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The inscribed circle of the triangle $A_{1} A_{2} A_{3}$ touches the sides $A_{2} A_{3}, A_{3} A_{1}$ and $A_{1} A_{2}$ at points $S_{1}, S_{2}, S_{3}$, respectively. Let $O_{1}, O_{2}, O_{3}$ be the centres of the inscribed circles of triangles $A_{1} S_{2} S_{3}, A_{2} S_{3} S_{1}$ and $A_{3} S_{1} S_{2}$, respectively. Prove that the straight lines $O_{1} S_{1}, O_{2} S_{2}$ and $O_{3} S_{3}$ intersect at one point.
|
We shall prove that the lines $S_{1} O_{1}, S_{2} O_{2}, S_{3} O_{3}$ are the bisectors of the angles of the triangle $S_{1} S_{2} S_{3}$. Let $O$ and $r$ be the centre and radius of the inscribed circle $C$ of the triangle $A_{1} A_{2} A_{3}$. Further, let $P_{1}$ and $H_{1}$ be the points where the inscribed circle of the triangle $A_{1} S_{2} S_{3}$ (with the centre $O_{1}$ and radius $r_{1}$ ) touches its sides $A_{1} S_{2}$ and $S_{2} S_{3}$, respectively (see Figure 2). To show that $S_{1} O_{1}$ is the bisector of the angle $\angle S_{3} S_{1} S_{2}$ it is sufficient to prove that $O_{1}$ lies on the circumference of circle $C$, for in this case the arcs $O_{1} S_{2}$ and $O_{1} S_{3}$ will obviously be equal. To prove this, first note that as $A_{1} S_{2} S_{3}$ is an isosceles triangle the point $H_{1}$, as well as $O_{1}$, lies on the straight line $A_{1} O$. Now, it suffices to show that $\left|O H_{1}\right|=r-r_{1}$. Indeed, we have
$$
\begin{aligned}
& \frac{r-r_{1}}{r}=1-\frac{r_{1}}{r}=1-\frac{\left|O_{1} P_{1}\right|}{\left|O S_{2}\right|}=1-\frac{\left|P_{1} A_{1}\right|}{\left|S_{2} A_{1}\right|}=\frac{\left|S_{2} A_{1}\right|-\left|P_{1} A_{1}\right|}{\left|S_{2} A_{1}\right|} \\
& =\frac{\left|S_{2} P_{1}\right|}{\left|S_{2} A_{1}\right|}=\frac{\left|S_{2} H_{1}\right|}{\left|S_{2} A_{1}\right|}=\frac{\left|O H_{1}\right|}{\left|O S_{2}\right|}=\frac{\left|O H_{1}\right|}{r} .
\end{aligned}
$$
Figure 2
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
The inscribed circle of the triangle $A_{1} A_{2} A_{3}$ touches the sides $A_{2} A_{3}, A_{3} A_{1}$ and $A_{1} A_{2}$ at points $S_{1}, S_{2}, S_{3}$, respectively. Let $O_{1}, O_{2}, O_{3}$ be the centres of the inscribed circles of triangles $A_{1} S_{2} S_{3}, A_{2} S_{3} S_{1}$ and $A_{3} S_{1} S_{2}$, respectively. Prove that the straight lines $O_{1} S_{1}, O_{2} S_{2}$ and $O_{3} S_{3}$ intersect at one point.
|
We shall prove that the lines $S_{1} O_{1}, S_{2} O_{2}, S_{3} O_{3}$ are the bisectors of the angles of the triangle $S_{1} S_{2} S_{3}$. Let $O$ and $r$ be the centre and radius of the inscribed circle $C$ of the triangle $A_{1} A_{2} A_{3}$. Further, let $P_{1}$ and $H_{1}$ be the points where the inscribed circle of the triangle $A_{1} S_{2} S_{3}$ (with the centre $O_{1}$ and radius $r_{1}$ ) touches its sides $A_{1} S_{2}$ and $S_{2} S_{3}$, respectively (see Figure 2). To show that $S_{1} O_{1}$ is the bisector of the angle $\angle S_{3} S_{1} S_{2}$ it is sufficient to prove that $O_{1}$ lies on the circumference of circle $C$, for in this case the arcs $O_{1} S_{2}$ and $O_{1} S_{3}$ will obviously be equal. To prove this, first note that as $A_{1} S_{2} S_{3}$ is an isosceles triangle the point $H_{1}$, as well as $O_{1}$, lies on the straight line $A_{1} O$. Now, it suffices to show that $\left|O H_{1}\right|=r-r_{1}$. Indeed, we have
$$
\begin{aligned}
& \frac{r-r_{1}}{r}=1-\frac{r_{1}}{r}=1-\frac{\left|O_{1} P_{1}\right|}{\left|O S_{2}\right|}=1-\frac{\left|P_{1} A_{1}\right|}{\left|S_{2} A_{1}\right|}=\frac{\left|S_{2} A_{1}\right|-\left|P_{1} A_{1}\right|}{\left|S_{2} A_{1}\right|} \\
& =\frac{\left|S_{2} P_{1}\right|}{\left|S_{2} A_{1}\right|}=\frac{\left|S_{2} H_{1}\right|}{\left|S_{2} A_{1}\right|}=\frac{\left|O H_{1}\right|}{\left|O S_{2}\right|}=\frac{\left|O H_{1}\right|}{r} .
\end{aligned}
$$
Figure 2
|
{
"exam": "BalticWay",
"problem_label": "12",
"problem_match": "\n12.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1994"
}
|
Find the smallest number $a$ such that a square of side $a$ can contain five disks of radius 1 so that no two of the disks have a common interior point.
|
Let $P Q R S$ be a square which has the property described in the problem. Clearly, $a>2$. Let $P^{\prime} Q^{\prime} R^{\prime} S^{\prime}$ be the square inside $P Q R S$ whose sides are at distance 1 from the sides of $P Q R S$, and, consequently, are of length $a-2$. Since all the five disks are inside $P Q R S$, their centres are inside $P^{\prime} Q^{\prime} R^{\prime} S^{\prime}$. Divide $P^{\prime} Q^{\prime} R^{\prime} S^{\prime}$ into four congruent squares of side length $\frac{a}{2}-1$. By the pigeonhole principle, at least two of the five centres are in the same small square. Their distance, then, is at most $\sqrt{2}\left(\frac{a}{2}-1\right)$. Since the distance has to be at least 2 , we have $a \geq 2+2 \sqrt{2}$. On the other hand, if $a=2+2 \sqrt{2}$, we can place the five disks in such a way that one is centred at the centre of $P Q R S$ and the other four have centres at $P^{\prime}, Q^{\prime}$, $R^{\prime}$ and $S^{\prime}$.
|
2+2\sqrt{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Find the smallest number $a$ such that a square of side $a$ can contain five disks of radius 1 so that no two of the disks have a common interior point.
|
Let $P Q R S$ be a square which has the property described in the problem. Clearly, $a>2$. Let $P^{\prime} Q^{\prime} R^{\prime} S^{\prime}$ be the square inside $P Q R S$ whose sides are at distance 1 from the sides of $P Q R S$, and, consequently, are of length $a-2$. Since all the five disks are inside $P Q R S$, their centres are inside $P^{\prime} Q^{\prime} R^{\prime} S^{\prime}$. Divide $P^{\prime} Q^{\prime} R^{\prime} S^{\prime}$ into four congruent squares of side length $\frac{a}{2}-1$. By the pigeonhole principle, at least two of the five centres are in the same small square. Their distance, then, is at most $\sqrt{2}\left(\frac{a}{2}-1\right)$. Since the distance has to be at least 2 , we have $a \geq 2+2 \sqrt{2}$. On the other hand, if $a=2+2 \sqrt{2}$, we can place the five disks in such a way that one is centred at the centre of $P Q R S$ and the other four have centres at $P^{\prime}, Q^{\prime}$, $R^{\prime}$ and $S^{\prime}$.
|
{
"exam": "BalticWay",
"problem_label": "13",
"problem_match": "\n13.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1994"
}
|
Let $\alpha, \beta, \gamma$ be the angles of a triangle opposite to its sides with lengths $a, b$ and $c$, respectively. Prove the inequality
$$
a \cdot\left(\frac{1}{\beta}+\frac{1}{\gamma}\right)+b \cdot\left(\frac{1}{\gamma}+\frac{1}{\alpha}\right)+c \cdot\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \geq 2 \cdot\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right) \cdot
$$
|
Clearly, the inequality $a>b$ implies $\alpha>\beta$ and similarly $a<b$ implies $\alpha<\beta$, hence $(a-b)(\alpha-\beta) \geq$ 0 and $a \alpha+b \beta \geq a \beta+b \alpha$. Dividing the last equality by $\alpha \beta$ we get
$$
\frac{a}{\beta}+\frac{b}{\alpha} \geq \frac{a}{\alpha}+\frac{b}{\beta}
$$
Similarly we get
$$
\frac{a}{\gamma}+\frac{c}{\alpha} \geq \frac{a}{\alpha}+\frac{c}{\gamma}
$$
and
$$
\frac{b}{\gamma}+\frac{c}{\beta} \geq \frac{b}{\beta}+\frac{c}{\gamma}
$$
To finish the proof it suffices to add the inequalities (6)-(8).
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $\alpha, \beta, \gamma$ be the angles of a triangle opposite to its sides with lengths $a, b$ and $c$, respectively. Prove the inequality
$$
a \cdot\left(\frac{1}{\beta}+\frac{1}{\gamma}\right)+b \cdot\left(\frac{1}{\gamma}+\frac{1}{\alpha}\right)+c \cdot\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \geq 2 \cdot\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right) \cdot
$$
|
Clearly, the inequality $a>b$ implies $\alpha>\beta$ and similarly $a<b$ implies $\alpha<\beta$, hence $(a-b)(\alpha-\beta) \geq$ 0 and $a \alpha+b \beta \geq a \beta+b \alpha$. Dividing the last equality by $\alpha \beta$ we get
$$
\frac{a}{\beta}+\frac{b}{\alpha} \geq \frac{a}{\alpha}+\frac{b}{\beta}
$$
Similarly we get
$$
\frac{a}{\gamma}+\frac{c}{\alpha} \geq \frac{a}{\alpha}+\frac{c}{\gamma}
$$
and
$$
\frac{b}{\gamma}+\frac{c}{\beta} \geq \frac{b}{\beta}+\frac{c}{\gamma}
$$
To finish the proof it suffices to add the inequalities (6)-(8).
|
{
"exam": "BalticWay",
"problem_label": "14",
"problem_match": "\n14.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1994"
}
|
Does there exist a triangle such that the lengths of all its sides and altitudes are integers and its perimeter is equal to 1995 ?
|
Consider a triangle $A B C$ with all its sides and heights having integer lengths. From the cosine theorem we conclude that $\cos \angle A, \cos \angle B$ and $\cos \angle C$ are rational numbers. Let $A H$ be one of the heights of the triangle $A B C$, with the point $H$ lying on the straight line determined by the side $B C$. Then $|B H|$ and $|C H|$ must be rational and hence integer (consider the Pythagorean theorem for the triangles $A B H$ and $A C H$ ). Now, if $|B H|$ and $|C H|$ have different parity then $|A B|$ and $|A C|$ also have different parity and $|B C|$ is odd. If $|B H|$ and $|C H|$ have the same parity then $|A B|$ and $|A C|$ also have the same parity and $|B C|$ is even. In both cases the perimeter of triangle $A B C$ is an even number and hence cannot be equal to 1995
Remark. In the solution we only used the fact that all three sides and one height of the triangle $A B C$ are integers.
Figure 3
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Does there exist a triangle such that the lengths of all its sides and altitudes are integers and its perimeter is equal to 1995 ?
|
Consider a triangle $A B C$ with all its sides and heights having integer lengths. From the cosine theorem we conclude that $\cos \angle A, \cos \angle B$ and $\cos \angle C$ are rational numbers. Let $A H$ be one of the heights of the triangle $A B C$, with the point $H$ lying on the straight line determined by the side $B C$. Then $|B H|$ and $|C H|$ must be rational and hence integer (consider the Pythagorean theorem for the triangles $A B H$ and $A C H$ ). Now, if $|B H|$ and $|C H|$ have different parity then $|A B|$ and $|A C|$ also have different parity and $|B C|$ is odd. If $|B H|$ and $|C H|$ have the same parity then $|A B|$ and $|A C|$ also have the same parity and $|B C|$ is even. In both cases the perimeter of triangle $A B C$ is an even number and hence cannot be equal to 1995
Remark. In the solution we only used the fact that all three sides and one height of the triangle $A B C$ are integers.
Figure 3
|
{
"exam": "BalticWay",
"problem_label": "15",
"problem_match": "\n15.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1994"
}
|
The Wonder Island is inhabited by Hedgehogs. Each Hedgehog consists of three segments of unit length having a common endpoint, with all three angles between them equal to $120^{\circ}$ (see Figure 3). Given that all Hedgehogs are lying flat on the island and no two of them touch each other, prove that there is a finite number of Hedgehogs on Wonder Island.
|
It suffices to prove that if the distance between the centres of two Hedgehogs is less than 0.2, then these Hedgehogs intersect. To show this, consider two Hedgehogs with their centres at points $O$ and $M$, respectively, such that $|O M|<0.2$. Let $A, B$ and $C$ be the endpoints of the needles of the first Hedgehog (see Figure 4) and draw a straight line $l$ parallel to $A C$ through the point $M$. As $|A C|=\sqrt{3}$ implies $|K L| \leq$ $\frac{0.2}{0.5}|A C|<1$ and the second Hedgehog has at least one of its needles pointing inside the triangle $O K L$, this needle intersects the first Hedgehog.
Remark. If the Hedgehogs can move their needles so that the angles between them can take any positive value then there can be an infinite number of Hedgehogs on the Wonder Island.
Figure 4
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
The Wonder Island is inhabited by Hedgehogs. Each Hedgehog consists of three segments of unit length having a common endpoint, with all three angles between them equal to $120^{\circ}$ (see Figure 3). Given that all Hedgehogs are lying flat on the island and no two of them touch each other, prove that there is a finite number of Hedgehogs on Wonder Island.
|
It suffices to prove that if the distance between the centres of two Hedgehogs is less than 0.2, then these Hedgehogs intersect. To show this, consider two Hedgehogs with their centres at points $O$ and $M$, respectively, such that $|O M|<0.2$. Let $A, B$ and $C$ be the endpoints of the needles of the first Hedgehog (see Figure 4) and draw a straight line $l$ parallel to $A C$ through the point $M$. As $|A C|=\sqrt{3}$ implies $|K L| \leq$ $\frac{0.2}{0.5}|A C|<1$ and the second Hedgehog has at least one of its needles pointing inside the triangle $O K L$, this needle intersects the first Hedgehog.
Remark. If the Hedgehogs can move their needles so that the angles between them can take any positive value then there can be an infinite number of Hedgehogs on the Wonder Island.
Figure 4
|
{
"exam": "BalticWay",
"problem_label": "16",
"problem_match": "\n16.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1994"
}
|
In a certain kingdom, the king has decided to build 25 new towns on 13 uninhabited islands so that on each island there will be at least one town. Direct ferry connections will be established between any pair of new towns which are on different islands. Determine the least possible number of these connections.
|
Let $a_{1}, \ldots, a_{13}$ be the numbers of towns on each island. Suppose there exist numbers $i$ and $j$ such that $a_{i} \geq a_{j}>1$ and consider an arbitrary town $A$ on the $j$-th island. The number of ferry connections from town $A$ is equal to $25-a_{j}$. On the other hand, if we "move" town $A$ to the $i$-th island then there will be $25-\left(a_{i}+1\right)$ connections from town $A$ while no other connections will be affected by this move. Hence, the smallest number of connections will be achieved if there are 13 towns on one island and one town on each of the other 12 islands. In this case there will be $13 \cdot 12+\frac{12 \cdot 11}{2}=222$ connections.
|
222
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In a certain kingdom, the king has decided to build 25 new towns on 13 uninhabited islands so that on each island there will be at least one town. Direct ferry connections will be established between any pair of new towns which are on different islands. Determine the least possible number of these connections.
|
Let $a_{1}, \ldots, a_{13}$ be the numbers of towns on each island. Suppose there exist numbers $i$ and $j$ such that $a_{i} \geq a_{j}>1$ and consider an arbitrary town $A$ on the $j$-th island. The number of ferry connections from town $A$ is equal to $25-a_{j}$. On the other hand, if we "move" town $A$ to the $i$-th island then there will be $25-\left(a_{i}+1\right)$ connections from town $A$ while no other connections will be affected by this move. Hence, the smallest number of connections will be achieved if there are 13 towns on one island and one town on each of the other 12 islands. In this case there will be $13 \cdot 12+\frac{12 \cdot 11}{2}=222$ connections.
|
{
"exam": "BalticWay",
"problem_label": "17",
"problem_match": "\n17.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1994"
}
|
There are $n$ lines $(n>2)$ given in the plane. No two of the lines are parallel and no three of them intersect at one point. Every point of intersection of these lines is labelled with a natural number between 1 and $n-1$. Prove that, if and only if $n$ is even, it is possible to assign the labels in such a way that every line has all the numbers from 1 to $n-1$ at its points of intersection with the other $n-1$ lines.
|
Suppose we have assigned the labels in the required manner. When a point has label 1 then there can be no more occurrences of label 1 on the two lines that intersect at that point. Therefore the number of intersection points labelled with 1 has to be exactly $\frac{n}{2}$, and so $n$ must be even. Now, let $n$ be an even number and denote the $n$ lines by $l_{1}, l_{2}, \ldots, l_{n}$. First write the lines $l_{i}$ in the following table:
$$
\begin{array}{llllll}
& & l_{3} & l_{4} & \ldots & l_{n / 2+1} \\
l_{1} & l_{2} & & & & \\
& & l_{n} & l_{n-1} & \ldots & l_{n / 2+2}
\end{array}
$$
and then rotate the picture $n-1$ times:
$$
\begin{array}{llllll}
& & l_{2} & l_{3} & \ldots & l_{n / 2} \\
l_{1} & l_{n} & l_{n-1} & l_{n-2} & \ldots & l_{n / 2+1} \\
& & & & & \\
& & l_{n} & l_{2} & \ldots & l_{n / 2-1} \\
l_{1} & l_{n-1} & & & & l_{n / 2}
\end{array}
$$
etc.
According to these tables, we can join the lines in pairs in $n-1$ different ways $-l_{1}$ with the line next to it and every other line with the line directly above or under it. Now we can assign the label $i$ to all the intersection points of the pairs of lines shown in the $i$ th table.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
There are $n$ lines $(n>2)$ given in the plane. No two of the lines are parallel and no three of them intersect at one point. Every point of intersection of these lines is labelled with a natural number between 1 and $n-1$. Prove that, if and only if $n$ is even, it is possible to assign the labels in such a way that every line has all the numbers from 1 to $n-1$ at its points of intersection with the other $n-1$ lines.
|
Suppose we have assigned the labels in the required manner. When a point has label 1 then there can be no more occurrences of label 1 on the two lines that intersect at that point. Therefore the number of intersection points labelled with 1 has to be exactly $\frac{n}{2}$, and so $n$ must be even. Now, let $n$ be an even number and denote the $n$ lines by $l_{1}, l_{2}, \ldots, l_{n}$. First write the lines $l_{i}$ in the following table:
$$
\begin{array}{llllll}
& & l_{3} & l_{4} & \ldots & l_{n / 2+1} \\
l_{1} & l_{2} & & & & \\
& & l_{n} & l_{n-1} & \ldots & l_{n / 2+2}
\end{array}
$$
and then rotate the picture $n-1$ times:
$$
\begin{array}{llllll}
& & l_{2} & l_{3} & \ldots & l_{n / 2} \\
l_{1} & l_{n} & l_{n-1} & l_{n-2} & \ldots & l_{n / 2+1} \\
& & & & & \\
& & l_{n} & l_{2} & \ldots & l_{n / 2-1} \\
l_{1} & l_{n-1} & & & & l_{n / 2}
\end{array}
$$
etc.
According to these tables, we can join the lines in pairs in $n-1$ different ways $-l_{1}$ with the line next to it and every other line with the line directly above or under it. Now we can assign the label $i$ to all the intersection points of the pairs of lines shown in the $i$ th table.
|
{
"exam": "BalticWay",
"problem_label": "18",
"problem_match": "\n18.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1994"
}
|
The Wonder Island Intelligence Service has 16 spies in Tartu. Each of them watches on some of his colleagues. It is known that if spy $A$ watches on spy $B$ then $B$ does not watch on $A$. Moreover, any 10 spies can be numbered in such a way that the first spy watches on the second, the second watches on the third, .., the tenth watches on the first. Prove that any 11 spies can also be numbered in a similar manner.
|
We call two spies $A$ and $B$ neutral to each other if neither $A$ watches on $B$ nor $B$ watches on $A$.
Denote the spies $A_{1}, A_{2}, \ldots, A_{16}$. Let $a_{i}, b_{i}$ and $c_{i}$ denote the number of spies that watch on $A_{i}$, the number of that are watched by $A_{i}$ and the number of spies neutral to $A_{i}$, respectively. Clearly, we have
$$
\begin{aligned}
a_{i}+b_{i}+c_{i} & =15, \\
a_{i}+c_{i} & \leq 8, \\
b_{i}+c_{i} & \leq 8
\end{aligned}
$$
for any $i=1, \ldots, 16$ (if any of the last two inequalities does not hold then there exist 10 spies who cannot be numbered in the required manner). Combining the relations above we find $c_{i} \leq 1$. Hence, for any spy, the number of his neutral colleagues is 0 or 1 .
Now suppose there is a group of 11 spies that cannot be numbered as required. Let $B$ be an arbitrary spy in this group. Number the other 10 spies as $C_{1}, C_{2}, \ldots, C_{10}$ so that $C_{1}$ watches on $C_{2}, \ldots, C_{10}$ watches on $C_{1}$. Suppose there is no spy neutral to $B$ among $C_{1}, \ldots, C_{10}$. Then, if $C_{1}$ watches on $B$ then $B$ cannot watch on $C_{2}$, as otherwise $C_{1}, B, C_{2}, \ldots, C_{10}$ would form an 11-cycle. So $C_{2}$ watches on $B$, etc. As some of the spies $C_{1}, C_{2}, \ldots, C_{10}$ must watch on $B$ we get all of them watching on $B$, a contradiction. Therefore, each of the 11 spies must have exactly one spy neutral to him among the other 10 - but this is impossible.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
The Wonder Island Intelligence Service has 16 spies in Tartu. Each of them watches on some of his colleagues. It is known that if spy $A$ watches on spy $B$ then $B$ does not watch on $A$. Moreover, any 10 spies can be numbered in such a way that the first spy watches on the second, the second watches on the third, .., the tenth watches on the first. Prove that any 11 spies can also be numbered in a similar manner.
|
We call two spies $A$ and $B$ neutral to each other if neither $A$ watches on $B$ nor $B$ watches on $A$.
