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Two circles, $\omega_{1}$ and $\omega_{2}$, of equal radius intersect at different points $X_{1}$ and $X_{2}$. Consider a circle $\omega$ externally tangent to $\omega_{1}$ at a point $T_{1}$, and internally tangent to $\omega_{2}$ at a point $T_{2}$. Prove that lines $X_{1} T_{1}$ and $X_{2} T_{2}$ intersect at a point lying on $\omega$.
|
Let the line $X_{k} T_{k}$ and $\omega$ meet again at $X_{k}^{\prime}, k=1,2$, and notice that the tangent $t_{k}$ to $\omega_{k}$ at $X_{k}$ and the tangent $t_{k}^{\prime}$ to $\omega$ at $X_{k}^{\prime}$ are parallel. Since the $\omega_{k}$ have equal radii, the $t_{k}$ are parallel, so the $t_{k}^{\prime}$ are parallel, and consequently the points $X_{1}^{\prime}$ and $X_{2}^{\prime}$ coincide (they are not antipodal, since they both lie on the same side of the line $T_{1} T_{2}$ ). The conclusion follows.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Two circles, $\omega_{1}$ and $\omega_{2}$, of equal radius intersect at different points $X_{1}$ and $X_{2}$. Consider a circle $\omega$ externally tangent to $\omega_{1}$ at a point $T_{1}$, and internally tangent to $\omega_{2}$ at a point $T_{2}$. Prove that lines $X_{1} T_{1}$ and $X_{2} T_{2}$ intersect at a point lying on $\omega$.
|
Let the line $X_{k} T_{k}$ and $\omega$ meet again at $X_{k}^{\prime}, k=1,2$, and notice that the tangent $t_{k}$ to $\omega_{k}$ at $X_{k}$ and the tangent $t_{k}^{\prime}$ to $\omega$ at $X_{k}^{\prime}$ are parallel. Since the $\omega_{k}$ have equal radii, the $t_{k}$ are parallel, so the $t_{k}^{\prime}$ are parallel, and consequently the points $X_{1}^{\prime}$ and $X_{2}^{\prime}$ coincide (they are not antipodal, since they both lie on the same side of the line $T_{1} T_{2}$ ). The conclusion follows.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "EGMO/segmented/en-2016-solutions.jsonl",
"solution_match": "\nSolution 1. ",
"tier": "T2",
"year": "2016"
}
|
Two circles, $\omega_{1}$ and $\omega_{2}$, of equal radius intersect at different points $X_{1}$ and $X_{2}$. Consider a circle $\omega$ externally tangent to $\omega_{1}$ at a point $T_{1}$, and internally tangent to $\omega_{2}$ at a point $T_{2}$. Prove that lines $X_{1} T_{1}$ and $X_{2} T_{2}$ intersect at a point lying on $\omega$.
|
The circle $\omega$ is the image of $\omega_{k}$ under a homothety $h_{k}$ centred at $T_{k}, k=1,2$. The tangent to $\omega$ at $X_{k}^{\prime}=h_{k}\left(X_{k}\right)$ is therefore parallel to the tangent $t_{k}$ to $\omega_{k}$ at $X_{k}$. Since the $\omega_{k}$ have equal radii, the $t_{k}$ are parallel, so $X_{1}^{\prime}=X_{2}^{\prime}$; and since the points $X_{k}, T_{k}$ and $X_{k}^{\prime}$ are collinear, the conclusion follows.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Two circles, $\omega_{1}$ and $\omega_{2}$, of equal radius intersect at different points $X_{1}$ and $X_{2}$. Consider a circle $\omega$ externally tangent to $\omega_{1}$ at a point $T_{1}$, and internally tangent to $\omega_{2}$ at a point $T_{2}$. Prove that lines $X_{1} T_{1}$ and $X_{2} T_{2}$ intersect at a point lying on $\omega$.
|
The circle $\omega$ is the image of $\omega_{k}$ under a homothety $h_{k}$ centred at $T_{k}, k=1,2$. The tangent to $\omega$ at $X_{k}^{\prime}=h_{k}\left(X_{k}\right)$ is therefore parallel to the tangent $t_{k}$ to $\omega_{k}$ at $X_{k}$. Since the $\omega_{k}$ have equal radii, the $t_{k}$ are parallel, so $X_{1}^{\prime}=X_{2}^{\prime}$; and since the points $X_{k}, T_{k}$ and $X_{k}^{\prime}$ are collinear, the conclusion follows.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "EGMO/segmented/en-2016-solutions.jsonl",
"solution_match": "\nSolution 2. ",
"tier": "T2",
"year": "2016"
}
|
Two circles, $\omega_{1}$ and $\omega_{2}$, of equal radius intersect at different points $X_{1}$ and $X_{2}$. Consider a circle $\omega$ externally tangent to $\omega_{1}$ at a point $T_{1}$, and internally tangent to $\omega_{2}$ at a point $T_{2}$. Prove that lines $X_{1} T_{1}$ and $X_{2} T_{2}$ intersect at a point lying on $\omega$.
|
Invert from $X_{1}$ and use an asterisk to denote images under this inversion. Notice that $\omega_{k}^{*}$ is the tangent from $X_{2}^{*}$ to $\omega^{*}$ at $T_{k}^{*}$, and the pole $X_{1}$ lies on the bisectrix of the angle formed by the $\omega_{k}^{*}$, not containing $\omega^{*}$. Letting $X_{1} T_{1}^{*}$ and $\omega^{*}$ meet again at $Y$, standard angle chase shows that $Y$ lies on the circle $X_{1} X_{2}^{*} T_{2}^{*}$, and the conclusion follows.
Remarks. The product $h_{1} h_{2}$ of the two homotheties in the first solution is reflexion across the midpoint of the segment $X_{1} X_{2}$, which lies on the line $T_{1} T_{2}$ 。
Various arguments, involving similarities, radical axes, and the like, work equally well to prove the required result.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Two circles, $\omega_{1}$ and $\omega_{2}$, of equal radius intersect at different points $X_{1}$ and $X_{2}$. Consider a circle $\omega$ externally tangent to $\omega_{1}$ at a point $T_{1}$, and internally tangent to $\omega_{2}$ at a point $T_{2}$. Prove that lines $X_{1} T_{1}$ and $X_{2} T_{2}$ intersect at a point lying on $\omega$.
|
Invert from $X_{1}$ and use an asterisk to denote images under this inversion. Notice that $\omega_{k}^{*}$ is the tangent from $X_{2}^{*}$ to $\omega^{*}$ at $T_{k}^{*}$, and the pole $X_{1}$ lies on the bisectrix of the angle formed by the $\omega_{k}^{*}$, not containing $\omega^{*}$. Letting $X_{1} T_{1}^{*}$ and $\omega^{*}$ meet again at $Y$, standard angle chase shows that $Y$ lies on the circle $X_{1} X_{2}^{*} T_{2}^{*}$, and the conclusion follows.
Remarks. The product $h_{1} h_{2}$ of the two homotheties in the first solution is reflexion across the midpoint of the segment $X_{1} X_{2}$, which lies on the line $T_{1} T_{2}$ 。
Various arguments, involving similarities, radical axes, and the like, work equally well to prove the required result.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "EGMO/segmented/en-2016-solutions.jsonl",
"solution_match": "\nSolution 3. ",
"tier": "T2",
"year": "2016"
}
|
Let $k$ and $n$ be integers such that $k \geq 2$ and $k \leq n \leq 2 k-1$. Place rectangular tiles, each of size $1 \times k$ or $k \times 1$, on an $n \times n$ chessboard so that each tile covers exactly $k$ cells, and no two tiles overlap. Do this until
no further tile can be placed in this way. For each such $k$ and $n$, determine the minimum number of tiles that such an arrangement may contain.
|
The required minimum is $n$ if $n=k$, and it is $\min (n, 2 n-2 k+2)$ if $k<n<2 k$.
The case $n=k$ being clear, assume henceforth $k<n<2 k$. Begin by describing maximal arrangements on the board $[0, n] \times[0, n]$, having the above mentioned cardinalities.
If $k<n<2 k-1$, then $\min (n, 2 n-2 k+2)=2 n-2 k+2$. To obtain a maximal arrangement of this cardinality, place four tiles, $[0, k] \times[0,1]$, $[0,1] \times[1, k+1],[1, k+1] \times[k, k+1]$ and $[k, k+1] \times[0, k]$ in the square $[0, k] \times[0, k]$, stack $n-k-1$ horizontal tiles in the rectangle $[1, k+1] \times[k+1, n]$, and erect $n-k-1$ vertical tiles in the rectangle $[k+1, n] \times[1, k+1]$.
If $n=2 k-1$, then $\min (n, 2 n-2 k+2)=n=2 k-1$. A maximal arrangement of $2 k-1$ tiles is obtained by stacking $k-1$ horizontal tiles in the rectangle $[0, k] \times[0, k-1]$, another $k-1$ horizontal tiles in the rectangle $[0, k] \times[k, 2 k-1]$, and adding the horizontal tile $[k-1,2 k-1] \times[k-1, k]$.
The above examples show that the required minimum does not exceed the mentioned values.
To prove the reverse inequality, consider a maximal arrangement and let $r$, respectively $c$, be the number of rows, respectively columns, not containing a tile.
If $r=0$ or $c=0$, the arrangement clearly contains at least $n$ tiles.
If $r$ and $c$ are both positive, we show that the arrangement contains at least $2 n-2 k+2$ tiles. To this end, we will prove that the rows, respectively columns, not containing a tile are consecutive. Assume this for the moment, to notice that these $r$ rows and $c$ columns cross to form an $r \times c$ rectangular array containing no tile at all, so $r<k$ and $c<k$ by maximality. Consequently, there are $n-r \geq n-k+1$ rows containing at least one horizontal tile each, and $n-c \geq n-k+1$ columns containing at least one vertical tile each, whence a total of at least $2 n-2 k+2$ tiles.
We now show that the rows not containing a tile are consecutive; columns are dealt with similarly. Consider a horizontal tile $T$. Since $n<2 k$, the nearest horizontal side of the board is at most $k-1$ rows away from the row containing $T$. These rows, if any, cross the $k$ columns $T$ crosses to form a rectangular array no vertical tile fits in. Maximality forces each of these rows to contain a horizontal tile and the claim follows.
Consequently, the cardinality of every maximal arrangement is at least $\min (n, 2 n-2 k+2)$, and the conclusion follows.
Remarks. (1) If $k \geq 3$ and $n=2 k$, the minimum is $n+1=2 k+1$
and is achieved, for instance, by the maximal arrangement consisting of the vertical tile $[0,1] \times[1, k+1]$ along with $k-1$ horizontal tiles stacked in $[1, k+1] \times[0, k-1]$, another $k-1$ horizontal tiles stacked in $[1, k+1] \times[k+1,2 k]$, and two horizontal tiles stacked in $[k, 2 k] \times[k-1, k+1]$. This example shows that the corresponding minimum does not exceed $n+1<2 n-2 k+2$. The argument in the solution also applies to the case $n=2 k$ to infer that for a maximal arrangement of minimal cardinality either $r=0$ or $c=0$, and the cardinality is at least $n$. Clearly, we may and will assume $r=0$. Suppose, if possible, such an arrangement contains exactly $n$ tiles. Then each row contains exactly one tile, and there are no vertical tiles. Since there is no room left for an additional tile, some tile $T$ must cover a cell of the leftmost column, so it covers the $k$ leftmost cells along its row, and there is then room for another tile along that row - a contradiction.
(2) For every pair $(r, c)$ of integers in the range $2 k-n, \ldots, k-1$, at least one of which is positive, say $c>0$, there exists a maximal arrangement of cardinality $2 n-r-c$.
Use again the board $[0, n] \times[0, n]$ to stack $k-r$ horizontal tiles in each of the rectangles $[0, k] \times[0, k-r]$ and $[k-c, 2 k-c] \times[k, 2 k-r]$, erect $k-c$ vertical tiles in each of the rectangles $[0, k-c] \times[k-r, 2 k-r]$ and $[k, 2 k-c] \times[0, k]$, then stack $n-2 k+r$ horizontal tiles in the rectangle $[k-c, 2 k-c] \times[2 k-r, n]$, and erect $n-2 k+c$ vertical tiles in the rectangle $[2 k-c, n] \times[1, k+1]$.
|
\min (n, 2 n-2 k+2)
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $k$ and $n$ be integers such that $k \geq 2$ and $k \leq n \leq 2 k-1$. Place rectangular tiles, each of size $1 \times k$ or $k \times 1$, on an $n \times n$ chessboard so that each tile covers exactly $k$ cells, and no two tiles overlap. Do this until
no further tile can be placed in this way. For each such $k$ and $n$, determine the minimum number of tiles that such an arrangement may contain.
|
The required minimum is $n$ if $n=k$, and it is $\min (n, 2 n-2 k+2)$ if $k<n<2 k$.
The case $n=k$ being clear, assume henceforth $k<n<2 k$. Begin by describing maximal arrangements on the board $[0, n] \times[0, n]$, having the above mentioned cardinalities.
If $k<n<2 k-1$, then $\min (n, 2 n-2 k+2)=2 n-2 k+2$. To obtain a maximal arrangement of this cardinality, place four tiles, $[0, k] \times[0,1]$, $[0,1] \times[1, k+1],[1, k+1] \times[k, k+1]$ and $[k, k+1] \times[0, k]$ in the square $[0, k] \times[0, k]$, stack $n-k-1$ horizontal tiles in the rectangle $[1, k+1] \times[k+1, n]$, and erect $n-k-1$ vertical tiles in the rectangle $[k+1, n] \times[1, k+1]$.
If $n=2 k-1$, then $\min (n, 2 n-2 k+2)=n=2 k-1$. A maximal arrangement of $2 k-1$ tiles is obtained by stacking $k-1$ horizontal tiles in the rectangle $[0, k] \times[0, k-1]$, another $k-1$ horizontal tiles in the rectangle $[0, k] \times[k, 2 k-1]$, and adding the horizontal tile $[k-1,2 k-1] \times[k-1, k]$.
The above examples show that the required minimum does not exceed the mentioned values.
To prove the reverse inequality, consider a maximal arrangement and let $r$, respectively $c$, be the number of rows, respectively columns, not containing a tile.
If $r=0$ or $c=0$, the arrangement clearly contains at least $n$ tiles.
If $r$ and $c$ are both positive, we show that the arrangement contains at least $2 n-2 k+2$ tiles. To this end, we will prove that the rows, respectively columns, not containing a tile are consecutive. Assume this for the moment, to notice that these $r$ rows and $c$ columns cross to form an $r \times c$ rectangular array containing no tile at all, so $r<k$ and $c<k$ by maximality. Consequently, there are $n-r \geq n-k+1$ rows containing at least one horizontal tile each, and $n-c \geq n-k+1$ columns containing at least one vertical tile each, whence a total of at least $2 n-2 k+2$ tiles.
We now show that the rows not containing a tile are consecutive; columns are dealt with similarly. Consider a horizontal tile $T$. Since $n<2 k$, the nearest horizontal side of the board is at most $k-1$ rows away from the row containing $T$. These rows, if any, cross the $k$ columns $T$ crosses to form a rectangular array no vertical tile fits in. Maximality forces each of these rows to contain a horizontal tile and the claim follows.
Consequently, the cardinality of every maximal arrangement is at least $\min (n, 2 n-2 k+2)$, and the conclusion follows.
Remarks. (1) If $k \geq 3$ and $n=2 k$, the minimum is $n+1=2 k+1$
and is achieved, for instance, by the maximal arrangement consisting of the vertical tile $[0,1] \times[1, k+1]$ along with $k-1$ horizontal tiles stacked in $[1, k+1] \times[0, k-1]$, another $k-1$ horizontal tiles stacked in $[1, k+1] \times[k+1,2 k]$, and two horizontal tiles stacked in $[k, 2 k] \times[k-1, k+1]$. This example shows that the corresponding minimum does not exceed $n+1<2 n-2 k+2$. The argument in the solution also applies to the case $n=2 k$ to infer that for a maximal arrangement of minimal cardinality either $r=0$ or $c=0$, and the cardinality is at least $n$. Clearly, we may and will assume $r=0$. Suppose, if possible, such an arrangement contains exactly $n$ tiles. Then each row contains exactly one tile, and there are no vertical tiles. Since there is no room left for an additional tile, some tile $T$ must cover a cell of the leftmost column, so it covers the $k$ leftmost cells along its row, and there is then room for another tile along that row - a contradiction.
(2) For every pair $(r, c)$ of integers in the range $2 k-n, \ldots, k-1$, at least one of which is positive, say $c>0$, there exists a maximal arrangement of cardinality $2 n-r-c$.
Use again the board $[0, n] \times[0, n]$ to stack $k-r$ horizontal tiles in each of the rectangles $[0, k] \times[0, k-r]$ and $[k-c, 2 k-c] \times[k, 2 k-r]$, erect $k-c$ vertical tiles in each of the rectangles $[0, k-c] \times[k-r, 2 k-r]$ and $[k, 2 k-c] \times[0, k]$, then stack $n-2 k+r$ horizontal tiles in the rectangle $[k-c, 2 k-c] \times[2 k-r, n]$, and erect $n-2 k+c$ vertical tiles in the rectangle $[2 k-c, n] \times[1, k+1]$.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "\nProblem 5.",
"resource_path": "EGMO/segmented/en-2016-solutions.jsonl",
"solution_match": "\nSolution.",
"tier": "T2",
"year": "2016"
}
|
Let $S$ be the set of all positive integers $n$ such that $n^{4}$ has a divisor in the range $n^{2}+1, n^{2}+2, \ldots, n^{2}+2 n$. Prove that there are infinitely many elements of $S$ of each of the forms $7 m, 7 m+1,7 m+2,7 m+5,7 m+6$ and no elements of $S$ of the form $7 m+3$ or $7 m+4$, where $m$ is an integer.
|
The conclusion is a consequence of the lemma below which actually provides a recursive description of $S$. The proof of the lemma is at the end of the solution.
Lemma. The fourth power of a positive integer $n$ has a divisor in the range $n^{2}+1, n^{2}+2, \ldots, n^{2}+2 n$ if and only if at least one of the numbers $2 n^{2}+1$ and $12 n^{2}+9$ is a perfect square.
Consequently, a positive integer $n$ is a member of $S$ if and only if $m^{2}-$ $2 n^{2}=1$ or $m^{2}-12 n^{2}=9$ for some positive integer $m$.
The former is a Pell equation whose solutions are $\left(m_{1}, n_{1}\right)=(3,2)$ and
$$
\left(m_{k+1}, n_{k+1}\right)=\left(3 m_{k}+4 n_{k}, 2 m_{k}+3 n_{k}\right), \quad k=1,2,3, \ldots
$$
In what follows, all congruences are modulo 7. Iteration shows that $\left(m_{k+3}, n_{k+3}\right) \equiv\left(m_{k}, n_{k}\right)$. Since $\left(m_{1}, n_{1}\right) \equiv(3,2),\left(m_{2}, n_{2}\right) \equiv(3,-2)$, and
$\left(m_{3}, n_{3}\right) \equiv(1,0)$, it follows that $S$ contains infinitely many integers from each of the residue classes 0 and $\pm 2$ modulo 7 .
The other equation is easily transformed into a Pell equation, $m^{\prime 2}-$ $12 n^{\prime 2}=1$, by noticing that $m$ and $n$ are both divisible by 3 , say $m=3 m^{\prime}$ and $n=3 n^{\prime}$. In this case, the solutions are $\left(m_{1}, n_{1}\right)=(21,6)$ and
$$
\left(m_{k+1}, n_{k+1}\right)=\left(7 m_{k}+24 n_{k}, 2 m_{k}+7 n_{k}\right), \quad k=1,2,3, \ldots
$$
This time iteration shows that $\left(m_{k+4}, n_{k+4}\right) \equiv\left(m_{k}, n_{k}\right)$. Since $\left(m_{1}, n_{1}\right) \equiv$ $(0,-1),\left(m_{2}, n_{2}\right) \equiv(-3,0),\left(m_{3}, n_{3}\right) \equiv(0,1)$, and $\left(m_{4}, n_{4}\right) \equiv(3,0)$, it follows that $S$ contains infinitely many integers from each of the residue classes 0 and $\pm 1$ modulo 7 .
Finally, since the $n_{k}$ from the two sets of formulae exhaust $S$, by the preceding no integer in the residue classes $\pm 3$ modulo 7 is a member of $S$.
We now turn to the proof of the lemma. Let $n$ be a member of $S$, and let $d=n^{2}+m$ be a divisor of $n^{4}$ in the range $n^{2}+1, n^{2}+2, \ldots, n^{2}+2 n$, so $1 \leq m \leq 2 n$. Consideration of the square of $n^{2}=d-m$ shows $m^{2}$ divisible by $d$, so $m^{2} / d$ is a positive integer. Since $n^{2}<d<(n+1)^{2}$, it follows that $d$ is not a square; in particular, $m^{2} / d \neq 1$, so $m^{2} / d \geq 2$. On the other hand, $1 \leq m \leq 2 n$, so $m^{2} / d=m^{2} /\left(n^{2}+m\right) \leq 4 n^{2} /\left(n^{2}+1\right)<4$. Consequently, $m^{2} / d=2$ or $m^{2} / d=3$; that is, $m^{2} /\left(n^{2}+m\right)=2$ or $m^{2} /\left(n^{2}+m\right)=3$. In the former case, $2 n^{2}+1=(m-1)^{2}$, and in the latter, $12 n^{2}+9=(2 m-3)^{2}$.
Conversely, if $2 n^{2}+1=m^{2}$ for some positive integer $m$, then $1<m^{2}<$ $4 n^{2}$, so $1<m<2 n$, and $n^{4}=\left(n^{2}+m+1\right)\left(n^{2}-m+1\right)$, so the first factor is the desired divisor.
Similarly, if $12 n^{2}+9=m^{2}$ for some positive integer $m$, then $m$ is odd, $n \geq 6$, and $n^{4}=\left(n^{2}+m / 2+3 / 2\right)\left(n^{2}-m / 2+3 / 2\right)$, and again the first factor is the desired divisor.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $S$ be the set of all positive integers $n$ such that $n^{4}$ has a divisor in the range $n^{2}+1, n^{2}+2, \ldots, n^{2}+2 n$. Prove that there are infinitely many elements of $S$ of each of the forms $7 m, 7 m+1,7 m+2,7 m+5,7 m+6$ and no elements of $S$ of the form $7 m+3$ or $7 m+4$, where $m$ is an integer.
|
The conclusion is a consequence of the lemma below which actually provides a recursive description of $S$. The proof of the lemma is at the end of the solution.
Lemma. The fourth power of a positive integer $n$ has a divisor in the range $n^{2}+1, n^{2}+2, \ldots, n^{2}+2 n$ if and only if at least one of the numbers $2 n^{2}+1$ and $12 n^{2}+9$ is a perfect square.
Consequently, a positive integer $n$ is a member of $S$ if and only if $m^{2}-$ $2 n^{2}=1$ or $m^{2}-12 n^{2}=9$ for some positive integer $m$.
The former is a Pell equation whose solutions are $\left(m_{1}, n_{1}\right)=(3,2)$ and
$$
\left(m_{k+1}, n_{k+1}\right)=\left(3 m_{k}+4 n_{k}, 2 m_{k}+3 n_{k}\right), \quad k=1,2,3, \ldots
$$
In what follows, all congruences are modulo 7. Iteration shows that $\left(m_{k+3}, n_{k+3}\right) \equiv\left(m_{k}, n_{k}\right)$. Since $\left(m_{1}, n_{1}\right) \equiv(3,2),\left(m_{2}, n_{2}\right) \equiv(3,-2)$, and
$\left(m_{3}, n_{3}\right) \equiv(1,0)$, it follows that $S$ contains infinitely many integers from each of the residue classes 0 and $\pm 2$ modulo 7 .
The other equation is easily transformed into a Pell equation, $m^{\prime 2}-$ $12 n^{\prime 2}=1$, by noticing that $m$ and $n$ are both divisible by 3 , say $m=3 m^{\prime}$ and $n=3 n^{\prime}$. In this case, the solutions are $\left(m_{1}, n_{1}\right)=(21,6)$ and
$$
\left(m_{k+1}, n_{k+1}\right)=\left(7 m_{k}+24 n_{k}, 2 m_{k}+7 n_{k}\right), \quad k=1,2,3, \ldots
$$
This time iteration shows that $\left(m_{k+4}, n_{k+4}\right) \equiv\left(m_{k}, n_{k}\right)$. Since $\left(m_{1}, n_{1}\right) \equiv$ $(0,-1),\left(m_{2}, n_{2}\right) \equiv(-3,0),\left(m_{3}, n_{3}\right) \equiv(0,1)$, and $\left(m_{4}, n_{4}\right) \equiv(3,0)$, it follows that $S$ contains infinitely many integers from each of the residue classes 0 and $\pm 1$ modulo 7 .
Finally, since the $n_{k}$ from the two sets of formulae exhaust $S$, by the preceding no integer in the residue classes $\pm 3$ modulo 7 is a member of $S$.
We now turn to the proof of the lemma. Let $n$ be a member of $S$, and let $d=n^{2}+m$ be a divisor of $n^{4}$ in the range $n^{2}+1, n^{2}+2, \ldots, n^{2}+2 n$, so $1 \leq m \leq 2 n$. Consideration of the square of $n^{2}=d-m$ shows $m^{2}$ divisible by $d$, so $m^{2} / d$ is a positive integer. Since $n^{2}<d<(n+1)^{2}$, it follows that $d$ is not a square; in particular, $m^{2} / d \neq 1$, so $m^{2} / d \geq 2$. On the other hand, $1 \leq m \leq 2 n$, so $m^{2} / d=m^{2} /\left(n^{2}+m\right) \leq 4 n^{2} /\left(n^{2}+1\right)<4$. Consequently, $m^{2} / d=2$ or $m^{2} / d=3$; that is, $m^{2} /\left(n^{2}+m\right)=2$ or $m^{2} /\left(n^{2}+m\right)=3$. In the former case, $2 n^{2}+1=(m-1)^{2}$, and in the latter, $12 n^{2}+9=(2 m-3)^{2}$.
Conversely, if $2 n^{2}+1=m^{2}$ for some positive integer $m$, then $1<m^{2}<$ $4 n^{2}$, so $1<m<2 n$, and $n^{4}=\left(n^{2}+m+1\right)\left(n^{2}-m+1\right)$, so the first factor is the desired divisor.
Similarly, if $12 n^{2}+9=m^{2}$ for some positive integer $m$, then $m$ is odd, $n \geq 6$, and $n^{4}=\left(n^{2}+m / 2+3 / 2\right)\left(n^{2}-m / 2+3 / 2\right)$, and again the first factor is the desired divisor.
|
{
"exam": "EGMO",
"problem_label": "6",
"problem_match": "\nProblem 6.",
"resource_path": "EGMO/segmented/en-2016-solutions.jsonl",
"solution_match": "\nSolution.",
"tier": "T2",
"year": "2016"
}
|
Let $A B C D$ be a convex quadrilateral with $\angle D A B=\angle B C D=90^{\circ}$ and $\angle A B C>\angle C D A$. Let $Q$ and $R$ be points on the segments $B C$ and $C D$, respectively, such that the line $Q R$ intersects lines $A B$ and $A D$ at points $P$ and $S$, respectively. It is given that $P Q=R S$. Let the midpoint of $B D$ be $M$ and the midpoint of $Q R$ be $N$. Prove that $M, N, A$ and $C$ lie on a circle.
Mark Mordechai Etkind, Israel
|
1
Note that $N$ is also the midpoint of $P S$. From right-angled triangles $P A S$ and $C Q R$ we obtain $\angle A N P=$ $2 \angle A S P, \angle C N Q=2 \angle C R Q$, hence $\angle A N C=\angle A N P+\angle C N Q=2(\angle A S P+\angle C R Q)=2(\angle R S D+\angle D R S)=$ $2 \angle A D C$.
Similarly, using right-angled triangles $B A D$ and $B C D$, we obtain $\angle A M C=2 \angle A D C$.
Thus $\angle A M C=\angle A N C$, and the required statement follows.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a convex quadrilateral with $\angle D A B=\angle B C D=90^{\circ}$ and $\angle A B C>\angle C D A$. Let $Q$ and $R$ be points on the segments $B C$ and $C D$, respectively, such that the line $Q R$ intersects lines $A B$ and $A D$ at points $P$ and $S$, respectively. It is given that $P Q=R S$. Let the midpoint of $B D$ be $M$ and the midpoint of $Q R$ be $N$. Prove that $M, N, A$ and $C$ lie on a circle.
Mark Mordechai Etkind, Israel
|
1
Note that $N$ is also the midpoint of $P S$. From right-angled triangles $P A S$ and $C Q R$ we obtain $\angle A N P=$ $2 \angle A S P, \angle C N Q=2 \angle C R Q$, hence $\angle A N C=\angle A N P+\angle C N Q=2(\angle A S P+\angle C R Q)=2(\angle R S D+\angle D R S)=$ $2 \angle A D C$.
Similarly, using right-angled triangles $B A D$ and $B C D$, we obtain $\angle A M C=2 \angle A D C$.
Thus $\angle A M C=\angle A N C$, and the required statement follows.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "# Problem 1",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "# Solution ",
"tier": "T2",
"year": "2017"
}
|
Let $A B C D$ be a convex quadrilateral with $\angle D A B=\angle B C D=90^{\circ}$ and $\angle A B C>\angle C D A$. Let $Q$ and $R$ be points on the segments $B C$ and $C D$, respectively, such that the line $Q R$ intersects lines $A B$ and $A D$ at points $P$ and $S$, respectively. It is given that $P Q=R S$. Let the midpoint of $B D$ be $M$ and the midpoint of $Q R$ be $N$. Prove that $M, N, A$ and $C$ lie on a circle.
Mark Mordechai Etkind, Israel
|
2
In this proof we show that we have $\angle N C M=\angle N A M$ instead. From right-angled triangles $B C D$ and $Q C R$ we get $\angle D R S=\angle C R Q=\angle R C N$ and $\angle B D C=\angle D C M$. Hence $\angle N C M=\angle D C M-\angle R C N$. From rightangled triangle $A P S$ we get $\angle P S A=\angle S A N$. From right-angled triangle $B A D$ we have $\angle M A D=\angle B D A$. Moreover, $\angle B D A=\angle D R S+\angle R S D-\angle R D B$.
Therefore $\angle N A M=\angle N A S-\angle M A D=\angle C D B-\angle D R S=\angle N C M$, and the required statement follows.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a convex quadrilateral with $\angle D A B=\angle B C D=90^{\circ}$ and $\angle A B C>\angle C D A$. Let $Q$ and $R$ be points on the segments $B C$ and $C D$, respectively, such that the line $Q R$ intersects lines $A B$ and $A D$ at points $P$ and $S$, respectively. It is given that $P Q=R S$. Let the midpoint of $B D$ be $M$ and the midpoint of $Q R$ be $N$. Prove that $M, N, A$ and $C$ lie on a circle.
Mark Mordechai Etkind, Israel
|
2
In this proof we show that we have $\angle N C M=\angle N A M$ instead. From right-angled triangles $B C D$ and $Q C R$ we get $\angle D R S=\angle C R Q=\angle R C N$ and $\angle B D C=\angle D C M$. Hence $\angle N C M=\angle D C M-\angle R C N$. From rightangled triangle $A P S$ we get $\angle P S A=\angle S A N$. From right-angled triangle $B A D$ we have $\angle M A D=\angle B D A$. Moreover, $\angle B D A=\angle D R S+\angle R S D-\angle R D B$.
Therefore $\angle N A M=\angle N A S-\angle M A D=\angle C D B-\angle D R S=\angle N C M$, and the required statement follows.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "# Problem 1",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "# Solution ",
"tier": "T2",
"year": "2017"
}
|
Let $A B C D$ be a convex quadrilateral with $\angle D A B=\angle B C D=90^{\circ}$ and $\angle A B C>\angle C D A$. Let $Q$ and $R$ be points on the segments $B C$ and $C D$, respectively, such that the line $Q R$ intersects lines $A B$ and $A D$ at points $P$ and $S$, respectively. It is given that $P Q=R S$. Let the midpoint of $B D$ be $M$ and the midpoint of $Q R$ be $N$. Prove that $M, N, A$ and $C$ lie on a circle.
Mark Mordechai Etkind, Israel
|
3
As $N$ is also the midpoint of $P S$, we can shrink triangle $A P S$ to a triangle $A_{0} Q R$ (where $P$ is sent to $Q$ and $S$ is sent to $R$ ). Then $A_{0}, Q, R$ and $C$ lie on a circle with center $N$. According to the shrinking the line $A_{0} R$ is parallel to the line $A D$. Therefore $\angle C N A=\angle C N A_{0}=2 \angle C R A_{0}=2 \angle C D A=\angle C M A$. The required statement follows.
|
proof
|
Yes
|
Problem not solved
|
proof
|
Geometry
|
Let $A B C D$ be a convex quadrilateral with $\angle D A B=\angle B C D=90^{\circ}$ and $\angle A B C>\angle C D A$. Let $Q$ and $R$ be points on the segments $B C$ and $C D$, respectively, such that the line $Q R$ intersects lines $A B$ and $A D$ at points $P$ and $S$, respectively. It is given that $P Q=R S$. Let the midpoint of $B D$ be $M$ and the midpoint of $Q R$ be $N$. Prove that $M, N, A$ and $C$ lie on a circle.
Mark Mordechai Etkind, Israel
|
3
As $N$ is also the midpoint of $P S$, we can shrink triangle $A P S$ to a triangle $A_{0} Q R$ (where $P$ is sent to $Q$ and $S$ is sent to $R$ ). Then $A_{0}, Q, R$ and $C$ lie on a circle with center $N$. According to the shrinking the line $A_{0} R$ is parallel to the line $A D$. Therefore $\angle C N A=\angle C N A_{0}=2 \angle C R A_{0}=2 \angle C D A=\angle C M A$. The required statement follows.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "# Problem 1",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "# Solution ",
"tier": "T2",
"year": "2017"
}
|
Find the smallest positive integer $k$ for which there exist a colouring of the positive integers $\mathbb{Z}_{>0}$ with $k$ colours and a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ with the following two properties:
(i) For all positive integers $m, n$ of the same colour, $f(m+n)=f(m)+f(n)$.
(ii) There are positive integers $m, n$ such that $f(m+n) \neq f(m)+f(n)$.
In a colouring of $\mathbb{Z}_{>0}$ with $k$ colours, every integer is coloured in exactly one of the $k$ colours. In both (i) and (ii) the positive integers $m, n$ are not necessarily different.
Merlijn Staps, the Netherlands
|
The answer is $k=3$.
First we show that there is such a function and coloring for $k=3$. Consider $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ given by $f(n)=n$ for all $n \equiv 1$ or 2 modulo 3 , and $f(n)=2 n$ for $n \equiv 0$ modulo 3 . Moreover, give a positive integer $n$ the $i$-th color if $n \equiv i$ (3).
By construction we have $f(1+2)=6 \neq 3=f(1)+f(2)$ and hence $f$ has property (ii).
Now let $n, m$ be positive integers with the same color $i$. If $i=0$, then $n+m$ has color 0 , so $f(n+m)=$ $2(n+m)=2 n+2 m=f(n)+f(m)$. If $i=1$, then $n+m$ has color 2 , so $f(n+m)=n+m=f(n)+f(m)$. Finally, if $i=2$, then $n+m$ has color 1 , so $f(n+m)=n+m=f(n)+f(m)$. Therefore $f$ also satisfies condition (i).
Next we show that there is no such function and coloring for $k=2$.
Consider any coloring of $\mathbb{Z}_{>0}$ with 2 colors and any function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ satisfying conditions (i) and (ii). Then there exist positive integers $m$ and $n$ such that $f(m+n) \neq f(m)+f(n)$. Choose $m$ and $n$ such that their sum is minimal among all such $m, n$ and define $a=m+n$. Then in particular for every $b<a$ we have $f(b)=b f(1)$ and $f(a) \neq a f(1)$.
If $a$ is even, then condition (i) for $m=n=\frac{a}{2}$ implies $f(a)=f\left(\frac{a}{2}\right)+f\left(\frac{a}{2}\right)=f(1) a$, a contradiction. Hence $a$ is odd. We will prove two lemmas.
Lemma 1. Any odd integer $b<a$ has a different color than $a$.
Proof. Suppose that $b<a$ is an odd integer, and that $a$ and $b$ have the same color. Then on the one hand, $f(a+b)=f(a)+b f(1)$. On the other hand, we also have $f(a+b)=f\left(\frac{a+b}{2}\right)+f\left(\frac{a+b}{2}\right)=(a+b) f(1)$, as $\frac{a+b}{2}$ is a positive integer smaller than $a$. Hence $f(a)=f(a+b)-b f(1)=(a+b) f(1)-b f(1)=a f(1)$, which is again a contradiction. Therefore all odd integers smaller than $a$ have a color different from that of $a$.
Lemma 2. Any even integer $b<a$ has the same color as $a$
Proof. Suppose $b<a$ is an even integer, and that $a$ and $b$ have different colors. Then $a-b$ is an odd integer smaller than $a$, so it has the same color as $b$. Thus $f(a)=f(a-b)+f(b)=(a-b) f(1)+b f(1)=a f(1)$, a contradiction. Hence all even integers smaller than $a$ have the same color as $a$.
Suppose now $a+1$ has the same color as $a$. As $a>1$, we have $\frac{a+1}{2}<a$ and therefore $f(a+1)=2 f\left(\frac{a+1}{2}\right)=$ $(a+1) f(1)$. As $a-1$ is an even integer smaller than $a$, we have by Lemma 2 that $a-1$ also has the same color as $a$. Hence $2 f(a)=f(2 a)=f(a+1)+f(a-1)=(a+1) f(1)+(a-1) f(1)=2 a f(1)$, which implies that $f(a)=a f(1)$, a contradiction. So $a$ and $a+1$ have different colors.
Since $a-2$ is an odd integer smaller than $a$, by Lemma 1 it has a color different from that of $a$, so $a-2$ and $a+1$ have the same color. Also, we have seen by Lemma 2 that $a-1$ and $a$ have the same color. So $f(a)+f(a-1)=f(2 a-1)=f(a+1)+f(a-2)=(a+1) f(1)+(a-2) f(1)=(2 a-1) f(1)$, from which it follows that $f(a)=(2 a-1) f(1)-f(a-1)=(2 a-1) f(1)-(a-1) f(1)=a f(1)$, which contradicts our choice of $a$ and finishes the proof.
|
3
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Find the smallest positive integer $k$ for which there exist a colouring of the positive integers $\mathbb{Z}_{>0}$ with $k$ colours and a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ with the following two properties:
(i) For all positive integers $m, n$ of the same colour, $f(m+n)=f(m)+f(n)$.
(ii) There are positive integers $m, n$ such that $f(m+n) \neq f(m)+f(n)$.
In a colouring of $\mathbb{Z}_{>0}$ with $k$ colours, every integer is coloured in exactly one of the $k$ colours. In both (i) and (ii) the positive integers $m, n$ are not necessarily different.
Merlijn Staps, the Netherlands
|
The answer is $k=3$.
First we show that there is such a function and coloring for $k=3$. Consider $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ given by $f(n)=n$ for all $n \equiv 1$ or 2 modulo 3 , and $f(n)=2 n$ for $n \equiv 0$ modulo 3 . Moreover, give a positive integer $n$ the $i$-th color if $n \equiv i$ (3).
By construction we have $f(1+2)=6 \neq 3=f(1)+f(2)$ and hence $f$ has property (ii).
Now let $n, m$ be positive integers with the same color $i$. If $i=0$, then $n+m$ has color 0 , so $f(n+m)=$ $2(n+m)=2 n+2 m=f(n)+f(m)$. If $i=1$, then $n+m$ has color 2 , so $f(n+m)=n+m=f(n)+f(m)$. Finally, if $i=2$, then $n+m$ has color 1 , so $f(n+m)=n+m=f(n)+f(m)$. Therefore $f$ also satisfies condition (i).
Next we show that there is no such function and coloring for $k=2$.
Consider any coloring of $\mathbb{Z}_{>0}$ with 2 colors and any function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ satisfying conditions (i) and (ii). Then there exist positive integers $m$ and $n$ such that $f(m+n) \neq f(m)+f(n)$. Choose $m$ and $n$ such that their sum is minimal among all such $m, n$ and define $a=m+n$. Then in particular for every $b<a$ we have $f(b)=b f(1)$ and $f(a) \neq a f(1)$.
If $a$ is even, then condition (i) for $m=n=\frac{a}{2}$ implies $f(a)=f\left(\frac{a}{2}\right)+f\left(\frac{a}{2}\right)=f(1) a$, a contradiction. Hence $a$ is odd. We will prove two lemmas.
Lemma 1. Any odd integer $b<a$ has a different color than $a$.
Proof. Suppose that $b<a$ is an odd integer, and that $a$ and $b$ have the same color. Then on the one hand, $f(a+b)=f(a)+b f(1)$. On the other hand, we also have $f(a+b)=f\left(\frac{a+b}{2}\right)+f\left(\frac{a+b}{2}\right)=(a+b) f(1)$, as $\frac{a+b}{2}$ is a positive integer smaller than $a$. Hence $f(a)=f(a+b)-b f(1)=(a+b) f(1)-b f(1)=a f(1)$, which is again a contradiction. Therefore all odd integers smaller than $a$ have a color different from that of $a$.
Lemma 2. Any even integer $b<a$ has the same color as $a$
Proof. Suppose $b<a$ is an even integer, and that $a$ and $b$ have different colors. Then $a-b$ is an odd integer smaller than $a$, so it has the same color as $b$. Thus $f(a)=f(a-b)+f(b)=(a-b) f(1)+b f(1)=a f(1)$, a contradiction. Hence all even integers smaller than $a$ have the same color as $a$.
Suppose now $a+1$ has the same color as $a$. As $a>1$, we have $\frac{a+1}{2}<a$ and therefore $f(a+1)=2 f\left(\frac{a+1}{2}\right)=$ $(a+1) f(1)$. As $a-1$ is an even integer smaller than $a$, we have by Lemma 2 that $a-1$ also has the same color as $a$. Hence $2 f(a)=f(2 a)=f(a+1)+f(a-1)=(a+1) f(1)+(a-1) f(1)=2 a f(1)$, which implies that $f(a)=a f(1)$, a contradiction. So $a$ and $a+1$ have different colors.
Since $a-2$ is an odd integer smaller than $a$, by Lemma 1 it has a color different from that of $a$, so $a-2$ and $a+1$ have the same color. Also, we have seen by Lemma 2 that $a-1$ and $a$ have the same color. So $f(a)+f(a-1)=f(2 a-1)=f(a+1)+f(a-2)=(a+1) f(1)+(a-2) f(1)=(2 a-1) f(1)$, from which it follows that $f(a)=(2 a-1) f(1)-f(a-1)=(2 a-1) f(1)-(a-1) f(1)=a f(1)$, which contradicts our choice of $a$ and finishes the proof.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "# Problem 2",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "# Solution 1:",
"tier": "T2",
"year": "2017"
}
|
Find the smallest positive integer $k$ for which there exist a colouring of the positive integers $\mathbb{Z}_{>0}$ with $k$ colours and a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ with the following two properties:
(i) For all positive integers $m, n$ of the same colour, $f(m+n)=f(m)+f(n)$.
(ii) There are positive integers $m, n$ such that $f(m+n) \neq f(m)+f(n)$.
In a colouring of $\mathbb{Z}_{>0}$ with $k$ colours, every integer is coloured in exactly one of the $k$ colours. In both (i) and (ii) the positive integers $m, n$ are not necessarily different.
Merlijn Staps, the Netherlands
|
We prove that $k \leq 3$ just as in first solution.
Next we show that there is no such function and coloring for $k=2$.
Consider any coloring of $\mathbb{Z}_{>0}$ with 2 colors and any function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ satisfying conditions (i) and (ii). We first notice with $m=n$ that $f(2 n)=2 f(n)$.
Lemma 3. For every $n \in \mathbb{Z}_{>0}, f(3 n)=3 f(n)$ holds.
Proof. Define $c=f(n), d=f(3 n)$. Then we have the relations
$$
f(2 n)=2 c, \quad f(4 n)=4 c, \quad f(6 n)=2 d
$$
- If $n$ and $2 n$ have the same color, then $f(3 n)=f(n)+f(2 n)=3 c=3 f(n)$.
- If $n$ and $3 n$ have the same color, then $4 c=f(4 n)=f(n)+f(3 n)=c+f(3 n)$, so $f(3 n)=3 f(n)$.
- If $2 n$ and $4 n$ have the same color, then $2 d=f(6 n)=f(2 n)+f(4 n)=2 c+4 c=6 c$, so $f(3 n)=d=3 c$.
- Otherwise $n$ and $4 n$ have the same color, and $2 n$ and $3 n$ both have the opposite color to $n$. Therefore we compute $5 c=f(n)+f(4 n)=f(5 n)=f(2 n)+f(3 n)=2 c+f(3 n)$ so $f(3 n)=3 f(n)$.
Consequently, for $k=2$ we necessarily have $f(3 n)=3 f(n)$.
Now let $a$ be the smallest integer such that $f(a) \neq a f(1)$. In particular $a$ is odd and $a>3$. Consider the three integers $a, \frac{a-3}{2}, \frac{a+3}{2}$. By pigeonhole principle two of them have the same color.
- If $\frac{a-3}{2}$ and $\frac{a+3}{2}$ have the same color, then $f(a)=\frac{a-3}{2} f(1)+\frac{a+3}{2} f(1)=a f(1)$.
- If $a$ and $\frac{a-3}{2}$ have the same color, then $3 \frac{a-1}{2} f(1)=3 f\left(\frac{a-1}{2}\right)=f\left(\frac{3 a-3}{2}\right)=f(a)+f\left(\frac{a-3}{2}\right)=f(a)+$ $\frac{a-3}{2} f(1)$, so $f(a)=a f(1)$.
- If $a$ and $\frac{a+3}{2}$ have the same color, then $3 \frac{a+1}{2} f(1)=3 f\left(\frac{a+1}{2}\right)=f\left(\frac{3 a+3}{2}\right)=f(a)+f\left(\frac{a+3}{2}\right)=f(a)+$ $\frac{a+3}{2} f(1)$, so $f(a)=a f(1)$.
In the three cases we find a contradiction with $f(a) \neq a f(1)$, so it finishes the proof.
|
3
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Find the smallest positive integer $k$ for which there exist a colouring of the positive integers $\mathbb{Z}_{>0}$ with $k$ colours and a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ with the following two properties:
(i) For all positive integers $m, n$ of the same colour, $f(m+n)=f(m)+f(n)$.
(ii) There are positive integers $m, n$ such that $f(m+n) \neq f(m)+f(n)$.
In a colouring of $\mathbb{Z}_{>0}$ with $k$ colours, every integer is coloured in exactly one of the $k$ colours. In both (i) and (ii) the positive integers $m, n$ are not necessarily different.
Merlijn Staps, the Netherlands
|
We prove that $k \leq 3$ just as in first solution.
Next we show that there is no such function and coloring for $k=2$.
Consider any coloring of $\mathbb{Z}_{>0}$ with 2 colors and any function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ satisfying conditions (i) and (ii). We first notice with $m=n$ that $f(2 n)=2 f(n)$.
Lemma 3. For every $n \in \mathbb{Z}_{>0}, f(3 n)=3 f(n)$ holds.
Proof. Define $c=f(n), d=f(3 n)$. Then we have the relations
$$
f(2 n)=2 c, \quad f(4 n)=4 c, \quad f(6 n)=2 d
$$
- If $n$ and $2 n$ have the same color, then $f(3 n)=f(n)+f(2 n)=3 c=3 f(n)$.
- If $n$ and $3 n$ have the same color, then $4 c=f(4 n)=f(n)+f(3 n)=c+f(3 n)$, so $f(3 n)=3 f(n)$.
- If $2 n$ and $4 n$ have the same color, then $2 d=f(6 n)=f(2 n)+f(4 n)=2 c+4 c=6 c$, so $f(3 n)=d=3 c$.
- Otherwise $n$ and $4 n$ have the same color, and $2 n$ and $3 n$ both have the opposite color to $n$. Therefore we compute $5 c=f(n)+f(4 n)=f(5 n)=f(2 n)+f(3 n)=2 c+f(3 n)$ so $f(3 n)=3 f(n)$.
Consequently, for $k=2$ we necessarily have $f(3 n)=3 f(n)$.
Now let $a$ be the smallest integer such that $f(a) \neq a f(1)$. In particular $a$ is odd and $a>3$. Consider the three integers $a, \frac{a-3}{2}, \frac{a+3}{2}$. By pigeonhole principle two of them have the same color.
- If $\frac{a-3}{2}$ and $\frac{a+3}{2}$ have the same color, then $f(a)=\frac{a-3}{2} f(1)+\frac{a+3}{2} f(1)=a f(1)$.
- If $a$ and $\frac{a-3}{2}$ have the same color, then $3 \frac{a-1}{2} f(1)=3 f\left(\frac{a-1}{2}\right)=f\left(\frac{3 a-3}{2}\right)=f(a)+f\left(\frac{a-3}{2}\right)=f(a)+$ $\frac{a-3}{2} f(1)$, so $f(a)=a f(1)$.
- If $a$ and $\frac{a+3}{2}$ have the same color, then $3 \frac{a+1}{2} f(1)=3 f\left(\frac{a+1}{2}\right)=f\left(\frac{3 a+3}{2}\right)=f(a)+f\left(\frac{a+3}{2}\right)=f(a)+$ $\frac{a+3}{2} f(1)$, so $f(a)=a f(1)$.
In the three cases we find a contradiction with $f(a) \neq a f(1)$, so it finishes the proof.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "# Problem 2",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "# Solution 2:",
"tier": "T2",
"year": "2017"
}
|
Find the smallest positive integer $k$ for which there exist a colouring of the positive integers $\mathbb{Z}_{>0}$ with $k$ colours and a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ with the following two properties:
(i) For all positive integers $m, n$ of the same colour, $f(m+n)=f(m)+f(n)$.
(ii) There are positive integers $m, n$ such that $f(m+n) \neq f(m)+f(n)$.
In a colouring of $\mathbb{Z}_{>0}$ with $k$ colours, every integer is coloured in exactly one of the $k$ colours. In both (i) and (ii) the positive integers $m, n$ are not necessarily different.
Merlijn Staps, the Netherlands
|
As before we prove that $k \leq 3$ and for any such function and colouring we have $f(2 n)=2 f(n)$.
Now we show that there is no such function and coloring for $k=2$.
Consider any coloring of $\mathbb{Z}_{>0}$ with 2 colors and any function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ satisfying conditions (i) and (ii). Say the two colors are white (W) and black (B). Pick $m, n$ any two integers such that $f(m+n)=f(m)+f(n)$. Without loss of generality we may assume that $m+n, m$ are black and $n$ is white.
Lemma 4. For all $l \in \mathbb{Z}_{>0}$ and every $x$ whose color is black, we have $x+\operatorname{lm}$ is black and $f(x+\operatorname{lm})=$ $f(x)+l f(m)$.
Proof. We proceed by induction. It is clearly true for $l=0$. If $x+\operatorname{lm}$ is black and satisfies $f(x+\operatorname{lm})=$ $f(x)+l f(m)$, then $f(x+(l+1) m)=f(x+l m)+f(m)=f(x)+(l+1) f(m)$ and $f(x+(l+1) m+n)=$ $f(x+l m)+f(m+n)=f(x)+l f(m)+f(m+n) \neq f(x)+(l+1) f(m)+f(n)=f(x+(l+1) m)+f(n)$, so $x+(l+1) m$ is not the same color of $n$, therefore $x+(l+1) m$ is black. Thjs completes the induction.
In particular we then must have that $2^{l} n$ is white for every $l$, because otherwise since $2^{l} m$ is black we would have $2^{l} f(m+n)=f\left(2^{l} m+2^{l} n\right)=f\left(2^{l} m\right)+f\left(2^{l} n\right)=2^{l}(f(m)+f(n))$, and consequently $f(m+n)=$ $f(m)+f(n)$.
Lemma 5. For every $l \geq 1,2^{l} m+2^{l-1} n$ is black.
Proof. On the one hand we have $2^{l} f(m+n)=f\left(2^{l} m+2^{l} n\right)=f\left(2^{l-1}(2 m+n)+2^{l-1} n\right)$. On the other hand we have
$\left.2^{l} f(m+n)=2^{l-1} \cdot 2 f(m+n) \neq 2^{l-1}(f(m+n)+f(m)+f(n))=2^{l-1}(f(2 m+n)+f(n))=f\left(2^{l} m+2^{l-1} n\right)\right)+f\left(2^{l-1} n\right)$.
Therefore $2^{l} m+2^{l-1} n$ and $2^{l-1} n$ have different color, which means $2^{l} m+2^{l-1} n$ is black.
Combining the two lemmas give $j m+2^{l-1} n$ is black for all $j \geq 2^{l}$ and every $l \geq 1$.
Now write $m=2^{l-1} m^{\prime}$ with $m^{\prime}$ odd. Let $t$ be a number such that $\frac{2^{t}-1}{m^{\prime}}$ is an integer and $j=\frac{2^{t}-1}{m^{\prime}} n \geq 2^{l}$, i.e. $t$ is some multiple of $\phi\left(m^{\prime}\right)$. Then we must have that $j m+2^{l-1} n$ is black, but by definition ${ }_{j m}^{m^{\prime}}+\overline{2}^{l-1} n=$ $\left(2^{t}-1\right) 2^{l-1} n+2^{l-1} n=2^{t+l-1} n$ is white. This is a contradiction, so $k=2$ is impossible.
|
3
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Find the smallest positive integer $k$ for which there exist a colouring of the positive integers $\mathbb{Z}_{>0}$ with $k$ colours and a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ with the following two properties:
(i) For all positive integers $m, n$ of the same colour, $f(m+n)=f(m)+f(n)$.
(ii) There are positive integers $m, n$ such that $f(m+n) \neq f(m)+f(n)$.
In a colouring of $\mathbb{Z}_{>0}$ with $k$ colours, every integer is coloured in exactly one of the $k$ colours. In both (i) and (ii) the positive integers $m, n$ are not necessarily different.
Merlijn Staps, the Netherlands
|
As before we prove that $k \leq 3$ and for any such function and colouring we have $f(2 n)=2 f(n)$.
Now we show that there is no such function and coloring for $k=2$.
Consider any coloring of $\mathbb{Z}_{>0}$ with 2 colors and any function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ satisfying conditions (i) and (ii). Say the two colors are white (W) and black (B). Pick $m, n$ any two integers such that $f(m+n)=f(m)+f(n)$. Without loss of generality we may assume that $m+n, m$ are black and $n$ is white.
Lemma 4. For all $l \in \mathbb{Z}_{>0}$ and every $x$ whose color is black, we have $x+\operatorname{lm}$ is black and $f(x+\operatorname{lm})=$ $f(x)+l f(m)$.
Proof. We proceed by induction. It is clearly true for $l=0$. If $x+\operatorname{lm}$ is black and satisfies $f(x+\operatorname{lm})=$ $f(x)+l f(m)$, then $f(x+(l+1) m)=f(x+l m)+f(m)=f(x)+(l+1) f(m)$ and $f(x+(l+1) m+n)=$ $f(x+l m)+f(m+n)=f(x)+l f(m)+f(m+n) \neq f(x)+(l+1) f(m)+f(n)=f(x+(l+1) m)+f(n)$, so $x+(l+1) m$ is not the same color of $n$, therefore $x+(l+1) m$ is black. Thjs completes the induction.
In particular we then must have that $2^{l} n$ is white for every $l$, because otherwise since $2^{l} m$ is black we would have $2^{l} f(m+n)=f\left(2^{l} m+2^{l} n\right)=f\left(2^{l} m\right)+f\left(2^{l} n\right)=2^{l}(f(m)+f(n))$, and consequently $f(m+n)=$ $f(m)+f(n)$.
Lemma 5. For every $l \geq 1,2^{l} m+2^{l-1} n$ is black.
Proof. On the one hand we have $2^{l} f(m+n)=f\left(2^{l} m+2^{l} n\right)=f\left(2^{l-1}(2 m+n)+2^{l-1} n\right)$. On the other hand we have
$\left.2^{l} f(m+n)=2^{l-1} \cdot 2 f(m+n) \neq 2^{l-1}(f(m+n)+f(m)+f(n))=2^{l-1}(f(2 m+n)+f(n))=f\left(2^{l} m+2^{l-1} n\right)\right)+f\left(2^{l-1} n\right)$.
Therefore $2^{l} m+2^{l-1} n$ and $2^{l-1} n$ have different color, which means $2^{l} m+2^{l-1} n$ is black.
Combining the two lemmas give $j m+2^{l-1} n$ is black for all $j \geq 2^{l}$ and every $l \geq 1$.
Now write $m=2^{l-1} m^{\prime}$ with $m^{\prime}$ odd. Let $t$ be a number such that $\frac{2^{t}-1}{m^{\prime}}$ is an integer and $j=\frac{2^{t}-1}{m^{\prime}} n \geq 2^{l}$, i.e. $t$ is some multiple of $\phi\left(m^{\prime}\right)$. Then we must have that $j m+2^{l-1} n$ is black, but by definition ${ }_{j m}^{m^{\prime}}+\overline{2}^{l-1} n=$ $\left(2^{t}-1\right) 2^{l-1} n+2^{l-1} n=2^{t+l-1} n$ is white. This is a contradiction, so $k=2$ is impossible.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "# Problem 2",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "# Solution 3:",
"tier": "T2",
"year": "2017"
}
|
There are 2017 lines in a plane such that no 3 of them go through the same point. Turbo the snail can slide along the lines in the following fashion: she initially moves on one of the lines and continues moving on a given line until she reaches an intersection of 2 lines. At the intersection, she follows her journey on the other line turning left or right, alternating the direction she chooses at each intersection point she passes. Can it happen that she slides through a line segment for a second time in her journey but in the opposite direction as she did for the first time?
Márk Di Giovanni, Hungary
|
We show that this is not possible.
The lines divide the plane into disjoint regions. We claim that there exists an alternating 2-coloring of these regions, that is each region can be colored in black or white, such that if two regions share a line segment, they have a different color. We show this inductively.
If there are no lines, this is obvious. Consider now an arrangement of $n$ lines in the plane and an alternating 2-coloring of the regions. If we add a line $g$, we can simply switch the colors of all regions in one of the half planes divided by $g$ from white to black and vice versa. Any line segment not in $g$ will still be between two regions of different color. Any line segment in $g$ cuts a region determined by the $n$ lines in two, and since we switched colors on one side of $g$ this segment will also lie between two regions of different color.
Now without loss of generality we may assume, that Turbo starts on a line segment with a white region on her left and a black one on her right. At any intersection, if she turns right, she will keep the black tile to her right. If she turns left, she will keep the white tile to her left. Thus wherever she goes, she will always have a white tile on her left and a black tile on her right. As a consequence, she can cross every line segment only in the direction where she has a white tile to her left, and never the opposite direction where she would have a black tile to the left.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
There are 2017 lines in a plane such that no 3 of them go through the same point. Turbo the snail can slide along the lines in the following fashion: she initially moves on one of the lines and continues moving on a given line until she reaches an intersection of 2 lines. At the intersection, she follows her journey on the other line turning left or right, alternating the direction she chooses at each intersection point she passes. Can it happen that she slides through a line segment for a second time in her journey but in the opposite direction as she did for the first time?
Márk Di Giovanni, Hungary
|
We show that this is not possible.
The lines divide the plane into disjoint regions. We claim that there exists an alternating 2-coloring of these regions, that is each region can be colored in black or white, such that if two regions share a line segment, they have a different color. We show this inductively.
If there are no lines, this is obvious. Consider now an arrangement of $n$ lines in the plane and an alternating 2-coloring of the regions. If we add a line $g$, we can simply switch the colors of all regions in one of the half planes divided by $g$ from white to black and vice versa. Any line segment not in $g$ will still be between two regions of different color. Any line segment in $g$ cuts a region determined by the $n$ lines in two, and since we switched colors on one side of $g$ this segment will also lie between two regions of different color.
Now without loss of generality we may assume, that Turbo starts on a line segment with a white region on her left and a black one on her right. At any intersection, if she turns right, she will keep the black tile to her right. If she turns left, she will keep the white tile to her left. Thus wherever she goes, she will always have a white tile on her left and a black tile on her right. As a consequence, she can cross every line segment only in the direction where she has a white tile to her left, and never the opposite direction where she would have a black tile to the left.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "# Problem 3",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "\n## 1. Solution",
"tier": "T2",
"year": "2017"
}
|
There are 2017 lines in a plane such that no 3 of them go through the same point. Turbo the snail can slide along the lines in the following fashion: she initially moves on one of the lines and continues moving on a given line until she reaches an intersection of 2 lines. At the intersection, she follows her journey on the other line turning left or right, alternating the direction she chooses at each intersection point she passes. Can it happen that she slides through a line segment for a second time in her journey but in the opposite direction as she did for the first time?
Márk Di Giovanni, Hungary
|
Suppose the assumption is true.
Let's label each segment in the snail's path with $\mathbf{L}$ or $\mathbf{R}$ depending on the direction that Turbo chose at the start point of this segment (one segment can have several labels if it has been visited several times).
Consider the first segment that has been visited twice in different directions, name this segment $s_{1}$. Assume without loss of generality that it is labeled with $\mathbf{L}$. Then next segment must be labeled with $\mathbf{R}$, name this one $s_{2}$.
Let's look at the label which $s_{1}$ can get on the second visit. If it gets $\mathbf{L}$ then the previous segment in the path must be $s_{2}$. But in this case $s_{1}$ is not the first segment that has been visited twice in different directions because $s_{2}$ has been visited earlier. So the second label of $s_{1}$ must be $\mathbf{R}$, and Turbo must have come from the opposite side of $s_{2}$.
Since Turbo alters directions at each point, labels in her path also alter. And because two labels of $s_{1}$ are different, the number of visited segments between these two visits must be even.
Now let's make the following observation: each segment in the path corresponds to exactly one line, and its previous and next segments are on opposite sides of this line.
Again consider the path between two visits of $s_{1}$.
Each line intersecting this path must be crossed an even number of times because Turbo has to return to the initial side of each line. Therefore, an even number of segments of Turbo's path are contained on each of these lines. But the line containing $s_{1}$ must be crossed an odd number times. Since each crossing corresponds to exactly one segment in the path, the number of segments must be odd.
Here we get the contradiction. Therefore, the assumption is false.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
There are 2017 lines in a plane such that no 3 of them go through the same point. Turbo the snail can slide along the lines in the following fashion: she initially moves on one of the lines and continues moving on a given line until she reaches an intersection of 2 lines. At the intersection, she follows her journey on the other line turning left or right, alternating the direction she chooses at each intersection point she passes. Can it happen that she slides through a line segment for a second time in her journey but in the opposite direction as she did for the first time?
Márk Di Giovanni, Hungary
|
Suppose the assumption is true.
Let's label each segment in the snail's path with $\mathbf{L}$ or $\mathbf{R}$ depending on the direction that Turbo chose at the start point of this segment (one segment can have several labels if it has been visited several times).
Consider the first segment that has been visited twice in different directions, name this segment $s_{1}$. Assume without loss of generality that it is labeled with $\mathbf{L}$. Then next segment must be labeled with $\mathbf{R}$, name this one $s_{2}$.
Let's look at the label which $s_{1}$ can get on the second visit. If it gets $\mathbf{L}$ then the previous segment in the path must be $s_{2}$. But in this case $s_{1}$ is not the first segment that has been visited twice in different directions because $s_{2}$ has been visited earlier. So the second label of $s_{1}$ must be $\mathbf{R}$, and Turbo must have come from the opposite side of $s_{2}$.
Since Turbo alters directions at each point, labels in her path also alter. And because two labels of $s_{1}$ are different, the number of visited segments between these two visits must be even.
Now let's make the following observation: each segment in the path corresponds to exactly one line, and its previous and next segments are on opposite sides of this line.
Again consider the path between two visits of $s_{1}$.
Each line intersecting this path must be crossed an even number of times because Turbo has to return to the initial side of each line. Therefore, an even number of segments of Turbo's path are contained on each of these lines. But the line containing $s_{1}$ must be crossed an odd number times. Since each crossing corresponds to exactly one segment in the path, the number of segments must be odd.
Here we get the contradiction. Therefore, the assumption is false.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "# Problem 3",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "\n## 2. Solution",
"tier": "T2",
"year": "2017"
}
|
There are 2017 lines in a plane such that no 3 of them go through the same point. Turbo the snail can slide along the lines in the following fashion: she initially moves on one of the lines and continues moving on a given line until she reaches an intersection of 2 lines. At the intersection, she follows her journey on the other line turning left or right, alternating the direction she chooses at each intersection point she passes. Can it happen that she slides through a line segment for a second time in her journey but in the opposite direction as she did for the first time?
Márk Di Giovanni, Hungary
|
Suppose that the snail always slides slightly to the right of the line segments on her path. When turning to the right, she does not cross any line, whereas when turning to the left, she crosses exactly two lines. This means that at any time of her journey, she has crossed an even number of lines.
Assuming that at some point she slides along a segment for the second time, but in the opposite direction, we argue that she needs to cross an odd number of lines. Let $\ell$ be the line on which the revisit happens. In order to get to the other side of $\ell$, the snail has to cross $\ell$ an odd number of times. To visit the same segment of $\ell$, she must cross every other line an even number of times.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
There are 2017 lines in a plane such that no 3 of them go through the same point. Turbo the snail can slide along the lines in the following fashion: she initially moves on one of the lines and continues moving on a given line until she reaches an intersection of 2 lines. At the intersection, she follows her journey on the other line turning left or right, alternating the direction she chooses at each intersection point she passes. Can it happen that she slides through a line segment for a second time in her journey but in the opposite direction as she did for the first time?
Márk Di Giovanni, Hungary
|
Suppose that the snail always slides slightly to the right of the line segments on her path. When turning to the right, she does not cross any line, whereas when turning to the left, she crosses exactly two lines. This means that at any time of her journey, she has crossed an even number of lines.
Assuming that at some point she slides along a segment for the second time, but in the opposite direction, we argue that she needs to cross an odd number of lines. Let $\ell$ be the line on which the revisit happens. In order to get to the other side of $\ell$, the snail has to cross $\ell$ an odd number of times. To visit the same segment of $\ell$, she must cross every other line an even number of times.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "# Problem 3",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "\n## 3. Solution",
"tier": "T2",
"year": "2017"
}
|
There are 2017 lines in a plane such that no 3 of them go through the same point. Turbo the snail can slide along the lines in the following fashion: she initially moves on one of the lines and continues moving on a given line until she reaches an intersection of 2 lines. At the intersection, she follows her journey on the other line turning left or right, alternating the direction she chooses at each intersection point she passes. Can it happen that she slides through a line segment for a second time in her journey but in the opposite direction as she did for the first time?
Márk Di Giovanni, Hungary
|
Let us color in red all intersection points of the given lines and let us choose one of two possible directions on each segment (draw an arrow on each segment). Consider a red point $R$ where two given lines $a$ and $b$ meet, and the four segments $a_{1}, a_{2}, b_{1}, b_{2}$ with endpoint $R$ (so that $a_{i} \subset a, b_{j} \subset b$ ). $R$ is called a saddle if on $a_{1}, a_{2}$ the arrows go out of $R$ while on $b_{1}, b_{2}$ the arrows enter $R$, or visa versa, on $b_{1}, b_{2}$ the arrows go out of $R$ while on $a_{1}, a_{2}$ the arrows enter $R$. The set of arrows (chosen on all segments) is said to be good if all red points are saddles. It is sufficient to prove that there exists a good set of arrows. Indeed, if initially Turbo is moving along (or opposite) the arrow, then this condition holds after she turns at a red point.
The given lines cut the plane into regions. Further we need the following property of the good set of arrows (this property directly follows from the definition): the boundary of any bounded region is a directed cycle of arrows; the boundary of any unbounded region is a directed chain of arrows.