Denote the spies $A_{1}, A_{2}, \ldots, A_{16}$. Let $a_{i}, b_{i}$ and $c_{i}$ denote the number of spies that watch on $A_{i}$, the number of that are watched by $A_{i}$ and the number of spies neutral to $A_{i}$, respectively. Clearly, we have
$$
\begin{aligned}
a_{i}+b_{i}+c_{i} & =15, \\
a_{i}+c_{i} & \leq 8, \\
b_{i}+c_{i} & \leq 8
\end{aligned}
$$
for any $i=1, \ldots, 16$ (if any of the last two inequalities does not hold then there exist 10 spies who cannot be numbered in the required manner). Combining the relations above we find $c_{i} \leq 1$. Hence, for any spy, the number of his neutral colleagues is 0 or 1 .
Now suppose there is a group of 11 spies that cannot be numbered as required. Let $B$ be an arbitrary spy in this group. Number the other 10 spies as $C_{1}, C_{2}, \ldots, C_{10}$ so that $C_{1}$ watches on $C_{2}, \ldots, C_{10}$ watches on $C_{1}$. Suppose there is no spy neutral to $B$ among $C_{1}, \ldots, C_{10}$. Then, if $C_{1}$ watches on $B$ then $B$ cannot watch on $C_{2}$, as otherwise $C_{1}, B, C_{2}, \ldots, C_{10}$ would form an 11-cycle. So $C_{2}$ watches on $B$, etc. As some of the spies $C_{1}, C_{2}, \ldots, C_{10}$ must watch on $B$ we get all of them watching on $B$, a contradiction. Therefore, each of the 11 spies must have exactly one spy neutral to him among the other 10 - but this is impossible.
|
{
"exam": "BalticWay",
"problem_label": "19",
"problem_match": "\n19.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1994"
}
|
An equilateral triangle is divided into 9000000 congruent equilateral triangles by lines parallel to its sides. Each vertex of the small triangles is coloured in one of three colours. Prove that there exist three points of the same colour being the vertices of a triangle with its sides parallel to the sides of the original triangle.
|
Consider the side $A B$ of the big triangle $A B C$ as "horizontal" and suppose the statement of the problem does not hold. The side $A B$ contains 3001 vertices $A=A_{0}, A_{1}, \ldots, A_{3000}=B$ of 3 colours. Hence, there are at least 1001 vertices of one colour, e.g., red. For any two red vertices $A_{k}$ and $A_{n}$ there exists a unique vertex $B_{k n}$ such that the triangle $B_{k n} A_{k} A_{n}$ is equilateral. That vertex $B_{k n}$ cannot be red. For different pairs $(k, n)$ the corresponding vertices $B_{k n}$ are different, so we have at least $\left(\begin{array}{c}1001 \\ 2\end{array}\right)>500000$ vertices of type $B_{k n}$ that cannot be red. As all these vertices are situated on 3000 horizontal lines, there exists a line $L$ which contains more than 160 vertices of type $B_{k n}$, each of them coloured in one of the two remaining colours. Hence there exist at least 81 vertices of the same colour, e.g., blue, on line $L$. For every two blue vertices $B_{k n}$ and $B_{m l}$ on line $L$ there exists a unique vertex $C_{k n m l}$ such that:
(i) $C_{k n m l}$ lies above the line $L$;
(ii) The triangle $C_{k n m l} B_{k n} B_{m l}$ is equilateral;
(iii) $C_{k n m l}=B_{p q}$ where $p=\min (k, m)$ and $q=\max (n, l)$.
Different pairs of vertices $B_{k n}$ belonging to line $L$ define different vertices $C_{k n m l}$. So we have at least $\left(\begin{array}{c}81 \\ 2\end{array}\right)>3200$ vertices of type $C_{k n m l}$ that can be neither blue nor red. As the number of these vertices exceeds the number of horizontal lines, there must be two vertices $C_{k n m l}$ and $C_{p q r s}$ on one horizontal line. Now, these two vertices define a new vertex $D_{\text {knmlpqrs }}$ that cannot have any of the three colours, a contradiction.
Remark. The minimal size of the big triangle that can be handled by this proof is 2557 .
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
An equilateral triangle is divided into 9000000 congruent equilateral triangles by lines parallel to its sides. Each vertex of the small triangles is coloured in one of three colours. Prove that there exist three points of the same colour being the vertices of a triangle with its sides parallel to the sides of the original triangle.
|
Consider the side $A B$ of the big triangle $A B C$ as "horizontal" and suppose the statement of the problem does not hold. The side $A B$ contains 3001 vertices $A=A_{0}, A_{1}, \ldots, A_{3000}=B$ of 3 colours. Hence, there are at least 1001 vertices of one colour, e.g., red. For any two red vertices $A_{k}$ and $A_{n}$ there exists a unique vertex $B_{k n}$ such that the triangle $B_{k n} A_{k} A_{n}$ is equilateral. That vertex $B_{k n}$ cannot be red. For different pairs $(k, n)$ the corresponding vertices $B_{k n}$ are different, so we have at least $\left(\begin{array}{c}1001 \\ 2\end{array}\right)>500000$ vertices of type $B_{k n}$ that cannot be red. As all these vertices are situated on 3000 horizontal lines, there exists a line $L$ which contains more than 160 vertices of type $B_{k n}$, each of them coloured in one of the two remaining colours. Hence there exist at least 81 vertices of the same colour, e.g., blue, on line $L$. For every two blue vertices $B_{k n}$ and $B_{m l}$ on line $L$ there exists a unique vertex $C_{k n m l}$ such that:
(i) $C_{k n m l}$ lies above the line $L$;
(ii) The triangle $C_{k n m l} B_{k n} B_{m l}$ is equilateral;
(iii) $C_{k n m l}=B_{p q}$ where $p=\min (k, m)$ and $q=\max (n, l)$.
Different pairs of vertices $B_{k n}$ belonging to line $L$ define different vertices $C_{k n m l}$. So we have at least $\left(\begin{array}{c}81 \\ 2\end{array}\right)>3200$ vertices of type $C_{k n m l}$ that can be neither blue nor red. As the number of these vertices exceeds the number of horizontal lines, there must be two vertices $C_{k n m l}$ and $C_{p q r s}$ on one horizontal line. Now, these two vertices define a new vertex $D_{\text {knmlpqrs }}$ that cannot have any of the three colours, a contradiction.
Remark. The minimal size of the big triangle that can be handled by this proof is 2557 .
|
{
"exam": "BalticWay",
"problem_label": "20",
"problem_match": "\n20.",
"resource_path": "BalticWay/segmented/en-bw94sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1994"
}
|
Find all triples $(x, y, z)$ of positive integers satisfying the system of equations
$$
\left\{\begin{array}{l}
x^{2}=2(y+z) \\
x^{6}=y^{6}+z^{6}+31\left(y^{2}+z^{2}\right)
\end{array}\right.
$$
|
From the first equation it follows that $x$ is even. The second equation implies $x>y$ and $x>z$. Hence $4 x>2(y+z)=x^{2}$, and therefore $x=2$ and $y+z=2$, so $y=z=1$. It is easy to check that the triple $(2,1,1)$ satisfies the given system of equations.
|
(2,1,1)
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all triples $(x, y, z)$ of positive integers satisfying the system of equations
$$
\left\{\begin{array}{l}
x^{2}=2(y+z) \\
x^{6}=y^{6}+z^{6}+31\left(y^{2}+z^{2}\right)
\end{array}\right.
$$
|
From the first equation it follows that $x$ is even. The second equation implies $x>y$ and $x>z$. Hence $4 x>2(y+z)=x^{2}$, and therefore $x=2$ and $y+z=2$, so $y=z=1$. It is easy to check that the triple $(2,1,1)$ satisfies the given system of equations.
|
{
"exam": "BalticWay",
"problem_label": "1",
"problem_match": "\n1.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
Let $a$ and $k$ be positive integers such that $a^{2}+k$ divides $(a-1) a(a+1)$. Prove that $k \geq a$.
|
We have $(a-1) a(a+1)=a\left(a^{2}+k\right)-(k+1) a$. Hence $a^{2}+k$ divides $(k+1) a$, and thus $k+1 \geq a$, or equivalently, $k \geq a$.
|
k \geq a
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a$ and $k$ be positive integers such that $a^{2}+k$ divides $(a-1) a(a+1)$. Prove that $k \geq a$.
|
We have $(a-1) a(a+1)=a\left(a^{2}+k\right)-(k+1) a$. Hence $a^{2}+k$ divides $(k+1) a$, and thus $k+1 \geq a$, or equivalently, $k \geq a$.
|
{
"exam": "BalticWay",
"problem_label": "2",
"problem_match": "\n2.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
The positive integers $a, b, c$ are pairwise relatively prime, $a$ and $c$ are odd and the numbers satisfy the equation $a^{2}+b^{2}=c^{2}$. Prove that $b+c$ is a square of an integer.
|
Since $a$ and $c$ are odd, $b$ must be even. We have $a^{2}=c^{2}-b^{2}=(c+b)(c-b)$. Let $d=\operatorname{gcd}(c+b, c-b)$. Then $d$ divides $(c+b)+(c-b)=2 c$ and $(c+b)-(c-b)=2 b$. Since $c+b$ and $c-b$ are odd, $d$ is odd, and hence $d$ divides both $b$ and $c$. But $b$ and $c$ are relatively prime, so $d=1$, i.e., $c+b$ and $c-b$ are also relatively prime. Since $(c+b)(c-b)=a^{2}$ is a square, it follows that $c+b$ and $c-b$ are also squares. In particular, $b+c$ is a square as required.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
The positive integers $a, b, c$ are pairwise relatively prime, $a$ and $c$ are odd and the numbers satisfy the equation $a^{2}+b^{2}=c^{2}$. Prove that $b+c$ is a square of an integer.
|
Since $a$ and $c$ are odd, $b$ must be even. We have $a^{2}=c^{2}-b^{2}=(c+b)(c-b)$. Let $d=\operatorname{gcd}(c+b, c-b)$. Then $d$ divides $(c+b)+(c-b)=2 c$ and $(c+b)-(c-b)=2 b$. Since $c+b$ and $c-b$ are odd, $d$ is odd, and hence $d$ divides both $b$ and $c$. But $b$ and $c$ are relatively prime, so $d=1$, i.e., $c+b$ and $c-b$ are also relatively prime. Since $(c+b)(c-b)=a^{2}$ is a square, it follows that $c+b$ and $c-b$ are also squares. In particular, $b+c$ is a square as required.
|
{
"exam": "BalticWay",
"problem_label": "3",
"problem_match": "\n3.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
John is older than Mary. He notices that if he switches the two digits of his age (an integer), he gets Mary's age. Moreover, the difference between the squares of their ages is the square of an integer. How old are Mary and John?
|
Let John's age be $10 a+b$ where $0 \leq a, b \leq 9$. Then Mary's age is $10 b+a$, and hence $a>b$. Now
$$
(10 a+b)^{2}-(10 b+a)^{2}=9 \cdot 11(a+b)(a-b) .
$$
Since this is the square of an integer, $a+b$ or $a-b$ must be divisible by 11 . The only possibility is clearly $a+b=11$. Hence $a-b$ must be a square. A case study yields the only possibility $a=6, b=5$. Thus John is 65 and Mary 56 years old.
|
65, 56
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
John is older than Mary. He notices that if he switches the two digits of his age (an integer), he gets Mary's age. Moreover, the difference between the squares of their ages is the square of an integer. How old are Mary and John?
|
Let John's age be $10 a+b$ where $0 \leq a, b \leq 9$. Then Mary's age is $10 b+a$, and hence $a>b$. Now
$$
(10 a+b)^{2}-(10 b+a)^{2}=9 \cdot 11(a+b)(a-b) .
$$
Since this is the square of an integer, $a+b$ or $a-b$ must be divisible by 11 . The only possibility is clearly $a+b=11$. Hence $a-b$ must be a square. A case study yields the only possibility $a=6, b=5$. Thus John is 65 and Mary 56 years old.
|
{
"exam": "BalticWay",
"problem_label": "4",
"problem_match": "\n4.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
Let $a<b<c$ be three positive integers. Prove that among any $2 c$ consecutive positive integers there exist three different numbers $x, y, z$ such that $a b c$ divides $x y z$.
|
First we show that among any $b$ consecutive numbers there are two different numbers $x$ and $y$ such that $a b$ divides $x y$. Among the $b$ consecutive numbers there is clearly a number $x^{\prime}$ divisible by $b$, and a number $y^{\prime}$ divisible by $a$. If $x^{\prime} \neq y^{\prime}$, we can take $x=x^{\prime}$ and $y=y^{\prime}$, and we are done. Now assume that $x^{\prime}=y^{\prime}$. Then $x^{\prime}$ is divisible by $e$, the least common multiple of $a$ and $b$. Let $d=\operatorname{gcd}(a, b)$. As $a<b$, we have $d \leq \frac{1}{2} b$. Hence there is a number $z^{\prime} \neq x^{\prime}$ among the $b$ consecutive numbers such that $z^{\prime}$ is divisible by $d$. Hence $x^{\prime} z^{\prime}$ is divisible by $d e$. But $d e=a b$, so we can take $x=x^{\prime}$ and $y=z^{\prime}$.
Now divide the $2 c$ consecutive numbers into two groups of $c$ consecutive numbers. In the first group, by the above reasoning, there exist distinct numbers $x$ and $y$ such that $a b$ divides $x y$. The second group contains a number $z$ divisible by $c$. Then $a b c$ divides $x y z$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a<b<c$ be three positive integers. Prove that among any $2 c$ consecutive positive integers there exist three different numbers $x, y, z$ such that $a b c$ divides $x y z$.
|
First we show that among any $b$ consecutive numbers there are two different numbers $x$ and $y$ such that $a b$ divides $x y$. Among the $b$ consecutive numbers there is clearly a number $x^{\prime}$ divisible by $b$, and a number $y^{\prime}$ divisible by $a$. If $x^{\prime} \neq y^{\prime}$, we can take $x=x^{\prime}$ and $y=y^{\prime}$, and we are done. Now assume that $x^{\prime}=y^{\prime}$. Then $x^{\prime}$ is divisible by $e$, the least common multiple of $a$ and $b$. Let $d=\operatorname{gcd}(a, b)$. As $a<b$, we have $d \leq \frac{1}{2} b$. Hence there is a number $z^{\prime} \neq x^{\prime}$ among the $b$ consecutive numbers such that $z^{\prime}$ is divisible by $d$. Hence $x^{\prime} z^{\prime}$ is divisible by $d e$. But $d e=a b$, so we can take $x=x^{\prime}$ and $y=z^{\prime}$.
Now divide the $2 c$ consecutive numbers into two groups of $c$ consecutive numbers. In the first group, by the above reasoning, there exist distinct numbers $x$ and $y$ such that $a b$ divides $x y$. The second group contains a number $z$ divisible by $c$. Then $a b c$ divides $x y z$.
|
{
"exam": "BalticWay",
"problem_label": "5",
"problem_match": "\n5.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
Prove that for positive $a, b, c, d$
$$
\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a} \geq 4 .
$$
|
The inequality between the arithmetic and harmonic mean gives
$$
\begin{aligned}
& \frac{a+c}{a+b}+\frac{c+a}{c+d} \geq \frac{4}{\frac{a+b}{a+c}+\frac{c+d}{c+a}}=4 \cdot \frac{a+c}{a+b+c+d} \\
& \frac{b+d}{b+c}+\frac{d+b}{d+a} \geq \frac{4}{\frac{b+c}{b+d}+\frac{d+a}{d+b}}=4 \cdot \frac{b+d}{a+b+c+d}
\end{aligned}
$$
and adding these inequalities yields the required inequality.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Prove that for positive $a, b, c, d$
$$
\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a} \geq 4 .
$$
|
The inequality between the arithmetic and harmonic mean gives
$$
\begin{aligned}
& \frac{a+c}{a+b}+\frac{c+a}{c+d} \geq \frac{4}{\frac{a+b}{a+c}+\frac{c+d}{c+a}}=4 \cdot \frac{a+c}{a+b+c+d} \\
& \frac{b+d}{b+c}+\frac{d+b}{d+a} \geq \frac{4}{\frac{b+c}{b+d}+\frac{d+a}{d+b}}=4 \cdot \frac{b+d}{a+b+c+d}
\end{aligned}
$$
and adding these inequalities yields the required inequality.
|
{
"exam": "BalticWay",
"problem_label": "6",
"problem_match": "\n6.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
Prove that $\sin ^{3} 18^{\circ}+\sin ^{2} 18^{\circ}=1 / 8$.
|
We have
$$
\begin{aligned}
\sin ^{3} 18^{\circ}+\sin ^{2} 18^{\circ} & =\sin ^{2} 18^{\circ}\left(\sin 18^{\circ}+\sin 90^{\circ}\right)=\sin ^{2} 18^{\circ} \cdot 2 \sin 54^{\circ} \cos 36^{\circ}=2 \sin ^{2} 18^{\circ} \cos ^{2} 36^{\circ} \\
& =\frac{2 \sin ^{2} 18^{\circ} \cos ^{2} 18^{\circ} \cos ^{2} 36^{\circ}}{\cos ^{2} 18^{\circ}}=\frac{\sin ^{2} 36^{\circ} \cos ^{2} 36^{\circ}}{2 \cos ^{2} 18^{\circ}}=\frac{\sin ^{2} 72^{\circ}}{8 \cos ^{2} 18^{\circ}}=\frac{1}{8} .
\end{aligned}
$$
|
\frac{1}{8}
|
Yes
|
Yes
|
proof
|
Algebra
|
Prove that $\sin ^{3} 18^{\circ}+\sin ^{2} 18^{\circ}=1 / 8$.
|
We have
$$
\begin{aligned}
\sin ^{3} 18^{\circ}+\sin ^{2} 18^{\circ} & =\sin ^{2} 18^{\circ}\left(\sin 18^{\circ}+\sin 90^{\circ}\right)=\sin ^{2} 18^{\circ} \cdot 2 \sin 54^{\circ} \cos 36^{\circ}=2 \sin ^{2} 18^{\circ} \cos ^{2} 36^{\circ} \\
& =\frac{2 \sin ^{2} 18^{\circ} \cos ^{2} 18^{\circ} \cos ^{2} 36^{\circ}}{\cos ^{2} 18^{\circ}}=\frac{\sin ^{2} 36^{\circ} \cos ^{2} 36^{\circ}}{2 \cos ^{2} 18^{\circ}}=\frac{\sin ^{2} 72^{\circ}}{8 \cos ^{2} 18^{\circ}}=\frac{1}{8} .
\end{aligned}
$$
|
{
"exam": "BalticWay",
"problem_label": "7",
"problem_match": "\n7.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
The real numbers $a, b$ and $c$ satisfy the inequalities $|a| \geq|b+c|,|b| \geq|c+a|$ and $|c| \geq|a+b|$. Prove that $a+b+c=0$.
|
Squaring both sides of the given inequalities we get
$$
\left\{\begin{array}{l}
a^{2} \geq(b+c)^{2} \\
b^{2} \geq(c+a)^{2} \\
c^{2} \geq(a+b)^{2}
\end{array}\right.
$$
Adding these three inequalities and rearranging, we get $(a+b+c)^{2} \leq 0$. Clearly equality must hold, and we have $a+b+c=0$.
|
a+b+c=0
|
Yes
|
Yes
|
proof
|
Inequalities
|
The real numbers $a, b$ and $c$ satisfy the inequalities $|a| \geq|b+c|,|b| \geq|c+a|$ and $|c| \geq|a+b|$. Prove that $a+b+c=0$.
|
Squaring both sides of the given inequalities we get
$$
\left\{\begin{array}{l}
a^{2} \geq(b+c)^{2} \\
b^{2} \geq(c+a)^{2} \\
c^{2} \geq(a+b)^{2}
\end{array}\right.
$$
Adding these three inequalities and rearranging, we get $(a+b+c)^{2} \leq 0$. Clearly equality must hold, and we have $a+b+c=0$.
|
{
"exam": "BalticWay",
"problem_label": "8",
"problem_match": "\n8.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
Prove that
$$
\frac{1995}{2}-\frac{1994}{3}+\frac{1993}{4}-\cdots-\frac{2}{1995}+\frac{1}{1996}=\frac{1}{999}+\frac{3}{1000}+\cdots+\frac{1995}{1996} .
$$
|
Denote the left-hand side of the equation by $L$, and the right-hand side by $R$. Then
$$
\begin{aligned}
L & =\sum_{k=1}^{1996}(-1)^{k+1}\left(\frac{1997}{k+1}-1\right)=1997 \cdot \sum_{k=1}^{1996}(-1)^{k+1} \cdot \frac{1}{k+1}=1997 \cdot \sum_{k=1}^{1996}(-1)^{k} \cdot \frac{1}{k}+1996, \\
R & =\sum_{k=1}^{998}\left(\frac{2 k+1996}{998+k}-\frac{1997}{998+k}\right)=1996-1997 \cdot \sum_{k=1}^{998} \frac{1}{k+998} .
\end{aligned}
$$
We must verify that $\sum_{k=1}^{1996}(-1)^{k-1} \cdot \frac{1}{k}=\sum_{k=1}^{998} \frac{1}{k+998}$. But this follows from the calculation
$$
\sum_{k=1}^{1996}(-1)^{k-1} \cdot \frac{1}{k}=\sum_{k=1}^{1996} \frac{1}{k}-2 \cdot \sum_{k=1}^{998} \frac{1}{2 k}=\sum_{k=1}^{998} \frac{1}{k+998}
$$
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Prove that
$$
\frac{1995}{2}-\frac{1994}{3}+\frac{1993}{4}-\cdots-\frac{2}{1995}+\frac{1}{1996}=\frac{1}{999}+\frac{3}{1000}+\cdots+\frac{1995}{1996} .
$$
|
Denote the left-hand side of the equation by $L$, and the right-hand side by $R$. Then
$$
\begin{aligned}
L & =\sum_{k=1}^{1996}(-1)^{k+1}\left(\frac{1997}{k+1}-1\right)=1997 \cdot \sum_{k=1}^{1996}(-1)^{k+1} \cdot \frac{1}{k+1}=1997 \cdot \sum_{k=1}^{1996}(-1)^{k} \cdot \frac{1}{k}+1996, \\
R & =\sum_{k=1}^{998}\left(\frac{2 k+1996}{998+k}-\frac{1997}{998+k}\right)=1996-1997 \cdot \sum_{k=1}^{998} \frac{1}{k+998} .
\end{aligned}
$$
We must verify that $\sum_{k=1}^{1996}(-1)^{k-1} \cdot \frac{1}{k}=\sum_{k=1}^{998} \frac{1}{k+998}$. But this follows from the calculation
$$
\sum_{k=1}^{1996}(-1)^{k-1} \cdot \frac{1}{k}=\sum_{k=1}^{1996} \frac{1}{k}-2 \cdot \sum_{k=1}^{998} \frac{1}{2 k}=\sum_{k=1}^{998} \frac{1}{k+998}
$$
|
{
"exam": "BalticWay",
"problem_label": "9",
"problem_match": "\n9.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
Find all real-valued functions $f$ defined on the set of all non-zero real numbers such that:
(i) $f(1)=1$,
(ii) $f\left(\frac{1}{x+y}\right)=f\left(\frac{1}{x}\right)+f\left(\frac{1}{y}\right)$ for all non-zero $x, y, x+y$,
(iii) $(x+y) f(x+y)=x y f(x) f(y)$ for all non-zero $x, y, x+y$.
|
Substituting $x=y=\frac{1}{2} z$ in (ii) we get
$$
f\left(\frac{1}{z}\right)=2 f\left(\frac{2}{z}\right)
$$
for all $z \neq 0$. Substituting $x=y=\frac{1}{z}$ in (iii) yields
$$
\frac{2}{z} f\left(\frac{2}{z}\right)=\frac{1}{z^{2}}\left(f\left(\frac{1}{z}\right)\right)^{2}
$$
for all $z \neq 0$, and hence
$$
2 f\left(\frac{2}{z}\right)=\frac{1}{z}\left(f\left(\frac{1}{z}\right)\right)^{2} .
$$
From (1) and (2) we get
$$
f\left(\frac{1}{z}\right)=\frac{1}{z}\left(f\left(\frac{1}{z}\right)\right)^{2},
$$
or, equivalently,
$$
f(x)=x(f(x))^{2}
$$
for all $x \neq 0$. If $f(x)=0$ for some $x$, then by (iii) we would have
$$
f(1)=(x+(1-x)) f(x+(1-x))=(1-x) f(x) f(1-x)=0
$$
which contradicts the condition (i). Hence $f(x) \neq 0$ for all $x$, and (3) implies $x f(x)=1$ for all $x$, and thus $f(x)=\frac{1}{x}$. It is easily verified that this function satisfies the given conditions.
|
f(x)=\frac{1}{x}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all real-valued functions $f$ defined on the set of all non-zero real numbers such that:
(i) $f(1)=1$,
(ii) $f\left(\frac{1}{x+y}\right)=f\left(\frac{1}{x}\right)+f\left(\frac{1}{y}\right)$ for all non-zero $x, y, x+y$,
(iii) $(x+y) f(x+y)=x y f(x) f(y)$ for all non-zero $x, y, x+y$.
|
Substituting $x=y=\frac{1}{2} z$ in (ii) we get
$$
f\left(\frac{1}{z}\right)=2 f\left(\frac{2}{z}\right)
$$
for all $z \neq 0$. Substituting $x=y=\frac{1}{z}$ in (iii) yields
$$
\frac{2}{z} f\left(\frac{2}{z}\right)=\frac{1}{z^{2}}\left(f\left(\frac{1}{z}\right)\right)^{2}
$$
for all $z \neq 0$, and hence
$$
2 f\left(\frac{2}{z}\right)=\frac{1}{z}\left(f\left(\frac{1}{z}\right)\right)^{2} .
$$
From (1) and (2) we get
$$
f\left(\frac{1}{z}\right)=\frac{1}{z}\left(f\left(\frac{1}{z}\right)\right)^{2},
$$
or, equivalently,
$$
f(x)=x(f(x))^{2}
$$
for all $x \neq 0$. If $f(x)=0$ for some $x$, then by (iii) we would have
$$
f(1)=(x+(1-x)) f(x+(1-x))=(1-x) f(x) f(1-x)=0
$$
which contradicts the condition (i). Hence $f(x) \neq 0$ for all $x$, and (3) implies $x f(x)=1$ for all $x$, and thus $f(x)=\frac{1}{x}$. It is easily verified that this function satisfies the given conditions.
|
{
"exam": "BalticWay",
"problem_label": "10",
"problem_match": "\n10.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
In how many ways can the set of integers $\{1,2, \ldots, 1995\}$ be partitioned into three nonempty sets so that none of these sets contains two consecutive integers?
|
We construct the three subsets by adding the numbers successively, and disregard at first the condition that the sets must be non-empty. The numbers 1 and 2 must belong to two different subsets, say $A$ and $B$. We then have two choices for each of the numbers $3,4, \ldots, 1995$, and different choices lead to different partitions. Hence there are $2^{1993}$ such partitions, one of which has an empty part. The number of partitions satisfying the requirements of the problem is therefore $2^{1993}-1$.
|
2^{1993}-1
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In how many ways can the set of integers $\{1,2, \ldots, 1995\}$ be partitioned into three nonempty sets so that none of these sets contains two consecutive integers?
|
We construct the three subsets by adding the numbers successively, and disregard at first the condition that the sets must be non-empty. The numbers 1 and 2 must belong to two different subsets, say $A$ and $B$. We then have two choices for each of the numbers $3,4, \ldots, 1995$, and different choices lead to different partitions. Hence there are $2^{1993}$ such partitions, one of which has an empty part. The number of partitions satisfying the requirements of the problem is therefore $2^{1993}-1$.
|
{
"exam": "BalticWay",
"problem_label": "11",
"problem_match": "\n11.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
Assume we have 95 boxes and 19 balls distributed in these boxes in an arbitrary manner. We take six new balls at a time and place them in six of the boxes, one ball in each of the six. Can we, by repeating this process a suitable number of times, achieve a situation in which each of the 95 boxes contains an equal number of balls?
|
Since $6 \cdot 16=96$, we can put 16 times 6 balls in the boxes so that the number of balls in one of the boxes increases by two, while in all other boxes it increases by one. Repeating this procedure, we can either diminish the difference between the number of balls in the box which has most balls and the number of balls in the box with the least number of balls, or diminish the number of boxes having the least number of balls, until all boxes have the same number of balls.
|
not found
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Assume we have 95 boxes and 19 balls distributed in these boxes in an arbitrary manner. We take six new balls at a time and place them in six of the boxes, one ball in each of the six. Can we, by repeating this process a suitable number of times, achieve a situation in which each of the 95 boxes contains an equal number of balls?
|
Since $6 \cdot 16=96$, we can put 16 times 6 balls in the boxes so that the number of balls in one of the boxes increases by two, while in all other boxes it increases by one. Repeating this procedure, we can either diminish the difference between the number of balls in the box which has most balls and the number of balls in the box with the least number of balls, or diminish the number of boxes having the least number of balls, until all boxes have the same number of balls.
|
{
"exam": "BalticWay",
"problem_label": "12",
"problem_match": "\n12.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
Consider the following two person game. A number of pebbles are situated on the table. Two players make their moves alternately. A move consists of taking off the table $x$ pebbles where $x$ is the square of any positive integer. The player who is unable to make a move loses. Prove that there are infinitely many initial situations in which the second player can win no matter how his opponent plays.
|
Suppose that there is an $n$ such that the first player always wins if there are initially more than $n$ pebbles. Consider the initial situation with $n^{2}+n+1$ pebbles. Since $(n+1)^{2}>n^{2}+n+1$, the first player can take at most $n^{2}$ pebbles, leaving at least $n+1$ pebbles on the table. By the assumption, the second player now wins. This contradiction proves that there are infinitely many situations in which the second player wins no matter how the first player plays.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Consider the following two person game. A number of pebbles are situated on the table. Two players make their moves alternately. A move consists of taking off the table $x$ pebbles where $x$ is the square of any positive integer. The player who is unable to make a move loses. Prove that there are infinitely many initial situations in which the second player can win no matter how his opponent plays.
|
Suppose that there is an $n$ such that the first player always wins if there are initially more than $n$ pebbles. Consider the initial situation with $n^{2}+n+1$ pebbles. Since $(n+1)^{2}>n^{2}+n+1$, the first player can take at most $n^{2}$ pebbles, leaving at least $n+1$ pebbles on the table. By the assumption, the second player now wins. This contradiction proves that there are infinitely many situations in which the second player wins no matter how the first player plays.
|
{
"exam": "BalticWay",
"problem_label": "13",
"problem_match": "\n13.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
There are $n$ fleas on an infinite sheet of triangulated paper. Initially the fleas are in different small triangles, all of which are inside some equilateral triangle consisting of $n^{2}$ small triangles (see Figure 1 for a possible initial configuration with $n=5$ ). Once a second each flea jumps from its triangle to one of the three small triangles as indicated in the figure. For which positive integers $n$ does there exist an initial configuration such that after a finite number of jumps all the $n$ fleas can meet in a single small triangle?
Figure 1
|
The small triangles can be coloured in four colours as shown in Figure 2. Then each flea can only reach triangles of a single colour. Moreover, number the horizontal rows are numbered as in Figure 2, and note that with each move a flea jumps from a triangle in an even-numbered row to a triangle in an odd-numbered row, or vice versa. Hence, if all the fleas are to meet in one small triangle, then they must initially be located in triangles of the same colour and in rows of the same parity. On the other hand, if these conditions are met, then the fleas can end up all in some designated triangle (of the right colour and parity). When a flea reaches this triangle, it can jump back and forth between the designated triangle and one of its neighbours until the other fleas arrive.
It remains to find the values of $n$ for which the big triangle contains at least $n$ small triangles of one colour, in rows of the same parity. For any odd $n$ there are at least $1+2+\cdots+\frac{n+1}{2}=\frac{1}{8}\left(n^{2}+4 n+3\right) \geq n$ such triangles. For even $n \geq 6$ we also have at least $1+2+\cdots+\frac{n}{2}=\frac{1}{8}\left(n^{2}+2 n\right) \geq n$ triangles of the required kind. Finally, it is easy to check that for $n=2$ and $n=4$ the necessary set of small triangles cannot be found.
Hence it is possible for the fleas to meet in one small triangle for all $n$ except 2 and 4 .
Figure 2
|
n \neq 2, 4
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
There are $n$ fleas on an infinite sheet of triangulated paper. Initially the fleas are in different small triangles, all of which are inside some equilateral triangle consisting of $n^{2}$ small triangles (see Figure 1 for a possible initial configuration with $n=5$ ). Once a second each flea jumps from its triangle to one of the three small triangles as indicated in the figure. For which positive integers $n$ does there exist an initial configuration such that after a finite number of jumps all the $n$ fleas can meet in a single small triangle?
Figure 1
|
The small triangles can be coloured in four colours as shown in Figure 2. Then each flea can only reach triangles of a single colour. Moreover, number the horizontal rows are numbered as in Figure 2, and note that with each move a flea jumps from a triangle in an even-numbered row to a triangle in an odd-numbered row, or vice versa. Hence, if all the fleas are to meet in one small triangle, then they must initially be located in triangles of the same colour and in rows of the same parity. On the other hand, if these conditions are met, then the fleas can end up all in some designated triangle (of the right colour and parity). When a flea reaches this triangle, it can jump back and forth between the designated triangle and one of its neighbours until the other fleas arrive.
It remains to find the values of $n$ for which the big triangle contains at least $n$ small triangles of one colour, in rows of the same parity. For any odd $n$ there are at least $1+2+\cdots+\frac{n+1}{2}=\frac{1}{8}\left(n^{2}+4 n+3\right) \geq n$ such triangles. For even $n \geq 6$ we also have at least $1+2+\cdots+\frac{n}{2}=\frac{1}{8}\left(n^{2}+2 n\right) \geq n$ triangles of the required kind. Finally, it is easy to check that for $n=2$ and $n=4$ the necessary set of small triangles cannot be found.
Hence it is possible for the fleas to meet in one small triangle for all $n$ except 2 and 4 .
Figure 2
|
{
"exam": "BalticWay",
"problem_label": "14",
"problem_match": "\n14.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
A polygon with $2 n+1$ vertices is given. Show that it is possible to label the vertices and midpoints of the sides of the polygon, using all the numbers $1,2, \ldots, 4 n+2$, so that the sums of the three numbers assigned to each side are all equal.
|
First, label the midpoints of the sides of the polygon with the numbers $1,2, \ldots, 2 n+1$, in clockwise order. Then, beginning with the vertex between the sides labelled by 1 and 2 , label every second vertex in clockwise order with the numbers $4 n+2,4 n+1, \ldots, 2 n+2$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
A polygon with $2 n+1$ vertices is given. Show that it is possible to label the vertices and midpoints of the sides of the polygon, using all the numbers $1,2, \ldots, 4 n+2$, so that the sums of the three numbers assigned to each side are all equal.
|
First, label the midpoints of the sides of the polygon with the numbers $1,2, \ldots, 2 n+1$, in clockwise order. Then, beginning with the vertex between the sides labelled by 1 and 2 , label every second vertex in clockwise order with the numbers $4 n+2,4 n+1, \ldots, 2 n+2$.
|
{
"exam": "BalticWay",
"problem_label": "15",
"problem_match": "\n15.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
In the triangle $A B C$, let $l$ be the bisector of the external angle at $C$. The line through the midpoint $O$ of the segment $A B$ parallel to $l$ meets the line $A C$ at $E$. Determine $|C E|$, if $|A C|=7$ and $|C B|=4$.
|
Let $F$ be the intersection point of $l$ and the line $A B$. Since $|A C|>|B C|$, the point $E$ lies on the segment $A C$, and $F$ lies on the ray $A B$. Let the line through $B$ parallel to $A C$ meet $C F$ at $G$. Then the triangles $A F C$ and $B F G$ are similar. Moreover, we have $\angle B G C=\angle B C G$, and hence the triangle $C B G$ is isosceles with $|B C|=|B G|$. Hence $\frac{|F A|}{|F B|}=\frac{|A C|}{|B G|}=\frac{|A C|}{|B C|}=\frac{7}{4}$. Therefore $\frac{|A O|}{|A F|}=\frac{3}{2} / 7=\frac{3}{14}$. Since the triangles $A C F$ and $A E O$ are similar, $\frac{|A E|}{|A C|}=\frac{|A O|}{|A F|}=\frac{3}{14}$, whence $|A E|=\frac{3}{2}$ and $|E C|=\frac{11}{2}$.
|
\frac{11}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
In the triangle $A B C$, let $l$ be the bisector of the external angle at $C$. The line through the midpoint $O$ of the segment $A B$ parallel to $l$ meets the line $A C$ at $E$. Determine $|C E|$, if $|A C|=7$ and $|C B|=4$.
|
Let $F$ be the intersection point of $l$ and the line $A B$. Since $|A C|>|B C|$, the point $E$ lies on the segment $A C$, and $F$ lies on the ray $A B$. Let the line through $B$ parallel to $A C$ meet $C F$ at $G$. Then the triangles $A F C$ and $B F G$ are similar. Moreover, we have $\angle B G C=\angle B C G$, and hence the triangle $C B G$ is isosceles with $|B C|=|B G|$. Hence $\frac{|F A|}{|F B|}=\frac{|A C|}{|B G|}=\frac{|A C|}{|B C|}=\frac{7}{4}$. Therefore $\frac{|A O|}{|A F|}=\frac{3}{2} / 7=\frac{3}{14}$. Since the triangles $A C F$ and $A E O$ are similar, $\frac{|A E|}{|A C|}=\frac{|A O|}{|A F|}=\frac{3}{14}$, whence $|A E|=\frac{3}{2}$ and $|E C|=\frac{11}{2}$.
|
{
"exam": "BalticWay",
"problem_label": "16",
"problem_match": "\n16.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
Prove that there exists a number $\alpha$ such that for any triangle $A B C$ the inequality
$$
\max \left(h_{A}, h_{B}, h_{C}\right) \leq \alpha \cdot \min \left(m_{A}, m_{B}, m_{C}\right)
$$
holds, where $h_{A}, h_{B}, h_{C}$ denote the lengths of the altitudes and $m_{A}, m_{B}, m_{C}$ denote the lengths of the medians. Find the smallest possible value of $\alpha$.
|
Let $h=\max \left(h_{A}, h_{B}, h_{C}\right)$ and $m=\min \left(m_{A}, m_{B}, m_{C}\right)$. If the longest height and the shortest median are drawn from the same vertex, then obviously $h \leq m$. Now let the longest height and shortest median be $A D$ and $B E$, respectively, with $|A D|=h$ and $|B E|=m$. Let $F$ be the point on the line $B C$ such that $E F$ is parallel to $A D$. Then $m=|E B| \geq|E F|=\frac{h}{2}$, whence $h \leq 2 m$. For an example with $h=2 m$, consider a triangle where $D$ lies on the ray $C B$ with $|C B|=|B D|$. Hence the smallest such value is $\alpha=2$.
|
2
|
Yes
|
Yes
|
proof
|
Inequalities
|
Prove that there exists a number $\alpha$ such that for any triangle $A B C$ the inequality
$$
\max \left(h_{A}, h_{B}, h_{C}\right) \leq \alpha \cdot \min \left(m_{A}, m_{B}, m_{C}\right)
$$
holds, where $h_{A}, h_{B}, h_{C}$ denote the lengths of the altitudes and $m_{A}, m_{B}, m_{C}$ denote the lengths of the medians. Find the smallest possible value of $\alpha$.
|
Let $h=\max \left(h_{A}, h_{B}, h_{C}\right)$ and $m=\min \left(m_{A}, m_{B}, m_{C}\right)$. If the longest height and the shortest median are drawn from the same vertex, then obviously $h \leq m$. Now let the longest height and shortest median be $A D$ and $B E$, respectively, with $|A D|=h$ and $|B E|=m$. Let $F$ be the point on the line $B C$ such that $E F$ is parallel to $A D$. Then $m=|E B| \geq|E F|=\frac{h}{2}$, whence $h \leq 2 m$. For an example with $h=2 m$, consider a triangle where $D$ lies on the ray $C B$ with $|C B|=|B D|$. Hence the smallest such value is $\alpha=2$.
|
{
"exam": "BalticWay",
"problem_label": "17",
"problem_match": "\n17.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
Let $M$ be the midpoint of the side $A C$ of a triangle $A B C$ and let $H$ be the foot point of the altitude from $B$. Let $P$ and $Q$ be the orthogonal projections of $A$ and $C$ on the bisector of angle $B$. Prove that the four points $M, H, P$ and $Q$ lie on the same circle.
|
If $|A B|=|B C|$, the points $M, H, P$ and $Q$ coincide and the circle degenerates to a point. We will assume that $|A B|<|B C|$, so that $P$ lies inside the triangle $A B C$, and $Q$ lies outside of it.
Let the line $A P$ intersect $B C$ at $P_{1}$, and let $C Q$ intersect $A B$ at $Q_{1}$. Then $|A P|=\left|P P_{1}\right|$ (since $\triangle A P B \cong$ $\left.\triangle P_{1} P B\right)$, and therefore $M P \| B C$. Similarly, $M Q \| A B$. Therefore $\angle A M Q=\angle B A C$. We have two cases:
(i) $\angle B A C \leq 90^{\circ}$. Then $A, H, P$ and $B$ lie on a circle in this order. Hence $\angle H P Q=180^{\circ}-\angle H P B=$ $\angle B A C=\angle H M Q$. Therefore $H, P, M$ and $Q$ lie on a circle.
(ii) $\angle B A C>90^{\circ}$. Then $A, H, B$ and $P$ lie on a circle in this order. Hence $\angle H P Q=180^{\circ}-\angle H P B=$ $180^{\circ}-\angle H A B=\angle B A C=\angle H M Q$, and therefore $H, P, M$ and $Q$ lie on a circle.
Figure 3
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $M$ be the midpoint of the side $A C$ of a triangle $A B C$ and let $H$ be the foot point of the altitude from $B$. Let $P$ and $Q$ be the orthogonal projections of $A$ and $C$ on the bisector of angle $B$. Prove that the four points $M, H, P$ and $Q$ lie on the same circle.
|
If $|A B|=|B C|$, the points $M, H, P$ and $Q$ coincide and the circle degenerates to a point. We will assume that $|A B|<|B C|$, so that $P$ lies inside the triangle $A B C$, and $Q$ lies outside of it.
Let the line $A P$ intersect $B C$ at $P_{1}$, and let $C Q$ intersect $A B$ at $Q_{1}$. Then $|A P|=\left|P P_{1}\right|$ (since $\triangle A P B \cong$ $\left.\triangle P_{1} P B\right)$, and therefore $M P \| B C$. Similarly, $M Q \| A B$. Therefore $\angle A M Q=\angle B A C$. We have two cases:
(i) $\angle B A C \leq 90^{\circ}$. Then $A, H, P$ and $B$ lie on a circle in this order. Hence $\angle H P Q=180^{\circ}-\angle H P B=$ $\angle B A C=\angle H M Q$. Therefore $H, P, M$ and $Q$ lie on a circle.
(ii) $\angle B A C>90^{\circ}$. Then $A, H, B$ and $P$ lie on a circle in this order. Hence $\angle H P Q=180^{\circ}-\angle H P B=$ $180^{\circ}-\angle H A B=\angle B A C=\angle H M Q$, and therefore $H, P, M$ and $Q$ lie on a circle.
Figure 3
|
{
"exam": "BalticWay",
"problem_label": "18",
"problem_match": "\n18.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
The following construction is used for training astronauts: A circle $C_{2}$ of radius $2 R$ rolls along the inside of another, fixed circle $C_{1}$ of radius $n R$, where $n$ is an integer greater than 2 . The astronaut is fastened to a third circle $C_{3}$ of radius $R$ which rolls along the inside of circle $C_{2}$ in such a way that the touching point of the circles $C_{2}$ and $C_{3}$ remains at maximum distance from the touching point of the circles $C_{1}$ and $C_{2}$ at all times (see Figure 3).
How many revolutions (relative to the ground) does the astronaut perform together with the circle $C_{3}$ while the circle $C_{2}$ completes one full lap around the inside of circle $C_{1}$ ?
|
Consider a circle $C_{4}$ with radius $R$ that rolls inside $C_{2}$ in such a way that the two circles always touch in the point opposite to the touching point of $C_{2}$ and $C_{3}$. Then the circles $C_{3}$ and $C_{4}$ follow each other and make the same number of revolutions, and so we will assume that the astronaut is inside the circle $C_{4}$ instead. But the touching point of $C_{2}$ and $C_{4}$ coincides with the touching point of $C_{1}$ and $C_{2}$. Hence the circles $C_{4}$ and $C_{1}$ always touch each other, and we can disregard the circle $C_{2}$ completely.
Suppose the circle $C_{4}$ rolls inside $C_{1}$ in counterclockwise direction. Then the astronaut revolves in clockwise direction. If the circle $C_{4}$ had rolled along a straight line of length $2 \pi n R$ (instead of the inside of $C_{1}$ ), the circle $C_{4}$ would have made $n$ revolutions during its movement. As the path of the circle $C_{4}$ makes a $360^{\circ}$ counterclockwise turn itself, the total number of revolutions of the astronaut relative to the ground is $n-1$.
Remark: The radius of the intermediate circle $C_{2}$ is irrelevant. Moreover, for any number of intermediate circles the answer remains the same, depending only on the radii of the outermost and innermost circles.
|
n-1
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
The following construction is used for training astronauts: A circle $C_{2}$ of radius $2 R$ rolls along the inside of another, fixed circle $C_{1}$ of radius $n R$, where $n$ is an integer greater than 2 . The astronaut is fastened to a third circle $C_{3}$ of radius $R$ which rolls along the inside of circle $C_{2}$ in such a way that the touching point of the circles $C_{2}$ and $C_{3}$ remains at maximum distance from the touching point of the circles $C_{1}$ and $C_{2}$ at all times (see Figure 3).