We construct a good set of arrows by induction on $n$ with trivial base $n=1$. Now erase one of $n$ given lines and assume we have a good set of arrows for remaining $n-1$ lines. Now restore the $n$-th line $\ell$, assume that $\ell$ is horizontal. Denote by $A_{1}, \ldots, A_{n-1}$ all new red points on $\ell$ from the left to the right. Each of $A_{i}$ belongs to some old segment $m_{i}$ of the line $\ell_{i}$. Let us call $A_{i}$ ascending if the arrow on $m_{i}$ goes up, and descending if the arrow on $m_{i}$ goes down. Consider the region containing the segment $A_{i} A_{i+1}$. By the property, $A_{i}$ and $A_{i+1}$ can not be both ascending or both descending. Thus we can choose arrows on all pieces of $\ell$ so that each arrow goes from a descending to an ascending vertex.
Each of points $A_{i}$ cuts $m_{i}$ into two new pieces; the direction of new pieces supposed to be the same as on $m_{i}$. Now simultaneously change the direction of arrows on all pieces below the line $\ell$. It is easy to see that $A_{1}, \ldots, A_{n-1}$ become saddles, while the other red points remain saddles. This completes the induction step.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
There are 2017 lines in a plane such that no 3 of them go through the same point. Turbo the snail can slide along the lines in the following fashion: she initially moves on one of the lines and continues moving on a given line until she reaches an intersection of 2 lines. At the intersection, she follows her journey on the other line turning left or right, alternating the direction she chooses at each intersection point she passes. Can it happen that she slides through a line segment for a second time in her journey but in the opposite direction as she did for the first time?
Márk Di Giovanni, Hungary
|
Let us color in red all intersection points of the given lines and let us choose one of two possible directions on each segment (draw an arrow on each segment). Consider a red point $R$ where two given lines $a$ and $b$ meet, and the four segments $a_{1}, a_{2}, b_{1}, b_{2}$ with endpoint $R$ (so that $a_{i} \subset a, b_{j} \subset b$ ). $R$ is called a saddle if on $a_{1}, a_{2}$ the arrows go out of $R$ while on $b_{1}, b_{2}$ the arrows enter $R$, or visa versa, on $b_{1}, b_{2}$ the arrows go out of $R$ while on $a_{1}, a_{2}$ the arrows enter $R$. The set of arrows (chosen on all segments) is said to be good if all red points are saddles. It is sufficient to prove that there exists a good set of arrows. Indeed, if initially Turbo is moving along (or opposite) the arrow, then this condition holds after she turns at a red point.
The given lines cut the plane into regions. Further we need the following property of the good set of arrows (this property directly follows from the definition): the boundary of any bounded region is a directed cycle of arrows; the boundary of any unbounded region is a directed chain of arrows.
We construct a good set of arrows by induction on $n$ with trivial base $n=1$. Now erase one of $n$ given lines and assume we have a good set of arrows for remaining $n-1$ lines. Now restore the $n$-th line $\ell$, assume that $\ell$ is horizontal. Denote by $A_{1}, \ldots, A_{n-1}$ all new red points on $\ell$ from the left to the right. Each of $A_{i}$ belongs to some old segment $m_{i}$ of the line $\ell_{i}$. Let us call $A_{i}$ ascending if the arrow on $m_{i}$ goes up, and descending if the arrow on $m_{i}$ goes down. Consider the region containing the segment $A_{i} A_{i+1}$. By the property, $A_{i}$ and $A_{i+1}$ can not be both ascending or both descending. Thus we can choose arrows on all pieces of $\ell$ so that each arrow goes from a descending to an ascending vertex.
Each of points $A_{i}$ cuts $m_{i}$ into two new pieces; the direction of new pieces supposed to be the same as on $m_{i}$. Now simultaneously change the direction of arrows on all pieces below the line $\ell$. It is easy to see that $A_{1}, \ldots, A_{n-1}$ become saddles, while the other red points remain saddles. This completes the induction step.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "# Problem 3",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "\n## 4. Solution",
"tier": "T2",
"year": "2017"
}
|
Let $n \geq 1$ be an integer and let $t_{1}<t_{2}<\ldots<t_{n}$ be positive integers. In a group of $t_{n}+1$ people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following conditions to hold at the same time:
i) The number of games played by each person is one of $t_{1}, t_{2}, \ldots, t_{n}$,
ii) For every $i$ with $1 \leq i \leq n$, there is someone who has played exactly $t_{i}$ games of chess.
Gerhard Wöginger, Luxembourg
## Comment
In graph theory terms the problem is to prove that for any finite nonempty set $\mathcal{T}$ of positive integers there exists a graph of size $\max \mathcal{T}+1$ such that the degree set of the graph is equal to $\mathcal{T}$.
Among graph theory specialists a generalization of this problem is known [1]. Nevertheless, the problem still suited the contest.
|
Let $\mathcal{T}=\left\{t_{1}, \ldots, t_{n}\right\}$. The proof proceeds by induction on $n=|\mathcal{T}|$. If $n=1$ and $\mathcal{T}=\{t\}$, choose a group of $t+1$ people and let every pair of two persons play against each other. Then every person has played $t$ games and the conditions of the problem are satisfied.
In the inductive step, suppose that $\mathcal{T}$ has $n \geq 2$ elements $t_{1}<t_{2}<\cdots<t_{n}$. Consider the set
$$
\mathcal{T}^{\prime}=\left\{t_{n}-t_{n-1}, t_{n}-t_{n-2}, \ldots, t_{n}-t_{1}\right\} .
$$
By the inductive hypothesis, there exists a group $G^{\prime}$ of $t_{n}-t_{1}+1$ people that satisfies the conditions of the problem for $T^{\prime}$.
Next construct a group $G^{\prime \prime}$ of $t_{n}+1$ people by adding $t_{1}$ people who do not know any of the other $t_{n}-t_{1}+1$ people in $G^{\prime}$. Finally, construct a group $G$ by complementing the knowledge relation in $G^{\prime \prime}$ : two persons play against each other in $G$ if and only if they do not play against each other in $G^{\prime \prime}$.
By construction $t \in \mathcal{T}$ if and only if there exists a person in $G^{\prime \prime}$ that played against exactly $t_{n}-t$ other people (if $t=t_{n}$, choose one of the $t_{1}$ people added to $\left.G^{\prime}\right)$. That person knows $t_{n}-\left(t_{n}-t\right)=t$ other students in $G$, completing the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n \geq 1$ be an integer and let $t_{1}<t_{2}<\ldots<t_{n}$ be positive integers. In a group of $t_{n}+1$ people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following conditions to hold at the same time:
i) The number of games played by each person is one of $t_{1}, t_{2}, \ldots, t_{n}$,
ii) For every $i$ with $1 \leq i \leq n$, there is someone who has played exactly $t_{i}$ games of chess.
Gerhard Wöginger, Luxembourg
## Comment
In graph theory terms the problem is to prove that for any finite nonempty set $\mathcal{T}$ of positive integers there exists a graph of size $\max \mathcal{T}+1$ such that the degree set of the graph is equal to $\mathcal{T}$.
Among graph theory specialists a generalization of this problem is known [1]. Nevertheless, the problem still suited the contest.
|
Let $\mathcal{T}=\left\{t_{1}, \ldots, t_{n}\right\}$. The proof proceeds by induction on $n=|\mathcal{T}|$. If $n=1$ and $\mathcal{T}=\{t\}$, choose a group of $t+1$ people and let every pair of two persons play against each other. Then every person has played $t$ games and the conditions of the problem are satisfied.
In the inductive step, suppose that $\mathcal{T}$ has $n \geq 2$ elements $t_{1}<t_{2}<\cdots<t_{n}$. Consider the set
$$
\mathcal{T}^{\prime}=\left\{t_{n}-t_{n-1}, t_{n}-t_{n-2}, \ldots, t_{n}-t_{1}\right\} .
$$
By the inductive hypothesis, there exists a group $G^{\prime}$ of $t_{n}-t_{1}+1$ people that satisfies the conditions of the problem for $T^{\prime}$.
Next construct a group $G^{\prime \prime}$ of $t_{n}+1$ people by adding $t_{1}$ people who do not know any of the other $t_{n}-t_{1}+1$ people in $G^{\prime}$. Finally, construct a group $G$ by complementing the knowledge relation in $G^{\prime \prime}$ : two persons play against each other in $G$ if and only if they do not play against each other in $G^{\prime \prime}$.
By construction $t \in \mathcal{T}$ if and only if there exists a person in $G^{\prime \prime}$ that played against exactly $t_{n}-t$ other people (if $t=t_{n}$, choose one of the $t_{1}$ people added to $\left.G^{\prime}\right)$. That person knows $t_{n}-\left(t_{n}-t\right)=t$ other students in $G$, completing the proof.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "# Problem 4",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "\n## 1. Solution (see also [2])",
"tier": "T2",
"year": "2017"
}
|
Let $n \geq 1$ be an integer and let $t_{1}<t_{2}<\ldots<t_{n}$ be positive integers. In a group of $t_{n}+1$ people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following conditions to hold at the same time:
i) The number of games played by each person is one of $t_{1}, t_{2}, \ldots, t_{n}$,
ii) For every $i$ with $1 \leq i \leq n$, there is someone who has played exactly $t_{i}$ games of chess.
Gerhard Wöginger, Luxembourg
## Comment
In graph theory terms the problem is to prove that for any finite nonempty set $\mathcal{T}$ of positive integers there exists a graph of size $\max \mathcal{T}+1$ such that the degree set of the graph is equal to $\mathcal{T}$.
Among graph theory specialists a generalization of this problem is known [1]. Nevertheless, the problem still suited the contest.
|
Let $\mathcal{T}=\left\{t_{1}, \ldots, t_{n}\right\}$. The proof proceeds by induction on $n=|\mathcal{T}|$. If $n=1$ and $\mathcal{T}=\{t\}$, we choose a group of $t+1$ people such that everyone plays with
everyone else. If $n=2$ and $\mathcal{T}=\left\{t_{1}, t_{2}\right\}$ with $t_{1}<t_{2}$, divide the $t_{2}+1$ people into groups $A$ resp. $B$ of size $t_{1}$ resp. $t_{2}-t_{1}+1$ such that everyone from group $A$ played with everyone else whereas people from group $B$ only played with the people from group $A$. Then the people from group $A$ resp. $B$ played with exactly $t_{2}$ resp. $t_{1}$ other people. In the inductive step, suppose that $T$ has $n>2$ elements $t_{1}<\ldots<t_{n}$. Consider the set
$$
\mathcal{T}^{\prime}=\left(\mathcal{T} \backslash\left\{t_{1}, t_{n}\right\}\right)-t_{1}=\left\{t_{n-1}-t_{1}, t_{n-1}-t_{1}, \ldots, t_{2}-t_{1}\right\}
$$
By the induction hypothesis there exists a group $C$ of $t_{n-1}-t_{1}+1$ people that satisfies the conditions of the problem for $T^{\prime}$. Next add groups $D$ resp. $E$ of $t_{1}$ resp. $t_{n}-t_{n-1}$ people such that people from group $D$ played with everyone else whereas people from group $E$ only played with the people from group $D$. Then the people from group $C$ played with $t$ other people if and only if they played with $t-t_{1}$ many people among $C$, i.e. if and only if $t \in\left\{t_{2}, \ldots, t_{n-1}\right\}=$ $\mathcal{T} \backslash\left\{t_{1}, t_{n}\right\}$. People from group $D$ resp. $E$ played with $t_{n}$ resp $t_{1}$ peoples, which completes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n \geq 1$ be an integer and let $t_{1}<t_{2}<\ldots<t_{n}$ be positive integers. In a group of $t_{n}+1$ people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following conditions to hold at the same time:
i) The number of games played by each person is one of $t_{1}, t_{2}, \ldots, t_{n}$,
ii) For every $i$ with $1 \leq i \leq n$, there is someone who has played exactly $t_{i}$ games of chess.
Gerhard Wöginger, Luxembourg
## Comment
In graph theory terms the problem is to prove that for any finite nonempty set $\mathcal{T}$ of positive integers there exists a graph of size $\max \mathcal{T}+1$ such that the degree set of the graph is equal to $\mathcal{T}$.
Among graph theory specialists a generalization of this problem is known [1]. Nevertheless, the problem still suited the contest.
|
Let $\mathcal{T}=\left\{t_{1}, \ldots, t_{n}\right\}$. The proof proceeds by induction on $n=|\mathcal{T}|$. If $n=1$ and $\mathcal{T}=\{t\}$, we choose a group of $t+1$ people such that everyone plays with
everyone else. If $n=2$ and $\mathcal{T}=\left\{t_{1}, t_{2}\right\}$ with $t_{1}<t_{2}$, divide the $t_{2}+1$ people into groups $A$ resp. $B$ of size $t_{1}$ resp. $t_{2}-t_{1}+1$ such that everyone from group $A$ played with everyone else whereas people from group $B$ only played with the people from group $A$. Then the people from group $A$ resp. $B$ played with exactly $t_{2}$ resp. $t_{1}$ other people. In the inductive step, suppose that $T$ has $n>2$ elements $t_{1}<\ldots<t_{n}$. Consider the set
$$
\mathcal{T}^{\prime}=\left(\mathcal{T} \backslash\left\{t_{1}, t_{n}\right\}\right)-t_{1}=\left\{t_{n-1}-t_{1}, t_{n-1}-t_{1}, \ldots, t_{2}-t_{1}\right\}
$$
By the induction hypothesis there exists a group $C$ of $t_{n-1}-t_{1}+1$ people that satisfies the conditions of the problem for $T^{\prime}$. Next add groups $D$ resp. $E$ of $t_{1}$ resp. $t_{n}-t_{n-1}$ people such that people from group $D$ played with everyone else whereas people from group $E$ only played with the people from group $D$. Then the people from group $C$ played with $t$ other people if and only if they played with $t-t_{1}$ many people among $C$, i.e. if and only if $t \in\left\{t_{2}, \ldots, t_{n-1}\right\}=$ $\mathcal{T} \backslash\left\{t_{1}, t_{n}\right\}$. People from group $D$ resp. $E$ played with $t_{n}$ resp $t_{1}$ peoples, which completes the proof.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "# Problem 4",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "\n## 2. Solution",
"tier": "T2",
"year": "2017"
}
|
Let $n \geq 1$ be an integer and let $t_{1}<t_{2}<\ldots<t_{n}$ be positive integers. In a group of $t_{n}+1$ people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following conditions to hold at the same time:
i) The number of games played by each person is one of $t_{1}, t_{2}, \ldots, t_{n}$,
ii) For every $i$ with $1 \leq i \leq n$, there is someone who has played exactly $t_{i}$ games of chess.
Gerhard Wöginger, Luxembourg
## Comment
In graph theory terms the problem is to prove that for any finite nonempty set $\mathcal{T}$ of positive integers there exists a graph of size $\max \mathcal{T}+1$ such that the degree set of the graph is equal to $\mathcal{T}$.
Among graph theory specialists a generalization of this problem is known [1]. Nevertheless, the problem still suited the contest.
|
The proof proceeds by induction on $\left|t_{n}\right|$. If $t_{n}=1$ we have $n=1$ and we can consider two persons that play against each other. Then every player has played 1 game and the conditions of the problem are satisfied. If $t_{n}>1$ we distinguish the two cases $t_{1}>1$ and $t_{1}=1$. If $t_{1}>1$ there exists, by the induction hypothesis, a group $A$ of size $t_{n}$ that satisfies the conditions of the problem for $t_{1}^{\prime}=t_{1}-1, \ldots, t_{n}^{\prime}=t_{n}-1$. Now add a new person to $A$ and let him/her play against everyone from $A$. The new group will be of size $t_{n}+1$ and there exists a person which has played $t$ games if and only if there exists a person that has played $t-1$ games within $A$, i.e. if and only if $t \in\left\{t_{1}, \ldots, t_{n}\right\}$. Hence the conditions of the problem are satisfied.
If $t_{1}=1$ there exists, by the induction hypothesis, a group $B$ of size $t_{n-1}$ that satisfies the conditions of the problem for $t_{1}^{\prime}=t_{2}-1, \ldots, t_{n-2}^{\prime}=t_{n-1}-1$. Now add a new person $P$ and let him/her play with everyone from group $B$ and a group $C$ of size $t_{n}-t_{n-1}>0$ and let them play with $P$. The new group will be of size $t_{n-1}+1+\left(t_{n}-t_{n-1}\right)+1=t_{n}+1$. Since person $P$ has played against everyone he will have played $t_{n}$ games. The people in $C$ will have played $1=t_{1}$ games. There exists a person in $B$ that has played $t$ games if and only if there exist a person in $B$ that has played $t-1$ games within $B$, i.e. if and only if $t \in\left\{t_{2}, \ldots, t_{n-1}\right\}$. Hence the conditions of the problem are satisfied.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n \geq 1$ be an integer and let $t_{1}<t_{2}<\ldots<t_{n}$ be positive integers. In a group of $t_{n}+1$ people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following conditions to hold at the same time:
i) The number of games played by each person is one of $t_{1}, t_{2}, \ldots, t_{n}$,
ii) For every $i$ with $1 \leq i \leq n$, there is someone who has played exactly $t_{i}$ games of chess.
Gerhard Wöginger, Luxembourg
## Comment
In graph theory terms the problem is to prove that for any finite nonempty set $\mathcal{T}$ of positive integers there exists a graph of size $\max \mathcal{T}+1$ such that the degree set of the graph is equal to $\mathcal{T}$.
Among graph theory specialists a generalization of this problem is known [1]. Nevertheless, the problem still suited the contest.
|
The proof proceeds by induction on $\left|t_{n}\right|$. If $t_{n}=1$ we have $n=1$ and we can consider two persons that play against each other. Then every player has played 1 game and the conditions of the problem are satisfied. If $t_{n}>1$ we distinguish the two cases $t_{1}>1$ and $t_{1}=1$. If $t_{1}>1$ there exists, by the induction hypothesis, a group $A$ of size $t_{n}$ that satisfies the conditions of the problem for $t_{1}^{\prime}=t_{1}-1, \ldots, t_{n}^{\prime}=t_{n}-1$. Now add a new person to $A$ and let him/her play against everyone from $A$. The new group will be of size $t_{n}+1$ and there exists a person which has played $t$ games if and only if there exists a person that has played $t-1$ games within $A$, i.e. if and only if $t \in\left\{t_{1}, \ldots, t_{n}\right\}$. Hence the conditions of the problem are satisfied.
If $t_{1}=1$ there exists, by the induction hypothesis, a group $B$ of size $t_{n-1}$ that satisfies the conditions of the problem for $t_{1}^{\prime}=t_{2}-1, \ldots, t_{n-2}^{\prime}=t_{n-1}-1$. Now add a new person $P$ and let him/her play with everyone from group $B$ and a group $C$ of size $t_{n}-t_{n-1}>0$ and let them play with $P$. The new group will be of size $t_{n-1}+1+\left(t_{n}-t_{n-1}\right)+1=t_{n}+1$. Since person $P$ has played against everyone he will have played $t_{n}$ games. The people in $C$ will have played $1=t_{1}$ games. There exists a person in $B$ that has played $t$ games if and only if there exist a person in $B$ that has played $t-1$ games within $B$, i.e. if and only if $t \in\left\{t_{2}, \ldots, t_{n-1}\right\}$. Hence the conditions of the problem are satisfied.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "# Problem 4",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "\n## 3. Solution",
"tier": "T2",
"year": "2017"
}
|
Let $n \geq 1$ be an integer and let $t_{1}<t_{2}<\ldots<t_{n}$ be positive integers. In a group of $t_{n}+1$ people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following conditions to hold at the same time:
i) The number of games played by each person is one of $t_{1}, t_{2}, \ldots, t_{n}$,
ii) For every $i$ with $1 \leq i \leq n$, there is someone who has played exactly $t_{i}$ games of chess.
Gerhard Wöginger, Luxembourg
## Comment
In graph theory terms the problem is to prove that for any finite nonempty set $\mathcal{T}$ of positive integers there exists a graph of size $\max \mathcal{T}+1$ such that the degree set of the graph is equal to $\mathcal{T}$.
Among graph theory specialists a generalization of this problem is known [1]. Nevertheless, the problem still suited the contest.
|
We generalize the construction for $\mathcal{T}=\{1, \ldots, n\}$
## Construction
Take sets of people $A_{1}, \ldots, A_{n}$. Let all people of $A_{i}$ play chess with all people in $A_{j}$ with $j \geq n-i+1$
Now the number of games played by anyone in $A_{i}$ is
$\left(\sum_{j \geq n-i+1}\left|A_{j}\right|\right)$ or $\left(\sum_{j \geq n-i+1}\left|A_{j}\right|\right)-1$ if $i \geq n-i+1$.
Now if we start with one person in each $A_{i}$ and two people in $A_{\left\lceil\frac{n}{2}\right\rceil}$. The number of played games for anyone in $A_{i}$ is equal to $i$. In particular this is a construction for $\mathcal{T}=\{1, . ., n\}$
Now to get to numbers of general sets $\mathcal{T}$ of size $n$ we can change the sizes of $A_{i}$ but keep the construction.
## Variant 1
Observation 1 Adding a person to a set $A_{i}$ increases the number of games played in $A_{j}$ for $j \geq n-i+1$, by exactly one.
Start with the construction above and then add $t_{1}-1$ people to group $A_{n}$, making the new set of games played equal to $\left\{t_{1}, t_{1}+1, \ldots, n+t_{1}-1\right\}$. Then add $t_{2}-t_{1}-1$ to $A_{n-1}$ to get set of games played to $\left\{t_{1}, t_{2}, t_{2}+1, \ldots, n+t_{2}-2\right\}$ and repeat until we get to the set $\mathcal{T}$ adding a total of $\sum_{j=1}^{n} t_{j}-t_{j-1}-1=t_{n}-n$ people (let $t_{0}=0$ ), so we get $t_{n}+1$ people in the end.
Clearly we can start by adding vertices to $A_{1}$ or any other set instead of $A_{n}$ first and obtain an equivalent construction with the same number of people.
## Variant 2
It is also possible to calculate the necessary sizes of $A_{i}$ 's all at once. We have by construction the number of games played in $A_{1}$ is less than the number of games played in $A_{2}$ etc. So we have that in the end we want the games played in $A_{i}$ to be exactly $t_{i}$.
So $\left(t_{t}, t_{2}, t_{3}, \ldots, t_{n}\right) \stackrel{!}{=}\left(\left|A_{n}\right|,\left|A_{n}\right|+\left|A_{n-1}\right|, \ldots,\left(\sum_{j=2}^{n}\left|A_{j}\right|\right)-1,\left(\sum_{j=1}^{n}\left|A_{j}\right|\right)-\right.$ 1).
This gives us by induction that $\left|A_{n}\right| \stackrel{!}{=} t_{1},\left|A_{n-1}\right| \stackrel{!}{=} t_{2}-t_{1}, \ldots,\left|A_{\left\lceil\frac{n}{2}\right\rceil}\right| \stackrel{!}{=}$ $t_{n-\left\lceil\frac{n}{2}\right\rceil+1}-t_{n-\left\lceil\frac{n}{2}\right\rceil}+1, \ldots,\left|A_{1}\right| \stackrel{!}{=} t_{n}-t_{n-1}$ and a quick calculation shows that the sum of all sets is exactly $1+\sum_{j=1}^{n}\left(t_{j}-t_{j-1}\right)=t_{n}+1$. (where the +1 comes from the set $A_{\left\lceil\frac{n}{2}\right\rceil}$ and $t_{0}=0$.)
## References
[1] Timothy A. Sipka. The orders of graphs with prescribed degree sets. Journal of Graph Theory, 4(3):301-307, 1980.
[2] Amitabha Tripathi and Sujith Vijay. A short proof of a theorem on degree sets of graphs. Discrete Appl. Math., 155(5):670-671, 2007.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n \geq 1$ be an integer and let $t_{1}<t_{2}<\ldots<t_{n}$ be positive integers. In a group of $t_{n}+1$ people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following conditions to hold at the same time:
i) The number of games played by each person is one of $t_{1}, t_{2}, \ldots, t_{n}$,
ii) For every $i$ with $1 \leq i \leq n$, there is someone who has played exactly $t_{i}$ games of chess.
Gerhard Wöginger, Luxembourg
## Comment
In graph theory terms the problem is to prove that for any finite nonempty set $\mathcal{T}$ of positive integers there exists a graph of size $\max \mathcal{T}+1$ such that the degree set of the graph is equal to $\mathcal{T}$.
Among graph theory specialists a generalization of this problem is known [1]. Nevertheless, the problem still suited the contest.
|
We generalize the construction for $\mathcal{T}=\{1, \ldots, n\}$
## Construction
Take sets of people $A_{1}, \ldots, A_{n}$. Let all people of $A_{i}$ play chess with all people in $A_{j}$ with $j \geq n-i+1$
Now the number of games played by anyone in $A_{i}$ is
$\left(\sum_{j \geq n-i+1}\left|A_{j}\right|\right)$ or $\left(\sum_{j \geq n-i+1}\left|A_{j}\right|\right)-1$ if $i \geq n-i+1$.
Now if we start with one person in each $A_{i}$ and two people in $A_{\left\lceil\frac{n}{2}\right\rceil}$. The number of played games for anyone in $A_{i}$ is equal to $i$. In particular this is a construction for $\mathcal{T}=\{1, . ., n\}$
Now to get to numbers of general sets $\mathcal{T}$ of size $n$ we can change the sizes of $A_{i}$ but keep the construction.
## Variant 1
Observation 1 Adding a person to a set $A_{i}$ increases the number of games played in $A_{j}$ for $j \geq n-i+1$, by exactly one.
Start with the construction above and then add $t_{1}-1$ people to group $A_{n}$, making the new set of games played equal to $\left\{t_{1}, t_{1}+1, \ldots, n+t_{1}-1\right\}$. Then add $t_{2}-t_{1}-1$ to $A_{n-1}$ to get set of games played to $\left\{t_{1}, t_{2}, t_{2}+1, \ldots, n+t_{2}-2\right\}$ and repeat until we get to the set $\mathcal{T}$ adding a total of $\sum_{j=1}^{n} t_{j}-t_{j-1}-1=t_{n}-n$ people (let $t_{0}=0$ ), so we get $t_{n}+1$ people in the end.
Clearly we can start by adding vertices to $A_{1}$ or any other set instead of $A_{n}$ first and obtain an equivalent construction with the same number of people.
## Variant 2
It is also possible to calculate the necessary sizes of $A_{i}$ 's all at once. We have by construction the number of games played in $A_{1}$ is less than the number of games played in $A_{2}$ etc. So we have that in the end we want the games played in $A_{i}$ to be exactly $t_{i}$.
So $\left(t_{t}, t_{2}, t_{3}, \ldots, t_{n}\right) \stackrel{!}{=}\left(\left|A_{n}\right|,\left|A_{n}\right|+\left|A_{n-1}\right|, \ldots,\left(\sum_{j=2}^{n}\left|A_{j}\right|\right)-1,\left(\sum_{j=1}^{n}\left|A_{j}\right|\right)-\right.$ 1).
This gives us by induction that $\left|A_{n}\right| \stackrel{!}{=} t_{1},\left|A_{n-1}\right| \stackrel{!}{=} t_{2}-t_{1}, \ldots,\left|A_{\left\lceil\frac{n}{2}\right\rceil}\right| \stackrel{!}{=}$ $t_{n-\left\lceil\frac{n}{2}\right\rceil+1}-t_{n-\left\lceil\frac{n}{2}\right\rceil}+1, \ldots,\left|A_{1}\right| \stackrel{!}{=} t_{n}-t_{n-1}$ and a quick calculation shows that the sum of all sets is exactly $1+\sum_{j=1}^{n}\left(t_{j}-t_{j-1}\right)=t_{n}+1$. (where the +1 comes from the set $A_{\left\lceil\frac{n}{2}\right\rceil}$ and $t_{0}=0$.)
## References
[1] Timothy A. Sipka. The orders of graphs with prescribed degree sets. Journal of Graph Theory, 4(3):301-307, 1980.
[2] Amitabha Tripathi and Sujith Vijay. A short proof of a theorem on degree sets of graphs. Discrete Appl. Math., 155(5):670-671, 2007.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "# Problem 4",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "\n## 4. Solution",
"tier": "T2",
"year": "2017"
}
|
Let $n \geq 2$ be an integer. An $n$-tuple $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of positive integers is expensive if there exists a positive integer $k$ such that
$$
\left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \cdots \cdots\left(a_{n-1}+a_{n}\right)\left(a_{n}+a_{1}\right)=2^{2 k-1} .
$$
a) Find all positive integers $n \geq 2$ for which there exists an expensive $n$-tuple.
b) Prove that for every positive integer $m$ there exists an integer $n \geq 2$ such that $m$ belongs to an expensive $n$-tuple.
There are exactly $n$ factors in the product on the left hand side.
Harun Hindija, Bosnia and Herzegovina
|
1
a) Notice that for odd integers $n>2$, the tuple $(1,1, \ldots, 1)$ is expensive. We will prove that there are no expensive $n$-tuples for even $n$.
Lemma 0.1. If an expensive $n$-tuple exists for some $n \geq 4$, then also an expensive $n-2$-tuple.
Proof. In what follows all indices are considered modulo $n$. Let $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ be an expensive $n$-tuple and $a_{t}$ the largest element of the tuple. We have the inequalities
$$
\begin{aligned}
a_{t-1}+a_{t} & \leq 2 a_{t}<2\left(a_{t}+a_{t+1}\right) \\
a_{t}+a_{t+1} & \leq 2 a_{t}<2\left(a_{t-1}+a_{t}\right) .
\end{aligned}
$$
Since both $a_{t-1}+a_{t}$ and $a_{t}+a_{t+1}$ are powers of 2 (they are divisors of a power of 2), we deduce from (1) and (2)
$$
a_{t-1}+a_{t}=a_{t}+a_{t+1}=2^{r}
$$
for some positive integer $r$, and in particular $a_{t-1}=a_{t+1}$.
Consider now the $n$-2-tuple $\left(b_{1}, \ldots, b_{n-2}\right)$ obtained by removing $a_{t}$ and $a_{t+1}$ from $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$. By what we just said we have
$$
\prod_{i=1}^{n-2}\left(b_{i}+b_{i+1}\right)=\frac{\prod_{i=1}^{n}\left(a_{i}+a_{i+1}\right)}{\left(a_{t-1}+a_{t}\right)\left(a_{t}+a_{t+1}\right)}=2^{2(k-r)-1}
$$
and hence $\left(b_{1}, \ldots, b_{n-2}\right)$ is again expensive.
From the lemma we now conclude that if there exists an expensive $n$-tuple for some even $n$, then also an expensive 2 -tuple i.e.
$$
\left(a_{1}+a_{2}\right)^{2}=2^{2 k-1}
$$
for some positive integers $a_{1}, a_{2}$, which is impossible since the right hand side is not a square.
b) We prove this by induction. In $a$ ) we saw that 1 belongs to an expensive $n$-tuple. Assume now that all odd positive integers less that $2^{k}$ belong to an expensive $n$-tuple, for some $k \geq 1$. Hence for any odd $r<2^{k}$ there is an integer $n$ and an expensive $n$-tuple $\left(a_{1}, \ldots, r, \ldots, a_{n}\right)$. We notice that then also ( $a_{1}, \ldots, r, 2^{k+1}-$ $\left.r, r, \ldots, a_{n}\right)$ is expensive. Since $2^{k+1}-r$ can take all odd values between $2^{k}$ and $2^{k+1}$ the induction step is complete.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $n \geq 2$ be an integer. An $n$-tuple $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of positive integers is expensive if there exists a positive integer $k$ such that
$$
\left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \cdots \cdots\left(a_{n-1}+a_{n}\right)\left(a_{n}+a_{1}\right)=2^{2 k-1} .
$$
a) Find all positive integers $n \geq 2$ for which there exists an expensive $n$-tuple.
b) Prove that for every positive integer $m$ there exists an integer $n \geq 2$ such that $m$ belongs to an expensive $n$-tuple.
There are exactly $n$ factors in the product on the left hand side.
Harun Hindija, Bosnia and Herzegovina
|
1
a) Notice that for odd integers $n>2$, the tuple $(1,1, \ldots, 1)$ is expensive. We will prove that there are no expensive $n$-tuples for even $n$.
Lemma 0.1. If an expensive $n$-tuple exists for some $n \geq 4$, then also an expensive $n-2$-tuple.
Proof. In what follows all indices are considered modulo $n$. Let $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ be an expensive $n$-tuple and $a_{t}$ the largest element of the tuple. We have the inequalities
$$
\begin{aligned}
a_{t-1}+a_{t} & \leq 2 a_{t}<2\left(a_{t}+a_{t+1}\right) \\
a_{t}+a_{t+1} & \leq 2 a_{t}<2\left(a_{t-1}+a_{t}\right) .
\end{aligned}
$$
Since both $a_{t-1}+a_{t}$ and $a_{t}+a_{t+1}$ are powers of 2 (they are divisors of a power of 2), we deduce from (1) and (2)
$$
a_{t-1}+a_{t}=a_{t}+a_{t+1}=2^{r}
$$
for some positive integer $r$, and in particular $a_{t-1}=a_{t+1}$.
Consider now the $n$-2-tuple $\left(b_{1}, \ldots, b_{n-2}\right)$ obtained by removing $a_{t}$ and $a_{t+1}$ from $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$. By what we just said we have
$$
\prod_{i=1}^{n-2}\left(b_{i}+b_{i+1}\right)=\frac{\prod_{i=1}^{n}\left(a_{i}+a_{i+1}\right)}{\left(a_{t-1}+a_{t}\right)\left(a_{t}+a_{t+1}\right)}=2^{2(k-r)-1}
$$
and hence $\left(b_{1}, \ldots, b_{n-2}\right)$ is again expensive.
From the lemma we now conclude that if there exists an expensive $n$-tuple for some even $n$, then also an expensive 2 -tuple i.e.
$$
\left(a_{1}+a_{2}\right)^{2}=2^{2 k-1}
$$
for some positive integers $a_{1}, a_{2}$, which is impossible since the right hand side is not a square.
b) We prove this by induction. In $a$ ) we saw that 1 belongs to an expensive $n$-tuple. Assume now that all odd positive integers less that $2^{k}$ belong to an expensive $n$-tuple, for some $k \geq 1$. Hence for any odd $r<2^{k}$ there is an integer $n$ and an expensive $n$-tuple $\left(a_{1}, \ldots, r, \ldots, a_{n}\right)$. We notice that then also ( $a_{1}, \ldots, r, 2^{k+1}-$ $\left.r, r, \ldots, a_{n}\right)$ is expensive. Since $2^{k+1}-r$ can take all odd values between $2^{k}$ and $2^{k+1}$ the induction step is complete.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "# Problem 5",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "# Solution ",
"tier": "T2",
"year": "2017"
}
|
Let $n \geq 2$ be an integer. An $n$-tuple $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of positive integers is expensive if there exists a positive integer $k$ such that
$$
\left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \cdots \cdots\left(a_{n-1}+a_{n}\right)\left(a_{n}+a_{1}\right)=2^{2 k-1} .
$$
a) Find all positive integers $n \geq 2$ for which there exists an expensive $n$-tuple.
b) Prove that for every positive integer $m$ there exists an integer $n \geq 2$ such that $m$ belongs to an expensive $n$-tuple.
There are exactly $n$ factors in the product on the left hand side.
Harun Hindija, Bosnia and Herzegovina
|
2
a) For odd $n$ the tuple $(1,1, \ldots, 1)$ is a solution.
Now consider $n$ even. Since the product $\prod\left(a_{i}+a_{i+1}\right)$ is a power of two, every factor needs to be a power of two. We are going to prove that for all tuples $\left(a_{1}, \ldots, a_{n}\right)$ such that $a_{i}+a_{i+1}$ is always a power of two , it is the case that $\prod\left(a_{i}+a_{i+1}\right)$ is equal to an even power of two. We are going to prove this with strong induction on $\sum a_{i}$. When all $a_{i}$ are equal to one this is certainly the case. Since $a_{i}+a_{i+1}>1$ it is even and we conclude that the $a_{i}$ are either all odd or all even. In the case they are all even, then consider the tuple $\left(b_{1}, \ldots, b_{n}\right)$ with $b_{i}=a_{i} / 2$. This tuple clearly satisfies the hypothesis as well and we have $\sum b_{i}<\sum a_{i}$. Furthermore we have $\prod\left(a_{i}+a_{i+1}\right)=2^{n} \prod\left(b_{i}+b_{i+1}\right)$ and since $n$ is even we are done in this case.
Now all $a_{i}$ are odd. Suppose none of are one, then consider the tuple $\left(b_{1}, \ldots, b_{n}\right)$ with $b_{i}=\left(a_{i}+(-1)^{i}\right) / 2$. Since all $a_{i}$ are odd and strictly larger than one, the $b_{i}$ are positive integers and satisfy $b_{i}+b_{i+1}=\left(a_{i}+a_{i+1}\right) / 2$, a power of two. Since $\sum b_{i}<\sum a_{i}$ and $\prod\left(a_{i}+a_{i+1}\right)=2^{n} \prod\left(b_{i}+b_{i+1}\right)$ we are done in this case again. Now there is at least one $a_{i}$ being one. We may assume $i=1$, because the condition is cyclic. Moreover we may also assume that $a_{2}>1$ since not all of the $a_{i}$ are equal to one. Let now $k$ be the smallest index larger than one such that $a_{k}$ is equal to one. We are not excluding the case $k=n+1$, yet. Now for $i=1, \ldots, k-1$ we have $a_{i}+a_{i+1}>2$ and thus divisible by four. By induction it easily follows that $a_{i} \equiv(-1)^{i+1} \bmod$ (4) for $i=1, \ldots, k-1$. In particular, since $a_{k}=1$ we find that $k$ is odd and at least three. Now consider the tuple $\left(b_{1}, \ldots, b_{n}\right)$ with $b_{i}=\left(a_{i}-(-1)^{i}\right) / 2$ for $i=1, \ldots, k$ and $b_{i}=a_{i}$ otherwise. This is again a tuple that satisfies the hypothesis, since $b_{1}=a_{1}=1=b_{k}=a_{k}$. Moreover $b_{2}<a_{2}$ and thus $\sum b_{i}<\sum a_{i}$. Finally we have $\prod\left(a_{i}+a_{i+1}\right)=2^{k-1} \prod\left(b_{i}+b_{i+1}\right)$ and since $k$ is odd we conclude the proof.
b) We use some of the ideas from a). Consider the operators $T_{ \pm}(n)=2 n \pm 1$. We claim that for every odd integer $m$ there is an integer $r$ and signs $\epsilon_{i} \in\{+,-\}$ for $i=1, \ldots r$ such that $T_{\epsilon_{r}} \circ \ldots \circ T_{\epsilon_{1}}(1)=m$. This is certainly true for $m=1$ and for $m>1$ we find that $m=T_{-}((m+1) / 2)$ if $m \equiv 1 \bmod (4)$ and $m=T_{+}((m-1) / 2)$ if $m \equiv 3 \bmod (4)$. Note that both $(m+1) / 2$ and $(m-1) / 2$ are odd integers in their respective cases and $(m-1) / 2 \leq(m+1) / 2<m$ for $m>1$. Therefore iterating the procedure will eventually terminate in one.
For the construction it is most convenient to set $n=2 l+1$ and label the tuple $\left(a_{-l}, a_{-l+1}, \ldots, a_{l}\right)$. For $m=1$ we have the expensive tuple $(1,1,1)$. For $m>1$ we will define operators $T_{ \pm}$on expensive tuples with the condition $a_{-l}=a_{l}=1$ that give rise to a new expensive tuple $\left(b_{-l^{\prime}}, \ldots, b_{l^{\prime}}\right)$ with $b_{-l^{\prime}}=b_{l^{\prime}}=1$ and $b_{0}=T_{ \pm} a_{0}$. It is then clear that $T_{\epsilon_{r}} \circ \cdots \circ T_{\epsilon_{1}}((1,1,1))$ is an expensive tuple containing $m$. We define $T_{ \pm}$as follows: set $l^{\prime}=l+1$ and $b_{-l^{\prime}}=b_{l^{\prime}}=1$ and $b_{i}=T_{ \pm(-1)^{i}}\left(a_{i}\right)$ for $i=-l, \ldots, l$. Here we identify + with +1 and - with -1 . We are left to prove that the new tuple is indeed expensive. If $\pm(-1)^{l}=-1$, then $\prod\left(b_{i}+b_{i+1}\right)=4 \cdot 2^{2 l} \prod\left(a_{i}+a_{i+1}\right)$, and if $\pm(-1)^{l}=+1$, then $\prod\left(b_{i}+b_{i+1}\right)=4 \cdot 2^{2 l+2} \prod\left(a_{i}+a_{i+1}\right)$. In both cases we end up with an expensive tuple again.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $n \geq 2$ be an integer. An $n$-tuple $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of positive integers is expensive if there exists a positive integer $k$ such that
$$
\left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \cdots \cdots\left(a_{n-1}+a_{n}\right)\left(a_{n}+a_{1}\right)=2^{2 k-1} .
$$
a) Find all positive integers $n \geq 2$ for which there exists an expensive $n$-tuple.
b) Prove that for every positive integer $m$ there exists an integer $n \geq 2$ such that $m$ belongs to an expensive $n$-tuple.
There are exactly $n$ factors in the product on the left hand side.
Harun Hindija, Bosnia and Herzegovina
|
2
a) For odd $n$ the tuple $(1,1, \ldots, 1)$ is a solution.
Now consider $n$ even. Since the product $\prod\left(a_{i}+a_{i+1}\right)$ is a power of two, every factor needs to be a power of two. We are going to prove that for all tuples $\left(a_{1}, \ldots, a_{n}\right)$ such that $a_{i}+a_{i+1}$ is always a power of two , it is the case that $\prod\left(a_{i}+a_{i+1}\right)$ is equal to an even power of two. We are going to prove this with strong induction on $\sum a_{i}$. When all $a_{i}$ are equal to one this is certainly the case. Since $a_{i}+a_{i+1}>1$ it is even and we conclude that the $a_{i}$ are either all odd or all even. In the case they are all even, then consider the tuple $\left(b_{1}, \ldots, b_{n}\right)$ with $b_{i}=a_{i} / 2$. This tuple clearly satisfies the hypothesis as well and we have $\sum b_{i}<\sum a_{i}$. Furthermore we have $\prod\left(a_{i}+a_{i+1}\right)=2^{n} \prod\left(b_{i}+b_{i+1}\right)$ and since $n$ is even we are done in this case.
Now all $a_{i}$ are odd. Suppose none of are one, then consider the tuple $\left(b_{1}, \ldots, b_{n}\right)$ with $b_{i}=\left(a_{i}+(-1)^{i}\right) / 2$. Since all $a_{i}$ are odd and strictly larger than one, the $b_{i}$ are positive integers and satisfy $b_{i}+b_{i+1}=\left(a_{i}+a_{i+1}\right) / 2$, a power of two. Since $\sum b_{i}<\sum a_{i}$ and $\prod\left(a_{i}+a_{i+1}\right)=2^{n} \prod\left(b_{i}+b_{i+1}\right)$ we are done in this case again. Now there is at least one $a_{i}$ being one. We may assume $i=1$, because the condition is cyclic. Moreover we may also assume that $a_{2}>1$ since not all of the $a_{i}$ are equal to one. Let now $k$ be the smallest index larger than one such that $a_{k}$ is equal to one. We are not excluding the case $k=n+1$, yet. Now for $i=1, \ldots, k-1$ we have $a_{i}+a_{i+1}>2$ and thus divisible by four. By induction it easily follows that $a_{i} \equiv(-1)^{i+1} \bmod$ (4) for $i=1, \ldots, k-1$. In particular, since $a_{k}=1$ we find that $k$ is odd and at least three. Now consider the tuple $\left(b_{1}, \ldots, b_{n}\right)$ with $b_{i}=\left(a_{i}-(-1)^{i}\right) / 2$ for $i=1, \ldots, k$ and $b_{i}=a_{i}$ otherwise. This is again a tuple that satisfies the hypothesis, since $b_{1}=a_{1}=1=b_{k}=a_{k}$. Moreover $b_{2}<a_{2}$ and thus $\sum b_{i}<\sum a_{i}$. Finally we have $\prod\left(a_{i}+a_{i+1}\right)=2^{k-1} \prod\left(b_{i}+b_{i+1}\right)$ and since $k$ is odd we conclude the proof.
b) We use some of the ideas from a). Consider the operators $T_{ \pm}(n)=2 n \pm 1$. We claim that for every odd integer $m$ there is an integer $r$ and signs $\epsilon_{i} \in\{+,-\}$ for $i=1, \ldots r$ such that $T_{\epsilon_{r}} \circ \ldots \circ T_{\epsilon_{1}}(1)=m$. This is certainly true for $m=1$ and for $m>1$ we find that $m=T_{-}((m+1) / 2)$ if $m \equiv 1 \bmod (4)$ and $m=T_{+}((m-1) / 2)$ if $m \equiv 3 \bmod (4)$. Note that both $(m+1) / 2$ and $(m-1) / 2$ are odd integers in their respective cases and $(m-1) / 2 \leq(m+1) / 2<m$ for $m>1$. Therefore iterating the procedure will eventually terminate in one.
For the construction it is most convenient to set $n=2 l+1$ and label the tuple $\left(a_{-l}, a_{-l+1}, \ldots, a_{l}\right)$. For $m=1$ we have the expensive tuple $(1,1,1)$. For $m>1$ we will define operators $T_{ \pm}$on expensive tuples with the condition $a_{-l}=a_{l}=1$ that give rise to a new expensive tuple $\left(b_{-l^{\prime}}, \ldots, b_{l^{\prime}}\right)$ with $b_{-l^{\prime}}=b_{l^{\prime}}=1$ and $b_{0}=T_{ \pm} a_{0}$. It is then clear that $T_{\epsilon_{r}} \circ \cdots \circ T_{\epsilon_{1}}((1,1,1))$ is an expensive tuple containing $m$. We define $T_{ \pm}$as follows: set $l^{\prime}=l+1$ and $b_{-l^{\prime}}=b_{l^{\prime}}=1$ and $b_{i}=T_{ \pm(-1)^{i}}\left(a_{i}\right)$ for $i=-l, \ldots, l$. Here we identify + with +1 and - with -1 . We are left to prove that the new tuple is indeed expensive. If $\pm(-1)^{l}=-1$, then $\prod\left(b_{i}+b_{i+1}\right)=4 \cdot 2^{2 l} \prod\left(a_{i}+a_{i+1}\right)$, and if $\pm(-1)^{l}=+1$, then $\prod\left(b_{i}+b_{i+1}\right)=4 \cdot 2^{2 l+2} \prod\left(a_{i}+a_{i+1}\right)$. In both cases we end up with an expensive tuple again.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "# Problem 5",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "# Solution ",
"tier": "T2",
"year": "2017"
}
|
Let $A B C$ be an acute-angled triangle in which no two sides have the same length. The reflections of the centroid $G$ and the circumcentre $O$ of $A B C$ in its sides $B C, C A, A B$ are denoted by $G_{1}, G_{2}, G_{3}$, and $O_{1}, O_{2}, O_{3}$, respectively. Show that the circumcircles of the triangles $G_{1} G_{2} C, G_{1} G_{3} B, G_{2} G_{3} A, O_{1} O_{2} C, O_{1} O_{3} B, O_{2} O_{3} A$ and $A B C$ have a common point.
The centroid of a triangle is the intersection point of the three medians. A median is a line connecting a vertex of the triangle to the midpoint of the opposite side.
Charles Leytem, Luxembourg
|
Let $H$ denote the orthocenter of $A B C$, and let $e$ denote its Euler line. Let $e_{1}, e_{2}, e_{3}$ denote the respective reflections of $e$ in $B C, C A, A B$. The proof naturally divides into two parts: we first show that pairwise intersections of the circles in question correspond to pairwise intersections of $e_{1}, e_{2}, e_{3}$, and then prove that $e_{1}, e_{2}, e_{3}$ intersect in a single point on the circumcircle of $A B C$.
Now consider for example the circumcircles of $O_{1} O_{2} C$ and $G_{1} G_{2} C$. By construction, it is clear that $\angle O_{2} C O_{1}=\angle G_{2} C G_{1}=2 \angle A C B$. Let $G_{1} O_{1}$ and $G_{2} O_{2}$ meet at $X$, and let $e$ meet $e_{1}, e_{2}$ at $E_{1}, E_{2}$, respectively, as shown in the diagram below. Chasing angles,
$$
\begin{aligned}
\angle G_{2} X G_{1}=\angle O_{2} X O_{1} & =\angle E_{2} X E_{1}=180^{\circ}-\angle E_{1} E_{2} X-\angle X E_{1} E_{2} \\
& =180^{\circ}-2 \angle E_{1} E_{2} C-\left(\angle C E_{1} E_{2}-\angle C E_{1} X\right) .
\end{aligned}
$$
But $\angle C E_{1} X=\angle B E_{1} O_{1}=\angle B E_{1} O=180^{\circ}-\angle C E_{1} E_{2}$, and thus
$$
\angle G_{2} X G_{1}=\angle O_{2} X O_{1}=2\left(180^{\circ}-\angle E_{1} E_{2} C-\angle C E_{1} E_{2}\right)=2 \angle A C B
$$
It follows from this that $X$ lies on the circumcircles of $G_{1} G_{2} C$ and $O_{1} O_{2} C$. In other words, the second point of intersection of the circumcircles of $G_{1} G_{2} C$ and $O_{1} O_{2} C$ is the intersection of $e_{1}$ and $e_{2}$. Similarly, the circumcircles of $G_{1} G_{3} B$ and $O_{1} O_{3} B$ meet again at the intersection of $e_{1}$ and $e_{3}$, and those of $G_{2} G_{3} A$ and $O_{2} O_{3} A$ meet again at the intersection of $e_{2}$ and $e_{3}$.
It thus remains to show that $e_{1}, e_{2}, e_{3}$ are concurrent, and intersect on the circumcircle of $A B C$. Let $e$ meet the circumcircles of the triangles $B C H, A C H, A B H$ at $X_{1}, X_{2}, X_{3}$, respectively. It is well known that the reflections of $H$ in the sides of $A B C$ lie on the circumcircle of $A B C$. For this reason, the circumcircles of $B C H, A C H, A B H$ have the same radius as the circumcircle of $A B C$, and hence the reflections $X_{1}^{\prime}, X_{2}^{\prime}, X_{3}^{\prime}$ of $X_{1}, X_{2}, X_{3}$ in the sides $[B C],[C A],[A B]$ lie on the circumcircle of $A B C$. By definition, $X_{1}^{\prime}, X_{2}^{\prime}, X_{3}^{\prime}$ lie on $e_{1}, e_{2}, e_{3}$, respectively. It thus remains to show that they coincide.
To show that, for example, $X_{1}^{\prime}=X_{2}^{\prime}$, it will be sufficient to show that $\angle X_{2} A C=\angle X_{1}^{\prime} A C$, since we have already shown that $X_{1}^{\prime}$ and $X_{2}^{\prime}$ lie on the circumcircle of $A B C$. But, chasing angles in the diagram above,
$$
\begin{aligned}
\angle X_{1}^{\prime} A C & =\angle X_{1}^{\prime} A B-\angle B A C=\left(180^{\circ}-\angle X_{1}^{\prime} C B\right)-\angle B A C \\
& =180^{\circ}-\angle X_{1} C B-\angle B A C=\left(180^{\circ}-\angle B A C\right)-\angle X_{1} H B \\
& =\angle B H C-\angle X_{1} H B=\angle X_{2} H C=\angle X_{2} A C
\end{aligned}
$$
where we have used the fact that $B H C X_{1}$ and $A H X_{2} C$ are cyclic by construction, and the fact that $\angle B H C=180^{\circ}-\angle B A C$. This shows that $X_{1}^{\prime}=X_{2}^{\prime}$. Similarly, $X_{1}^{\prime}=X_{3}^{\prime}$, which completes the proof.
Remark. The statement of the problem remains true if $O$ and $G$ are replaced with two points that are aligned with the orthocenter $H$ of the triangle, and indeed, the proof above did not require any property of $G$ and $O$ other than the fact that they lie on a line through $H$, the Euler line.
|
proof
|
Yes
|
Incomplete
|
proof
|
Geometry
|
Let $A B C$ be an acute-angled triangle in which no two sides have the same length. The reflections of the centroid $G$ and the circumcentre $O$ of $A B C$ in its sides $B C, C A, A B$ are denoted by $G_{1}, G_{2}, G_{3}$, and $O_{1}, O_{2}, O_{3}$, respectively. Show that the circumcircles of the triangles $G_{1} G_{2} C, G_{1} G_{3} B, G_{2} G_{3} A, O_{1} O_{2} C, O_{1} O_{3} B, O_{2} O_{3} A$ and $A B C$ have a common point.
The centroid of a triangle is the intersection point of the three medians. A median is a line connecting a vertex of the triangle to the midpoint of the opposite side.
Charles Leytem, Luxembourg
|
Let $H$ denote the orthocenter of $A B C$, and let $e$ denote its Euler line. Let $e_{1}, e_{2}, e_{3}$ denote the respective reflections of $e$ in $B C, C A, A B$. The proof naturally divides into two parts: we first show that pairwise intersections of the circles in question correspond to pairwise intersections of $e_{1}, e_{2}, e_{3}$, and then prove that $e_{1}, e_{2}, e_{3}$ intersect in a single point on the circumcircle of $A B C$.
Now consider for example the circumcircles of $O_{1} O_{2} C$ and $G_{1} G_{2} C$. By construction, it is clear that $\angle O_{2} C O_{1}=\angle G_{2} C G_{1}=2 \angle A C B$. Let $G_{1} O_{1}$ and $G_{2} O_{2}$ meet at $X$, and let $e$ meet $e_{1}, e_{2}$ at $E_{1}, E_{2}$, respectively, as shown in the diagram below. Chasing angles,
$$
\begin{aligned}
\angle G_{2} X G_{1}=\angle O_{2} X O_{1} & =\angle E_{2} X E_{1}=180^{\circ}-\angle E_{1} E_{2} X-\angle X E_{1} E_{2} \\
& =180^{\circ}-2 \angle E_{1} E_{2} C-\left(\angle C E_{1} E_{2}-\angle C E_{1} X\right) .
\end{aligned}
$$
But $\angle C E_{1} X=\angle B E_{1} O_{1}=\angle B E_{1} O=180^{\circ}-\angle C E_{1} E_{2}$, and thus
$$
\angle G_{2} X G_{1}=\angle O_{2} X O_{1}=2\left(180^{\circ}-\angle E_{1} E_{2} C-\angle C E_{1} E_{2}\right)=2 \angle A C B
$$
It follows from this that $X$ lies on the circumcircles of $G_{1} G_{2} C$ and $O_{1} O_{2} C$. In other words, the second point of intersection of the circumcircles of $G_{1} G_{2} C$ and $O_{1} O_{2} C$ is the intersection of $e_{1}$ and $e_{2}$. Similarly, the circumcircles of $G_{1} G_{3} B$ and $O_{1} O_{3} B$ meet again at the intersection of $e_{1}$ and $e_{3}$, and those of $G_{2} G_{3} A$ and $O_{2} O_{3} A$ meet again at the intersection of $e_{2}$ and $e_{3}$.
It thus remains to show that $e_{1}, e_{2}, e_{3}$ are concurrent, and intersect on the circumcircle of $A B C$. Let $e$ meet the circumcircles of the triangles $B C H, A C H, A B H$ at $X_{1}, X_{2}, X_{3}$, respectively. It is well known that the reflections of $H$ in the sides of $A B C$ lie on the circumcircle of $A B C$. For this reason, the circumcircles of $B C H, A C H, A B H$ have the same radius as the circumcircle of $A B C$, and hence the reflections $X_{1}^{\prime}, X_{2}^{\prime}, X_{3}^{\prime}$ of $X_{1}, X_{2}, X_{3}$ in the sides $[B C],[C A],[A B]$ lie on the circumcircle of $A B C$. By definition, $X_{1}^{\prime}, X_{2}^{\prime}, X_{3}^{\prime}$ lie on $e_{1}, e_{2}, e_{3}$, respectively. It thus remains to show that they coincide.
To show that, for example, $X_{1}^{\prime}=X_{2}^{\prime}$, it will be sufficient to show that $\angle X_{2} A C=\angle X_{1}^{\prime} A C$, since we have already shown that $X_{1}^{\prime}$ and $X_{2}^{\prime}$ lie on the circumcircle of $A B C$. But, chasing angles in the diagram above,
$$
\begin{aligned}
\angle X_{1}^{\prime} A C & =\angle X_{1}^{\prime} A B-\angle B A C=\left(180^{\circ}-\angle X_{1}^{\prime} C B\right)-\angle B A C \\
& =180^{\circ}-\angle X_{1} C B-\angle B A C=\left(180^{\circ}-\angle B A C\right)-\angle X_{1} H B \\
& =\angle B H C-\angle X_{1} H B=\angle X_{2} H C=\angle X_{2} A C
\end{aligned}
$$
where we have used the fact that $B H C X_{1}$ and $A H X_{2} C$ are cyclic by construction, and the fact that $\angle B H C=180^{\circ}-\angle B A C$. This shows that $X_{1}^{\prime}=X_{2}^{\prime}$. Similarly, $X_{1}^{\prime}=X_{3}^{\prime}$, which completes the proof.
Remark. The statement of the problem remains true if $O$ and $G$ are replaced with two points that are aligned with the orthocenter $H$ of the triangle, and indeed, the proof above did not require any property of $G$ and $O$ other than the fact that they lie on a line through $H$, the Euler line.
|
{
"exam": "EGMO",
"problem_label": "6",
"problem_match": "# Problem 6",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "# Solution 1 (Euler lines)",
"tier": "T2",
"year": "2017"
}
|
Let $A B C$ be an acute-angled triangle in which no two sides have the same length. The reflections of the centroid $G$ and the circumcentre $O$ of $A B C$ in its sides $B C, C A, A B$ are denoted by $G_{1}, G_{2}, G_{3}$, and $O_{1}, O_{2}, O_{3}$, respectively. Show that the circumcircles of the triangles $G_{1} G_{2} C, G_{1} G_{3} B, G_{2} G_{3} A, O_{1} O_{2} C, O_{1} O_{3} B, O_{2} O_{3} A$ and $A B C$ have a common point.
The centroid of a triangle is the intersection point of the three medians. A median is a line connecting a vertex of the triangle to the midpoint of the opposite side.
Charles Leytem, Luxembourg
|
2
The proof consists of two parts. First, we show that if $P$ is any point inside the triangle $A B C$ and $P_{1}, P_{2}, P_{3}$ are its reflections in the sides $B C, C A, A B$, then the circumcircles of the triangles $P_{1} P_{2} C, P_{1} P_{3} B, P_{2} P_{3} A$ intersect in a point $T_{P}$ on the circumcircle of the triangle $A B C$. In the second part, we show that $T_{G}$ coincides with $T_{O}$.
Now let $P$ be any point inside the triangle $A B C$ and let $P_{1}, P_{2}, P_{3}$ be the reflections in the sides as above. Let $T_{P}$ be the second intersection of the circumcircles of the triangles $P_{1} P_{2} C$ and $A B C$. We want to show that $T_{P}$ lies on the circumcircles of the triangles $P_{1} P_{3} B$ and $P_{2} P_{3} A$.
By construction, we have $P_{1} C=P_{2} C$, hence
$$
\angle C P_{1} P_{2}=90^{\circ}-\frac{1}{2} \angle P_{2} C P_{1}=90^{\circ}-\angle A C B
$$
Similarly, $\angle P_{2} P_{3} A=90^{\circ}-\angle B A C$. This gives us
$$
\begin{aligned}
\angle P_{2} T_{P} A & =\angle C T_{P} A-\angle C T_{P} P_{2}=\angle C B A-\angle C P_{1} P_{2} \\
& =\angle C B A-90^{\circ}+\angle A C B=90^{\circ}-\angle B A C=\angle P_{2} P_{3} A,
\end{aligned}
$$
so $T_{P}$ lies on the circumcircle of the triangle $P_{2} P_{3} A$. Similarly, $T_{P}$ lies on the circumcircle of the triangle $P_{1} P_{3} B$ which completes the first part.
Note that if $P_{2}$ is given, then $T_{P}$ is the unique point on the circumcircle of the triangle $A B C$ with $\angle C T_{P} P_{2}=90^{\circ}-\angle A C B$. In the second part, we will use this as follows: If we can find a point $T$ on the circumcircle of the triangle $A B C$ with $\angle C T G_{2}=\angle C T O_{2}=90^{\circ}-\angle A C B$,
then $T=T_{G}=T_{O}$ and we are done.
Let $H$ be the orthocenter of the triangle $A B C$ and let $H_{2}$ be the reflection in the side $A C$. It is known that $H_{2}$ lies on the circumcircle of the triangle $A B C . G, O, H$ lie on the Euler line, so $G_{2}, O_{2}, H_{2}$ are collinear as well. Let $T$ be the second intersection of $G_{2} H_{2}$ and the circumcircle of the triangle $A B C$. We can now complete the proof by seeing that
$$
\angle C T G_{2}=\angle C T O_{2}=\angle C T H_{2}=\angle C B H_{2}=90^{\circ}-\angle A C B .
$$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute-angled triangle in which no two sides have the same length. The reflections of the centroid $G$ and the circumcentre $O$ of $A B C$ in its sides $B C, C A, A B$ are denoted by $G_{1}, G_{2}, G_{3}$, and $O_{1}, O_{2}, O_{3}$, respectively. Show that the circumcircles of the triangles $G_{1} G_{2} C, G_{1} G_{3} B, G_{2} G_{3} A, O_{1} O_{2} C, O_{1} O_{3} B, O_{2} O_{3} A$ and $A B C$ have a common point.
The centroid of a triangle is the intersection point of the three medians. A median is a line connecting a vertex of the triangle to the midpoint of the opposite side.
Charles Leytem, Luxembourg
|
2
The proof consists of two parts. First, we show that if $P$ is any point inside the triangle $A B C$ and $P_{1}, P_{2}, P_{3}$ are its reflections in the sides $B C, C A, A B$, then the circumcircles of the triangles $P_{1} P_{2} C, P_{1} P_{3} B, P_{2} P_{3} A$ intersect in a point $T_{P}$ on the circumcircle of the triangle $A B C$. In the second part, we show that $T_{G}$ coincides with $T_{O}$.
Now let $P$ be any point inside the triangle $A B C$ and let $P_{1}, P_{2}, P_{3}$ be the reflections in the sides as above. Let $T_{P}$ be the second intersection of the circumcircles of the triangles $P_{1} P_{2} C$ and $A B C$. We want to show that $T_{P}$ lies on the circumcircles of the triangles $P_{1} P_{3} B$ and $P_{2} P_{3} A$.
By construction, we have $P_{1} C=P_{2} C$, hence
$$
\angle C P_{1} P_{2}=90^{\circ}-\frac{1}{2} \angle P_{2} C P_{1}=90^{\circ}-\angle A C B
$$
Similarly, $\angle P_{2} P_{3} A=90^{\circ}-\angle B A C$. This gives us
$$
\begin{aligned}
\angle P_{2} T_{P} A & =\angle C T_{P} A-\angle C T_{P} P_{2}=\angle C B A-\angle C P_{1} P_{2} \\
& =\angle C B A-90^{\circ}+\angle A C B=90^{\circ}-\angle B A C=\angle P_{2} P_{3} A,
\end{aligned}
$$
so $T_{P}$ lies on the circumcircle of the triangle $P_{2} P_{3} A$. Similarly, $T_{P}$ lies on the circumcircle of the triangle $P_{1} P_{3} B$ which completes the first part.
Note that if $P_{2}$ is given, then $T_{P}$ is the unique point on the circumcircle of the triangle $A B C$ with $\angle C T_{P} P_{2}=90^{\circ}-\angle A C B$. In the second part, we will use this as follows: If we can find a point $T$ on the circumcircle of the triangle $A B C$ with $\angle C T G_{2}=\angle C T O_{2}=90^{\circ}-\angle A C B$,
then $T=T_{G}=T_{O}$ and we are done.
Let $H$ be the orthocenter of the triangle $A B C$ and let $H_{2}$ be the reflection in the side $A C$. It is known that $H_{2}$ lies on the circumcircle of the triangle $A B C . G, O, H$ lie on the Euler line, so $G_{2}, O_{2}, H_{2}$ are collinear as well. Let $T$ be the second intersection of $G_{2} H_{2}$ and the circumcircle of the triangle $A B C$. We can now complete the proof by seeing that
$$
\angle C T G_{2}=\angle C T O_{2}=\angle C T H_{2}=\angle C B H_{2}=90^{\circ}-\angle A C B .
$$
|
{
"exam": "EGMO",
"problem_label": "6",
"problem_match": "# Problem 6",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "# Solution ",
"tier": "T2",
"year": "2017"
}
|
Let $A B C$ be an acute-angled triangle in which no two sides have the same length. The reflections of the centroid $G$ and the circumcentre $O$ of $A B C$ in its sides $B C, C A, A B$ are denoted by $G_{1}, G_{2}, G_{3}$, and $O_{1}, O_{2}, O_{3}$, respectively. Show that the circumcircles of the triangles $G_{1} G_{2} C, G_{1} G_{3} B, G_{2} G_{3} A, O_{1} O_{2} C, O_{1} O_{3} B, O_{2} O_{3} A$ and $A B C$ have a common point.
The centroid of a triangle is the intersection point of the three medians. A median is a line connecting a vertex of the triangle to the midpoint of the opposite side.
Charles Leytem, Luxembourg
|
For every point $P$, let $p$ denote the corresponding complex number. Set $O$ to be the origin, so $o=0$, and without loss of generality we can assume that $a, b$ and $c$ lie on the unit circle. Then the centroid can be expressed as $g=\frac{a+b+c}{3}$.
The segments $O o_{1}$ and $b c$ have a common midpoint, so $o_{1}+o=b+c$, and then $o_{1}=b+c$. Similarly $o_{2}=a+c$ and $o_{3}=a+b$. In order to compute $g_{1}$, define $y$ to be the projection of $g$ onto $b c$. Since $b$ and $c$ are on the unit circle, it is well known that $y$ can be expressed as
$$
y=\frac{1}{2}(b+c+g-b c \bar{g}) .
$$
By using $\bar{a}=\frac{1}{a}, \bar{b}=\frac{1}{b}$ and $\bar{c}=\frac{1}{c}$ (points on the unit circle), we obtain
$$
g_{1}=b+c-\frac{a b+b c+c a}{3 a} .
$$
Similarly, we get $g_{2}=a+c-\frac{a b+b c+c a}{3 b}$ and $g_{3}=a+b-\frac{a b+b c+c a}{3 c}$.
1) Proof that circumcircles of triangles $a b c, o_{1} O_{2} c, o_{1} O_{3} b$ and $o_{2} O_{3} a$ have common point.
Let $x$ be the point of intersection of circumcircles of triangles $o_{1} O_{2} c$ and $a b c(x \neq c)$. We know that $x, o_{1}, o_{2}$ and $c$ are concyclic if and only if $\frac{x-c}{o_{1}-c}: \frac{x-o_{2}}{o_{1}-o_{2}}$ is real number, which is equivalent to
$$
\frac{x-c}{\bar{x}-\bar{c}} \cdot \frac{o_{1}-o_{2}}{\overline{o_{1}}-\overline{o_{2}}}=\frac{o_{1}-c}{\overline{o_{1}}-\bar{c}} \cdot \frac{x-o_{2}}{\bar{x}-\overline{o_{2}}}
$$
Since $x$ and $c$ are on the unit circle $\frac{x-c}{\bar{x}-\bar{c}}=-x c$. Also, $\frac{o_{1}-o_{2}}{\overline{o_{1}-\overline{o_{2}}}}=\frac{b-a}{\bar{b}-\bar{a}}=-a b$, and $\frac{o_{1}-c}{\overline{o_{1}}-\bar{c}}=\frac{b}{\bar{b}}=b^{2}$. Since $\bar{x}=\frac{1}{x}$, from (1) and previous relations, we have:
$$
x=\frac{a b+b c+c a}{a+b+c}
$$
This formula is symmetric, so we conclude that $x$ also belongs to circumcircles of $o_{1} O_{3} b$ and $o_{2} O_{3} a$.