How many revolutions (relative to the ground) does the astronaut perform together with the circle $C_{3}$ while the circle $C_{2}$ completes one full lap around the inside of circle $C_{1}$ ?
|
Consider a circle $C_{4}$ with radius $R$ that rolls inside $C_{2}$ in such a way that the two circles always touch in the point opposite to the touching point of $C_{2}$ and $C_{3}$. Then the circles $C_{3}$ and $C_{4}$ follow each other and make the same number of revolutions, and so we will assume that the astronaut is inside the circle $C_{4}$ instead. But the touching point of $C_{2}$ and $C_{4}$ coincides with the touching point of $C_{1}$ and $C_{2}$. Hence the circles $C_{4}$ and $C_{1}$ always touch each other, and we can disregard the circle $C_{2}$ completely.
Suppose the circle $C_{4}$ rolls inside $C_{1}$ in counterclockwise direction. Then the astronaut revolves in clockwise direction. If the circle $C_{4}$ had rolled along a straight line of length $2 \pi n R$ (instead of the inside of $C_{1}$ ), the circle $C_{4}$ would have made $n$ revolutions during its movement. As the path of the circle $C_{4}$ makes a $360^{\circ}$ counterclockwise turn itself, the total number of revolutions of the astronaut relative to the ground is $n-1$.
Remark: The radius of the intermediate circle $C_{2}$ is irrelevant. Moreover, for any number of intermediate circles the answer remains the same, depending only on the radii of the outermost and innermost circles.
|
{
"exam": "BalticWay",
"problem_label": "19",
"problem_match": "\n19.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
Prove that if both coordinates of every vertex of a convex pentagon are integers, then the area of this pentagon is not less than $\frac{5}{2}$.
|
There are two vertices $A_{1}$ and $A_{2}$ of the pentagon that have their first coordinates of the parity, and their second coordinates of the same parity. Therefore the midpoint $M$ of $A_{1} A_{2}$ has integer coordinates. There are two possibilities:
(i) The considered vertices are not consecutive. Then $M$ lies inside the pentagon (because it is convex) and is the common vertex of five triangles having as their bases the sides of the pentagon. The area of any one of these triangles is not less than $\frac{1}{2}$, so the area of the pentagon is at least $\frac{5}{2}$.
(ii) The considered vertices are consecutive. Since the pentagon is convex, the side $A_{1} A_{2}$ is not simultaneously parallel to $A_{3} A_{4}$ and $A_{4} A_{5}$. Suppose that the segments $A_{1} A_{2}$ and $A_{3} A_{4}$ are not parallel. Then the triangles $A_{2} A_{3} A_{4}, M A_{3} A_{4}$ and $A-1 A_{3} A_{4}$ have different areas, since their altitudes dropped onto the side $A_{3} A_{4}$ form a monotone sequence. At least one of these triangles has area not less than $\frac{3}{2}$, and the pentagon has area not less than $\frac{5}{2}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Prove that if both coordinates of every vertex of a convex pentagon are integers, then the area of this pentagon is not less than $\frac{5}{2}$.
|
There are two vertices $A_{1}$ and $A_{2}$ of the pentagon that have their first coordinates of the parity, and their second coordinates of the same parity. Therefore the midpoint $M$ of $A_{1} A_{2}$ has integer coordinates. There are two possibilities:
(i) The considered vertices are not consecutive. Then $M$ lies inside the pentagon (because it is convex) and is the common vertex of five triangles having as their bases the sides of the pentagon. The area of any one of these triangles is not less than $\frac{1}{2}$, so the area of the pentagon is at least $\frac{5}{2}$.
(ii) The considered vertices are consecutive. Since the pentagon is convex, the side $A_{1} A_{2}$ is not simultaneously parallel to $A_{3} A_{4}$ and $A_{4} A_{5}$. Suppose that the segments $A_{1} A_{2}$ and $A_{3} A_{4}$ are not parallel. Then the triangles $A_{2} A_{3} A_{4}, M A_{3} A_{4}$ and $A-1 A_{3} A_{4}$ have different areas, since their altitudes dropped onto the side $A_{3} A_{4}$ form a monotone sequence. At least one of these triangles has area not less than $\frac{3}{2}$, and the pentagon has area not less than $\frac{5}{2}$.
|
{
"exam": "BalticWay",
"problem_label": "20",
"problem_match": "\n20.",
"resource_path": "BalticWay/segmented/en-bw95sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1995"
}
|
Let $\alpha$ be the angle between two lines containing the diagonals of a regular 1996-gon, and let $\beta \neq 0$ be another such angle. Prove that $\alpha / \beta$ is a rational number.
|
Let $O$ be the circumcentre of the 1996-gon. Consider two diagonals $A B$ and $C D$. There is a rotation around $O$ that takes the point $C$ to $A$ and $D$ to a point $D^{\prime}$. Clearly the angle of this rotation is a multiple of $2 \varphi=2 \pi / 1996$.
The angle $B A D^{\prime}$ is the inscribed angle on the $\operatorname{arc} B D^{\prime}$, and hence is an integral multiple of $\varphi$, the inscribed angle on the arc between any two adjacent vertices of the 1996-gon. Hence the angle between $A B$ and $C D$ is also an integral multiple of $\varphi$.
Since both $\alpha$ and $\beta$ are integral multiples of $\varphi, \alpha / \beta$ is a rational number.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $\alpha$ be the angle between two lines containing the diagonals of a regular 1996-gon, and let $\beta \neq 0$ be another such angle. Prove that $\alpha / \beta$ is a rational number.
|
Let $O$ be the circumcentre of the 1996-gon. Consider two diagonals $A B$ and $C D$. There is a rotation around $O$ that takes the point $C$ to $A$ and $D$ to a point $D^{\prime}$. Clearly the angle of this rotation is a multiple of $2 \varphi=2 \pi / 1996$.
The angle $B A D^{\prime}$ is the inscribed angle on the $\operatorname{arc} B D^{\prime}$, and hence is an integral multiple of $\varphi$, the inscribed angle on the arc between any two adjacent vertices of the 1996-gon. Hence the angle between $A B$ and $C D$ is also an integral multiple of $\varphi$.
Since both $\alpha$ and $\beta$ are integral multiples of $\varphi, \alpha / \beta$ is a rational number.
|
{
"exam": "BalticWay",
"problem_label": "1",
"problem_match": "\n1.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
In the figure below, you see three half-circles. The circle $C$ is tangent to two of the half-circles and to the line $P Q$ perpendicular to the diameter $A B$. The area of the shaded region is $39 \pi$, and the area of the circle $C$ is $9 \pi$. Find the length of the diameter $A B$.
Figure 1
|
Let $r$ and $s$ be the radii of the half-circles with diameters $A P$ and $B P$. Then we have
$$
39 \pi=\frac{\pi}{2}\left((r+s)^{2}-r^{2}-s^{2}\right)-9 \pi
$$
hence $r s=48$. Let $M$ be the midpoint of the diameter $A B, N$ be the midpoint of $P B, O$ be the centre of the circle $C$, and let $F$ be the orthogonal projection of $O$ on $A B$. Since the radius of $C$ is 3 , we have $|M O|=r+s-3,|M F|=r-s+3,|O N|=s+3$, and $|F N|=s-3$.
Applying the Pythagorean theorem to the triangles $M F O$ and $N F O$ yields
$$
(r+s-3)^{2}-(r-s+3)^{2}=|O F|^{2}=(s+3)^{2}-(s-3)^{2},
$$
which implies $r(s-3)=3 s$, so that $3(r+s)=r s=48$. Hence $|A B|=2(r+s)=32$.
|
32
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
In the figure below, you see three half-circles. The circle $C$ is tangent to two of the half-circles and to the line $P Q$ perpendicular to the diameter $A B$. The area of the shaded region is $39 \pi$, and the area of the circle $C$ is $9 \pi$. Find the length of the diameter $A B$.
Figure 1
|
Let $r$ and $s$ be the radii of the half-circles with diameters $A P$ and $B P$. Then we have
$$
39 \pi=\frac{\pi}{2}\left((r+s)^{2}-r^{2}-s^{2}\right)-9 \pi
$$
hence $r s=48$. Let $M$ be the midpoint of the diameter $A B, N$ be the midpoint of $P B, O$ be the centre of the circle $C$, and let $F$ be the orthogonal projection of $O$ on $A B$. Since the radius of $C$ is 3 , we have $|M O|=r+s-3,|M F|=r-s+3,|O N|=s+3$, and $|F N|=s-3$.
Applying the Pythagorean theorem to the triangles $M F O$ and $N F O$ yields
$$
(r+s-3)^{2}-(r-s+3)^{2}=|O F|^{2}=(s+3)^{2}-(s-3)^{2},
$$
which implies $r(s-3)=3 s$, so that $3(r+s)=r s=48$. Hence $|A B|=2(r+s)=32$.
|
{
"exam": "BalticWay",
"problem_label": "2",
"problem_match": "\n2.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
Let $A B C D$ be a unit square and let $P$ and $Q$ be points in the plane such that $Q$ is the circumcentre of triangle $B P C$ and $D$ is the circumcentre of triangle $P Q A$. Find all possible values of the length of segment $P Q$.
|
As $Q$ is the circumcentre of triangle $B P C$, we have $|P Q|=|Q C|$ and $Q$ lies on the perpendicular bisector $s$ of $B C$. On the other hand, as $D$ is the circumcentre of triangle $P Q A, Q$ lies on the circle centred at $D$ and passing through $A$. Thus $Q$ must be one of the two intersection points $Q_{1}$ and $Q_{2}$ of this circle and the line $s$. We may choose $Q_{1}$ to lie inside, and $Q_{2}$ outside of the square $A B C D$.
Let $E$ and $F$ be the midpoints of $A D$ and $B C$, respectively. We have $\left|A Q_{1}\right|=\left|D Q_{1}\right|=|D A|=1$. Hence $\left|E Q_{1}\right|=\frac{\sqrt{3}}{2}$ and $\left|F Q_{1}\right|=1-\frac{\sqrt{3}}{2}$. The Pythagorean theorem applied to the triangle $C F Q_{1}$ now yields
$$
\left|C Q_{1}\right|^{2}=|C F|^{2}+\left|F Q_{1}\right|^{2}=\left(\frac{1}{2}\right)^{2}+\left(1-\frac{\sqrt{3}}{2}\right)^{2}=2-\sqrt{3}
$$
and hence $\left|C Q_{1}\right|=\sqrt{2-\sqrt{3}}$. Similarly, $\left|Q_{2} E\right|=\frac{\sqrt{3}}{2}$, and the Pythagorean theorem applied to the triangle $C F Q_{2}$ now yields
$$
\left|C Q_{2}\right|^{2}=|C F|^{2}+\left|F Q_{2}\right|^{2}=\left(\frac{1}{2}\right)^{2}+\left(1+\frac{\sqrt{3}}{2}\right)^{2}=2+\sqrt{3}
$$
and hence $\left|C Q_{2}\right|=\sqrt{2+\sqrt{3}}$. Hence the possible values of the length of the segment $P Q$ are $\sqrt{2-\sqrt{3}}$ and $\sqrt{2+\sqrt{3}}$.
Remark. The actual location of the point $P$ is unimportant for us. Note however that the point $P$ exists because $P$ and $C$ are the two intersection points of the circle centred at $D$ passing through $A$ and the circle centred at $Q$ passing through $C$.
|
\sqrt{2-\sqrt{3}} \text{ and } \sqrt{2+\sqrt{3}}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C D$ be a unit square and let $P$ and $Q$ be points in the plane such that $Q$ is the circumcentre of triangle $B P C$ and $D$ is the circumcentre of triangle $P Q A$. Find all possible values of the length of segment $P Q$.
|
As $Q$ is the circumcentre of triangle $B P C$, we have $|P Q|=|Q C|$ and $Q$ lies on the perpendicular bisector $s$ of $B C$. On the other hand, as $D$ is the circumcentre of triangle $P Q A, Q$ lies on the circle centred at $D$ and passing through $A$. Thus $Q$ must be one of the two intersection points $Q_{1}$ and $Q_{2}$ of this circle and the line $s$. We may choose $Q_{1}$ to lie inside, and $Q_{2}$ outside of the square $A B C D$.
Let $E$ and $F$ be the midpoints of $A D$ and $B C$, respectively. We have $\left|A Q_{1}\right|=\left|D Q_{1}\right|=|D A|=1$. Hence $\left|E Q_{1}\right|=\frac{\sqrt{3}}{2}$ and $\left|F Q_{1}\right|=1-\frac{\sqrt{3}}{2}$. The Pythagorean theorem applied to the triangle $C F Q_{1}$ now yields
$$
\left|C Q_{1}\right|^{2}=|C F|^{2}+\left|F Q_{1}\right|^{2}=\left(\frac{1}{2}\right)^{2}+\left(1-\frac{\sqrt{3}}{2}\right)^{2}=2-\sqrt{3}
$$
and hence $\left|C Q_{1}\right|=\sqrt{2-\sqrt{3}}$. Similarly, $\left|Q_{2} E\right|=\frac{\sqrt{3}}{2}$, and the Pythagorean theorem applied to the triangle $C F Q_{2}$ now yields
$$
\left|C Q_{2}\right|^{2}=|C F|^{2}+\left|F Q_{2}\right|^{2}=\left(\frac{1}{2}\right)^{2}+\left(1+\frac{\sqrt{3}}{2}\right)^{2}=2+\sqrt{3}
$$
and hence $\left|C Q_{2}\right|=\sqrt{2+\sqrt{3}}$. Hence the possible values of the length of the segment $P Q$ are $\sqrt{2-\sqrt{3}}$ and $\sqrt{2+\sqrt{3}}$.
Remark. The actual location of the point $P$ is unimportant for us. Note however that the point $P$ exists because $P$ and $C$ are the two intersection points of the circle centred at $D$ passing through $A$ and the circle centred at $Q$ passing through $C$.
|
{
"exam": "BalticWay",
"problem_label": "3",
"problem_match": "\n3.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
$A B C D$ is a trapezium $(A D \| B C) . P$ is the point on the line $A B$ such that $\angle C P D$ is maximal. $Q$ is the point on the line $C D$ such that $\angle B Q A$ is maximal. Given that $P$ lies on the segment $A B$, prove that $\angle C P D=\angle B Q A$.
|
The property that $\angle C P D$ is maximal is equivalent to the property that the circle $C P D$ touches the line $A B$ (at $P$ ). Let $O$ be the intersection point of the lines $A B$ and $C D$, and let $\ell$ be the bisector of $\angle A O D$. Let $A^{\prime}, B^{\prime}$ and $Q^{\prime}$ be the points symmetrical to $A, B$ and $Q$, respectively, relative to the line $\ell$. Then the circle $A Q B$ is symmetrical to the circle $A^{\prime} Q^{\prime} B^{\prime}$ that touches the line $A B$ at $Q^{\prime}$. We have
$$
\frac{|O D|}{\left|O A^{\prime}\right|}=\frac{|O D|}{|O A|}=\frac{|O C|}{|O B|}=\frac{|O C|}{\left|O B^{\prime}\right|}
$$
Hence the homothety with centre $O$ and coefficient $|O D| /|O A|$ takes $A^{\prime}$ to $D, B^{\prime}$ to $C$, and $Q^{\prime}$ to a point $Q^{\prime \prime}$ such that the circle $C Q^{\prime \prime} D$ touches the line $A B$, and thus $Q^{\prime \prime}$ coincides with $P$. Therefore $\angle A Q B=\angle A^{\prime} Q^{\prime} B^{\prime}=\angle C Q^{\prime \prime} D=\angle C P D$ as required.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
$A B C D$ is a trapezium $(A D \| B C) . P$ is the point on the line $A B$ such that $\angle C P D$ is maximal. $Q$ is the point on the line $C D$ such that $\angle B Q A$ is maximal. Given that $P$ lies on the segment $A B$, prove that $\angle C P D=\angle B Q A$.
|
The property that $\angle C P D$ is maximal is equivalent to the property that the circle $C P D$ touches the line $A B$ (at $P$ ). Let $O$ be the intersection point of the lines $A B$ and $C D$, and let $\ell$ be the bisector of $\angle A O D$. Let $A^{\prime}, B^{\prime}$ and $Q^{\prime}$ be the points symmetrical to $A, B$ and $Q$, respectively, relative to the line $\ell$. Then the circle $A Q B$ is symmetrical to the circle $A^{\prime} Q^{\prime} B^{\prime}$ that touches the line $A B$ at $Q^{\prime}$. We have
$$
\frac{|O D|}{\left|O A^{\prime}\right|}=\frac{|O D|}{|O A|}=\frac{|O C|}{|O B|}=\frac{|O C|}{\left|O B^{\prime}\right|}
$$
Hence the homothety with centre $O$ and coefficient $|O D| /|O A|$ takes $A^{\prime}$ to $D, B^{\prime}$ to $C$, and $Q^{\prime}$ to a point $Q^{\prime \prime}$ such that the circle $C Q^{\prime \prime} D$ touches the line $A B$, and thus $Q^{\prime \prime}$ coincides with $P$. Therefore $\angle A Q B=\angle A^{\prime} Q^{\prime} B^{\prime}=\angle C Q^{\prime \prime} D=\angle C P D$ as required.
|
{
"exam": "BalticWay",
"problem_label": "4",
"problem_match": "\n4.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
Let $A B C D$ be a cyclic convex quadrilateral and let $r_{a}, r_{b}, r_{c}, r_{d}$ be the radii of the circles inscribed in the triangles $B C D, A C D, A B D, A B C$ respectively. Prove that $r_{a}+r_{c}=r_{b}+r_{d}$.
|
For a triangle $M N K$ with in-radius $r$ and circumradius $R$, the equality
$$
\cos \angle M+\cos \angle N+\cos \angle K=1+\frac{r}{R}
$$
hold; this follows from the cosine theorem and formulas for $r$ and $R$.
We have $\angle A C B=\angle A D B, \angle B D C=\angle B A C, \angle C A D=\angle C B D$ and $\angle D B A=\angle D C A$. Denoting these angles by $\alpha, \beta, \gamma$ and $\delta$, respectively, we get $r_{a}=(\cos \beta+\cos \gamma+\cos (\alpha+\delta)-1) R$ and $r_{c}=(\cos \alpha+\cos \delta+$ $\cos (\beta+\gamma)-1) R$. Since $\cos (\alpha+\delta)=-\cos (\beta+\gamma)$, we get
$$
r_{a}+r_{c}=(\cos \alpha+\cos \beta+\cos \gamma+\cos \delta-2) R .
$$
Similarly,
$$
r_{b}+r_{d}=(\cos \alpha+\cos \beta+\cos \gamma+\cos \delta-2) R,
$$
where $R$ is the circumradius of the quadrangle $A B C D$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a cyclic convex quadrilateral and let $r_{a}, r_{b}, r_{c}, r_{d}$ be the radii of the circles inscribed in the triangles $B C D, A C D, A B D, A B C$ respectively. Prove that $r_{a}+r_{c}=r_{b}+r_{d}$.
|
For a triangle $M N K$ with in-radius $r$ and circumradius $R$, the equality
$$
\cos \angle M+\cos \angle N+\cos \angle K=1+\frac{r}{R}
$$
hold; this follows from the cosine theorem and formulas for $r$ and $R$.
We have $\angle A C B=\angle A D B, \angle B D C=\angle B A C, \angle C A D=\angle C B D$ and $\angle D B A=\angle D C A$. Denoting these angles by $\alpha, \beta, \gamma$ and $\delta$, respectively, we get $r_{a}=(\cos \beta+\cos \gamma+\cos (\alpha+\delta)-1) R$ and $r_{c}=(\cos \alpha+\cos \delta+$ $\cos (\beta+\gamma)-1) R$. Since $\cos (\alpha+\delta)=-\cos (\beta+\gamma)$, we get
$$
r_{a}+r_{c}=(\cos \alpha+\cos \beta+\cos \gamma+\cos \delta-2) R .
$$
Similarly,
$$
r_{b}+r_{d}=(\cos \alpha+\cos \beta+\cos \gamma+\cos \delta-2) R,
$$
where $R$ is the circumradius of the quadrangle $A B C D$.
|
{
"exam": "BalticWay",
"problem_label": "5",
"problem_match": "\n5.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
Let $a, b, c, d$ be positive integers such that $a b=c d$. Prove that $a+b+c+d$ is not prime.
|
. As $a b=c d$, we get $a(a+b+c+d)=(a+c)(a+d)$. If $a+b+c+d$ were a prime, then it would be a factor in either $a+c$ or $a+d$, which are both smaller than $a+b+c+d$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a, b, c, d$ be positive integers such that $a b=c d$. Prove that $a+b+c+d$ is not prime.
|
. As $a b=c d$, we get $a(a+b+c+d)=(a+c)(a+d)$. If $a+b+c+d$ were a prime, then it would be a factor in either $a+c$ or $a+d$, which are both smaller than $a+b+c+d$.
|
{
"exam": "BalticWay",
"problem_label": "6",
"problem_match": "\n6.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution 1",
"tier": "T3",
"year": "1996"
}
|
Let $a, b, c, d$ be positive integers such that $a b=c d$. Prove that $a+b+c+d$ is not prime.
|
. Let $r=\operatorname{gcd}(a, c)$ and $s=\operatorname{gcd}(b, d)$. Let $a=a^{\prime} r, b=b^{\prime} s, c=c^{\prime} r$ and $d=d^{\prime} s$. Then $a^{\prime} b^{\prime}=c^{\prime} d^{\prime}$. But $\operatorname{gcd}\left(a^{\prime}, c^{\prime}\right)=1$ and $\operatorname{gcd}\left(b^{\prime}, d^{\prime}\right)=1$, so we must have $a^{\prime}=d^{\prime}$ and $b^{\prime}=c^{\prime}$. This gives
$$
a+b+c+d=a^{\prime} r+b^{\prime} s+c^{\prime} r+d^{\prime} s=a^{\prime} r+b^{\prime} s+b^{\prime} r+a^{\prime} s=\left(a^{\prime}+b^{\prime}\right)(r+s) \text {. }
$$
Since $a^{\prime}, b^{\prime}, r$ and $s$ are positive integers, $a+b+c+d$ is not a prime.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a, b, c, d$ be positive integers such that $a b=c d$. Prove that $a+b+c+d$ is not prime.
|
. Let $r=\operatorname{gcd}(a, c)$ and $s=\operatorname{gcd}(b, d)$. Let $a=a^{\prime} r, b=b^{\prime} s, c=c^{\prime} r$ and $d=d^{\prime} s$. Then $a^{\prime} b^{\prime}=c^{\prime} d^{\prime}$. But $\operatorname{gcd}\left(a^{\prime}, c^{\prime}\right)=1$ and $\operatorname{gcd}\left(b^{\prime}, d^{\prime}\right)=1$, so we must have $a^{\prime}=d^{\prime}$ and $b^{\prime}=c^{\prime}$. This gives
$$
a+b+c+d=a^{\prime} r+b^{\prime} s+c^{\prime} r+d^{\prime} s=a^{\prime} r+b^{\prime} s+b^{\prime} r+a^{\prime} s=\left(a^{\prime}+b^{\prime}\right)(r+s) \text {. }
$$
Since $a^{\prime}, b^{\prime}, r$ and $s$ are positive integers, $a+b+c+d$ is not a prime.
|
{
"exam": "BalticWay",
"problem_label": "6",
"problem_match": "\n6.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution 2",
"tier": "T3",
"year": "1996"
}
|
A sequence of integers $a_{1}, a_{2}, \ldots$, is such that $a_{1}=1, a_{2}=2$ and for $n \geq 1$
$$
a_{n+2}= \begin{cases}5 a_{n+1}-3 a_{n} & \text { if } a_{n} \cdot a_{n+1} \text { is even, } \\ a_{n+1}-a_{n} & \text { if } a_{n} \cdot a_{n+1} \text { is odd. }\end{cases}
$$
Prove that $a_{n} \neq 0$ for all $n$.
|
Considering the sequence modulo 6 we obtain 1, 2, 1, 5, 4, 5, 1, 2, ... The conclusion follows.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
A sequence of integers $a_{1}, a_{2}, \ldots$, is such that $a_{1}=1, a_{2}=2$ and for $n \geq 1$
$$
a_{n+2}= \begin{cases}5 a_{n+1}-3 a_{n} & \text { if } a_{n} \cdot a_{n+1} \text { is even, } \\ a_{n+1}-a_{n} & \text { if } a_{n} \cdot a_{n+1} \text { is odd. }\end{cases}
$$
Prove that $a_{n} \neq 0$ for all $n$.
|
Considering the sequence modulo 6 we obtain 1, 2, 1, 5, 4, 5, 1, 2, ... The conclusion follows.
|
{
"exam": "BalticWay",
"problem_label": "7",
"problem_match": "\n7.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
Consider the sequence
$$
\begin{aligned}
x_{1} & =19, \\
x_{2} & =95, \\
x_{n+2} & =\operatorname{lcm}\left(x_{n+1}, x_{n}\right)+x_{n},
\end{aligned}
$$
for $n>1$, where $\operatorname{lcm}(a, b)$ means the least common multiple of $a$ and $b$. Find the greatest common divisor of $x_{1995}$ and $x_{1996}$.
|
Let $d=\operatorname{gcd}\left(x_{k}, x_{k+1}\right)$. Then $\operatorname{lcm}\left(x_{k}, x_{k+1}\right)=x_{k} x_{k+1} / d$, and
$$
\operatorname{gcd}\left(x_{k+1}, x_{k+2}\right)=\operatorname{gcd}\left(x_{k+1}, \frac{x_{k} x_{k+1}}{d}+x_{k}\right)=\operatorname{gcd}\left(x_{k+1}, \frac{x_{k}}{d}\left(x_{k+1}+d\right)\right) .
$$
Since $x_{k+1}$ and $x_{k} / d$ are relatively prime, this equals $\operatorname{gcd}\left(x_{k+1}, x_{k+1}+d\right)=d$. It follows by induction that $\operatorname{gcd}\left(x_{n}, x_{n+1}\right)=\operatorname{gcd}\left(x_{1}, x_{2}\right)=19$ for all $n \geq 1$. Hence $\operatorname{gcd}\left(x_{1995}, x_{1996}\right)=19$.
|
19
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Consider the sequence
$$
\begin{aligned}
x_{1} & =19, \\
x_{2} & =95, \\
x_{n+2} & =\operatorname{lcm}\left(x_{n+1}, x_{n}\right)+x_{n},
\end{aligned}
$$
for $n>1$, where $\operatorname{lcm}(a, b)$ means the least common multiple of $a$ and $b$. Find the greatest common divisor of $x_{1995}$ and $x_{1996}$.
|
Let $d=\operatorname{gcd}\left(x_{k}, x_{k+1}\right)$. Then $\operatorname{lcm}\left(x_{k}, x_{k+1}\right)=x_{k} x_{k+1} / d$, and
$$
\operatorname{gcd}\left(x_{k+1}, x_{k+2}\right)=\operatorname{gcd}\left(x_{k+1}, \frac{x_{k} x_{k+1}}{d}+x_{k}\right)=\operatorname{gcd}\left(x_{k+1}, \frac{x_{k}}{d}\left(x_{k+1}+d\right)\right) .
$$
Since $x_{k+1}$ and $x_{k} / d$ are relatively prime, this equals $\operatorname{gcd}\left(x_{k+1}, x_{k+1}+d\right)=d$. It follows by induction that $\operatorname{gcd}\left(x_{n}, x_{n+1}\right)=\operatorname{gcd}\left(x_{1}, x_{2}\right)=19$ for all $n \geq 1$. Hence $\operatorname{gcd}\left(x_{1995}, x_{1996}\right)=19$.
|
{
"exam": "BalticWay",
"problem_label": "8",
"problem_match": "\n8.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
Let $n$ and $k$ be integers, $1<k \leq n$. Find an integer $b$ and a set $A$ of $n$ integers satisfying the following conditions:
(i) No product of $k-1$ distinct elements of $A$ is divisible by $b$.
(ii) Every product of $k$ distinct elements of $A$ is divisible by $b$.
(iii) For all distinct $a, a^{\prime}$ in $A$, $a$ does not divide $a^{\prime}$.
|
Let $p_{1}, \ldots, p_{n}$ be the first $n$ odd primes. Then we can take $A=\left\{2 p_{1}, 2 p_{2}, \ldots, 2 p_{n}\right\}$ and $b=2^{k}$. It is easily seen that the conditions are satisfied.
|
A=\left\{2 p_{1}, 2 p_{2}, \ldots, 2 p_{n}\right\}, b=2^{k}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $n$ and $k$ be integers, $1<k \leq n$. Find an integer $b$ and a set $A$ of $n$ integers satisfying the following conditions:
(i) No product of $k-1$ distinct elements of $A$ is divisible by $b$.
(ii) Every product of $k$ distinct elements of $A$ is divisible by $b$.
(iii) For all distinct $a, a^{\prime}$ in $A$, $a$ does not divide $a^{\prime}$.
|
Let $p_{1}, \ldots, p_{n}$ be the first $n$ odd primes. Then we can take $A=\left\{2 p_{1}, 2 p_{2}, \ldots, 2 p_{n}\right\}$ and $b=2^{k}$. It is easily seen that the conditions are satisfied.
|
{
"exam": "BalticWay",
"problem_label": "9",
"problem_match": "\n9.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
Denote by $d(n)$ the number of distinct positive divisors of a positive integer $n$ (including 1 and $n$ ). Let $a>1$ and $n>0$ be integers such that $a^{n}+1$ is a prime. Prove that
$$
d\left(a^{n}-1\right) \geq n .
$$
|
First we show that $n=2^{s}$ for some integer $s \geq 0$. Indeed, if $n=m p$ where $p$ is an odd prime, then $a^{n}+1=a^{m p}+1=\left(a^{m}+1\right)\left(a^{m(p-1)}-a^{m(p-2)}+\cdots-a+1\right)$, a contradiction.
Now we use induction on $s$ to prove that $d\left(a^{2^{s}}-1\right) \geq 2^{s}$. The case $s=0$ is obvious. As $a^{2^{s}}-1=$ $\left(a^{2^{s-1}}-1\right)\left(a^{2^{s-1}}+1\right)$, then for any divisor $q$ of $a^{2^{s-1}}-1$, both $q$ and $q\left(a^{2^{s-1}}+1\right)$ are divisors of $a^{2^{s}}-1$. Since the divisors of the form $q\left(a^{2^{s-1}}+1\right)$ are all larger than $a^{2^{s-1}}-1$ we have $d\left(a^{2^{s}}-1\right) \geq 2 \cdot d\left(a^{2^{s-1}}-1\right)=2^{s}$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Denote by $d(n)$ the number of distinct positive divisors of a positive integer $n$ (including 1 and $n$ ). Let $a>1$ and $n>0$ be integers such that $a^{n}+1$ is a prime. Prove that
$$
d\left(a^{n}-1\right) \geq n .
$$
|
First we show that $n=2^{s}$ for some integer $s \geq 0$. Indeed, if $n=m p$ where $p$ is an odd prime, then $a^{n}+1=a^{m p}+1=\left(a^{m}+1\right)\left(a^{m(p-1)}-a^{m(p-2)}+\cdots-a+1\right)$, a contradiction.
Now we use induction on $s$ to prove that $d\left(a^{2^{s}}-1\right) \geq 2^{s}$. The case $s=0$ is obvious. As $a^{2^{s}}-1=$ $\left(a^{2^{s-1}}-1\right)\left(a^{2^{s-1}}+1\right)$, then for any divisor $q$ of $a^{2^{s-1}}-1$, both $q$ and $q\left(a^{2^{s-1}}+1\right)$ are divisors of $a^{2^{s}}-1$. Since the divisors of the form $q\left(a^{2^{s-1}}+1\right)$ are all larger than $a^{2^{s-1}}-1$ we have $d\left(a^{2^{s}}-1\right) \geq 2 \cdot d\left(a^{2^{s-1}}-1\right)=2^{s}$.
|
{
"exam": "BalticWay",
"problem_label": "10",
"problem_match": "\n10.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
The real numbers $x_{1}, x_{2}, \ldots, x_{1996}$ have the following property: for any polynomial $W$ of degree 2 at least three of the numbers $W\left(x_{1}\right), W\left(x_{2}\right), \ldots, W\left(x_{1996}\right)$ are equal. Prove that at least three of the numbers $x_{1}, x_{2}, \ldots, x_{1996}$ are equal.
|
Let $m=\min \left\{x_{1}, \ldots, x_{1996}\right\}$. Then the polynomial $W(x)=(x-m)^{2}$ is strictly increasing for $x \geq m$. Hence if $W\left(x_{i}\right)=W\left(x_{j}\right)$ we must have $x_{i}=x_{j}$, and the conclusion follows.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
The real numbers $x_{1}, x_{2}, \ldots, x_{1996}$ have the following property: for any polynomial $W$ of degree 2 at least three of the numbers $W\left(x_{1}\right), W\left(x_{2}\right), \ldots, W\left(x_{1996}\right)$ are equal. Prove that at least three of the numbers $x_{1}, x_{2}, \ldots, x_{1996}$ are equal.
|
Let $m=\min \left\{x_{1}, \ldots, x_{1996}\right\}$. Then the polynomial $W(x)=(x-m)^{2}$ is strictly increasing for $x \geq m$. Hence if $W\left(x_{i}\right)=W\left(x_{j}\right)$ we must have $x_{i}=x_{j}$, and the conclusion follows.
|
{
"exam": "BalticWay",
"problem_label": "11",
"problem_match": "\n11.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
Let $S$ be a set of integers containing the numbers 0 and 1996. Suppose further that any integer root of any non-zero polynomial with coefficients in $S$ also belongs to $S$. Prove that -2 belongs to $S$.
|
Consider the polynomial $W(x)=1996 x+1996$. As $W(-1)=0$ we conclude that $-1 \in S$. Now consider the polynomial $U(x)=-x^{1996}-x^{1995}-\cdots-x^{2}-x+1996$. As $U(1)=0$ we have $1 \in S$. Finally, let $T(x)=-x^{10}+x^{9}-x^{8}+x^{7}-x^{6}+x^{3}-x^{2}+1996$. Then $-2 \in S$ since $T(-2)=0$.
|
-2 \in S
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $S$ be a set of integers containing the numbers 0 and 1996. Suppose further that any integer root of any non-zero polynomial with coefficients in $S$ also belongs to $S$. Prove that -2 belongs to $S$.
|
Consider the polynomial $W(x)=1996 x+1996$. As $W(-1)=0$ we conclude that $-1 \in S$. Now consider the polynomial $U(x)=-x^{1996}-x^{1995}-\cdots-x^{2}-x+1996$. As $U(1)=0$ we have $1 \in S$. Finally, let $T(x)=-x^{10}+x^{9}-x^{8}+x^{7}-x^{6}+x^{3}-x^{2}+1996$. Then $-2 \in S$ since $T(-2)=0$.
|
{
"exam": "BalticWay",
"problem_label": "12",
"problem_match": "\n12.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
Consider the functions $f$ defined on the set of integers such that
$$
f(x)=f\left(x^{2}+x+1\right),
$$
for all integers $x$. Find
(a) all even functions,
(b) all odd functions of this kind.
|
(a) For $f$ even, we have $f(x-1)=f\left((x-1)^{2}+(x-1)+1\right)=f\left(x^{2}-x+1\right)=f\left((-x)^{2}-x+1\right)=f(-x)=f(x)$ for any $x \in \mathbb{Z}$. Hence $f$ has a constant value; any constant will do.
(b) For $f$ odd, a similar computation yields $f(x-1)=-f(x)$. Since $f(0)=0$, we see that $f(x)=0$ for all $x \in \mathbb{Z}$.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Consider the functions $f$ defined on the set of integers such that
$$
f(x)=f\left(x^{2}+x+1\right),
$$
for all integers $x$. Find
(a) all even functions,
(b) all odd functions of this kind.
|
(a) For $f$ even, we have $f(x-1)=f\left((x-1)^{2}+(x-1)+1\right)=f\left(x^{2}-x+1\right)=f\left((-x)^{2}-x+1\right)=f(-x)=f(x)$ for any $x \in \mathbb{Z}$. Hence $f$ has a constant value; any constant will do.
(b) For $f$ odd, a similar computation yields $f(x-1)=-f(x)$. Since $f(0)=0$, we see that $f(x)=0$ for all $x \in \mathbb{Z}$.
|
{
"exam": "BalticWay",
"problem_label": "13",
"problem_match": "\n13.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "# Solution.",
"tier": "T3",
"year": "1996"
}
|
The graph of the function $f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ (where $n>1$ ), intersects the line $y=b$ at the points $B_{1}, B_{2}, \ldots, B_{n}$ (from left to right), and the line $y=c(c \neq b)$ at the points $C_{1}, C_{2}, \ldots, C_{n}$ (from left to right). Let $P$ be a point on the line $y=c$, to the right to the point $C_{n}$. Find the sum $\cot \angle B_{1} C_{1} P+\cdots+\cot \angle B_{n} C_{n} P$.
|
Let the points $B_{i}$ and $C_{i}$ have the coordinates $\left(b_{i}, b\right)$ and $\left(c_{i}, c\right)$, respectively, for $i=1,2, \ldots, n$. Then we have
$$
\cot \angle B_{1} C_{1} P+\cdots+\cot \angle B_{n} C_{n} P=\frac{1}{b-c} \sum_{i=1}^{n}\left(b_{i}-c_{i}\right)
$$
The numbers $b_{i}$ and $c_{i}$ are the solutions of $f(x)-b=0$ and $f(x)-c=0$, respectively. As $n \geq 2$, it follows from the relationships between the roots and coefficients of a polynomial (Viète's relations) that $\sum_{i=1}^{n} b_{i}=$ $\sum_{i=1}^{n} c_{i}=-a_{n-1}$ regardless of the values of $b$ and $c$, and hence $\cot \angle B_{1} C_{1} P+\cdots+\cot \angle B_{n} C_{n} P=0$.
|
0
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
The graph of the function $f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ (where $n>1$ ), intersects the line $y=b$ at the points $B_{1}, B_{2}, \ldots, B_{n}$ (from left to right), and the line $y=c(c \neq b)$ at the points $C_{1}, C_{2}, \ldots, C_{n}$ (from left to right). Let $P$ be a point on the line $y=c$, to the right to the point $C_{n}$. Find the sum $\cot \angle B_{1} C_{1} P+\cdots+\cot \angle B_{n} C_{n} P$.
|
Let the points $B_{i}$ and $C_{i}$ have the coordinates $\left(b_{i}, b\right)$ and $\left(c_{i}, c\right)$, respectively, for $i=1,2, \ldots, n$. Then we have
$$
\cot \angle B_{1} C_{1} P+\cdots+\cot \angle B_{n} C_{n} P=\frac{1}{b-c} \sum_{i=1}^{n}\left(b_{i}-c_{i}\right)
$$
The numbers $b_{i}$ and $c_{i}$ are the solutions of $f(x)-b=0$ and $f(x)-c=0$, respectively. As $n \geq 2$, it follows from the relationships between the roots and coefficients of a polynomial (Viète's relations) that $\sum_{i=1}^{n} b_{i}=$ $\sum_{i=1}^{n} c_{i}=-a_{n-1}$ regardless of the values of $b$ and $c$, and hence $\cot \angle B_{1} C_{1} P+\cdots+\cot \angle B_{n} C_{n} P=0$.
|
{
"exam": "BalticWay",
"problem_label": "14",
"problem_match": "\n14.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
For which positive real numbers $a, b$ does the inequality
$$
x_{1} \cdot x_{2}+x_{2} \cdot x_{3}+\cdots+x_{n-1} \cdot x_{n}+x_{n} \cdot x_{1} \geq x_{1}^{a} \cdot x_{2}^{b} \cdot x_{3}^{a}+x_{2}^{a} \cdot x_{3}^{b} \cdot x_{4}^{a}+\cdots+x_{n}^{a} \cdot x_{1}^{b} \cdot x_{2}^{a}
$$
hold for all integers $n>2$ and positive real numbers $x_{1}, x_{2}, \ldots, x_{n}$ ?
|
Substituting $x_{i}=x$ easily yields that $2 a+b=2$. Now take $n=4, x_{1}=x_{3}=x$ and $x_{2}=x_{4}=1$. This gives $2 x \geq x^{2 a}+x^{b}$. But the inequality between the arithmetic and geometric mean yields $x^{2 a}+x^{b} \geq$ $2 \sqrt{x^{2 a} x^{b}}=2 x$. Here equality must hold, and this implies that $x^{2 a}=x^{b}$, which gives $2 a=b=1$.
On the other hand, if $b=1$ and $a=\frac{1}{2}$, we let $y_{i}=\sqrt{x_{i} x_{i+1}}$ for $1 \leq i \leq n$, with $x_{n+1}=x_{1}$. The inequality then takes the form
$$
y_{1}^{2}+\cdots+y_{n}^{2} \geq y_{1} y_{2}+y_{2} y_{3}+\cdots+y_{n} y_{1} \text {. }
$$
But the inequality between the arithmetic and geometric mean yields
$$
\frac{1}{2}\left(y_{i}^{2}+y_{i+1}^{2}\right) \geq y_{i} y_{i+1}, \quad 1 \leq i \leq n,
$$
where $y_{n+1}=y_{n}$. Adding these $n$ inequalities yields the inequality (1).
The inequality (1) can also be obtained from the Cauchy-Schwarz inequality, which implies that $\sum_{i=1}^{n} y_{i}^{2} \sum_{i=1}^{n} y_{i+1}^{2} \geq\left(\sum_{i=1}^{n} y_{i} y_{i+1}\right)^{2}$, which is exactly the stated inequality.
|
b=1, a=\frac{1}{2}
|
Yes
|
Yes
|
math-word-problem
|
Inequalities
|
For which positive real numbers $a, b$ does the inequality
$$
x_{1} \cdot x_{2}+x_{2} \cdot x_{3}+\cdots+x_{n-1} \cdot x_{n}+x_{n} \cdot x_{1} \geq x_{1}^{a} \cdot x_{2}^{b} \cdot x_{3}^{a}+x_{2}^{a} \cdot x_{3}^{b} \cdot x_{4}^{a}+\cdots+x_{n}^{a} \cdot x_{1}^{b} \cdot x_{2}^{a}
$$
hold for all integers $n>2$ and positive real numbers $x_{1}, x_{2}, \ldots, x_{n}$ ?
|
Substituting $x_{i}=x$ easily yields that $2 a+b=2$. Now take $n=4, x_{1}=x_{3}=x$ and $x_{2}=x_{4}=1$. This gives $2 x \geq x^{2 a}+x^{b}$. But the inequality between the arithmetic and geometric mean yields $x^{2 a}+x^{b} \geq$ $2 \sqrt{x^{2 a} x^{b}}=2 x$. Here equality must hold, and this implies that $x^{2 a}=x^{b}$, which gives $2 a=b=1$.
On the other hand, if $b=1$ and $a=\frac{1}{2}$, we let $y_{i}=\sqrt{x_{i} x_{i+1}}$ for $1 \leq i \leq n$, with $x_{n+1}=x_{1}$. The inequality then takes the form
$$
y_{1}^{2}+\cdots+y_{n}^{2} \geq y_{1} y_{2}+y_{2} y_{3}+\cdots+y_{n} y_{1} \text {. }
$$
But the inequality between the arithmetic and geometric mean yields
$$
\frac{1}{2}\left(y_{i}^{2}+y_{i+1}^{2}\right) \geq y_{i} y_{i+1}, \quad 1 \leq i \leq n,
$$
where $y_{n+1}=y_{n}$. Adding these $n$ inequalities yields the inequality (1).
The inequality (1) can also be obtained from the Cauchy-Schwarz inequality, which implies that $\sum_{i=1}^{n} y_{i}^{2} \sum_{i=1}^{n} y_{i+1}^{2} \geq\left(\sum_{i=1}^{n} y_{i} y_{i+1}\right)^{2}$, which is exactly the stated inequality.
|
{
"exam": "BalticWay",
"problem_label": "15",
"problem_match": "\n15.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
On an infinite checkerboard, two players alternately mark one unmarked cell. One of them uses $\times$, the other $o$. The first who fills a $2 \times 2$ square with his symbols wins. Can the player who starts always win?
|
Divide the plane into dominoes in the way indicated by the thick lines in Figure 2. The second player can respond by marking the other cell of the same domino where the first player placed his mark. Since every $2 \times 2$ square contains one whole domino, the first player cannot win.
Figure 2
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
On an infinite checkerboard, two players alternately mark one unmarked cell. One of them uses $\times$, the other $o$. The first who fills a $2 \times 2$ square with his symbols wins. Can the player who starts always win?
|
Divide the plane into dominoes in the way indicated by the thick lines in Figure 2. The second player can respond by marking the other cell of the same domino where the first player placed his mark. Since every $2 \times 2$ square contains one whole domino, the first player cannot win.
Figure 2
|
{
"exam": "BalticWay",
"problem_label": "16",
"problem_match": "\n16.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
Using each of the eight digits $1,3,4,5,6,7,8$ and 9 exactly once, a three-digit number $A$, two twodigit numbers $B$ and $C, B<C$, and a one-digit number $D$ are formed. The numbers are such that $A+D=B+C=143$. In how many ways can this be done?
|
From $A=143-D$ and $1 \leq D \leq 9$, it follows that $134 \leq A \leq 142$. The hundreds digit of $A$ is therefore 1 , and the tens digit is either 3 or 4 . If the tens digit of $A$ is 4 , then the sum of the units digits of
$A$ and $D$ must be 3 , which is impossible, as the digits 0 and 2 are not among the eight digits given. Hence the first two digits of $A$ are uniquely determined as 1 and 3 . The sum of the units digits of $A$ and $D$ must be 13. This can be achieved in six different ways as $13=4+9=5+8=6+7=7+6=8+5=9+4$.