2) Proof that $x$ belongs to circumcircles of $g_{1} g_{2} c, g_{1} g_{3} b$ and $g_{2} g_{3} a$.
Because of symmetry, it is enough to prove that $x$ belongs to circumcircle of $g_{1} g_{2} c$, i.e. to prove the following:
$$
\frac{x-c}{\bar{x}-\bar{c}} \cdot \frac{g_{1}-g_{2}}{\overline{g_{1}}-\overline{g_{2}}}=\frac{g_{1}-c}{\overline{g_{1}}-\bar{c}} \cdot \frac{x-g_{2}}{\bar{x}-\overline{g_{2}}}
$$
Easy computations give that
$$
g_{1}-g_{2}=(b-a) \frac{2 a b-b c-a c}{3 a b}, \quad \overline{g_{1}}-\overline{g_{2}}=(\bar{b}-\bar{a}) \frac{2-\frac{a}{c}-\frac{b}{c}}{3}
$$
and then
$$
\frac{g_{1}-g_{2}}{\overline{g_{1}}-\overline{g_{2}}}=\frac{c(b c+a c-2 a b)}{2 c-a-b}
$$
On the other hand we have
$$
g_{1}-c=\frac{2 a b-b c-a c}{3 a}, \quad \overline{g_{1}}-\bar{c}=\frac{2 c-a-b}{3 b c} .
$$
This implies
$$
\frac{g_{1}-c}{\overline{g_{1}-\bar{c}}}=\frac{2 a b-b c-a c}{2 c-a-b} \cdot \frac{b c}{a}
$$
Then (2) is equivalent to
$$
\begin{aligned}
-x c \cdot \frac{c(b c+a c-2 a b)}{2 c-a-b} & =\frac{2 a b-b c-a c}{2 c-a-b} \cdot \frac{b c}{a} \cdot \frac{x-g_{2}}{\bar{x}-\overline{g_{2}}} \\
\Longleftrightarrow x c a\left(\bar{x}-\bar{g}_{2}\right) & =b\left(x-g_{2}\right)
\end{aligned}
$$
which is also equivalent to
$\frac{a b+b c+c a}{a+b+c} \cdot c a\left(\frac{a+b+c}{a b+b c+c a}-\frac{1}{a}-\frac{1}{c}+\frac{a+b+c}{3 a c}\right)=b \cdot\left(\frac{a b+b c+c a}{a+b+c}-a-c+\frac{a b+b c+c a}{3 b}\right)$.
The last equality can easily be verified, which implies that $x$ belongs to circumcircle of triangle $g_{1} g_{2} c$. This concludes our proof.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute-angled triangle in which no two sides have the same length. The reflections of the centroid $G$ and the circumcentre $O$ of $A B C$ in its sides $B C, C A, A B$ are denoted by $G_{1}, G_{2}, G_{3}$, and $O_{1}, O_{2}, O_{3}$, respectively. Show that the circumcircles of the triangles $G_{1} G_{2} C, G_{1} G_{3} B, G_{2} G_{3} A, O_{1} O_{2} C, O_{1} O_{3} B, O_{2} O_{3} A$ and $A B C$ have a common point.
The centroid of a triangle is the intersection point of the three medians. A median is a line connecting a vertex of the triangle to the midpoint of the opposite side.
Charles Leytem, Luxembourg
|
For every point $P$, let $p$ denote the corresponding complex number. Set $O$ to be the origin, so $o=0$, and without loss of generality we can assume that $a, b$ and $c$ lie on the unit circle. Then the centroid can be expressed as $g=\frac{a+b+c}{3}$.
The segments $O o_{1}$ and $b c$ have a common midpoint, so $o_{1}+o=b+c$, and then $o_{1}=b+c$. Similarly $o_{2}=a+c$ and $o_{3}=a+b$. In order to compute $g_{1}$, define $y$ to be the projection of $g$ onto $b c$. Since $b$ and $c$ are on the unit circle, it is well known that $y$ can be expressed as
$$
y=\frac{1}{2}(b+c+g-b c \bar{g}) .
$$
By using $\bar{a}=\frac{1}{a}, \bar{b}=\frac{1}{b}$ and $\bar{c}=\frac{1}{c}$ (points on the unit circle), we obtain
$$
g_{1}=b+c-\frac{a b+b c+c a}{3 a} .
$$
Similarly, we get $g_{2}=a+c-\frac{a b+b c+c a}{3 b}$ and $g_{3}=a+b-\frac{a b+b c+c a}{3 c}$.
1) Proof that circumcircles of triangles $a b c, o_{1} O_{2} c, o_{1} O_{3} b$ and $o_{2} O_{3} a$ have common point.
Let $x$ be the point of intersection of circumcircles of triangles $o_{1} O_{2} c$ and $a b c(x \neq c)$. We know that $x, o_{1}, o_{2}$ and $c$ are concyclic if and only if $\frac{x-c}{o_{1}-c}: \frac{x-o_{2}}{o_{1}-o_{2}}$ is real number, which is equivalent to
$$
\frac{x-c}{\bar{x}-\bar{c}} \cdot \frac{o_{1}-o_{2}}{\overline{o_{1}}-\overline{o_{2}}}=\frac{o_{1}-c}{\overline{o_{1}}-\bar{c}} \cdot \frac{x-o_{2}}{\bar{x}-\overline{o_{2}}}
$$
Since $x$ and $c$ are on the unit circle $\frac{x-c}{\bar{x}-\bar{c}}=-x c$. Also, $\frac{o_{1}-o_{2}}{\overline{o_{1}-\overline{o_{2}}}}=\frac{b-a}{\bar{b}-\bar{a}}=-a b$, and $\frac{o_{1}-c}{\overline{o_{1}}-\bar{c}}=\frac{b}{\bar{b}}=b^{2}$. Since $\bar{x}=\frac{1}{x}$, from (1) and previous relations, we have:
$$
x=\frac{a b+b c+c a}{a+b+c}
$$
This formula is symmetric, so we conclude that $x$ also belongs to circumcircles of $o_{1} O_{3} b$ and $o_{2} O_{3} a$.
2) Proof that $x$ belongs to circumcircles of $g_{1} g_{2} c, g_{1} g_{3} b$ and $g_{2} g_{3} a$.
Because of symmetry, it is enough to prove that $x$ belongs to circumcircle of $g_{1} g_{2} c$, i.e. to prove the following:
$$
\frac{x-c}{\bar{x}-\bar{c}} \cdot \frac{g_{1}-g_{2}}{\overline{g_{1}}-\overline{g_{2}}}=\frac{g_{1}-c}{\overline{g_{1}}-\bar{c}} \cdot \frac{x-g_{2}}{\bar{x}-\overline{g_{2}}}
$$
Easy computations give that
$$
g_{1}-g_{2}=(b-a) \frac{2 a b-b c-a c}{3 a b}, \quad \overline{g_{1}}-\overline{g_{2}}=(\bar{b}-\bar{a}) \frac{2-\frac{a}{c}-\frac{b}{c}}{3}
$$
and then
$$
\frac{g_{1}-g_{2}}{\overline{g_{1}}-\overline{g_{2}}}=\frac{c(b c+a c-2 a b)}{2 c-a-b}
$$
On the other hand we have
$$
g_{1}-c=\frac{2 a b-b c-a c}{3 a}, \quad \overline{g_{1}}-\bar{c}=\frac{2 c-a-b}{3 b c} .
$$
This implies
$$
\frac{g_{1}-c}{\overline{g_{1}-\bar{c}}}=\frac{2 a b-b c-a c}{2 c-a-b} \cdot \frac{b c}{a}
$$
Then (2) is equivalent to
$$
\begin{aligned}
-x c \cdot \frac{c(b c+a c-2 a b)}{2 c-a-b} & =\frac{2 a b-b c-a c}{2 c-a-b} \cdot \frac{b c}{a} \cdot \frac{x-g_{2}}{\bar{x}-\overline{g_{2}}} \\
\Longleftrightarrow x c a\left(\bar{x}-\bar{g}_{2}\right) & =b\left(x-g_{2}\right)
\end{aligned}
$$
which is also equivalent to
$\frac{a b+b c+c a}{a+b+c} \cdot c a\left(\frac{a+b+c}{a b+b c+c a}-\frac{1}{a}-\frac{1}{c}+\frac{a+b+c}{3 a c}\right)=b \cdot\left(\frac{a b+b c+c a}{a+b+c}-a-c+\frac{a b+b c+c a}{3 b}\right)$.
The last equality can easily be verified, which implies that $x$ belongs to circumcircle of triangle $g_{1} g_{2} c$. This concludes our proof.
|
{
"exam": "EGMO",
"problem_label": "6",
"problem_match": "# Problem 6",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "# Solution 3 (complex numbers)",
"tier": "T2",
"year": "2017"
}
|
Let $A B C$ be an acute-angled triangle in which no two sides have the same length. The reflections of the centroid $G$ and the circumcentre $O$ of $A B C$ in its sides $B C, C A, A B$ are denoted by $G_{1}, G_{2}, G_{3}$, and $O_{1}, O_{2}, O_{3}$, respectively. Show that the circumcircles of the triangles $G_{1} G_{2} C, G_{1} G_{3} B, G_{2} G_{3} A, O_{1} O_{2} C, O_{1} O_{3} B, O_{2} O_{3} A$ and $A B C$ have a common point.
The centroid of a triangle is the intersection point of the three medians. A median is a line connecting a vertex of the triangle to the midpoint of the opposite side.
Charles Leytem, Luxembourg
|
The first part of the first solution and the second part of the second solution can also be done by the following rotation argument: The rotation through $2 \angle B A C$ about $A$ takes $O_{3}$ to $O_{2}, G_{3}$ to $G_{2}$ and $H_{3}$ to $H_{2}$ (again, $H_{2}, H_{3}$ are the reflections of the orthocenter $H$ in the sides $C A, A B$ ). Let $X$ be the intersection of the Euler line reflections $e_{2}$ (going through $O_{2}, G_{2}, H_{2}$ ) and $e_{3}$ (going through $O_{3}, G_{3}, H_{3}$ ). We now use the well-known fact that if a rotation about a point $A$ takes a line $l$ and a point $P$ on $l$ to the line $l^{\prime}$ and the point $P^{\prime}$, then the quadrilateral $A P P^{\prime} X$ is cyclic, where $X$ is the intersection of $l$ and $l^{\prime}$. For this reason, $A O_{3} O_{2} X, A G_{3} G_{2} X$ and $A H_{3} H_{2} X$ are cyclic quadrilaterals. Since $H_{2}$ and $H_{3}$ lie on the circumcircle of the triangle $A B C$, the circumcircles of the triangles $A H_{3} H_{2}$ and $A B C$ are the same, hence $X$ lies on the circumcircle of $A B C$.
This proves the first part of the first solution as well as the second part of the second solution.
Remark: Problem 6 is a special case of Corollary 3 in Darij Grinberg's paper Anti-Steiner points with respect to a triangle.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute-angled triangle in which no two sides have the same length. The reflections of the centroid $G$ and the circumcentre $O$ of $A B C$ in its sides $B C, C A, A B$ are denoted by $G_{1}, G_{2}, G_{3}$, and $O_{1}, O_{2}, O_{3}$, respectively. Show that the circumcircles of the triangles $G_{1} G_{2} C, G_{1} G_{3} B, G_{2} G_{3} A, O_{1} O_{2} C, O_{1} O_{3} B, O_{2} O_{3} A$ and $A B C$ have a common point.
The centroid of a triangle is the intersection point of the three medians. A median is a line connecting a vertex of the triangle to the midpoint of the opposite side.
Charles Leytem, Luxembourg
|
The first part of the first solution and the second part of the second solution can also be done by the following rotation argument: The rotation through $2 \angle B A C$ about $A$ takes $O_{3}$ to $O_{2}, G_{3}$ to $G_{2}$ and $H_{3}$ to $H_{2}$ (again, $H_{2}, H_{3}$ are the reflections of the orthocenter $H$ in the sides $C A, A B$ ). Let $X$ be the intersection of the Euler line reflections $e_{2}$ (going through $O_{2}, G_{2}, H_{2}$ ) and $e_{3}$ (going through $O_{3}, G_{3}, H_{3}$ ). We now use the well-known fact that if a rotation about a point $A$ takes a line $l$ and a point $P$ on $l$ to the line $l^{\prime}$ and the point $P^{\prime}$, then the quadrilateral $A P P^{\prime} X$ is cyclic, where $X$ is the intersection of $l$ and $l^{\prime}$. For this reason, $A O_{3} O_{2} X, A G_{3} G_{2} X$ and $A H_{3} H_{2} X$ are cyclic quadrilaterals. Since $H_{2}$ and $H_{3}$ lie on the circumcircle of the triangle $A B C$, the circumcircles of the triangles $A H_{3} H_{2}$ and $A B C$ are the same, hence $X$ lies on the circumcircle of $A B C$.
This proves the first part of the first solution as well as the second part of the second solution.
Remark: Problem 6 is a special case of Corollary 3 in Darij Grinberg's paper Anti-Steiner points with respect to a triangle.
|
{
"exam": "EGMO",
"problem_label": "6",
"problem_match": "# Problem 6",
"resource_path": "EGMO/segmented/en-2017-solutions.jsonl",
"solution_match": "# Solution 4 (rotation)",
"tier": "T2",
"year": "2017"
}
|
Let $A B C$ be a triangle with $C A=C B$ and $\angle A C B=120^{\circ}$, and let $M$ be the midpoint of $A B$. Let $P$ be a variable point on the circumcircle of $A B C$, and let $Q$ be the point on the segment $C P$ such that $Q P=2 Q C$. It is given that the line through $P$ and perpendicular to $A B$ intersects the line $M Q$ at a unique point $N$.
Prove that there exists a fixed circle such that $N$ lies on this circle for all possible positions of $P$.
(Velina Ivanova, Bulgaria)
|
Let $O$ be the circumcenter of $A B C$. From the assumption that $\angle A C B=120^{\circ}$ it follows that $M$ is the midpoint of $C O$.
Let $\omega$ denote the circle with center in $C$ and radius $C O$. This circle in the image of the circumcircle of $A B C$ through the translation that sends $O$ to $C$. We claim that $N$ lies on $\omega$.
Let us consider the triangles $Q N P$ and $Q M C$. The angles in $Q$ are equal. Since $N P$ is parallel to $M C$ (both lines are perpendicular to $A B$ ), it turns out that $\angle Q N P=\angle Q M C$, and hence the two triangles are similar. Since $Q P=2 Q C$, it follows that
$$
N P=2 M C=C O
$$
which proves that $N$ lies on $\omega$.
Comment The possible positions of $N$ are all the points of $\omega$ with the exception of the two points lying on the line $C O$. Indeed, $P$ does not lie on the line $C O$ because otherwise the point $N$ is not well-defined, and therefore also $N$ does not lie on the same line.
Conversely, let $N$ be any point on $\omega$ and not lying on the line $C O$. Let $P$ be the corresponding point on the circumcircle of $A B C$, namely such that $N P$ is parallel and equal to $C O$. Let $Q$ be
the intersection of $C P$ and $N M$. As before, the triangles $Q N P$ and $Q M C$ are similar, and now from the relation $N P=2 M C$ we deduce that $Q P=2 Q C$. This proves that $N$ can be obtained from $P$ through the construction described in the statement of the problem.
Alternative solution Let $M^{\prime}$ denote the symmetric of $M$ with respect to $O$.
Let us consider the quadrilateral $M M^{\prime} P N$. The lines $M M^{\prime}$ and $N P$ are parallel by construction. Also the lines $P M^{\prime}$ and $N M$ are parallel (homothety from $C$ with coefficient 3). It follows that $M M^{\prime} P N$ is a parallelogram, and hence $P N=M M^{\prime}=O C$.
Computational solution There are many computation approaches to this problem. For example, we can set Cartesian coordinates so that
$$
A=\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right), \quad B=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right), \quad C=(0,1), \quad M=\left(0, \frac{1}{2}\right)
$$
Setting $P=(a, b)$, we obtain that $Q=(a / 3,(2+b) / 3)$. The equation of the line through $P$ and perpendicular to $A B$ is $x=a$. The equation of the line $M Q$ (if $a \neq 0$ ) is
$$
y-\frac{1}{2}=\frac{x}{a}\left(\frac{1}{2}+b\right) .
$$
The intersection of the two lines is therefore
$$
N=(a, 1+b)=P+(0,1)
$$
This shows that the map $P \rightarrow N$ in the translation by the vector $(0,1)$. This result is independent of the position of $P$ (provided that $a \neq 0$, because otherwise $N$ is not well-defined).
When $P$ lies on the circumcircle of $A B C$, with the exception of the two points with $a=0$, then necessarily $N$ lies on the translated circle (which is the circle with center in $C$ and radius 1).
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $C A=C B$ and $\angle A C B=120^{\circ}$, and let $M$ be the midpoint of $A B$. Let $P$ be a variable point on the circumcircle of $A B C$, and let $Q$ be the point on the segment $C P$ such that $Q P=2 Q C$. It is given that the line through $P$ and perpendicular to $A B$ intersects the line $M Q$ at a unique point $N$.
Prove that there exists a fixed circle such that $N$ lies on this circle for all possible positions of $P$.
(Velina Ivanova, Bulgaria)
|
Let $O$ be the circumcenter of $A B C$. From the assumption that $\angle A C B=120^{\circ}$ it follows that $M$ is the midpoint of $C O$.
Let $\omega$ denote the circle with center in $C$ and radius $C O$. This circle in the image of the circumcircle of $A B C$ through the translation that sends $O$ to $C$. We claim that $N$ lies on $\omega$.
Let us consider the triangles $Q N P$ and $Q M C$. The angles in $Q$ are equal. Since $N P$ is parallel to $M C$ (both lines are perpendicular to $A B$ ), it turns out that $\angle Q N P=\angle Q M C$, and hence the two triangles are similar. Since $Q P=2 Q C$, it follows that
$$
N P=2 M C=C O
$$
which proves that $N$ lies on $\omega$.
Comment The possible positions of $N$ are all the points of $\omega$ with the exception of the two points lying on the line $C O$. Indeed, $P$ does not lie on the line $C O$ because otherwise the point $N$ is not well-defined, and therefore also $N$ does not lie on the same line.
Conversely, let $N$ be any point on $\omega$ and not lying on the line $C O$. Let $P$ be the corresponding point on the circumcircle of $A B C$, namely such that $N P$ is parallel and equal to $C O$. Let $Q$ be
the intersection of $C P$ and $N M$. As before, the triangles $Q N P$ and $Q M C$ are similar, and now from the relation $N P=2 M C$ we deduce that $Q P=2 Q C$. This proves that $N$ can be obtained from $P$ through the construction described in the statement of the problem.
Alternative solution Let $M^{\prime}$ denote the symmetric of $M$ with respect to $O$.
Let us consider the quadrilateral $M M^{\prime} P N$. The lines $M M^{\prime}$ and $N P$ are parallel by construction. Also the lines $P M^{\prime}$ and $N M$ are parallel (homothety from $C$ with coefficient 3). It follows that $M M^{\prime} P N$ is a parallelogram, and hence $P N=M M^{\prime}=O C$.
Computational solution There are many computation approaches to this problem. For example, we can set Cartesian coordinates so that
$$
A=\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right), \quad B=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right), \quad C=(0,1), \quad M=\left(0, \frac{1}{2}\right)
$$
Setting $P=(a, b)$, we obtain that $Q=(a / 3,(2+b) / 3)$. The equation of the line through $P$ and perpendicular to $A B$ is $x=a$. The equation of the line $M Q$ (if $a \neq 0$ ) is
$$
y-\frac{1}{2}=\frac{x}{a}\left(\frac{1}{2}+b\right) .
$$
The intersection of the two lines is therefore
$$
N=(a, 1+b)=P+(0,1)
$$
This shows that the map $P \rightarrow N$ in the translation by the vector $(0,1)$. This result is independent of the position of $P$ (provided that $a \neq 0$, because otherwise $N$ is not well-defined).
When $P$ lies on the circumcircle of $A B C$, with the exception of the two points with $a=0$, then necessarily $N$ lies on the translated circle (which is the circle with center in $C$ and radius 1).
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1",
"resource_path": "EGMO/segmented/en-2018-solutions.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2018"
}
|
Consider the set
$$
A=\left\{1+\frac{1}{k}: k=1,2,3, \ldots\right\}
$$
(a) Prove that every integer $x \geq 2$ can be written as the product of one or more elements of $A$, which are not necessarily different.
(b) For every integer $x \geq 2$, let $f(x)$ denote the minimum integer such that $x$ can be written as the product of $f(x)$ elements of $A$, which are not necessarily different.
Prove that there exist infinitely many pairs $(x, y)$ of integers with $x \geq 2, y \geq 2$, and
$$
f(x y)<f(x)+f(y) .
$$
(Pairs $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ are different if $x_{1} \neq x_{2}$ or $\left.y_{1} \neq y_{2}\right)$.
(Mihail Baluna, Romania)
|
Every integer $x \geq 2$ can be written as the telescopic product of $x-1$ elements of $A$ as
$$
x=\left(1+\frac{1}{x-1}\right) \cdot\left(1+\frac{1}{x-2}\right) \cdot \ldots \cdot\left(1+\frac{1}{2}\right) \cdot\left(1+\frac{1}{1}\right)
$$
which is enough to establish part (a). We now consider part (b). Notice that for any positive integer $k$ we have
$$
f\left(2^{k}+1\right) \leq k+1
$$
because $2^{k}+1=\left(1+\frac{1}{2^{k}}\right) \cdot 2^{k}$ is a representation of $2^{k}+1$ as a product of $k+1$ elements of $A$. We claim that all the pairs $(x, y)$ of the form
$$
x=5, \quad y=\frac{2^{4 k+2}+1}{5}
$$
satisfy the required inequality. Notice that $y$ is an integer for any positive value of $k$, because $2^{4 k+2}+1 \equiv 16^{k} \cdot 4+1 \equiv 5 \equiv 0(\bmod 5)$. Furthermore, $f(x y)=f\left(2^{4 k+2}+1\right) \leq 4 k+3$ (and $f(x)=f\left(2^{2}+1\right) \leq 3$ ) by the above. We now need some lower bounds on the values of $f$. Notice that no element of $A$ exceeds 2, and therefore the product of at most $k$ elements of $A$ does not exceed $2^{k}$ : it follows that
$$
f(n) \geq\left\lceil\log _{2}(n)\right\rceil
$$
and in particular that
$$
f(5)=f\left(2^{2}+1\right) \geq\left\lceil\log _{2}(5)\right\rceil=3
$$
We have thus proven $f(x)=f(5)=3$. We want to show $f(x y)<f(x)+f(y)$, and since we know $f(x y) \leq 4 k+3$ and $f(x)=3$ we are reduced to showing $f(y)>4 k$. Since $y>2^{4 k-1}$, from (Q2.1) we already know that $f(y) \geq 4 k$, and hence we just need to exclude that $f(y)=4 k$. Let us assume that we can represent $y$ in the form $a_{1} \cdot \ldots a_{4 k}$ with every $a_{i}$ in $A$. At least one of the $a_{i}$ is not 2 (otherwise the product would be a power of 2 , while $y$ is odd), and hence it is less than or equal to $3 / 2$. It follows that
$$
a_{1} \cdot \ldots \cdot a_{4 k} \leq 2^{4 k-1} \cdot \frac{3}{2}=15 \cdot \frac{2^{4 k-2}}{5}<\frac{2^{4 k+2}}{5}<y
$$
which contradicts the fact that $a_{1} \cdot \ldots \cdot a_{4 k}$ is a representation of $y$.
Note. Using a similar approach one can also prove that all pairs of the form
$$
\left(3, \frac{2^{2 k+1}+1}{3}\right) \quad \text { and } \quad\left(11, \frac{2^{10 k+5}+1}{11}\right)
$$
satisfy the required inequality.
Second solution As in the previous solution we obtain the lower bound (Q2.1).
Now we claim that all the pairs of the form
$$
x=2^{k}+1, \quad y=4^{k}-2^{k}+1
$$
satisfy the required inequality when $k$ is large enough. To begin with, it is easy to see that
$$
2^{k}+1=\frac{2^{k}+1}{2^{k}} \cdot \underbrace{2 \cdot \ldots \cdot 2}_{k \text { terms }} \quad \text { and } \quad 2^{3 k}+1=\frac{2^{3 k}+1}{2^{3 k}} \cdot \underbrace{2 \cdot \ldots \cdot 2}_{3 k \text { terms }}
$$
which shows that $f\left(2^{k}+1\right) \leq k+1$ and $f\left(2^{3 k}+1\right) \leq 3 k+1$. On the other hand, from (Q2.1) we deduce that the previous inequalities are actually equalities, and therefore
$$
f(x)=k+1 \quad \text { and } \quad f(x y)=3 k+1
$$
Therefore, it remain to show that $f(y)>2 k$. Since $y>2^{2 k-1}$ (for $k \geq 1$ ), from (Q2.1) we already know that $f(y) \geq 2 k$, and hence we just need to exclude that $f(y)=2 k$. Let us assume that we can represent $y$ in the form $a_{1} \cdot \ldots \cdot a_{2 k}$. At least one of the factors is not 2 , and hence it is less than or equal to $3 / 2$. Thus when $k$ is large enough it follows that
$$
a_{1} \cdot \ldots \cdot a_{2 k} \leq 2^{2 k-1} \cdot \frac{3}{2}=\frac{3}{4} \cdot 2^{2 k}<2^{2 k}-2^{k}<y
$$
which contradicts the fact that $a_{1} \cdot \ldots \cdot a_{2 k}$ is a representation of $y$.
Third solution Let's start by showing that $(x, y)=(7,7)$ satisfies $f(x y)<f(x)+f(y)$. We have $f(7) \geq 4$ since 7 cannot be written as the product of 3 or fewer elements of $A$ : indeed $2^{3}>7$, and any other product of at most three elements of $A$ does not exceed $2^{2} \cdot \frac{3}{2}=6<7$. On the other hand, $f(49) \leq 7$ since $49=2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot \frac{3}{2} \cdot \frac{49}{48}$.
Suppose by contradiction that there exist only finitely many pairs $(x, y)$ that satisfy $f(x y)<$ $f(x)+f(y)$. This implies that there exists $M$ large enough so that whenever $a>M$ or $b>M$ holds we have $f(a b)=f(a)+f(b)$ (indeed, it is clear that the reverse inequality $f(a b) \leq f(a)+f(b)$ is always satisfied).
Now take any pair $(x, y)$ that satisfies $f(x y)<f(x)+f(y)$ and let $n>M$ be any integer. We obtain
$$
f(n)+f(x y)=f(n x y)=f(n x)+f(y)=f(n)+f(x)+f(y),
$$
which contradicts $f(x y)<f(x)+f(y)$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Consider the set
$$
A=\left\{1+\frac{1}{k}: k=1,2,3, \ldots\right\}
$$
(a) Prove that every integer $x \geq 2$ can be written as the product of one or more elements of $A$, which are not necessarily different.
(b) For every integer $x \geq 2$, let $f(x)$ denote the minimum integer such that $x$ can be written as the product of $f(x)$ elements of $A$, which are not necessarily different.
Prove that there exist infinitely many pairs $(x, y)$ of integers with $x \geq 2, y \geq 2$, and
$$
f(x y)<f(x)+f(y) .
$$
(Pairs $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ are different if $x_{1} \neq x_{2}$ or $\left.y_{1} \neq y_{2}\right)$.
(Mihail Baluna, Romania)
|
Every integer $x \geq 2$ can be written as the telescopic product of $x-1$ elements of $A$ as
$$
x=\left(1+\frac{1}{x-1}\right) \cdot\left(1+\frac{1}{x-2}\right) \cdot \ldots \cdot\left(1+\frac{1}{2}\right) \cdot\left(1+\frac{1}{1}\right)
$$
which is enough to establish part (a). We now consider part (b). Notice that for any positive integer $k$ we have
$$
f\left(2^{k}+1\right) \leq k+1
$$
because $2^{k}+1=\left(1+\frac{1}{2^{k}}\right) \cdot 2^{k}$ is a representation of $2^{k}+1$ as a product of $k+1$ elements of $A$. We claim that all the pairs $(x, y)$ of the form
$$
x=5, \quad y=\frac{2^{4 k+2}+1}{5}
$$
satisfy the required inequality. Notice that $y$ is an integer for any positive value of $k$, because $2^{4 k+2}+1 \equiv 16^{k} \cdot 4+1 \equiv 5 \equiv 0(\bmod 5)$. Furthermore, $f(x y)=f\left(2^{4 k+2}+1\right) \leq 4 k+3$ (and $f(x)=f\left(2^{2}+1\right) \leq 3$ ) by the above. We now need some lower bounds on the values of $f$. Notice that no element of $A$ exceeds 2, and therefore the product of at most $k$ elements of $A$ does not exceed $2^{k}$ : it follows that
$$
f(n) \geq\left\lceil\log _{2}(n)\right\rceil
$$
and in particular that
$$
f(5)=f\left(2^{2}+1\right) \geq\left\lceil\log _{2}(5)\right\rceil=3
$$
We have thus proven $f(x)=f(5)=3$. We want to show $f(x y)<f(x)+f(y)$, and since we know $f(x y) \leq 4 k+3$ and $f(x)=3$ we are reduced to showing $f(y)>4 k$. Since $y>2^{4 k-1}$, from (Q2.1) we already know that $f(y) \geq 4 k$, and hence we just need to exclude that $f(y)=4 k$. Let us assume that we can represent $y$ in the form $a_{1} \cdot \ldots a_{4 k}$ with every $a_{i}$ in $A$. At least one of the $a_{i}$ is not 2 (otherwise the product would be a power of 2 , while $y$ is odd), and hence it is less than or equal to $3 / 2$. It follows that
$$
a_{1} \cdot \ldots \cdot a_{4 k} \leq 2^{4 k-1} \cdot \frac{3}{2}=15 \cdot \frac{2^{4 k-2}}{5}<\frac{2^{4 k+2}}{5}<y
$$
which contradicts the fact that $a_{1} \cdot \ldots \cdot a_{4 k}$ is a representation of $y$.
Note. Using a similar approach one can also prove that all pairs of the form
$$
\left(3, \frac{2^{2 k+1}+1}{3}\right) \quad \text { and } \quad\left(11, \frac{2^{10 k+5}+1}{11}\right)
$$
satisfy the required inequality.
Second solution As in the previous solution we obtain the lower bound (Q2.1).
Now we claim that all the pairs of the form
$$
x=2^{k}+1, \quad y=4^{k}-2^{k}+1
$$
satisfy the required inequality when $k$ is large enough. To begin with, it is easy to see that
$$
2^{k}+1=\frac{2^{k}+1}{2^{k}} \cdot \underbrace{2 \cdot \ldots \cdot 2}_{k \text { terms }} \quad \text { and } \quad 2^{3 k}+1=\frac{2^{3 k}+1}{2^{3 k}} \cdot \underbrace{2 \cdot \ldots \cdot 2}_{3 k \text { terms }}
$$
which shows that $f\left(2^{k}+1\right) \leq k+1$ and $f\left(2^{3 k}+1\right) \leq 3 k+1$. On the other hand, from (Q2.1) we deduce that the previous inequalities are actually equalities, and therefore
$$
f(x)=k+1 \quad \text { and } \quad f(x y)=3 k+1
$$
Therefore, it remain to show that $f(y)>2 k$. Since $y>2^{2 k-1}$ (for $k \geq 1$ ), from (Q2.1) we already know that $f(y) \geq 2 k$, and hence we just need to exclude that $f(y)=2 k$. Let us assume that we can represent $y$ in the form $a_{1} \cdot \ldots \cdot a_{2 k}$. At least one of the factors is not 2 , and hence it is less than or equal to $3 / 2$. Thus when $k$ is large enough it follows that
$$
a_{1} \cdot \ldots \cdot a_{2 k} \leq 2^{2 k-1} \cdot \frac{3}{2}=\frac{3}{4} \cdot 2^{2 k}<2^{2 k}-2^{k}<y
$$
which contradicts the fact that $a_{1} \cdot \ldots \cdot a_{2 k}$ is a representation of $y$.
Third solution Let's start by showing that $(x, y)=(7,7)$ satisfies $f(x y)<f(x)+f(y)$. We have $f(7) \geq 4$ since 7 cannot be written as the product of 3 or fewer elements of $A$ : indeed $2^{3}>7$, and any other product of at most three elements of $A$ does not exceed $2^{2} \cdot \frac{3}{2}=6<7$. On the other hand, $f(49) \leq 7$ since $49=2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot \frac{3}{2} \cdot \frac{49}{48}$.
Suppose by contradiction that there exist only finitely many pairs $(x, y)$ that satisfy $f(x y)<$ $f(x)+f(y)$. This implies that there exists $M$ large enough so that whenever $a>M$ or $b>M$ holds we have $f(a b)=f(a)+f(b)$ (indeed, it is clear that the reverse inequality $f(a b) \leq f(a)+f(b)$ is always satisfied).
Now take any pair $(x, y)$ that satisfies $f(x y)<f(x)+f(y)$ and let $n>M$ be any integer. We obtain
$$
f(n)+f(x y)=f(n x y)=f(n x)+f(y)=f(n)+f(x)+f(y),
$$
which contradicts $f(x y)<f(x)+f(y)$.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "\nProblem 2",
"resource_path": "EGMO/segmented/en-2018-solutions.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2018"
}
|
The $n$ contestants of an EGMO are named $C_{1}, \ldots, C_{n}$. After the competition they queue in front of the restaurant according to the following rules.
- The Jury chooses the initial order of the contestants in the queue.
- Every minute, the Jury chooses an integer $i$ with $1 \leq i \leq n$.
- If contestant $C_{i}$ has at least $i$ other contestants in front of her, she pays one euro to the Jury and moves forward in the queue by exactly $i$ positions.
- If contestant $C_{i}$ has fewer than $i$ other contestants in front of her, the restaurant opens and the process ends.
(a) Prove that the process cannot continue indefinitely, regardless of the Jury's choices.
(b) Determine for every $n$ the maximum number of euros that the Jury can collect by cunningly choosing the initial order and the sequence of moves.
(Hungary)
|
The maximal number of euros is $2^{n}-n-1$.
To begin with, we show that it is possible for the Jury to collect this number of euros. We argue by induction. Let us assume that the Jury can collect $M_{n}$ euros in a configuration with $n$ contestants. Then we show that the Jury can collect at least $2 M_{n}+n$ moves in a configuration with $n+1$ contestants. Indeed, let us begin with all the contestants lined up in reverse order. In the first $M_{n}$ moves the Jury keeps $C_{n+1}$ in first position and reverses the order of the remaining contestants, then in the next $n$ moves all contestants $C_{1}, \ldots, C_{n}$ (in this order) jump over $C_{n+1}$ and end up in the first $n$ positions of the line in reverse order, and finally in the last $M_{n}$ moves the Jury rearranges the first $n$ positions.
Since $M_{1}=0$ and $M_{n+1} \geq 2 M_{n}+n$, an easy induction shows that $M_{n} \geq 2^{n}-n-1$.
Let us show now that at most $2^{n}-n-1$ moves are possible. To this end, let us identify a line of contestants with a permutation $\sigma$ of $\{1, \ldots, n\}$. To each permutation we associate the set of reverse pairs
$$
R(\sigma):=\{(i, j): 1 \leq i<j \leq n \text { and } \sigma(i)>\sigma(j)\}
$$
and the nonnegative integer
$$
W(\sigma):=\sum_{(i, j) \in R(\sigma)} 2^{i}
$$
which we call the total weight of the permutation. We claim that the total weight decreases after any move of the contestants. Indeed, let us assume that $C_{i}$ moves forward in the queue, let $\sigma$ be the permutation before the move, and let $\sigma^{\prime}$ denote the permutation after the move. Since $C_{i}$ jumps over exactly $i$ contestants, necessarily she jumps over at least one contestant $C_{j}$ with index
$j>i$. This means that the pair $(i, j)$ is reverse with respect to $\sigma$ but not with respect to $\sigma^{\prime}$, and this yields a reduction of $2^{i}$ in the total weight. On the other hand, the move by $C_{i}$ can create new reverse pairs of the form $(k, i)$ with $k<i$, but their total contribution is at most
$$
2^{0}+2^{1}+\ldots+2^{i-1}=2^{i}-1
$$
In conclusion, when passing from $\sigma$ to $\sigma^{\prime}$, at least one term $2^{i}$ disappears from the computation of the total weight, and the sum of all the new terms that might have been created is at most $2^{i}-1$. This shows that $W\left(\sigma^{\prime}\right) \leq W(\sigma)-1$.
We conclude by observing that the maximum possible value of $W(\sigma)$ is realized when all pairs are reverse, in which case
$$
W(\sigma)=\sum_{i=1}^{n}(i-1) 2^{i}=2^{n}-n-1 .
$$
This proves that the number of moves is less than or equal to $2^{n}-n-1$, and in particular it is finite.
Alternative solution As in the previous solution, the fundamental observation is again that, when a contestant $C_{i}$ moves forward, necessarily she has to jump over at least one contestant $C_{j}$ with $j>i$.
Let us show now that the process ends after a finite number of moves. Let us assume that this is not the case. Then at least one contestant moves infinitely many times. Let $i_{0}$ be the largest index such that $C_{i_{0}}$ moves infinitely many times. Then necessarily $C_{i_{0}}$ jumps infinitely many times over some fixed $C_{j_{0}}$ with $j_{0}>i_{0}$. On the other hand, we know that $C_{j_{0}}$ makes only a finite number of moves, and therefore she can precede $C_{i_{0}}$ in the line only a finite number of times, which is absurd.
In order to estimate from above the maximal number of moves, we show that the contestant $C_{i}$ can make at most $2^{n-i}-1$ moves. Indeed, let us argue by "backward extended induction". To begin with, we observe that the estimate is trivially true for $C_{n}$ because she has no legal move.
Let us assume now that the estimate has been proved for $C_{i}, C_{i+1}, \ldots, C_{n}$, and let us prove it for $C_{i-1}$. When $C_{i-1}$ moves, at least one contestant $C_{j}$ with $j>i-1$ must precede her in the line. The initial configuration can provide at most $n-i$ contestants with larger index in front of $C_{i-1}$, which means at most $n-i$ moves for $C_{i-1}$. All other moves are possible only if some contestant in the range $C_{i}, C_{i+1}, \ldots, C_{n}$ jumps over $C_{i-1}$ during her moves. As a consequence, the total number of moves of $C_{i-1}$ is at most
$$
n-i+\sum_{k=i}^{n}\left(2^{n-k}-1\right)=2^{n-i+1}-1
$$
Summing over all indices we obtain that
$$
\sum_{i=1}^{n}\left(2^{n-i}-1\right)=2^{n}-n-1
$$
which gives an estimate for the total number of moves.
The same example of the first solution shows that this upper bound can actually be achieved.
Comment In every move of the example, the moving contestant jumps over exactly one contestant with larger index (and as a consequence over all contestants with smaller index).
|
2^{n}-n-1
|
Yes
|
Yes
|
proof
|
Combinatorics
|
The $n$ contestants of an EGMO are named $C_{1}, \ldots, C_{n}$. After the competition they queue in front of the restaurant according to the following rules.
- The Jury chooses the initial order of the contestants in the queue.
- Every minute, the Jury chooses an integer $i$ with $1 \leq i \leq n$.
- If contestant $C_{i}$ has at least $i$ other contestants in front of her, she pays one euro to the Jury and moves forward in the queue by exactly $i$ positions.
- If contestant $C_{i}$ has fewer than $i$ other contestants in front of her, the restaurant opens and the process ends.
(a) Prove that the process cannot continue indefinitely, regardless of the Jury's choices.
(b) Determine for every $n$ the maximum number of euros that the Jury can collect by cunningly choosing the initial order and the sequence of moves.
(Hungary)
|
The maximal number of euros is $2^{n}-n-1$.
To begin with, we show that it is possible for the Jury to collect this number of euros. We argue by induction. Let us assume that the Jury can collect $M_{n}$ euros in a configuration with $n$ contestants. Then we show that the Jury can collect at least $2 M_{n}+n$ moves in a configuration with $n+1$ contestants. Indeed, let us begin with all the contestants lined up in reverse order. In the first $M_{n}$ moves the Jury keeps $C_{n+1}$ in first position and reverses the order of the remaining contestants, then in the next $n$ moves all contestants $C_{1}, \ldots, C_{n}$ (in this order) jump over $C_{n+1}$ and end up in the first $n$ positions of the line in reverse order, and finally in the last $M_{n}$ moves the Jury rearranges the first $n$ positions.
Since $M_{1}=0$ and $M_{n+1} \geq 2 M_{n}+n$, an easy induction shows that $M_{n} \geq 2^{n}-n-1$.
Let us show now that at most $2^{n}-n-1$ moves are possible. To this end, let us identify a line of contestants with a permutation $\sigma$ of $\{1, \ldots, n\}$. To each permutation we associate the set of reverse pairs
$$
R(\sigma):=\{(i, j): 1 \leq i<j \leq n \text { and } \sigma(i)>\sigma(j)\}
$$
and the nonnegative integer
$$
W(\sigma):=\sum_{(i, j) \in R(\sigma)} 2^{i}
$$
which we call the total weight of the permutation. We claim that the total weight decreases after any move of the contestants. Indeed, let us assume that $C_{i}$ moves forward in the queue, let $\sigma$ be the permutation before the move, and let $\sigma^{\prime}$ denote the permutation after the move. Since $C_{i}$ jumps over exactly $i$ contestants, necessarily she jumps over at least one contestant $C_{j}$ with index
$j>i$. This means that the pair $(i, j)$ is reverse with respect to $\sigma$ but not with respect to $\sigma^{\prime}$, and this yields a reduction of $2^{i}$ in the total weight. On the other hand, the move by $C_{i}$ can create new reverse pairs of the form $(k, i)$ with $k<i$, but their total contribution is at most
$$
2^{0}+2^{1}+\ldots+2^{i-1}=2^{i}-1
$$
In conclusion, when passing from $\sigma$ to $\sigma^{\prime}$, at least one term $2^{i}$ disappears from the computation of the total weight, and the sum of all the new terms that might have been created is at most $2^{i}-1$. This shows that $W\left(\sigma^{\prime}\right) \leq W(\sigma)-1$.
We conclude by observing that the maximum possible value of $W(\sigma)$ is realized when all pairs are reverse, in which case
$$
W(\sigma)=\sum_{i=1}^{n}(i-1) 2^{i}=2^{n}-n-1 .
$$
This proves that the number of moves is less than or equal to $2^{n}-n-1$, and in particular it is finite.
Alternative solution As in the previous solution, the fundamental observation is again that, when a contestant $C_{i}$ moves forward, necessarily she has to jump over at least one contestant $C_{j}$ with $j>i$.
Let us show now that the process ends after a finite number of moves. Let us assume that this is not the case. Then at least one contestant moves infinitely many times. Let $i_{0}$ be the largest index such that $C_{i_{0}}$ moves infinitely many times. Then necessarily $C_{i_{0}}$ jumps infinitely many times over some fixed $C_{j_{0}}$ with $j_{0}>i_{0}$. On the other hand, we know that $C_{j_{0}}$ makes only a finite number of moves, and therefore she can precede $C_{i_{0}}$ in the line only a finite number of times, which is absurd.
In order to estimate from above the maximal number of moves, we show that the contestant $C_{i}$ can make at most $2^{n-i}-1$ moves. Indeed, let us argue by "backward extended induction". To begin with, we observe that the estimate is trivially true for $C_{n}$ because she has no legal move.
Let us assume now that the estimate has been proved for $C_{i}, C_{i+1}, \ldots, C_{n}$, and let us prove it for $C_{i-1}$. When $C_{i-1}$ moves, at least one contestant $C_{j}$ with $j>i-1$ must precede her in the line. The initial configuration can provide at most $n-i$ contestants with larger index in front of $C_{i-1}$, which means at most $n-i$ moves for $C_{i-1}$. All other moves are possible only if some contestant in the range $C_{i}, C_{i+1}, \ldots, C_{n}$ jumps over $C_{i-1}$ during her moves. As a consequence, the total number of moves of $C_{i-1}$ is at most
$$
n-i+\sum_{k=i}^{n}\left(2^{n-k}-1\right)=2^{n-i+1}-1
$$
Summing over all indices we obtain that
$$
\sum_{i=1}^{n}\left(2^{n-i}-1\right)=2^{n}-n-1
$$
which gives an estimate for the total number of moves.
The same example of the first solution shows that this upper bound can actually be achieved.
Comment In every move of the example, the moving contestant jumps over exactly one contestant with larger index (and as a consequence over all contestants with smaller index).
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "\nProblem 3",
"resource_path": "EGMO/segmented/en-2018-solutions.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2018"
}
|
A domino is a $1 \times 2$ or $2 \times 1$ tile.
Let $n \geq 3$ be an integer. Dominoes are placed on an $n \times n$ board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap.
The value of a row or column is the number of dominoes that cover at least one cell of this row or column. The configuration is called balanced if there exists some $k \geq 1$ such that each row and each column has a value of $k$.
Prove that a balanced configuration exists for every $n \geq 3$, and find the minimum number of dominoes needed in such a configuration.
(Merlijn Staps, The Netherlands)
|
The minimal number of dominoes required in a balanced configuration is $2 n / 3$ if $n$ is a multiple of 3 , and $2 n$ otherwise.
In order to show that this number is necessary, we count in two different ways the number of elements of the set $S$ of all pairs $(\ell, d)$, where $\ell$ is a row or a column of the board, and $d$ is a domino that covers at least one cell of that row or column. On the one hand, since each row or column intersects the same number $k$ of dominoes, the set $S$ has $2 n k$ elements. On the other hand, since each domino intersects 3 rows/columns, the set $S$ has $3 D$ elements, where $D$ is the total number of dominoes on the board. This leads to the equality
$$
2 n k=3 D
$$
If $n$ is a multiple of 3 , from the trivial inequality $k \geq 1$ we obtain that $D \geq 2 n / 3$. If $n$ is not a multiple of 3 , then $k$ is a multiple of 3 , which means that $k \geq 3$ and hence $D \geq 2 n$.
Now we need to exhibit a balanced configuration with this number of dominoes. The following diagram shows a balanced configuration with $n=3$ and $k=1$.
If $n$ is any multiple of 3 , we can obtain a balanced configuration with $k=1$ by using $n / 3$ of these $3 \times 3$ blocks along the principal diagonal of the board.
The following diagrams show balanced configurations with $k=3$ and $n \in\{4,5,6,7\}$.
Any $n \geq 8$ can be written in the form $4 A+r$ where $A$ is a positive integer and $r \in\{4,5,6,7\}$. Therefore, we can obtain a balanced configuration with $n \geq 8$ and $k=3$ by using one block with size $r \times r$, and $A$ blocks with size $4 \times 4$ along the principal diagonal of the board. In particular, this construction covers all the cases where $n$ is not a multiple of 3 .
|
2 n / 3 \text{ if } n \text{ is a multiple of 3, and } 2 n \text{ otherwise}
|
Yes
|
Yes
|
proof
|
Combinatorics
|
A domino is a $1 \times 2$ or $2 \times 1$ tile.
Let $n \geq 3$ be an integer. Dominoes are placed on an $n \times n$ board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap.
The value of a row or column is the number of dominoes that cover at least one cell of this row or column. The configuration is called balanced if there exists some $k \geq 1$ such that each row and each column has a value of $k$.
Prove that a balanced configuration exists for every $n \geq 3$, and find the minimum number of dominoes needed in such a configuration.
(Merlijn Staps, The Netherlands)
|
The minimal number of dominoes required in a balanced configuration is $2 n / 3$ if $n$ is a multiple of 3 , and $2 n$ otherwise.
In order to show that this number is necessary, we count in two different ways the number of elements of the set $S$ of all pairs $(\ell, d)$, where $\ell$ is a row or a column of the board, and $d$ is a domino that covers at least one cell of that row or column. On the one hand, since each row or column intersects the same number $k$ of dominoes, the set $S$ has $2 n k$ elements. On the other hand, since each domino intersects 3 rows/columns, the set $S$ has $3 D$ elements, where $D$ is the total number of dominoes on the board. This leads to the equality
$$
2 n k=3 D
$$
If $n$ is a multiple of 3 , from the trivial inequality $k \geq 1$ we obtain that $D \geq 2 n / 3$. If $n$ is not a multiple of 3 , then $k$ is a multiple of 3 , which means that $k \geq 3$ and hence $D \geq 2 n$.
Now we need to exhibit a balanced configuration with this number of dominoes. The following diagram shows a balanced configuration with $n=3$ and $k=1$.
If $n$ is any multiple of 3 , we can obtain a balanced configuration with $k=1$ by using $n / 3$ of these $3 \times 3$ blocks along the principal diagonal of the board.
The following diagrams show balanced configurations with $k=3$ and $n \in\{4,5,6,7\}$.
Any $n \geq 8$ can be written in the form $4 A+r$ where $A$ is a positive integer and $r \in\{4,5,6,7\}$. Therefore, we can obtain a balanced configuration with $n \geq 8$ and $k=3$ by using one block with size $r \times r$, and $A$ blocks with size $4 \times 4$ along the principal diagonal of the board. In particular, this construction covers all the cases where $n$ is not a multiple of 3 .
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "\nProblem 4",
"resource_path": "EGMO/segmented/en-2018-solutions.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2018"
}
|
Let $\Gamma$ be the circumcircle of triangle $A B C$. A circle $\Omega$ is tangent to the line segment $A B$ and is tangent to $\Gamma$ at a point lying on the same side of the line $A B$ as $C$. The angle bisector of $\angle B C A$ intersects $\Omega$ at two different points $P$ and $Q$.
Prove that $\angle A B P=\angle Q B C$.
(Dominika Regiec, Poland)
|
1 Let $M$ be the midpoint of the arc $A B$ that does not contain $C$, let $V$ be the intersection of $\Omega$ and $\Gamma$, and let $U$ be the intersection of $\Omega$ and $A B$.
The proof can be divided in two steps:
1). Proving that $M P \cdot M Q=M B^{2}$.
It is well-known that $V, U$ and $M$ are collinear (indeed the homothety with center in $V$ that sends $\Omega$ to $\Gamma$ sends $U$ to the point of $\Gamma$ where the tangent to $\Gamma$ is parallel to $A B$, and this point is $M$ ), and
$$
M V \cdot M U=M A^{2}=M B^{2}
$$
This follows from the similitude between the triangles $\triangle M A V$ and $\triangle M U A$. Alternatively, it is a consequence of the following well-known lemma: Given a circle $\Gamma$ with a chord $A B$, let $M$ be the middle point of one of the two arcs $A B$. Take a line through $M$ which intersects $\Gamma$ again at $X$ and $A B$ at $Y$. Then $M X \cdot M Y$ is independent of the choice of the line.
Computing the power of $M$ with respect to $\Omega$ we obtain that
$$
M P \cdot M Q=M U \cdot M V=M B^{2}
$$
2). Conclude the proof given that $M P \cdot M Q=M B^{2}$.
The relation $M P \cdot M Q=M B^{2}$ in turn implies that triangle $\triangle M B P$ is similar to triangle $\triangle M Q B$, and in particular $\angle M B P=\angle M Q B$. Keeping into account that $\angle M C B=$ $\angle M B A$, we finally conclude that
$$
\angle Q B C=\angle M Q B-\angle M C B=\angle M B P-\angle M B A=\angle P B A
$$
as required.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $\Gamma$ be the circumcircle of triangle $A B C$. A circle $\Omega$ is tangent to the line segment $A B$ and is tangent to $\Gamma$ at a point lying on the same side of the line $A B$ as $C$. The angle bisector of $\angle B C A$ intersects $\Omega$ at two different points $P$ and $Q$.
Prove that $\angle A B P=\angle Q B C$.
(Dominika Regiec, Poland)
|
1 Let $M$ be the midpoint of the arc $A B$ that does not contain $C$, let $V$ be the intersection of $\Omega$ and $\Gamma$, and let $U$ be the intersection of $\Omega$ and $A B$.
The proof can be divided in two steps:
1). Proving that $M P \cdot M Q=M B^{2}$.
It is well-known that $V, U$ and $M$ are collinear (indeed the homothety with center in $V$ that sends $\Omega$ to $\Gamma$ sends $U$ to the point of $\Gamma$ where the tangent to $\Gamma$ is parallel to $A B$, and this point is $M$ ), and
$$
M V \cdot M U=M A^{2}=M B^{2}
$$
This follows from the similitude between the triangles $\triangle M A V$ and $\triangle M U A$. Alternatively, it is a consequence of the following well-known lemma: Given a circle $\Gamma$ with a chord $A B$, let $M$ be the middle point of one of the two arcs $A B$. Take a line through $M$ which intersects $\Gamma$ again at $X$ and $A B$ at $Y$. Then $M X \cdot M Y$ is independent of the choice of the line.
Computing the power of $M$ with respect to $\Omega$ we obtain that
$$
M P \cdot M Q=M U \cdot M V=M B^{2}
$$
2). Conclude the proof given that $M P \cdot M Q=M B^{2}$.
The relation $M P \cdot M Q=M B^{2}$ in turn implies that triangle $\triangle M B P$ is similar to triangle $\triangle M Q B$, and in particular $\angle M B P=\angle M Q B$. Keeping into account that $\angle M C B=$ $\angle M B A$, we finally conclude that
$$
\angle Q B C=\angle M Q B-\angle M C B=\angle M B P-\angle M B A=\angle P B A
$$
as required.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "\nProblem 5",
"resource_path": "EGMO/segmented/en-2018-solutions.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2018"
}
|
Let $\Gamma$ be the circumcircle of triangle $A B C$. A circle $\Omega$ is tangent to the line segment $A B$ and is tangent to $\Gamma$ at a point lying on the same side of the line $A B$ as $C$. The angle bisector of $\angle B C A$ intersects $\Omega$ at two different points $P$ and $Q$.
Prove that $\angle A B P=\angle Q B C$.
(Dominika Regiec, Poland)
|
2 The second solution is in fact a different proof of the first part of Solution 1.
Let us consider the inversion with respect to circle with center $M$ and radius $M A=M B$. This inversion switches $A B$ and $\Gamma$, and fixes the line passing through $M, U, V$. As a consequence, it keeps $\Omega$ fixed, and therefore it switches $P$ and $Q$. This is because they are the intersections between the fixed line $M C$ and $\Omega$, and the only fixed point on the segment $M C$ is its intersection with the inversion circle (thus $P$ and $Q$ are switched). This implies that $M P \cdot M Q=M B^{2}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $\Gamma$ be the circumcircle of triangle $A B C$. A circle $\Omega$ is tangent to the line segment $A B$ and is tangent to $\Gamma$ at a point lying on the same side of the line $A B$ as $C$. The angle bisector of $\angle B C A$ intersects $\Omega$ at two different points $P$ and $Q$.
Prove that $\angle A B P=\angle Q B C$.
(Dominika Regiec, Poland)
|
2 The second solution is in fact a different proof of the first part of Solution 1.
Let us consider the inversion with respect to circle with center $M$ and radius $M A=M B$. This inversion switches $A B$ and $\Gamma$, and fixes the line passing through $M, U, V$. As a consequence, it keeps $\Omega$ fixed, and therefore it switches $P$ and $Q$. This is because they are the intersections between the fixed line $M C$ and $\Omega$, and the only fixed point on the segment $M C$ is its intersection with the inversion circle (thus $P$ and $Q$ are switched). This implies that $M P \cdot M Q=M B^{2}$.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "\nProblem 5",
"resource_path": "EGMO/segmented/en-2018-solutions.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2018"
}
|
Let $\Gamma$ be the circumcircle of triangle $A B C$. A circle $\Omega$ is tangent to the line segment $A B$ and is tangent to $\Gamma$ at a point lying on the same side of the line $A B$ as $C$. The angle bisector of $\angle B C A$ intersects $\Omega$ at two different points $P$ and $Q$.
Prove that $\angle A B P=\angle Q B C$.
(Dominika Regiec, Poland)
|
3 This solution is instead a different proof of the second step of Solution 1.
Let $I$ and $J$ be the incenter and the $C$-excenter of $\triangle A B C$ respectively. It is well-known that $M A=M I=M J$, therefore the relation $M P \cdot M Q=M A^{2}$ implies that $(P, Q, I, J)=-1$.
Now observe that $\angle I B J=90^{\circ}$, thus $B I$ is the angle bisector of $\angle P B Q$ as it is well-known from the theory of harmonic pencils, and this leads easily to the conclusion.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $\Gamma$ be the circumcircle of triangle $A B C$. A circle $\Omega$ is tangent to the line segment $A B$ and is tangent to $\Gamma$ at a point lying on the same side of the line $A B$ as $C$. The angle bisector of $\angle B C A$ intersects $\Omega$ at two different points $P$ and $Q$.
Prove that $\angle A B P=\angle Q B C$.
(Dominika Regiec, Poland)
|
3 This solution is instead a different proof of the second step of Solution 1.
Let $I$ and $J$ be the incenter and the $C$-excenter of $\triangle A B C$ respectively. It is well-known that $M A=M I=M J$, therefore the relation $M P \cdot M Q=M A^{2}$ implies that $(P, Q, I, J)=-1$.
Now observe that $\angle I B J=90^{\circ}$, thus $B I$ is the angle bisector of $\angle P B Q$ as it is well-known from the theory of harmonic pencils, and this leads easily to the conclusion.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "\nProblem 5",
"resource_path": "EGMO/segmented/en-2018-solutions.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2018"
}
|
Let $\Gamma$ be the circumcircle of triangle $A B C$. A circle $\Omega$ is tangent to the line segment $A B$ and is tangent to $\Gamma$ at a point lying on the same side of the line $A B$ as $C$. The angle bisector of $\angle B C A$ intersects $\Omega$ at two different points $P$ and $Q$.
Prove that $\angle A B P=\angle Q B C$.
(Dominika Regiec, Poland)
|
4 Let $D$ denote the intersection of $A B$ and $C M$. Let us consider an inversion with respect to $B$, and let us use primes to denote corresponding points in the transformed diagram, with the gentlemen agreement that $B^{\prime}=B$.
Since inversion preserves angles, it turns out that
$$
\angle A^{\prime} B^{\prime} M^{\prime}=\angle A^{\prime} M^{\prime} B^{\prime}=\angle A C B,
$$
and in particular triangle $A^{\prime} B^{\prime} M^{\prime}$ is isosceles with basis $B^{\prime} M^{\prime}$.
The image of $C M$ is the circumcircle of $B^{\prime} C^{\prime} M^{\prime}$, which we denote by $\omega^{\prime}$. It follows that the centers of both $\omega^{\prime}$ and the image $\Omega^{\prime}$ of $\Omega$ lie on the perpendicular bisector of $B^{\prime} M^{\prime}$. Therefore, the whole transformed diagram is symmetric with respect to the perpendicular bisector of $B^{\prime} M^{\prime}$, and in particular the $\operatorname{arcs} D^{\prime} P^{\prime}$ and $Q^{\prime} C^{\prime}$ of $\omega^{\prime}$ are equal.
This is enough to conclude that $\angle D^{\prime} B^{\prime} P^{\prime}=\angle Q^{\prime} B^{\prime} C^{\prime}$, which implies the conclusion.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $\Gamma$ be the circumcircle of triangle $A B C$. A circle $\Omega$ is tangent to the line segment $A B$ and is tangent to $\Gamma$ at a point lying on the same side of the line $A B$ as $C$. The angle bisector of $\angle B C A$ intersects $\Omega$ at two different points $P$ and $Q$.
Prove that $\angle A B P=\angle Q B C$.
(Dominika Regiec, Poland)
|
4 Let $D$ denote the intersection of $A B$ and $C M$. Let us consider an inversion with respect to $B$, and let us use primes to denote corresponding points in the transformed diagram, with the gentlemen agreement that $B^{\prime}=B$.
Since inversion preserves angles, it turns out that
$$
\angle A^{\prime} B^{\prime} M^{\prime}=\angle A^{\prime} M^{\prime} B^{\prime}=\angle A C B,
$$
and in particular triangle $A^{\prime} B^{\prime} M^{\prime}$ is isosceles with basis $B^{\prime} M^{\prime}$.
The image of $C M$ is the circumcircle of $B^{\prime} C^{\prime} M^{\prime}$, which we denote by $\omega^{\prime}$. It follows that the centers of both $\omega^{\prime}$ and the image $\Omega^{\prime}$ of $\Omega$ lie on the perpendicular bisector of $B^{\prime} M^{\prime}$. Therefore, the whole transformed diagram is symmetric with respect to the perpendicular bisector of $B^{\prime} M^{\prime}$, and in particular the $\operatorname{arcs} D^{\prime} P^{\prime}$ and $Q^{\prime} C^{\prime}$ of $\omega^{\prime}$ are equal.
This is enough to conclude that $\angle D^{\prime} B^{\prime} P^{\prime}=\angle Q^{\prime} B^{\prime} C^{\prime}$, which implies the conclusion.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "\nProblem 5",
"resource_path": "EGMO/segmented/en-2018-solutions.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2018"
}
|
(a) Prove that for every real number $t$ such that $0<t<\frac{1}{2}$ there exists a positive integer $n$ with the following property: for every set $S$ of $n$ positive integers there exist two different elements $x$ and $y$ of $S$, and a non-negative integer $m$ (i.e. $m \geq 0$ ), such that
$$
|x-m y| \leq t y
$$
(b) Determine whether for every real number $t$ such that $0<t<\frac{1}{2}$ there exists an infinite set $S$ of positive integers such that
$$
|x-m y|>t y
$$
for every pair of different elements $x$ and $y$ of $S$ and every positive integer $m$ (i.e. $m>0$ ).
(Merlijn Staps, The Netherlands)
|
Part (a) Let $n$ be any positive integer such that
$$
(1+t)^{n-1} \geq \frac{1}{t}
$$
(this inequality is actually true for every large enough $n$ due to Bernoulli's inequality).
Let $S$ be any set of $n$ distinct positive integers, which we denote by
$$
s_{1}<s_{2}<\ldots<s_{n}
$$
We distinguish two cases.
- If $s_{i+1} \leq(1+t) s_{i}$ for some $i \in\{1, \ldots, n-1\}$, then
$$
\left|s_{i+1}-s_{i}\right|=s_{i+1}-s_{i} \leq t s_{i}
$$
and therefore the required inequality is satisfied with $x=s_{i+1}, y=s_{i}$, and $m=1$.
- If $s_{i+1}>(1+t) s_{i}$ for every $i \in\{1, \ldots, n-1\}$, then by induction we obtain that
$$
s_{n}>(1+t)^{n-1} s_{1} .
$$
As a consequence, from (Q6.1) it follows that
$$
\left|s_{1}\right|=s_{1}<\frac{1}{(1+t)^{n-1}} \cdot s_{n} \leq t s_{n}
$$
and therefore the required inequality is satisfied with $x=s_{1}, y=s_{n}$, and $m=0$.
Part (b) (Explicit formula) We claim that an infinite set with the required property exists. To this end, we rewrite the required condition in the form
$$
\left|\frac{x}{y}-m\right|>t
$$
This is equivalent to saying that the distance between the ratio $x / y$ and the set of positive integers is greater than $t$.
Now we construct an increasing sequence $s_{n}$ of odd coprime positive integers satisfying
$$
\frac{1}{2}-\frac{1}{2 s_{n}}>t \quad \forall n \geq 1
$$
and such that for every $j>i$ it turns out that
$$
\frac{s_{i}}{s_{j}}<\frac{1}{2} \quad \text { and } \quad t<\left\{\frac{s_{j}}{s_{i}}\right\}<\frac{1}{2}
$$
where $\{\alpha\}$ denotes the fractional part of $\alpha$. This is enough to show that the set $S:=\left\{s_{n}: n \geq 1\right\}$ has the required property.
To this end, we consider the sequence defined recursively by
$$
s_{n+1}=\frac{\left(s_{1} \cdot \ldots \cdot s_{n}\right)^{2}+1}{2}
$$
with $s_{1}$ large enough. An easy induction shows that this is an increasing sequence of odd positive integers. For every $i \in\{1, \ldots, n\}$ it turns out that
$$
\frac{s_{i}}{s_{n+1}} \leq \frac{2}{s_{i}} \leq \frac{2}{s_{1}}<\frac{1}{2}
$$
because $s_{1}$ is large enough, which proves the first relation in (Q6.3). Moreover, it turns out that
$$
\frac{s_{n+1}}{s_{i}}=\frac{\left(s_{1} \cdot \ldots \cdot s_{n}\right)^{2}}{2 s_{i}}+\frac{1}{2 s_{i}}
$$
The first term is a positive integer plus $1 / 2$, from which it follows that the distance of $s_{n+1} / s_{i}$ from the positive integers is greater than or equal to
$$
\frac{1}{2}-\frac{1}{2 s_{i}} \geq \frac{1}{2}-\frac{1}{2 s_{1}}
$$
which is greater than $t$ if $s_{1}$ is large enough. This proves the second relation in (Q6.3).
Part (b) (Arithmetic approach) We produce an increasing sequence $s_{n}$ of odd and coprime positive integers that satisfies (Q6.3) every $j>i$. As in the previous solution, this is enough to conclude.
We argue by induction. To begin with, we choose $s_{1}$ to be any odd integer satisfying the inequality in (Q6.2). Let us assume now that $s_{1}, \ldots, s_{n}$ have already been chosen, and let us choose $s_{n+1}$ in such a way that
$$
s_{n+1} \equiv \frac{s_{i}-1}{2} \quad\left(\bmod s_{i}\right) \quad \forall i \in\{1, \ldots, n\}
$$
We can solve this system because the previously chosen integers are odd and coprime. Moreover, any solution of this system is coprime with $s_{1}, \ldots, s_{n}$. Indeed, for every $1 \leq i \leq n$ it turns out that
$$
s_{n+1}=\frac{s_{i}-1}{2}+k_{i} s_{i}
$$
for some positive integer $k_{i}$. Therefore, any prime $p$ that divides both $s_{n+1}$ and $s_{i}$ divides also $\left(2 k_{i}+1\right) s_{i}-2 s_{n+1}=1$, which is absurd. Finally, we observe that we can assume that $s_{n+1}$ is odd and large enough. In this way we can guarantee that
$$
\frac{s_{i}}{s_{n+1}}<\frac{1}{2} \quad \forall i \in\{1, \ldots, n\}
$$
which is the first requirement in (Q6.3), and
$$
k_{i}+t<k_{i}+\frac{1}{2}-\frac{1}{2 s_{i}}=\frac{s_{n+1}}{s_{i}}<k_{i}+\frac{1}{2} \quad \forall i \in\{1, \ldots, n\}
$$
which implies the second requirement in (Q6.3).
Part (b) (Algebraic approach) Again we produce an increasing sequence $s_{n}$ of positive integers that satisfies (Q6.3) every $j>i$.
To this end, for every positive integer $x$, we define its security region
$$
S(x):=\bigcup_{n \geq 1}\left((n+t) x,\left(n+\frac{1}{2}\right) x\right) .
$$
The security region $S(x)$ is a periodic countable union of intervals of length $\left(\frac{1}{2}-t\right) x$, whose left-hand or right-hand endpoints form an arithmetic sequence. It has the property that
$$
t<\left\{\frac{y}{x}\right\}<\frac{1}{2} \quad \forall y \in S(x)
$$
Now we prove by induction that we can choose a sequence $s_{n}$ of positive integers satisfying (Q6.3) and in addition the fact that every interval of the security region $S\left(s_{n}\right)$ contains at least one interval of $S\left(s_{n-1}\right)$.
To begin with, we choose $s_{1}$ large enough so that the length of the intervals of $S\left(s_{1}\right)$ is larger than 1. This guarantees that any interval of $S\left(s_{1}\right)$ contains at least a positive integer. Now let us choose a positive integer $s_{2} \in S\left(s_{1}\right)$ that is large enough. This guarantees that $s_{1} / s_{2}$ is small enough, that the fractional part of $s_{2} / s_{1}$ is in $(t, 1 / 2)$, and that every interval of the security region $S\left(s_{2}\right)$ contains at least one interval of $S\left(s_{1}\right)$, and hence at least one positive integer.