The sum of the units digits of $B$ and $C$ must again be 13, and as $B+C=143$, this must also be true for the tens digits. For each choice of the numbers $A$ and $D$, the remaining four digits form two pairs, both with the sum 13. The units digits of $B$ and $C$ may then be chosen in four ways. The tens digits are then uniquely determined by the remaining pair and the relation $B<C$. The total number of possibilities is therefore $6 \cdot 4=24$.
|
24
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Using each of the eight digits $1,3,4,5,6,7,8$ and 9 exactly once, a three-digit number $A$, two twodigit numbers $B$ and $C, B<C$, and a one-digit number $D$ are formed. The numbers are such that $A+D=B+C=143$. In how many ways can this be done?
|
From $A=143-D$ and $1 \leq D \leq 9$, it follows that $134 \leq A \leq 142$. The hundreds digit of $A$ is therefore 1 , and the tens digit is either 3 or 4 . If the tens digit of $A$ is 4 , then the sum of the units digits of
$A$ and $D$ must be 3 , which is impossible, as the digits 0 and 2 are not among the eight digits given. Hence the first two digits of $A$ are uniquely determined as 1 and 3 . The sum of the units digits of $A$ and $D$ must be 13. This can be achieved in six different ways as $13=4+9=5+8=6+7=7+6=8+5=9+4$.
The sum of the units digits of $B$ and $C$ must again be 13, and as $B+C=143$, this must also be true for the tens digits. For each choice of the numbers $A$ and $D$, the remaining four digits form two pairs, both with the sum 13. The units digits of $B$ and $C$ may then be chosen in four ways. The tens digits are then uniquely determined by the remaining pair and the relation $B<C$. The total number of possibilities is therefore $6 \cdot 4=24$.
|
{
"exam": "BalticWay",
"problem_label": "17",
"problem_match": "\n17.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
The jury of an olympiad has 30 members in the beginning. Each member of the jury thinks that some of his colleagues are competent, while all the others are not, and these opinions do not change. At the beginning of every session a voting takes place, and those members who are not competent in the opinion of more than one half of the voters are excluded from the jury for the rest of the olympiad. Prove that after at most 15 sessions there will be no more exclusions. (Note that nobody votes about his own competence.)
|
First we note that if nobody is excluded in some session, then the situation becomes stable and nobody can be excluded in any later session.
We use induction to prove the slightly more general claim that if the jury has $2 n$ members, $n \geq 2$, then after at most $n$ sessions nobody will be excluded anymore. For $n=2$ the claim is obvious, since if some members are excluded in the first two sessions, there are at most two members left, and hence nobody is excluded in the third session.
Now assuming that the claim is true for $n \leq k-1$, suppose the jury has $2 k$ members, and consider the first session. If nobody is excluded, we are done. If a positive and even number of members are excluded, there will be $2 r$ members left with $r<k$, and by the induction hypotheses the jury will stabilize after at most $r$ more sessions, giving a total of at most $r+1 \leq k$ sessions, as required.
Finally suppose that an odd number of members are excluded in the first session. There are three alternatives:
(i) An even number of members are excluded in each of the next $m$ sessions, after which nobody is excluded. Then the number of members left is at most $2 k-1-2 m$. Hence $2 k-1-2 m \geq 1$, so that $k \geq m+1$. Hence the number of sessions is at most $k$.
(ii) An even number of members are excluded in each of the next $m$ sessions, after which an odd number of members greater than 1 are excluded. Then there are at most $2 k-1-2 m-3$ members left, and by the induction hypotheses, the jury will stabilize in no more than $k-m-2$ sessions. The total number of sessions is therefore $1+m+1+(k-m-2)=k$.
(iii) An even number of members are excluded in each of the next $m$ sessions, followed by a session where precisely one member $M$ is excluded. In this session, there were $2 r+1$ members present for some $r$, and $r+1$ of these voted for the exclusion of $M$. But then any member other than $M$ was thought to be incompetent by at most $r$ others. In the next session the jury will have $2 r$ members, and since the members do not change their sympathies, nobody can be excluded. Hence the situation is stable after $m+2$ sessions, and at least $1+2 m+1=2 m+2$ members have been excluded. But there must be at least 3 members left, for one member cannot be excluded from a jury of 2 members. Hence $2 m+2 \leq 2 k-3$, whence $m+2 \leq k$.
Thus the claim holds for $n=k$ also. We conclude that the claim holds for all $n \geq 2$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
The jury of an olympiad has 30 members in the beginning. Each member of the jury thinks that some of his colleagues are competent, while all the others are not, and these opinions do not change. At the beginning of every session a voting takes place, and those members who are not competent in the opinion of more than one half of the voters are excluded from the jury for the rest of the olympiad. Prove that after at most 15 sessions there will be no more exclusions. (Note that nobody votes about his own competence.)
|
First we note that if nobody is excluded in some session, then the situation becomes stable and nobody can be excluded in any later session.
We use induction to prove the slightly more general claim that if the jury has $2 n$ members, $n \geq 2$, then after at most $n$ sessions nobody will be excluded anymore. For $n=2$ the claim is obvious, since if some members are excluded in the first two sessions, there are at most two members left, and hence nobody is excluded in the third session.
Now assuming that the claim is true for $n \leq k-1$, suppose the jury has $2 k$ members, and consider the first session. If nobody is excluded, we are done. If a positive and even number of members are excluded, there will be $2 r$ members left with $r<k$, and by the induction hypotheses the jury will stabilize after at most $r$ more sessions, giving a total of at most $r+1 \leq k$ sessions, as required.
Finally suppose that an odd number of members are excluded in the first session. There are three alternatives:
(i) An even number of members are excluded in each of the next $m$ sessions, after which nobody is excluded. Then the number of members left is at most $2 k-1-2 m$. Hence $2 k-1-2 m \geq 1$, so that $k \geq m+1$. Hence the number of sessions is at most $k$.
(ii) An even number of members are excluded in each of the next $m$ sessions, after which an odd number of members greater than 1 are excluded. Then there are at most $2 k-1-2 m-3$ members left, and by the induction hypotheses, the jury will stabilize in no more than $k-m-2$ sessions. The total number of sessions is therefore $1+m+1+(k-m-2)=k$.
(iii) An even number of members are excluded in each of the next $m$ sessions, followed by a session where precisely one member $M$ is excluded. In this session, there were $2 r+1$ members present for some $r$, and $r+1$ of these voted for the exclusion of $M$. But then any member other than $M$ was thought to be incompetent by at most $r$ others. In the next session the jury will have $2 r$ members, and since the members do not change their sympathies, nobody can be excluded. Hence the situation is stable after $m+2$ sessions, and at least $1+2 m+1=2 m+2$ members have been excluded. But there must be at least 3 members left, for one member cannot be excluded from a jury of 2 members. Hence $2 m+2 \leq 2 k-3$, whence $m+2 \leq k$.
Thus the claim holds for $n=k$ also. We conclude that the claim holds for all $n \geq 2$.
|
{
"exam": "BalticWay",
"problem_label": "18",
"problem_match": "\n18.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
Four heaps contain 38, 45, 61, and 70 matches respectively. Two players take turns choosing any two of the heaps and take some non-zero number of matches from one heap and some non-zero number of matches from the other heap. The player who cannot make a move, loses. Which one of the players has a winning strategy?
|
The first player wins by making moves so that the opponent must face positions of the form $(a, a, a, b)$, where $a \leq b$.
|
not found
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Four heaps contain 38, 45, 61, and 70 matches respectively. Two players take turns choosing any two of the heaps and take some non-zero number of matches from one heap and some non-zero number of matches from the other heap. The player who cannot make a move, loses. Which one of the players has a winning strategy?
|
The first player wins by making moves so that the opponent must face positions of the form $(a, a, a, b)$, where $a \leq b$.
|
{
"exam": "BalticWay",
"problem_label": "19",
"problem_match": "\n19.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
Is it possible to partition all positive integers into disjoint sets $A$ and $B$ such that
(i) no three numbers of $A$ form arithmetic progression,
(ii) no infinite non-constant arithmetic progression can be formed by numbers of $B$ ?
|
Let $\mathbb{N}$ denote the set of positive integers. There is a bijective function $f: \mathbb{N} \rightarrow \mathbb{N} \times \mathbb{N}$. Let $a_{0}=1$, and for $k \geq 1$, let $a_{k}$ be the least integer of the form $m+t n$ for some integer $t \geq 0$ where $f(k)=(m, n)$, such that $a_{k} \geq 2 a_{k-1}$. Let $A=\left\{a_{0}, a_{1}, \ldots\right\}$ and let $B=\mathbb{N} \backslash A$. We now show that $A$ and $B$ satisfy the given conditions.
(i) For any non-negative integers $i<j<k$, we have $a_{k} \geq a_{j+1} \geq 2 a_{j}$, and hence $a_{k}-a_{j} \geq a_{j}>a_{j}-a_{i}$. Thus $a_{i}, a_{j}$ and $a_{k}$ do not form an arithmetic progression, since this would mean that $a_{k}-a_{j}=a_{j}-a_{i}$. Hence no three numbers in $A$ form an arithmetic progression.
(ii) Consider an infinite arithmetic progression $m, m+n, m+2 n, \ldots$, with $m, n \in \mathbb{N}$. Then $m+n t=a_{k}$ for some integer $t \geq 0$, where $k=f^{-1}(m, n)$. Thus $a_{k}$ belongs to the arithmetic progression, but $a_{k} \notin B$. Hence $B$ does not contain any infinite non-constant arithmetic progression.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Is it possible to partition all positive integers into disjoint sets $A$ and $B$ such that
(i) no three numbers of $A$ form arithmetic progression,
(ii) no infinite non-constant arithmetic progression can be formed by numbers of $B$ ?
|
Let $\mathbb{N}$ denote the set of positive integers. There is a bijective function $f: \mathbb{N} \rightarrow \mathbb{N} \times \mathbb{N}$. Let $a_{0}=1$, and for $k \geq 1$, let $a_{k}$ be the least integer of the form $m+t n$ for some integer $t \geq 0$ where $f(k)=(m, n)$, such that $a_{k} \geq 2 a_{k-1}$. Let $A=\left\{a_{0}, a_{1}, \ldots\right\}$ and let $B=\mathbb{N} \backslash A$. We now show that $A$ and $B$ satisfy the given conditions.
(i) For any non-negative integers $i<j<k$, we have $a_{k} \geq a_{j+1} \geq 2 a_{j}$, and hence $a_{k}-a_{j} \geq a_{j}>a_{j}-a_{i}$. Thus $a_{i}, a_{j}$ and $a_{k}$ do not form an arithmetic progression, since this would mean that $a_{k}-a_{j}=a_{j}-a_{i}$. Hence no three numbers in $A$ form an arithmetic progression.
(ii) Consider an infinite arithmetic progression $m, m+n, m+2 n, \ldots$, with $m, n \in \mathbb{N}$. Then $m+n t=a_{k}$ for some integer $t \geq 0$, where $k=f^{-1}(m, n)$. Thus $a_{k}$ belongs to the arithmetic progression, but $a_{k} \notin B$. Hence $B$ does not contain any infinite non-constant arithmetic progression.
|
{
"exam": "BalticWay",
"problem_label": "20",
"problem_match": "\n20.",
"resource_path": "BalticWay/segmented/en-bw96sol.jsonl",
"solution_match": "\nSolution.",
"tier": "T3",
"year": "1996"
}
|
Determine all functions $f$ from the real numbers to the real numbers, different from the zero function, such that $f(x) f(y)=f(x-y)$ for all real numbers $x$ and $y$.
|
Answer: $f(x) \equiv 1$ is the only such function.
Since $f$ is not the zero function, there is an $x_{0}$ such that $f\left(x_{0}\right) \neq 0$. From $f\left(x_{0}\right) f(0)=f\left(x_{0}-0\right)=f\left(x_{0}\right)$ we then get $f(0)=1$. Then by $f(x)^{2}=f(x) f(x)=f(x-x)=f(0)$ we have $f(x) \neq 0$ for any real $x$. Finally from $f(x) f\left(\frac{x}{2}\right)=f\left(x-\frac{x}{2}\right)=f\left(\frac{x}{2}\right)$ we get $f(x)=1$ for any real $x$. It is readily verified that this function satisfies the equation.
|
f(x) \equiv 1
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Determine all functions $f$ from the real numbers to the real numbers, different from the zero function, such that $f(x) f(y)=f(x-y)$ for all real numbers $x$ and $y$.
|
Answer: $f(x) \equiv 1$ is the only such function.
Since $f$ is not the zero function, there is an $x_{0}$ such that $f\left(x_{0}\right) \neq 0$. From $f\left(x_{0}\right) f(0)=f\left(x_{0}-0\right)=f\left(x_{0}\right)$ we then get $f(0)=1$. Then by $f(x)^{2}=f(x) f(x)=f(x-x)=f(0)$ we have $f(x) \neq 0$ for any real $x$. Finally from $f(x) f\left(\frac{x}{2}\right)=f\left(x-\frac{x}{2}\right)=f\left(\frac{x}{2}\right)$ we get $f(x)=1$ for any real $x$. It is readily verified that this function satisfies the equation.
|
{
"exam": "BalticWay",
"problem_label": "1",
"problem_match": "\n1.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n1.",
"tier": "T3",
"year": "1997"
}
|
Given a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers in which every positive integer occurs exactly once. Prove that there exist integers $\ell$ and $m$, $1<\ell<m$, such that $a_{1}+a_{m}=2 a_{\ell}$.
|
Let $\ell$ be the least index such that $a_{\ell}>a_{1}$. Since $2 a_{\ell}-a_{1}$ is a positive integer larger than $a_{1}$, it occurs in the given sequence beyond $a_{\ell}$. In other words, there exists an index $m>\ell$ such that $a_{m}=2 a_{\ell}-a_{1}$. This completes the proof.
Remarks. The problem was proposed in the slightly more general form where the first term of the arithmetic progression has an arbitary index. The remarks below refer to this version. The problem committee felt that no essential new aspects would arise from the generalization.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Given a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers in which every positive integer occurs exactly once. Prove that there exist integers $\ell$ and $m$, $1<\ell<m$, such that $a_{1}+a_{m}=2 a_{\ell}$.
|
Let $\ell$ be the least index such that $a_{\ell}>a_{1}$. Since $2 a_{\ell}-a_{1}$ is a positive integer larger than $a_{1}$, it occurs in the given sequence beyond $a_{\ell}$. In other words, there exists an index $m>\ell$ such that $a_{m}=2 a_{\ell}-a_{1}$. This completes the proof.
Remarks. The problem was proposed in the slightly more general form where the first term of the arithmetic progression has an arbitary index. The remarks below refer to this version. The problem committee felt that no essential new aspects would arise from the generalization.
|
{
"exam": "BalticWay",
"problem_label": "2",
"problem_match": "\n2.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n2.",
"tier": "T3",
"year": "1997"
}
|
Determine all functions $f$ from the real numbers to the real numbers, different from the zero function, such that $f(x) f(y)=f(x-y)$ for all real numbers $x$ and $y$.
|
A generalization of this problem is to ask about an existence of an $s$-term arithmetic subsequence of the sequence $\left(a_{n}\right)$ (such a subsequence
always exists for $s=3$, as shown above). It turns out that for $s=5$ such a subsequence may not exist. The proof can be found in [1]. The same problem for $s=4$ is still open!
|
not found
|
Yes
|
Problem not solved
|
math-word-problem
|
Algebra
|
Determine all functions $f$ from the real numbers to the real numbers, different from the zero function, such that $f(x) f(y)=f(x-y)$ for all real numbers $x$ and $y$.
|
A generalization of this problem is to ask about an existence of an $s$-term arithmetic subsequence of the sequence $\left(a_{n}\right)$ (such a subsequence
always exists for $s=3$, as shown above). It turns out that for $s=5$ such a subsequence may not exist. The proof can be found in [1]. The same problem for $s=4$ is still open!
|
{
"exam": "BalticWay",
"problem_label": "1",
"problem_match": "\n1.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n1.",
"tier": "T3",
"year": "1997"
}
|
Given a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers in which every positive integer occurs exactly once. Prove that there exist integers $\ell$ and $m$, $1<\ell<m$, such that $a_{1}+a_{m}=2 a_{\ell}$.
|
The present problem $(s=3)$ and the above solution is also taken from [1].
Reference. [1] J. A. Davis, R. C. Entringer, R. L. Graham and G. J. Simmons, On permutations containing no long arithmetic progressions, Acta Arithmetica 1(1977), pp. 81-90.
|
proof
|
Yes
|
Incomplete
|
proof
|
Number Theory
|
Given a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers in which every positive integer occurs exactly once. Prove that there exist integers $\ell$ and $m$, $1<\ell<m$, such that $a_{1}+a_{m}=2 a_{\ell}$.
|
The present problem $(s=3)$ and the above solution is also taken from [1].
Reference. [1] J. A. Davis, R. C. Entringer, R. L. Graham and G. J. Simmons, On permutations containing no long arithmetic progressions, Acta Arithmetica 1(1977), pp. 81-90.
|
{
"exam": "BalticWay",
"problem_label": "2",
"problem_match": "\n2.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n2.",
"tier": "T3",
"year": "1997"
}
|
Let $x_{1}=1$ and $x_{n+1}=x_{n}+\left\lfloor\frac{x_{n}}{n}\right\rfloor+2$ for $n=1,2,3, \ldots$, where $\lfloor x\rfloor$ denotes the largest integer not greater than $x$. Determine $x_{1997}$.
|
Answer: $x_{1997}=23913$.
Note that if $x_{n}=a n+b$ with $0 \leqslant b<n$, then
$$
x_{n+1}=x_{n}+a+2=a(n+1)+b+2 \text {. }
$$
Hence if $x_{N}=A N$ for some positive integers $A$ and $N$, then for $i=0,1, \ldots, N$ we have $x_{N+i}=A(N+i)+2 i$, and $x_{2 N}=(A+1) \cdot 2 N$. Since for $N=1$ the condition $x_{N}=A N$ holds with $A=1$, then for $N=2^{k}$ (where $k$ is any non-negative integer) it also holds with $A=k+1$. Now for $N=2^{10}=1024$ we have $A=11$ and $x_{N+i}=A(N+i)+2 i$, which for $i=973$ makes $x_{1997}=11 \cdot 1997+2 \cdot 973=23913$.
|
23913
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $x_{1}=1$ and $x_{n+1}=x_{n}+\left\lfloor\frac{x_{n}}{n}\right\rfloor+2$ for $n=1,2,3, \ldots$, where $\lfloor x\rfloor$ denotes the largest integer not greater than $x$. Determine $x_{1997}$.
|
Answer: $x_{1997}=23913$.
Note that if $x_{n}=a n+b$ with $0 \leqslant b<n$, then
$$
x_{n+1}=x_{n}+a+2=a(n+1)+b+2 \text {. }
$$
Hence if $x_{N}=A N$ for some positive integers $A$ and $N$, then for $i=0,1, \ldots, N$ we have $x_{N+i}=A(N+i)+2 i$, and $x_{2 N}=(A+1) \cdot 2 N$. Since for $N=1$ the condition $x_{N}=A N$ holds with $A=1$, then for $N=2^{k}$ (where $k$ is any non-negative integer) it also holds with $A=k+1$. Now for $N=2^{10}=1024$ we have $A=11$ and $x_{N+i}=A(N+i)+2 i$, which for $i=973$ makes $x_{1997}=11 \cdot 1997+2 \cdot 973=23913$.
|
{
"exam": "BalticWay",
"problem_label": "3",
"problem_match": "\n3.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n3.",
"tier": "T3",
"year": "1997"
}
|
Prove that the arithmetic mean $a$ of $x_{1}, \ldots, x_{n}$ satisfies
$$
\left(x_{1}-a\right)^{2}+\cdots+\left(x_{n}-a\right)^{2} \leqslant \frac{1}{2}\left(\left|x_{1}-a\right|+\cdots+\left|x_{n}-a\right|\right)^{2} .
$$
|
Denote $y_{i}=x_{i}-a$. Then $y_{1}+y_{2}+\cdots+y_{n}=0$. We can assume $y_{1} \leqslant y_{2} \leqslant \cdots \leqslant y_{k} \leqslant 0 \leqslant y_{k+1} \leqslant \cdots \leqslant y_{n}$. Let $y_{1}+y_{2}+\cdots+y_{k}=-z$, then $y_{k+1}+\cdots+y_{n}=z$ and
$$
\begin{aligned}
y_{1}^{2}+y_{2}^{2}+\cdots+y_{n}^{2} & =y_{1}^{2}+y_{2}^{2}+\cdots+y_{k}^{2}+y_{k+1}^{2}+\cdots+y_{n}^{2} \leqslant \\
& \leqslant\left(y_{1}+y_{2}+\cdots+y_{k}\right)^{2}+\left(y_{k+1}+\cdots+y_{n}\right)^{2}=2 z^{2}= \\
& =\frac{1}{2}(2 z)^{2}=\frac{1}{2}\left(\left|y_{1}\right|+\left|y_{2}\right|+\cdots+\left|y_{n}\right|\right)^{2} .
\end{aligned}
$$
Alternative solution. The case $n=1$ is trivial (then $x_{1}-a=0$ and we get the inequality $0 \leqslant 0$ ). Suppose now that $n \geqslant 2$. Consider a square of side length $\left|x_{1}-a\right|+\left|x_{2}-a\right|+\ldots+\left|x_{n}-a\right|$ and construct squares of side lengths $\left|x_{1}-a\right|,\left|x_{2}-a\right|, \ldots,\left|x_{n}-a\right|$ side by side inside it as shown on Figure 1. Since none of the side lengths of the small squares exceeds half of the side length of the large square, then all the small squares are contained within the upper half of the large square, i.e. the sum of their areas does not exceed half of the area of the large square, q.e.d.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Prove that the arithmetic mean $a$ of $x_{1}, \ldots, x_{n}$ satisfies
$$
\left(x_{1}-a\right)^{2}+\cdots+\left(x_{n}-a\right)^{2} \leqslant \frac{1}{2}\left(\left|x_{1}-a\right|+\cdots+\left|x_{n}-a\right|\right)^{2} .
$$
|
Denote $y_{i}=x_{i}-a$. Then $y_{1}+y_{2}+\cdots+y_{n}=0$. We can assume $y_{1} \leqslant y_{2} \leqslant \cdots \leqslant y_{k} \leqslant 0 \leqslant y_{k+1} \leqslant \cdots \leqslant y_{n}$. Let $y_{1}+y_{2}+\cdots+y_{k}=-z$, then $y_{k+1}+\cdots+y_{n}=z$ and
$$
\begin{aligned}
y_{1}^{2}+y_{2}^{2}+\cdots+y_{n}^{2} & =y_{1}^{2}+y_{2}^{2}+\cdots+y_{k}^{2}+y_{k+1}^{2}+\cdots+y_{n}^{2} \leqslant \\
& \leqslant\left(y_{1}+y_{2}+\cdots+y_{k}\right)^{2}+\left(y_{k+1}+\cdots+y_{n}\right)^{2}=2 z^{2}= \\
& =\frac{1}{2}(2 z)^{2}=\frac{1}{2}\left(\left|y_{1}\right|+\left|y_{2}\right|+\cdots+\left|y_{n}\right|\right)^{2} .