Let us now assume that $s_{1}, \ldots, s_{n}$ have been already chosen with the required properties. We know that every interval of $S\left(s_{n}\right)$ contains at least one interval of $S\left(s_{n-1}\right)$, which in turn contains an interval in $S\left(s_{n-2}\right)$, and so on up to $S\left(s_{1}\right)$. As a consequence, we can choose a large enough positive integer $s_{n+1}$ that lies in $S\left(s_{k}\right)$ for every $k \in\{1, \ldots, n\}$. Since $s_{n+1}$ is large enough, we are sure that
$$
\frac{s_{k}}{s_{n+1}}<t \quad \forall k \in\{1, \ldots, n\}
$$
Moreover, we are sure also that all the intervals of $S\left(s_{n+1}\right)$ are large enough, and therefore they contain at least one interval of $S\left(s_{n}\right)$, which in turn contains at least one interval of $S\left(s_{n-1}\right)$, and so on. Finally, the condition
$$
t<\left\{\frac{s_{n-1}}{s_{n}}\right\}<\frac{1}{2}
$$
is guaranteed by the fact that $s_{n+1}$ was chosen in an interval that is contained in $S\left(s_{k}\right)$ for every $k \in\{1, \ldots, n\}$. This completes the induction.
|
proof
|
Yes
|
Incomplete
|
proof
|
Number Theory
|
(a) Prove that for every real number $t$ such that $0<t<\frac{1}{2}$ there exists a positive integer $n$ with the following property: for every set $S$ of $n$ positive integers there exist two different elements $x$ and $y$ of $S$, and a non-negative integer $m$ (i.e. $m \geq 0$ ), such that
$$
|x-m y| \leq t y
$$
(b) Determine whether for every real number $t$ such that $0<t<\frac{1}{2}$ there exists an infinite set $S$ of positive integers such that
$$
|x-m y|>t y
$$
for every pair of different elements $x$ and $y$ of $S$ and every positive integer $m$ (i.e. $m>0$ ).
(Merlijn Staps, The Netherlands)
|
Part (a) Let $n$ be any positive integer such that
$$
(1+t)^{n-1} \geq \frac{1}{t}
$$
(this inequality is actually true for every large enough $n$ due to Bernoulli's inequality).
Let $S$ be any set of $n$ distinct positive integers, which we denote by
$$
s_{1}<s_{2}<\ldots<s_{n}
$$
We distinguish two cases.
- If $s_{i+1} \leq(1+t) s_{i}$ for some $i \in\{1, \ldots, n-1\}$, then
$$
\left|s_{i+1}-s_{i}\right|=s_{i+1}-s_{i} \leq t s_{i}
$$
and therefore the required inequality is satisfied with $x=s_{i+1}, y=s_{i}$, and $m=1$.
- If $s_{i+1}>(1+t) s_{i}$ for every $i \in\{1, \ldots, n-1\}$, then by induction we obtain that
$$
s_{n}>(1+t)^{n-1} s_{1} .
$$
As a consequence, from (Q6.1) it follows that
$$
\left|s_{1}\right|=s_{1}<\frac{1}{(1+t)^{n-1}} \cdot s_{n} \leq t s_{n}
$$
and therefore the required inequality is satisfied with $x=s_{1}, y=s_{n}$, and $m=0$.
Part (b) (Explicit formula) We claim that an infinite set with the required property exists. To this end, we rewrite the required condition in the form
$$
\left|\frac{x}{y}-m\right|>t
$$
This is equivalent to saying that the distance between the ratio $x / y$ and the set of positive integers is greater than $t$.
Now we construct an increasing sequence $s_{n}$ of odd coprime positive integers satisfying
$$
\frac{1}{2}-\frac{1}{2 s_{n}}>t \quad \forall n \geq 1
$$
and such that for every $j>i$ it turns out that
$$
\frac{s_{i}}{s_{j}}<\frac{1}{2} \quad \text { and } \quad t<\left\{\frac{s_{j}}{s_{i}}\right\}<\frac{1}{2}
$$
where $\{\alpha\}$ denotes the fractional part of $\alpha$. This is enough to show that the set $S:=\left\{s_{n}: n \geq 1\right\}$ has the required property.
To this end, we consider the sequence defined recursively by
$$
s_{n+1}=\frac{\left(s_{1} \cdot \ldots \cdot s_{n}\right)^{2}+1}{2}
$$
with $s_{1}$ large enough. An easy induction shows that this is an increasing sequence of odd positive integers. For every $i \in\{1, \ldots, n\}$ it turns out that
$$
\frac{s_{i}}{s_{n+1}} \leq \frac{2}{s_{i}} \leq \frac{2}{s_{1}}<\frac{1}{2}
$$
because $s_{1}$ is large enough, which proves the first relation in (Q6.3). Moreover, it turns out that
$$
\frac{s_{n+1}}{s_{i}}=\frac{\left(s_{1} \cdot \ldots \cdot s_{n}\right)^{2}}{2 s_{i}}+\frac{1}{2 s_{i}}
$$
The first term is a positive integer plus $1 / 2$, from which it follows that the distance of $s_{n+1} / s_{i}$ from the positive integers is greater than or equal to
$$
\frac{1}{2}-\frac{1}{2 s_{i}} \geq \frac{1}{2}-\frac{1}{2 s_{1}}
$$
which is greater than $t$ if $s_{1}$ is large enough. This proves the second relation in (Q6.3).
Part (b) (Arithmetic approach) We produce an increasing sequence $s_{n}$ of odd and coprime positive integers that satisfies (Q6.3) every $j>i$. As in the previous solution, this is enough to conclude.
We argue by induction. To begin with, we choose $s_{1}$ to be any odd integer satisfying the inequality in (Q6.2). Let us assume now that $s_{1}, \ldots, s_{n}$ have already been chosen, and let us choose $s_{n+1}$ in such a way that
$$
s_{n+1} \equiv \frac{s_{i}-1}{2} \quad\left(\bmod s_{i}\right) \quad \forall i \in\{1, \ldots, n\}
$$
We can solve this system because the previously chosen integers are odd and coprime. Moreover, any solution of this system is coprime with $s_{1}, \ldots, s_{n}$. Indeed, for every $1 \leq i \leq n$ it turns out that
$$
s_{n+1}=\frac{s_{i}-1}{2}+k_{i} s_{i}
$$
for some positive integer $k_{i}$. Therefore, any prime $p$ that divides both $s_{n+1}$ and $s_{i}$ divides also $\left(2 k_{i}+1\right) s_{i}-2 s_{n+1}=1$, which is absurd. Finally, we observe that we can assume that $s_{n+1}$ is odd and large enough. In this way we can guarantee that
$$
\frac{s_{i}}{s_{n+1}}<\frac{1}{2} \quad \forall i \in\{1, \ldots, n\}
$$
which is the first requirement in (Q6.3), and
$$
k_{i}+t<k_{i}+\frac{1}{2}-\frac{1}{2 s_{i}}=\frac{s_{n+1}}{s_{i}}<k_{i}+\frac{1}{2} \quad \forall i \in\{1, \ldots, n\}
$$
which implies the second requirement in (Q6.3).
Part (b) (Algebraic approach) Again we produce an increasing sequence $s_{n}$ of positive integers that satisfies (Q6.3) every $j>i$.
To this end, for every positive integer $x$, we define its security region
$$
S(x):=\bigcup_{n \geq 1}\left((n+t) x,\left(n+\frac{1}{2}\right) x\right) .
$$
The security region $S(x)$ is a periodic countable union of intervals of length $\left(\frac{1}{2}-t\right) x$, whose left-hand or right-hand endpoints form an arithmetic sequence. It has the property that
$$
t<\left\{\frac{y}{x}\right\}<\frac{1}{2} \quad \forall y \in S(x)
$$
Now we prove by induction that we can choose a sequence $s_{n}$ of positive integers satisfying (Q6.3) and in addition the fact that every interval of the security region $S\left(s_{n}\right)$ contains at least one interval of $S\left(s_{n-1}\right)$.
To begin with, we choose $s_{1}$ large enough so that the length of the intervals of $S\left(s_{1}\right)$ is larger than 1. This guarantees that any interval of $S\left(s_{1}\right)$ contains at least a positive integer. Now let us choose a positive integer $s_{2} \in S\left(s_{1}\right)$ that is large enough. This guarantees that $s_{1} / s_{2}$ is small enough, that the fractional part of $s_{2} / s_{1}$ is in $(t, 1 / 2)$, and that every interval of the security region $S\left(s_{2}\right)$ contains at least one interval of $S\left(s_{1}\right)$, and hence at least one positive integer.
Let us now assume that $s_{1}, \ldots, s_{n}$ have been already chosen with the required properties. We know that every interval of $S\left(s_{n}\right)$ contains at least one interval of $S\left(s_{n-1}\right)$, which in turn contains an interval in $S\left(s_{n-2}\right)$, and so on up to $S\left(s_{1}\right)$. As a consequence, we can choose a large enough positive integer $s_{n+1}$ that lies in $S\left(s_{k}\right)$ for every $k \in\{1, \ldots, n\}$. Since $s_{n+1}$ is large enough, we are sure that
$$
\frac{s_{k}}{s_{n+1}}<t \quad \forall k \in\{1, \ldots, n\}
$$
Moreover, we are sure also that all the intervals of $S\left(s_{n+1}\right)$ are large enough, and therefore they contain at least one interval of $S\left(s_{n}\right)$, which in turn contains at least one interval of $S\left(s_{n-1}\right)$, and so on. Finally, the condition
$$
t<\left\{\frac{s_{n-1}}{s_{n}}\right\}<\frac{1}{2}
$$
is guaranteed by the fact that $s_{n+1}$ was chosen in an interval that is contained in $S\left(s_{k}\right)$ for every $k \in\{1, \ldots, n\}$. This completes the induction.
|
{
"exam": "EGMO",
"problem_label": "6",
"problem_match": "# Problem 6",
"resource_path": "EGMO/segmented/en-2018-solutions.jsonl",
"solution_match": "# Solution",
"tier": "T2",
"year": "2018"
}
|
(Netherlands). Find all triples $(a, b, c)$ of real numbers such that $a b+b c+$ $c a=1$ and
$$
a^{2} b+c=b^{2} c+a=c^{2} a+b
$$
|
First suppose that $a=0$. Then we have $b c=1$ and $c=b^{2} c=b$. So $b=c$, which implies $b^{2}=1$ and hence $b= \pm 1$. This leads to the solutions $(a, b, c)=(0,1,1)$ and $(a, b, c)=(0,-1,-1)$. Similarly, $b=0$ gives the solutions $(a, b, c)=(1,0,1)$ and $(a, b, c)=(-1,0,-1)$, while $c=0$ gives $(a, b, c)=(1,1,0)$ and $(a, b, c)=(-1,-1,0)$.
Now we may assume that $a, b, c \neq=0$. We multiply $a b+b c+c a=1$ by $a$ to find $a^{2} b+a b c+c a^{2}=a$, hence $a^{2} b=a-a b c-a^{2} c$. Substituting this in $a^{2} b+c=b^{2} c+a$ yields $a-a b c-a^{2} c+c=b^{2} c+a$, so $b^{2} c+a b c+a^{2} c=c$. As $c \neq=0$, we find $b^{2}+a b+a^{2}=1$.
Analogously we have $b^{2}+b c+c^{2}=1$ and $a^{2}+a c+c^{2}=1$. Adding these three equations yields $2\left(a^{2}+b^{2}+c^{2}\right)+a b+b c+c a=3$, which implies $a^{2}+b^{2}+c^{2}=1$. Combining this result with $b^{2}+a b+a^{2}=1$, we get $1-a b=1-c^{2}$, so $c^{2}=a b$.
Analogously we also have $b^{2}=a c$ and $a^{2}=b c$. In particular we now have that $a b, b c$ and $c a$ are all positive. This means that $a, b$ and $c$ must all be positive or all be negative. Now assume that $|c|$ is the largest among $|a|,|b|$ and $|c|$, then $c^{2} \geq|a b|=a b=c^{2}$, so we must have equality. This means that $|c|=|a|$ and $|c|=|b|$. Since $(a, b, c)$ must all have the same sign, we find $a=b=c$. Now we have $3 a^{2}=1$, hence $a= \pm \frac{1}{3} \sqrt{3}$. We find the solutions $(a, b, c)=\left(\frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}\right)$ and $(a, b, c)=\left(-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3}\right)$.
We conclude that all possible triples $(a, b, c)$ are $(0,1,1),(0,-1,-1),(1,0,1),(-1,0,-1)$, $(1,1,0),(-1,-1,0),\left(\frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}\right)$ and $\left(-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3}\right)$.
| null |
Yes
|
Yes
|
math-word-problem
|
Algebra
|
(Netherlands). Find all triples $(a, b, c)$ of real numbers such that $a b+b c+$ $c a=1$ and
$$
a^{2} b+c=b^{2} c+a=c^{2} a+b
$$
|
First suppose that $a=0$. Then we have $b c=1$ and $c=b^{2} c=b$. So $b=c$, which implies $b^{2}=1$ and hence $b= \pm 1$. This leads to the solutions $(a, b, c)=(0,1,1)$ and $(a, b, c)=(0,-1,-1)$. Similarly, $b=0$ gives the solutions $(a, b, c)=(1,0,1)$ and $(a, b, c)=(-1,0,-1)$, while $c=0$ gives $(a, b, c)=(1,1,0)$ and $(a, b, c)=(-1,-1,0)$.
Now we may assume that $a, b, c \neq=0$. We multiply $a b+b c+c a=1$ by $a$ to find $a^{2} b+a b c+c a^{2}=a$, hence $a^{2} b=a-a b c-a^{2} c$. Substituting this in $a^{2} b+c=b^{2} c+a$ yields $a-a b c-a^{2} c+c=b^{2} c+a$, so $b^{2} c+a b c+a^{2} c=c$. As $c \neq=0$, we find $b^{2}+a b+a^{2}=1$.
Analogously we have $b^{2}+b c+c^{2}=1$ and $a^{2}+a c+c^{2}=1$. Adding these three equations yields $2\left(a^{2}+b^{2}+c^{2}\right)+a b+b c+c a=3$, which implies $a^{2}+b^{2}+c^{2}=1$. Combining this result with $b^{2}+a b+a^{2}=1$, we get $1-a b=1-c^{2}$, so $c^{2}=a b$.
Analogously we also have $b^{2}=a c$ and $a^{2}=b c$. In particular we now have that $a b, b c$ and $c a$ are all positive. This means that $a, b$ and $c$ must all be positive or all be negative. Now assume that $|c|$ is the largest among $|a|,|b|$ and $|c|$, then $c^{2} \geq|a b|=a b=c^{2}$, so we must have equality. This means that $|c|=|a|$ and $|c|=|b|$. Since $(a, b, c)$ must all have the same sign, we find $a=b=c$. Now we have $3 a^{2}=1$, hence $a= \pm \frac{1}{3} \sqrt{3}$. We find the solutions $(a, b, c)=\left(\frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}\right)$ and $(a, b, c)=\left(-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3}\right)$.
We conclude that all possible triples $(a, b, c)$ are $(0,1,1),(0,-1,-1),(1,0,1),(-1,0,-1)$, $(1,1,0),(-1,-1,0),\left(\frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}\right)$ and $\left(-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3}\right)$.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution 1. ",
"tier": "T2",
"year": "2019"
}
|
(Netherlands). Find all triples $(a, b, c)$ of real numbers such that $a b+b c+$ $c a=1$ and
$$
a^{2} b+c=b^{2} c+a=c^{2} a+b
$$
|
From the problem statement $a b=1-b c-c a$ and thus $b^{2} c+a=a^{2} b+c=$ $a-a b c-a^{2} c+c, c\left(b^{2}+a^{2}+a b-1\right)=0$. If $c=0$ then $a b=1$ and $a^{2} b=b$, which implies $a=b= \pm 1$. Otherwise $b^{2}+a^{2}+a b=1$. Cases $a=0$ and $b=0$ are completely analogous to $c=0$, so we may suppose that $a, b, c \neq 0$. In this case we end up with
$$
\left\{\begin{array}{l}
a^{2}+b^{2}+a b=1 \\
b^{2}+c^{2}+b c=1 \\
c^{2}+a^{2}+c a=1 \\
a b+b c+c a=1
\end{array}\right.
$$
Adding first three equations and subtracting the fourth yields $2\left(a^{2}+b^{2}+c^{2}\right)=2=$ $2(a b+b c+c a)$. Consequently, $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$. Now we can easily conclude that $a=b=c= \pm \frac{1}{\sqrt{3}}$.
|
a=b=c= \pm \frac{1}{\sqrt{3}}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
(Netherlands). Find all triples $(a, b, c)$ of real numbers such that $a b+b c+$ $c a=1$ and
$$
a^{2} b+c=b^{2} c+a=c^{2} a+b
$$
|
From the problem statement $a b=1-b c-c a$ and thus $b^{2} c+a=a^{2} b+c=$ $a-a b c-a^{2} c+c, c\left(b^{2}+a^{2}+a b-1\right)=0$. If $c=0$ then $a b=1$ and $a^{2} b=b$, which implies $a=b= \pm 1$. Otherwise $b^{2}+a^{2}+a b=1$. Cases $a=0$ and $b=0$ are completely analogous to $c=0$, so we may suppose that $a, b, c \neq 0$. In this case we end up with
$$
\left\{\begin{array}{l}
a^{2}+b^{2}+a b=1 \\
b^{2}+c^{2}+b c=1 \\
c^{2}+a^{2}+c a=1 \\
a b+b c+c a=1
\end{array}\right.
$$
Adding first three equations and subtracting the fourth yields $2\left(a^{2}+b^{2}+c^{2}\right)=2=$ $2(a b+b c+c a)$. Consequently, $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$. Now we can easily conclude that $a=b=c= \pm \frac{1}{\sqrt{3}}$.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution 2. ",
"tier": "T2",
"year": "2019"
}
|
(Netherlands). Find all triples $(a, b, c)$ of real numbers such that $a b+b c+$ $c a=1$ and
$$
a^{2} b+c=b^{2} c+a=c^{2} a+b
$$
|
by Achilleas Sinefakopoulos, Greece. We have
$$
c\left(1-b^{2}\right)=a(1-a b)=a(b c+c a)=c\left(a b+a^{2}\right)
$$
and so
$$
c\left(a^{2}+a b+b^{2}-1\right)=0
$$
Similarly, we have
$$
b\left(a^{2}+a c+c^{2}-1\right)=0 \quad \text { and } \quad a\left(b^{2}+b c+c^{2}-1\right)=0
$$
If $c=0$, then we get $a b=1$ and $a^{2} b=a=b$, which give us $a=b=1$, or $a=b=-1$. Similarly, if $a=0$, then $b=c=1$, or $b=c=-1$, while if $b=0$, then $a=c=1$, or $a=c=-1$.
So assume that $a b c \neq 0$. Then
$$
a^{2}+a b+b^{2}=b^{2}+b c+c^{2}=c^{2}+c a+a^{2}=1
$$
Adding these gives us
$$
2\left(a^{2}+b^{2}+c^{2}\right)+a b+b c+c a=3
$$
and using the fact that $a b+b c+c a=1$, we get
$$
a^{2}+b^{2}+c^{2}=1=a b+b c+c a
$$
Hence
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=2\left(a^{2}+b^{2}+c^{2}\right)-2(a b+b c+c a)=0
$$
and so $a=b=c= \pm \frac{1}{\sqrt{3}}$.
Therefore, the solutions $(a, b, c)$ are $(0,1,1),(0,-1,-1),(1,0,1),(-1,0,-1),(1,1,0)$, $(-1,-1,0),\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$
|
(0,1,1),(0,-1,-1),(1,0,1),(-1,0,-1),(1,1,0),(-1,-1,0),\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
(Netherlands). Find all triples $(a, b, c)$ of real numbers such that $a b+b c+$ $c a=1$ and
$$
a^{2} b+c=b^{2} c+a=c^{2} a+b
$$
|
by Achilleas Sinefakopoulos, Greece. We have
$$
c\left(1-b^{2}\right)=a(1-a b)=a(b c+c a)=c\left(a b+a^{2}\right)
$$
and so
$$
c\left(a^{2}+a b+b^{2}-1\right)=0
$$
Similarly, we have
$$
b\left(a^{2}+a c+c^{2}-1\right)=0 \quad \text { and } \quad a\left(b^{2}+b c+c^{2}-1\right)=0
$$
If $c=0$, then we get $a b=1$ and $a^{2} b=a=b$, which give us $a=b=1$, or $a=b=-1$. Similarly, if $a=0$, then $b=c=1$, or $b=c=-1$, while if $b=0$, then $a=c=1$, or $a=c=-1$.
So assume that $a b c \neq 0$. Then
$$
a^{2}+a b+b^{2}=b^{2}+b c+c^{2}=c^{2}+c a+a^{2}=1
$$
Adding these gives us
$$
2\left(a^{2}+b^{2}+c^{2}\right)+a b+b c+c a=3
$$
and using the fact that $a b+b c+c a=1$, we get
$$
a^{2}+b^{2}+c^{2}=1=a b+b c+c a
$$
Hence
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=2\left(a^{2}+b^{2}+c^{2}\right)-2(a b+b c+c a)=0
$$
and so $a=b=c= \pm \frac{1}{\sqrt{3}}$.
Therefore, the solutions $(a, b, c)$ are $(0,1,1),(0,-1,-1),(1,0,1),(-1,0,-1),(1,1,0)$, $(-1,-1,0),\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2019"
}
|
(Netherlands). Find all triples $(a, b, c)$ of real numbers such that $a b+b c+$ $c a=1$ and
$$
a^{2} b+c=b^{2} c+a=c^{2} a+b
$$
|
by Eirini Miliori (HEL2). It is $a b+b c+c a=1$ and
$$
a^{2} b+c=b^{2} c+a=c^{2} a+b
$$
We have
$$
\begin{aligned}
a^{2} b+c=b^{2} c+a & \Longleftrightarrow a^{2} b-a=b^{2} c-c \\
& \Longleftrightarrow a(a b-1)=c\left(b^{2}-1\right) \\
& \Longleftrightarrow a(-b c-a c)=c\left(b^{2}-1\right) \\
& \Longleftrightarrow-a c(a+b)=c\left(b^{2}-1\right)
\end{aligned}
$$
First, consider the case where one of $a, b, c$ is equal to 0 . Without loss of generality, assume that $a=0$. Then $b c=1$ and $b=c$ from (1), and so $b^{2}=1$ giving us $b=1$ or -1 . Hence $b=c=1$ or $b=c=-1$.
Therefore, $(a, b, c)$ equals one of the triples $(0,1,1),(0,-1,-1)$, as well as their rearrangements $(1,0,1)$ and $(-1,0,-1)$ when $b=0$, or $(1,1,0)$ and $(-1,-1,0)$ when $c=0$.
Now consider the case where $a \neq 0, b \neq 0$ and $c \neq 0$. Then (2) gives us
$$
-a(a+b)=b^{2}-1 \Longleftrightarrow-a^{2}-a b=b^{2}-1 \Longleftrightarrow a^{2}+a b+b^{2}-1=0
$$
The quadratic $P(x)=x^{2}+b x+b^{2}-1$ has $x=a$ as a root. Let $x_{1}$ be its second root (which could be equal to $a$ in the case where the discriminant is 0 ). From Vieta's formulas we get
$$
\left\{\begin{aligned}
x_{1}+a=-b & \Longleftrightarrow x_{1}=-b-a, \text { and } \\
x_{1} a=b^{2}-1 & \Longleftrightarrow x_{1}=\frac{b^{2}-1}{a} .
\end{aligned}\right.
$$
Using $a^{2} b+c=c^{2} a+b$ we obtain $b\left(a^{2}-1\right)=c(a c-1)$ yielding $a^{2}+a c+c^{2}-1=0$ in a similar way. The quadratic $Q(x)=x^{2}+c x+c^{2}-1$ has $x=a$ as a root. Let $x_{2}$ be its second root (which could be equal to $a$ in the case where the discriminant is 0 ). From Vieta's formulas we get
$$
\left\{\begin{aligned}
x_{2}+a=-c & \Longleftrightarrow x_{2}=-c-a, \text { and } \\
x_{2} a=c^{2}-1 & \Longleftrightarrow x_{2}=\frac{c^{2}-1}{a} .
\end{aligned}\right.
$$
Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-b-a-c-a, \text { and } \\
x_{1}+x_{2}=\frac{b^{2}-1}{a}+\frac{c^{2}-1}{a}
\end{array}\right.
$$
which give us
$$
\begin{aligned}
-(2 a+b+c)=\frac{b^{2}-1}{a}+\frac{c^{2}-1}{a} & \Longleftrightarrow-2 a^{2}-b a-c a=b^{2}+c^{2}-2 \\
& \Longleftrightarrow b c-1-2 a^{2}=b^{2}+c^{2}-2 \\
& \Longleftrightarrow 2 a^{2}+b^{2}+c^{2}=1+b c .
\end{aligned}
$$
By symmetry, we get
$$
\begin{aligned}
& 2 b^{2}+a^{2}+c^{2}=1+a c, \text { and } \\
& 2 c^{2}+a^{2}+b^{2}=1+b c
\end{aligned}
$$
Adding equations (3), (4), and (5), we get
$$
4\left(a^{2}+b^{2}+c^{2}\right)=3+a b+b c+c a \Longleftrightarrow 4\left(a^{2}+b^{2}+c^{2}\right)=4 \Longleftrightarrow a^{2}+b^{2}+c^{2}=1
$$
From this and (3), since $a b+b c+c a=1$, we get
$$
a^{2}=b c=1-a b-a c \Longleftrightarrow a(a+b+c)=1
$$
Similarly, from (4) we get
$$
b(a+b+c)=1
$$
and from (4),
$$
c(a+b+c)=1
$$
Clearly, it is $a+b+c \neq 0$ (for otherwise it would be $0=1$, a contradiction). Therefore,
$$
a=b=c=\frac{1}{a+b+c},
$$
and so $3 a^{2}=1$ giving us $a=b=c= \pm \frac{1}{\sqrt{3}}$.
In conclusion, the solutions $(a, b, c)$ are $(0,1,1),(0,-1,-1),(1,0,1),(-1,0,-1),(1,1,0)$, $(-1,-1,0),\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$, and $\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$.
|
(0,1,1),(0,-1,-1),(1,0,1),(-1,0,-1),(1,1,0),(-1,-1,0),\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
(Netherlands). Find all triples $(a, b, c)$ of real numbers such that $a b+b c+$ $c a=1$ and
$$
a^{2} b+c=b^{2} c+a=c^{2} a+b
$$
|
by Eirini Miliori (HEL2). It is $a b+b c+c a=1$ and
$$
a^{2} b+c=b^{2} c+a=c^{2} a+b
$$
We have
$$
\begin{aligned}
a^{2} b+c=b^{2} c+a & \Longleftrightarrow a^{2} b-a=b^{2} c-c \\
& \Longleftrightarrow a(a b-1)=c\left(b^{2}-1\right) \\
& \Longleftrightarrow a(-b c-a c)=c\left(b^{2}-1\right) \\
& \Longleftrightarrow-a c(a+b)=c\left(b^{2}-1\right)
\end{aligned}
$$
First, consider the case where one of $a, b, c$ is equal to 0 . Without loss of generality, assume that $a=0$. Then $b c=1$ and $b=c$ from (1), and so $b^{2}=1$ giving us $b=1$ or -1 . Hence $b=c=1$ or $b=c=-1$.
Therefore, $(a, b, c)$ equals one of the triples $(0,1,1),(0,-1,-1)$, as well as their rearrangements $(1,0,1)$ and $(-1,0,-1)$ when $b=0$, or $(1,1,0)$ and $(-1,-1,0)$ when $c=0$.
Now consider the case where $a \neq 0, b \neq 0$ and $c \neq 0$. Then (2) gives us
$$
-a(a+b)=b^{2}-1 \Longleftrightarrow-a^{2}-a b=b^{2}-1 \Longleftrightarrow a^{2}+a b+b^{2}-1=0
$$
The quadratic $P(x)=x^{2}+b x+b^{2}-1$ has $x=a$ as a root. Let $x_{1}$ be its second root (which could be equal to $a$ in the case where the discriminant is 0 ). From Vieta's formulas we get
$$
\left\{\begin{aligned}
x_{1}+a=-b & \Longleftrightarrow x_{1}=-b-a, \text { and } \\
x_{1} a=b^{2}-1 & \Longleftrightarrow x_{1}=\frac{b^{2}-1}{a} .
\end{aligned}\right.
$$
Using $a^{2} b+c=c^{2} a+b$ we obtain $b\left(a^{2}-1\right)=c(a c-1)$ yielding $a^{2}+a c+c^{2}-1=0$ in a similar way. The quadratic $Q(x)=x^{2}+c x+c^{2}-1$ has $x=a$ as a root. Let $x_{2}$ be its second root (which could be equal to $a$ in the case where the discriminant is 0 ). From Vieta's formulas we get
$$
\left\{\begin{aligned}
x_{2}+a=-c & \Longleftrightarrow x_{2}=-c-a, \text { and } \\
x_{2} a=c^{2}-1 & \Longleftrightarrow x_{2}=\frac{c^{2}-1}{a} .
\end{aligned}\right.
$$
Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-b-a-c-a, \text { and } \\
x_{1}+x_{2}=\frac{b^{2}-1}{a}+\frac{c^{2}-1}{a}
\end{array}\right.
$$
which give us
$$
\begin{aligned}
-(2 a+b+c)=\frac{b^{2}-1}{a}+\frac{c^{2}-1}{a} & \Longleftrightarrow-2 a^{2}-b a-c a=b^{2}+c^{2}-2 \\
& \Longleftrightarrow b c-1-2 a^{2}=b^{2}+c^{2}-2 \\
& \Longleftrightarrow 2 a^{2}+b^{2}+c^{2}=1+b c .
\end{aligned}
$$
By symmetry, we get
$$
\begin{aligned}
& 2 b^{2}+a^{2}+c^{2}=1+a c, \text { and } \\
& 2 c^{2}+a^{2}+b^{2}=1+b c
\end{aligned}
$$
Adding equations (3), (4), and (5), we get
$$
4\left(a^{2}+b^{2}+c^{2}\right)=3+a b+b c+c a \Longleftrightarrow 4\left(a^{2}+b^{2}+c^{2}\right)=4 \Longleftrightarrow a^{2}+b^{2}+c^{2}=1
$$
From this and (3), since $a b+b c+c a=1$, we get
$$
a^{2}=b c=1-a b-a c \Longleftrightarrow a(a+b+c)=1
$$
Similarly, from (4) we get
$$
b(a+b+c)=1
$$
and from (4),
$$
c(a+b+c)=1
$$
Clearly, it is $a+b+c \neq 0$ (for otherwise it would be $0=1$, a contradiction). Therefore,
$$
a=b=c=\frac{1}{a+b+c},
$$
and so $3 a^{2}=1$ giving us $a=b=c= \pm \frac{1}{\sqrt{3}}$.
In conclusion, the solutions $(a, b, c)$ are $(0,1,1),(0,-1,-1),(1,0,1),(-1,0,-1),(1,1,0)$, $(-1,-1,0),\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$, and $\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2019"
}
|
(Netherlands). Find all triples $(a, b, c)$ of real numbers such that $a b+b c+$ $c a=1$ and
$$
a^{2} b+c=b^{2} c+a=c^{2} a+b
$$
|
by ISR5. First, homogenize the condition $a^{2} b+c=b^{2} c+a=c^{2} a+b$ by replacing $c$ by $c(a b+b c+c a)$ (etc.), yielding
$$
a^{2} b+c=a^{2} b+a b c+b c^{2}+c^{2} a=a b c+\sum_{c y c} a^{2} b+\left(c^{2} b-b^{2} c\right)=a b c+\sum_{c y c} a^{2} b+b c(c-b) .
$$
Thus, after substracting the cyclicly symmetric part $a b c+\sum_{c y c} a^{2} b$ we find the condition is eqivalent to
$$
D:=b c(c-b)=c a(a-c)=a b(b-a)
$$
Ending 1. It is easy to see that if e.g. $a=0$ then $b=c= \pm 1$, and if e.g. $a=b$ then either $a=b=c= \pm \frac{1}{\sqrt{3}}$ or $a=b= \pm 1, c=0$, and these are indeed solutions. So, to show that these are all solutions (up to symmetries), we may assume by contradiction that $a, b, c$ are pairwise different and non-zero. All conditions are preserved under cyclic shifts and under simultaenously switching signs on all $a, b, c$, and by applying these operations as necessary we may assume $a<b<c$. It follows that $D^{3}=a^{2} b^{2} c^{2}(c-b)(a-c)(b-a)$ must be negative (the only negative term is $a-c$, hence $D$ is negative, i.e. $b c, a b<0<a c$. But this means that $a, c$ have the same sign and $b$ has a different one, which clearly contradicts $a<b<c$ ! So, such configurations are impossible.
Ending 2. Note that $3 D=\sum c^{2} b-\sum b^{2} c=(c-b)(c-a)(b-a)$ and $D^{3}=a^{2} b^{2} c^{2}(c-$ $b)(a-c)(b-a)=-3 a^{2} b^{2} c^{2} D$. Since $3 D$ and $D^{3}$ must have the same sign, and $-3 a^{2} b^{2} c^{2}$ is non-positive, necessarily $D=0$. Thus (up to cyclic permutation) $a=b$ and from there we immediately find either $a=b= \pm 1, c=0$ or $a=b=c= \pm \frac{1}{\sqrt{3}}$.
|
a=b=c= \pm \frac{1}{\sqrt{3}} \text{ or } a=b= \pm 1, c=0
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
(Netherlands). Find all triples $(a, b, c)$ of real numbers such that $a b+b c+$ $c a=1$ and
$$
a^{2} b+c=b^{2} c+a=c^{2} a+b
$$
|
by ISR5. First, homogenize the condition $a^{2} b+c=b^{2} c+a=c^{2} a+b$ by replacing $c$ by $c(a b+b c+c a)$ (etc.), yielding
$$
a^{2} b+c=a^{2} b+a b c+b c^{2}+c^{2} a=a b c+\sum_{c y c} a^{2} b+\left(c^{2} b-b^{2} c\right)=a b c+\sum_{c y c} a^{2} b+b c(c-b) .
$$
Thus, after substracting the cyclicly symmetric part $a b c+\sum_{c y c} a^{2} b$ we find the condition is eqivalent to
$$
D:=b c(c-b)=c a(a-c)=a b(b-a)
$$
Ending 1. It is easy to see that if e.g. $a=0$ then $b=c= \pm 1$, and if e.g. $a=b$ then either $a=b=c= \pm \frac{1}{\sqrt{3}}$ or $a=b= \pm 1, c=0$, and these are indeed solutions. So, to show that these are all solutions (up to symmetries), we may assume by contradiction that $a, b, c$ are pairwise different and non-zero. All conditions are preserved under cyclic shifts and under simultaenously switching signs on all $a, b, c$, and by applying these operations as necessary we may assume $a<b<c$. It follows that $D^{3}=a^{2} b^{2} c^{2}(c-b)(a-c)(b-a)$ must be negative (the only negative term is $a-c$, hence $D$ is negative, i.e. $b c, a b<0<a c$. But this means that $a, c$ have the same sign and $b$ has a different one, which clearly contradicts $a<b<c$ ! So, such configurations are impossible.
Ending 2. Note that $3 D=\sum c^{2} b-\sum b^{2} c=(c-b)(c-a)(b-a)$ and $D^{3}=a^{2} b^{2} c^{2}(c-$ $b)(a-c)(b-a)=-3 a^{2} b^{2} c^{2} D$. Since $3 D$ and $D^{3}$ must have the same sign, and $-3 a^{2} b^{2} c^{2}$ is non-positive, necessarily $D=0$. Thus (up to cyclic permutation) $a=b$ and from there we immediately find either $a=b= \pm 1, c=0$ or $a=b=c= \pm \frac{1}{\sqrt{3}}$.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2019"
}
|
(Luxembourg). Let $n$ be a positive integer. Dominoes are placed on a $2 n \times 2 n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
|
Let $M$ denote the maximal number of dominoes that can be placed on the chessboard. We claim that $M=n(n+1) / 2$. The proof naturally splits into two parts: we first prove that $n(n+1) / 2$ dominoes can be placed on the board, and then show that $M \leq n(n+1) / 2$ to complete the proof.
We construct placings of the dominoes by induction. The base cases $n=1$ and $n=2$ correspond to the placings
Next, we add dominoes to the border of a $2 n \times 2 n$ chessboard to obtain a placing of dominoes for the $2(n+2) \times 2(n+2)$ board,
depending on whether $n$ is odd or even. In these constructions, the interior square is filled with the placing for the $2 n \times 2 n$ board. This construction adds $2 n+3$ dominoes, and therefore places, in total,
$$
\frac{n(n+1)}{2}+(2 n+3)=\frac{(n+2)(n+3)}{2}
$$
dominoes on the board. Noticing that the contour and the interior mesh together appropriately, this proves, by induction, that $n(n+1) / 2$ dominoes can be placed on the $2 n n$ board.
To prove that $M \leq n(n+1) / 2$, we border the $2 n \times 2 n$ square board up to a $(2 n+2) \times(2 n+2)$ square board; this adds $8 n+4$ cells to the $4 n^{2}$ cells that we have started with. Calling a cell covered if it belongs to a domino or is adjacent to a domino, each domino on the $2 n \times 2 n$ board is seen to cover exactly 8 cells of the $(2 n+2) \times(2 n+2)$ board (some of which may belong to the border). By construction, each of the $4 n^{2}$ cells of the $2 n \times 2 n$ board is covered by precisely one domino.
If two adjacent cells on the border, away from a corner, are covered, then there will be at least two uncovered cells on both sides of them; if one covered cell lies between uncovered cells, then again, on both sides of it there will be at least two uncovered cells; three or more adjacent cells cannot be all covered. The following diagrams, in which the borders are shaded, $\times$ marks an uncovered cell on the border, + marks a covered cell not belonging to a domino, and - marks a cell which cannot belong to a domino, summarize the two possible situations,
| $\cdots$ | $\times$ | $\times$ | + | + | $\times$ | $\times$ | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | - | + | | | + | - | |
| | | - | + | + | - | | |
| | | | - | - | | | |
| $\vdots$ | | | | | | | $\vdots$ |
or
Close to a corner of the board, either the corner belongs to some domino,
| $\times$ | + | + | $\times$ | $\times$ | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| + | | | + | - | |
| $\times$ | + | + | - | | |
| $\times$ | - | - | | | |
| $\vdots$ | | | | | |
or one of the following situations, in which the corner cell of the original board is not covered by a domino, may occur:
It is thus seen that at most half of the cells on the border, i.e. $4 n+2$ cells, may be covered, and hence
$$
M \leq\left[\frac{4 n^{2}+(4 n+2)}{8}\right]=\left[\frac{n(n+1)}{2}+\frac{1}{2}\right]=\frac{n(n+1)}{2}
$$
which completes the proof of our claim.
|
\frac{n(n+1)}{2}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
(Luxembourg). Let $n$ be a positive integer. Dominoes are placed on a $2 n \times 2 n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
|
Let $M$ denote the maximal number of dominoes that can be placed on the chessboard. We claim that $M=n(n+1) / 2$. The proof naturally splits into two parts: we first prove that $n(n+1) / 2$ dominoes can be placed on the board, and then show that $M \leq n(n+1) / 2$ to complete the proof.
We construct placings of the dominoes by induction. The base cases $n=1$ and $n=2$ correspond to the placings
Next, we add dominoes to the border of a $2 n \times 2 n$ chessboard to obtain a placing of dominoes for the $2(n+2) \times 2(n+2)$ board,
depending on whether $n$ is odd or even. In these constructions, the interior square is filled with the placing for the $2 n \times 2 n$ board. This construction adds $2 n+3$ dominoes, and therefore places, in total,
$$
\frac{n(n+1)}{2}+(2 n+3)=\frac{(n+2)(n+3)}{2}
$$
dominoes on the board. Noticing that the contour and the interior mesh together appropriately, this proves, by induction, that $n(n+1) / 2$ dominoes can be placed on the $2 n n$ board.
To prove that $M \leq n(n+1) / 2$, we border the $2 n \times 2 n$ square board up to a $(2 n+2) \times(2 n+2)$ square board; this adds $8 n+4$ cells to the $4 n^{2}$ cells that we have started with. Calling a cell covered if it belongs to a domino or is adjacent to a domino, each domino on the $2 n \times 2 n$ board is seen to cover exactly 8 cells of the $(2 n+2) \times(2 n+2)$ board (some of which may belong to the border). By construction, each of the $4 n^{2}$ cells of the $2 n \times 2 n$ board is covered by precisely one domino.
If two adjacent cells on the border, away from a corner, are covered, then there will be at least two uncovered cells on both sides of them; if one covered cell lies between uncovered cells, then again, on both sides of it there will be at least two uncovered cells; three or more adjacent cells cannot be all covered. The following diagrams, in which the borders are shaded, $\times$ marks an uncovered cell on the border, + marks a covered cell not belonging to a domino, and - marks a cell which cannot belong to a domino, summarize the two possible situations,
| $\cdots$ | $\times$ | $\times$ | + | + | $\times$ | $\times$ | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | - | + | | | + | - | |
| | | - | + | + | - | | |
| | | | - | - | | | |
| $\vdots$ | | | | | | | $\vdots$ |
or
Close to a corner of the board, either the corner belongs to some domino,
| $\times$ | + | + | $\times$ | $\times$ | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| + | | | + | - | |
| $\times$ | + | + | - | | |
| $\times$ | - | - | | | |
| $\vdots$ | | | | | |
or one of the following situations, in which the corner cell of the original board is not covered by a domino, may occur:
It is thus seen that at most half of the cells on the border, i.e. $4 n+2$ cells, may be covered, and hence
$$
M \leq\left[\frac{4 n^{2}+(4 n+2)}{8}\right]=\left[\frac{n(n+1)}{2}+\frac{1}{2}\right]=\frac{n(n+1)}{2}
$$
which completes the proof of our claim.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "\nProblem 2",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution 1. ",
"tier": "T2",
"year": "2019"
}
|
(Luxembourg). Let $n$ be a positive integer. Dominoes are placed on a $2 n \times 2 n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
|
We use the same example as in Solution 1. Let $M$ denote the maximum number of dominoes which satisfy the condition of the problem. To prove that $M \leq$ $n(n+1) / 2$, we again border the $2 n \times 2 n$ square board up to a $(2 n+2) \times(2 n+2)$ square board. In fact, we shall ignore the corner border cells as they cannot be covered anyway and consider only the $2 n$ border cells along each side. We prove that out of each four border cells next to each other at most two can be covered. Suppose three out of four cells $A, B, C, D$ are covered. Then there are two possibilities below:
The first option is that $A, B$ and $D$ are covered (marked with + in top row). Then the cells inside the starting square next to $A, B$ and $D$ are covered by the dominoes, but the cell in between them has now two adjacent cells with dominoes, contradiction. The second option is that $A, B$ and $C$ are covered. Then the cells inside the given square next to $A, B$ and $C$ are covered by the dominoes. But then the cell next to B has two adjacent cells with dominoes, contradiction.
Now we can split the border cells along one side in groups of 4 (leaving one group of 2 if $n$ is odd). So when $n$ is even, at most $n$ of the $2 n$ border cells along one side can be covered, and when $n$ is odd, at most $n+1$ out of the $2 n$ border cells can be covered. For all four borders together, this gives a contribution of $4 n$ when $n$ is even and $4 n+4$ when $n$ is odd. Adding $4 n^{2}$ and dividing by 8 we get the desired result.
|
M \leq \frac{n(n+1)}{2}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
(Luxembourg). Let $n$ be a positive integer. Dominoes are placed on a $2 n \times 2 n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
|
We use the same example as in Solution 1. Let $M$ denote the maximum number of dominoes which satisfy the condition of the problem. To prove that $M \leq$ $n(n+1) / 2$, we again border the $2 n \times 2 n$ square board up to a $(2 n+2) \times(2 n+2)$ square board. In fact, we shall ignore the corner border cells as they cannot be covered anyway and consider only the $2 n$ border cells along each side. We prove that out of each four border cells next to each other at most two can be covered. Suppose three out of four cells $A, B, C, D$ are covered. Then there are two possibilities below:
The first option is that $A, B$ and $D$ are covered (marked with + in top row). Then the cells inside the starting square next to $A, B$ and $D$ are covered by the dominoes, but the cell in between them has now two adjacent cells with dominoes, contradiction. The second option is that $A, B$ and $C$ are covered. Then the cells inside the given square next to $A, B$ and $C$ are covered by the dominoes. But then the cell next to B has two adjacent cells with dominoes, contradiction.
Now we can split the border cells along one side in groups of 4 (leaving one group of 2 if $n$ is odd). So when $n$ is even, at most $n$ of the $2 n$ border cells along one side can be covered, and when $n$ is odd, at most $n+1$ out of the $2 n$ border cells can be covered. For all four borders together, this gives a contribution of $4 n$ when $n$ is even and $4 n+4$ when $n$ is odd. Adding $4 n^{2}$ and dividing by 8 we get the desired result.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "\nProblem 2",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution 2. ",
"tier": "T2",
"year": "2019"
}
|
(Luxembourg). Let $n$ be a positive integer. Dominoes are placed on a $2 n \times 2 n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
|
by ISR5. Consider the number of pairs of adjacent cells, such that one of them is covered by a domino. Since each cell is adjacent to one covered cell, the number of such pairs is exactly $4 n^{2}$. On the other hand, let $n_{2}$ be the number of covered corner cells, $n_{3}$ the number of covered edge cells (cells with 3 neighbours), and $n_{4}$ be the number of covered interior cells (cells with 4 neighbours). Thus the number of pairs is $2 n_{2}+3 n_{3}+4 n_{4}=4 n^{2}$, whereas the number of dominoes is $m=\frac{n_{2}+n_{3}+n_{4}}{2}$.
Considering only the outer frame (of corner and edge cells), observe that every covered cell dominates two others, so at most half of the cells are ccovered. The frame has a total of $4(2 n-1)$ cells, i.e. $n_{2}+n_{3} \leq 4 n-2$. Additionally $n_{2} \leq 4$ since there are only 4 corners, thus
$8 m=4 n_{2}+4 n_{3}+4 n_{4}=\left(2 n_{2}+3 n_{3}+4 n_{4}\right)+\left(n_{2}+n_{3}\right)+n_{2} \leq 4 n^{2}+(4 n-2)+4=4 n(n+1)+2$
Thus $m \leq \frac{n(n+1)}{2}+\frac{1}{4}$, so in fact $m \leq \frac{n(n+1)}{2}$.
|
m \leq \frac{n(n+1)}{2}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
(Luxembourg). Let $n$ be a positive integer. Dominoes are placed on a $2 n \times 2 n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
|
by ISR5. Consider the number of pairs of adjacent cells, such that one of them is covered by a domino. Since each cell is adjacent to one covered cell, the number of such pairs is exactly $4 n^{2}$. On the other hand, let $n_{2}$ be the number of covered corner cells, $n_{3}$ the number of covered edge cells (cells with 3 neighbours), and $n_{4}$ be the number of covered interior cells (cells with 4 neighbours). Thus the number of pairs is $2 n_{2}+3 n_{3}+4 n_{4}=4 n^{2}$, whereas the number of dominoes is $m=\frac{n_{2}+n_{3}+n_{4}}{2}$.
Considering only the outer frame (of corner and edge cells), observe that every covered cell dominates two others, so at most half of the cells are ccovered. The frame has a total of $4(2 n-1)$ cells, i.e. $n_{2}+n_{3} \leq 4 n-2$. Additionally $n_{2} \leq 4$ since there are only 4 corners, thus
$8 m=4 n_{2}+4 n_{3}+4 n_{4}=\left(2 n_{2}+3 n_{3}+4 n_{4}\right)+\left(n_{2}+n_{3}\right)+n_{2} \leq 4 n^{2}+(4 n-2)+4=4 n(n+1)+2$
Thus $m \leq \frac{n(n+1)}{2}+\frac{1}{4}$, so in fact $m \leq \frac{n(n+1)}{2}$.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "\nProblem 2",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution (upper bound) ",
"tier": "T2",
"year": "2019"
}
|
(Luxembourg). Let $n$ be a positive integer. Dominoes are placed on a $2 n \times 2 n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
|
by ISR5. We prove that this is the upper bound (and also the lower bound!) by proving that any two configurations, say $A$ and $B$, must contain exactly the same number of dominoes.
Colour the board in a black and white checkboard colouring. Let $W$ be the set of white cells covered by dominoes of tiling $A$. For each cell $w \in W$ let $N_{w}$ be the set of its adjacent (necessarily black) cells. Since each black cell has exactly one neighbour (necessarily white) covered by a domino of tiling $A$, it follows that each black cell is contained in exactly one $N_{w}$, i.e. the $N_{w}$ form a partition of the black cells. Since each white cell has exactly one (necessarily black) neighbour covered by a tile of $B$, each $B_{w}$ contains exactly one black tile covered by a domino of $B$. But, since each domino covers exactly one white and one black cell, we have
$$
|A|=|W|=\left|\left\{N_{w}: w \in W\right\}\right|=|B|
$$
as claimed.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
(Luxembourg). Let $n$ be a positive integer. Dominoes are placed on a $2 n \times 2 n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
|
by ISR5. We prove that this is the upper bound (and also the lower bound!) by proving that any two configurations, say $A$ and $B$, must contain exactly the same number of dominoes.
Colour the board in a black and white checkboard colouring. Let $W$ be the set of white cells covered by dominoes of tiling $A$. For each cell $w \in W$ let $N_{w}$ be the set of its adjacent (necessarily black) cells. Since each black cell has exactly one neighbour (necessarily white) covered by a domino of tiling $A$, it follows that each black cell is contained in exactly one $N_{w}$, i.e. the $N_{w}$ form a partition of the black cells. Since each white cell has exactly one (necessarily black) neighbour covered by a tile of $B$, each $B_{w}$ contains exactly one black tile covered by a domino of $B$. But, since each domino covers exactly one white and one black cell, we have
$$
|A|=|W|=\left|\left\{N_{w}: w \in W\right\}\right|=|B|
$$
as claimed.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "\nProblem 2",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution (upper and lower bound) ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
Let $S$ be the intersection point of $B C$ and the angle bisector of $\angle B A D$, and let $T$ be the intersection point of $B C$ and the angle bisector of $\angle B X C$. We will prove that both quadruples $A, I, B, S$ and $A, I, B, T$ are concyclic, which yields $S=T$.
Firstly denote by $M$ the middle of $\operatorname{arc} A B$ of the circumcenter of $A B C$ which does not contain $C$. Consider the circle centered at $M$ passing through $A, I$ and $B$ (it is well-known that $M A=M I=M B$ ); let it intersect $B C$ at $B$ and $S^{\prime}$. Since $\angle B A C>\angle C B A$ it is easy to check that $S^{\prime}$ lies on side $B C$. Denoting the angles in $A B C$ by $\alpha, \beta, \gamma$ we get
$$
\angle B A D=\angle B A C-\angle D A C=\alpha-\beta .
$$
Moreover since $\angle M B C=\angle M B A+\angle A B C=\frac{\gamma}{2}+\beta$, then
$$
\angle B M S^{\prime}=180^{\circ}-2 \angle M B C=180^{\circ}-\gamma-2 \beta=\alpha-\beta
$$
It follows that $\angle B A S^{\prime}=2 \angle B M S^{\prime}=2 \angle B A D$ which gives us $S=S^{\prime}$.
Secondly let $N$ be the middle of arc $B C$ of the circumcenter of $A B C$ which does not contain $A$. From $\angle B A C>\angle C B A$ we conclude that $X$ lies on the $\operatorname{arc} A B$ of circumcircle of $A B C$ not containing $C$. Obviously both $A I$ and $X T$ are passing through $N$. Since $\angle N B T=\frac{\alpha}{2}=\angle B X N$ we obtain $\triangle N B T \sim \triangle N X B$, therefore
$$
N T \cdot N X=N B^{2}=N I^{2}
$$
It follows that $\triangle N T I \sim \triangle N I X$. Keeping in mind that $\angle N B C=\angle N A C=\angle I X A$ we get
$$
\angle T I N=\angle I X N=\angle N X A-\angle I X A=\angle N B A-\angle N B C=\angle T B A .
$$
It means that $A, I, B, T$ are concyclic which ends the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
Let $S$ be the intersection point of $B C$ and the angle bisector of $\angle B A D$, and let $T$ be the intersection point of $B C$ and the angle bisector of $\angle B X C$. We will prove that both quadruples $A, I, B, S$ and $A, I, B, T$ are concyclic, which yields $S=T$.
Firstly denote by $M$ the middle of $\operatorname{arc} A B$ of the circumcenter of $A B C$ which does not contain $C$. Consider the circle centered at $M$ passing through $A, I$ and $B$ (it is well-known that $M A=M I=M B$ ); let it intersect $B C$ at $B$ and $S^{\prime}$. Since $\angle B A C>\angle C B A$ it is easy to check that $S^{\prime}$ lies on side $B C$. Denoting the angles in $A B C$ by $\alpha, \beta, \gamma$ we get
$$
\angle B A D=\angle B A C-\angle D A C=\alpha-\beta .
$$
Moreover since $\angle M B C=\angle M B A+\angle A B C=\frac{\gamma}{2}+\beta$, then
$$
\angle B M S^{\prime}=180^{\circ}-2 \angle M B C=180^{\circ}-\gamma-2 \beta=\alpha-\beta
$$
It follows that $\angle B A S^{\prime}=2 \angle B M S^{\prime}=2 \angle B A D$ which gives us $S=S^{\prime}$.
Secondly let $N$ be the middle of arc $B C$ of the circumcenter of $A B C$ which does not contain $A$. From $\angle B A C>\angle C B A$ we conclude that $X$ lies on the $\operatorname{arc} A B$ of circumcircle of $A B C$ not containing $C$. Obviously both $A I$ and $X T$ are passing through $N$. Since $\angle N B T=\frac{\alpha}{2}=\angle B X N$ we obtain $\triangle N B T \sim \triangle N X B$, therefore
$$
N T \cdot N X=N B^{2}=N I^{2}
$$
It follows that $\triangle N T I \sim \triangle N I X$. Keeping in mind that $\angle N B C=\angle N A C=\angle I X A$ we get
$$
\angle T I N=\angle I X N=\angle N X A-\angle I X A=\angle N B A-\angle N B C=\angle T B A .
$$
It means that $A, I, B, T$ are concyclic which ends the proof.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "\nProblem 3",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution 1. ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
Let $\angle B A C=\alpha, \angle A B C=\beta, \angle B C A=\gamma \angle A C X=\phi$. Denote by $W_{1}$ and $W_{2}$ the intersections of segment $B C$ with the angle bisectors of $\angle B X C$ and $\angle B A D$ respectively. Then $B W_{1} / W_{1} C=B X / X C$ and $B W_{2} / W_{2} D=B A / A D$. We shall show that $B W_{1}=B W_{2}$.
Since $\angle D A C=\angle C B A$, triangles $A D C$ and $B A C$ are similar and therefore
$$
\frac{D C}{A C}=\frac{A C}{B C}
$$
By the Law of sines
$$
\frac{B W_{2}}{W_{2} D}=\frac{B A}{A D}=\frac{B C}{A C}=\frac{\sin \alpha}{\sin \beta} .
$$
Consequently
$$
\begin{gathered}
\frac{B D}{B W_{2}}=\frac{W_{2} D}{B W_{2}}+1=\frac{\sin \beta}{\sin \alpha}+1, \\
\frac{B C}{B W_{2}}=\frac{B C}{B D} \cdot \frac{B D}{B W_{2}}=\frac{1}{1-D C / B C} \cdot \frac{B D}{B W_{2}}=\frac{1}{1-A C^{2} / B C^{2}} \cdot \frac{B D}{B W_{2}}= \\
\frac{\sin ^{2} \alpha}{\sin ^{2} \alpha-\sin ^{2} \beta} \cdot \frac{\sin \beta+\sin \alpha}{\sin \alpha}=\frac{\sin \alpha}{\sin \alpha-\sin \beta} .
\end{gathered}
$$
Note that $A X B C$ is cyclic and so $\angle B X C=\angle B A C=\alpha$. Hence, $\angle X B C=180^{\circ}-$ $\angle B X C-\angle B C X=180^{\circ}-\alpha-\phi$. By the Law of sines for the triangle $B X C$, we have
$$
\begin{gathered}
\frac{B C}{W_{1} B}=\frac{W_{1} C}{W_{1} B}+1=\frac{C X}{B X}+1=\frac{\sin \angle C B X}{\sin \phi}+1= \\
\frac{\sin (\alpha+\phi)}{\sin \phi}+1=\sin \alpha \cot \phi+\cos \alpha+1 .
\end{gathered}
$$
So, it's enough to prove that
$$
\frac{\sin \alpha}{\sin \alpha-\sin \beta}=\sin \alpha \cot \phi+\cos \alpha
$$
Since $A C$ is tangent to the circle $A I X$, we have $\angle A X I=\angle I A C=\alpha / 2$. Moreover $\angle X A I=\angle X A B+\angle B A I=\phi+\alpha / 2$ and $\angle X I A=180^{\circ}-\angle X A I-\angle A X I=180^{\circ}-\alpha-\phi$. Applying the Law of sines again $X A C, X A I, I A C$ we obtain
$$
\begin{gathered}
\frac{A X}{\sin (\alpha+\phi)}=\frac{A I}{\sin \alpha / 2}, \\
\frac{A X}{\sin (\gamma-\phi)}=\frac{A C}{\sin \angle A X C}=\frac{A C}{\sin \beta}, \\
\frac{A I}{\sin \gamma / 2}=\frac{A C}{\sin (\alpha / 2+\gamma / 2)} .
\end{gathered}
$$
Combining the last three equalities we end up with
$$
\begin{gathered}
\frac{\sin (\gamma-\phi)}{\sin (\alpha+\phi)}=\frac{A I}{A C} \cdot \frac{\sin \beta}{\sin \alpha / 2}=\frac{\sin \beta}{\sin \alpha / 2} \cdot \frac{\sin \gamma / 2}{\sin (\alpha / 2+\gamma / 2)} \\
\frac{\sin (\gamma-\phi)}{\sin (\alpha+\phi)}=\frac{\sin \gamma \cot \phi-\cos \gamma}{\sin \alpha \cot \phi+\cos \alpha}=\frac{2 \sin \beta / 2 \sin \gamma / 2}{\sin \alpha / 2}
\end{gathered}
$$
$$
\frac{\sin \alpha \sin \gamma \cot \phi-\sin \alpha \cos \gamma}{\sin \gamma \sin \alpha \cot \phi+\sin \gamma \cos \alpha}=\frac{2 \sin \beta / 2 \cos \alpha / 2}{\cos \gamma / 2}
$$
Subtracting 1 from both sides yields
$$
\begin{gathered}
\frac{-\sin \alpha \cos \gamma-\sin \gamma \cos \alpha}{\sin \gamma \sin \alpha \cot \phi+\sin \gamma \cos \alpha}=\frac{2 \sin \beta / 2 \cos \alpha / 2}{\cos \gamma / 2}-1= \\
\frac{2 \sin \beta / 2 \cos \alpha / 2-\sin (\alpha / 2+\beta / 2)}{\cos \gamma / 2}=\frac{\sin \beta / 2 \cos \alpha / 2-\sin \alpha / 2 \cos \beta / 2}{\cos \gamma / 2}, \\
\frac{-\sin (\alpha+\gamma)}{\sin \gamma \sin \alpha \cot \phi+\sin \gamma \cos \alpha}=\frac{\sin (\beta / 2-\alpha / 2)}{\cos \gamma / 2}, \\
\frac{-\sin \beta}{\sin \alpha \cot \phi+\cos \alpha}=2 \sin \gamma / 2 \sin (\beta / 2-\alpha / 2)= \\
2 \cos (\beta / 2+\alpha / 2) \sin (\beta / 2-\alpha / 2)=\sin \beta-\sin \alpha
\end{gathered}
$$
and the result follows. We are left to note that none of the denominators can vanish.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
Let $\angle B A C=\alpha, \angle A B C=\beta, \angle B C A=\gamma \angle A C X=\phi$. Denote by $W_{1}$ and $W_{2}$ the intersections of segment $B C$ with the angle bisectors of $\angle B X C$ and $\angle B A D$ respectively. Then $B W_{1} / W_{1} C=B X / X C$ and $B W_{2} / W_{2} D=B A / A D$. We shall show that $B W_{1}=B W_{2}$.
Since $\angle D A C=\angle C B A$, triangles $A D C$ and $B A C$ are similar and therefore
$$
\frac{D C}{A C}=\frac{A C}{B C}
$$
By the Law of sines
$$
\frac{B W_{2}}{W_{2} D}=\frac{B A}{A D}=\frac{B C}{A C}=\frac{\sin \alpha}{\sin \beta} .
$$
Consequently
$$
\begin{gathered}
\frac{B D}{B W_{2}}=\frac{W_{2} D}{B W_{2}}+1=\frac{\sin \beta}{\sin \alpha}+1, \\
\frac{B C}{B W_{2}}=\frac{B C}{B D} \cdot \frac{B D}{B W_{2}}=\frac{1}{1-D C / B C} \cdot \frac{B D}{B W_{2}}=\frac{1}{1-A C^{2} / B C^{2}} \cdot \frac{B D}{B W_{2}}= \\
\frac{\sin ^{2} \alpha}{\sin ^{2} \alpha-\sin ^{2} \beta} \cdot \frac{\sin \beta+\sin \alpha}{\sin \alpha}=\frac{\sin \alpha}{\sin \alpha-\sin \beta} .
\end{gathered}
$$
Note that $A X B C$ is cyclic and so $\angle B X C=\angle B A C=\alpha$. Hence, $\angle X B C=180^{\circ}-$ $\angle B X C-\angle B C X=180^{\circ}-\alpha-\phi$. By the Law of sines for the triangle $B X C$, we have
$$
\begin{gathered}
\frac{B C}{W_{1} B}=\frac{W_{1} C}{W_{1} B}+1=\frac{C X}{B X}+1=\frac{\sin \angle C B X}{\sin \phi}+1= \\
\frac{\sin (\alpha+\phi)}{\sin \phi}+1=\sin \alpha \cot \phi+\cos \alpha+1 .
\end{gathered}
$$
So, it's enough to prove that
$$
\frac{\sin \alpha}{\sin \alpha-\sin \beta}=\sin \alpha \cot \phi+\cos \alpha
$$
Since $A C$ is tangent to the circle $A I X$, we have $\angle A X I=\angle I A C=\alpha / 2$. Moreover $\angle X A I=\angle X A B+\angle B A I=\phi+\alpha / 2$ and $\angle X I A=180^{\circ}-\angle X A I-\angle A X I=180^{\circ}-\alpha-\phi$. Applying the Law of sines again $X A C, X A I, I A C$ we obtain
$$
\begin{gathered}
\frac{A X}{\sin (\alpha+\phi)}=\frac{A I}{\sin \alpha / 2}, \\
\frac{A X}{\sin (\gamma-\phi)}=\frac{A C}{\sin \angle A X C}=\frac{A C}{\sin \beta}, \\
\frac{A I}{\sin \gamma / 2}=\frac{A C}{\sin (\alpha / 2+\gamma / 2)} .
\end{gathered}
$$
Combining the last three equalities we end up with
$$
\begin{gathered}
\frac{\sin (\gamma-\phi)}{\sin (\alpha+\phi)}=\frac{A I}{A C} \cdot \frac{\sin \beta}{\sin \alpha / 2}=\frac{\sin \beta}{\sin \alpha / 2} \cdot \frac{\sin \gamma / 2}{\sin (\alpha / 2+\gamma / 2)} \\
\frac{\sin (\gamma-\phi)}{\sin (\alpha+\phi)}=\frac{\sin \gamma \cot \phi-\cos \gamma}{\sin \alpha \cot \phi+\cos \alpha}=\frac{2 \sin \beta / 2 \sin \gamma / 2}{\sin \alpha / 2}
\end{gathered}
$$
$$
\frac{\sin \alpha \sin \gamma \cot \phi-\sin \alpha \cos \gamma}{\sin \gamma \sin \alpha \cot \phi+\sin \gamma \cos \alpha}=\frac{2 \sin \beta / 2 \cos \alpha / 2}{\cos \gamma / 2}
$$
Subtracting 1 from both sides yields
$$
\begin{gathered}
\frac{-\sin \alpha \cos \gamma-\sin \gamma \cos \alpha}{\sin \gamma \sin \alpha \cot \phi+\sin \gamma \cos \alpha}=\frac{2 \sin \beta / 2 \cos \alpha / 2}{\cos \gamma / 2}-1= \\
\frac{2 \sin \beta / 2 \cos \alpha / 2-\sin (\alpha / 2+\beta / 2)}{\cos \gamma / 2}=\frac{\sin \beta / 2 \cos \alpha / 2-\sin \alpha / 2 \cos \beta / 2}{\cos \gamma / 2}, \\
\frac{-\sin (\alpha+\gamma)}{\sin \gamma \sin \alpha \cot \phi+\sin \gamma \cos \alpha}=\frac{\sin (\beta / 2-\alpha / 2)}{\cos \gamma / 2}, \\
\frac{-\sin \beta}{\sin \alpha \cot \phi+\cos \alpha}=2 \sin \gamma / 2 \sin (\beta / 2-\alpha / 2)= \\
2 \cos (\beta / 2+\alpha / 2) \sin (\beta / 2-\alpha / 2)=\sin \beta-\sin \alpha
\end{gathered}
$$
and the result follows. We are left to note that none of the denominators can vanish.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "\nProblem 3",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution 2. ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
by Achilleas Sinefakopoulos, Greece. We first note that
$$
\angle B A D=\angle B A C-\angle D A C=\angle A-\angle B .
$$
Let $C X$ and $A D$ meet at $K$. Then $\angle C X A=\angle A B C=\angle K A C$. Also, we have $\angle I X A=$ $\angle A / 2$, since $\omega$ is tangent to $A C$ at $A$. Therefore,
$$
\angle D A I=|\angle B-\angle A / 2|=|\angle K X A-\angle I X A|=\angle K X I,
$$
(the absolute value depends on whether $\angle B \geq \angle A / 2$ or not) which means that $X K I A$ is cyclic, i.e. $K$ lies also on $\omega$.
Let $I K$ meet $B C$ at $E$. (If $\angle B=\angle A / 2$, then $I K$ degenerates to the tangent line to $\omega$ at I.) Note that BEIA is cyclic, because
$$
\angle E I A=180^{\circ}-\angle K X A=180^{\circ}-\angle A B E .
$$
We have $\angle E K A=180^{\circ}-\angle A X I=180^{\circ}-\angle A / 2$ and $\angle A E I=\angle A B I=\angle B / 2$. Hence
$$
\begin{aligned}
\angle E A K & =180^{\circ}-\angle E K A-\angle A E I \\
& =180^{\circ}-\left(180^{\circ}-\angle A / 2\right)-\angle B / 2 \\
& =(\angle A-\angle B) / 2 \\
& =\angle B A D / 2 .
\end{aligned}
$$
This means that $A E$ is the angle bisector of $\angle B A D$. Next, let $M$ be the point of intersection of $A E$ and $B I$. Then
$$
\angle E M I=180^{\circ}-\angle B / 2-\angle B A D / 2=180^{\circ}-\angle A / 2
$$
and so, its supplement is
$$
\angle A M I=\angle A / 2=\angle A X I
$$
so $X, M, K, I, A$ all lie on $\omega$. Next, we have
$$
\begin{aligned}
\angle X M A & =\angle X K A \\
& =180^{\circ}-\angle A D C-\angle X C B \\
& =180^{\circ}-\angle A-\angle X C B \\
& =\angle B+\angle X C A \\
& =\angle B+\angle X B A \\
& =\angle X B E,
\end{aligned}
$$
and so $X, B, E, M$ are concyclic. Hence
$$
\begin{aligned}
\angle E X C & =\angle E X M+\angle M X C \\
& =\angle M B E+\angle M A K \\
& =\angle B / 2+\angle B A D / 2 \\
& =\angle A / 2 \\
& =\angle B X C / 2
\end{aligned}
$$
This means that $X E$ is the angle bisector of $\angle B X C$ and so we are done!
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
by Achilleas Sinefakopoulos, Greece. We first note that
$$
\angle B A D=\angle B A C-\angle D A C=\angle A-\angle B .
$$
Let $C X$ and $A D$ meet at $K$. Then $\angle C X A=\angle A B C=\angle K A C$. Also, we have $\angle I X A=$ $\angle A / 2$, since $\omega$ is tangent to $A C$ at $A$. Therefore,
$$
\angle D A I=|\angle B-\angle A / 2|=|\angle K X A-\angle I X A|=\angle K X I,
$$
(the absolute value depends on whether $\angle B \geq \angle A / 2$ or not) which means that $X K I A$ is cyclic, i.e. $K$ lies also on $\omega$.
Let $I K$ meet $B C$ at $E$. (If $\angle B=\angle A / 2$, then $I K$ degenerates to the tangent line to $\omega$ at I.) Note that BEIA is cyclic, because
$$
\angle E I A=180^{\circ}-\angle K X A=180^{\circ}-\angle A B E .
$$
We have $\angle E K A=180^{\circ}-\angle A X I=180^{\circ}-\angle A / 2$ and $\angle A E I=\angle A B I=\angle B / 2$. Hence
$$
\begin{aligned}
\angle E A K & =180^{\circ}-\angle E K A-\angle A E I \\
& =180^{\circ}-\left(180^{\circ}-\angle A / 2\right)-\angle B / 2 \\
& =(\angle A-\angle B) / 2 \\
& =\angle B A D / 2 .
\end{aligned}
$$
This means that $A E$ is the angle bisector of $\angle B A D$. Next, let $M$ be the point of intersection of $A E$ and $B I$. Then
$$
\angle E M I=180^{\circ}-\angle B / 2-\angle B A D / 2=180^{\circ}-\angle A / 2
$$
and so, its supplement is
$$
\angle A M I=\angle A / 2=\angle A X I
$$
so $X, M, K, I, A$ all lie on $\omega$. Next, we have
$$
\begin{aligned}
\angle X M A & =\angle X K A \\
& =180^{\circ}-\angle A D C-\angle X C B \\
& =180^{\circ}-\angle A-\angle X C B \\
& =\angle B+\angle X C A \\
& =\angle B+\angle X B A \\
& =\angle X B E,
\end{aligned}
$$
and so $X, B, E, M$ are concyclic. Hence
$$
\begin{aligned}
\angle E X C & =\angle E X M+\angle M X C \\
& =\angle M B E+\angle M A K \\
& =\angle B / 2+\angle B A D / 2 \\
& =\angle A / 2 \\
& =\angle B X C / 2
\end{aligned}
$$
This means that $X E$ is the angle bisector of $\angle B X C$ and so we are done!
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "\nProblem 3",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
based on that by Eirini Miliori (HEL2), edited by A. Sinefakopoulos, Greece. It is $\angle A B D=\angle D A C$, and so $\overline{A C}$ is tangent to the circumcircle of $\triangle B A D$ at $A$. Hence $C A^{2}=C D \cdot C B$.
Triangle $\triangle A B C$ is similar to triangle $\triangle C A D$, because $\angle C$ is a common angle and $\angle C A D=\angle A B C$, and so $\angle A D C=\angle B A C=2 \varphi$.
Let $Q$ be the point of intersection of $\overline{A D}$ and $\overline{C X}$. Since $\angle B X C=\angle B A C=2 \varphi$, it follows that $B D Q X$ is cyclic. Therefore, $C D \cdot C B=C Q \cdot C X=C A^{2}$ which implies that $Q$ lies on $\omega$.
Next let $P$ be the point of intersection of $\overline{A D}$ with the circumcircle of triangle $\triangle A B C$. Then $\angle P B C=\angle P A C=\angle A B C=\angle A P C$ yielding $C A=C P$. So, let $T$ be on the side $\overline{B C}$ such that $C T=C A=C P$. Then
$$
\angle T A D=\angle T A C-\angle D A C=\left(90^{\circ}-\frac{\angle C}{2}\right)-\angle B=\frac{\angle A-\angle B}{2}=\frac{\angle B A D}{2}
$$
that is, line $\overline{A T}$ is the angle bisector of $\angle B A D$. We want to show that $\overline{X T}$ is the angle bisector of $\angle B X C$. To this end, it suffices to show that $\angle T X C=\varphi$.
It is $C T^{2}=C A^{2}=C Q \cdot C X$, and so $\overline{C T}$ is tangent to the circumcircle of $\triangle X T Q$ at $T$. Since $\angle T X Q=\angle Q T C$ and $\angle Q D C=2 \varphi$, it suffices to show that $\angle T Q D=\varphi$, or, in other words, that $I, Q$, and $T$ are collinear.
Let $T^{\prime}$ is the point of intersection of $\overline{I Q}$ and $\overline{B C}$. Then $\triangle A I C$ is congruent to $\triangle T^{\prime} I C$, since they share $\overline{C I}$ as a common side, $\angle A C I=\angle T^{\prime} C I$, and
$$
\angle I T^{\prime} D=2 \varphi-\angle T^{\prime} Q D=2 \varphi-\angle I Q A=2 \varphi-\angle I X A=\varphi=\angle I A C .
$$
Therefore, $C T^{\prime}=C A=C T$, which means that $T$ coincides with $T^{\prime}$ and completes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
based on that by Eirini Miliori (HEL2), edited by A. Sinefakopoulos, Greece. It is $\angle A B D=\angle D A C$, and so $\overline{A C}$ is tangent to the circumcircle of $\triangle B A D$ at $A$. Hence $C A^{2}=C D \cdot C B$.
Triangle $\triangle A B C$ is similar to triangle $\triangle C A D$, because $\angle C$ is a common angle and $\angle C A D=\angle A B C$, and so $\angle A D C=\angle B A C=2 \varphi$.
Let $Q$ be the point of intersection of $\overline{A D}$ and $\overline{C X}$. Since $\angle B X C=\angle B A C=2 \varphi$, it follows that $B D Q X$ is cyclic. Therefore, $C D \cdot C B=C Q \cdot C X=C A^{2}$ which implies that $Q$ lies on $\omega$.
Next let $P$ be the point of intersection of $\overline{A D}$ with the circumcircle of triangle $\triangle A B C$. Then $\angle P B C=\angle P A C=\angle A B C=\angle A P C$ yielding $C A=C P$. So, let $T$ be on the side $\overline{B C}$ such that $C T=C A=C P$. Then
$$
\angle T A D=\angle T A C-\angle D A C=\left(90^{\circ}-\frac{\angle C}{2}\right)-\angle B=\frac{\angle A-\angle B}{2}=\frac{\angle B A D}{2}
$$
that is, line $\overline{A T}$ is the angle bisector of $\angle B A D$. We want to show that $\overline{X T}$ is the angle bisector of $\angle B X C$. To this end, it suffices to show that $\angle T X C=\varphi$.
It is $C T^{2}=C A^{2}=C Q \cdot C X$, and so $\overline{C T}$ is tangent to the circumcircle of $\triangle X T Q$ at $T$. Since $\angle T X Q=\angle Q T C$ and $\angle Q D C=2 \varphi$, it suffices to show that $\angle T Q D=\varphi$, or, in other words, that $I, Q$, and $T$ are collinear.
Let $T^{\prime}$ is the point of intersection of $\overline{I Q}$ and $\overline{B C}$. Then $\triangle A I C$ is congruent to $\triangle T^{\prime} I C$, since they share $\overline{C I}$ as a common side, $\angle A C I=\angle T^{\prime} C I$, and
$$
\angle I T^{\prime} D=2 \varphi-\angle T^{\prime} Q D=2 \varphi-\angle I Q A=2 \varphi-\angle I X A=\varphi=\angle I A C .
$$
Therefore, $C T^{\prime}=C A=C T$, which means that $T$ coincides with $T^{\prime}$ and completes the proof.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "\nProblem 3",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
based on the work of Artemis-Chrysanthi Savva (HEL4), completed by A. Sinefakopoulos, Greece. Let $G$ be the point of intersection of $\overline{A D}$ and $\overline{C X}$. Since the quadrilateral $A X B C$ is cyclic, it is $\angle A X C=\angle A B C$.
Let the line $\overline{A D}$ meet $\omega$ at $K$. Then it is $\angle A X K=\angle C A D=\angle A B C$, because the angle that is formed by a chord and a tangent to the circle at an endpoint of the chord equals the inscribed angle to that chord. Therefore, $\angle A X K=\angle A X C=\angle A X G$. This means that the point $G$ coincides with the point $K$ and so $G$ belongs to the circle $\omega$.
Let $E$ be the point of intersection of the angle bisector of $\angle D A B$ with $\overline{B C}$. It suffices to show that
$$
\frac{C E}{B E}=\frac{X C}{X B} .
$$
Let $F$ be the second point of intersection of $\omega$ with $\overline{A B}$. Then we have $\angle I A F=\frac{\angle C A B}{2}=$ $\angle I X F$, where $I$ is the incenter of $\triangle A B C$, because $\angle I A F$ and $\angle I X F$ are inscribed in the same arc of $\omega$. Thus $\triangle A I F$ is isosceles with $A I=I F$. Since $I$ is the incenter of $\triangle A B C$, we have $A F=2(s-a)$, where $s=(a+b+c) / 2$ is the semiperimeter of $\triangle A B C$. Also, it is $C E=A C=b$ because in triangle $\triangle A C E$, we have
$$
\begin{aligned}
\angle A E C & =\angle A B C+\angle B A E \\
& =\angle A B C+\frac{\angle B A D}{2} \\
& =\angle A B C+\frac{\angle B A C-\angle A B C}{2} \\
& =90^{\circ}-\frac{\angle A C E}{2}
\end{aligned}
$$
and so $\angle C A E=180^{\circ}-\angle A E C-\angle A C E=90^{\circ}-\frac{\angle A C E}{2}=\angle A E C$. Hence
$$
B F=B A-A F=c-2(s-a)=a-b=C B-C E=B E .
$$
Moreover, triangle $\triangle C A X$ is similar to triangle $\triangle B F X$, because $\angle A C X=\angle F B X$ and
$$
\angle X F B=\angle X A F+\angle A X F=\angle X A F+\angle C A F=\angle C A X
$$
Therefore
$$
\frac{C E}{B E}=\frac{A C}{B F}=\frac{X C}{X B},
$$
as desired. The proof is complete.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
based on the work of Artemis-Chrysanthi Savva (HEL4), completed by A. Sinefakopoulos, Greece. Let $G$ be the point of intersection of $\overline{A D}$ and $\overline{C X}$. Since the quadrilateral $A X B C$ is cyclic, it is $\angle A X C=\angle A B C$.
Let the line $\overline{A D}$ meet $\omega$ at $K$. Then it is $\angle A X K=\angle C A D=\angle A B C$, because the angle that is formed by a chord and a tangent to the circle at an endpoint of the chord equals the inscribed angle to that chord. Therefore, $\angle A X K=\angle A X C=\angle A X G$. This means that the point $G$ coincides with the point $K$ and so $G$ belongs to the circle $\omega$.
Let $E$ be the point of intersection of the angle bisector of $\angle D A B$ with $\overline{B C}$. It suffices to show that
$$
\frac{C E}{B E}=\frac{X C}{X B} .
$$
Let $F$ be the second point of intersection of $\omega$ with $\overline{A B}$. Then we have $\angle I A F=\frac{\angle C A B}{2}=$ $\angle I X F$, where $I$ is the incenter of $\triangle A B C$, because $\angle I A F$ and $\angle I X F$ are inscribed in the same arc of $\omega$. Thus $\triangle A I F$ is isosceles with $A I=I F$. Since $I$ is the incenter of $\triangle A B C$, we have $A F=2(s-a)$, where $s=(a+b+c) / 2$ is the semiperimeter of $\triangle A B C$. Also, it is $C E=A C=b$ because in triangle $\triangle A C E$, we have
$$
\begin{aligned}
\angle A E C & =\angle A B C+\angle B A E \\
& =\angle A B C+\frac{\angle B A D}{2} \\
& =\angle A B C+\frac{\angle B A C-\angle A B C}{2} \\
& =90^{\circ}-\frac{\angle A C E}{2}
\end{aligned}
$$
and so $\angle C A E=180^{\circ}-\angle A E C-\angle A C E=90^{\circ}-\frac{\angle A C E}{2}=\angle A E C$. Hence
$$
B F=B A-A F=c-2(s-a)=a-b=C B-C E=B E .
$$
Moreover, triangle $\triangle C A X$ is similar to triangle $\triangle B F X$, because $\angle A C X=\angle F B X$ and
$$
\angle X F B=\angle X A F+\angle A X F=\angle X A F+\angle C A F=\angle C A X
$$
Therefore
$$
\frac{C E}{B E}=\frac{A C}{B F}=\frac{X C}{X B},
$$
as desired. The proof is complete.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "\nProblem 3",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
by IRL1 and IRL 5. Let $\omega$ denote the circle through $A$ and $I$ tangent to $A C$. Let $Y$ be the second point of intersection of the circle $\omega$ with the line $A D$. Let $L$
be the intersection of $B C$ with the angle bisector of $\angle B A D$. We will prove $\angle L X C=$ $1 / 2 \angle B A C=1 / 2 \angle B X C$.
We will refer to the angles of $\triangle A B C$ as $\angle A, \angle B, \angle C$. Thus $\angle B A D=\angle A-\angle B$.
On the circumcircle of $\triangle A B C$, we have $\angle A X C=\angle A B C=\angle C A D$, and since $A C$ is tangent to $\omega$, we have $\angle C A D=\angle C A Y=\angle A X Y$. Hence $C, X, Y$ are collinear.
Also note that $\triangle C A L$ is isosceles with $\angle C A L=\angle C L A=\frac{1}{2}(\angle B A D)+\angle A B C=\frac{1}{2}(\angle A+$ $\angle B)$ hence $A C=C L$. Moreover, $C I$ is angle bisector to $\angle A C L$ so it's the symmetry axis for the triangle, hence $\angle I L C=\angle I A C=1 / 2 \angle A$ and $\angle A L I=\angle L I A=1 / 2 \angle B$. Since $A C$ is tangent to $\omega$, we have $\angle A Y I=\angle I A C=1 / 2 \angle A=\angle L A Y+\angle A L I$. Hence $L, Y, I$ are collinear.
Since $A C$ is tangent to $\omega$, we have $\triangle C A Y \sim \triangle C X A$ hence $C A^{2}=C X \cdot C Y$. However we proved $C A=C L$ hence $C L^{2}=C X \cdot C Y$. Hence $\triangle C L Y \sim \triangle C X L$ and hence $\angle C X L=\angle C L Y=\angle C A I=1 / 2 \angle A$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
by IRL1 and IRL 5. Let $\omega$ denote the circle through $A$ and $I$ tangent to $A C$. Let $Y$ be the second point of intersection of the circle $\omega$ with the line $A D$. Let $L$
be the intersection of $B C$ with the angle bisector of $\angle B A D$. We will prove $\angle L X C=$ $1 / 2 \angle B A C=1 / 2 \angle B X C$.
We will refer to the angles of $\triangle A B C$ as $\angle A, \angle B, \angle C$. Thus $\angle B A D=\angle A-\angle B$.
On the circumcircle of $\triangle A B C$, we have $\angle A X C=\angle A B C=\angle C A D$, and since $A C$ is tangent to $\omega$, we have $\angle C A D=\angle C A Y=\angle A X Y$. Hence $C, X, Y$ are collinear.
Also note that $\triangle C A L$ is isosceles with $\angle C A L=\angle C L A=\frac{1}{2}(\angle B A D)+\angle A B C=\frac{1}{2}(\angle A+$ $\angle B)$ hence $A C=C L$. Moreover, $C I$ is angle bisector to $\angle A C L$ so it's the symmetry axis for the triangle, hence $\angle I L C=\angle I A C=1 / 2 \angle A$ and $\angle A L I=\angle L I A=1 / 2 \angle B$. Since $A C$ is tangent to $\omega$, we have $\angle A Y I=\angle I A C=1 / 2 \angle A=\angle L A Y+\angle A L I$. Hence $L, Y, I$ are collinear.
Since $A C$ is tangent to $\omega$, we have $\triangle C A Y \sim \triangle C X A$ hence $C A^{2}=C X \cdot C Y$. However we proved $C A=C L$ hence $C L^{2}=C X \cdot C Y$. Hence $\triangle C L Y \sim \triangle C X L$ and hence $\angle C X L=\angle C L Y=\angle C A I=1 / 2 \angle A$.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "\nProblem 3",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
by IRL 5. Let $M$ be the midpoint of the $\operatorname{arc} B C$. Let $\omega$ denote the circle through $A$ and $I$ tangent to $A C$. Let $N$ be the second point of intersection of $\omega$ with $A B$ and $L$ the intersection of $B C$ with the angle bisector of $\angle B A D$. We know $\frac{D L}{L B}=\frac{A D}{A B}$ and want to prove $\frac{X B}{X C}=\frac{L B}{L C}$.
First note that $\triangle C A L$ is isosceles with $\angle C A L=\angle C L A=\frac{1}{2}(\angle B A D)+\angle A B C$ hence $A C=C L$ and $\frac{L B}{L C}=\frac{L B}{A C}$.
Now we calculate $\frac{X B}{X C}$ :
Comparing angles on the circles $\omega$ and the circumcircle of $\triangle A B C$ we get $\triangle X I N \sim$ $\triangle X M B$ and hence also $\triangle X I M \sim \triangle X N B$ (having equal angles at $X$ and proportional adjoint sides). Hence $\frac{X B}{X M}=\frac{N B}{I M}$.
Also comparing angles on the circles $\omega$ and the circumcircle of $\triangle A B C$ and using the tangent $A C$ we get $\triangle X A I \sim \triangle X C M$ and hence also $\triangle X A C \sim \triangle X I M$. Hence $\frac{X C}{X M}=$ $\frac{A C}{I M}$.
Comparing the last two equations we get $\frac{X B}{X C}=\frac{N B}{A C}$. Comparing with $\frac{L B}{L C}=\frac{L B}{A C}$, it remains to prove $N B=L B$.
We prove $\triangle I N B \equiv \triangle I L B$ as follows:
First, we note that $I$ is the circumcentre of $\triangle A L N$. Indeed, $C I$ is angle bisector in the isosceles triangle $A C L$ so it's perpendicular bisector for $A L$. As well, $\triangle I A N$ is isosceles with $\angle I N A=\angle C A I=\angle I A B$ hence $I$ is also on the perpendicular bisector of $A N$.
Hence $I N=I L$ and also $\angle N I L=2 \angle N A L=\angle A-\angle B=2 \angle N I B$ (the last angle is calculated using that the exterior angle of $\triangle N I B$ is $\angle I N A=\angle A / 2$. Hence $\angle N I B=$ $\angle L I B$ and $\triangle I N B \equiv \triangle I L B$ by SAS.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
by IRL 5. Let $M$ be the midpoint of the $\operatorname{arc} B C$. Let $\omega$ denote the circle through $A$ and $I$ tangent to $A C$. Let $N$ be the second point of intersection of $\omega$ with $A B$ and $L$ the intersection of $B C$ with the angle bisector of $\angle B A D$. We know $\frac{D L}{L B}=\frac{A D}{A B}$ and want to prove $\frac{X B}{X C}=\frac{L B}{L C}$.
First note that $\triangle C A L$ is isosceles with $\angle C A L=\angle C L A=\frac{1}{2}(\angle B A D)+\angle A B C$ hence $A C=C L$ and $\frac{L B}{L C}=\frac{L B}{A C}$.
Now we calculate $\frac{X B}{X C}$ :
Comparing angles on the circles $\omega$ and the circumcircle of $\triangle A B C$ we get $\triangle X I N \sim$ $\triangle X M B$ and hence also $\triangle X I M \sim \triangle X N B$ (having equal angles at $X$ and proportional adjoint sides). Hence $\frac{X B}{X M}=\frac{N B}{I M}$.
Also comparing angles on the circles $\omega$ and the circumcircle of $\triangle A B C$ and using the tangent $A C$ we get $\triangle X A I \sim \triangle X C M$ and hence also $\triangle X A C \sim \triangle X I M$. Hence $\frac{X C}{X M}=$ $\frac{A C}{I M}$.
Comparing the last two equations we get $\frac{X B}{X C}=\frac{N B}{A C}$. Comparing with $\frac{L B}{L C}=\frac{L B}{A C}$, it remains to prove $N B=L B$.
We prove $\triangle I N B \equiv \triangle I L B$ as follows:
First, we note that $I$ is the circumcentre of $\triangle A L N$. Indeed, $C I$ is angle bisector in the isosceles triangle $A C L$ so it's perpendicular bisector for $A L$. As well, $\triangle I A N$ is isosceles with $\angle I N A=\angle C A I=\angle I A B$ hence $I$ is also on the perpendicular bisector of $A N$.
Hence $I N=I L$ and also $\angle N I L=2 \angle N A L=\angle A-\angle B=2 \angle N I B$ (the last angle is calculated using that the exterior angle of $\triangle N I B$ is $\angle I N A=\angle A / 2$. Hence $\angle N I B=$ $\angle L I B$ and $\triangle I N B \equiv \triangle I L B$ by SAS.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "\nProblem 3",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
by ISR5 (with help from IRL5). Let $M, N$ be the midpoints of arcs $B C, B A$ of the circumcircle $A B C$, respectively. Let $Y$ be the second intersection of $A D$ and circle $A B C$. Let $E$ be the incenter of triangle $A B Y$ and note that $E$ lies on the angle bisectors of the triangle, which are the lines $Y N$ (immediate), $B C$ (since $\angle C B Y=\angle C A Y=\angle C A D=\angle A B C)$ and the angle bisector of $\angle D A B$; so the question reduces to showing that $E$ is also on $X M$, which is the angle bisector of $\angle C X B$.
We claim that the three lines $C X, A D Y, I E$ are concurrent at a point $D^{\prime}$. We will complete the proof using this fact, and the proof will appear at the end (and see the solution by HEL5 for an alternative proof of this fact).
To show that XEM are collinear, we construct a projective transformation which projects $M$ to $X$ through center $E$. We produce it as a composition of three other projections. Let $O$ be the intersection of lines $A D^{\prime} D Y$ and $C I N$. Projecting the points $Y N C M$ on the circle $A B C$ through the (concyclic) point $A$ to the line $C N$ yields the points $O N C I$. Projecting these points through $E$ to the line $A Y$ yields $O Y D D^{\prime}$ (here we use the facts that $D^{\prime}$ lies on $I E$ and $A Y$ ). Projecting these points to the circle $A B C$ through $C$ yields $N Y B X$ (here we use the fact that $D^{\prime}$ lies on $C X$ ). Composing, we observe that we found a projection of the circle $A B C$ to itself sending $Y N C M$ to $N Y B X$. Since the projection of the circle through $E$ also sends $Y N C$ to $N Y B$, and three points determine a projective transformation, the projection through $E$ also sends $M$ to $X$, as claimed.
Let $B^{\prime}, D^{\prime}$ be the intersections of $A B, A D$ with the circle $A X I$, respectively. We wish to show that this $D^{\prime}$ is the concurrency point defined above, i.e. that $C D^{\prime} X$ and $I D^{\prime} E$ are collinear. Additionally, we will show that $I$ is the circumcenter of $A B^{\prime} E$.
Consider the inversion with center $C$ and radius $C A$. The circles $A X I$ and $A B D$ are tangent to $C A$ at $A$ (the former by definition, the latter since $\angle C A D=\angle A B C$ ), so they are preserved under the inversion. In particular, the inversion transposes $D$ and $B$ and preserves $A$, so sends the circle $C A B$ to the line $A D$. Thus $X$, which is the second intersection of circles $A B C$ and $A X I$, is sent by the inversion to the second intersection of $A D$ and circle $A X I$, which is $D^{\prime}$. In particular $C D^{\prime} X$ are collinear.
In the circle $A I B^{\prime}, A I$ is the angle bisector of $B^{\prime} A$ and the tangent at $A$, so $I$ is the midpoint of the arc $A B^{\prime}$, and in particular $A I=I B^{\prime}$. By angle chasing, we find that $A C E$ is an isosceles triangle:
$\angle C A E=\angle C A D+\angle D A E=\angle A B C+\angle E A B=\angle A B E+\angle E A B=\angle A E B=\angle A E C$,
thus the angle bisector $C I$ is the perpendicular bisector of $A E$ and $A I=I E$. Thus $I$ is the circumcenter of $A B^{\prime} E$.
We can now show that $I D^{\prime} E$ are collinear by angle chasing:
$$
\angle E I B^{\prime}=2 \angle E A B^{\prime}=2 \angle E A B=\angle D A B=\angle D^{\prime} A B^{\prime}=\angle D^{\prime} I B^{\prime} .
$$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
by ISR5 (with help from IRL5). Let $M, N$ be the midpoints of arcs $B C, B A$ of the circumcircle $A B C$, respectively. Let $Y$ be the second intersection of $A D$ and circle $A B C$. Let $E$ be the incenter of triangle $A B Y$ and note that $E$ lies on the angle bisectors of the triangle, which are the lines $Y N$ (immediate), $B C$ (since $\angle C B Y=\angle C A Y=\angle C A D=\angle A B C)$ and the angle bisector of $\angle D A B$; so the question reduces to showing that $E$ is also on $X M$, which is the angle bisector of $\angle C X B$.
We claim that the three lines $C X, A D Y, I E$ are concurrent at a point $D^{\prime}$. We will complete the proof using this fact, and the proof will appear at the end (and see the solution by HEL5 for an alternative proof of this fact).
To show that XEM are collinear, we construct a projective transformation which projects $M$ to $X$ through center $E$. We produce it as a composition of three other projections. Let $O$ be the intersection of lines $A D^{\prime} D Y$ and $C I N$. Projecting the points $Y N C M$ on the circle $A B C$ through the (concyclic) point $A$ to the line $C N$ yields the points $O N C I$. Projecting these points through $E$ to the line $A Y$ yields $O Y D D^{\prime}$ (here we use the facts that $D^{\prime}$ lies on $I E$ and $A Y$ ). Projecting these points to the circle $A B C$ through $C$ yields $N Y B X$ (here we use the fact that $D^{\prime}$ lies on $C X$ ). Composing, we observe that we found a projection of the circle $A B C$ to itself sending $Y N C M$ to $N Y B X$. Since the projection of the circle through $E$ also sends $Y N C$ to $N Y B$, and three points determine a projective transformation, the projection through $E$ also sends $M$ to $X$, as claimed.
Let $B^{\prime}, D^{\prime}$ be the intersections of $A B, A D$ with the circle $A X I$, respectively. We wish to show that this $D^{\prime}$ is the concurrency point defined above, i.e. that $C D^{\prime} X$ and $I D^{\prime} E$ are collinear. Additionally, we will show that $I$ is the circumcenter of $A B^{\prime} E$.
Consider the inversion with center $C$ and radius $C A$. The circles $A X I$ and $A B D$ are tangent to $C A$ at $A$ (the former by definition, the latter since $\angle C A D=\angle A B C$ ), so they are preserved under the inversion. In particular, the inversion transposes $D$ and $B$ and preserves $A$, so sends the circle $C A B$ to the line $A D$. Thus $X$, which is the second intersection of circles $A B C$ and $A X I$, is sent by the inversion to the second intersection of $A D$ and circle $A X I$, which is $D^{\prime}$. In particular $C D^{\prime} X$ are collinear.
In the circle $A I B^{\prime}, A I$ is the angle bisector of $B^{\prime} A$ and the tangent at $A$, so $I$ is the midpoint of the arc $A B^{\prime}$, and in particular $A I=I B^{\prime}$. By angle chasing, we find that $A C E$ is an isosceles triangle:
$\angle C A E=\angle C A D+\angle D A E=\angle A B C+\angle E A B=\angle A B E+\angle E A B=\angle A E B=\angle A E C$,
thus the angle bisector $C I$ is the perpendicular bisector of $A E$ and $A I=I E$. Thus $I$ is the circumcenter of $A B^{\prime} E$.
We can now show that $I D^{\prime} E$ are collinear by angle chasing:
$$
\angle E I B^{\prime}=2 \angle E A B^{\prime}=2 \angle E A B=\angle D A B=\angle D^{\prime} A B^{\prime}=\angle D^{\prime} I B^{\prime} .
$$
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "\nProblem 3",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
inspired by ISR2. Let $W$ be the midpoint of arc $B C$, let $D^{\prime}$ be the second intersection point of $A D$ and the circle $A B C$. Let $P$ be the intersection of the angle bisector $X W$ of $\angle C X B$ with $B C$; we wish to prove that $A P$ is the angle bisector of $D A B$. Denote $\alpha=\frac{\angle C A B}{2}, \beta=\angle A B C$.
Let $M$ be the intersection of $A D$ and $X C$. Angle chasing finds:
$$
\begin{aligned}
\angle M X I & =\angle A X I-\angle A X M=\angle C A I-\angle A X C=\angle C A I-\angle A B C=\alpha-\beta \\
& =\angle C A I-\angle C A D=\angle D A I=\angle M A I
\end{aligned}
$$
And in particular $M$ is on $\omega$. By angle chasing we find
$$
\angle X I A=\angle I X A+\angle X A I=\angle I C A+\angle X A I=\angle X A C=\angle X B C=\angle X B P
$$
and $\angle P X B=\alpha=\angle C A I=\angle A X I$, and it follows that $\triangle X I A \sim \triangle X B P$. Let $S$ be the second intersection point of the cirumcircles of $X I A$ and $X B P$. Then by the spiral map lemma (or by the equivalent angle chasing) it follows that $I S B$ and $A S P$ are collinear.
Let $L$ be the second intersection of $\omega$ and $A B$. We want to prove that $A S P$ is the angle bisector of $\angle D A B=\angle M A L$, i.e. that $S$ is the midpoint of the $\operatorname{arc} M L$ of $\omega$. And this follows easily from chasing angular arc lengths in $\omega$ :
$$
\begin{aligned}
& \overparen{A I}=\angle C A I=\alpha \\
& \widehat{I L}=\angle I A L=\alpha \\
& \widehat{M I}=\angle M X I=\alpha-\beta \\
& \widehat{A I}-\widehat{S L}=\angle A B I=\frac{\beta}{2}
\end{aligned}
$$
And thus
$$
\widehat{M L}=\widehat{M I}+\widehat{I L}=2 \alpha-\beta=2\left(\widehat{A I}-\frac{\beta}{2}\right)=2 \widehat{S L}
$$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
inspired by ISR2. Let $W$ be the midpoint of arc $B C$, let $D^{\prime}$ be the second intersection point of $A D$ and the circle $A B C$. Let $P$ be the intersection of the angle bisector $X W$ of $\angle C X B$ with $B C$; we wish to prove that $A P$ is the angle bisector of $D A B$. Denote $\alpha=\frac{\angle C A B}{2}, \beta=\angle A B C$.
Let $M$ be the intersection of $A D$ and $X C$. Angle chasing finds:
$$
\begin{aligned}
\angle M X I & =\angle A X I-\angle A X M=\angle C A I-\angle A X C=\angle C A I-\angle A B C=\alpha-\beta \\
& =\angle C A I-\angle C A D=\angle D A I=\angle M A I
\end{aligned}
$$
And in particular $M$ is on $\omega$. By angle chasing we find
$$
\angle X I A=\angle I X A+\angle X A I=\angle I C A+\angle X A I=\angle X A C=\angle X B C=\angle X B P
$$
and $\angle P X B=\alpha=\angle C A I=\angle A X I$, and it follows that $\triangle X I A \sim \triangle X B P$. Let $S$ be the second intersection point of the cirumcircles of $X I A$ and $X B P$. Then by the spiral map lemma (or by the equivalent angle chasing) it follows that $I S B$ and $A S P$ are collinear.
Let $L$ be the second intersection of $\omega$ and $A B$. We want to prove that $A S P$ is the angle bisector of $\angle D A B=\angle M A L$, i.e. that $S$ is the midpoint of the $\operatorname{arc} M L$ of $\omega$. And this follows easily from chasing angular arc lengths in $\omega$ :
$$
\begin{aligned}
& \overparen{A I}=\angle C A I=\alpha \\
& \widehat{I L}=\angle I A L=\alpha \\
& \widehat{M I}=\angle M X I=\alpha-\beta \\
& \widehat{A I}-\widehat{S L}=\angle A B I=\frac{\beta}{2}
\end{aligned}
$$
And thus
$$
\widehat{M L}=\widehat{M I}+\widehat{I L}=2 \alpha-\beta=2\left(\widehat{A I}-\frac{\beta}{2}\right)=2 \widehat{S L}
$$
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "\nProblem 3",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
by inversion, by JPN Observer A, Satoshi Hayakawa. Let $E$ be the intersection of the bisector of $\angle B A D$ and $B C$, and $N$ be the middle point of arc $B C$ of the circumcircle of $A B C$. Then it suffices to show that $E$ is on line $X N$.
We consider the inversion at $A$. Let $P^{*}$ be the image of a point denoted by $P$. Then $A, B^{*}, C^{*}, E^{*}$ are concyclic, $X^{*}, B^{*}, C^{*}$ are colinear, and $X^{*} I^{*}$ and $A C^{*}$ are parallel. Now it suffices to show that $A, X^{*}, E^{*}, N^{*}$ are concyclic. Let $Y$ be the intersection of $B^{*} C^{*}$ and $A E^{*}$. Then, by the power of a point, we get
$$
\begin{aligned}
A, X^{*}, E^{*}, N^{*} \text { are concyclic } & \Longleftrightarrow Y X^{*} \cdot Y N^{*}=Y A \cdot Y E^{*} \\
& \Longleftrightarrow Y X^{*} \cdot Y N^{*}=Y B^{*} \cdot Y C^{*} \\
& \left(A, B^{*}, C^{*}, E^{*} \text { are concyclic }\right)
\end{aligned}
$$
Here, by the property of inversion, we have
$$
\angle A I^{*} B^{*}=\angle A B I=\frac{1}{2} \angle A B C=\frac{1}{2} \angle C^{*} A D^{*} .
$$
Define $Q, R$ as described in the figure, and we get by simple angle chasing
$$
\angle Q A I^{*}=\angle Q I^{*} A, \quad \angle R A I^{*}=\angle B^{*} I^{*} A
$$
Especially, $B^{*} R$ and $A I^{*}$ are parallel, so that we have
$$
\frac{Y B^{*}}{Y N^{*}}=\frac{Y R}{Y A}=\frac{Y X^{*}}{Y C^{*}}
$$
and the proof is completed.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.
|
by inversion, by JPN Observer A, Satoshi Hayakawa. Let $E$ be the intersection of the bisector of $\angle B A D$ and $B C$, and $N$ be the middle point of arc $B C$ of the circumcircle of $A B C$. Then it suffices to show that $E$ is on line $X N$.
We consider the inversion at $A$. Let $P^{*}$ be the image of a point denoted by $P$. Then $A, B^{*}, C^{*}, E^{*}$ are concyclic, $X^{*}, B^{*}, C^{*}$ are colinear, and $X^{*} I^{*}$ and $A C^{*}$ are parallel. Now it suffices to show that $A, X^{*}, E^{*}, N^{*}$ are concyclic. Let $Y$ be the intersection of $B^{*} C^{*}$ and $A E^{*}$. Then, by the power of a point, we get
$$
\begin{aligned}
A, X^{*}, E^{*}, N^{*} \text { are concyclic } & \Longleftrightarrow Y X^{*} \cdot Y N^{*}=Y A \cdot Y E^{*} \\
& \Longleftrightarrow Y X^{*} \cdot Y N^{*}=Y B^{*} \cdot Y C^{*} \\
& \left(A, B^{*}, C^{*}, E^{*} \text { are concyclic }\right)
\end{aligned}
$$
Here, by the property of inversion, we have
$$
\angle A I^{*} B^{*}=\angle A B I=\frac{1}{2} \angle A B C=\frac{1}{2} \angle C^{*} A D^{*} .
$$
Define $Q, R$ as described in the figure, and we get by simple angle chasing
$$
\angle Q A I^{*}=\angle Q I^{*} A, \quad \angle R A I^{*}=\angle B^{*} I^{*} A
$$
Especially, $B^{*} R$ and $A I^{*}$ are parallel, so that we have
$$
\frac{Y B^{*}}{Y N^{*}}=\frac{Y R}{Y A}=\frac{Y X^{*}}{Y C^{*}}
$$
and the proof is completed.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "\nProblem 3",
"resource_path": "EGMO/segmented/en-2019-solutions-day1.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle with incentre $I$. The circle through $B$ tangent to $A I$ at $I$ meets side $A B$ again at $P$. The circle through $C$ tangent to $A I$ at $I$ meets side $A C$ again at $Q$. Prove that $P Q$ is tangent to the incircle of $A B C$.
|
Let $Q X, P Y$ be tangent to the incircle of $A B C$, where $X, Y$ lie on the incircle and do not lie on $A C, A B$. Denote $\angle B A C=\alpha, \angle C B A=\beta, \angle A C B=\gamma$.
Since $A I$ is tangent to the circumcircle of $C Q I$ we get $\angle Q I A=\angle Q C I=\frac{\gamma}{2}$. Thus
$$
\angle I Q C=\angle I A Q+\angle Q I A=\frac{\alpha}{2}+\frac{\gamma}{2} .
$$
By the definition of $X$ we have $\angle I Q C=\angle X Q I$, therefore
$$
\angle A Q X=180^{\circ}-\angle X Q C=180^{\circ}-\alpha-\gamma=\beta
$$
Similarly one can prove that $\angle A P Y=\gamma$. This means that $Q, P, X, Y$ are collinear which leads us to the conclusion that $X=Y$ and $Q P$ is tangent to the incircle at $X$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle with incentre $I$. The circle through $B$ tangent to $A I$ at $I$ meets side $A B$ again at $P$. The circle through $C$ tangent to $A I$ at $I$ meets side $A C$ again at $Q$. Prove that $P Q$ is tangent to the incircle of $A B C$.
|
Let $Q X, P Y$ be tangent to the incircle of $A B C$, where $X, Y$ lie on the incircle and do not lie on $A C, A B$. Denote $\angle B A C=\alpha, \angle C B A=\beta, \angle A C B=\gamma$.
Since $A I$ is tangent to the circumcircle of $C Q I$ we get $\angle Q I A=\angle Q C I=\frac{\gamma}{2}$. Thus
$$
\angle I Q C=\angle I A Q+\angle Q I A=\frac{\alpha}{2}+\frac{\gamma}{2} .
$$
By the definition of $X$ we have $\angle I Q C=\angle X Q I$, therefore
$$
\angle A Q X=180^{\circ}-\angle X Q C=180^{\circ}-\alpha-\gamma=\beta
$$
Similarly one can prove that $\angle A P Y=\gamma$. This means that $Q, P, X, Y$ are collinear which leads us to the conclusion that $X=Y$ and $Q P$ is tangent to the incircle at $X$.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "\nProblem 4",
"resource_path": "EGMO/segmented/en-2019-solutions-day2.jsonl",
"solution_match": "\nSolution 1. ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle with incentre $I$. The circle through $B$ tangent to $A I$ at $I$ meets side $A B$ again at $P$. The circle through $C$ tangent to $A I$ at $I$ meets side $A C$ again at $Q$. Prove that $P Q$ is tangent to the incircle of $A B C$.
|
By the power of a point we have
$$
A D \cdot A C=A I^{2}=A P \cdot A B, \quad \text { which means that } \quad \frac{A Q}{A P}=\frac{A B}{A C}
$$
and therefore triangles $A D P, A B C$ are similar. Let $J$ be the incenter of $A Q P$. We obtain
$$
\angle J P Q=\angle I C B=\angle Q C I=\angle Q I J
$$
thus $J, P, I, Q$ are concyclic. Let $S$ be the intersection of $A I$ and $B C$. It follows that
$$
\angle I Q P=\angle I J P=\angle S I C=\angle I Q C .
$$
This means that $I Q$ is the angle bisector of $\angle C Q P$, so $Q P$ is indeed tangent to the incircle of $A B C$.
Comment. The final angle chasing from the Solution 2 may simply be replaced by the observation that since $J, P, I, Q$ are concyclic, then $I$ is the $A$-excenter of triangle $A P Q$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle with incentre $I$. The circle through $B$ tangent to $A I$ at $I$ meets side $A B$ again at $P$. The circle through $C$ tangent to $A I$ at $I$ meets side $A C$ again at $Q$. Prove that $P Q$ is tangent to the incircle of $A B C$.
|
By the power of a point we have
$$
A D \cdot A C=A I^{2}=A P \cdot A B, \quad \text { which means that } \quad \frac{A Q}{A P}=\frac{A B}{A C}
$$
and therefore triangles $A D P, A B C$ are similar. Let $J$ be the incenter of $A Q P$. We obtain
$$
\angle J P Q=\angle I C B=\angle Q C I=\angle Q I J
$$
thus $J, P, I, Q$ are concyclic. Let $S$ be the intersection of $A I$ and $B C$. It follows that
$$
\angle I Q P=\angle I J P=\angle S I C=\angle I Q C .
$$
This means that $I Q$ is the angle bisector of $\angle C Q P$, so $Q P$ is indeed tangent to the incircle of $A B C$.
Comment. The final angle chasing from the Solution 2 may simply be replaced by the observation that since $J, P, I, Q$ are concyclic, then $I$ is the $A$-excenter of triangle $A P Q$.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "\nProblem 4",
"resource_path": "EGMO/segmented/en-2019-solutions-day2.jsonl",
"solution_match": "\nSolution 2. ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle with incentre $I$. The circle through $B$ tangent to $A I$ at $I$ meets side $A B$ again at $P$. The circle through $C$ tangent to $A I$ at $I$ meets side $A C$ again at $Q$. Prove that $P Q$ is tangent to the incircle of $A B C$.
|
Like before, notice that $A Q \cdot A C=A P \cdot A B=A I^{2}$. Consider the positive inversion $\Psi$ with center $A$ and power $A I^{2}$. This maps $P$ to $B$ (and vice-versa), $Q$ to $C$
(and vice-versa), and keeps the incenter $I$ fixed. The problem statement will follow from the fact that the image of the incircle of triangle $A B C$ under $\Psi$ is the so-called mixtilinear incircle of $A B C$, which is defined to be the circle tangent to the lines $A B, A C$, and the circumcircle of $A B C$. Indeed, since the image of the line $Q P$ is the circumcircle of $A B C$, and inversion preserves tangencies, this implies that $Q P$ is tangent to the incircle of $A B C$.
We justify the claim as follows: let $\gamma$ be the incircle of $A B C$ and let $\Gamma_{A}$ be the $A$-mixtilinear incircle of $A B C$. Let $K$ and $L$ be the tangency points of $\gamma$ with the sides $A B$ and $A C$, and let $U$ and $V$ be the tangency points of $\Gamma_{A}$ with the sides $A B$ and $A C$, respectively. It is well-known that the incenter $I$ is the midpoint of segment $U V$. In particular, since also $A I \perp U V$, this implies that $A U=A V=\frac{A I}{\cos \frac{A}{2}}$. Note that $A K=A L=A I \cdot \cos \frac{A}{2}$. Therefore, $A U \cdot A K=A V \cdot A L=A I^{2}$, which means that $U$ and $V$ are the images of $K$ and $L$ under $\Psi$. Since $\Gamma_{A}$ is the unique circle simultaneously tangent to $A B$ at $U$ and to $A C$ at $V$, it follows that the image of $\gamma$ under $\Psi$ must be precisely $\Gamma_{A}$, as claimed.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle with incentre $I$. The circle through $B$ tangent to $A I$ at $I$ meets side $A B$ again at $P$. The circle through $C$ tangent to $A I$ at $I$ meets side $A C$ again at $Q$. Prove that $P Q$ is tangent to the incircle of $A B C$.
|
Like before, notice that $A Q \cdot A C=A P \cdot A B=A I^{2}$. Consider the positive inversion $\Psi$ with center $A$ and power $A I^{2}$. This maps $P$ to $B$ (and vice-versa), $Q$ to $C$
(and vice-versa), and keeps the incenter $I$ fixed. The problem statement will follow from the fact that the image of the incircle of triangle $A B C$ under $\Psi$ is the so-called mixtilinear incircle of $A B C$, which is defined to be the circle tangent to the lines $A B, A C$, and the circumcircle of $A B C$. Indeed, since the image of the line $Q P$ is the circumcircle of $A B C$, and inversion preserves tangencies, this implies that $Q P$ is tangent to the incircle of $A B C$.
We justify the claim as follows: let $\gamma$ be the incircle of $A B C$ and let $\Gamma_{A}$ be the $A$-mixtilinear incircle of $A B C$. Let $K$ and $L$ be the tangency points of $\gamma$ with the sides $A B$ and $A C$, and let $U$ and $V$ be the tangency points of $\Gamma_{A}$ with the sides $A B$ and $A C$, respectively. It is well-known that the incenter $I$ is the midpoint of segment $U V$. In particular, since also $A I \perp U V$, this implies that $A U=A V=\frac{A I}{\cos \frac{A}{2}}$. Note that $A K=A L=A I \cdot \cos \frac{A}{2}$. Therefore, $A U \cdot A K=A V \cdot A L=A I^{2}$, which means that $U$ and $V$ are the images of $K$ and $L$ under $\Psi$. Since $\Gamma_{A}$ is the unique circle simultaneously tangent to $A B$ at $U$ and to $A C$ at $V$, it follows that the image of $\gamma$ under $\Psi$ must be precisely $\Gamma_{A}$, as claimed.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "\nProblem 4",
"resource_path": "EGMO/segmented/en-2019-solutions-day2.jsonl",
"solution_match": "\nSolution 3. ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle with incentre $I$. The circle through $B$ tangent to $A I$ at $I$ meets side $A B$ again at $P$. The circle through $C$ tangent to $A I$ at $I$ meets side $A C$ again at $Q$. Prove that $P Q$ is tangent to the incircle of $A B C$.
|
by Achilleas Sinefakopoulos, Greece. From the power of a point theorem, we have
$$
A P \cdot A B=A I^{2}=A Q \cdot A C
$$
Hence $P B C Q$ is cyclic, and so, $\angle A P Q=\angle B C A$. Let $K$ be the circumcenter of $\triangle B I P$ and let $L$ be the circumcenter of $\triangle Q I C$. Then $\overline{K L}$ is perpendicular to $\overline{A I}$ at $I$.
Let $N$ be the point of intersection of line $\overline{K L}$ with $\overline{A B}$.Then in the right triangle $\triangle N I A$, we have $\angle A N I=90^{\circ}-\frac{\angle B A C}{2}$ and from the external angle theorem for triangle $\triangle B N I$, we have $\angle A N I=\frac{\angle A B C}{2}+\angle N I B$. Hence
$$
\angle N I B=\angle A N I-\frac{\angle A B C}{2}=\left(90^{\circ}-\frac{\angle B A C}{2}\right)-\frac{\angle A B C}{2}=\frac{\angle B C A}{2} .
$$
Since $M I$ is tangent to the circumcircle of $\triangle B I P$ at $I$, we have
$$
\angle B P I=\angle B I M=\angle N I M-\angle N I B=90^{\circ}-\frac{\angle B C A}{2} .
$$
Also, since $\angle A P Q=\angle B C A$, we have
$$
\angle Q P I=180^{\circ}-\angle A P Q-\angle B P I=180^{\circ}-\angle B C A-\left(90^{\circ}-\frac{\angle B C A}{2}\right)=90^{\circ}-\frac{\angle B C A}{2}
$$
as well. Hence $I$ lies on the angle bisector of $\angle B P Q$, and so it is equidistant from its sides $\overline{P Q}$ and $\overline{P B}$. Therefore, the distance of $I$ from $\overline{P Q}$ equals the inradius of $\triangle A B C$, as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle with incentre $I$. The circle through $B$ tangent to $A I$ at $I$ meets side $A B$ again at $P$. The circle through $C$ tangent to $A I$ at $I$ meets side $A C$ again at $Q$. Prove that $P Q$ is tangent to the incircle of $A B C$.
|
by Achilleas Sinefakopoulos, Greece. From the power of a point theorem, we have
$$
A P \cdot A B=A I^{2}=A Q \cdot A C
$$
Hence $P B C Q$ is cyclic, and so, $\angle A P Q=\angle B C A$. Let $K$ be the circumcenter of $\triangle B I P$ and let $L$ be the circumcenter of $\triangle Q I C$. Then $\overline{K L}$ is perpendicular to $\overline{A I}$ at $I$.
Let $N$ be the point of intersection of line $\overline{K L}$ with $\overline{A B}$.Then in the right triangle $\triangle N I A$, we have $\angle A N I=90^{\circ}-\frac{\angle B A C}{2}$ and from the external angle theorem for triangle $\triangle B N I$, we have $\angle A N I=\frac{\angle A B C}{2}+\angle N I B$. Hence
$$
\angle N I B=\angle A N I-\frac{\angle A B C}{2}=\left(90^{\circ}-\frac{\angle B A C}{2}\right)-\frac{\angle A B C}{2}=\frac{\angle B C A}{2} .
$$
Since $M I$ is tangent to the circumcircle of $\triangle B I P$ at $I$, we have
$$
\angle B P I=\angle B I M=\angle N I M-\angle N I B=90^{\circ}-\frac{\angle B C A}{2} .
$$
Also, since $\angle A P Q=\angle B C A$, we have
$$
\angle Q P I=180^{\circ}-\angle A P Q-\angle B P I=180^{\circ}-\angle B C A-\left(90^{\circ}-\frac{\angle B C A}{2}\right)=90^{\circ}-\frac{\angle B C A}{2}
$$
as well. Hence $I$ lies on the angle bisector of $\angle B P Q$, and so it is equidistant from its sides $\overline{P Q}$ and $\overline{P B}$. Therefore, the distance of $I$ from $\overline{P Q}$ equals the inradius of $\triangle A B C$, as desired.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "\nProblem 4",
"resource_path": "EGMO/segmented/en-2019-solutions-day2.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2019"
}
|
(Poland). Let $A B C$ be a triangle with incentre $I$. The circle through $B$ tangent to $A I$ at $I$ meets side $A B$ again at $P$. The circle through $C$ tangent to $A I$ at $I$ meets side $A C$ again at $Q$. Prove that $P Q$ is tangent to the incircle of $A B C$.
|
by Eirini Miliori (HEL2). Let $D$ be the point of intersection of $\overline{A I}$ and $\overline{B C}$ and let $R$ be the point of intersection of $\overline{A I}$ and $\overline{P Q}$. We have $\angle R I P=\angle P B I=\frac{\angle B}{2}$, $\angle R I Q=\angle I C Q=\frac{\angle C}{2}, \angle I Q C=\angle D I C=x$ and $\angle B P I=\angle B I D=\varphi$, since $\overline{A I}^{2}$ is tangent to both circles.
From the angle bisector theorem, we have
$$
\frac{R Q}{R P}=\frac{A Q}{A P} \quad \text { and } \quad \frac{A C}{A B}=\frac{D C}{B D}
$$
Since $\overline{A I}$ is tangent to both circles at $I$, we have $A I^{2}=A Q \cdot A C$ and $A I^{2}=A P \cdot A B$. Therefore,
$$
\frac{R Q}{R P} \cdot \frac{D C}{B D}=\frac{A Q \cdot A C}{A B \cdot A P}=1
$$
From the sine law in triangles $\triangle Q R I$ and $\triangle P R I$, it follows that $\frac{R Q}{\sin \frac{\angle C}{2}}=\frac{R I}{\sin y}$ and $\frac{R P}{\sin \frac{\angle B}{2}}=\frac{R I}{\sin \omega}$, respectively. Hence
$$
\frac{R Q}{R P} \cdot \frac{\sin \frac{\angle B}{2}}{\sin \frac{\angle C}{2}}=\frac{\sin \omega}{\sin y}
$$
Similarly, from the sine law in triangles $\triangle I D C$ and $\triangle I D B$, it is $\frac{D C}{\sin x}=\frac{I D}{\sin \frac{\angle C}{2}}$ and $\frac{B D}{\sin \varphi}=\frac{I D}{\sin \frac{\angle B}{2}}$, and so
$$
\frac{D C}{B D} \cdot \frac{\sin \varphi}{\sin x}=\frac{\sin \frac{\angle B}{2}}{\sin \frac{\angle C}{2}}
$$
By multiplying equations (2) with (3), we obtain $\frac{R Q}{R P} \cdot \frac{D C}{B D} \cdot \frac{\sin \varphi}{\sin x}=\frac{\sin \omega}{\sin y}$, which combined with (1) and cross-multiplying yields
$$
\sin \varphi \cdot \sin y=\sin \omega \cdot \sin x
$$
Let $\theta=90^{\circ}+\frac{\angle A}{2}$. Since $I$ is the incenter of $\triangle A B C$, we have $x=90^{\circ}+\frac{\angle A}{2}-\varphi=\theta-\phi$. Also, in triangle $\triangle P I Q$, we see that $\omega+y+\frac{\angle B}{2}+\frac{\angle C}{2}=180^{\circ}$, and so $y=\theta-\omega$.
Therefore, equation (4) yields
$$
\sin \varphi \cdot \sin (\theta-\omega)=\sin \omega \cdot \sin (\theta-\varphi)
$$
or
$$
\frac{1}{2}(\cos (\varphi-\theta+\omega)-\cos (\varphi+\theta-\omega))=\frac{1}{2}(\cos (\omega-\theta+\varphi)-\cos (\omega+\theta-\varphi))
$$
which is equivalent to
$$
\cos (\varphi+\theta-\omega)=\cos (\omega+\theta-\varphi)
$$
So
$$
\varphi+\theta-\omega=2 k \cdot 180^{\circ} \pm(\omega+\theta-\varphi), \quad(k \in \mathbb{Z} .)
$$
If $\varphi+\theta-\omega=2 k \cdot 180^{\circ}+(\omega+\theta-\varphi)$, then $2(\varphi-\omega)=2 k \cdot 180^{\circ}$, with $|\varphi-\omega|<180^{\circ}$ forcing $k=0$ and $\varphi=\omega$. If $\varphi+\theta-\omega=2 k \cdot 180^{\circ}-(\omega+\theta-\varphi)$, then $2 \theta=2 k \cdot 180^{\circ}$, which contradicts the fact that $0^{\circ}<\theta<180^{\circ}$. Hence $\varphi=\omega$, and so $P I$ is the angle bisector of $\angle Q P B$.
Therefore the distance of $I$ from $\overline{P Q}$ is the same with the distance of $I$ from $A B$, which is equal to the inradius of $\triangle A B C$. Consequently, $\overline{P Q}$ is tangent to the incircle of $\triangle A B C$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
(Poland). Let $A B C$ be a triangle with incentre $I$. The circle through $B$ tangent to $A I$ at $I$ meets side $A B$ again at $P$. The circle through $C$ tangent to $A I$ at $I$ meets side $A C$ again at $Q$. Prove that $P Q$ is tangent to the incircle of $A B C$.
|
by Eirini Miliori (HEL2). Let $D$ be the point of intersection of $\overline{A I}$ and $\overline{B C}$ and let $R$ be the point of intersection of $\overline{A I}$ and $\overline{P Q}$. We have $\angle R I P=\angle P B I=\frac{\angle B}{2}$, $\angle R I Q=\angle I C Q=\frac{\angle C}{2}, \angle I Q C=\angle D I C=x$ and $\angle B P I=\angle B I D=\varphi$, since $\overline{A I}^{2}$ is tangent to both circles.
From the angle bisector theorem, we have
$$
\frac{R Q}{R P}=\frac{A Q}{A P} \quad \text { and } \quad \frac{A C}{A B}=\frac{D C}{B D}
$$
Since $\overline{A I}$ is tangent to both circles at $I$, we have $A I^{2}=A Q \cdot A C$ and $A I^{2}=A P \cdot A B$. Therefore,
$$
\frac{R Q}{R P} \cdot \frac{D C}{B D}=\frac{A Q \cdot A C}{A B \cdot A P}=1
$$
From the sine law in triangles $\triangle Q R I$ and $\triangle P R I$, it follows that $\frac{R Q}{\sin \frac{\angle C}{2}}=\frac{R I}{\sin y}$ and $\frac{R P}{\sin \frac{\angle B}{2}}=\frac{R I}{\sin \omega}$, respectively. Hence
$$
\frac{R Q}{R P} \cdot \frac{\sin \frac{\angle B}{2}}{\sin \frac{\angle C}{2}}=\frac{\sin \omega}{\sin y}
$$
Similarly, from the sine law in triangles $\triangle I D C$ and $\triangle I D B$, it is $\frac{D C}{\sin x}=\frac{I D}{\sin \frac{\angle C}{2}}$ and $\frac{B D}{\sin \varphi}=\frac{I D}{\sin \frac{\angle B}{2}}$, and so
$$
\frac{D C}{B D} \cdot \frac{\sin \varphi}{\sin x}=\frac{\sin \frac{\angle B}{2}}{\sin \frac{\angle C}{2}}
$$
By multiplying equations (2) with (3), we obtain $\frac{R Q}{R P} \cdot \frac{D C}{B D} \cdot \frac{\sin \varphi}{\sin x}=\frac{\sin \omega}{\sin y}$, which combined with (1) and cross-multiplying yields
$$
\sin \varphi \cdot \sin y=\sin \omega \cdot \sin x
$$
Let $\theta=90^{\circ}+\frac{\angle A}{2}$. Since $I$ is the incenter of $\triangle A B C$, we have $x=90^{\circ}+\frac{\angle A}{2}-\varphi=\theta-\phi$. Also, in triangle $\triangle P I Q$, we see that $\omega+y+\frac{\angle B}{2}+\frac{\angle C}{2}=180^{\circ}$, and so $y=\theta-\omega$.
Therefore, equation (4) yields
$$
\sin \varphi \cdot \sin (\theta-\omega)=\sin \omega \cdot \sin (\theta-\varphi)
$$
or
$$
\frac{1}{2}(\cos (\varphi-\theta+\omega)-\cos (\varphi+\theta-\omega))=\frac{1}{2}(\cos (\omega-\theta+\varphi)-\cos (\omega+\theta-\varphi))
$$
which is equivalent to
$$
\cos (\varphi+\theta-\omega)=\cos (\omega+\theta-\varphi)
$$
So
$$
\varphi+\theta-\omega=2 k \cdot 180^{\circ} \pm(\omega+\theta-\varphi), \quad(k \in \mathbb{Z} .)
$$
If $\varphi+\theta-\omega=2 k \cdot 180^{\circ}+(\omega+\theta-\varphi)$, then $2(\varphi-\omega)=2 k \cdot 180^{\circ}$, with $|\varphi-\omega|<180^{\circ}$ forcing $k=0$ and $\varphi=\omega$. If $\varphi+\theta-\omega=2 k \cdot 180^{\circ}-(\omega+\theta-\varphi)$, then $2 \theta=2 k \cdot 180^{\circ}$, which contradicts the fact that $0^{\circ}<\theta<180^{\circ}$. Hence $\varphi=\omega$, and so $P I$ is the angle bisector of $\angle Q P B$.
Therefore the distance of $I$ from $\overline{P Q}$ is the same with the distance of $I$ from $A B$, which is equal to the inradius of $\triangle A B C$. Consequently, $\overline{P Q}$ is tangent to the incircle of $\triangle A B C$.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "\nProblem 4",
"resource_path": "EGMO/segmented/en-2019-solutions-day2.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2019"
}
|
(Netherlands).
Let $n \geq 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive integers. Show that there exist positive integers $b_{1}, b_{2}, \ldots, b_{n}$ satisfying the following three conditions:
1). $a_{i} \leq b_{i}$ for $i=1,2, \ldots, n$;
2). the remainders of $b_{1}, b_{2}, \ldots, b_{n}$ on division by $n$ are pairwise different; and
3). $b_{1}+\cdots+b_{n} \leq n\left(\frac{n-1}{2}+\left\lfloor\frac{a_{1}+\cdots+a_{n}}{n}\right\rfloor\right)$.
(Here, $\lfloor x\rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
|
We define the $b_{i}$ recursively by letting $b_{i}$ be the smallest integer such that $b_{i} \geq a_{i}$ and such that $b_{i}$ is not congruent to any of $b_{1}, \ldots, b_{i-1}$ modulo $n$. Then $b_{i}-a_{i} \leq i-1$, since of the $i$ consecutive integers $a_{i}, a_{i}+1, \ldots, a_{i}+i-1$, at most $i-1$ are congruent to one of $b_{1}, \ldots, b_{i-1}$ modulo $n$. Since all $b_{i}$ are distinct modulo $n$, we have $\sum_{i=1}^{n} b_{i} \equiv \sum_{i=1}^{n}(i-1)=\frac{1}{2} n(n-1)$ modulo $n$, so $n$ divides $\sum_{i=1}^{n} b_{i}-\frac{1}{2} n(n-1)$. Moreover, we have $\sum_{i=1}^{n} b_{i}-\sum_{i=1}^{n} a_{i} \leq \sum_{i=1}^{n}(i-1)=\frac{1}{2} n(n-1)$, hence $\sum_{i=1}^{n} b_{i}-\frac{1}{2} n(n-1) \leq \sum_{i=1}^{n}$. As the left hand side is divisible by $n$, we have
$$
\frac{1}{n}\left(\sum_{i=1}^{n} b_{i}-\frac{1}{2} n(n-1)\right) \leq\left[\frac{1}{n} \sum_{i=1}^{n} a_{i}\right]
$$
which we can rewrite as
$$
\sum_{i=1}^{n} b_{i} \leq n\left(\frac{n-1}{2}+\left[\frac{1}{n} \sum_{i=1}^{n} a_{i}\right]\right)
$$
as required.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
(Netherlands).
Let $n \geq 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive integers. Show that there exist positive integers $b_{1}, b_{2}, \ldots, b_{n}$ satisfying the following three conditions:
1). $a_{i} \leq b_{i}$ for $i=1,2, \ldots, n$;
2). the remainders of $b_{1}, b_{2}, \ldots, b_{n}$ on division by $n$ are pairwise different; and
3). $b_{1}+\cdots+b_{n} \leq n\left(\frac{n-1}{2}+\left\lfloor\frac{a_{1}+\cdots+a_{n}}{n}\right\rfloor\right)$.
(Here, $\lfloor x\rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
|
We define the $b_{i}$ recursively by letting $b_{i}$ be the smallest integer such that $b_{i} \geq a_{i}$ and such that $b_{i}$ is not congruent to any of $b_{1}, \ldots, b_{i-1}$ modulo $n$. Then $b_{i}-a_{i} \leq i-1$, since of the $i$ consecutive integers $a_{i}, a_{i}+1, \ldots, a_{i}+i-1$, at most $i-1$ are congruent to one of $b_{1}, \ldots, b_{i-1}$ modulo $n$. Since all $b_{i}$ are distinct modulo $n$, we have $\sum_{i=1}^{n} b_{i} \equiv \sum_{i=1}^{n}(i-1)=\frac{1}{2} n(n-1)$ modulo $n$, so $n$ divides $\sum_{i=1}^{n} b_{i}-\frac{1}{2} n(n-1)$. Moreover, we have $\sum_{i=1}^{n} b_{i}-\sum_{i=1}^{n} a_{i} \leq \sum_{i=1}^{n}(i-1)=\frac{1}{2} n(n-1)$, hence $\sum_{i=1}^{n} b_{i}-\frac{1}{2} n(n-1) \leq \sum_{i=1}^{n}$. As the left hand side is divisible by $n$, we have
$$
\frac{1}{n}\left(\sum_{i=1}^{n} b_{i}-\frac{1}{2} n(n-1)\right) \leq\left[\frac{1}{n} \sum_{i=1}^{n} a_{i}\right]
$$
which we can rewrite as
$$
\sum_{i=1}^{n} b_{i} \leq n\left(\frac{n-1}{2}+\left[\frac{1}{n} \sum_{i=1}^{n} a_{i}\right]\right)
$$
as required.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "# Problem 5",
"resource_path": "EGMO/segmented/en-2019-solutions-day2.jsonl",
"solution_match": "\nSolution 1. ",
"tier": "T2",
"year": "2019"
}
|
(Netherlands).
Let $n \geq 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive integers. Show that there exist positive integers $b_{1}, b_{2}, \ldots, b_{n}$ satisfying the following three conditions:
1). $a_{i} \leq b_{i}$ for $i=1,2, \ldots, n$;
2). the remainders of $b_{1}, b_{2}, \ldots, b_{n}$ on division by $n$ are pairwise different; and
3). $b_{1}+\cdots+b_{n} \leq n\left(\frac{n-1}{2}+\left\lfloor\frac{a_{1}+\cdots+a_{n}}{n}\right\rfloor\right)$.
(Here, $\lfloor x\rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
|
Note that the problem is invariant under each of the following operations:
- adding a multiple of $n$ to some $a_{i}$ (and the corresponding $b_{i}$ );
- adding the same integer to all $a_{i}$ (and all $b_{i}$ );
- permuting the index set $1,2, \ldots, n$.
We may therefore remove the restriction that our $a_{i}$ and $b_{i}$ be positive.
For each congruence class $\bar{k}$ modulo $n(\bar{k}=\overline{0}, \ldots, \overline{n-1})$, let $h(k)$ be the number of $i$ such that $a_{i}$ belongs to $\bar{k}$. We will now show that the problem is solved if we can find a $t \in \mathbb{Z}$ such that
$$
\begin{array}{cl}
h(t) & \geq 1, \\
h(t)+h(t+1) & \geq 2 \\
h(t)+h(t+1)+h(t+2) & \geq 3
\end{array}
$$
Indeed, these inequalities guarantee the existence of elements $a_{i_{1}} \in \bar{t}, a_{i_{2}} \in \bar{t} \cup \overline{t+1}$, $a_{i_{3}} \in \bar{t} \cup \overline{t+1} \cup \overline{t+2}$, et cetera, where all $i_{k}$ are different. Subtracting appropriate
multiples of $n$ and reordering our elements, we may assume $a_{1}=t, a_{2} \in\{t, t+1\}$, $a_{3} \in\{t, t+1, t+2\}$, et cetera. Finally subtracting $t$ from the complete sequence, we may assume $a_{1}=0, a_{2} \in\{0,1\}, a_{3} \in\{0,1,2\}$ et cetera. Now simply setting $b_{i}=i-1$ for all $i$ suffices, since $a_{i} \leq b_{i}$ for all $i$, the $b_{i}$ are all different modulo $n$, and
$$
\sum_{i=1}^{n} b_{i}=\frac{n(n-1)}{2} \leq \frac{n(n-1)}{2}+n\left[\frac{\sum_{i=1}^{n} a_{i}}{n}\right] .
$$
Put $x_{i}=h(i)-1$ for all $i=0, \ldots, n-1$. Note that $x_{i} \geq-1$, because $h(i) \geq 0$. If we have $x_{i} \geq 0$ for all $i=0, \ldots, n-1$, then taking $t=0$ completes the proof. Otherwise, we can pick some index $j$ such that $x_{j}=-1$. Let $y_{i}=x_{i}$ where $i=0, \ldots, j-1, j+1, \ldots, n-1$ and $y_{j}=0$. For sequence $\left\{y_{i}\right\}$ we have
$$
\sum_{i=0}^{n-1} y_{i}=\sum_{i=0}^{n-1} x_{i}+1=\sum_{i=0}^{n-1} h(i)-n+1=1
$$
so from Raney's lemma there exists index $k$ such that $\sum_{i=k}^{k+j} y_{i}>0$ for all $j=0, \ldots, n-1$ where $y_{n+j}=y_{j}$ for $j=0, \ldots, k-1$. Taking $t=k$ we will have
$$
\sum_{t=k}^{k+i} h(t)-(i+1)=\sum_{t=k}^{k+i} x(t) \geq \sum_{t=k}^{k+i} y(t)-1 \geq 0
$$
for all $i=0, \ldots, n-1$ and we are done.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
(Netherlands).
Let $n \geq 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive integers. Show that there exist positive integers $b_{1}, b_{2}, \ldots, b_{n}$ satisfying the following three conditions:
1). $a_{i} \leq b_{i}$ for $i=1,2, \ldots, n$;
2). the remainders of $b_{1}, b_{2}, \ldots, b_{n}$ on division by $n$ are pairwise different; and
3). $b_{1}+\cdots+b_{n} \leq n\left(\frac{n-1}{2}+\left\lfloor\frac{a_{1}+\cdots+a_{n}}{n}\right\rfloor\right)$.