\end{aligned}
$$
Alternative solution. The case $n=1$ is trivial (then $x_{1}-a=0$ and we get the inequality $0 \leqslant 0$ ). Suppose now that $n \geqslant 2$. Consider a square of side length $\left|x_{1}-a\right|+\left|x_{2}-a\right|+\ldots+\left|x_{n}-a\right|$ and construct squares of side lengths $\left|x_{1}-a\right|,\left|x_{2}-a\right|, \ldots,\left|x_{n}-a\right|$ side by side inside it as shown on Figure 1. Since none of the side lengths of the small squares exceeds half of the side length of the large square, then all the small squares are contained within the upper half of the large square, i.e. the sum of their areas does not exceed half of the area of the large square, q.e.d.
|
{
"exam": "BalticWay",
"problem_label": "4",
"problem_match": "\n4.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n4.",
"tier": "T3",
"year": "1997"
}
|
In a sequence $u_{0}, u_{1}, \ldots$ of positive integers, $u_{0}$ is arbitrary, and for any non-negative integer $n$,
$$
u_{n+1}= \begin{cases}\frac{1}{2} u_{n} & \text { for even } u_{n} \\ a+u_{n} & \text { for odd } u_{n}\end{cases}
$$
where $a$ is a fixed odd positive integer. Prove that the sequence is periodic from a certain step.
|
Suppose $u_{n}>a$. Then, if $u_{n}$ is even we have $u_{n+1}=\frac{1}{2} u_{n}<u_{n}$, and if
$u_{n}$ is odd we have $u_{n+1}=a+u_{n}<2 u_{n}$ and $u_{n+2}=\frac{1}{2} u_{n+1}<u_{n}$. Hence the iteration results in $u_{n} \leqslant a$ in a finite number of steps. Thus for any non-negative integer $m$, some non-negative integer $n>m$ satisfies $u_{n} \leqslant a$, and there must be an infinite set of such integers $n$.
Since the set of natural numbers not exceeding $a$ is finite and such values arise in the sequence $\left(u_{n}\right)$ an infinite number of times, there exist nonnegative integers $m$ and $n$ with $n>m$ such that $u_{n}=u_{m}$. Starting from $u_{m}$ the sequence is then periodic with a period dividing $n-m$.
Figure 1
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
In a sequence $u_{0}, u_{1}, \ldots$ of positive integers, $u_{0}$ is arbitrary, and for any non-negative integer $n$,
$$
u_{n+1}= \begin{cases}\frac{1}{2} u_{n} & \text { for even } u_{n} \\ a+u_{n} & \text { for odd } u_{n}\end{cases}
$$
where $a$ is a fixed odd positive integer. Prove that the sequence is periodic from a certain step.
|
Suppose $u_{n}>a$. Then, if $u_{n}$ is even we have $u_{n+1}=\frac{1}{2} u_{n}<u_{n}$, and if
$u_{n}$ is odd we have $u_{n+1}=a+u_{n}<2 u_{n}$ and $u_{n+2}=\frac{1}{2} u_{n+1}<u_{n}$. Hence the iteration results in $u_{n} \leqslant a$ in a finite number of steps. Thus for any non-negative integer $m$, some non-negative integer $n>m$ satisfies $u_{n} \leqslant a$, and there must be an infinite set of such integers $n$.
Since the set of natural numbers not exceeding $a$ is finite and such values arise in the sequence $\left(u_{n}\right)$ an infinite number of times, there exist nonnegative integers $m$ and $n$ with $n>m$ such that $u_{n}=u_{m}$. Starting from $u_{m}$ the sequence is then periodic with a period dividing $n-m$.
Figure 1
|
{
"exam": "BalticWay",
"problem_label": "5",
"problem_match": "\n5.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n5.",
"tier": "T3",
"year": "1997"
}
|
Find all triples $(a, b, c)$ of non-negative integers satisfying $a \geqslant b \geqslant c$ and $1 \cdot a^{3}+9 \cdot b^{2}+9 \cdot c+7=1997$.
|
Answer: $(10,10,10)$ is the only such triple.
The equality immediately implies $a^{3}+9 b^{2}+9 c=1990 \equiv 1(\bmod 9)$. Hence $a^{3} \equiv 1(\bmod 9)$ and $a \equiv 1(\bmod 3)$. Since $13^{3}=2197>1990$ then the possible values for $a$ are $1,4,7,10$.
On the other hand, if $a \leqslant 7$ then by $a \geqslant b \geqslant c$ we have
$$
a^{3}+9 b^{2}+9 c^{2} \leqslant 7^{3}+9 \cdot 7^{2}+9 \cdot 7=847<1990
$$
a contradiction. Hence $a=10$ and $9 b^{2}+9 c=990$, whence by $c \leqslant b \leqslant 10$ we have $c=b=10$.
|
(10,10,10)
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find all triples $(a, b, c)$ of non-negative integers satisfying $a \geqslant b \geqslant c$ and $1 \cdot a^{3}+9 \cdot b^{2}+9 \cdot c+7=1997$.
|
Answer: $(10,10,10)$ is the only such triple.
The equality immediately implies $a^{3}+9 b^{2}+9 c=1990 \equiv 1(\bmod 9)$. Hence $a^{3} \equiv 1(\bmod 9)$ and $a \equiv 1(\bmod 3)$. Since $13^{3}=2197>1990$ then the possible values for $a$ are $1,4,7,10$.
On the other hand, if $a \leqslant 7$ then by $a \geqslant b \geqslant c$ we have
$$
a^{3}+9 b^{2}+9 c^{2} \leqslant 7^{3}+9 \cdot 7^{2}+9 \cdot 7=847<1990
$$
a contradiction. Hence $a=10$ and $9 b^{2}+9 c=990$, whence by $c \leqslant b \leqslant 10$ we have $c=b=10$.
|
{
"exam": "BalticWay",
"problem_label": "6",
"problem_match": "\n6.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n6.",
"tier": "T3",
"year": "1997"
}
|
Let $P$ and $Q$ be polynomials with integer coefficients. Suppose that the integers $a$ and $a+1997$ are roots of $P$, and that $Q(1998)=2000$. Prove that the equation $Q(P(x))=1$ has no integer solutions.
|
Suppose $b$ is an integer such that $Q(P(b))=1$. Since $a$ and $a+1997$ are roots of $P$ we have $P(x)=(x-a)(x-a-1997) R(x)$ where $R$ is a polynomial with integer coefficients. For any integer $b$ the integers $b-a$ and
$b-a-1997$ are of different parity and hence $P(b)=(b-a)(b-a-1997) R(b)$ is even. Since $Q(1998)=2000$ then the constant term in the expansion of $Q(x)$ is even (otherwise $Q(x)$ would be odd for any even integer $x$ ), and $Q(c)$ is even for any even integer $c$. Hence $Q(P(b))$ is also even and cannot be equal to 1 .
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $P$ and $Q$ be polynomials with integer coefficients. Suppose that the integers $a$ and $a+1997$ are roots of $P$, and that $Q(1998)=2000$. Prove that the equation $Q(P(x))=1$ has no integer solutions.
|
Suppose $b$ is an integer such that $Q(P(b))=1$. Since $a$ and $a+1997$ are roots of $P$ we have $P(x)=(x-a)(x-a-1997) R(x)$ where $R$ is a polynomial with integer coefficients. For any integer $b$ the integers $b-a$ and
$b-a-1997$ are of different parity and hence $P(b)=(b-a)(b-a-1997) R(b)$ is even. Since $Q(1998)=2000$ then the constant term in the expansion of $Q(x)$ is even (otherwise $Q(x)$ would be odd for any even integer $x$ ), and $Q(c)$ is even for any even integer $c$. Hence $Q(P(b))$ is also even and cannot be equal to 1 .
|
{
"exam": "BalticWay",
"problem_label": "7",
"problem_match": "\n7.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n7.",
"tier": "T3",
"year": "1997"
}
|
If we add 1996 and 1997, we first add the unit digits 6 and 7. Obtaining 13, we write down 3 and "carry" 1 to the next column. Thus we make a carry. Continuing, we see that we are to make three carries in total:
$$
\begin{array}{r}
111 \\
1996 \\
+1997 \\
\hline 3993
\end{array}
$$
Does there exist a positive integer $k$ such that adding $1996 \cdot k$ to $1997 \cdot k$ no carry arises during the whole calculation?
|
Answer: yes.
The key to the proof is noting that if we add two positive integers and the result is an integer consisting only of digits 9 then the process of addition must have gone without any carries. Therefore it is enough to prove that there exists an integer $k$ such that $3993 k$ is of the form $999 \ldots 9$.
Consider the first 3994 positive integers consisting only of digits 9:
$$
9,99,999, \ldots, \underbrace{999 \ldots 9}_{3994} .
$$
By the pigeonhole principle some two of these give the same remainder upon division by 3993 , so their difference
$$
\underbrace{99 \ldots 9}_{n} \underbrace{00 \ldots 0}_{r}=\underbrace{99 \ldots 9}_{n} \cdot 10^{r}
$$
is divisible by 3993 . Since 10 and 3993 are coprime we get an integer consisting only of digits 9 and divisible by 3993 .
## Remarks.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
If we add 1996 and 1997, we first add the unit digits 6 and 7. Obtaining 13, we write down 3 and "carry" 1 to the next column. Thus we make a carry. Continuing, we see that we are to make three carries in total:
$$
\begin{array}{r}
111 \\
1996 \\
+1997 \\
\hline 3993
\end{array}
$$
Does there exist a positive integer $k$ such that adding $1996 \cdot k$ to $1997 \cdot k$ no carry arises during the whole calculation?
|
Answer: yes.
The key to the proof is noting that if we add two positive integers and the result is an integer consisting only of digits 9 then the process of addition must have gone without any carries. Therefore it is enough to prove that there exists an integer $k$ such that $3993 k$ is of the form $999 \ldots 9$.
Consider the first 3994 positive integers consisting only of digits 9:
$$
9,99,999, \ldots, \underbrace{999 \ldots 9}_{3994} .
$$
By the pigeonhole principle some two of these give the same remainder upon division by 3993 , so their difference
$$
\underbrace{99 \ldots 9}_{n} \underbrace{00 \ldots 0}_{r}=\underbrace{99 \ldots 9}_{n} \cdot 10^{r}
$$
is divisible by 3993 . Since 10 and 3993 are coprime we get an integer consisting only of digits 9 and divisible by 3993 .
## Remarks.
|
{
"exam": "BalticWay",
"problem_label": "8",
"problem_match": "\n8.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n8.",
"tier": "T3",
"year": "1997"
}
|
Determine all functions $f$ from the real numbers to the real numbers, different from the zero function, such that $f(x) f(y)=f(x-y)$ for all real numbers $x$ and $y$.
|
The existence of an integer $10^{\ell}-1$ consisting only of digits 9 and divisible by 3993 may also be demonstrated quite elegantly by means of Euler's Theorem. The numbers 10 and 3993 are coprime, so $10^{\varphi(3993)}-1$ is divisible by 3993 . Thus we may take $\ell=\varphi(3993)$.
2 . By a computer search it can be found that the smallest integer $k$ satisfying the condition of the problem is $k=162$. Then $1996 \cdot 162=323352$; $1997 \cdot 162=323514$ and
$$
\begin{array}{r}
323352 \\
+323514 \\
\hline 646866
\end{array}
$$
|
not found
|
Yes
|
Problem not solved
|
math-word-problem
|
Algebra
|
Determine all functions $f$ from the real numbers to the real numbers, different from the zero function, such that $f(x) f(y)=f(x-y)$ for all real numbers $x$ and $y$.
|
The existence of an integer $10^{\ell}-1$ consisting only of digits 9 and divisible by 3993 may also be demonstrated quite elegantly by means of Euler's Theorem. The numbers 10 and 3993 are coprime, so $10^{\varphi(3993)}-1$ is divisible by 3993 . Thus we may take $\ell=\varphi(3993)$.
2 . By a computer search it can be found that the smallest integer $k$ satisfying the condition of the problem is $k=162$. Then $1996 \cdot 162=323352$; $1997 \cdot 162=323514$ and
$$
\begin{array}{r}
323352 \\
+323514 \\
\hline 646866
\end{array}
$$
|
{
"exam": "BalticWay",
"problem_label": "1",
"problem_match": "\n1.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n1.",
"tier": "T3",
"year": "1997"
}
|
The worlds in the Worlds' Sphere are numbered $1,2,3, \ldots$ and connected so that for any integer $n \geqslant 1$, Gandalf the Wizard can move in both directions between any worlds with numbers $n, 2 n$ and $3 n+1$. Starting his travel from an arbitrary world, can Gandalf reach every other world?
|
Answer: yes.
For any two given worlds, Gandalf can move between them either in both
directions or none. Hence, it suffices to show that Gandalf can move to the world 1 from any given world $n$. For that, it is sufficient for him to be able to move from any world $n>1$ to some world $m$ such that $m<n$. We consider three possible cases:
a) If $n=3 k+1$, then Gandalf can move directly from the world $n$ to the world $k<n$.
b) If $n=3 k+2$, then Gandalf can move from the world $n$ to the world $2 n=6 k+4=3 \cdot(2 k+1)+1$ and further to the world $2 k+1<n$. c) If $n=3 k$ then Gandalf can move from the world $n$ to the world $3 n+1=9 k+1$, further from there to the worlds $2 \cdot(9 k+1)=18 k+2$, $2 \cdot(18 k+2)=36 k+4=3 \cdot(12 k+1)+1,12 k+1,4 k$ and finally to the world $2 k<n$.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
The worlds in the Worlds' Sphere are numbered $1,2,3, \ldots$ and connected so that for any integer $n \geqslant 1$, Gandalf the Wizard can move in both directions between any worlds with numbers $n, 2 n$ and $3 n+1$. Starting his travel from an arbitrary world, can Gandalf reach every other world?
|
Answer: yes.
For any two given worlds, Gandalf can move between them either in both
directions or none. Hence, it suffices to show that Gandalf can move to the world 1 from any given world $n$. For that, it is sufficient for him to be able to move from any world $n>1$ to some world $m$ such that $m<n$. We consider three possible cases:
a) If $n=3 k+1$, then Gandalf can move directly from the world $n$ to the world $k<n$.
b) If $n=3 k+2$, then Gandalf can move from the world $n$ to the world $2 n=6 k+4=3 \cdot(2 k+1)+1$ and further to the world $2 k+1<n$. c) If $n=3 k$ then Gandalf can move from the world $n$ to the world $3 n+1=9 k+1$, further from there to the worlds $2 \cdot(9 k+1)=18 k+2$, $2 \cdot(18 k+2)=36 k+4=3 \cdot(12 k+1)+1,12 k+1,4 k$ and finally to the world $2 k<n$.
|
{
"exam": "BalticWay",
"problem_label": "9",
"problem_match": "\n9.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n9.",
"tier": "T3",
"year": "1997"
}
|
Prove that in every sequence of 79 consecutive positive integers written in the decimal system, there is a positive integer whose sum of digits is divisible by 13 .
|
Among the first 40 numbers in the sequence, four are divisible by 10 and at least one of these has its second digit from the right less than or equal to 6 . Let this number be $x$ and let $y$ be its sum of digits. Then the numbers $x, x+1, x+2, \ldots, x+39$ all belong to the sequence, and each of $y, y+1, \ldots, y+12$ appears at least once among their sums of decimal digits. One of these is divisible by 13 .
Remark: there exist 78 consecutive natural numbers, none of which has its sum of digits divisible by 13 - e.g. 859999999961 through 860000000038 .
Figure 2
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Prove that in every sequence of 79 consecutive positive integers written in the decimal system, there is a positive integer whose sum of digits is divisible by 13 .
|
Among the first 40 numbers in the sequence, four are divisible by 10 and at least one of these has its second digit from the right less than or equal to 6 . Let this number be $x$ and let $y$ be its sum of digits. Then the numbers $x, x+1, x+2, \ldots, x+39$ all belong to the sequence, and each of $y, y+1, \ldots, y+12$ appears at least once among their sums of decimal digits. One of these is divisible by 13 .
Remark: there exist 78 consecutive natural numbers, none of which has its sum of digits divisible by 13 - e.g. 859999999961 through 860000000038 .
Figure 2
|
{
"exam": "BalticWay",
"problem_label": "10",
"problem_match": "\n10.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n10.",
"tier": "T3",
"year": "1997"
}
|
On two parallel lines, the distinct points $A_{1}, A_{2}, A_{3}, \ldots$ respectively $B_{1}$, $B_{2}, B_{3}, \ldots$ are marked in such a way that $\left|A_{i} A_{i+1}\right|=1$ and $\left|B_{i} B_{i+1}\right|=2$ for $i=1,2, \ldots$ (see Figure). Provided that $\angle A_{1} A_{2} B_{1}=\alpha$, find the infinite sum $\angle A_{1} B_{1} A_{2}+\angle A_{2} B_{2} A_{3}+\angle A_{3} B_{3} A_{4}+\ldots$
|
Answer: $\pi-\alpha$.
Let $C_{1}, C_{2}, C_{3}, \ldots$ be points on the upper line such that $\left|C_{i} C_{i+1}\right|=1$ and $B_{i}=C_{2 i}$ for each $i=1,2, \ldots$ (see Figure 2 ). Then for any $i=1,2, \ldots$ we have
$$
\angle A_{i} B_{i} A_{i+1}=\angle A_{i} C_{2 i} A_{i+1}=\angle A_{1} C_{i+1} A_{2}=\angle C_{i+1} A_{2} C_{i+2}
$$
Hence
$$
\begin{aligned}
& \angle A_{1} B_{1} A_{2}+\angle A_{2} B_{2} A_{3}+\angle A_{3} B_{3} A_{4}+\ldots= \\
& \quad=\angle C_{2} A_{2} C_{3}+\angle C_{3} A_{2} C_{4}+\angle C_{4} A_{2} C_{5}+\ldots=\pi-\alpha .
\end{aligned}
$$
|
\pi-\alpha
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
On two parallel lines, the distinct points $A_{1}, A_{2}, A_{3}, \ldots$ respectively $B_{1}$, $B_{2}, B_{3}, \ldots$ are marked in such a way that $\left|A_{i} A_{i+1}\right|=1$ and $\left|B_{i} B_{i+1}\right|=2$ for $i=1,2, \ldots$ (see Figure). Provided that $\angle A_{1} A_{2} B_{1}=\alpha$, find the infinite sum $\angle A_{1} B_{1} A_{2}+\angle A_{2} B_{2} A_{3}+\angle A_{3} B_{3} A_{4}+\ldots$
|
Answer: $\pi-\alpha$.
Let $C_{1}, C_{2}, C_{3}, \ldots$ be points on the upper line such that $\left|C_{i} C_{i+1}\right|=1$ and $B_{i}=C_{2 i}$ for each $i=1,2, \ldots$ (see Figure 2 ). Then for any $i=1,2, \ldots$ we have
$$
\angle A_{i} B_{i} A_{i+1}=\angle A_{i} C_{2 i} A_{i+1}=\angle A_{1} C_{i+1} A_{2}=\angle C_{i+1} A_{2} C_{i+2}
$$
Hence
$$
\begin{aligned}
& \angle A_{1} B_{1} A_{2}+\angle A_{2} B_{2} A_{3}+\angle A_{3} B_{3} A_{4}+\ldots= \\
& \quad=\angle C_{2} A_{2} C_{3}+\angle C_{3} A_{2} C_{4}+\angle C_{4} A_{2} C_{5}+\ldots=\pi-\alpha .
\end{aligned}
$$
|
{
"exam": "BalticWay",
"problem_label": "11",
"problem_match": "\n11.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n11.",
"tier": "T3",
"year": "1997"
}
|
Two circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect in $P$ and $Q$. A line through $P$ intersects $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ again in $A$ and $B$, respectively, and $X$ is the midpoint of $A B$. The line through $Q$ and $X$ intersects $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ again in $Y$ and $Z$, respectively. Prove that $X$ is the midpoint of $Y Z$.
|
Depending on the radii of the circles, the distance between their centres and the choice of the line through $P$ we have several possible arrangements of the points $A, B, P$ and $Y, Z, Q$. We shall show that in each case the triangles $A X Y$ and $B X Z$ are congruent, whence $|Y X|=|X Z|$.
(a) Point $P$ lies within segment $A B$ and point $Q$ lies within segment $Y Z$ (see Figure 3). Then
$$
\angle A Y X=\angle A Y Q=\pi-\angle A P Q=\angle B P Q=\angle B Z Q=\angle B Z X .
$$
Since also $\angle A X Y=\angle B X Z$ and $|A X|=|X B|$, triangles $A X Y$ and $B X Z$ are congruent.
(b) Point $P$ lies outside of segment $A B$ and point $Q$ lies within segment $Y Z$ (see Figure 4). Then
$$
\angle A Y X=\angle A Y Q=\angle A P Q=\angle B P Q=\angle B Z Q=\angle B Z X .
$$
Figure 3
Figure 4
(c) Point $P$ lies outside of segment $A B$ and point $Q$ lies outside of segment $Y Z$ (see Figure 5). Then
$$
\angle A Y X=\pi-\angle A Y Q=\angle A P Q=\angle B P Q=\angle B Z Q=\angle B Z X .
$$
(d) Point $P$ lies within segment $A B$ and point $Q$ lies outside of segment $Y Z$. This case is similar to (b): exchange the roles of points $P$ and $Q, A$ and $Y, B$ and $Z$.
Figure 5
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Two circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect in $P$ and $Q$. A line through $P$ intersects $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ again in $A$ and $B$, respectively, and $X$ is the midpoint of $A B$. The line through $Q$ and $X$ intersects $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ again in $Y$ and $Z$, respectively. Prove that $X$ is the midpoint of $Y Z$.
|
Depending on the radii of the circles, the distance between their centres and the choice of the line through $P$ we have several possible arrangements of the points $A, B, P$ and $Y, Z, Q$. We shall show that in each case the triangles $A X Y$ and $B X Z$ are congruent, whence $|Y X|=|X Z|$.
(a) Point $P$ lies within segment $A B$ and point $Q$ lies within segment $Y Z$ (see Figure 3). Then
$$
\angle A Y X=\angle A Y Q=\pi-\angle A P Q=\angle B P Q=\angle B Z Q=\angle B Z X .
$$
Since also $\angle A X Y=\angle B X Z$ and $|A X|=|X B|$, triangles $A X Y$ and $B X Z$ are congruent.