(Here, $\lfloor x\rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
|
Note that the problem is invariant under each of the following operations:
- adding a multiple of $n$ to some $a_{i}$ (and the corresponding $b_{i}$ );
- adding the same integer to all $a_{i}$ (and all $b_{i}$ );
- permuting the index set $1,2, \ldots, n$.
We may therefore remove the restriction that our $a_{i}$ and $b_{i}$ be positive.
For each congruence class $\bar{k}$ modulo $n(\bar{k}=\overline{0}, \ldots, \overline{n-1})$, let $h(k)$ be the number of $i$ such that $a_{i}$ belongs to $\bar{k}$. We will now show that the problem is solved if we can find a $t \in \mathbb{Z}$ such that
$$
\begin{array}{cl}
h(t) & \geq 1, \\
h(t)+h(t+1) & \geq 2 \\
h(t)+h(t+1)+h(t+2) & \geq 3
\end{array}
$$
Indeed, these inequalities guarantee the existence of elements $a_{i_{1}} \in \bar{t}, a_{i_{2}} \in \bar{t} \cup \overline{t+1}$, $a_{i_{3}} \in \bar{t} \cup \overline{t+1} \cup \overline{t+2}$, et cetera, where all $i_{k}$ are different. Subtracting appropriate
multiples of $n$ and reordering our elements, we may assume $a_{1}=t, a_{2} \in\{t, t+1\}$, $a_{3} \in\{t, t+1, t+2\}$, et cetera. Finally subtracting $t$ from the complete sequence, we may assume $a_{1}=0, a_{2} \in\{0,1\}, a_{3} \in\{0,1,2\}$ et cetera. Now simply setting $b_{i}=i-1$ for all $i$ suffices, since $a_{i} \leq b_{i}$ for all $i$, the $b_{i}$ are all different modulo $n$, and
$$
\sum_{i=1}^{n} b_{i}=\frac{n(n-1)}{2} \leq \frac{n(n-1)}{2}+n\left[\frac{\sum_{i=1}^{n} a_{i}}{n}\right] .
$$
Put $x_{i}=h(i)-1$ for all $i=0, \ldots, n-1$. Note that $x_{i} \geq-1$, because $h(i) \geq 0$. If we have $x_{i} \geq 0$ for all $i=0, \ldots, n-1$, then taking $t=0$ completes the proof. Otherwise, we can pick some index $j$ such that $x_{j}=-1$. Let $y_{i}=x_{i}$ where $i=0, \ldots, j-1, j+1, \ldots, n-1$ and $y_{j}=0$. For sequence $\left\{y_{i}\right\}$ we have
$$
\sum_{i=0}^{n-1} y_{i}=\sum_{i=0}^{n-1} x_{i}+1=\sum_{i=0}^{n-1} h(i)-n+1=1
$$
so from Raney's lemma there exists index $k$ such that $\sum_{i=k}^{k+j} y_{i}>0$ for all $j=0, \ldots, n-1$ where $y_{n+j}=y_{j}$ for $j=0, \ldots, k-1$. Taking $t=k$ we will have
$$
\sum_{t=k}^{k+i} h(t)-(i+1)=\sum_{t=k}^{k+i} x(t) \geq \sum_{t=k}^{k+i} y(t)-1 \geq 0
$$
for all $i=0, \ldots, n-1$ and we are done.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "# Problem 5",
"resource_path": "EGMO/segmented/en-2019-solutions-day2.jsonl",
"solution_match": "\nSolution 2. ",
"tier": "T2",
"year": "2019"
}
|
(Netherlands).
Let $n \geq 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive integers. Show that there exist positive integers $b_{1}, b_{2}, \ldots, b_{n}$ satisfying the following three conditions:
1). $a_{i} \leq b_{i}$ for $i=1,2, \ldots, n$;
2). the remainders of $b_{1}, b_{2}, \ldots, b_{n}$ on division by $n$ are pairwise different; and
3). $b_{1}+\cdots+b_{n} \leq n\left(\frac{n-1}{2}+\left\lfloor\frac{a_{1}+\cdots+a_{n}}{n}\right\rfloor\right)$.
(Here, $\lfloor x\rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
|
Choose a random permutation $c_{1}, \ldots, c_{n}$ of the integers $1,2, \ldots, n$. Let $b_{i}=a_{i}+f\left(c_{i}-a_{i}\right)$, where $f(x) \in\{0, \ldots, n-1\}$ denotes a remainder of $x$ modulo $n$. Observe, that for such defined sequence the first two conditions hold. The expected value of $B:=b_{1}+\ldots+b_{n}$ is easily seen to be equal to $a_{1}+\ldots+a_{n}+n(n-1) / 2$. Indeed, for each $i$ the random number $c_{i}-a_{i}$ has uniform distribution modulo $n$, thus the expected value of $f\left(c_{i}-a_{i}\right)$ is $(0+\ldots+(n-1)) / n=(n-1) / 2$. Therefore we may find such $c$ that $B \leq a_{1}+\ldots+a_{n}+n(n-1) / 2$. But $B-n(n-1) / 2$ is divisible by $n$ and therefore $B \leq n\left[\left(a_{1}+\ldots+a_{n}\right) / n\right]+n(n-1) / 2$ as needed.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
(Netherlands).
Let $n \geq 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive integers. Show that there exist positive integers $b_{1}, b_{2}, \ldots, b_{n}$ satisfying the following three conditions:
1). $a_{i} \leq b_{i}$ for $i=1,2, \ldots, n$;
2). the remainders of $b_{1}, b_{2}, \ldots, b_{n}$ on division by $n$ are pairwise different; and
3). $b_{1}+\cdots+b_{n} \leq n\left(\frac{n-1}{2}+\left\lfloor\frac{a_{1}+\cdots+a_{n}}{n}\right\rfloor\right)$.
(Here, $\lfloor x\rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
|
Choose a random permutation $c_{1}, \ldots, c_{n}$ of the integers $1,2, \ldots, n$. Let $b_{i}=a_{i}+f\left(c_{i}-a_{i}\right)$, where $f(x) \in\{0, \ldots, n-1\}$ denotes a remainder of $x$ modulo $n$. Observe, that for such defined sequence the first two conditions hold. The expected value of $B:=b_{1}+\ldots+b_{n}$ is easily seen to be equal to $a_{1}+\ldots+a_{n}+n(n-1) / 2$. Indeed, for each $i$ the random number $c_{i}-a_{i}$ has uniform distribution modulo $n$, thus the expected value of $f\left(c_{i}-a_{i}\right)$ is $(0+\ldots+(n-1)) / n=(n-1) / 2$. Therefore we may find such $c$ that $B \leq a_{1}+\ldots+a_{n}+n(n-1) / 2$. But $B-n(n-1) / 2$ is divisible by $n$ and therefore $B \leq n\left[\left(a_{1}+\ldots+a_{n}\right) / n\right]+n(n-1) / 2$ as needed.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "# Problem 5",
"resource_path": "EGMO/segmented/en-2019-solutions-day2.jsonl",
"solution_match": "\nSolution 3. ",
"tier": "T2",
"year": "2019"
}
|
(Netherlands).
Let $n \geq 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive integers. Show that there exist positive integers $b_{1}, b_{2}, \ldots, b_{n}$ satisfying the following three conditions:
1). $a_{i} \leq b_{i}$ for $i=1,2, \ldots, n$;
2). the remainders of $b_{1}, b_{2}, \ldots, b_{n}$ on division by $n$ are pairwise different; and
3). $b_{1}+\cdots+b_{n} \leq n\left(\frac{n-1}{2}+\left\lfloor\frac{a_{1}+\cdots+a_{n}}{n}\right\rfloor\right)$.
(Here, $\lfloor x\rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
|
We will prove the required statement for all sequences of non-negative integers $a_{i}$ by induction on $n$.
Case $n=1$ is obvious, just set $b_{1}=a_{1}$.
Now suppose that the statement is true for some $n \geq 1$; we shall prove it for $n+1$.
First note that, by subtracting a multiple of $n+1$ to each $a_{i}$ and possibly rearranging indices we can reduce the problem to the case where $0 \leq a_{1} \leq a_{2} \leq \cdots \leq a_{n} \leq a_{n+1}<$ $n+1$.
Now, by the induction hypothesis there exists a sequence $d_{1}, d_{2}, \ldots, d_{n}$ which satisfies the properties required by the statement in relation to the numbers $a_{1}, \ldots, a_{n}$. Set $I=\{i \mid 1 \leq$ $i \leq n$ and $\left.d_{i} \bmod n \geq a_{i}\right\}$ and construct $b_{i}$, for $i=1, \ldots, n+1$, as follows:
$$
b_{i}=\left\{\begin{array}{l}
d_{i} \bmod n, \text { when } i \in I, \\
n+1+\left(d_{i} \bmod n\right), \text { when } i \in\{1, \ldots, n\} \backslash I, \\
n, \text { for } i=n+1
\end{array}\right.
$$
Now, $a_{i} \leq d_{i} \bmod n \leq b_{i}$ for $i \in I$, while for $i \notin I$ we have $a_{i} \leq n \leq b_{i}$. Thus the sequence $\left(b_{i}\right)_{i=1}^{n+1}$ satisfies the first condition from the problem statement.
By the induction hypothesis, the numbers $d_{i} \bmod n$ are distinct for $i \in\{1, \ldots, n\}$, so the values $b_{i} \bmod (n+1)$ are distinct elements of $\{0, \ldots, n-1\}$ for $i \in\{1, \ldots, n\}$. Since $b_{n+1}=n$, the second condition is also satisfied.
Denote $k=|I|$. We have
$$
\begin{gathered}
\sum_{i=1}^{n+1} b_{i}=\sum_{i=1}^{n} b_{i}+n=\sum_{i=1}^{n} d_{i} \bmod n+(n-k)(n+1)+n= \\
\frac{n(n+1)}{2}+(n-k)(n+1)
\end{gathered}
$$
hence we need to show that
$$
\frac{n(n+1)}{2}+(n-k)(n+1) \leq \frac{n(n+1)}{2}+(n+1)\left[\frac{\sum_{i=1}^{n+1} a_{i}}{n+1}\right]
$$
equivalently, that
$$
n-k \leq\left[\frac{\sum_{i=1}^{n+1} a_{i}}{n+1}\right]
$$
Next, from the induction hypothesis we have
$$
\begin{gathered}
\frac{n(n-1)}{2}+n\left[\frac{\sum_{i=1}^{n} a_{i}}{n}\right] \geq \sum_{i=1}^{n} d_{i}=\sum_{i \in I} d_{i}+\sum_{i \notin I} d_{i} \geq \\
\sum_{i \in I} d_{i} \bmod n+\sum_{i \notin I}\left(n+d_{i} \bmod n\right)=\frac{n(n-1)}{2}+(n-k) n
\end{gathered}
$$
or
$$
n-k \leq\left[\frac{\sum_{i=1}^{n} a_{i}}{n}\right]
$$
Thus, it's enough to show that
$$
\frac{\sum_{i=1}^{n} a_{i}}{n} \leq \frac{\sum_{i=1}^{n+1} a_{i}}{n+1}
$$
because then
$$
n-k \leq\left[\frac{\sum_{i=1}^{n} a_{i}}{n}\right] \leq\left[\frac{\sum_{i=1}^{n+1} a_{i}}{n+1}\right]
$$
But the required inequality is equivalent to $\sum_{i=1}^{n} a_{i} \leq n a_{n+1}$, which is obvious.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
(Netherlands).
Let $n \geq 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive integers. Show that there exist positive integers $b_{1}, b_{2}, \ldots, b_{n}$ satisfying the following three conditions:
1). $a_{i} \leq b_{i}$ for $i=1,2, \ldots, n$;
2). the remainders of $b_{1}, b_{2}, \ldots, b_{n}$ on division by $n$ are pairwise different; and
3). $b_{1}+\cdots+b_{n} \leq n\left(\frac{n-1}{2}+\left\lfloor\frac{a_{1}+\cdots+a_{n}}{n}\right\rfloor\right)$.
(Here, $\lfloor x\rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
|
We will prove the required statement for all sequences of non-negative integers $a_{i}$ by induction on $n$.
Case $n=1$ is obvious, just set $b_{1}=a_{1}$.
Now suppose that the statement is true for some $n \geq 1$; we shall prove it for $n+1$.
First note that, by subtracting a multiple of $n+1$ to each $a_{i}$ and possibly rearranging indices we can reduce the problem to the case where $0 \leq a_{1} \leq a_{2} \leq \cdots \leq a_{n} \leq a_{n+1}<$ $n+1$.
Now, by the induction hypothesis there exists a sequence $d_{1}, d_{2}, \ldots, d_{n}$ which satisfies the properties required by the statement in relation to the numbers $a_{1}, \ldots, a_{n}$. Set $I=\{i \mid 1 \leq$ $i \leq n$ and $\left.d_{i} \bmod n \geq a_{i}\right\}$ and construct $b_{i}$, for $i=1, \ldots, n+1$, as follows:
$$
b_{i}=\left\{\begin{array}{l}
d_{i} \bmod n, \text { when } i \in I, \\
n+1+\left(d_{i} \bmod n\right), \text { when } i \in\{1, \ldots, n\} \backslash I, \\
n, \text { for } i=n+1
\end{array}\right.
$$
Now, $a_{i} \leq d_{i} \bmod n \leq b_{i}$ for $i \in I$, while for $i \notin I$ we have $a_{i} \leq n \leq b_{i}$. Thus the sequence $\left(b_{i}\right)_{i=1}^{n+1}$ satisfies the first condition from the problem statement.
By the induction hypothesis, the numbers $d_{i} \bmod n$ are distinct for $i \in\{1, \ldots, n\}$, so the values $b_{i} \bmod (n+1)$ are distinct elements of $\{0, \ldots, n-1\}$ for $i \in\{1, \ldots, n\}$. Since $b_{n+1}=n$, the second condition is also satisfied.
Denote $k=|I|$. We have
$$
\begin{gathered}
\sum_{i=1}^{n+1} b_{i}=\sum_{i=1}^{n} b_{i}+n=\sum_{i=1}^{n} d_{i} \bmod n+(n-k)(n+1)+n= \\
\frac{n(n+1)}{2}+(n-k)(n+1)
\end{gathered}
$$
hence we need to show that
$$
\frac{n(n+1)}{2}+(n-k)(n+1) \leq \frac{n(n+1)}{2}+(n+1)\left[\frac{\sum_{i=1}^{n+1} a_{i}}{n+1}\right]
$$
equivalently, that
$$
n-k \leq\left[\frac{\sum_{i=1}^{n+1} a_{i}}{n+1}\right]
$$
Next, from the induction hypothesis we have
$$
\begin{gathered}
\frac{n(n-1)}{2}+n\left[\frac{\sum_{i=1}^{n} a_{i}}{n}\right] \geq \sum_{i=1}^{n} d_{i}=\sum_{i \in I} d_{i}+\sum_{i \notin I} d_{i} \geq \\
\sum_{i \in I} d_{i} \bmod n+\sum_{i \notin I}\left(n+d_{i} \bmod n\right)=\frac{n(n-1)}{2}+(n-k) n
\end{gathered}
$$
or
$$
n-k \leq\left[\frac{\sum_{i=1}^{n} a_{i}}{n}\right]
$$
Thus, it's enough to show that
$$
\frac{\sum_{i=1}^{n} a_{i}}{n} \leq \frac{\sum_{i=1}^{n+1} a_{i}}{n+1}
$$
because then
$$
n-k \leq\left[\frac{\sum_{i=1}^{n} a_{i}}{n}\right] \leq\left[\frac{\sum_{i=1}^{n+1} a_{i}}{n+1}\right]
$$
But the required inequality is equivalent to $\sum_{i=1}^{n} a_{i} \leq n a_{n+1}$, which is obvious.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "# Problem 5",
"resource_path": "EGMO/segmented/en-2019-solutions-day2.jsonl",
"solution_match": "\nSolution 4. ",
"tier": "T2",
"year": "2019"
}
|
(Netherlands).
Let $n \geq 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive integers. Show that there exist positive integers $b_{1}, b_{2}, \ldots, b_{n}$ satisfying the following three conditions:
1). $a_{i} \leq b_{i}$ for $i=1,2, \ldots, n$;
2). the remainders of $b_{1}, b_{2}, \ldots, b_{n}$ on division by $n$ are pairwise different; and
3). $b_{1}+\cdots+b_{n} \leq n\left(\frac{n-1}{2}+\left\lfloor\frac{a_{1}+\cdots+a_{n}}{n}\right\rfloor\right)$.
(Here, $\lfloor x\rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
|
We can assume that all $a_{i} \in\{0,1, \ldots, n-1\}$, as we can deduct $n$ from both $a_{i}$ and $b_{i}$ for arbitrary $i$ without violating any of the three conditions from the problem statement. We shall also assume that $a_{1} \leq \ldots \leq a_{n}$.
Now let us provide an algorithm for constructing $b_{1}, \ldots, b_{n}$.
We start at step 1 by choosing $f(1)$ to be the maximum $i$ in $\{1, \ldots, n\}$ such that $a_{i} \leq n-1$, that is $f(1)=n$. We set $b_{f(1)}=n-1$.
Having performed steps 1 through $j$, at step $j+1$ we set $f(j+1)$ to be the maximum $i$ in $\{1, \ldots, n\} \backslash\{f(1), \ldots, f(j)\}$ such that $a_{i} \leq n-j-1$, if such an index exists. If it does, we set $b_{f(j+1)}=n-j-1$. If there is no such index, then we define $T=j$ and assign to the terms $b_{i}$, where $i \notin f(\{1, \ldots, j\})$, the values $n, n+1 \ldots, 2 n-j-1$, in any order, thus concluding the run of our algorithm.
Notice that the sequence $\left(b_{i}\right)_{i=1}^{n}$ satisfies the first and second required conditions by construction. We wish to show that it also satisfies the third.
Notice that, since the values chosen for the $b_{i}$ 's are those from $n-T$ to $2 n-T-1$, we have
$$
\sum_{i=1}^{n} b_{i}=\frac{n(n-1)}{2}+(n-T) n
$$
It therefore suffices to show that
$$
\left[\frac{a_{1}+\ldots+a_{n}}{n}\right] \geq n-T
$$
or (since the RHS is obviously an integer) $a_{1}+\ldots+a_{n} \geq(n-T) n$.
First, we show that there exists $1 \leq i \leq T$ such that $n-i=b_{f(i)}=a_{f(i)}$.
Indeed, this is true if $a_{n}=n-1$, so we may suppose $a_{n}<n-1$ and therefore $a_{n-1} \leq n-2$, so that $T \geq 2$. If $a_{n-1}=n-2$, we are done. If not, then $a_{n-1}<n-2$ and therefore $a_{n-2} \leq n-3$ and $T \geq 3$. Inductively, we actually obtain $T=n$ and necessarily $f(n)=1$ and $a_{1}=b_{1}=0$, which gives the desired result.
Now let $t$ be the largest such index $i$. We know that $n-t=b_{f(t)}=a_{f(t)}$ and therefore $a_{1} \leq \ldots \leq a_{f(t)} \leq n-t$. If we have $a_{1}=\ldots=a_{f(t)}=n-t$, then $T=t$ and we have $a_{i} \geq n-T$ for all $i$, hence $\sum_{i} a_{i} \geq n(n-T)$. Otherwise, $T>t$ and in fact one can show $T=t+f(t+1)$ by proceeding inductively and using the fact that $t$ is the last time for which $a_{f(t)}=b_{f(t)}$.
Now we get that, since $a_{f(t+1)+1} \geq n-t$, then $\sum_{i} a_{i} \geq(n-t)(n-f(t+1))=(n-T+f(t+$ 1) $)(n-f(t+1))=n(n-T)+n f(t+1)-f(t+1)(n-T+f(t+1))=n(n-T)+t f(t+1) \geq$ $n(n-T)$.
Greedy algorithm variant $\mathbf{1}$ (ISR). Consider the residues $0, \ldots, n-1$ modulo $n$ arranged in a circle clockwise, and place each $a_{i}$ on its corresponding residue; so that on each residue there is a stack of all $a_{i}$ s congruent to it modulo $n$, and the sum of the sizes of all stacks is exactly $n$. We iteratively flatten and spread the stacks forward, in such a way that the $a_{i}$ s are placed in the nearest available space on the circle clockwise (skipping over any already flattened residue or still standing stack). We may choose the order in which the stacks are flattened. Since the total amount of numbers equals the total number of spaces, there is always an available space and at the end all spaces are covered. The $b_{i} \mathrm{~s}$ are then defined by adding to each $a_{i}$ the number of places it was moved forward, which clearly satifies (i) and (ii), and we must prove that they satisfy (iii) as well.
Suppose that we flatten a stack of $k$ numbers at a residue $i$, causing it to overtake a stack of $l$ numbers at residue $j \in(i, i+k)$ (we can allow $j$ to be larger than $n$ and identify it
with its residue modulo $n$ ). Then in fact in fact in whichever order we would flatten the two stacks, the total number of forward steps would be the same, and the total sum of the corresponding $b_{t}$ (such that $a_{t} \bmod n \in\{i, j\}$ ) would be the same. Moreover, we can merge the stacks to a single stack of $k+l$ numbers at residue $i$, by replacing each $a_{t} \equiv j$ $(\bmod n)$ by $a_{t}^{\prime}=a_{t}-(j-i)$, and this stack would be flattened forward into the same positions as the separate stacks would have been, so applying our algorithm to the new stacks will yield the same total sum of $\sum b_{i}-$ but the $a_{i}$ s are strictly decreased, so $\sum a_{i}$ is decreased, so $\left\lfloor\frac{\sum a_{i}}{n}\right\rfloor$ is not increased - so by merging the stacks, we can only make the inequality we wish to prove tighter.
Thus, as long as there is some stack that when flattened will overtake another stack, we may merge stacks and only make the inequality tighter. Since the amount of numbers equals the amount of places, the merging process terminates with stacks of sizes $k_{1}, \ldots, k_{m}$, such that the stack $j$, when flattened, will exactly cover the interval to the next stack. Clearly the numbers in each such stack were advanced by a total of $\sum_{t=1}^{k_{j}-1}=\frac{k_{j}\left(k_{j}-1\right)}{2}$, thus $\sum b_{i}=\sum a_{i}+\sum_{j} \frac{k_{j}\left(k_{j}-1\right)}{2}$. Writing $\sum a_{i}=n \cdot r+s$ with $0 \leq s<n$, we must therefore show
$$
s+\sum_{j} \frac{k_{j}\left(k_{j}-1\right)}{2} \leq \frac{n(n-1)}{2}
$$
Ending 1. Observing that both sides of the last inequality are congruent modulo $n$ (both are congruent to the sum of all different residues), and that $0 \leq s<n$, the inequality is eqivalent to the simpler $\sum_{j} \frac{k_{j}\left(k_{j}-1\right)}{2} \leq \frac{n(n-1)}{2}$. Since $x(x-1)$ is convex, and $k_{j}$ are nonnegative integers with $\sum_{j} k_{j}=n$, the left hand side is maximal when $k_{j^{\prime}}=n$ and the rest are 0 , and then eqaulity is achieved. (Alternatively it follows easily for any non-negative reals from AM-GM.)
Ending 2. If $m=1$ (and $k_{1}=n$ ), then all numbers are in a single stack and have the same residue, so $s=0$ and equality is attained. If $m \geq 2$, then by convexity $\sum_{j} \frac{k_{j}\left(k_{j}-1\right)}{2}$ is maximal for $m=2$ and $\left(k_{1}, k_{2}\right)=(n-1,1)$, where it equals $\frac{(n-1)(n-2)}{2}$. Since we always have $s \leq n-1$, we find
$$
s+\sum_{j} \frac{k_{j}\left(k_{j}-1\right)}{2} \leq(n-1)+\frac{(n-1)(n-2)}{2}=\frac{n(n-1)}{2}
$$
as required.
Greedy algorithm variant 1' (ISR). We apply the same algorithm as in the previous solution. However, this time we note that we may merge stacks not only when they overlap after flattening, but also when they merely touch front-to-back: That is, we relax the condition $j \in(i, i+k)$ to $j \in(i, i+k]$; the argument for why such merges are allowed is exactly the same (But note that this is now sharp, as merging non-touching stacks can cause the sum of $b_{i}$ s to decrease).
We now observe that as long as there at least two stacks left, at least one will spread to touch (or overtake) the next stack, so we can perform merges until there is only one stack left. We are left with verifying that the inequality indeed holds for the case of only one stack which is spread forward, and this is indeed immediate (and in fact equality is achieved).
Greedy algorithm variant 2 (ISR). Let $c_{i}=a_{i} \bmod n$. Iteratively define $b_{i}=a_{i}+l_{i}$ greedily, write $d_{i}=c_{i}+l_{i}$, and observe that $l_{i} \leq n-1$ (since all residues are present in $\left.a_{i}, \ldots, a_{i}+n-1\right)$, hence $0 \leq d_{i} \leq 2 n-2$. Let $I=\left\{i \in I: d_{i} \geq n\right\}$, and note that $d_{i}=b_{i}$ $\bmod n$ if $i \notin I$ and $d_{i}=\left(b_{i} \bmod n\right)+n$ if $i \in I$. Then we must show
$$
\begin{aligned}
& \sum\left(a_{i}+l_{i}\right)=\sum b_{i} \leq \frac{n(n-1)}{2}+n\left\lfloor\frac{\sum a_{i}}{n}\right\rfloor \\
\Longleftrightarrow & \sum\left(c_{i}+l_{i}\right) \leq \sum\left(b_{i} \bmod n\right)+n\left\lfloor\frac{\sum c_{i}}{n}\right\rfloor \\
\Longleftrightarrow & \left.n|I| \leq n\left\lfloor\frac{\sum c_{i}}{n}\right\rfloor \Longleftrightarrow|I| \leq \left\lvert\, \frac{\sum c_{i}}{n}\right.\right\rfloor \Longleftrightarrow|I| \leq \frac{\sum c_{i}}{n}
\end{aligned}
$$
Let $k=|I|$, and for each $0 \leq m<n$ let $J_{m}=\left\{i: c_{i} \geq n-m\right\}$. We claim that there must be some $m$ for which $\left|J_{m}\right| \geq m+k$ (clearly for such $m$, at least $k$ of the sums $d_{j}$ with $j \in J_{m}$ must exceed $n$, i.e. at least $k$ of the elements of $J_{m}$ must also be in $I$, so this $m$ is a "witness" to the fact $|I| \geq k$ ). Once we find such an $m$, then we clearly have
$$
\sum c_{i} \geq(n-m)\left|J_{m}\right| \geq(n-m)(k+m)=n k+m(n-(k+m)) \geq n k=n|I|
$$
as required. We now construct such an $m$ explicitly.
If $k=0$, then clearly $m=n$ works (and also the original inequality is trivial). Otherwise, there are some $d_{i}$ s greater than $n$, and let $r+n=\max d_{i}$, and suppose $d_{t}=r+n$ and let $s=c_{t}$. Note that $r<s<r+n$ since $l_{t}<n$. Let $m \geq 0$ be the smallest number such that $n-m-1$ is not in $\left\{d_{1}, \ldots, d_{t}\right\}$, or equivalently $m$ is the largest such that $[n-m, n) \subset\left\{d_{1}, \ldots, d_{t}\right\}$. We claim that this $m$ satisfies the required property. More specifically, we claim that $J_{m}^{\prime}=\left\{i \leq t: d_{i} \geq n-m\right\}$ contains exactly $m+k$ elements and is a subset of $J_{m}$.
Note that by the greediness of the algorithm, it is impossible that for $\left[c_{i}, d_{i}\right)$ to contain numbers congruent to $d_{j} \bmod n$ with $j>i$ (otherwise, the greedy choice would prefer $d_{j}$ to $d_{i}$ at stage $i$ ). We call this the greedy property. In particular, it follows that all $i$ such that $d_{i} \in\left[s, d_{t}\right)=\left[c_{t}, d_{t}\right)$ must satisfy $i<t$. Additionally, $\left\{d_{i}\right\}$ is disjoint from $[n+r+1,2 n)$ (by maximality of $d_{t}$ ), but does intersect every residue class, so it contains $[r+1, n)$ and in particular also $[s, n)$. By the greedy property the latter can only be attained by $d_{i}$ with $i<t$, thus $[s, n) \subset\left\{d_{1}, \ldots, d_{t}\right\}$, and in particular $n-m \leq s$ (and in particular $m \geq 1$ ).
On the other hand $n-m>r$ (since $r \notin\left\{d_{i}\right\}$ at all), so $n-m-1 \geq r$. It follows that there is a time $t^{\prime} \geq t$ for which $d_{t^{\prime}} \equiv n-m-1(\bmod n)$ : If $n-m-1=r$ then this is true for $t^{\prime}=t$ with $d_{t}=n+r=2 n-m-1$; whereas if $n-m-1 \in[r+1, n)$ then there is some $t^{\prime}$ for which $d_{t^{\prime}}=n-m-1$, and by the definition of $m$ it satisfies $t^{\prime}>t$.
Therefore for all $i<t \leq t^{\prime}$ for which $d_{i} \geq n-m$, necessarily also $c_{i} \geq n-m$, since otherwise $d_{t^{\prime}} \in\left[c_{i}, d_{i}\right)$, in contradiction to the greedy property. This is also true for $i=t$, since $c_{t}=s \geq n-m$ as previously shown. Thus, $J_{m}^{\prime} \subset J_{m}$ as claimed.
Finally, since by definition of $m$ and greediness we have $[n-m, n) \cup\left\{d_{i}: i \in I\right\} \subset$ $\left\{d_{1}, \ldots, d_{t}\right\}$, we find that $\left\{d_{j}: j \in J_{m}^{\prime}\right\}=[n-m, n) \cup\left\{d_{i}: i \in I\right\}$ and thus $\left|J_{m}^{\prime}\right|=$ $|[n-m, n)|+|I|=m+k$ as claimed.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
(Netherlands).
Let $n \geq 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive integers. Show that there exist positive integers $b_{1}, b_{2}, \ldots, b_{n}$ satisfying the following three conditions:
1). $a_{i} \leq b_{i}$ for $i=1,2, \ldots, n$;
2). the remainders of $b_{1}, b_{2}, \ldots, b_{n}$ on division by $n$ are pairwise different; and
3). $b_{1}+\cdots+b_{n} \leq n\left(\frac{n-1}{2}+\left\lfloor\frac{a_{1}+\cdots+a_{n}}{n}\right\rfloor\right)$.
(Here, $\lfloor x\rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
|
We can assume that all $a_{i} \in\{0,1, \ldots, n-1\}$, as we can deduct $n$ from both $a_{i}$ and $b_{i}$ for arbitrary $i$ without violating any of the three conditions from the problem statement. We shall also assume that $a_{1} \leq \ldots \leq a_{n}$.
Now let us provide an algorithm for constructing $b_{1}, \ldots, b_{n}$.
We start at step 1 by choosing $f(1)$ to be the maximum $i$ in $\{1, \ldots, n\}$ such that $a_{i} \leq n-1$, that is $f(1)=n$. We set $b_{f(1)}=n-1$.
Having performed steps 1 through $j$, at step $j+1$ we set $f(j+1)$ to be the maximum $i$ in $\{1, \ldots, n\} \backslash\{f(1), \ldots, f(j)\}$ such that $a_{i} \leq n-j-1$, if such an index exists. If it does, we set $b_{f(j+1)}=n-j-1$. If there is no such index, then we define $T=j$ and assign to the terms $b_{i}$, where $i \notin f(\{1, \ldots, j\})$, the values $n, n+1 \ldots, 2 n-j-1$, in any order, thus concluding the run of our algorithm.
Notice that the sequence $\left(b_{i}\right)_{i=1}^{n}$ satisfies the first and second required conditions by construction. We wish to show that it also satisfies the third.
Notice that, since the values chosen for the $b_{i}$ 's are those from $n-T$ to $2 n-T-1$, we have
$$
\sum_{i=1}^{n} b_{i}=\frac{n(n-1)}{2}+(n-T) n
$$
It therefore suffices to show that
$$
\left[\frac{a_{1}+\ldots+a_{n}}{n}\right] \geq n-T
$$
or (since the RHS is obviously an integer) $a_{1}+\ldots+a_{n} \geq(n-T) n$.
First, we show that there exists $1 \leq i \leq T$ such that $n-i=b_{f(i)}=a_{f(i)}$.
Indeed, this is true if $a_{n}=n-1$, so we may suppose $a_{n}<n-1$ and therefore $a_{n-1} \leq n-2$, so that $T \geq 2$. If $a_{n-1}=n-2$, we are done. If not, then $a_{n-1}<n-2$ and therefore $a_{n-2} \leq n-3$ and $T \geq 3$. Inductively, we actually obtain $T=n$ and necessarily $f(n)=1$ and $a_{1}=b_{1}=0$, which gives the desired result.
Now let $t$ be the largest such index $i$. We know that $n-t=b_{f(t)}=a_{f(t)}$ and therefore $a_{1} \leq \ldots \leq a_{f(t)} \leq n-t$. If we have $a_{1}=\ldots=a_{f(t)}=n-t$, then $T=t$ and we have $a_{i} \geq n-T$ for all $i$, hence $\sum_{i} a_{i} \geq n(n-T)$. Otherwise, $T>t$ and in fact one can show $T=t+f(t+1)$ by proceeding inductively and using the fact that $t$ is the last time for which $a_{f(t)}=b_{f(t)}$.
Now we get that, since $a_{f(t+1)+1} \geq n-t$, then $\sum_{i} a_{i} \geq(n-t)(n-f(t+1))=(n-T+f(t+$ 1) $)(n-f(t+1))=n(n-T)+n f(t+1)-f(t+1)(n-T+f(t+1))=n(n-T)+t f(t+1) \geq$ $n(n-T)$.
Greedy algorithm variant $\mathbf{1}$ (ISR). Consider the residues $0, \ldots, n-1$ modulo $n$ arranged in a circle clockwise, and place each $a_{i}$ on its corresponding residue; so that on each residue there is a stack of all $a_{i}$ s congruent to it modulo $n$, and the sum of the sizes of all stacks is exactly $n$. We iteratively flatten and spread the stacks forward, in such a way that the $a_{i}$ s are placed in the nearest available space on the circle clockwise (skipping over any already flattened residue or still standing stack). We may choose the order in which the stacks are flattened. Since the total amount of numbers equals the total number of spaces, there is always an available space and at the end all spaces are covered. The $b_{i} \mathrm{~s}$ are then defined by adding to each $a_{i}$ the number of places it was moved forward, which clearly satifies (i) and (ii), and we must prove that they satisfy (iii) as well.
Suppose that we flatten a stack of $k$ numbers at a residue $i$, causing it to overtake a stack of $l$ numbers at residue $j \in(i, i+k)$ (we can allow $j$ to be larger than $n$ and identify it
with its residue modulo $n$ ). Then in fact in fact in whichever order we would flatten the two stacks, the total number of forward steps would be the same, and the total sum of the corresponding $b_{t}$ (such that $a_{t} \bmod n \in\{i, j\}$ ) would be the same. Moreover, we can merge the stacks to a single stack of $k+l$ numbers at residue $i$, by replacing each $a_{t} \equiv j$ $(\bmod n)$ by $a_{t}^{\prime}=a_{t}-(j-i)$, and this stack would be flattened forward into the same positions as the separate stacks would have been, so applying our algorithm to the new stacks will yield the same total sum of $\sum b_{i}-$ but the $a_{i}$ s are strictly decreased, so $\sum a_{i}$ is decreased, so $\left\lfloor\frac{\sum a_{i}}{n}\right\rfloor$ is not increased - so by merging the stacks, we can only make the inequality we wish to prove tighter.
Thus, as long as there is some stack that when flattened will overtake another stack, we may merge stacks and only make the inequality tighter. Since the amount of numbers equals the amount of places, the merging process terminates with stacks of sizes $k_{1}, \ldots, k_{m}$, such that the stack $j$, when flattened, will exactly cover the interval to the next stack. Clearly the numbers in each such stack were advanced by a total of $\sum_{t=1}^{k_{j}-1}=\frac{k_{j}\left(k_{j}-1\right)}{2}$, thus $\sum b_{i}=\sum a_{i}+\sum_{j} \frac{k_{j}\left(k_{j}-1\right)}{2}$. Writing $\sum a_{i}=n \cdot r+s$ with $0 \leq s<n$, we must therefore show
$$
s+\sum_{j} \frac{k_{j}\left(k_{j}-1\right)}{2} \leq \frac{n(n-1)}{2}
$$
Ending 1. Observing that both sides of the last inequality are congruent modulo $n$ (both are congruent to the sum of all different residues), and that $0 \leq s<n$, the inequality is eqivalent to the simpler $\sum_{j} \frac{k_{j}\left(k_{j}-1\right)}{2} \leq \frac{n(n-1)}{2}$. Since $x(x-1)$ is convex, and $k_{j}$ are nonnegative integers with $\sum_{j} k_{j}=n$, the left hand side is maximal when $k_{j^{\prime}}=n$ and the rest are 0 , and then eqaulity is achieved. (Alternatively it follows easily for any non-negative reals from AM-GM.)
Ending 2. If $m=1$ (and $k_{1}=n$ ), then all numbers are in a single stack and have the same residue, so $s=0$ and equality is attained. If $m \geq 2$, then by convexity $\sum_{j} \frac{k_{j}\left(k_{j}-1\right)}{2}$ is maximal for $m=2$ and $\left(k_{1}, k_{2}\right)=(n-1,1)$, where it equals $\frac{(n-1)(n-2)}{2}$. Since we always have $s \leq n-1$, we find
$$
s+\sum_{j} \frac{k_{j}\left(k_{j}-1\right)}{2} \leq(n-1)+\frac{(n-1)(n-2)}{2}=\frac{n(n-1)}{2}
$$
as required.
Greedy algorithm variant 1' (ISR). We apply the same algorithm as in the previous solution. However, this time we note that we may merge stacks not only when they overlap after flattening, but also when they merely touch front-to-back: That is, we relax the condition $j \in(i, i+k)$ to $j \in(i, i+k]$; the argument for why such merges are allowed is exactly the same (But note that this is now sharp, as merging non-touching stacks can cause the sum of $b_{i}$ s to decrease).
We now observe that as long as there at least two stacks left, at least one will spread to touch (or overtake) the next stack, so we can perform merges until there is only one stack left. We are left with verifying that the inequality indeed holds for the case of only one stack which is spread forward, and this is indeed immediate (and in fact equality is achieved).
Greedy algorithm variant 2 (ISR). Let $c_{i}=a_{i} \bmod n$. Iteratively define $b_{i}=a_{i}+l_{i}$ greedily, write $d_{i}=c_{i}+l_{i}$, and observe that $l_{i} \leq n-1$ (since all residues are present in $\left.a_{i}, \ldots, a_{i}+n-1\right)$, hence $0 \leq d_{i} \leq 2 n-2$. Let $I=\left\{i \in I: d_{i} \geq n\right\}$, and note that $d_{i}=b_{i}$ $\bmod n$ if $i \notin I$ and $d_{i}=\left(b_{i} \bmod n\right)+n$ if $i \in I$. Then we must show
$$
\begin{aligned}
& \sum\left(a_{i}+l_{i}\right)=\sum b_{i} \leq \frac{n(n-1)}{2}+n\left\lfloor\frac{\sum a_{i}}{n}\right\rfloor \\
\Longleftrightarrow & \sum\left(c_{i}+l_{i}\right) \leq \sum\left(b_{i} \bmod n\right)+n\left\lfloor\frac{\sum c_{i}}{n}\right\rfloor \\
\Longleftrightarrow & \left.n|I| \leq n\left\lfloor\frac{\sum c_{i}}{n}\right\rfloor \Longleftrightarrow|I| \leq \left\lvert\, \frac{\sum c_{i}}{n}\right.\right\rfloor \Longleftrightarrow|I| \leq \frac{\sum c_{i}}{n}
\end{aligned}
$$
Let $k=|I|$, and for each $0 \leq m<n$ let $J_{m}=\left\{i: c_{i} \geq n-m\right\}$. We claim that there must be some $m$ for which $\left|J_{m}\right| \geq m+k$ (clearly for such $m$, at least $k$ of the sums $d_{j}$ with $j \in J_{m}$ must exceed $n$, i.e. at least $k$ of the elements of $J_{m}$ must also be in $I$, so this $m$ is a "witness" to the fact $|I| \geq k$ ). Once we find such an $m$, then we clearly have
$$
\sum c_{i} \geq(n-m)\left|J_{m}\right| \geq(n-m)(k+m)=n k+m(n-(k+m)) \geq n k=n|I|
$$
as required. We now construct such an $m$ explicitly.
If $k=0$, then clearly $m=n$ works (and also the original inequality is trivial). Otherwise, there are some $d_{i}$ s greater than $n$, and let $r+n=\max d_{i}$, and suppose $d_{t}=r+n$ and let $s=c_{t}$. Note that $r<s<r+n$ since $l_{t}<n$. Let $m \geq 0$ be the smallest number such that $n-m-1$ is not in $\left\{d_{1}, \ldots, d_{t}\right\}$, or equivalently $m$ is the largest such that $[n-m, n) \subset\left\{d_{1}, \ldots, d_{t}\right\}$. We claim that this $m$ satisfies the required property. More specifically, we claim that $J_{m}^{\prime}=\left\{i \leq t: d_{i} \geq n-m\right\}$ contains exactly $m+k$ elements and is a subset of $J_{m}$.
Note that by the greediness of the algorithm, it is impossible that for $\left[c_{i}, d_{i}\right)$ to contain numbers congruent to $d_{j} \bmod n$ with $j>i$ (otherwise, the greedy choice would prefer $d_{j}$ to $d_{i}$ at stage $i$ ). We call this the greedy property. In particular, it follows that all $i$ such that $d_{i} \in\left[s, d_{t}\right)=\left[c_{t}, d_{t}\right)$ must satisfy $i<t$. Additionally, $\left\{d_{i}\right\}$ is disjoint from $[n+r+1,2 n)$ (by maximality of $d_{t}$ ), but does intersect every residue class, so it contains $[r+1, n)$ and in particular also $[s, n)$. By the greedy property the latter can only be attained by $d_{i}$ with $i<t$, thus $[s, n) \subset\left\{d_{1}, \ldots, d_{t}\right\}$, and in particular $n-m \leq s$ (and in particular $m \geq 1$ ).
On the other hand $n-m>r$ (since $r \notin\left\{d_{i}\right\}$ at all), so $n-m-1 \geq r$. It follows that there is a time $t^{\prime} \geq t$ for which $d_{t^{\prime}} \equiv n-m-1(\bmod n)$ : If $n-m-1=r$ then this is true for $t^{\prime}=t$ with $d_{t}=n+r=2 n-m-1$; whereas if $n-m-1 \in[r+1, n)$ then there is some $t^{\prime}$ for which $d_{t^{\prime}}=n-m-1$, and by the definition of $m$ it satisfies $t^{\prime}>t$.
Therefore for all $i<t \leq t^{\prime}$ for which $d_{i} \geq n-m$, necessarily also $c_{i} \geq n-m$, since otherwise $d_{t^{\prime}} \in\left[c_{i}, d_{i}\right)$, in contradiction to the greedy property. This is also true for $i=t$, since $c_{t}=s \geq n-m$ as previously shown. Thus, $J_{m}^{\prime} \subset J_{m}$ as claimed.
Finally, since by definition of $m$ and greediness we have $[n-m, n) \cup\left\{d_{i}: i \in I\right\} \subset$ $\left\{d_{1}, \ldots, d_{t}\right\}$, we find that $\left\{d_{j}: j \in J_{m}^{\prime}\right\}=[n-m, n) \cup\left\{d_{i}: i \in I\right\}$ and thus $\left|J_{m}^{\prime}\right|=$ $|[n-m, n)|+|I|=m+k$ as claimed.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "# Problem 5",
"resource_path": "EGMO/segmented/en-2019-solutions-day2.jsonl",
"solution_match": "\nSolution 5. ",
"tier": "T2",
"year": "2019"
}
|
(United Kingdom).
On a circle, Alina draws 2019 chords, the endpoints of which are all different. A point is considered marked if it is either
(i) one of the 4038 endpoints of a chord; or
(ii) an intersection point of at least two chords.
Alina labels each marked point. Of the 4038 points meeting criterion (i), Alina labels 2019 points with a 0 and the other 2019 points with a 1 . She labels each point meeting criterion (ii) with an arbitrary integer (not necessarily positive).
Along each chord, Alina considers the segments connecting two consecutive marked points. (A chord with $k$ marked points has $k-1$ such segments.) She labels each such segment in yellow with the sum of the labels of its two endpoints and in blue with the absolute value of their difference.
Alina finds that the $N+1$ yellow labels take each value $0,1, \ldots, N$ exactly once. Show that at least one blue label is a multiple of 3 .
(A chord is a line segment joining two different points on a circle.)
|
First we prove the following:
Lemma: if we color all of the points white or black, then the number of white-black edges, which we denote $E_{W B}$, is equal modulo 2 to the number of white (or black) points on the circumference, which we denote $C_{W}$, resp. $C_{B}$.
Observe that changing the colour of any interior point does not change the parity of $E_{W B}$, as each interior point has even degree, so it suffices to show the statement holds when all interior points are black. But then $E_{W B}=C_{W}$ so certainly the parities are equal.
Now returning to the original problem, assume that no two adjacent vertex labels differ by a multiple of three, and three-colour the vertices according to the residue class of the labels modulo 3. Let $E_{01}$ denote the number of edges between 0 -vertices and 1 -vertices, and $C_{0}$ denote the number of 0 -vertices on the boundary, and so on.
Then, consider the two-coloring obtained by combining the 1-vertices and 2-vertices. By applying the lemma, we see that $E_{01}+E_{02} \equiv C_{0} \bmod 2$.
$$
\text { Similarly } E_{01}+E_{12} \equiv C_{1}, \quad \text { and } E_{02}+E_{12} \equiv C_{2}, \quad \bmod 2
$$
Using the fact that $C_{0}=C_{1}=2019$ and $C_{2}=0$, we deduce that either $E_{02}$ and $E_{12}$ are even and $E_{01}$ is odd; or $E_{02}$ and $E_{12}$ are odd and $E_{01}$ is even.
But if the edge labels are the first $N$ non-negative integers, then $E_{01}=E_{12}$ unless $N \equiv 0$ modulo 3 , in which case $E_{01}=E_{02}$. So however Alina chooses the vertex labels, it is not possible that the multiset of edge labels is $\{0, \ldots, N\}$.
Hence in fact two vertex labels must differ by a multiple of 3 .
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
(United Kingdom).
On a circle, Alina draws 2019 chords, the endpoints of which are all different. A point is considered marked if it is either
(i) one of the 4038 endpoints of a chord; or
(ii) an intersection point of at least two chords.
Alina labels each marked point. Of the 4038 points meeting criterion (i), Alina labels 2019 points with a 0 and the other 2019 points with a 1 . She labels each point meeting criterion (ii) with an arbitrary integer (not necessarily positive).
Along each chord, Alina considers the segments connecting two consecutive marked points. (A chord with $k$ marked points has $k-1$ such segments.) She labels each such segment in yellow with the sum of the labels of its two endpoints and in blue with the absolute value of their difference.
Alina finds that the $N+1$ yellow labels take each value $0,1, \ldots, N$ exactly once. Show that at least one blue label is a multiple of 3 .
(A chord is a line segment joining two different points on a circle.)
|
First we prove the following:
Lemma: if we color all of the points white or black, then the number of white-black edges, which we denote $E_{W B}$, is equal modulo 2 to the number of white (or black) points on the circumference, which we denote $C_{W}$, resp. $C_{B}$.
Observe that changing the colour of any interior point does not change the parity of $E_{W B}$, as each interior point has even degree, so it suffices to show the statement holds when all interior points are black. But then $E_{W B}=C_{W}$ so certainly the parities are equal.
Now returning to the original problem, assume that no two adjacent vertex labels differ by a multiple of three, and three-colour the vertices according to the residue class of the labels modulo 3. Let $E_{01}$ denote the number of edges between 0 -vertices and 1 -vertices, and $C_{0}$ denote the number of 0 -vertices on the boundary, and so on.
Then, consider the two-coloring obtained by combining the 1-vertices and 2-vertices. By applying the lemma, we see that $E_{01}+E_{02} \equiv C_{0} \bmod 2$.
$$
\text { Similarly } E_{01}+E_{12} \equiv C_{1}, \quad \text { and } E_{02}+E_{12} \equiv C_{2}, \quad \bmod 2
$$
Using the fact that $C_{0}=C_{1}=2019$ and $C_{2}=0$, we deduce that either $E_{02}$ and $E_{12}$ are even and $E_{01}$ is odd; or $E_{02}$ and $E_{12}$ are odd and $E_{01}$ is even.
But if the edge labels are the first $N$ non-negative integers, then $E_{01}=E_{12}$ unless $N \equiv 0$ modulo 3 , in which case $E_{01}=E_{02}$. So however Alina chooses the vertex labels, it is not possible that the multiset of edge labels is $\{0, \ldots, N\}$.
Hence in fact two vertex labels must differ by a multiple of 3 .
|
{
"exam": "EGMO",
"problem_label": "6",
"problem_match": "# Problem 6",
"resource_path": "EGMO/segmented/en-2019-solutions-day2.jsonl",
"solution_match": "\nSolution 1. ",
"tier": "T2",
"year": "2019"
}
|
(United Kingdom).
On a circle, Alina draws 2019 chords, the endpoints of which are all different. A point is considered marked if it is either
(i) one of the 4038 endpoints of a chord; or
(ii) an intersection point of at least two chords.
Alina labels each marked point. Of the 4038 points meeting criterion (i), Alina labels 2019 points with a 0 and the other 2019 points with a 1 . She labels each point meeting criterion (ii) with an arbitrary integer (not necessarily positive).
Along each chord, Alina considers the segments connecting two consecutive marked points. (A chord with $k$ marked points has $k-1$ such segments.) She labels each such segment in yellow with the sum of the labels of its two endpoints and in blue with the absolute value of their difference.
Alina finds that the $N+1$ yellow labels take each value $0,1, \ldots, N$ exactly once. Show that at least one blue label is a multiple of 3 .
(A chord is a line segment joining two different points on a circle.)
|
As before, colour vertices based on their label modulo 3.
Suppose this gives a valid 3-colouring of the graph with 2019 0s and 2019 1s on the
circumference. Identify pairs of 0-labelled vertices and pairs of 1-labelled vertices on the circumference, with one 0 and one 1 left over. The resulting graph has even degrees except these two leaves. So the connected component $\mathcal{C}$ containing these leaves has an Eulerian path, and any other component has an Eulerian cycle.
Let $E_{01}^{*}$ denote the number of edges between 0 -vertices and 1 -vertices in $\mathcal{C}$, and let $E_{01}^{\prime}$ denote the number of such edges in the other components, and so on. By studying whether a given vertex has label congruent to 0 modulo 3 or not as we go along the Eulerian path in $\mathcal{C}$, we find $E_{01}^{*}+E_{02}^{*}$ is odd, and similarly $E_{01}^{*}+E_{12}^{*}$ is odd. Since neither start nor end vertex is a 2 -vertex, $E_{02}^{*}+E_{12}^{*}$ must be even.
Applying the same argument for the Eulerian cycle in each other component and adding up, we find that $E_{01}^{\prime}+E_{02}^{\prime}, E_{01}^{\prime}+E_{12}^{\prime}, E_{02}^{\prime}+E_{12}^{\prime}$ are all even. So, again we find $E_{01}+E_{02}$, $E_{01}+E_{12}$ are odd, and $E_{02}+E_{12}$ is even, and we finish as in the original solution.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
(United Kingdom).
On a circle, Alina draws 2019 chords, the endpoints of which are all different. A point is considered marked if it is either
(i) one of the 4038 endpoints of a chord; or
(ii) an intersection point of at least two chords.
Alina labels each marked point. Of the 4038 points meeting criterion (i), Alina labels 2019 points with a 0 and the other 2019 points with a 1 . She labels each point meeting criterion (ii) with an arbitrary integer (not necessarily positive).
Along each chord, Alina considers the segments connecting two consecutive marked points. (A chord with $k$ marked points has $k-1$ such segments.) She labels each such segment in yellow with the sum of the labels of its two endpoints and in blue with the absolute value of their difference.
Alina finds that the $N+1$ yellow labels take each value $0,1, \ldots, N$ exactly once. Show that at least one blue label is a multiple of 3 .
(A chord is a line segment joining two different points on a circle.)
|
As before, colour vertices based on their label modulo 3.
Suppose this gives a valid 3-colouring of the graph with 2019 0s and 2019 1s on the
circumference. Identify pairs of 0-labelled vertices and pairs of 1-labelled vertices on the circumference, with one 0 and one 1 left over. The resulting graph has even degrees except these two leaves. So the connected component $\mathcal{C}$ containing these leaves has an Eulerian path, and any other component has an Eulerian cycle.
Let $E_{01}^{*}$ denote the number of edges between 0 -vertices and 1 -vertices in $\mathcal{C}$, and let $E_{01}^{\prime}$ denote the number of such edges in the other components, and so on. By studying whether a given vertex has label congruent to 0 modulo 3 or not as we go along the Eulerian path in $\mathcal{C}$, we find $E_{01}^{*}+E_{02}^{*}$ is odd, and similarly $E_{01}^{*}+E_{12}^{*}$ is odd. Since neither start nor end vertex is a 2 -vertex, $E_{02}^{*}+E_{12}^{*}$ must be even.
Applying the same argument for the Eulerian cycle in each other component and adding up, we find that $E_{01}^{\prime}+E_{02}^{\prime}, E_{01}^{\prime}+E_{12}^{\prime}, E_{02}^{\prime}+E_{12}^{\prime}$ are all even. So, again we find $E_{01}+E_{02}$, $E_{01}+E_{12}$ are odd, and $E_{02}+E_{12}$ is even, and we finish as in the original solution.
|
{
"exam": "EGMO",
"problem_label": "6",
"problem_match": "# Problem 6",
"resource_path": "EGMO/segmented/en-2019-solutions-day2.jsonl",
"solution_match": "\nSolution 2. ",
"tier": "T2",
"year": "2019"
}
|
According to Anna, the number 2021 is fantabulous. She states that if any element of the set $\{m, 2 m+1,3 m\}$ is fantabulousfor a positive integer $m$, then they are all fantabulous. Is the number 2021 ${ }^{2021}$ fantabulous?
(Australia, Angelo Di Pasquale)
Answer: Yes
|
Consider the sequence of positive integers $m, 3 m, 6 m+1,12 m+3,4 m+1,2 m$. Since each number in the sequence is fantabulous if and only if the next one is, we deduce that $m$ is fantabulous if and only if $2 m$ is fantabulous.
Combined with the fact that $m$ is fantabulous if and only if $2 m+1$ is fantabulous, this implies that $m>1$ is fantabulous if and only if $f(m)=\left[\frac{m}{2}\right]$ is fantabulous. We can apply $f$ sufficiently many times to any positive integer $n$ to conclude that $n$ is fantabulous if and only if 1 is fantabulous. Therefore, the fact that 2021 is fantabulous implies that 1 is fantabulous, which in turn implies that $2021^{2021}$ is fantabulous.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
According to Anna, the number 2021 is fantabulous. She states that if any element of the set $\{m, 2 m+1,3 m\}$ is fantabulousfor a positive integer $m$, then they are all fantabulous. Is the number 2021 ${ }^{2021}$ fantabulous?
(Australia, Angelo Di Pasquale)
Answer: Yes
|
Consider the sequence of positive integers $m, 3 m, 6 m+1,12 m+3,4 m+1,2 m$. Since each number in the sequence is fantabulous if and only if the next one is, we deduce that $m$ is fantabulous if and only if $2 m$ is fantabulous.
Combined with the fact that $m$ is fantabulous if and only if $2 m+1$ is fantabulous, this implies that $m>1$ is fantabulous if and only if $f(m)=\left[\frac{m}{2}\right]$ is fantabulous. We can apply $f$ sufficiently many times to any positive integer $n$ to conclude that $n$ is fantabulous if and only if 1 is fantabulous. Therefore, the fact that 2021 is fantabulous implies that 1 is fantabulous, which in turn implies that $2021^{2021}$ is fantabulous.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "EGMO/segmented/en-2021-solutions.jsonl",
"solution_match": "# Solution 1.",
"tier": "T2",
"year": "2021"
}
|
According to Anna, the number 2021 is fantabulous. She states that if any element of the set $\{m, 2 m+1,3 m\}$ is fantabulousfor a positive integer $m$, then they are all fantabulous. Is the number 2021 ${ }^{2021}$ fantabulous?
(Australia, Angelo Di Pasquale)
Answer: Yes
|
Let $m>1$ be a fantabulous number. Note that at least one of the following four cases must hold.
- Case 1. The number $m$ is odd;
We have $m=2 a+1$ for some positive integer $a$, so $a<m$ is also fantabulous.
- Case 2. The number $m$ is a multiple of 3;
We have $m=3 a$ for some positive integer $a$, so $a<m$ is also fantabulous.
- Case 3. The number $m$ is 4 modulo 6;
We have $m=6 a-2$ for some positive integer $a$. We have the sequence of fantabulous numbers
$$
(6 a-2) \rightarrow(12 a-3) \rightarrow(4 a-1)
$$
so $4 a-1<m$ is also fantabulous.
## - Case 4. The number $m$ is 2 modulo 6;
We have $m=6 a+2$ for some positive integer $a$. We have the sequence of fantabulous numbers
$$
(6 a+2) \rightarrow(12 a+5) \rightarrow(36 a+15) \rightarrow(18 a+7) \rightarrow(9 a+3) \rightarrow(3 a+1)
$$
so $3 a+1<m$ is also fantabulous.
In all cases, we see that there is another fantabulous number less than $m$. Since 2021 is fantabulous, it follows that 1 is fantabulous.
Observe that a number $m$ is not fantabulous if and only if all of the elements of the set $\{m, 2 m+1,3 m\}$ are not fantabulous. So, the argument above shows that if there exists a positive integer that is not fantabulous, then 1 would not be fantabulous either. This is a contradiction, so all positive integers are fantabulous and, in particular, $2021^{2021}$ is fantabulous.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
According to Anna, the number 2021 is fantabulous. She states that if any element of the set $\{m, 2 m+1,3 m\}$ is fantabulousfor a positive integer $m$, then they are all fantabulous. Is the number 2021 ${ }^{2021}$ fantabulous?
(Australia, Angelo Di Pasquale)
Answer: Yes
|
Let $m>1$ be a fantabulous number. Note that at least one of the following four cases must hold.
- Case 1. The number $m$ is odd;
We have $m=2 a+1$ for some positive integer $a$, so $a<m$ is also fantabulous.
- Case 2. The number $m$ is a multiple of 3;
We have $m=3 a$ for some positive integer $a$, so $a<m$ is also fantabulous.
- Case 3. The number $m$ is 4 modulo 6;
We have $m=6 a-2$ for some positive integer $a$. We have the sequence of fantabulous numbers
$$
(6 a-2) \rightarrow(12 a-3) \rightarrow(4 a-1)
$$
so $4 a-1<m$ is also fantabulous.
## - Case 4. The number $m$ is 2 modulo 6;
We have $m=6 a+2$ for some positive integer $a$. We have the sequence of fantabulous numbers
$$
(6 a+2) \rightarrow(12 a+5) \rightarrow(36 a+15) \rightarrow(18 a+7) \rightarrow(9 a+3) \rightarrow(3 a+1)
$$
so $3 a+1<m$ is also fantabulous.
In all cases, we see that there is another fantabulous number less than $m$. Since 2021 is fantabulous, it follows that 1 is fantabulous.
Observe that a number $m$ is not fantabulous if and only if all of the elements of the set $\{m, 2 m+1,3 m\}$ are not fantabulous. So, the argument above shows that if there exists a positive integer that is not fantabulous, then 1 would not be fantabulous either. This is a contradiction, so all positive integers are fantabulous and, in particular, $2021^{2021}$ is fantabulous.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "EGMO/segmented/en-2021-solutions.jsonl",
"solution_match": "# Solution 2.",
"tier": "T2",
"year": "2021"
}
|
According to Anna, the number 2021 is fantabulous. She states that if any element of the set $\{m, 2 m+1,3 m\}$ is fantabulousfor a positive integer $m$, then they are all fantabulous. Is the number 2021 ${ }^{2021}$ fantabulous?
(Australia, Angelo Di Pasquale)
Answer: Yes
|
The following transformations show that $a$ is fantabulous if and only if $3 a, 3 a+1$ or $3 a+2$ are fantabulous.
$$
\begin{aligned}
& a \rightarrow 3 a \\
& a \rightarrow 2 a+1 \rightarrow 6 a+3 \rightarrow 3 a+1 \\
& a \rightarrow 2 a+1 \rightarrow 4 a+3 \rightarrow 12 a+9 \rightarrow 36 a+27 \rightarrow 18 a+13 \rightarrow 9 a+6 \rightarrow 3 a+2
\end{aligned}
$$
This implies that $a \geq 3$ is fantabulous if and only if $f(a)=\left[\frac{a}{3}\right]$ is fantabulous. We can use this to deduce that 1 and 2 are fantabulous from the fact that 2021 is fantabulous in the following way:
$$
2021 \rightarrow 673 \rightarrow 224 \rightarrow 74 \rightarrow 24 \rightarrow 8 \rightarrow 2 \rightarrow 5 \rightarrow 1
$$
We can apply $f$ sufficiently many times to any positive integer $n$ to arrive at the number 1 or 2 . It follows that every positive integer is fantabulous, so $2021^{2021}$ is fantabulous.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
According to Anna, the number 2021 is fantabulous. She states that if any element of the set $\{m, 2 m+1,3 m\}$ is fantabulousfor a positive integer $m$, then they are all fantabulous. Is the number 2021 ${ }^{2021}$ fantabulous?
(Australia, Angelo Di Pasquale)
Answer: Yes
|
The following transformations show that $a$ is fantabulous if and only if $3 a, 3 a+1$ or $3 a+2$ are fantabulous.
$$
\begin{aligned}
& a \rightarrow 3 a \\
& a \rightarrow 2 a+1 \rightarrow 6 a+3 \rightarrow 3 a+1 \\
& a \rightarrow 2 a+1 \rightarrow 4 a+3 \rightarrow 12 a+9 \rightarrow 36 a+27 \rightarrow 18 a+13 \rightarrow 9 a+6 \rightarrow 3 a+2
\end{aligned}
$$
This implies that $a \geq 3$ is fantabulous if and only if $f(a)=\left[\frac{a}{3}\right]$ is fantabulous. We can use this to deduce that 1 and 2 are fantabulous from the fact that 2021 is fantabulous in the following way:
$$
2021 \rightarrow 673 \rightarrow 224 \rightarrow 74 \rightarrow 24 \rightarrow 8 \rightarrow 2 \rightarrow 5 \rightarrow 1
$$
We can apply $f$ sufficiently many times to any positive integer $n$ to arrive at the number 1 or 2 . It follows that every positive integer is fantabulous, so $2021^{2021}$ is fantabulous.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "EGMO/segmented/en-2021-solutions.jsonl",
"solution_match": "# Solution 3.",
"tier": "T2",
"year": "2021"
}
|
Find all functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$ such that the equation
$$
f(x f(x)+y)=f(y)+x^{2}
$$
holds for all rational numbers $x$ and $y$.
Here, $\mathbb{Q}$ denotes the set of rational numbers.
(Slovakia, Patrik Bak)
Answer: $f(x)=x$ and $f(x)=-x$.
|
Denote the equation from the statement by (1). Let $x f(x)=A$ and $x^{2}=B$. The equation (1) is of the form
$$
f(A+y)=f(y)+B
$$
Also, if we put $y \rightarrow-A+y$, we have $f(A-A+y)=f(-A+y)+B$. Therefore
$$
f(-A+y)=f(y)-B
$$
We can easily show that for any integer $n$ we even have
$$
f(n A+y)=f(y)+n B
$$
Indeed, it's trivially true for $n=0$ and if this holds true for some integer $n$, then
$$
f((n+1) A+y)=f(A+y+n A)=f(n y+A)+B=f(y)+n B+B=f(y)+(n+1) B
$$
and
$$
f((n-1) A+y)=f(-A+n A+y)=f(n A+y)-B=f(y)+n B-B=f(y)+(n-1) B .
$$
So, equation (2) follows from the induction on $n$.
Now we can say that for any integer $k$ it holds
$$
f(n x f(x)+y)=f(y)+n x^{2}
$$
If $y$ is given, then $f(y)+n x^{2}$ can be any rational number, since $n x^{2}$ can be any rational number. If it is supposed to be $\frac{p}{q}$, where $q \neq 0$, then we may take $n=p q$, and $x=\frac{1}{q}$. Therefore $f$ is surjective on $\mathbb{Q}$. So there's a rational number $c$ such that $f(c)=0$. Be putting $x=c$ into (1) we immediately get $c=0$, i.e. $f(0)=0$. Therefore, $f(x)=0$ if and only if $x=0$.
For any integer $n$ and for any rational $x, y$ it holds
$$
f\left(n^{2} x f(x)+y\right)=f(y)+n^{2} x^{2}=f(y)+(n x)^{2}=f(n x f(n x)+y)
$$
After taking $y=-n x f(n x)$ in (4), the right-hand side becomes 0 , therefore
$$
n^{2} x f(x)-n x f(n x)=0
$$
This simplifies into $n f(x)=f(n x)$ for $x \neq 0$, but it also holds for $x=0$. Therefore, for any rational number $x=\frac{p}{q}$ we have,
$$
f(x)=f\left(\frac{p}{q}\right)=f\left(p \cdot \frac{1}{q}\right)=p \cdot f\left(\frac{1}{p}\right)=p \cdot \frac{f\left(q \cdot \frac{1}{q}\right)}{q}=\frac{p}{q} \cdot f(1)=x f(1)
$$
So, we have $f(x)=k x$, for some rational number $k$. Let's put this answer in (1) and we get $k(x k x+y)=k y+x^{2}$, thus $k^{2}=1$. Therefore $f(x)=x$ and $f(x)=-x$ are solutions.
|
f(x)=x \text{ and } f(x)=-x
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$ such that the equation
$$
f(x f(x)+y)=f(y)+x^{2}
$$
holds for all rational numbers $x$ and $y$.
Here, $\mathbb{Q}$ denotes the set of rational numbers.
(Slovakia, Patrik Bak)
Answer: $f(x)=x$ and $f(x)=-x$.
|
Denote the equation from the statement by (1). Let $x f(x)=A$ and $x^{2}=B$. The equation (1) is of the form
$$
f(A+y)=f(y)+B
$$
Also, if we put $y \rightarrow-A+y$, we have $f(A-A+y)=f(-A+y)+B$. Therefore
$$
f(-A+y)=f(y)-B
$$
We can easily show that for any integer $n$ we even have
$$
f(n A+y)=f(y)+n B
$$
Indeed, it's trivially true for $n=0$ and if this holds true for some integer $n$, then
$$
f((n+1) A+y)=f(A+y+n A)=f(n y+A)+B=f(y)+n B+B=f(y)+(n+1) B
$$
and
$$
f((n-1) A+y)=f(-A+n A+y)=f(n A+y)-B=f(y)+n B-B=f(y)+(n-1) B .
$$
So, equation (2) follows from the induction on $n$.
Now we can say that for any integer $k$ it holds
$$
f(n x f(x)+y)=f(y)+n x^{2}
$$
If $y$ is given, then $f(y)+n x^{2}$ can be any rational number, since $n x^{2}$ can be any rational number. If it is supposed to be $\frac{p}{q}$, where $q \neq 0$, then we may take $n=p q$, and $x=\frac{1}{q}$. Therefore $f$ is surjective on $\mathbb{Q}$. So there's a rational number $c$ such that $f(c)=0$. Be putting $x=c$ into (1) we immediately get $c=0$, i.e. $f(0)=0$. Therefore, $f(x)=0$ if and only if $x=0$.
For any integer $n$ and for any rational $x, y$ it holds
$$
f\left(n^{2} x f(x)+y\right)=f(y)+n^{2} x^{2}=f(y)+(n x)^{2}=f(n x f(n x)+y)
$$
After taking $y=-n x f(n x)$ in (4), the right-hand side becomes 0 , therefore
$$
n^{2} x f(x)-n x f(n x)=0
$$
This simplifies into $n f(x)=f(n x)$ for $x \neq 0$, but it also holds for $x=0$. Therefore, for any rational number $x=\frac{p}{q}$ we have,
$$
f(x)=f\left(\frac{p}{q}\right)=f\left(p \cdot \frac{1}{q}\right)=p \cdot f\left(\frac{1}{p}\right)=p \cdot \frac{f\left(q \cdot \frac{1}{q}\right)}{q}=\frac{p}{q} \cdot f(1)=x f(1)
$$
So, we have $f(x)=k x$, for some rational number $k$. Let's put this answer in (1) and we get $k(x k x+y)=k y+x^{2}$, thus $k^{2}=1$. Therefore $f(x)=x$ and $f(x)=-x$ are solutions.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "\nProblem 2.",
"resource_path": "EGMO/segmented/en-2021-solutions.jsonl",
"solution_match": "\nSolution.",
"tier": "T2",
"year": "2021"
}
|
Let $A B C$ be a triangle with an obtuse angle at $A$. Let $E$ and $F$ be the intersections of the external bisector of angle $A$ with the altitudes of $A B C$ through $B$ and $C$ respectively. Let $M$ and $N$ be the points on the segments $E C$ and $F B$ respectively such that $\angle E M A=\angle B C A$ and $\angle A N F=\angle A B C$. Prove that the points $E, F, N, M$ lie on a circle.
|
The first solution is based on the main Lemma. We present this Lemma with two different proofs.
Lemma: Let $A B C$ be an acute triangle with $A B=B C$. Let $P$ be any point on $A C$. Line passing through $P$ perpendicular to $A B$, intersects ray $B C$ in point $T$. If the line $A T$ intersects the circumscribed circle of the triangle $A B C$ the second time at point $K$, then $\angle A K P=\angle A B P$.
## Proof 1:
Let $H$ be the orthocenter of the triangle $A B P$. Then
$$
\angle B H P=180^{\circ}-\angle B A C=180^{\circ}-\angle B C P
$$
So $B H P C$ is cyclic. Then we get
$$
T K \cdot T A=T C \cdot T B=T P \cdot T H
$$
So, $A H P K$ is also cyclic. But then
$$
\angle A K P=180^{\circ}-\angle A H P=\angle A B P
$$
Proof 2:
Consider the symmetric points $B^{\prime}$ and $C^{\prime}$ of $B$ and $C$, respectively, with respect to the line $P T$. It is clear that
$$
\mathrm{TC}^{\prime} \cdot T B^{\prime}=T C \cdot T B=T K \cdot T A
$$
So $B^{\prime} C^{\prime} K A$ is cyclic. Also, because of the symmetry we have
$$
\angle P C^{\prime} B^{\prime}=\angle P C B=\angle P A B
$$
So $B^{\prime} C^{\prime} P A$ is also cyclic. Therefore, the points $B^{\prime}, C^{\prime}, K, P$ and $A$ all lie on the common circle. Because of this fact and because of the symmetry again we have
$$
\angle P K A=\angle P B^{\prime} A=\angle P B A .
$$
So, lemma is proved and now return to the problem.
Let $H$ be intersection point of the altitudes at $B$ and $C$. Denote by $M^{\prime}$ and $N^{\prime}$ the intersection points of the circumcircle of the triangle $H E F$ with the segments $E C$ and $F B$, respectively. We are going to show that $M=M^{\prime}$ and $N=N^{\prime}$ and it will prove the points $E, F, N, M$ lie on a common circle.
Of course, $A$ is an orthocenter of the triangle $B C H$. Therefore $\angle B H A=\angle B C A, \angle C H A=\angle C B A$ and $\angle H B A=\angle H C A$. Thus
$$
\angle H E F=\angle H B A+\angle E A B=\angle H C A+\angle F A C=\angle H F E
$$
So, the triangle $H E F$ is isosceles, $H E=H F$.
By using lemma, we get
$$
\angle A M^{\prime} E=\angle A H E=\angle A C B
$$
and
$$
\angle A N^{\prime} F=\angle A H F=\angle A B C \text {. }
$$
Therefore $M=M^{\prime}$ and $N=N^{\prime}$ and we are done.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with an obtuse angle at $A$. Let $E$ and $F$ be the intersections of the external bisector of angle $A$ with the altitudes of $A B C$ through $B$ and $C$ respectively. Let $M$ and $N$ be the points on the segments $E C$ and $F B$ respectively such that $\angle E M A=\angle B C A$ and $\angle A N F=\angle A B C$. Prove that the points $E, F, N, M$ lie on a circle.
|
The first solution is based on the main Lemma. We present this Lemma with two different proofs.
Lemma: Let $A B C$ be an acute triangle with $A B=B C$. Let $P$ be any point on $A C$. Line passing through $P$ perpendicular to $A B$, intersects ray $B C$ in point $T$. If the line $A T$ intersects the circumscribed circle of the triangle $A B C$ the second time at point $K$, then $\angle A K P=\angle A B P$.
## Proof 1:
Let $H$ be the orthocenter of the triangle $A B P$. Then
$$
\angle B H P=180^{\circ}-\angle B A C=180^{\circ}-\angle B C P
$$
So $B H P C$ is cyclic. Then we get
$$
T K \cdot T A=T C \cdot T B=T P \cdot T H
$$
So, $A H P K$ is also cyclic. But then
$$
\angle A K P=180^{\circ}-\angle A H P=\angle A B P
$$
Proof 2:
Consider the symmetric points $B^{\prime}$ and $C^{\prime}$ of $B$ and $C$, respectively, with respect to the line $P T$. It is clear that
$$
\mathrm{TC}^{\prime} \cdot T B^{\prime}=T C \cdot T B=T K \cdot T A
$$
So $B^{\prime} C^{\prime} K A$ is cyclic. Also, because of the symmetry we have
$$
\angle P C^{\prime} B^{\prime}=\angle P C B=\angle P A B
$$
So $B^{\prime} C^{\prime} P A$ is also cyclic. Therefore, the points $B^{\prime}, C^{\prime}, K, P$ and $A$ all lie on the common circle. Because of this fact and because of the symmetry again we have
$$
\angle P K A=\angle P B^{\prime} A=\angle P B A .
$$
So, lemma is proved and now return to the problem.
Let $H$ be intersection point of the altitudes at $B$ and $C$. Denote by $M^{\prime}$ and $N^{\prime}$ the intersection points of the circumcircle of the triangle $H E F$ with the segments $E C$ and $F B$, respectively. We are going to show that $M=M^{\prime}$ and $N=N^{\prime}$ and it will prove the points $E, F, N, M$ lie on a common circle.
Of course, $A$ is an orthocenter of the triangle $B C H$. Therefore $\angle B H A=\angle B C A, \angle C H A=\angle C B A$ and $\angle H B A=\angle H C A$. Thus
$$
\angle H E F=\angle H B A+\angle E A B=\angle H C A+\angle F A C=\angle H F E
$$
So, the triangle $H E F$ is isosceles, $H E=H F$.
By using lemma, we get
$$
\angle A M^{\prime} E=\angle A H E=\angle A C B
$$
and
$$
\angle A N^{\prime} F=\angle A H F=\angle A B C \text {. }
$$
Therefore $M=M^{\prime}$ and $N=N^{\prime}$ and we are done.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "EGMO/segmented/en-2021-solutions.jsonl",
"solution_match": "# Solution 1.",
"tier": "T2",
"year": "2021"
}
|
Let $A B C$ be a triangle with an obtuse angle at $A$. Let $E$ and $F$ be the intersections of the external bisector of angle $A$ with the altitudes of $A B C$ through $B$ and $C$ respectively. Let $M$ and $N$ be the points on the segments $E C$ and $F B$ respectively such that $\angle E M A=\angle B C A$ and $\angle A N F=\angle A B C$. Prove that the points $E, F, N, M$ lie on a circle.
|
Let $X, Y$ be projections of $B$ on $A C$, and $C$ on $A B$, respectively. Let $\omega$ be circumcircle of $B X Y C$. Let $Z$ be intersection of $E C$ and $\omega$ and $D$ be projection of $E$ on $B A$.
$$
\angle M A C=\angle A M E-\angle M C A=\angle X C B-\angle X C E=\angle Z C B=\angle Z X B
$$
Since $B X Y C$ is cyclic $\angle A C Y=\angle X B A$, and since $D E X A$ is cyclic
$$
\angle E X D=\angle E A D=\angle F A C
$$
Therefore, we get that the quadrangles $B Z X D$ and $C M A F$ are similar. Hence $\angle F M C=\angle D Z B$. Since $Z E D B$ is cyclic,
$$
\angle D Z B=\angle D E B=\angle X A B .
$$
Thus $\angle F M C=\angle X A B$. Similarly, $\angle E N B=\angle Y A C$. We get that $\angle F M C=\angle E N B$ and it implies that the points $E, F, N, M$ lie on a circle.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with an obtuse angle at $A$. Let $E$ and $F$ be the intersections of the external bisector of angle $A$ with the altitudes of $A B C$ through $B$ and $C$ respectively. Let $M$ and $N$ be the points on the segments $E C$ and $F B$ respectively such that $\angle E M A=\angle B C A$ and $\angle A N F=\angle A B C$. Prove that the points $E, F, N, M$ lie on a circle.
|
Let $X, Y$ be projections of $B$ on $A C$, and $C$ on $A B$, respectively. Let $\omega$ be circumcircle of $B X Y C$. Let $Z$ be intersection of $E C$ and $\omega$ and $D$ be projection of $E$ on $B A$.
$$
\angle M A C=\angle A M E-\angle M C A=\angle X C B-\angle X C E=\angle Z C B=\angle Z X B
$$
Since $B X Y C$ is cyclic $\angle A C Y=\angle X B A$, and since $D E X A$ is cyclic
$$
\angle E X D=\angle E A D=\angle F A C
$$
Therefore, we get that the quadrangles $B Z X D$ and $C M A F$ are similar. Hence $\angle F M C=\angle D Z B$. Since $Z E D B$ is cyclic,
$$
\angle D Z B=\angle D E B=\angle X A B .
$$
Thus $\angle F M C=\angle X A B$. Similarly, $\angle E N B=\angle Y A C$. We get that $\angle F M C=\angle E N B$ and it implies that the points $E, F, N, M$ lie on a circle.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "\nProblem 3.",
"resource_path": "EGMO/segmented/en-2021-solutions.jsonl",
"solution_match": "# Solution 2.",
"tier": "T2",
"year": "2021"
}
|
Let $A B C$ be a triangle with incentre Iand let $D$ be an arbitrary point on the side $B C$. Let the line through $D$ perpendicular to $B I$ intersect $C I$ at $E$. Let the line through $D$ perpendicular to $C I$ intersect $B I$ at $F$. Prove that the reflection of $A$ in the line $E F$ lies on the line $B C$.
(Australia, Sampson Wong)
|
Let us consider the case when $I$ lies inside of triangle $E F D$. For the other cases the proof is almost the same only with the slight difference.
We are going to prove that the intersection point of the circumcircles of $A E C$ and $A F B$ (denote it by $T$ ) lies on the line $B C$ and this point is the symmetric point of $A$ with respect to $E F$. First of all we prove that $A E I F$ is cyclic, which implies that $T$ lies on the line $B C$, because
$$
\angle A T B+\angle A T C=\angle A F B+\angle A E C=\angle A F I+\angle A E I=180^{\circ} .
$$
Denote by $N$ an intersection point of the lines $D F$ and $A C$. Of course $N$ is the symmetric point with respect to $C I$. Thus, $\angle I N A=\angle I D B$. Also,
$$
\angle I F D=\angle N D C-\angle I B C=90^{\circ}-\angle I C B-\angle I B C=\angle I A N .
$$
So, we get that $A, I, N$ and $F$ lie on a common circle. Therefore, we have $\angle A F I=\angle I N A=\angle I D B$. Analogously, $\angle A E I=\angle I D C$ and we have
$$
\angle A F I+\angle A E I=\angle I D B+\angle I D C
$$
So $\angle A F I+\angle A E I=180^{\circ}$, thus $A E I F$ is cyclic and $T$ lies on the line $B C$.
Because $E C$ bisects the angle $A C B$ and $A E T C$ is cyclic we get $E A=E T$. Because of the similar reasons we have $F A=F T$. Therefore $T$ is the symmetric point of $A$ with respect to the line $E F$ and it lies on the line $B C$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with incentre Iand let $D$ be an arbitrary point on the side $B C$. Let the line through $D$ perpendicular to $B I$ intersect $C I$ at $E$. Let the line through $D$ perpendicular to $C I$ intersect $B I$ at $F$. Prove that the reflection of $A$ in the line $E F$ lies on the line $B C$.
(Australia, Sampson Wong)
|
Let us consider the case when $I$ lies inside of triangle $E F D$. For the other cases the proof is almost the same only with the slight difference.
We are going to prove that the intersection point of the circumcircles of $A E C$ and $A F B$ (denote it by $T$ ) lies on the line $B C$ and this point is the symmetric point of $A$ with respect to $E F$. First of all we prove that $A E I F$ is cyclic, which implies that $T$ lies on the line $B C$, because
$$
\angle A T B+\angle A T C=\angle A F B+\angle A E C=\angle A F I+\angle A E I=180^{\circ} .
$$
Denote by $N$ an intersection point of the lines $D F$ and $A C$. Of course $N$ is the symmetric point with respect to $C I$. Thus, $\angle I N A=\angle I D B$. Also,
$$
\angle I F D=\angle N D C-\angle I B C=90^{\circ}-\angle I C B-\angle I B C=\angle I A N .
$$
So, we get that $A, I, N$ and $F$ lie on a common circle. Therefore, we have $\angle A F I=\angle I N A=\angle I D B$. Analogously, $\angle A E I=\angle I D C$ and we have
$$
\angle A F I+\angle A E I=\angle I D B+\angle I D C
$$
So $\angle A F I+\angle A E I=180^{\circ}$, thus $A E I F$ is cyclic and $T$ lies on the line $B C$.
Because $E C$ bisects the angle $A C B$ and $A E T C$ is cyclic we get $E A=E T$. Because of the similar reasons we have $F A=F T$. Therefore $T$ is the symmetric point of $A$ with respect to the line $E F$ and it lies on the line $B C$.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "EGMO/segmented/en-2021-solutions.jsonl",
"solution_match": "# Solution 1.",
"tier": "T2",
"year": "2021"
}
|
Let $A B C$ be a triangle with incentre Iand let $D$ be an arbitrary point on the side $B C$. Let the line through $D$ perpendicular to $B I$ intersect $C I$ at $E$. Let the line through $D$ perpendicular to $C I$ intersect $B I$ at $F$. Prove that the reflection of $A$ in the line $E F$ lies on the line $B C$.
(Australia, Sampson Wong)
|
Like to the first solution, consider the case when $I$ lies inside of triangle $E F D$. we need to prove that $A E I F$ is cyclic. The finish of the proof is the same.
first note that $\triangle F D B \sim \triangle A I B$, because $\angle F B D=\angle A B I$, and
$$
\angle B F D=\angle F D C-\angle I B C=90^{\circ}-\angle I C D-\angle I B C=\angle I A B .
$$
Because of the similarity we have $\frac{A B}{A F}=\frac{B I}{B D}$. This equality of the length ratios with $\angle I B D=\angle A B F$ implies that $\triangle A B F \sim \triangle I B D$. Therefore, we have $\angle I D B=\angle A F B$. Analogously, we can get $\angle I D C=\angle A E C$, thus
$$
\angle A F I+\angle A E I=\angle I D B+\angle I D C=180^{\circ} .
$$
So, $A E I F$ is cyclic and we are done.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with incentre Iand let $D$ be an arbitrary point on the side $B C$. Let the line through $D$ perpendicular to $B I$ intersect $C I$ at $E$. Let the line through $D$ perpendicular to $C I$ intersect $B I$ at $F$. Prove that the reflection of $A$ in the line $E F$ lies on the line $B C$.
(Australia, Sampson Wong)
|
Like to the first solution, consider the case when $I$ lies inside of triangle $E F D$. we need to prove that $A E I F$ is cyclic. The finish of the proof is the same.
first note that $\triangle F D B \sim \triangle A I B$, because $\angle F B D=\angle A B I$, and
$$
\angle B F D=\angle F D C-\angle I B C=90^{\circ}-\angle I C D-\angle I B C=\angle I A B .
$$
Because of the similarity we have $\frac{A B}{A F}=\frac{B I}{B D}$. This equality of the length ratios with $\angle I B D=\angle A B F$ implies that $\triangle A B F \sim \triangle I B D$. Therefore, we have $\angle I D B=\angle A F B$. Analogously, we can get $\angle I D C=\angle A E C$, thus
$$
\angle A F I+\angle A E I=\angle I D B+\angle I D C=180^{\circ} .
$$
So, $A E I F$ is cyclic and we are done.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "\nProblem 4.",
"resource_path": "EGMO/segmented/en-2021-solutions.jsonl",
"solution_match": "# Solution 2.",
"tier": "T2",
"year": "2021"
}
|
A plane has a special point $O$ called the origin. Let $P$ be a set of 2021 points in the plane, such that
(i) no three points in $P$ lie on a line and
(ii) no two points in $P$ lie on a line through the origin.
A triangle with vertices in $P$ is fat, if $O$ is strictly inside the triangle. Find the maximum number of fat triangles.
(Austria, Veronika Schreitter)
Answer: $2021 \cdot 505 \cdot 337$
|
We will count minimal number of triangles that are not fat. Let $F$ set of fat triangles, and S set of triangles that are not fat. If triangle $X Y Z \in S$, we call $X$ and $Z$ good vertices if $O Y$ is located between $O X$ and $O Z$. For $A \in P$ let $S_{A} \subseteq S$ be set of triangles in $S$ for which $A$ is one of the good vertex.
It is easy to see that
$$
2|S|=\sum_{A \in P}\left|S_{A}\right|
$$
For $A \in P$, let $R_{A} \subset P$ and $L_{A} \subset P$ be parts of $P \backslash\{A\}$ divided by $A O$. Suppose for $A X Y \in S$ vertex $A$ is good, then clearly $X, Y \in R_{A}$ or $X, Y \in L_{A}$. On the other hand, if $X, Y \in R_{A}$ or $X, Y \in L_{A}$ then clearly $A X Y \in S$ and $A$ is its good vertex. Therefore,
$$
\left|S_{A}\right|=\binom{\left|R_{A}\right|}{2}+\binom{\left|L_{A}\right|}{2}
$$
It is easy to show following identity:
$$
\frac{x(x-1)}{2}+\frac{y(y-1)}{2}-2 \cdot \frac{\frac{x+y}{2}\left(\frac{x+y}{2}-1\right)}{2}=\frac{(x-y)^{2}}{4}
$$
By using (2) and (3) we get
$$
\left|S_{A}\right| \geq 2 \cdot\binom{\frac{\left|R_{A}\right|+\left|L_{A}\right|}{2}}{2}=2 \cdot\binom{1010}{2}=1010 \cdot 1009
$$
and the equality holds when $\left|R_{A}\right|=\left|L_{A}\right|=1010$. Hence
$$
|S|=\frac{\sum_{A \in P}\left|S_{A}\right|}{2} \geq \frac{2021 \cdot 1010 \cdot 1009}{2}=2021 \cdot 505 \cdot 1009
$$
Therefore,
$$
|F|=\binom{2021}{3}-|S| \leq 2021 \cdot 1010 \cdot 673-2021 \cdot 505 \cdot 1009=2021 \cdot 505 \cdot 337
$$
For configuration of points on regular 2021-gon which is centered at $O$, in equalities in (4), (5), (6) become equalities. Hence $2021 \cdot 505 \cdot 337$ is indeed the answer.
|
2021 \cdot 505 \cdot 337
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A plane has a special point $O$ called the origin. Let $P$ be a set of 2021 points in the plane, such that
(i) no three points in $P$ lie on a line and
(ii) no two points in $P$ lie on a line through the origin.
A triangle with vertices in $P$ is fat, if $O$ is strictly inside the triangle. Find the maximum number of fat triangles.
(Austria, Veronika Schreitter)
Answer: $2021 \cdot 505 \cdot 337$
|
We will count minimal number of triangles that are not fat. Let $F$ set of fat triangles, and S set of triangles that are not fat. If triangle $X Y Z \in S$, we call $X$ and $Z$ good vertices if $O Y$ is located between $O X$ and $O Z$. For $A \in P$ let $S_{A} \subseteq S$ be set of triangles in $S$ for which $A$ is one of the good vertex.
It is easy to see that
$$
2|S|=\sum_{A \in P}\left|S_{A}\right|
$$
For $A \in P$, let $R_{A} \subset P$ and $L_{A} \subset P$ be parts of $P \backslash\{A\}$ divided by $A O$. Suppose for $A X Y \in S$ vertex $A$ is good, then clearly $X, Y \in R_{A}$ or $X, Y \in L_{A}$. On the other hand, if $X, Y \in R_{A}$ or $X, Y \in L_{A}$ then clearly $A X Y \in S$ and $A$ is its good vertex. Therefore,
$$
\left|S_{A}\right|=\binom{\left|R_{A}\right|}{2}+\binom{\left|L_{A}\right|}{2}
$$
It is easy to show following identity:
$$
\frac{x(x-1)}{2}+\frac{y(y-1)}{2}-2 \cdot \frac{\frac{x+y}{2}\left(\frac{x+y}{2}-1\right)}{2}=\frac{(x-y)^{2}}{4}
$$
By using (2) and (3) we get
$$
\left|S_{A}\right| \geq 2 \cdot\binom{\frac{\left|R_{A}\right|+\left|L_{A}\right|}{2}}{2}=2 \cdot\binom{1010}{2}=1010 \cdot 1009
$$
and the equality holds when $\left|R_{A}\right|=\left|L_{A}\right|=1010$. Hence
$$
|S|=\frac{\sum_{A \in P}\left|S_{A}\right|}{2} \geq \frac{2021 \cdot 1010 \cdot 1009}{2}=2021 \cdot 505 \cdot 1009
$$
Therefore,
$$
|F|=\binom{2021}{3}-|S| \leq 2021 \cdot 1010 \cdot 673-2021 \cdot 505 \cdot 1009=2021 \cdot 505 \cdot 337
$$
For configuration of points on regular 2021-gon which is centered at $O$, in equalities in (4), (5), (6) become equalities. Hence $2021 \cdot 505 \cdot 337$ is indeed the answer.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "\nProblems 5.",
"resource_path": "EGMO/segmented/en-2021-solutions.jsonl",
"solution_match": "# Solution",
"tier": "T2",
"year": "2021"
}
|
Does there exist a nonnegative integer $a$ for which the equation
$$
\left\lfloor\frac{m}{1}\right\rfloor+\left\lfloor\frac{m}{2}\right\rfloor+\left\lfloor\frac{m}{3}\right\rfloor+\cdots+\left\lfloor\frac{m}{m}\right\rfloor=n^{2}+a
$$
has more than one million different solutions $(m, n)$ where $m$ and $n$ are positive integers? (The expression $\lfloor x\rfloor$ denotesthe integer part(or floor) of the real numberx. Thus $\lfloor\sqrt{2}\rfloor=1,\lfloor\pi\rfloor=\left\lfloor\frac{22}{7}\right\rfloor=$ $3,\lfloor 42\rfloor=42$ and $\lfloor 0\rfloor=0)$
Answer: Yes.
|
Denote the equation from the statement by (1). The left hand side of (1) depends only on $m$, and will throughout be denoted by $L(m)$. Fix an integer $q>10^{7}$ and note that for $m=q^{3}$
$$
L\left(q^{3}\right)=\sum_{k=1}^{q^{3}}\left[\frac{q^{3}}{k}\right] \leq \sum_{k=1}^{q^{3}} \frac{q^{3}}{k} \leq q^{3} \cdot \sum_{k=1}^{q^{3}} \frac{1}{k} \leq q^{3} \cdot q=q^{4}
$$
Indeed, the first inequality results from $[x] \leq x$. The second inequality can be seen (for instance) as follows. We divide the terms in the sum $\sum_{k=1}^{q^{3}} \frac{1}{k}$ into several groups: For $j \geq 0$, the $j$-th group contains the $2^{j}$ consecutive terms $\frac{1}{2^{j}}, \ldots, \frac{1}{2^{j+1}-1}$. Since every term in the $j$-th group is bounded by $\frac{1}{2^{j}}$, the overall contribution of the $j$-th group to the sum is at most 1 . Since the first $q$ groups together would contain $2^{q}-1>q^{3}$ terms, the number of groups does not exceed $q$, and hence the value of the sum under consideration is indeed bounded by $q$.
Call an integer $m$ special, if it satisfies $1 \leq L(m) \leq q^{4}$. Denote by $g(m) \geq 1$ the largest integer whose square is bounded by $L(m)$; in other words $g^{2}(m) \leq L(m)<(g(m)+1)^{2}$. Notethat $g(m) \leq$ $q^{2}$ for all special $m$, which implies
$$
0 \leq L(m)-g^{2}(m)<(g(m)+1)^{2}-g^{2}(m)=2 g(m)+1 \leq 2 q^{2}+1
$$
Finally, we do some counting. Inequality (2) and the monotonicity of $L(m)$ imply that there exist at least $q^{3}$ special integers. Because of (3), every special integer $m$ has $0 \leq L(m)-g^{2}(m) \leq 2 q^{2}+1$. By averaging, at least $\frac{q^{3}}{2 q^{2}+2}>10^{6}$ special integers must yield the same value $L(m)-g^{2}(m)$. This frequently occurring value is our choice for $\alpha$, which yields more than $10^{6}$ solutions $(m, g(m))$ to equation (1). Hence, the answer to the problem is YES.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Does there exist a nonnegative integer $a$ for which the equation
$$
\left\lfloor\frac{m}{1}\right\rfloor+\left\lfloor\frac{m}{2}\right\rfloor+\left\lfloor\frac{m}{3}\right\rfloor+\cdots+\left\lfloor\frac{m}{m}\right\rfloor=n^{2}+a
$$
has more than one million different solutions $(m, n)$ where $m$ and $n$ are positive integers? (The expression $\lfloor x\rfloor$ denotesthe integer part(or floor) of the real numberx. Thus $\lfloor\sqrt{2}\rfloor=1,\lfloor\pi\rfloor=\left\lfloor\frac{22}{7}\right\rfloor=$ $3,\lfloor 42\rfloor=42$ and $\lfloor 0\rfloor=0)$
Answer: Yes.
|
Denote the equation from the statement by (1). The left hand side of (1) depends only on $m$, and will throughout be denoted by $L(m)$. Fix an integer $q>10^{7}$ and note that for $m=q^{3}$
$$
L\left(q^{3}\right)=\sum_{k=1}^{q^{3}}\left[\frac{q^{3}}{k}\right] \leq \sum_{k=1}^{q^{3}} \frac{q^{3}}{k} \leq q^{3} \cdot \sum_{k=1}^{q^{3}} \frac{1}{k} \leq q^{3} \cdot q=q^{4}
$$
Indeed, the first inequality results from $[x] \leq x$. The second inequality can be seen (for instance) as follows. We divide the terms in the sum $\sum_{k=1}^{q^{3}} \frac{1}{k}$ into several groups: For $j \geq 0$, the $j$-th group contains the $2^{j}$ consecutive terms $\frac{1}{2^{j}}, \ldots, \frac{1}{2^{j+1}-1}$. Since every term in the $j$-th group is bounded by $\frac{1}{2^{j}}$, the overall contribution of the $j$-th group to the sum is at most 1 . Since the first $q$ groups together would contain $2^{q}-1>q^{3}$ terms, the number of groups does not exceed $q$, and hence the value of the sum under consideration is indeed bounded by $q$.
Call an integer $m$ special, if it satisfies $1 \leq L(m) \leq q^{4}$. Denote by $g(m) \geq 1$ the largest integer whose square is bounded by $L(m)$; in other words $g^{2}(m) \leq L(m)<(g(m)+1)^{2}$. Notethat $g(m) \leq$ $q^{2}$ for all special $m$, which implies
$$
0 \leq L(m)-g^{2}(m)<(g(m)+1)^{2}-g^{2}(m)=2 g(m)+1 \leq 2 q^{2}+1
$$
Finally, we do some counting. Inequality (2) and the monotonicity of $L(m)$ imply that there exist at least $q^{3}$ special integers. Because of (3), every special integer $m$ has $0 \leq L(m)-g^{2}(m) \leq 2 q^{2}+1$. By averaging, at least $\frac{q^{3}}{2 q^{2}+2}>10^{6}$ special integers must yield the same value $L(m)-g^{2}(m)$. This frequently occurring value is our choice for $\alpha$, which yields more than $10^{6}$ solutions $(m, g(m))$ to equation (1). Hence, the answer to the problem is YES.
|
{
"exam": "EGMO",
"problem_label": "6",
"problem_match": "\nProblems 6.",
"resource_path": "EGMO/segmented/en-2021-solutions.jsonl",
"solution_match": "# Solution.",
"tier": "T2",
"year": "2021"
}
|
Let $A B C$ be an acute-angled triangle in which $B C<A B$ and $B C<C A$. Let point $P$ lie on segment $A B$ and point $Q$ lie on segment $A C$ such that $P \neq B, Q \neq C$ and $B Q=B C=C P$. Let $T$ be the circumcentre of triangle $A P Q, H$ the orthocentre of triangle $A B C$, and $S$ the point of intersection of the lines $B Q$ and $C P$. Prove that $T, H$ and $S$ are collinear.
Proposed by: Netherlands
|
We show that $T$ and $H$ are both on the angle bisector $\ell$ of $\angle B S C$.
We first prove that $H \in \ell$. The altitude $C H$ in triangle $A B C$ is also the altitude in isosceles triangle $P B C$ with $C P=C B$. Therefore, $C H$ is also the angle bisector of $\angle P C B$ and hence also of $\angle S C B$. Analogously, $B H$ is the angle bisector of $\angle S B C$. We conclude that $H$, as the intersection of two angle bisectors of $\triangle B S C$, is also on the third angle bisector, which is $\ell$.
We now prove that $T \in \ell$.
Variant 1. In the isosceles triangles $\triangle B C P$ and $\triangle C B Q$ we see that $\angle B C P=180^{\circ}-2 \angle B$ and $\angle C B Q=180^{\circ}-2 \angle C$. This yields $\angle P S Q=\angle B S C=180^{\circ}-\left(180^{\circ}-2 \angle B\right)-\left(180^{\circ}-2 \angle C\right)=$ $180^{\circ}-2 \angle A$. Furthermore, $\angle P T Q=2 \angle P A Q=2 \angle A$ ( $T$ being circumcentre of $\left.\triangle A P Q\right)$. Now $\angle P T Q+\angle P S Q=180^{\circ}$, so $P T Q S$ is a cyclic quadrilateral. From $P T=T Q$ we then obtain that $\angle P S T=\angle P Q T=\angle Q P T=\angle Q S T$, so $T$ is on the angle bisector $\angle P S Q$, which is also $\ell$.
We conclude that $T, S$ and $H$ are collinear.
Variant 2. Let $R$ be the second intersection of $B Q$ and $\odot A P Q$.
$A P R Q$ is cyclic quadrilateral, so $\angle P R S=\angle A, \angle A P R=\angle B Q C=\angle C$. On the other hand $\angle B P C=\angle B$. Therefore $\angle R P S=180^{\circ}-\angle B-\angle C=\angle A$. Hence, the triangle $P R S$ is isosceles with $S P=S R$; then $\ell$ is the perpendicular bisector of the chord $P R$ in the circle that passes through $T$.
Note that if $B Q$ is tangent to $\odot A P Q$, then $C P$ is also tangent to $\odot A P Q$ and triangle $A B C$ is isosceles, so $T, S$ and $H$ lie on the altitude from $A$.
Remark. The fact that $C P$ is also tangent to $\odot A P Q$ could be shown with $P T Q S$ beeing a cyclic quadrilateral like in the first variant of the solution. Otherwise we can consider $R$ the second intersection of $C P$ and $\odot A P Q$ and prove that triangle $Q R S$ is isosceles.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute-angled triangle in which $B C<A B$ and $B C<C A$. Let point $P$ lie on segment $A B$ and point $Q$ lie on segment $A C$ such that $P \neq B, Q \neq C$ and $B Q=B C=C P$. Let $T$ be the circumcentre of triangle $A P Q, H$ the orthocentre of triangle $A B C$, and $S$ the point of intersection of the lines $B Q$ and $C P$. Prove that $T, H$ and $S$ are collinear.
Proposed by: Netherlands
|
We show that $T$ and $H$ are both on the angle bisector $\ell$ of $\angle B S C$.
We first prove that $H \in \ell$. The altitude $C H$ in triangle $A B C$ is also the altitude in isosceles triangle $P B C$ with $C P=C B$. Therefore, $C H$ is also the angle bisector of $\angle P C B$ and hence also of $\angle S C B$. Analogously, $B H$ is the angle bisector of $\angle S B C$. We conclude that $H$, as the intersection of two angle bisectors of $\triangle B S C$, is also on the third angle bisector, which is $\ell$.
We now prove that $T \in \ell$.
Variant 1. In the isosceles triangles $\triangle B C P$ and $\triangle C B Q$ we see that $\angle B C P=180^{\circ}-2 \angle B$ and $\angle C B Q=180^{\circ}-2 \angle C$. This yields $\angle P S Q=\angle B S C=180^{\circ}-\left(180^{\circ}-2 \angle B\right)-\left(180^{\circ}-2 \angle C\right)=$ $180^{\circ}-2 \angle A$. Furthermore, $\angle P T Q=2 \angle P A Q=2 \angle A$ ( $T$ being circumcentre of $\left.\triangle A P Q\right)$. Now $\angle P T Q+\angle P S Q=180^{\circ}$, so $P T Q S$ is a cyclic quadrilateral. From $P T=T Q$ we then obtain that $\angle P S T=\angle P Q T=\angle Q P T=\angle Q S T$, so $T$ is on the angle bisector $\angle P S Q$, which is also $\ell$.
We conclude that $T, S$ and $H$ are collinear.
Variant 2. Let $R$ be the second intersection of $B Q$ and $\odot A P Q$.
$A P R Q$ is cyclic quadrilateral, so $\angle P R S=\angle A, \angle A P R=\angle B Q C=\angle C$. On the other hand $\angle B P C=\angle B$. Therefore $\angle R P S=180^{\circ}-\angle B-\angle C=\angle A$. Hence, the triangle $P R S$ is isosceles with $S P=S R$; then $\ell$ is the perpendicular bisector of the chord $P R$ in the circle that passes through $T$.
Note that if $B Q$ is tangent to $\odot A P Q$, then $C P$ is also tangent to $\odot A P Q$ and triangle $A B C$ is isosceles, so $T, S$ and $H$ lie on the altitude from $A$.
Remark. The fact that $C P$ is also tangent to $\odot A P Q$ could be shown with $P T Q S$ beeing a cyclic quadrilateral like in the first variant of the solution. Otherwise we can consider $R$ the second intersection of $C P$ and $\odot A P Q$ and prove that triangle $Q R S$ is isosceles.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "# Problem 1.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "# Solution 1.",
"tier": "T2",
"year": "2022"
}
|
Let $A B C$ be an acute-angled triangle in which $B C<A B$ and $B C<C A$. Let point $P$ lie on segment $A B$ and point $Q$ lie on segment $A C$ such that $P \neq B, Q \neq C$ and $B Q=B C=C P$. Let $T$ be the circumcentre of triangle $A P Q, H$ the orthocentre of triangle $A B C$, and $S$ the point of intersection of the lines $B Q$ and $C P$. Prove that $T, H$ and $S$ are collinear.
Proposed by: Netherlands
|
In the same way as in the previous solution, we see that $\angle P S Q=180^{\circ}-2 \angle A$, so $\angle C S Q=2 \angle A$. From the cyclic quadrilateral $A E H D$ (with $E$ and $D$ feet of the altitudes $C H$ and $B H$ ) we see that $\angle D H C=\angle D A E=\angle A$. Since $B H$ is the perpendicular bisector of $C Q$, we have $\angle D H Q=\angle A$ as well, so $\angle C H Q=2 \angle A$. From $\angle C H Q=2 \angle A=\angle C S Q$, we see $C H S Q$ is a cyclic quadrilateral. This means $\angle Q H S=\angle Q C S$.
Since triangles $P T Q$ and $C H Q$ are both isosceles with apex $2 \angle A$, we get $\triangle P T Q \sim \triangle C H Q$. We see that one can be obtained from the other by a spiral similarity centered at $Q$, so we also obtain $\triangle Q T H \sim \triangle Q P C$. This means that $\angle Q H T=\angle Q C P$. Combining this with $\angle Q H S=\angle Q C S$, we see that $\angle Q H T=\angle Q C P=\angle Q C S=\angle Q H S$. So $\angle Q H T=\angle Q H S$, which means that $T, S$ and $H$ are collinear.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute-angled triangle in which $B C<A B$ and $B C<C A$. Let point $P$ lie on segment $A B$ and point $Q$ lie on segment $A C$ such that $P \neq B, Q \neq C$ and $B Q=B C=C P$. Let $T$ be the circumcentre of triangle $A P Q, H$ the orthocentre of triangle $A B C$, and $S$ the point of intersection of the lines $B Q$ and $C P$. Prove that $T, H$ and $S$ are collinear.
Proposed by: Netherlands
|
In the same way as in the previous solution, we see that $\angle P S Q=180^{\circ}-2 \angle A$, so $\angle C S Q=2 \angle A$. From the cyclic quadrilateral $A E H D$ (with $E$ and $D$ feet of the altitudes $C H$ and $B H$ ) we see that $\angle D H C=\angle D A E=\angle A$. Since $B H$ is the perpendicular bisector of $C Q$, we have $\angle D H Q=\angle A$ as well, so $\angle C H Q=2 \angle A$. From $\angle C H Q=2 \angle A=\angle C S Q$, we see $C H S Q$ is a cyclic quadrilateral. This means $\angle Q H S=\angle Q C S$.
Since triangles $P T Q$ and $C H Q$ are both isosceles with apex $2 \angle A$, we get $\triangle P T Q \sim \triangle C H Q$. We see that one can be obtained from the other by a spiral similarity centered at $Q$, so we also obtain $\triangle Q T H \sim \triangle Q P C$. This means that $\angle Q H T=\angle Q C P$. Combining this with $\angle Q H S=\angle Q C S$, we see that $\angle Q H T=\angle Q C P=\angle Q C S=\angle Q H S$. So $\angle Q H T=\angle Q H S$, which means that $T, S$ and $H$ are collinear.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "# Problem 1.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "# Solution 2.",
"tier": "T2",
"year": "2022"
}
|
Let $A B C$ be an acute-angled triangle in which $B C<A B$ and $B C<C A$. Let point $P$ lie on segment $A B$ and point $Q$ lie on segment $A C$ such that $P \neq B, Q \neq C$ and $B Q=B C=C P$. Let $T$ be the circumcentre of triangle $A P Q, H$ the orthocentre of triangle $A B C$, and $S$ the point of intersection of the lines $B Q$ and $C P$. Prove that $T, H$ and $S$ are collinear.
Proposed by: Netherlands
|
Let us draw a parallel $f$ to $B C$ through $A$. Let $B^{\prime}=B Q \cap f, C^{\prime}=C P \cap f$. Then $A Q B^{\prime} \sim C Q B$ and $A P C^{\prime} \sim B P C$, therefore both $Q A B^{\prime}$ and $A P C^{\prime}$ will be isosceles.
Also, $B C S \sim B^{\prime} C^{\prime} S$ with respect to the similarity with center $S$, therefore if we take the image of the line $B H$ (which is the perpendicular bisector of the segment $C Q$ ) through this transformation, it will go through the point $B^{\prime}$, and be perpendicular to $A Q$ (as the image of $C Q$ is parallel to $C Q$ ). As $A Q B^{\prime}$ is isosceles, this line is the perpendicular bisector of the segment $A Q$. This means it goes through $T$, the circumcenter of $A P Q$. Similarly on the other side the image of $C H$ also goes through $T$. This means that the image of $H$ with respect to the similarity though $S$ will be $T$, so $T, S, H$ are collinear.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute-angled triangle in which $B C<A B$ and $B C<C A$. Let point $P$ lie on segment $A B$ and point $Q$ lie on segment $A C$ such that $P \neq B, Q \neq C$ and $B Q=B C=C P$. Let $T$ be the circumcentre of triangle $A P Q, H$ the orthocentre of triangle $A B C$, and $S$ the point of intersection of the lines $B Q$ and $C P$. Prove that $T, H$ and $S$ are collinear.
Proposed by: Netherlands
|
Let us draw a parallel $f$ to $B C$ through $A$. Let $B^{\prime}=B Q \cap f, C^{\prime}=C P \cap f$. Then $A Q B^{\prime} \sim C Q B$ and $A P C^{\prime} \sim B P C$, therefore both $Q A B^{\prime}$ and $A P C^{\prime}$ will be isosceles.
Also, $B C S \sim B^{\prime} C^{\prime} S$ with respect to the similarity with center $S$, therefore if we take the image of the line $B H$ (which is the perpendicular bisector of the segment $C Q$ ) through this transformation, it will go through the point $B^{\prime}$, and be perpendicular to $A Q$ (as the image of $C Q$ is parallel to $C Q$ ). As $A Q B^{\prime}$ is isosceles, this line is the perpendicular bisector of the segment $A Q$. This means it goes through $T$, the circumcenter of $A P Q$. Similarly on the other side the image of $C H$ also goes through $T$. This means that the image of $H$ with respect to the similarity though $S$ will be $T$, so $T, S, H$ are collinear.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "# Problem 1.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "# Solution 3.",
"tier": "T2",
"year": "2022"
}
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
First, all such functions $f$ satisfy the conditions, as $v_{p}(a) \neq v_{p}(b)$ implies $v_{p}(a+b)=$ $\min \left(v_{p}(a), v_{p}(b)\right)$. Plugging $a, b=1$ into (1) gives $f(1)=1$. Also, a simple induction gives that $f\left(\prod_{i=1}^{k} p_{i}^{a_{i}}\right)=\prod_{i=1}^{k} f\left(p_{i}\right)^{a_{i}}$. Let $S$ be the set of all primes $p$ such that $f(p) \neq 1$. If $|S|=1$, then $S=\{p\}$ and $f(p)=a$ for some $a \neq 1$, and thus $f(n)=a^{v_{p}(n)}$ for all $n$. If $S=\emptyset$ then $f(n)=1$ for all $n$, which can be also written in the above form for $a=1$.
Now suppose that $S$ contains at least two primes. Let $p<q$ be the two smallest elements of $S$. Since all prime divisors of $q-p$ are smaller than $q$ and not equal to $p, f(q-p)=1$. Using (2) for $p$ and $q-p$ gives us that some two of $f(p) \neq 1, f(p-q)=1$ and $f(q) \neq 1$ are equal, meaning that $f(p)=f(q)$. Let $t \geq 2$ be the smallest integer such that $p^{t}>q$, and let $p^{t}=a q+b$ with $0 \leq b<q$. Since $q \geq p^{t-1}$ and $q$ is not divisible by $p$, we have $q>p^{t-1}$, so $a<p$, and thus $f(a)=1$. Therefore $a q$ is not divisible by $p$, thus neither is $b$, and since $b<q, f(b)=1$. Now using (2) for $a q$ and $b$ we get that some two of $f(a q)=f(a) f(q)=f(q)=f(p), f(b)=1$ and $f(a q+b)=f\left(p^{t}\right)=f(p)^{t}$ are equal, which is a contradiction.
|
f(n)=a^{v_{p}(n)}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
First, all such functions $f$ satisfy the conditions, as $v_{p}(a) \neq v_{p}(b)$ implies $v_{p}(a+b)=$ $\min \left(v_{p}(a), v_{p}(b)\right)$. Plugging $a, b=1$ into (1) gives $f(1)=1$. Also, a simple induction gives that $f\left(\prod_{i=1}^{k} p_{i}^{a_{i}}\right)=\prod_{i=1}^{k} f\left(p_{i}\right)^{a_{i}}$. Let $S$ be the set of all primes $p$ such that $f(p) \neq 1$. If $|S|=1$, then $S=\{p\}$ and $f(p)=a$ for some $a \neq 1$, and thus $f(n)=a^{v_{p}(n)}$ for all $n$. If $S=\emptyset$ then $f(n)=1$ for all $n$, which can be also written in the above form for $a=1$.
Now suppose that $S$ contains at least two primes. Let $p<q$ be the two smallest elements of $S$. Since all prime divisors of $q-p$ are smaller than $q$ and not equal to $p, f(q-p)=1$. Using (2) for $p$ and $q-p$ gives us that some two of $f(p) \neq 1, f(p-q)=1$ and $f(q) \neq 1$ are equal, meaning that $f(p)=f(q)$. Let $t \geq 2$ be the smallest integer such that $p^{t}>q$, and let $p^{t}=a q+b$ with $0 \leq b<q$. Since $q \geq p^{t-1}$ and $q$ is not divisible by $p$, we have $q>p^{t-1}$, so $a<p$, and thus $f(a)=1$. Therefore $a q$ is not divisible by $p$, thus neither is $b$, and since $b<q, f(b)=1$. Now using (2) for $a q$ and $b$ we get that some two of $f(a q)=f(a) f(q)=f(q)=f(p), f(b)=1$ and $f(a q+b)=f\left(p^{t}\right)=f(p)^{t}$ are equal, which is a contradiction.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "# Problem 2.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "\nSolution 1. ",
"tier": "T2",
"year": "2022"
}
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
1B (for the main part).
Suppose that there are two primes, $p<q$ such that $f(p)>1$ and $f(q)>1$, and these are the two smallest such primes.
$1<p-q<q$ and it is not divisible by $p$ and $q$, therefore its prime factors are smaller than $q$, so we conclude $f(q-p)=1$. By (2), two of $f(q-p)=1, f(q-p)>1$ and $f(q)>1$ are equal, so $f(q)=f(p)=a$.
If $p^{2}<q$ then $q-p^{2}$ is not divisible by $p$ and $q$, so we conclude $f\left(q-p^{2}\right)=1$ in the same way. Then $f\left(p^{2}\right)=f(p)^{2}=a^{2}, f\left(q-p^{2}\right)=1$ and $f(q)=a$ are distinct, contradicting (2). Hence, $q<p^{2}$.
Let $c=\left\lfloor p^{2} / q\right\rfloor$; then $0<c<p$, so $f(c)=1$. By $0<p^{2}-c q<q$, we get $f\left(p^{2}-c q\right)=1$. Now $f\left(p^{2}-c q\right)=1, f\left(p^{2}\right)=a^{2}$ and $f(c q)=a$ are distinct, contradicting (2) again.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
1B (for the main part).
Suppose that there are two primes, $p<q$ such that $f(p)>1$ and $f(q)>1$, and these are the two smallest such primes.
$1<p-q<q$ and it is not divisible by $p$ and $q$, therefore its prime factors are smaller than $q$, so we conclude $f(q-p)=1$. By (2), two of $f(q-p)=1, f(q-p)>1$ and $f(q)>1$ are equal, so $f(q)=f(p)=a$.
If $p^{2}<q$ then $q-p^{2}$ is not divisible by $p$ and $q$, so we conclude $f\left(q-p^{2}\right)=1$ in the same way. Then $f\left(p^{2}\right)=f(p)^{2}=a^{2}, f\left(q-p^{2}\right)=1$ and $f(q)=a$ are distinct, contradicting (2). Hence, $q<p^{2}$.
Let $c=\left\lfloor p^{2} / q\right\rfloor$; then $0<c<p$, so $f(c)=1$. By $0<p^{2}-c q<q$, we get $f\left(p^{2}-c q\right)=1$. Now $f\left(p^{2}-c q\right)=1, f\left(p^{2}\right)=a^{2}$ and $f(c q)=a$ are distinct, contradicting (2) again.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "# Problem 2.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "# Solution ",
"tier": "T2",
"year": "2022"
}
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
Suppose there exists a positive integer $n$ such that $f(n) \neq 1$ and $f(n+1) \neq 1$. We know that $f(1)=1$, so by $(2)$ two of $f(n) \neq 1, f(n+1) \neq 1$ and $f(1)=1$ are equal, therefore $f(n)=f(n+1)$.
$f\left(n^{2}\right)=f(n)^{2} \neq 1$ and $f\left(n^{2}-1\right)=f(n+1) f(n-1)=f(n) f(n-1) \neq 1$, but by (2) two of $f(1)=1, f\left(n^{2}-1\right) \neq 1$ and $f\left(n^{2}\right) \neq 1$ are equal, so $f\left(n^{2}-1\right)=f\left(n^{2}\right)$ and thus $f(n) f(n-1)=f(n)^{2}$, therefore $f(n-1)=f(n)$.
Then for $n-1$, it is also true that $f(n-1) \neq 1$ and $f(n-1+1) \neq 1$, so the same proof applies, $f(n-1-1)=f(n-1)$ and so on. By induction we get that $f(n)=f(1)$, which is a contradiction, as $f(1)=1$ and we assumed that $f(n) \neq 1$. Therefore, there can't be any $n$ for which $f(n) \neq 1$ and $f(n+1) \neq 1$.
Now suppose there are two primes, $p \neq q \in S$ such that $f(p) \neq 1$ and $f(q) \neq 1$. Then by the Chinese remainder theorem, there exists a number $n$ such that $p \mid n$ and $q \mid n+1$, so $f(p) \mid f(n)$ and $f(q) \mid f(n+1)$ thus $f(n) \neq 1$ and $f(n+1) \neq 1$. This is a contradiction, so $S$ can't have two distinct elements.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
Suppose there exists a positive integer $n$ such that $f(n) \neq 1$ and $f(n+1) \neq 1$. We know that $f(1)=1$, so by $(2)$ two of $f(n) \neq 1, f(n+1) \neq 1$ and $f(1)=1$ are equal, therefore $f(n)=f(n+1)$.
$f\left(n^{2}\right)=f(n)^{2} \neq 1$ and $f\left(n^{2}-1\right)=f(n+1) f(n-1)=f(n) f(n-1) \neq 1$, but by (2) two of $f(1)=1, f\left(n^{2}-1\right) \neq 1$ and $f\left(n^{2}\right) \neq 1$ are equal, so $f\left(n^{2}-1\right)=f\left(n^{2}\right)$ and thus $f(n) f(n-1)=f(n)^{2}$, therefore $f(n-1)=f(n)$.
Then for $n-1$, it is also true that $f(n-1) \neq 1$ and $f(n-1+1) \neq 1$, so the same proof applies, $f(n-1-1)=f(n-1)$ and so on. By induction we get that $f(n)=f(1)$, which is a contradiction, as $f(1)=1$ and we assumed that $f(n) \neq 1$. Therefore, there can't be any $n$ for which $f(n) \neq 1$ and $f(n+1) \neq 1$.
Now suppose there are two primes, $p \neq q \in S$ such that $f(p) \neq 1$ and $f(q) \neq 1$. Then by the Chinese remainder theorem, there exists a number $n$ such that $p \mid n$ and $q \mid n+1$, so $f(p) \mid f(n)$ and $f(q) \mid f(n+1)$ thus $f(n) \neq 1$ and $f(n+1) \neq 1$. This is a contradiction, so $S$ can't have two distinct elements.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "# Problem 2.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "# Solution 2 (for the main part).",
"tier": "T2",
"year": "2022"
}
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
(for the second part).
Claim 3.1. If $f(m)>1$ for some $m \in \mathbb{N}$, then there are less than $m$ different prime numbers $p_{i}$ with $f\left(p_{i}\right)>1$.
Proof. On the one hand, if $f(m)=c>1$, then for any $n \in \mathbb{N}$, there is some $k \in\{n+1, n+2 \ldots, n+m\}$ such that $f(k) \in\{1, c\}$. To show this, it is enough to see that condition (2) implies that for any $n \in \mathbb{N}$
- if $f(n)=c$, then $f(n+1) \in\{1, c\}$;
- if $f(n)=1$, then $f(n+m) \in\{1, c\}$.
On the other hand, if we suppose that there are $m$ different prime numbers $p_{1}, p_{2}, \ldots, p_{m}$, then we know (by the Chinese Remainder Theorem) that for any number $t \in \mathbb{N}$, there exists a number $n_{t} \in \mathbb{N}$ such that $p_{k}^{t} \mid n_{t}+k$ for each $k \in\{1,2, \ldots, m\}$. Hence for each $k$, we have
$$
f\left(n_{t}+k\right) \geq f\left(p_{i}^{t}\right)=f\left(p_{i}\right)^{t} \geq 2^{t} .
$$
If $2^{t}>c$, then it means that $f(k)>c$ for all $k \in\left\{n_{t}+1, n_{t}+2 \ldots, n_{t}+m\right\}$.
Now we can suppose that for some $m \geq 2$, the set of primes with $f(p)>1$ is $S=\left\{p_{1}, p_{2}, \ldots, p_{m}\right\}$. Now let $S_{1}$ and $S_{2}$ be two disjoint nonempty subsets such that $S_{1} \cup S_{2}=S$. Let $a=\prod_{p_{i} \in S_{1}} p_{i}$ and $b=\prod_{p_{j} \in S_{2}} p_{j}$.
Now for any $k, \ell \in \mathbb{N}, a^{k}+b^{\ell}$ is coprime to all primes in $S$, hence $f\left(a^{k}+b^{\ell}\right)=1$. It is easy to see that we can choose $k$ and $\ell$ such that $f\left(a^{k}\right) \neq f\left(b^{\ell}\right)$, which contradicts condition (2).
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
(for the second part).
Claim 3.1. If $f(m)>1$ for some $m \in \mathbb{N}$, then there are less than $m$ different prime numbers $p_{i}$ with $f\left(p_{i}\right)>1$.
Proof. On the one hand, if $f(m)=c>1$, then for any $n \in \mathbb{N}$, there is some $k \in\{n+1, n+2 \ldots, n+m\}$ such that $f(k) \in\{1, c\}$. To show this, it is enough to see that condition (2) implies that for any $n \in \mathbb{N}$
- if $f(n)=c$, then $f(n+1) \in\{1, c\}$;
- if $f(n)=1$, then $f(n+m) \in\{1, c\}$.
On the other hand, if we suppose that there are $m$ different prime numbers $p_{1}, p_{2}, \ldots, p_{m}$, then we know (by the Chinese Remainder Theorem) that for any number $t \in \mathbb{N}$, there exists a number $n_{t} \in \mathbb{N}$ such that $p_{k}^{t} \mid n_{t}+k$ for each $k \in\{1,2, \ldots, m\}$. Hence for each $k$, we have
$$
f\left(n_{t}+k\right) \geq f\left(p_{i}^{t}\right)=f\left(p_{i}\right)^{t} \geq 2^{t} .
$$
If $2^{t}>c$, then it means that $f(k)>c$ for all $k \in\left\{n_{t}+1, n_{t}+2 \ldots, n_{t}+m\right\}$.
Now we can suppose that for some $m \geq 2$, the set of primes with $f(p)>1$ is $S=\left\{p_{1}, p_{2}, \ldots, p_{m}\right\}$. Now let $S_{1}$ and $S_{2}$ be two disjoint nonempty subsets such that $S_{1} \cup S_{2}=S$. Let $a=\prod_{p_{i} \in S_{1}} p_{i}$ and $b=\prod_{p_{j} \in S_{2}} p_{j}$.
Now for any $k, \ell \in \mathbb{N}, a^{k}+b^{\ell}$ is coprime to all primes in $S$, hence $f\left(a^{k}+b^{\ell}\right)=1$. It is easy to see that we can choose $k$ and $\ell$ such that $f\left(a^{k}\right) \neq f\left(b^{\ell}\right)$, which contradicts condition (2).
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "# Problem 2.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "# Solution 3 (Joseph Myers) ",
"tier": "T2",
"year": "2022"
}
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
Let $p$ the minimal (prime) number with $f(p)=c>1$.
Claim 4.1. For any $k \in \mathbb{N}, f\left(1+p+p^{2}+\ldots+p^{k}\right) \in\left\{1, c, c^{2}, \ldots, c^{k}\right\}$.
Proof of the claim. Use condition (2) recursively.
Now suppose that there exists some prime $q>p$ with $f(q)>1$, and consider the following three values of $f$ :
- $f(1)=1$.
- $f\left(p^{q-1}-1\right)=f(p-1) f\left(p^{q-2}+p^{q-3}+\ldots+p+1\right) \in\left\{1, c, c^{2}, \ldots, c^{q-2}\right\}$ using our claim, (note that we know $f(p-1)=1$ by the minimality of $p$ ).
But we also know (by Fermat's little theorem) that $q \mid p^{q-1}-1$, hence $f(q) \mid f\left(p^{q-1}-1\right)$. Therefore we have that: $f\left(p^{q-1}-1\right) \in\left\{c, c^{2}, \ldots, c^{q-2}\right\}$.
- $f\left(p^{q-1}\right)=c^{q-1}$.
These are three different values, hence we have a contradiction by condition (2).
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
Let $p$ the minimal (prime) number with $f(p)=c>1$.
Claim 4.1. For any $k \in \mathbb{N}, f\left(1+p+p^{2}+\ldots+p^{k}\right) \in\left\{1, c, c^{2}, \ldots, c^{k}\right\}$.
Proof of the claim. Use condition (2) recursively.
Now suppose that there exists some prime $q>p$ with $f(q)>1$, and consider the following three values of $f$ :
- $f(1)=1$.
- $f\left(p^{q-1}-1\right)=f(p-1) f\left(p^{q-2}+p^{q-3}+\ldots+p+1\right) \in\left\{1, c, c^{2}, \ldots, c^{q-2}\right\}$ using our claim, (note that we know $f(p-1)=1$ by the minimality of $p$ ).
But we also know (by Fermat's little theorem) that $q \mid p^{q-1}-1$, hence $f(q) \mid f\left(p^{q-1}-1\right)$. Therefore we have that: $f\left(p^{q-1}-1\right) \in\left\{c, c^{2}, \ldots, c^{q-2}\right\}$.
- $f\left(p^{q-1}\right)=c^{q-1}$.
These are three different values, hence we have a contradiction by condition (2).
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "# Problem 2.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "\nSolution 4 (for the main part).",
"tier": "T2",
"year": "2022"
}
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
4B (for the main part). Let $p$ the minimal (prime) number with $f(p)=c>1$.
Lemma 4B.1. For any $k \in \mathbb{N}, f\left(1+p+p^{2}+\ldots+p^{k}\right)=1$.
Proof of the Lemma, by induction. $k=0$ is trivial.
$k=1$ is also easy, as $p+1$ cannot be a prime, except in the case of $p=2$, which is also easy. (If $f(2)=c>1$, then $f(3) \in\{1, c\}$ by $3=1+2$ and $f(3) \in\left\{1, c^{2}\right\}$ by $2^{2}=1+3$.)
From now we suppose $k \geq 2$ and $f\left(1+p+\ldots+p^{k-1}\right)=1$. By condition (2), we have that:
- As $f\left(1+p+\ldots+p^{k-1}\right)=1$ and $f\left(p^{k}\right)=c^{k}$, we have that $f\left(1+p+\ldots+p^{k-1}\right) \in\left\{1, c^{k}\right\}$
- As $f(1)=1$ and $f\left(p+\ldots+p^{k}\right)=f(p) f\left(1+p+\ldots+p^{k-1}\right)=c$, we also have that $f\left(1+p+\ldots+p^{k-1}\right) \in\{1, c\}$.
If $k>1, c \neq c^{k}$, therefore $f\left(1+p+\ldots+p^{k}\right)=1$.
Now suppose that there exists some prime $q>p$ with $f(q)>1$. Then by condition (1),
$$
f\left(p^{q-1}-1\right)=f(p-1) f\left(1+p+\ldots+p^{q-2}\right)
$$
here $f(p-1)=1$ (by the minimality of $p$ ) and $f\left(1+p+\ldots+p^{q-2}\right)=1$ by the lemma. But we also know (by Fermat's little theorem) that $q \mid p^{q-1}-1$, hence $1<f(q) \leq f\left(p^{q-1}-1\right.$ ), which is a contradiction.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
4B (for the main part). Let $p$ the minimal (prime) number with $f(p)=c>1$.
Lemma 4B.1. For any $k \in \mathbb{N}, f\left(1+p+p^{2}+\ldots+p^{k}\right)=1$.
Proof of the Lemma, by induction. $k=0$ is trivial.
$k=1$ is also easy, as $p+1$ cannot be a prime, except in the case of $p=2$, which is also easy. (If $f(2)=c>1$, then $f(3) \in\{1, c\}$ by $3=1+2$ and $f(3) \in\left\{1, c^{2}\right\}$ by $2^{2}=1+3$.)
From now we suppose $k \geq 2$ and $f\left(1+p+\ldots+p^{k-1}\right)=1$. By condition (2), we have that:
- As $f\left(1+p+\ldots+p^{k-1}\right)=1$ and $f\left(p^{k}\right)=c^{k}$, we have that $f\left(1+p+\ldots+p^{k-1}\right) \in\left\{1, c^{k}\right\}$
- As $f(1)=1$ and $f\left(p+\ldots+p^{k}\right)=f(p) f\left(1+p+\ldots+p^{k-1}\right)=c$, we also have that $f\left(1+p+\ldots+p^{k-1}\right) \in\{1, c\}$.
If $k>1, c \neq c^{k}$, therefore $f\left(1+p+\ldots+p^{k}\right)=1$.
Now suppose that there exists some prime $q>p$ with $f(q)>1$. Then by condition (1),
$$
f\left(p^{q-1}-1\right)=f(p-1) f\left(1+p+\ldots+p^{q-2}\right)
$$
here $f(p-1)=1$ (by the minimality of $p$ ) and $f\left(1+p+\ldots+p^{q-2}\right)=1$ by the lemma. But we also know (by Fermat's little theorem) that $q \mid p^{q-1}-1$, hence $1<f(q) \leq f\left(p^{q-1}-1\right.$ ), which is a contradiction.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "# Problem 2.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "\nSolution ",
"tier": "T2",
"year": "2022"
}
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
Lemma 5.1. If $f(m) \neq f(n)$, then $f(n+m)=f(n-m)$ (for all $n>m \in \mathbb{N}$ ).
Proof of the Lemma. By condition (2), $\{f(n+m), f(n-m)\} \subseteq\{f(n), f(m)\}$. We also have that:
$$
f(n-m) f(n+m)=f\left(n^{2}-m^{2}\right) \in\left\{f(n)^{2}, f(m)^{2}\right\}
$$
which is only possible if $f(n-m)=f(n+m)=f(n)$ or $f(n-m)=f(n+m)=f(m)$.
Now let $p$ be the smallest (prime) number with $f(p)>1$.
Claim 5.2. If $p$ does not divide $n$, then $f(n)=1$.
Proof of the Claim. Let $n$ be the minimal counterexample, and and let $n=d p+r$ with $0<r<p$. Here $d \geq 1$ by the minimality of $p$.
Now $1<f(d) f(p)=f(d p) \neq f(r)=1$, hence by our lemma, $f(d p+r)=f(d p-r)=1$, contradiction.
|
f(n)=a^{v_{p}(n)}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(a b)=f(a) f(b)$, and
(2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.
Proposed by: Fedir Yudin, Ukraine
Answer: $f(n)=a^{v_{p}(n)}$, where $a \in \mathbb{N}, p$ is a prime and $v_{p}(n)$ is the largest exponent of $p$ dividing $n$.
|
Lemma 5.1. If $f(m) \neq f(n)$, then $f(n+m)=f(n-m)$ (for all $n>m \in \mathbb{N}$ ).
Proof of the Lemma. By condition (2), $\{f(n+m), f(n-m)\} \subseteq\{f(n), f(m)\}$. We also have that:
$$
f(n-m) f(n+m)=f\left(n^{2}-m^{2}\right) \in\left\{f(n)^{2}, f(m)^{2}\right\}
$$
which is only possible if $f(n-m)=f(n+m)=f(n)$ or $f(n-m)=f(n+m)=f(m)$.
Now let $p$ be the smallest (prime) number with $f(p)>1$.
Claim 5.2. If $p$ does not divide $n$, then $f(n)=1$.
Proof of the Claim. Let $n$ be the minimal counterexample, and and let $n=d p+r$ with $0<r<p$. Here $d \geq 1$ by the minimality of $p$.
Now $1<f(d) f(p)=f(d p) \neq f(r)=1$, hence by our lemma, $f(d p+r)=f(d p-r)=1$, contradiction.
|
{
"exam": "EGMO",
"problem_label": "2",
"problem_match": "# Problem 2.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "# Solution 5 (for the main part).",
"tier": "T2",
"year": "2022"
}
|
An infinite sequence of positive integers $a_{1}, a_{2}, \ldots$ is called good if
(1) $a_{1}$ is a perfect square, and
(2) for any integer $n \geq 2, a_{n}$ is the smallest positive integer such that
$$
n a_{1}+(n-1) a_{2}+\ldots+2 a_{n-1}+a_{n}
$$
is a perfect square.
Prove that for any good sequence $a_{1}, a_{2}, \ldots$, there exists a positive integer $k$ such that $a_{n}=a_{k}$ for all integers $n \geq k$.
Proposed by: Joe Benton and Dominic Yeo, United Kingdom
|
Define the following auxiliary sequences:
$$
\begin{array}{ll}
b_{1}=a_{1}, & b_{n}=a_{1}+a_{2}+\cdots+a_{n} \\
c_{1}=b_{1}, & c_{n}=b_{1}+b_{2}+\cdots+b_{n}
\end{array}
$$
Observe that
$$
\begin{gathered}
c_{n}-c_{n-1}=b_{n} \\
b_{n}-b_{n-1}=\left(c_{n}-c_{n-1}\right)-\left(c_{n-1}-c_{n-2}\right)=a_{n}
\end{gathered}
$$
Therefore,
$$
c_{n}=\left(2 c_{n-1}-c_{n-2}\right)+a_{n}
$$
is the smallest square greater than $2 c_{n-1}-c_{n-2}$.
Claim. $\sqrt{c_{n}}-\sqrt{c_{n-1}} \leq \sqrt{c_{n-1}}-\sqrt{c_{n-2}}$
Proof 1. Let $c_{n}=m_{n}^{2}, c_{n-1}=m_{n-1}^{2}, c_{n-2}=m_{n-2}^{2}$, where the value of $c_{n}$ is momentarily permitted to exceed its minimum. Any value is permitted for which
$$
b_{n}=c_{n}-c_{n-1}>b_{n-1}=c_{n-1}-c_{n-2}
$$
Factoring,
$$
\left(m_{n}-m_{n-1}\right)\left(m_{n}+m_{n-1}\right)>\left(m_{n-1}-m_{n-2}\right)\left(m_{n-1}+m_{n-2}\right)
$$
In the case that $m_{n}-m_{n-1}=m_{n-1}-m_{n-2}$, this inequality clearly holds, as $m_{n}+m_{n-1}>m_{n-1}+m_{n-2}$. Thus, the minimal possible value of $m_{n}$ satisfies the claimed inequality.
Proof 2. Denote $c_{n-1}=x^{2}, c_{n-2}=(x-d)^{2}$. Then
$$
2 c_{n-1}-c_{n-2}=2 x^{2}-(x-d)^{2}=x^{2}+2 d x-d^{2}=(x+d)^{2}-2 d^{2}<(x+d)^{2}
$$
It follows that $c_{n} \leq(x+d)^{2}$.
And so the sequence of positive integers $\sqrt{c_{n}}-\sqrt{c_{n-1}}$ is decreasing. Any such sequence is eventually constant.
As a corollary, $c_{n}=(x+n d)^{2}$ for sufficiently large $n$, with fixed integers $x$ and $d$. Along the lines above, it becomes clear that $a_{n}=c_{n}-2 c_{n-1}+c_{n-2}=2 d^{2}$, so the sequence $\left(a_{n}\right)$ is constant.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
An infinite sequence of positive integers $a_{1}, a_{2}, \ldots$ is called good if
(1) $a_{1}$ is a perfect square, and
(2) for any integer $n \geq 2, a_{n}$ is the smallest positive integer such that
$$
n a_{1}+(n-1) a_{2}+\ldots+2 a_{n-1}+a_{n}
$$
is a perfect square.
Prove that for any good sequence $a_{1}, a_{2}, \ldots$, there exists a positive integer $k$ such that $a_{n}=a_{k}$ for all integers $n \geq k$.
Proposed by: Joe Benton and Dominic Yeo, United Kingdom
|
Define the following auxiliary sequences:
$$
\begin{array}{ll}
b_{1}=a_{1}, & b_{n}=a_{1}+a_{2}+\cdots+a_{n} \\
c_{1}=b_{1}, & c_{n}=b_{1}+b_{2}+\cdots+b_{n}
\end{array}
$$
Observe that
$$
\begin{gathered}
c_{n}-c_{n-1}=b_{n} \\
b_{n}-b_{n-1}=\left(c_{n}-c_{n-1}\right)-\left(c_{n-1}-c_{n-2}\right)=a_{n}
\end{gathered}
$$
Therefore,
$$
c_{n}=\left(2 c_{n-1}-c_{n-2}\right)+a_{n}
$$
is the smallest square greater than $2 c_{n-1}-c_{n-2}$.
Claim. $\sqrt{c_{n}}-\sqrt{c_{n-1}} \leq \sqrt{c_{n-1}}-\sqrt{c_{n-2}}$
Proof 1. Let $c_{n}=m_{n}^{2}, c_{n-1}=m_{n-1}^{2}, c_{n-2}=m_{n-2}^{2}$, where the value of $c_{n}$ is momentarily permitted to exceed its minimum. Any value is permitted for which
$$
b_{n}=c_{n}-c_{n-1}>b_{n-1}=c_{n-1}-c_{n-2}
$$
Factoring,
$$
\left(m_{n}-m_{n-1}\right)\left(m_{n}+m_{n-1}\right)>\left(m_{n-1}-m_{n-2}\right)\left(m_{n-1}+m_{n-2}\right)
$$
In the case that $m_{n}-m_{n-1}=m_{n-1}-m_{n-2}$, this inequality clearly holds, as $m_{n}+m_{n-1}>m_{n-1}+m_{n-2}$. Thus, the minimal possible value of $m_{n}$ satisfies the claimed inequality.
Proof 2. Denote $c_{n-1}=x^{2}, c_{n-2}=(x-d)^{2}$. Then
$$
2 c_{n-1}-c_{n-2}=2 x^{2}-(x-d)^{2}=x^{2}+2 d x-d^{2}=(x+d)^{2}-2 d^{2}<(x+d)^{2}
$$
It follows that $c_{n} \leq(x+d)^{2}$.
And so the sequence of positive integers $\sqrt{c_{n}}-\sqrt{c_{n-1}}$ is decreasing. Any such sequence is eventually constant.
As a corollary, $c_{n}=(x+n d)^{2}$ for sufficiently large $n$, with fixed integers $x$ and $d$. Along the lines above, it becomes clear that $a_{n}=c_{n}-2 c_{n-1}+c_{n-2}=2 d^{2}$, so the sequence $\left(a_{n}\right)$ is constant.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "# Problem 3.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "\nSolution.",
"tier": "T2",
"year": "2022"
}
|
An infinite sequence of positive integers $a_{1}, a_{2}, \ldots$ is called good if
(1) $a_{1}$ is a perfect square, and
(2) for any integer $n \geq 2, a_{n}$ is the smallest positive integer such that
$$
n a_{1}+(n-1) a_{2}+\ldots+2 a_{n-1}+a_{n}
$$
is a perfect square.
Prove that for any good sequence $a_{1}, a_{2}, \ldots$, there exists a positive integer $k$ such that $a_{n}=a_{k}$ for all integers $n \geq k$.
Proposed by: Joe Benton and Dominic Yeo, United Kingdom
|
We write:
$$
s_{n}^{2}=S_{n}=a_{1}+\left(a_{1}+a_{2}\right)+\ldots+\left(a_{1}+\ldots+a_{n}\right)
$$
So, setting $b_{n}:=a_{1}+\ldots+a_{n}$, we have $S_{n}=b_{1}+b_{2}+\ldots+b_{n}$ and, in particular $S_{n+1}=S_{n}+b_{n+1}$. Now, we study the quantity $S_{n} b_{n}=+b_{1}+b_{2}+\ldots+b_{n}+b_{n}$ in two different ways. Since $b_{n+1}$ is the smallest integer strictly greater than $b_{n}$ such that $b_{1}+\ldots+b_{n}+b_{n+1}$ is a perfect square, we must have
$$
b_{1}+b_{2}+\ldots+b_{n}+b_{n} \geq\left(s_{n+1}-1\right)^{2}
$$
However, we also have
$$
b_{1}+b_{2}+\ldots+b_{n}+b_{n}=S_{n}+b_{n}=2 S_{n}-S_{n-1}
$$
Combining, we obtain
$$
s_{n}^{2} \geq \frac{s_{n-1}^{2}+\left(s_{n+1}-1\right)^{2}}{2}>\left(\frac{s_{n-1}+s_{n+1}-1}{2}\right)^{2}
$$
where the final inequality is strict since the sequence $\left(s_{k}\right)$ is strictly increasing. Taking a square root, and noting that all the $\left(s_{n}\right)$ are integers, one obtains $s_{n+1}-s_{n} \leq s_{n}-s_{n-1}$.
Now we focus on the sequence $d_{n}=s_{n+1}-s_{n} .\left(s_{k}\right)$ is strictly increasing thus $\left(d_{k}\right)$ is positive. However we proved that $d_{n+1} \leq d_{n}$, so the sequence $\left(d_{k}\right)$ is eventually constant, so eventually $s_{n}=b n+c$ and $S_{n}=(b n+c)^{2}$ with some numbers $b, c$; then
$$
a_{n+2}=S_{n+2}-2 S_{n+1}+S_{n}=(b(n+2)+c)^{2}-2(b(n+1)+c)^{2}+(b n+c)^{2}=2 b^{2}
$$
Remark. The key idea is to use the $D$ difference operator on the space of sequences. That is a function which takes a sequence as an input and outputs another sequence the following way: if $(x)$ is the input sequence, then
$$
D(x)=D\left(\left(x_{1}, x_{2}, x_{3}, x_{4}, \ldots\right)\right)=\left(x_{2}-x_{1}, x_{3}-x_{2}, x_{4}-x_{3}, \ldots\right)
$$
Here $D D(S)=(a)$ (with some shift), $D(s)=(d)$. This $D$ operator has some nice properties; if someone studied calculus, these properties will be familiar. $D(x+y)=D(x)+D(y) ; D(\lambda x)=\lambda D(x) ;(x)$ constant iff $D x$ zero ; $(x)$ is linear iff $D x$ constant ; in general $(x)$ is a polynomial with degree $l+1$ iff $D(x)$ is a polynomial with degree $l$.
First we proved that sequence $(d)$ is eventually constant, thus $D(d)=D D(s)$ is eventually zero. Therefore the sequence $(s)$ is eventually a linear function, hence $s^{2}=S$ is eventually a quadratic polynomial, therefore $0=D D D(S)=D(a)$, so $(a)$ is constant eventually.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
An infinite sequence of positive integers $a_{1}, a_{2}, \ldots$ is called good if
(1) $a_{1}$ is a perfect square, and
(2) for any integer $n \geq 2, a_{n}$ is the smallest positive integer such that
$$
n a_{1}+(n-1) a_{2}+\ldots+2 a_{n-1}+a_{n}
$$
is a perfect square.
Prove that for any good sequence $a_{1}, a_{2}, \ldots$, there exists a positive integer $k$ such that $a_{n}=a_{k}$ for all integers $n \geq k$.
Proposed by: Joe Benton and Dominic Yeo, United Kingdom
|
We write:
$$
s_{n}^{2}=S_{n}=a_{1}+\left(a_{1}+a_{2}\right)+\ldots+\left(a_{1}+\ldots+a_{n}\right)
$$
So, setting $b_{n}:=a_{1}+\ldots+a_{n}$, we have $S_{n}=b_{1}+b_{2}+\ldots+b_{n}$ and, in particular $S_{n+1}=S_{n}+b_{n+1}$. Now, we study the quantity $S_{n} b_{n}=+b_{1}+b_{2}+\ldots+b_{n}+b_{n}$ in two different ways. Since $b_{n+1}$ is the smallest integer strictly greater than $b_{n}$ such that $b_{1}+\ldots+b_{n}+b_{n+1}$ is a perfect square, we must have
$$
b_{1}+b_{2}+\ldots+b_{n}+b_{n} \geq\left(s_{n+1}-1\right)^{2}
$$
However, we also have
$$
b_{1}+b_{2}+\ldots+b_{n}+b_{n}=S_{n}+b_{n}=2 S_{n}-S_{n-1}
$$
Combining, we obtain
$$
s_{n}^{2} \geq \frac{s_{n-1}^{2}+\left(s_{n+1}-1\right)^{2}}{2}>\left(\frac{s_{n-1}+s_{n+1}-1}{2}\right)^{2}
$$
where the final inequality is strict since the sequence $\left(s_{k}\right)$ is strictly increasing. Taking a square root, and noting that all the $\left(s_{n}\right)$ are integers, one obtains $s_{n+1}-s_{n} \leq s_{n}-s_{n-1}$.
Now we focus on the sequence $d_{n}=s_{n+1}-s_{n} .\left(s_{k}\right)$ is strictly increasing thus $\left(d_{k}\right)$ is positive. However we proved that $d_{n+1} \leq d_{n}$, so the sequence $\left(d_{k}\right)$ is eventually constant, so eventually $s_{n}=b n+c$ and $S_{n}=(b n+c)^{2}$ with some numbers $b, c$; then
$$
a_{n+2}=S_{n+2}-2 S_{n+1}+S_{n}=(b(n+2)+c)^{2}-2(b(n+1)+c)^{2}+(b n+c)^{2}=2 b^{2}
$$
Remark. The key idea is to use the $D$ difference operator on the space of sequences. That is a function which takes a sequence as an input and outputs another sequence the following way: if $(x)$ is the input sequence, then
$$
D(x)=D\left(\left(x_{1}, x_{2}, x_{3}, x_{4}, \ldots\right)\right)=\left(x_{2}-x_{1}, x_{3}-x_{2}, x_{4}-x_{3}, \ldots\right)
$$
Here $D D(S)=(a)$ (with some shift), $D(s)=(d)$. This $D$ operator has some nice properties; if someone studied calculus, these properties will be familiar. $D(x+y)=D(x)+D(y) ; D(\lambda x)=\lambda D(x) ;(x)$ constant iff $D x$ zero ; $(x)$ is linear iff $D x$ constant ; in general $(x)$ is a polynomial with degree $l+1$ iff $D(x)$ is a polynomial with degree $l$.
First we proved that sequence $(d)$ is eventually constant, thus $D(d)=D D(s)$ is eventually zero. Therefore the sequence $(s)$ is eventually a linear function, hence $s^{2}=S$ is eventually a quadratic polynomial, therefore $0=D D D(S)=D(a)$, so $(a)$ is constant eventually.
|
{
"exam": "EGMO",
"problem_label": "3",
"problem_match": "# Problem 3.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "\nSolution.",
"tier": "T2",
"year": "2022"
}
|
Given a positive integer $n \geq 2$, determine the largest positive integer $N$ for which there exist $N+1$ real numbers $a_{0}, a_{1}, \ldots, a_{N}$ such that
(1) $a_{0}+a_{1}=-\frac{1}{n}$, and
(2) $\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ for $1 \leq k \leq N-1$.
Proposed by: Romania
|
$\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ is equivalent to $\left(a_{k}+a_{k-1}+1\right)\left(a_{k}+a_{k+1}-1\right)=-1$. Let $b_{k}=a_{k}+a_{k+1}$. Thus we need $b_{0}, b_{1}, \ldots$ the following way: $b_{0}=-\frac{1}{n}$ and $\left(b_{k-1}+1\right)\left(b_{k}-1\right)=-1$. There is a proper sequence $b_{0}, b_{1}, \ldots, b_{N-1}$ if and only if there is proper sequence $a_{0}, a_{1}, \ldots, a_{N}$, because from a a proper $\left(a_{k}\right)$ sequence we can get a proper $\left(b_{k}\right)$ sequence with $b_{k}=a_{k}+a_{k+1}$ for $k=0,1, \ldots, N-1$ and from a proper $\left(b_{k}\right)$ sequence we can get a proper $\left(a_{k}\right)$ sequence by arbitrarily setting $a_{0}$ and then inductively defining $a_{k}=b_{k-1}-a_{k-1}$ for $k=1,2, \ldots, N$.
We prove by induction that $b_{k}=-\frac{1}{n-k}$ for $k<n$. This is true for $k=0$, as $b_{0}=-\frac{1}{n}$ and
$$
b_{k}=1-\frac{1}{b_{k-1}+1}=1-\frac{1}{1-\frac{1}{n-k+1}}=-\frac{1}{n-k}
$$
for $k<n$. Thus there is a proper sequence $b_{0}, b_{1}, \ldots, b_{n-1}$, but it can't be continued, because $b_{n-1}+1=$ 0 so there is no $b_{n}$ for which $\left(b_{n-1}+1\right)\left(b_{n}-1\right)=-1$.
Therefore the longest proper sequence $\left(b_{k}\right)$ is $n$-long, so the longest proper sequence $\left(a_{k}\right)$ is $n+1$ long, so $N=n$.
|
n
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Given a positive integer $n \geq 2$, determine the largest positive integer $N$ for which there exist $N+1$ real numbers $a_{0}, a_{1}, \ldots, a_{N}$ such that
(1) $a_{0}+a_{1}=-\frac{1}{n}$, and
(2) $\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ for $1 \leq k \leq N-1$.
Proposed by: Romania
|
$\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ is equivalent to $\left(a_{k}+a_{k-1}+1\right)\left(a_{k}+a_{k+1}-1\right)=-1$. Let $b_{k}=a_{k}+a_{k+1}$. Thus we need $b_{0}, b_{1}, \ldots$ the following way: $b_{0}=-\frac{1}{n}$ and $\left(b_{k-1}+1\right)\left(b_{k}-1\right)=-1$. There is a proper sequence $b_{0}, b_{1}, \ldots, b_{N-1}$ if and only if there is proper sequence $a_{0}, a_{1}, \ldots, a_{N}$, because from a a proper $\left(a_{k}\right)$ sequence we can get a proper $\left(b_{k}\right)$ sequence with $b_{k}=a_{k}+a_{k+1}$ for $k=0,1, \ldots, N-1$ and from a proper $\left(b_{k}\right)$ sequence we can get a proper $\left(a_{k}\right)$ sequence by arbitrarily setting $a_{0}$ and then inductively defining $a_{k}=b_{k-1}-a_{k-1}$ for $k=1,2, \ldots, N$.
We prove by induction that $b_{k}=-\frac{1}{n-k}$ for $k<n$. This is true for $k=0$, as $b_{0}=-\frac{1}{n}$ and
$$
b_{k}=1-\frac{1}{b_{k-1}+1}=1-\frac{1}{1-\frac{1}{n-k+1}}=-\frac{1}{n-k}
$$
for $k<n$. Thus there is a proper sequence $b_{0}, b_{1}, \ldots, b_{n-1}$, but it can't be continued, because $b_{n-1}+1=$ 0 so there is no $b_{n}$ for which $\left(b_{n-1}+1\right)\left(b_{n}-1\right)=-1$.
Therefore the longest proper sequence $\left(b_{k}\right)$ is $n$-long, so the longest proper sequence $\left(a_{k}\right)$ is $n+1$ long, so $N=n$.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "# Problem 4.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "\nSolution 1. ",
"tier": "T2",
"year": "2022"
}
|
Given a positive integer $n \geq 2$, determine the largest positive integer $N$ for which there exist $N+1$ real numbers $a_{0}, a_{1}, \ldots, a_{N}$ such that
(1) $a_{0}+a_{1}=-\frac{1}{n}$, and
(2) $\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ for $1 \leq k \leq N-1$.
Proposed by: Romania
|
The required maximum is $N=n$.
To rule out the case $N \geq n+1$, it is clearly sufficient to rule out the case $N=n+1$.
Assume for contradiction that $a_{0}, a_{1}, \ldots, a_{n+1}$ are real numbers satisfying both conditions in the statement. It is sufficient to show that $a_{k}+a_{k+1}=0$ for some $k \leq n$, because then $a_{k-1}-a_{k+1}=0$ so $a_{k+1}=a_{k-1}$, therefore $a_{k-1}+a_{k}=0$ and so on, by backwards recursion we get that $a_{j}+a_{j+1}=0$ for all $0 \leq j \leq k$, but this is a contradiction with $a_{0}+a_{1}=-\frac{1}{n}$.
To prove that $a_{k}+a_{k+1}=0$ for some $k \leq n$, assume that $a_{k}+a_{k+1} \neq 0$ for all $k \leq n$, to rewrite the second condition in the statement in the form
$$
\frac{1}{a_{k}+a_{k+1}}-\frac{1}{a_{k-1}+a_{k}}=1, \quad k=1, \ldots, n
$$
and sum both sides over the full range from $k=1$ to $n$. This gives
$$
\frac{1}{a_{n}+a_{n+1}}-\frac{1}{a_{0}+a_{1}}=n \text {. }
$$
As $a_{0}+a_{1}=-\frac{1}{n}$, this means that $\frac{1}{a_{n}+a_{n+1}}=0$, which is a contradiction. Consequently, $N \leq n$.
To provide $n+1$ real numbers satisfying both conditions in the statement, fix $a_{0}$ and go through the telescoping procedure above to obtain
$$
a_{k}=(-1)^{k} a_{0}+\sum_{j=1}^{k} \frac{(-1)^{k-j+1}}{n-j+1}, \quad k=1, \ldots, n
$$
This concludes the proof.
|
N=n
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Given a positive integer $n \geq 2$, determine the largest positive integer $N$ for which there exist $N+1$ real numbers $a_{0}, a_{1}, \ldots, a_{N}$ such that
(1) $a_{0}+a_{1}=-\frac{1}{n}$, and
(2) $\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ for $1 \leq k \leq N-1$.
Proposed by: Romania
|
The required maximum is $N=n$.
To rule out the case $N \geq n+1$, it is clearly sufficient to rule out the case $N=n+1$.
Assume for contradiction that $a_{0}, a_{1}, \ldots, a_{n+1}$ are real numbers satisfying both conditions in the statement. It is sufficient to show that $a_{k}+a_{k+1}=0$ for some $k \leq n$, because then $a_{k-1}-a_{k+1}=0$ so $a_{k+1}=a_{k-1}$, therefore $a_{k-1}+a_{k}=0$ and so on, by backwards recursion we get that $a_{j}+a_{j+1}=0$ for all $0 \leq j \leq k$, but this is a contradiction with $a_{0}+a_{1}=-\frac{1}{n}$.
To prove that $a_{k}+a_{k+1}=0$ for some $k \leq n$, assume that $a_{k}+a_{k+1} \neq 0$ for all $k \leq n$, to rewrite the second condition in the statement in the form
$$
\frac{1}{a_{k}+a_{k+1}}-\frac{1}{a_{k-1}+a_{k}}=1, \quad k=1, \ldots, n
$$
and sum both sides over the full range from $k=1$ to $n$. This gives
$$
\frac{1}{a_{n}+a_{n+1}}-\frac{1}{a_{0}+a_{1}}=n \text {. }
$$
As $a_{0}+a_{1}=-\frac{1}{n}$, this means that $\frac{1}{a_{n}+a_{n+1}}=0$, which is a contradiction. Consequently, $N \leq n$.
To provide $n+1$ real numbers satisfying both conditions in the statement, fix $a_{0}$ and go through the telescoping procedure above to obtain
$$
a_{k}=(-1)^{k} a_{0}+\sum_{j=1}^{k} \frac{(-1)^{k-j+1}}{n-j+1}, \quad k=1, \ldots, n
$$
This concludes the proof.
|
{
"exam": "EGMO",
"problem_label": "4",
"problem_match": "# Problem 4.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "\nSolution 2. ",
"tier": "T2",
"year": "2022"
}
|
For all positive integers $n, k$, let $f(n, 2 k)$ be the number of ways an $n \times 2 k$ board can be fully covered by $n k$ dominoes of size $2 \times 1$. (For example, $f(2,2)=2$ and $f(3,2)=3$.)
Find all positive integers $n$ such that for every positive integer $k$, the number $f(n, 2 k)$ is odd.
Proposed by: U.S.A.
|
The integers $n$ with the requested property are exactly the ones of the form $2^{k}-1$.
In what follows, let $f(m, n)$ denote the number of domino tilings of an $m \times n$ grid. (For convenience, we also allow $m$ and $n$ to be 0 , in which case $f(m, n)=1$.)
Claim. $f(m, 2 n+1) \equiv f(m, n)(\bmod 2)$ for all $n$ and even $m$
Proof. Consider reflecting $m \times(2 n+1)$ tilings across the central column. In this way, the tilings are grouped into singletons and pairs, so modulo $2, f(m, 2 n+1)$ is congruent to the number of singletons.
If a tiling is invariant under reflection in the central column, then every domino with one cell in that column must have both cells in that column.
In other words, the central column is filled with $\frac{m}{2}$ vertical dominoes, splitting the remainder of the grid into two $m \times n$ grids. Obeying the symmetry constraint, there are $f(m, n)$ tilings: each of the tilings of the left grid fixes the tiling of the right grid.
Claim. $f(n, n) \equiv 0(\bmod 2)$ for all even $n \geq 2$. (Recall that $f(0,0)=1$ is odd.)
Proof. Consider reflecting $n \times n$ tilings across the diagonal. This groups the tilings into pairs, (no tiling is grouped with itself). Hence the number of $n \times n$ tilings is even.
We are ready to complete the solution.
- If $n$ is odd, the first claim shows that $n$ satisfies the property if and only if $\frac{1}{2}(n-1)$ does.
- If $n \geq 2$ is even, the second claim shows that $n$ does not satisfy the property.
- If $n=0$, then $n$ satisfies the property, as $f(m, 0)=1$ always.
This concludes the proof that the sought numbers are the ones of the form $2^{k}-1$.
|
2^{k}-1
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
For all positive integers $n, k$, let $f(n, 2 k)$ be the number of ways an $n \times 2 k$ board can be fully covered by $n k$ dominoes of size $2 \times 1$. (For example, $f(2,2)=2$ and $f(3,2)=3$.)
Find all positive integers $n$ such that for every positive integer $k$, the number $f(n, 2 k)$ is odd.
Proposed by: U.S.A.
|
The integers $n$ with the requested property are exactly the ones of the form $2^{k}-1$.
In what follows, let $f(m, n)$ denote the number of domino tilings of an $m \times n$ grid. (For convenience, we also allow $m$ and $n$ to be 0 , in which case $f(m, n)=1$.)
Claim. $f(m, 2 n+1) \equiv f(m, n)(\bmod 2)$ for all $n$ and even $m$
Proof. Consider reflecting $m \times(2 n+1)$ tilings across the central column. In this way, the tilings are grouped into singletons and pairs, so modulo $2, f(m, 2 n+1)$ is congruent to the number of singletons.
If a tiling is invariant under reflection in the central column, then every domino with one cell in that column must have both cells in that column.
In other words, the central column is filled with $\frac{m}{2}$ vertical dominoes, splitting the remainder of the grid into two $m \times n$ grids. Obeying the symmetry constraint, there are $f(m, n)$ tilings: each of the tilings of the left grid fixes the tiling of the right grid.
Claim. $f(n, n) \equiv 0(\bmod 2)$ for all even $n \geq 2$. (Recall that $f(0,0)=1$ is odd.)
Proof. Consider reflecting $n \times n$ tilings across the diagonal. This groups the tilings into pairs, (no tiling is grouped with itself). Hence the number of $n \times n$ tilings is even.
We are ready to complete the solution.
- If $n$ is odd, the first claim shows that $n$ satisfies the property if and only if $\frac{1}{2}(n-1)$ does.
- If $n \geq 2$ is even, the second claim shows that $n$ does not satisfy the property.
- If $n=0$, then $n$ satisfies the property, as $f(m, 0)=1$ always.
This concludes the proof that the sought numbers are the ones of the form $2^{k}-1$.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "# Problem 5.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "\nSolution.",
"tier": "T2",
"year": "2022"
}
|
For all positive integers $n, k$, let $f(n, 2 k)$ be the number of ways an $n \times 2 k$ board can be fully covered by $n k$ dominoes of size $2 \times 1$. (For example, $f(2,2)=2$ and $f(3,2)=3$.)
Find all positive integers $n$ such that for every positive integer $k$, the number $f(n, 2 k)$ is odd.
Proposed by: U.S.A.
|
Color the board as a chessboard. Consider the bipartite graph whose vertices are the squares and the neighbors are connected by an edge. Notice that a domino tiling described in the problem corresponds to a perfect matching in this bipartite graph, so we are interested in the number of perfect matchings.
And that is the permanent of the (bipartite) adjacency matrix. However we need to compute it $\bmod 2\left(\right.$ over $\left.F_{2}\right)$, so that is the determinant of that matrix.
So our goal is to decide whether the determinant is 0 or $1 \bmod 2$, or in other world, that matrix is singular or not. From now we compute everything mod 2. Now we understand what does it mean, that a vector $v$ is in the nullspace (has eigenvalue 0 ). But each entry of $v$ is $0-1$, so in other words we choose a subset of the black squares ( $v$ is the characteristic vector of that subset). So in the language of the board, there is a nontrivial $v$ in the nullspace iff we have a subspace of black square such a way that each white square has even number of black chosen neighbors.
If $n$ is even, then we can choose a subset such a way in the $n \times n$ square: the diagonal.
Now we prove similar connection between $n$ and $2 n+1$ : there is a construction for $n$ iff we have one for $2 n+1 . k$ is always even. If we have a construction in $n \times k$ then we have one in $(2 n+1) \times k$ : we mirror the construction to the middle row.
Now we prove that from a construction in $(2 n+1) \times k$ we can make a construction in $n \times k$. First, we
don't have any chosen black square in the middle row. Then the chosen black squares in the upper (or lower) $n \times k$ rectangle is a construction for $n \times k$. Second, we have some chosen black square in the middle row. Because we can find a white square in the middle row such that is has only one chosen black neighbor in the middle row. Thus this construction isn't symmetric to the horizontal line the symmetric difference of the construction and its reflection (to the vertical line) is a nontrivial vector in the nullspace, thus we can use the first case.
|
not found
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
For all positive integers $n, k$, let $f(n, 2 k)$ be the number of ways an $n \times 2 k$ board can be fully covered by $n k$ dominoes of size $2 \times 1$. (For example, $f(2,2)=2$ and $f(3,2)=3$.)
Find all positive integers $n$ such that for every positive integer $k$, the number $f(n, 2 k)$ is odd.
Proposed by: U.S.A.
|
Color the board as a chessboard. Consider the bipartite graph whose vertices are the squares and the neighbors are connected by an edge. Notice that a domino tiling described in the problem corresponds to a perfect matching in this bipartite graph, so we are interested in the number of perfect matchings.
And that is the permanent of the (bipartite) adjacency matrix. However we need to compute it $\bmod 2\left(\right.$ over $\left.F_{2}\right)$, so that is the determinant of that matrix.
So our goal is to decide whether the determinant is 0 or $1 \bmod 2$, or in other world, that matrix is singular or not. From now we compute everything mod 2. Now we understand what does it mean, that a vector $v$ is in the nullspace (has eigenvalue 0 ). But each entry of $v$ is $0-1$, so in other words we choose a subset of the black squares ( $v$ is the characteristic vector of that subset). So in the language of the board, there is a nontrivial $v$ in the nullspace iff we have a subspace of black square such a way that each white square has even number of black chosen neighbors.
If $n$ is even, then we can choose a subset such a way in the $n \times n$ square: the diagonal.
Now we prove similar connection between $n$ and $2 n+1$ : there is a construction for $n$ iff we have one for $2 n+1 . k$ is always even. If we have a construction in $n \times k$ then we have one in $(2 n+1) \times k$ : we mirror the construction to the middle row.
Now we prove that from a construction in $(2 n+1) \times k$ we can make a construction in $n \times k$. First, we
don't have any chosen black square in the middle row. Then the chosen black squares in the upper (or lower) $n \times k$ rectangle is a construction for $n \times k$. Second, we have some chosen black square in the middle row. Because we can find a white square in the middle row such that is has only one chosen black neighbor in the middle row. Thus this construction isn't symmetric to the horizontal line the symmetric difference of the construction and its reflection (to the vertical line) is a nontrivial vector in the nullspace, thus we can use the first case.
|
{
"exam": "EGMO",
"problem_label": "5",
"problem_match": "# Problem 5.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "# Solution 2.",
"tier": "T2",
"year": "2022"
}
|
Let $A B C D$ be a cyclic quadrilateral with circumcentre $O$. Let the internal angle bisectors at $A$ and $B$ meet at $X$, the internal angle bisectors at $B$ and $C$ meet at $Y$, the internal angle bisectors at $C$ and $D$ meet at $Z$, and the internal angle bisectors at $D$ and $A$ meet at $W$. Further, let $A C$ and $B D$ meet at $P$. Suppose that the points $X, Y, Z, W, O$ and $P$ are distinct.
Prove that $O, X, Y, Z$ and $W$ lie on the same circle if and only if $P, X, Y, Z$ and $W$ lie on the same circle.
Proposed by: Ethan Tan, Australia
|
Let $\Omega$ be the circumcircle of the quadrilateral $A B C D$ and let $r$ be its radius.
First, notice that the points $X, Y, Z, W$ are concyclic. Indeed, using oriented (modulo $180^{\circ}$ ) angles,
$$
\angle(X W, X Y)+\angle(Z Y, Z W)=\angle(X A, X B)+\angle(Z C, Z D)=-\frac{\angle A+\angle B}{2}-\frac{\angle C+\angle D}{2}=0 .
$$
Let $\omega$ be the circle passing through these four points. Our goal is to prove that $O \in \omega$ occurs if and only if $P \in \omega$.
Next, we rule out the case when $A B C D$ is a trapezium. Suppose that if $A B C D$ is an isosceles trapezium; without loss of generality say $A B \| C D$. By symmetry, points $X, Z, O$ and $P$ lie on the symmetry axis (that is, the common perpendicular bisector of $A B$ and $C D$ ), therefore they are collinear. By the conditions of the problem, these four points are distinct and $X$ and $Z$ are on $\omega$, so neither $O$, nor $P$ can lie on $\omega$; therefore the problem statement is obvious. From now on we assume that $A B \nVdash C D$ and $B C \nVdash A D$.
Let $A B$ and $C D$ meet at $Q$, and let $B C$ and $A D$ meet at $R$. Without loss of generality, suppose that $B$ lies between $A$ and $Q$, and $D$ lies between $A$ and $R$. Now we show that line $Q R$ is the radical axis between circles $\Omega$ and $\omega$.
Point $W$ is the intersection point of the bisectors of $\angle A$ and $\angle D$, so in triangle $A D Q$, point $W$ is the incentre. Similary, $B Y$ and $C Y$ are the external bisectors of $\angle Q B C$ and $\angle B C Q$, so in triangle $B C Q$, point $Y$ is the excentre opposite to $Q$. Hence, both $Y$ and $W$ lie on the bisector of $\angle D Q A$.
By $\angle D Q A=180^{\circ}-\angle D-\angle A=\angle B-\angle A$ we get
$$
\angle B Y Q=\angle Y B A-\angle Y Q B=\frac{\angle B}{2}-\frac{\angle D Q A}{2}=\frac{\angle B}{2}-\frac{\angle B-\angle A}{2}=\frac{\angle A}{2}=\angle B A W
$$
so the points $A, B, Y, W$ are concyclic, and therefore $Q A \cdot Q B=Q Y \cdot Q W$; hence, $Q$ has equal power with respect to $\Omega$ and $\omega$. It can be seen similarly that $R$ has equal power with respect to the circles $\Omega$ and $\omega$. Hence, the radical axis of $\Omega$ and $\omega$ is line $Q R$.
Let lines $O P$ and $Q R$ meet at point $S$. It is well-known that with respect to circle $A B C D$, the diagonal triangle $P Q R$ is autopolar. As consequences we have $O P \perp Q R$, and the points $P$ and $S$ are symmetric to circle $A B C D$, so $O S \cdot O P=r^{2}$.
Notice that $P$ and $O$ lie inside $\Omega$, so the polar line $Q R$ lies entirely outside, so $S$ is different from $P$ and $O$. Moreover,
$$
S O \cdot S P=O S \cdot(O S-O P)=O S^{2}-O S \cdot O P=S O^{2}-r^{2}
$$
so $S O \cdot S P$ is equal to the power of $S$ with respect to $\Omega$. Since $S$ lies on the radical axis $Q R$, it has equal power with respect to the two circles; therefore, $S O \cdot S P$ is equal to the power of $S$ with respect to $\omega$. From this, it follows that $O \in \omega$ if and only if $P \in \omega$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a cyclic quadrilateral with circumcentre $O$. Let the internal angle bisectors at $A$ and $B$ meet at $X$, the internal angle bisectors at $B$ and $C$ meet at $Y$, the internal angle bisectors at $C$ and $D$ meet at $Z$, and the internal angle bisectors at $D$ and $A$ meet at $W$. Further, let $A C$ and $B D$ meet at $P$. Suppose that the points $X, Y, Z, W, O$ and $P$ are distinct.
Prove that $O, X, Y, Z$ and $W$ lie on the same circle if and only if $P, X, Y, Z$ and $W$ lie on the same circle.
Proposed by: Ethan Tan, Australia
|
Let $\Omega$ be the circumcircle of the quadrilateral $A B C D$ and let $r$ be its radius.
First, notice that the points $X, Y, Z, W$ are concyclic. Indeed, using oriented (modulo $180^{\circ}$ ) angles,
$$
\angle(X W, X Y)+\angle(Z Y, Z W)=\angle(X A, X B)+\angle(Z C, Z D)=-\frac{\angle A+\angle B}{2}-\frac{\angle C+\angle D}{2}=0 .
$$
Let $\omega$ be the circle passing through these four points. Our goal is to prove that $O \in \omega$ occurs if and only if $P \in \omega$.
Next, we rule out the case when $A B C D$ is a trapezium. Suppose that if $A B C D$ is an isosceles trapezium; without loss of generality say $A B \| C D$. By symmetry, points $X, Z, O$ and $P$ lie on the symmetry axis (that is, the common perpendicular bisector of $A B$ and $C D$ ), therefore they are collinear. By the conditions of the problem, these four points are distinct and $X$ and $Z$ are on $\omega$, so neither $O$, nor $P$ can lie on $\omega$; therefore the problem statement is obvious. From now on we assume that $A B \nVdash C D$ and $B C \nVdash A D$.
Let $A B$ and $C D$ meet at $Q$, and let $B C$ and $A D$ meet at $R$. Without loss of generality, suppose that $B$ lies between $A$ and $Q$, and $D$ lies between $A$ and $R$. Now we show that line $Q R$ is the radical axis between circles $\Omega$ and $\omega$.
Point $W$ is the intersection point of the bisectors of $\angle A$ and $\angle D$, so in triangle $A D Q$, point $W$ is the incentre. Similary, $B Y$ and $C Y$ are the external bisectors of $\angle Q B C$ and $\angle B C Q$, so in triangle $B C Q$, point $Y$ is the excentre opposite to $Q$. Hence, both $Y$ and $W$ lie on the bisector of $\angle D Q A$.
By $\angle D Q A=180^{\circ}-\angle D-\angle A=\angle B-\angle A$ we get
$$
\angle B Y Q=\angle Y B A-\angle Y Q B=\frac{\angle B}{2}-\frac{\angle D Q A}{2}=\frac{\angle B}{2}-\frac{\angle B-\angle A}{2}=\frac{\angle A}{2}=\angle B A W
$$
so the points $A, B, Y, W$ are concyclic, and therefore $Q A \cdot Q B=Q Y \cdot Q W$; hence, $Q$ has equal power with respect to $\Omega$ and $\omega$. It can be seen similarly that $R$ has equal power with respect to the circles $\Omega$ and $\omega$. Hence, the radical axis of $\Omega$ and $\omega$ is line $Q R$.
Let lines $O P$ and $Q R$ meet at point $S$. It is well-known that with respect to circle $A B C D$, the diagonal triangle $P Q R$ is autopolar. As consequences we have $O P \perp Q R$, and the points $P$ and $S$ are symmetric to circle $A B C D$, so $O S \cdot O P=r^{2}$.
Notice that $P$ and $O$ lie inside $\Omega$, so the polar line $Q R$ lies entirely outside, so $S$ is different from $P$ and $O$. Moreover,
$$
S O \cdot S P=O S \cdot(O S-O P)=O S^{2}-O S \cdot O P=S O^{2}-r^{2}
$$
so $S O \cdot S P$ is equal to the power of $S$ with respect to $\Omega$. Since $S$ lies on the radical axis $Q R$, it has equal power with respect to the two circles; therefore, $S O \cdot S P$ is equal to the power of $S$ with respect to $\omega$. From this, it follows that $O \in \omega$ if and only if $P \in \omega$.
|
{
"exam": "EGMO",
"problem_label": "6",
"problem_match": "# Problem 6.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "\nSolution 1. ",
"tier": "T2",
"year": "2022"
}
|
Let $A B C D$ be a cyclic quadrilateral with circumcentre $O$. Let the internal angle bisectors at $A$ and $B$ meet at $X$, the internal angle bisectors at $B$ and $C$ meet at $Y$, the internal angle bisectors at $C$ and $D$ meet at $Z$, and the internal angle bisectors at $D$ and $A$ meet at $W$. Further, let $A C$ and $B D$ meet at $P$. Suppose that the points $X, Y, Z, W, O$ and $P$ are distinct.
Prove that $O, X, Y, Z$ and $W$ lie on the same circle if and only if $P, X, Y, Z$ and $W$ lie on the same circle.
Proposed by: Ethan Tan, Australia
|
We will prove that the points $X, Y, Z, W, O, P$ lie on a conic section $\Gamma$. Five distinct points uniquely determine a conic section, therefore
$$
X, Y, Z, W, O \text { are concyclic } \Leftrightarrow \Gamma \text { is a circle } \Leftrightarrow X, Y, Z, W, P \text { are concyclic. }
$$
Let $\alpha=\angle A C B=\angle A D B, \beta=\angle B D C=\angle B A C, \gamma=\angle C A D=\angle C B D$ and $\delta=\angle D B A=$ $\angle D C A$, where of course $\alpha+\beta+\gamma+\delta=180^{\circ}$. For every point $\xi$ in the plane, let $a(\xi), b(\xi), c(\xi)$ and $d(\xi)$ be the signed distances between $\xi$ and the lines $A B, B C, C D$ and $D A$, respectively, such that the quadrilateral lies on the positive sides of these lines.
The equations of the bisectors of $\angle A X W, \angle B X Y, \angle C Y Z$ and $\angle D Z W$ are $a(\xi)-d(\xi)=0$, $b(\xi)-a(\xi)=0, c(\xi)-b(\xi)=0$ and $d(\xi)-c(\xi)=0$, respectively. Notice that the points $X, Y, Z, W$ satisfy the quadratic equations $(a(\xi)-d(\xi))(b(\xi)-c(\xi))=0$ and $(a(\xi)-b(\xi))(c(\xi)-d(\xi))=0$, so
$$
a(\xi) b(\xi)+c(\xi) d(\xi)=a(\xi) c(\xi)+b(\xi) d(\xi)=a(\xi) d(\xi)+b(\xi) c(\xi) \quad \text { for } \xi=X, Y, Z, W
$$
Without loss of generality, suppose that $A B C D$ is inscribed in a unit circle. Then
$$
a(O)=\cos \alpha, \quad b(O)=\cos \beta, \quad c(O)=\cos \gamma, \quad d(O)=\cos \delta
$$
By $\frac{a(P)}{b(P)}=\frac{B P \cdot \sin \delta}{B P \cdot \sin \gamma}=\frac{1 / \sin \gamma}{1 / \sin \delta}$ and the analogous relations we can see that
$$
a(P)=\frac{k}{\sin \gamma}, \quad b(P)=\frac{k}{\sin \delta}, \quad c(P)=\frac{k}{\sin \alpha}, \quad d(P)=\frac{k}{\sin \beta}
$$
with some positive number $k$.
Now let $s=\frac{\sin \alpha \sin \beta \sin \gamma \sin \delta}{k^{2}}$; then, by (2a) and (2b),
$$
\begin{aligned}
a(O) b(O)-s \cdot a(P) b(P) & =\cos \alpha \cos \beta-\sin \alpha \sin \beta=\cos (\alpha+\beta) \\
c(O) d(O)-s \cdot c(P) d(P) & =\cos (\gamma+\delta)=-\cos (\alpha+\beta) \\
a(O) b(O)+c(O) d(O) & =s \cdot(a(P) b(P)+c(P) d(P))
\end{aligned}
$$
analogous calculation provides
$$
\begin{aligned}
& a(O) c(O)+b(O) d(O)=s \cdot(a(P) c(P)+b(P) d(P)) \\
& a(O) d(O)+b(O) c(O)=s \cdot(a(P) d(P)+b(P) c(P))
\end{aligned}
$$
In order to find the equation of the curve $\Gamma$, choose real numbers $u, v, w$, not all zero, such that
$$
\begin{gathered}
u+v+w=0 \\
u \cdot(a(P) b(P)+c(P) d(P))+v \cdot(a(P) c(P)+b(P) d(P))+w \cdot(a(P) d(P)+b(P) c(P))=0
\end{gathered}
$$
This is always possible, because this a system of two homogeneous linear equations with three variables. Then the equation of $\Gamma$ will be
$$
f(\xi)=u \cdot(a(\xi) b(\xi)+c(\xi) d(\xi))+v \cdot(a(\xi) c(\xi)+b(\xi) d(\xi))+w \cdot(a(\xi) d(\xi)+b(\xi) c(\xi))=0
$$
As can be seen, $f(X)=f(Y)=f(Z)=f(W)=0$ follows from (1) and (4), $f(P)=0$ follows from (5), then $f(O)=s \cdot f(P)=0$ follows from $(3 a-3 c)$. So, the points $X, Y, Z, W, O, P$ all satisfy the equation $f(\xi)=0$.
Notice that $f(B)=u \cdot c(B) d(B)$ and $f(A)=w \cdot b(A) c(A)$; since at least one of $u$ and $w$ is nonzero, either $A$ or $B$ does not satisfy $(*)$. Therefore, the equation cannot degenerate to an identity.
Hence, the equation $f(\xi)=0$ is at most quadratic, it is not an identity, but satisfied by $X, Y, Z, W, O, P$, so these six points lie on a conic section.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a cyclic quadrilateral with circumcentre $O$. Let the internal angle bisectors at $A$ and $B$ meet at $X$, the internal angle bisectors at $B$ and $C$ meet at $Y$, the internal angle bisectors at $C$ and $D$ meet at $Z$, and the internal angle bisectors at $D$ and $A$ meet at $W$. Further, let $A C$ and $B D$ meet at $P$. Suppose that the points $X, Y, Z, W, O$ and $P$ are distinct.
Prove that $O, X, Y, Z$ and $W$ lie on the same circle if and only if $P, X, Y, Z$ and $W$ lie on the same circle.
Proposed by: Ethan Tan, Australia
|
We will prove that the points $X, Y, Z, W, O, P$ lie on a conic section $\Gamma$. Five distinct points uniquely determine a conic section, therefore
$$
X, Y, Z, W, O \text { are concyclic } \Leftrightarrow \Gamma \text { is a circle } \Leftrightarrow X, Y, Z, W, P \text { are concyclic. }
$$
Let $\alpha=\angle A C B=\angle A D B, \beta=\angle B D C=\angle B A C, \gamma=\angle C A D=\angle C B D$ and $\delta=\angle D B A=$ $\angle D C A$, where of course $\alpha+\beta+\gamma+\delta=180^{\circ}$. For every point $\xi$ in the plane, let $a(\xi), b(\xi), c(\xi)$ and $d(\xi)$ be the signed distances between $\xi$ and the lines $A B, B C, C D$ and $D A$, respectively, such that the quadrilateral lies on the positive sides of these lines.
The equations of the bisectors of $\angle A X W, \angle B X Y, \angle C Y Z$ and $\angle D Z W$ are $a(\xi)-d(\xi)=0$, $b(\xi)-a(\xi)=0, c(\xi)-b(\xi)=0$ and $d(\xi)-c(\xi)=0$, respectively. Notice that the points $X, Y, Z, W$ satisfy the quadratic equations $(a(\xi)-d(\xi))(b(\xi)-c(\xi))=0$ and $(a(\xi)-b(\xi))(c(\xi)-d(\xi))=0$, so
$$
a(\xi) b(\xi)+c(\xi) d(\xi)=a(\xi) c(\xi)+b(\xi) d(\xi)=a(\xi) d(\xi)+b(\xi) c(\xi) \quad \text { for } \xi=X, Y, Z, W
$$
Without loss of generality, suppose that $A B C D$ is inscribed in a unit circle. Then
$$
a(O)=\cos \alpha, \quad b(O)=\cos \beta, \quad c(O)=\cos \gamma, \quad d(O)=\cos \delta
$$
By $\frac{a(P)}{b(P)}=\frac{B P \cdot \sin \delta}{B P \cdot \sin \gamma}=\frac{1 / \sin \gamma}{1 / \sin \delta}$ and the analogous relations we can see that
$$
a(P)=\frac{k}{\sin \gamma}, \quad b(P)=\frac{k}{\sin \delta}, \quad c(P)=\frac{k}{\sin \alpha}, \quad d(P)=\frac{k}{\sin \beta}
$$
with some positive number $k$.
Now let $s=\frac{\sin \alpha \sin \beta \sin \gamma \sin \delta}{k^{2}}$; then, by (2a) and (2b),
$$
\begin{aligned}
a(O) b(O)-s \cdot a(P) b(P) & =\cos \alpha \cos \beta-\sin \alpha \sin \beta=\cos (\alpha+\beta) \\
c(O) d(O)-s \cdot c(P) d(P) & =\cos (\gamma+\delta)=-\cos (\alpha+\beta) \\
a(O) b(O)+c(O) d(O) & =s \cdot(a(P) b(P)+c(P) d(P))
\end{aligned}
$$
analogous calculation provides
$$
\begin{aligned}
& a(O) c(O)+b(O) d(O)=s \cdot(a(P) c(P)+b(P) d(P)) \\
& a(O) d(O)+b(O) c(O)=s \cdot(a(P) d(P)+b(P) c(P))
\end{aligned}
$$
In order to find the equation of the curve $\Gamma$, choose real numbers $u, v, w$, not all zero, such that
$$
\begin{gathered}
u+v+w=0 \\
u \cdot(a(P) b(P)+c(P) d(P))+v \cdot(a(P) c(P)+b(P) d(P))+w \cdot(a(P) d(P)+b(P) c(P))=0
\end{gathered}
$$
This is always possible, because this a system of two homogeneous linear equations with three variables. Then the equation of $\Gamma$ will be
$$
f(\xi)=u \cdot(a(\xi) b(\xi)+c(\xi) d(\xi))+v \cdot(a(\xi) c(\xi)+b(\xi) d(\xi))+w \cdot(a(\xi) d(\xi)+b(\xi) c(\xi))=0
$$
As can be seen, $f(X)=f(Y)=f(Z)=f(W)=0$ follows from (1) and (4), $f(P)=0$ follows from (5), then $f(O)=s \cdot f(P)=0$ follows from $(3 a-3 c)$. So, the points $X, Y, Z, W, O, P$ all satisfy the equation $f(\xi)=0$.
Notice that $f(B)=u \cdot c(B) d(B)$ and $f(A)=w \cdot b(A) c(A)$; since at least one of $u$ and $w$ is nonzero, either $A$ or $B$ does not satisfy $(*)$. Therefore, the equation cannot degenerate to an identity.
Hence, the equation $f(\xi)=0$ is at most quadratic, it is not an identity, but satisfied by $X, Y, Z, W, O, P$, so these six points lie on a conic section.
|
{
"exam": "EGMO",
"problem_label": "6",
"problem_match": "# Problem 6.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "\nSolution 2. ",
"tier": "T2",
"year": "2022"
}
|
Let $A B C D$ be a cyclic quadrilateral with circumcentre $O$. Let the internal angle bisectors at $A$ and $B$ meet at $X$, the internal angle bisectors at $B$ and $C$ meet at $Y$, the internal angle bisectors at $C$ and $D$ meet at $Z$, and the internal angle bisectors at $D$ and $A$ meet at $W$. Further, let $A C$ and $B D$ meet at $P$. Suppose that the points $X, Y, Z, W, O$ and $P$ are distinct.
Prove that $O, X, Y, Z$ and $W$ lie on the same circle if and only if $P, X, Y, Z$ and $W$ lie on the same circle.
Proposed by: Ethan Tan, Australia
|
Let $A^{\prime}, B^{\prime}, C^{\prime}$ and $D^{\prime}$ be the second intersection of the angle bisectors of $\angle D A C$, $\angle A B C, \angle B C D$ and $\angle C D A$ with the circumcircle of $A B C D$, respectively. Notice that $A^{\prime}$ and $C^{\prime}$ are the midpoints of the two arcs of $B D$ and $B^{\prime}$ and $D^{\prime}$ are the midpoints of arcs corresponding to $A C$ so the lines $A^{\prime} C^{\prime}$ and $B^{\prime} D^{\prime}$ intersect in $O$.
We will prove that the points $X, Y, Z, W, P$ and $O$ lie on a conic section. This is a more general statement because five points uniquely determine a conic sections so if $X, Y, Z, W$ and $P$ lie on a circle then this circle is the conic section so $O$ also lies on this circle. In the same way if $X, Y, Z$, $W$ and $O$ are on a circle then it also contains $P$.
Let $E$ be the intersection of the lines $A^{\prime} B$ and $C D^{\prime}$. Let $F$ be the intersection of the lines $A^{\prime} D$ and $B^{\prime} C$. Finally, let $G$ be the intesection of $X Y$ and $Z W$.
Now we use Pascal's theorem five times:
|
proof
|
Yes
|
Incomplete
|
proof
|
Geometry
|
Let $A B C D$ be a cyclic quadrilateral with circumcentre $O$. Let the internal angle bisectors at $A$ and $B$ meet at $X$, the internal angle bisectors at $B$ and $C$ meet at $Y$, the internal angle bisectors at $C$ and $D$ meet at $Z$, and the internal angle bisectors at $D$ and $A$ meet at $W$. Further, let $A C$ and $B D$ meet at $P$. Suppose that the points $X, Y, Z, W, O$ and $P$ are distinct.
Prove that $O, X, Y, Z$ and $W$ lie on the same circle if and only if $P, X, Y, Z$ and $W$ lie on the same circle.
Proposed by: Ethan Tan, Australia
|
Let $A^{\prime}, B^{\prime}, C^{\prime}$ and $D^{\prime}$ be the second intersection of the angle bisectors of $\angle D A C$, $\angle A B C, \angle B C D$ and $\angle C D A$ with the circumcircle of $A B C D$, respectively. Notice that $A^{\prime}$ and $C^{\prime}$ are the midpoints of the two arcs of $B D$ and $B^{\prime}$ and $D^{\prime}$ are the midpoints of arcs corresponding to $A C$ so the lines $A^{\prime} C^{\prime}$ and $B^{\prime} D^{\prime}$ intersect in $O$.
We will prove that the points $X, Y, Z, W, P$ and $O$ lie on a conic section. This is a more general statement because five points uniquely determine a conic sections so if $X, Y, Z, W$ and $P$ lie on a circle then this circle is the conic section so $O$ also lies on this circle. In the same way if $X, Y, Z$, $W$ and $O$ are on a circle then it also contains $P$.
Let $E$ be the intersection of the lines $A^{\prime} B$ and $C D^{\prime}$. Let $F$ be the intersection of the lines $A^{\prime} D$ and $B^{\prime} C$. Finally, let $G$ be the intesection of $X Y$ and $Z W$.
Now we use Pascal's theorem five times:
|
{
"exam": "EGMO",
"problem_label": "6",
"problem_match": "# Problem 6.",
"resource_path": "EGMO/segmented/en-2022-solutions.jsonl",
"solution_match": "\nSolution 3. ",
"tier": "T2",
"year": "2022"
}
|
There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$.
Prove that $a_{1}=a_{2}=\cdots=a_{n}$.
|
Suppose that not all $a_{i}$ are equal. Consider an index $i$ such that $a_{i}$ is maximal and $a_{i+1}<a_{i}$. Then
$$
b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}<\frac{2 a_{i}}{a_{i}}=2 .
$$
But since $a_{i}$ is maximal, $b_{i}$ is also maximal, so we must have $b_{j}<2$ for all $j \in\{1,2, \ldots, n\}$. However, consider the product $b_{1} b_{2} \ldots b_{n}$. We have
$$
\begin{aligned}
b_{1} b_{2} \ldots b_{n} & =\frac{a_{n}+a_{2}}{a_{1}} \cdot \frac{a_{1}+a_{3}}{a_{2}} \cdot \ldots \cdot \frac{a_{n-1}+a_{1}}{a_{n}} \\
& \geqslant 2^{n} \frac{\sqrt{a_{n} a_{2}} \sqrt{a_{1} a_{3}} \ldots \sqrt{a_{n-1} a_{1}}}{a_{1} a_{2} \ldots a_{n}} \\
& =2^{n}
\end{aligned}
$$
where we used the inequality $x+y \geqslant 2 \sqrt{x y}$ for $x=a_{i-1}, y=a_{i+1}$ for all $i \in\{1,2, \ldots, n\}$ in the second row.
Since the product of all $b_{i}$ is at least $2^{n}$, at least one of them must be greater than 2 , which is a contradiction with the previous conclusion.
Thus, all $a_{i}$ must be equal.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$.
Prove that $a_{1}=a_{2}=\cdots=a_{n}$.
|
Suppose that not all $a_{i}$ are equal. Consider an index $i$ such that $a_{i}$ is maximal and $a_{i+1}<a_{i}$. Then
$$
b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}<\frac{2 a_{i}}{a_{i}}=2 .
$$
But since $a_{i}$ is maximal, $b_{i}$ is also maximal, so we must have $b_{j}<2$ for all $j \in\{1,2, \ldots, n\}$. However, consider the product $b_{1} b_{2} \ldots b_{n}$. We have
$$
\begin{aligned}
b_{1} b_{2} \ldots b_{n} & =\frac{a_{n}+a_{2}}{a_{1}} \cdot \frac{a_{1}+a_{3}}{a_{2}} \cdot \ldots \cdot \frac{a_{n-1}+a_{1}}{a_{n}} \\
& \geqslant 2^{n} \frac{\sqrt{a_{n} a_{2}} \sqrt{a_{1} a_{3}} \ldots \sqrt{a_{n-1} a_{1}}}{a_{1} a_{2} \ldots a_{n}} \\
& =2^{n}
\end{aligned}
$$
where we used the inequality $x+y \geqslant 2 \sqrt{x y}$ for $x=a_{i-1}, y=a_{i+1}$ for all $i \in\{1,2, \ldots, n\}$ in the second row.
Since the product of all $b_{i}$ is at least $2^{n}$, at least one of them must be greater than 2 , which is a contradiction with the previous conclusion.
Thus, all $a_{i}$ must be equal.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "EGMO/segmented/en-2023-solutions.jsonl",
"solution_match": "\nSolution 1. ",
"tier": "T2",
"year": "2023"
}
|
There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$.
Prove that $a_{1}=a_{2}=\cdots=a_{n}$.
|
This is a version of Solution 1 without use of proof by contradiction.
Taking $a_{i}$ such that it is maximal among $a_{1}, \ldots, a_{n}$, we obtain $b_{i} \leqslant 2$. Thus $b_{i} \leqslant 2$ for all $j \in\{1,2, \ldots, n\}$.
The second part of Solution 1 then gives $2^{n} \geqslant b_{1} \cdots b_{n} \geqslant 2^{n}$, which together with $b_{i} \leqslant 2$ for all $j \in\{1,2, \ldots, n\}$ implies that $b_{j}=2$ for all $j \in\{1,2, \ldots, n\}$. Since we have $b_{1}=$ $b_{2}=\cdots=b_{n}$, the condition that $a_{i} \leqslant a_{j} \Longleftrightarrow b_{i} \leqslant b_{j}$ gives that $a_{1}=a_{2}=\cdots=a_{n}$.
|
a_{1}=a_{2}=\cdots=a_{n}
|
Yes
|
Yes
|
proof
|
Algebra
|
There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$.
Prove that $a_{1}=a_{2}=\cdots=a_{n}$.
|
This is a version of Solution 1 without use of proof by contradiction.
Taking $a_{i}$ such that it is maximal among $a_{1}, \ldots, a_{n}$, we obtain $b_{i} \leqslant 2$. Thus $b_{i} \leqslant 2$ for all $j \in\{1,2, \ldots, n\}$.
The second part of Solution 1 then gives $2^{n} \geqslant b_{1} \cdots b_{n} \geqslant 2^{n}$, which together with $b_{i} \leqslant 2$ for all $j \in\{1,2, \ldots, n\}$ implies that $b_{j}=2$ for all $j \in\{1,2, \ldots, n\}$. Since we have $b_{1}=$ $b_{2}=\cdots=b_{n}$, the condition that $a_{i} \leqslant a_{j} \Longleftrightarrow b_{i} \leqslant b_{j}$ gives that $a_{1}=a_{2}=\cdots=a_{n}$.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "EGMO/segmented/en-2023-solutions.jsonl",
"solution_match": "\nSolution 2. ",
"tier": "T2",
"year": "2023"
}
|
There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$.
Prove that $a_{1}=a_{2}=\cdots=a_{n}$.
|
We first show that $b_{j} \leqslant 2$ for all $j$ as in Solution 2. Then
$$
\begin{aligned}
2 n & \geqslant b_{1}+\cdots+b_{n}=\frac{a_{n}}{a_{1}}+\frac{a_{2}}{a_{1}}+\frac{a_{1}}{a_{2}}+\frac{a_{3}}{a_{2}}+\cdots+\frac{a_{n-1}}{a_{n}}+\frac{a_{1}}{a_{n}} \\
& \geqslant 2 n \sqrt[n]{\frac{a_{n}}{a_{1}} \cdot \frac{a_{2}}{a_{1}} \cdot \frac{a_{1}}{a_{2}} \cdot \frac{a_{3}}{a_{2}} \cdots \frac{a_{n-1}}{a_{n}} \cdot \frac{a_{1}}{a_{n}}=2 n \cdot 1=2 n,}
\end{aligned}
$$
where we used the AM-GM inequality.
It follows that all $b_{j}$ 's are equal which as in Solution 2 gives $a_{1}=a_{2}=\cdots=a_{n}$.
|
a_{1}=a_{2}=\cdots=a_{n}
|
Yes
|
Yes
|
proof
|
Algebra
|
There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$.
Prove that $a_{1}=a_{2}=\cdots=a_{n}$.
|
We first show that $b_{j} \leqslant 2$ for all $j$ as in Solution 2. Then
$$
\begin{aligned}
2 n & \geqslant b_{1}+\cdots+b_{n}=\frac{a_{n}}{a_{1}}+\frac{a_{2}}{a_{1}}+\frac{a_{1}}{a_{2}}+\frac{a_{3}}{a_{2}}+\cdots+\frac{a_{n-1}}{a_{n}}+\frac{a_{1}}{a_{n}} \\
& \geqslant 2 n \sqrt[n]{\frac{a_{n}}{a_{1}} \cdot \frac{a_{2}}{a_{1}} \cdot \frac{a_{1}}{a_{2}} \cdot \frac{a_{3}}{a_{2}} \cdots \frac{a_{n-1}}{a_{n}} \cdot \frac{a_{1}}{a_{n}}=2 n \cdot 1=2 n,}
\end{aligned}
$$
where we used the AM-GM inequality.
It follows that all $b_{j}$ 's are equal which as in Solution 2 gives $a_{1}=a_{2}=\cdots=a_{n}$.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "EGMO/segmented/en-2023-solutions.jsonl",
"solution_match": "\nSolution 3. ",
"tier": "T2",
"year": "2023"
}
|
There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$.
Prove that $a_{1}=a_{2}=\cdots=a_{n}$.
|
By assumption $a_{i} b_{i}=a_{i-1}+a_{i+1}$ for $i=\{1,2, \ldots, n\}$, hence,
$$
\sum_{i=1}^{n} a_{i} b_{i}=2 \sum_{i=1}^{n} a_{i} .
$$
Since $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$, the Chebyshev's inequality implies
$$
\left(\sum_{i=1}^{n} a_{i}\right) \cdot\left(\sum_{i=1}^{n} b_{i}\right) \leqslant n \cdot \sum_{i=1}^{n} a_{i} b_{i}=2 n \cdot \sum_{i=1}^{n} a_{i}
$$
and so $\sum_{i=1}^{n} b_{i} \leqslant 2 n$. On the other hand, we have
$$
\sum_{i=1}^{n} b_{i}=\sum_{i=1}^{n} \frac{a_{i-1}}{a_{i}}+\sum_{i=1}^{n} \frac{a_{i+1}}{a_{i}}=\sum_{i=1}^{n} \frac{a_{i-1}}{a_{i}}+\sum_{i=1}^{n} \frac{a_{i}}{a_{i-1}}=\sum_{i=1}^{n}\left(\frac{a_{i-1}}{a_{i}}+\frac{a_{i}}{a_{i-1}}\right)
$$
so we can use the AM-GM inequality to estimate
$$
\sum_{i=1}^{n} b_{i} \geqslant \sum_{i=1}^{n} 2 \sqrt{\frac{a_{i-1}}{a_{i}} \cdot \frac{a_{i}}{a_{i-1}}}=2 n
$$
We conclude that we must have equalities in all the above, which implies $\frac{a_{i-1}}{a_{i}}=\frac{a_{i}}{a_{i-1}}$ and consequently $a_{i}=a_{i-1}$ for all positive integers $i$. Hence, all $a$ 's are equal.
|
a_{1}=a_{2}=\cdots=a_{n}
|
Yes
|
Yes
|
proof
|
Algebra
|
There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$.
Prove that $a_{1}=a_{2}=\cdots=a_{n}$.
|
By assumption $a_{i} b_{i}=a_{i-1}+a_{i+1}$ for $i=\{1,2, \ldots, n\}$, hence,
$$
\sum_{i=1}^{n} a_{i} b_{i}=2 \sum_{i=1}^{n} a_{i} .
$$
Since $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$, the Chebyshev's inequality implies
$$
\left(\sum_{i=1}^{n} a_{i}\right) \cdot\left(\sum_{i=1}^{n} b_{i}\right) \leqslant n \cdot \sum_{i=1}^{n} a_{i} b_{i}=2 n \cdot \sum_{i=1}^{n} a_{i}
$$
and so $\sum_{i=1}^{n} b_{i} \leqslant 2 n$. On the other hand, we have
$$
\sum_{i=1}^{n} b_{i}=\sum_{i=1}^{n} \frac{a_{i-1}}{a_{i}}+\sum_{i=1}^{n} \frac{a_{i+1}}{a_{i}}=\sum_{i=1}^{n} \frac{a_{i-1}}{a_{i}}+\sum_{i=1}^{n} \frac{a_{i}}{a_{i-1}}=\sum_{i=1}^{n}\left(\frac{a_{i-1}}{a_{i}}+\frac{a_{i}}{a_{i-1}}\right)
$$
so we can use the AM-GM inequality to estimate
$$
\sum_{i=1}^{n} b_{i} \geqslant \sum_{i=1}^{n} 2 \sqrt{\frac{a_{i-1}}{a_{i}} \cdot \frac{a_{i}}{a_{i-1}}}=2 n
$$
We conclude that we must have equalities in all the above, which implies $\frac{a_{i-1}}{a_{i}}=\frac{a_{i}}{a_{i-1}}$ and consequently $a_{i}=a_{i-1}$ for all positive integers $i$. Hence, all $a$ 's are equal.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "EGMO/segmented/en-2023-solutions.jsonl",
"solution_match": "\nSolution 4. ",
"tier": "T2",
"year": "2023"
}
|
There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$.
Prove that $a_{1}=a_{2}=\cdots=a_{n}$.
|
As in Solution 4 we show that $\sum_{i=1}^{n} b_{i} \leqslant 2 n$ and as in Solution 1 we show that $\prod_{i=1}^{n} b_{i} \geqslant 2^{n}$. We now use the AM-GM inequality and the first inequality to get
$$
\prod_{i=1}^{n} b_{i} \leqslant\left(\frac{1}{n} \sum_{i=1}^{n} b_{i}\right)^{n} \leqslant\left(\frac{1}{n} \cdot 2 n\right)^{n}=2^{n}
$$
This implies that we must have equalities in all the above. In particular, we have equality in the AM-GM inequality, so all $b$ 's are equal and as in Solution 2 then all $a$ 's are equal.
|
a_{1}=a_{2}=\cdots=a_{n}
|
Yes
|
Yes
|
proof
|
Algebra
|
There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$.
Prove that $a_{1}=a_{2}=\cdots=a_{n}$.
|
As in Solution 4 we show that $\sum_{i=1}^{n} b_{i} \leqslant 2 n$ and as in Solution 1 we show that $\prod_{i=1}^{n} b_{i} \geqslant 2^{n}$. We now use the AM-GM inequality and the first inequality to get
$$
\prod_{i=1}^{n} b_{i} \leqslant\left(\frac{1}{n} \sum_{i=1}^{n} b_{i}\right)^{n} \leqslant\left(\frac{1}{n} \cdot 2 n\right)^{n}=2^{n}
$$
This implies that we must have equalities in all the above. In particular, we have equality in the AM-GM inequality, so all $b$ 's are equal and as in Solution 2 then all $a$ 's are equal.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "EGMO/segmented/en-2023-solutions.jsonl",
"solution_match": "\nSolution 5. ",
"tier": "T2",
"year": "2023"
}
|
There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$.
Prove that $a_{1}=a_{2}=\cdots=a_{n}$.
|
Let $a_{i}$ to be minimal and $a_{j}$ maximal among all $a$ 's. Then
$$
b_{j}=\frac{a_{j-1}+a_{j+1}}{a_{j}} \leqslant \frac{2 a_{j}}{a_{j}}=2=\frac{2 a_{i}}{a_{i}} \leqslant \frac{a_{i-1}+a_{i+1}}{a_{i}}=b_{i}
$$
and by assumption $b_{i} \leqslant b_{j}$. Hence, we have equalities in the above so $b_{j}=2$ so $a_{j-1}+$ $a_{j+1}=2 a_{j}$ and therefore $a_{j-1}=a_{j}=a_{j+1}$. We have thus shown that the two neighbors of a maximal $a$ are also maximal. By an inductive argument all $a$ 's are maximal, hence equal.
|
a_{1}=a_{2}=\cdots=a_{n}
|
Yes
|
Yes
|
proof
|
Algebra
|
There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$.
Prove that $a_{1}=a_{2}=\cdots=a_{n}$.
|
Let $a_{i}$ to be minimal and $a_{j}$ maximal among all $a$ 's. Then
$$
b_{j}=\frac{a_{j-1}+a_{j+1}}{a_{j}} \leqslant \frac{2 a_{j}}{a_{j}}=2=\frac{2 a_{i}}{a_{i}} \leqslant \frac{a_{i-1}+a_{i+1}}{a_{i}}=b_{i}
$$
and by assumption $b_{i} \leqslant b_{j}$. Hence, we have equalities in the above so $b_{j}=2$ so $a_{j-1}+$ $a_{j+1}=2 a_{j}$ and therefore $a_{j-1}=a_{j}=a_{j+1}$. We have thus shown that the two neighbors of a maximal $a$ are also maximal. By an inductive argument all $a$ 's are maximal, hence equal.
|
{
"exam": "EGMO",
"problem_label": "1",
"problem_match": "\nProblem 1.",
"resource_path": "EGMO/segmented/en-2023-solutions.jsonl",
"solution_match": "\nSolution 6. ",
"tier": "T2",
"year": "2023"
}
|
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