(b) Point $P$ lies outside of segment $A B$ and point $Q$ lies within segment $Y Z$ (see Figure 4). Then
$$
\angle A Y X=\angle A Y Q=\angle A P Q=\angle B P Q=\angle B Z Q=\angle B Z X .
$$
Figure 3
Figure 4
(c) Point $P$ lies outside of segment $A B$ and point $Q$ lies outside of segment $Y Z$ (see Figure 5). Then
$$
\angle A Y X=\pi-\angle A Y Q=\angle A P Q=\angle B P Q=\angle B Z Q=\angle B Z X .
$$
(d) Point $P$ lies within segment $A B$ and point $Q$ lies outside of segment $Y Z$. This case is similar to (b): exchange the roles of points $P$ and $Q, A$ and $Y, B$ and $Z$.
Figure 5
|
{
"exam": "BalticWay",
"problem_label": "12",
"problem_match": "\n12.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n12.",
"tier": "T3",
"year": "1997"
}
|
Five distinct points $A, B, C, D$ and $E$ lie on a line with
$$
|A B|=|B C|=|C D|=|D E| \text {. }
$$
The point $F$ lies outside the line. Let $G$ be the circumcentre of triangle $A D F$ and $H$ be the circumcentre of triangle $B E F$. Show that lines $G H$ and $F C$ are perpendicular.
|
Let $O, H^{\prime}$ and $G^{\prime}$ be the circumcentres of the triangles $B D F, B C F$ and $C D F$, respectively (see Figure 6). Then $O, G$ and $G^{\prime}$ lie on the perpendicular bisector of the segment $D F$, while $O, H$ and $H^{\prime}$ lie on the perpendicular bisector of the segment $B F$. Moreover, $G$ and $H^{\prime}$ lie on the perpendicular bisector of $B C, O$ lies on the perpendicular bisector of $B D$, $H$ and $G^{\prime}$ lie on the perpendicular bisector of $C D$ and $C$ is the midpoint of $B D$. Hence $H^{\prime}$ and $G^{\prime}$ are symmetric to $H$ and $G$, respectively, relative to point $O$. Hence triangles $O G H^{\prime}$ and $O G^{\prime} H$ are congruent, and $G H G^{\prime} H^{\prime}$ is a parallelogram.
Since $C F$ is the common side of triangles $B C F$ and $C D F$, the line $G^{\prime} H^{\prime}$ connecting their circumcentres is perpendicular to $C F$. Therefore $G H$ is also perpendicular to $C F$.
Figure 6
Alternative solution. Note that the diagonals of a quadrangle $X Y Z W$ are perpendicular to each other if and only if $|X Y|^{2}-|Z Y|^{2}=|X W|^{2}-|Z W|^{2}$. Applying this to the quadrangle $G F H C$ it is sufficient to prove that $|G F|^{2}-|H F|^{2}=|G C|^{2}-|H C|^{2}$. Denote $|A B|=|B C|=|C D|=|D E|=a$, $\angle G A C=\alpha$ and $\angle H E C=\beta$, and let $R_{1}, R_{2}$ be the circumradii of triangles $A D F$ and $B E F$, respectively (see Figure 7). Applying the cosine law to triangles $C G A$ and $C H E$, we have $|G C|^{2}=R_{1}^{2}+4 a^{2}-4 a R_{1} \cos \alpha$
and $|H C|^{2}=R_{2}^{2}+4 a^{2}-4 a R_{2} \cos \beta$. Together with $\cos \alpha=\frac{3 a}{2 R_{1}}$ and $\cos \beta=\frac{3 a}{2 R_{2}}$ this yields $|G C|^{2}-|H C|^{2}=R_{1}^{2}-R_{2}^{2}$. Since $|G F|=R_{1}$ and $|H F|=R_{2}$, we also have $|G F|^{2}-|H F|^{2}=R_{1}^{2}-R_{2}^{2}$.
Figure 7
Figure 8
Another solution. We shall use the following fact that can easily be derived from the properties of the power of a point: Let a line $s$ intersect two circles at points $K, L$ and $M, N$, respectively, and let these circles intersect each other at $P$ and $Q . A$ point $X$ on the line $s$ lies also on the line $P Q$ (i.e. is the intersection point of the lines $s$ and $P Q$ ) if and only if $|K X| \cdot|L X|=|M X| \cdot|N X|$.
The line $A E$ intersects the circumcircles of triangles $A D F$ and $B E F$ at $A, D$ and $B, E$, respectively. Since point $C$ lies on line $A E$ and $|A C| \cdot|D C|=|B C| \cdot|E C|$, then line $C F$ passes through the second intersection point of these circles (see Figure 8) and hence is perpendicular to the segment $G H$ connecting the centres of these circles.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Five distinct points $A, B, C, D$ and $E$ lie on a line with
$$
|A B|=|B C|=|C D|=|D E| \text {. }
$$
The point $F$ lies outside the line. Let $G$ be the circumcentre of triangle $A D F$ and $H$ be the circumcentre of triangle $B E F$. Show that lines $G H$ and $F C$ are perpendicular.
|
Let $O, H^{\prime}$ and $G^{\prime}$ be the circumcentres of the triangles $B D F, B C F$ and $C D F$, respectively (see Figure 6). Then $O, G$ and $G^{\prime}$ lie on the perpendicular bisector of the segment $D F$, while $O, H$ and $H^{\prime}$ lie on the perpendicular bisector of the segment $B F$. Moreover, $G$ and $H^{\prime}$ lie on the perpendicular bisector of $B C, O$ lies on the perpendicular bisector of $B D$, $H$ and $G^{\prime}$ lie on the perpendicular bisector of $C D$ and $C$ is the midpoint of $B D$. Hence $H^{\prime}$ and $G^{\prime}$ are symmetric to $H$ and $G$, respectively, relative to point $O$. Hence triangles $O G H^{\prime}$ and $O G^{\prime} H$ are congruent, and $G H G^{\prime} H^{\prime}$ is a parallelogram.
Since $C F$ is the common side of triangles $B C F$ and $C D F$, the line $G^{\prime} H^{\prime}$ connecting their circumcentres is perpendicular to $C F$. Therefore $G H$ is also perpendicular to $C F$.
Figure 6
Alternative solution. Note that the diagonals of a quadrangle $X Y Z W$ are perpendicular to each other if and only if $|X Y|^{2}-|Z Y|^{2}=|X W|^{2}-|Z W|^{2}$. Applying this to the quadrangle $G F H C$ it is sufficient to prove that $|G F|^{2}-|H F|^{2}=|G C|^{2}-|H C|^{2}$. Denote $|A B|=|B C|=|C D|=|D E|=a$, $\angle G A C=\alpha$ and $\angle H E C=\beta$, and let $R_{1}, R_{2}$ be the circumradii of triangles $A D F$ and $B E F$, respectively (see Figure 7). Applying the cosine law to triangles $C G A$ and $C H E$, we have $|G C|^{2}=R_{1}^{2}+4 a^{2}-4 a R_{1} \cos \alpha$
and $|H C|^{2}=R_{2}^{2}+4 a^{2}-4 a R_{2} \cos \beta$. Together with $\cos \alpha=\frac{3 a}{2 R_{1}}$ and $\cos \beta=\frac{3 a}{2 R_{2}}$ this yields $|G C|^{2}-|H C|^{2}=R_{1}^{2}-R_{2}^{2}$. Since $|G F|=R_{1}$ and $|H F|=R_{2}$, we also have $|G F|^{2}-|H F|^{2}=R_{1}^{2}-R_{2}^{2}$.
Figure 7
Figure 8
Another solution. We shall use the following fact that can easily be derived from the properties of the power of a point: Let a line $s$ intersect two circles at points $K, L$ and $M, N$, respectively, and let these circles intersect each other at $P$ and $Q . A$ point $X$ on the line $s$ lies also on the line $P Q$ (i.e. is the intersection point of the lines $s$ and $P Q$ ) if and only if $|K X| \cdot|L X|=|M X| \cdot|N X|$.
The line $A E$ intersects the circumcircles of triangles $A D F$ and $B E F$ at $A, D$ and $B, E$, respectively. Since point $C$ lies on line $A E$ and $|A C| \cdot|D C|=|B C| \cdot|E C|$, then line $C F$ passes through the second intersection point of these circles (see Figure 8) and hence is perpendicular to the segment $G H$ connecting the centres of these circles.
|
{
"exam": "BalticWay",
"problem_label": "13",
"problem_match": "\n13.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n13.",
"tier": "T3",
"year": "1997"
}
|
In the triangle $A B C,|A C|^{2}$ is the arithmetic mean of $|B C|^{2}$ and $|A B|^{2}$. Show that $\cot ^{2} B \geqslant \cot A \cot C$.
|
Denote $|B C|=a,|C A|=b$ and $|A B|=c$, then we have $2 b^{2}=a^{2}+c^{2}$. Applying the cosine and sine laws to triangle $A B C$ we have:
$$
\begin{aligned}
& \cot B=\frac{\cos B}{\sin B}=\frac{\left(a^{2}+c^{2}-b^{2}\right) \cdot 2 R}{2 a c \cdot b}=\frac{\left(a^{2}+c^{2}-b^{2}\right) \cdot R}{a b c}, \\
& \cot A=\frac{\cos A}{\sin A}=\frac{\left(b^{2}+c^{2}-a^{2}\right) \cdot R}{a b c},
\end{aligned}
$$
$$
\cot C=\frac{\cos C}{\sin C}=\frac{\left(a^{2}+b^{2}-c^{2}\right) \cdot R}{a b c}
$$
where $R$ is the circumradius of triangle $A B C$. To finish the proof it hence suffices to show that $\left(a^{2}+c^{2}-b^{2}\right)^{2} \geqslant\left(b^{2}+c^{2}-a^{2}\right)\left(a^{2}+b^{2}-c^{2}\right)$. Indeed, from the AM-GM inequality we get
$$
\begin{aligned}
\left(b^{2}+c^{2}-a^{2}\right)\left(a^{2}+b^{2}-c^{2}\right) & \leqslant \frac{\left(b^{2}+c^{2}-a^{2}+a^{2}+b^{2}-c^{2}\right)^{2}}{4}=b^{4}= \\
& =\left(2 b^{2}-b^{2}\right)^{2}=\left(a^{2}+c^{2}-b^{2}\right)^{2} .
\end{aligned}
$$
Figure 9
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In the triangle $A B C,|A C|^{2}$ is the arithmetic mean of $|B C|^{2}$ and $|A B|^{2}$. Show that $\cot ^{2} B \geqslant \cot A \cot C$.
|
Denote $|B C|=a,|C A|=b$ and $|A B|=c$, then we have $2 b^{2}=a^{2}+c^{2}$. Applying the cosine and sine laws to triangle $A B C$ we have:
$$
\begin{aligned}
& \cot B=\frac{\cos B}{\sin B}=\frac{\left(a^{2}+c^{2}-b^{2}\right) \cdot 2 R}{2 a c \cdot b}=\frac{\left(a^{2}+c^{2}-b^{2}\right) \cdot R}{a b c}, \\
& \cot A=\frac{\cos A}{\sin A}=\frac{\left(b^{2}+c^{2}-a^{2}\right) \cdot R}{a b c},
\end{aligned}
$$
$$
\cot C=\frac{\cos C}{\sin C}=\frac{\left(a^{2}+b^{2}-c^{2}\right) \cdot R}{a b c}
$$
where $R$ is the circumradius of triangle $A B C$. To finish the proof it hence suffices to show that $\left(a^{2}+c^{2}-b^{2}\right)^{2} \geqslant\left(b^{2}+c^{2}-a^{2}\right)\left(a^{2}+b^{2}-c^{2}\right)$. Indeed, from the AM-GM inequality we get
$$
\begin{aligned}
\left(b^{2}+c^{2}-a^{2}\right)\left(a^{2}+b^{2}-c^{2}\right) & \leqslant \frac{\left(b^{2}+c^{2}-a^{2}+a^{2}+b^{2}-c^{2}\right)^{2}}{4}=b^{4}= \\
& =\left(2 b^{2}-b^{2}\right)^{2}=\left(a^{2}+c^{2}-b^{2}\right)^{2} .
\end{aligned}
$$
Figure 9
|
{
"exam": "BalticWay",
"problem_label": "14",
"problem_match": "\n14.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n14.",
"tier": "T3",
"year": "1997"
}
|
In the acute triangle $A B C$, the bisectors of $\angle A, \angle B$ and $\angle C$ intersect the circumcircle again in $A_{1}, B_{1}$ and $C_{1}$, respectively. Let $M$ be the point of intersection of $A B$ and $B_{1} C_{1}$, and let $N$ be the point of intersection of $B C$ and $A_{1} B_{1}$. Prove that $M N$ passes through the incentre of triangle $A B C$.
|
Let $I$ be the incenter of triangle $A B C$ (the intersection point of the angle bisectors $A A_{1}, B B_{1}$ and $C C_{1}$ ), and let $B_{1} C_{1}$ intersect the side $A C$ and the angle bisector $A A_{1}$ at $P$ and $Q$, respectively (see Figure 9 ). Then
$$
\angle A Q C_{1}=\frac{1}{2}\left(\overparen{A C}_{1}+\overparen{A_{1} B_{1}}\right)=\frac{1}{2} \cdot\left(\frac{1}{2} \overparen{A B}+\frac{1}{2} \overparen{B C}+\frac{1}{2} \overparen{C A}\right)=90^{\circ}
$$
Since $\angle A C_{1} B_{1}=\angle B_{1} C_{1} C$ (as their supporting arcs are of equal size), then $C_{1} B_{1}$ is the bisector of angle $A C_{1} I$. Moreover, since $A I$ and $C_{1} B_{1}$ are perpendicular, then $C_{1} B_{1}$ is also the bisector of angle $A M I$. Similarly we can show that $B_{1} C_{1}$ bisects the angles $A B_{1} I$ ja $A P I$. Hence the diagonals of the quadrangle $A M I P$ are perpendicular and bisect its angles, i.e.
$A M I P$ is a rhombus and $M I$ is parrallel to $A C$. Similarly we can prove that $N I$ is parallel to $A C$, i.e. points $M, I$ and $N$ are collinear, q.e.d.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In the acute triangle $A B C$, the bisectors of $\angle A, \angle B$ and $\angle C$ intersect the circumcircle again in $A_{1}, B_{1}$ and $C_{1}$, respectively. Let $M$ be the point of intersection of $A B$ and $B_{1} C_{1}$, and let $N$ be the point of intersection of $B C$ and $A_{1} B_{1}$. Prove that $M N$ passes through the incentre of triangle $A B C$.
|
Let $I$ be the incenter of triangle $A B C$ (the intersection point of the angle bisectors $A A_{1}, B B_{1}$ and $C C_{1}$ ), and let $B_{1} C_{1}$ intersect the side $A C$ and the angle bisector $A A_{1}$ at $P$ and $Q$, respectively (see Figure 9 ). Then
$$
\angle A Q C_{1}=\frac{1}{2}\left(\overparen{A C}_{1}+\overparen{A_{1} B_{1}}\right)=\frac{1}{2} \cdot\left(\frac{1}{2} \overparen{A B}+\frac{1}{2} \overparen{B C}+\frac{1}{2} \overparen{C A}\right)=90^{\circ}
$$
Since $\angle A C_{1} B_{1}=\angle B_{1} C_{1} C$ (as their supporting arcs are of equal size), then $C_{1} B_{1}$ is the bisector of angle $A C_{1} I$. Moreover, since $A I$ and $C_{1} B_{1}$ are perpendicular, then $C_{1} B_{1}$ is also the bisector of angle $A M I$. Similarly we can show that $B_{1} C_{1}$ bisects the angles $A B_{1} I$ ja $A P I$. Hence the diagonals of the quadrangle $A M I P$ are perpendicular and bisect its angles, i.e.
$A M I P$ is a rhombus and $M I$ is parrallel to $A C$. Similarly we can prove that $N I$ is parallel to $A C$, i.e. points $M, I$ and $N$ are collinear, q.e.d.
|
{
"exam": "BalticWay",
"problem_label": "15",
"problem_match": "\n15.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n15.",
"tier": "T3",
"year": "1997"
}
|
On a $5 \times 5$ chessboard, two players play the following game. The first player places a knight on some square. Then the players alternately move the knight according to the rules of chess, starting with the second player. It is not allowed to move the knight to a square that has been visited previously. The player who cannot move loses. Which of the two players has a winning strategy?
|
Answer: the first player has a winning strategy.
Divide all the squares of the board except one in pairs so that the squares of each pair are accessible from each other by one move of the knight (see Figure 10 where the squares of each pair are marked with the same number, and the remaining square is marked by $X$ ). The winning strategy for the first player will be to place the knight on the square $X$ in the beginning and further make each move from a square to the other square paired with it.
| $X$ | 12 | 8 | 3 | 11 |
| :---: | :---: | :---: | :---: | :---: |
| 5 | 3 | 11 | 1 | 7 |
| 12 | 8 | 6 | 10 | 4 |
| 2 | 5 | 9 | 7 | 1 |
| 9 | 6 | 2 | 4 | 10 |
Figure 10
| 7 | | | | 1 |
| :--- | :--- | :--- | :--- | :--- |
| | | 8 | | |
| | 6 | | 2 | |
| | | 4 | 9 | |
| 5 | | | | 3 |
Figure 11
| 7 | 12 | 23 | 18 | 1 |
| :---: | :---: | :---: | :---: | :---: |
| 22 | 17 | 8 | 13 | 24 |
| 11 | 6 | 25 | 2 | 19 |
| 16 | 21 | 4 | 9 | 14 |
| 5 | 10 | 15 | 20 | 3 |
Figure 12
Alternative solution. If the first player places the knight on the square marked by 1 on Figure 11, then the second player will have two possible moves which are symmetric to each other relative to a diagonal of the board. Suppose w.l.o.g. that he makes a move to the square marked by 2 , then the first player can make his move to the square marked by 3 . At this point, the second player can only make a move to the square marked by 4 , and the first player can make his next move to the square marked by 5 ; then the second player can only make a move to the square marked by 6 , etc., until the first player will make a move to the square marked by 9 . Now the second player will again have two possible moves, but since these two squares are symmetric relative to a diagonal of the board (and the set of squares already used is symmetric to that diagonal as well) we can assume w.l.o.g. that he makes a move to the square marked by 10 . Now the first player can make his moves until the end of the game so that the second player will have no choice for his subsequent moves (these moves will be to the squares marked by 11 through 25 , in this order). We see that the first
player will be the one to make the last move, and hence the winner.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
On a $5 \times 5$ chessboard, two players play the following game. The first player places a knight on some square. Then the players alternately move the knight according to the rules of chess, starting with the second player. It is not allowed to move the knight to a square that has been visited previously. The player who cannot move loses. Which of the two players has a winning strategy?
|
Answer: the first player has a winning strategy.
Divide all the squares of the board except one in pairs so that the squares of each pair are accessible from each other by one move of the knight (see Figure 10 where the squares of each pair are marked with the same number, and the remaining square is marked by $X$ ). The winning strategy for the first player will be to place the knight on the square $X$ in the beginning and further make each move from a square to the other square paired with it.
| $X$ | 12 | 8 | 3 | 11 |
| :---: | :---: | :---: | :---: | :---: |
| 5 | 3 | 11 | 1 | 7 |
| 12 | 8 | 6 | 10 | 4 |
| 2 | 5 | 9 | 7 | 1 |
| 9 | 6 | 2 | 4 | 10 |
Figure 10
| 7 | | | | 1 |
| :--- | :--- | :--- | :--- | :--- |
| | | 8 | | |
| | 6 | | 2 | |
| | | 4 | 9 | |
| 5 | | | | 3 |
Figure 11
| 7 | 12 | 23 | 18 | 1 |
| :---: | :---: | :---: | :---: | :---: |
| 22 | 17 | 8 | 13 | 24 |
| 11 | 6 | 25 | 2 | 19 |
| 16 | 21 | 4 | 9 | 14 |
| 5 | 10 | 15 | 20 | 3 |
Figure 12
Alternative solution. If the first player places the knight on the square marked by 1 on Figure 11, then the second player will have two possible moves which are symmetric to each other relative to a diagonal of the board. Suppose w.l.o.g. that he makes a move to the square marked by 2 , then the first player can make his move to the square marked by 3 . At this point, the second player can only make a move to the square marked by 4 , and the first player can make his next move to the square marked by 5 ; then the second player can only make a move to the square marked by 6 , etc., until the first player will make a move to the square marked by 9 . Now the second player will again have two possible moves, but since these two squares are symmetric relative to a diagonal of the board (and the set of squares already used is symmetric to that diagonal as well) we can assume w.l.o.g. that he makes a move to the square marked by 10 . Now the first player can make his moves until the end of the game so that the second player will have no choice for his subsequent moves (these moves will be to the squares marked by 11 through 25 , in this order). We see that the first
player will be the one to make the last move, and hence the winner.
|
{
"exam": "BalticWay",
"problem_label": "16",
"problem_match": "\n16.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n16.",
"tier": "T3",
"year": "1997"
}
|
A rectangle can be divided into $n$ equal squares. The same rectangle can also be divided into $n+76$ equal squares. Find all possible values of $n$.
|
Answer: $n=324$.
Let $a b=n$ and $c d=n+76$, where $a, b$ and $c, d$ are the numbers of squares in each direction for the partitioning of the rectangle into $n$ and $n+76$ squares, respectively. Then $\frac{a}{c}=\frac{b}{d}$, or $a d=b c$. Denote $u=\operatorname{gcd}(a, c)$ ja $v=\operatorname{gcd}(b, d)$, then there exist positive integers $x$ and $y$ such that $\operatorname{gcd}(x, y)=1, a=u x, c=u y$ and $b=v x, d=v y$. Hence we have
$$
c d-a b=u v\left(y^{2}-x^{2}\right)=u v(y-x)(y+x)=76=2^{2} \cdot 19 .
$$
Since $y-x$ and $y+x$ are positive integers of the same parity and $\operatorname{gcd}(x, y)=1$, we have $y-x=1$ and $y+x=19$ as the only possibility, yielding $y=10, x=9$ and $u v=4$. Finally we have $n=a b=x^{2} u v=324$.
|
324
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
A rectangle can be divided into $n$ equal squares. The same rectangle can also be divided into $n+76$ equal squares. Find all possible values of $n$.
|
Answer: $n=324$.
Let $a b=n$ and $c d=n+76$, where $a, b$ and $c, d$ are the numbers of squares in each direction for the partitioning of the rectangle into $n$ and $n+76$ squares, respectively. Then $\frac{a}{c}=\frac{b}{d}$, or $a d=b c$. Denote $u=\operatorname{gcd}(a, c)$ ja $v=\operatorname{gcd}(b, d)$, then there exist positive integers $x$ and $y$ such that $\operatorname{gcd}(x, y)=1, a=u x, c=u y$ and $b=v x, d=v y$. Hence we have
$$
c d-a b=u v\left(y^{2}-x^{2}\right)=u v(y-x)(y+x)=76=2^{2} \cdot 19 .
$$
Since $y-x$ and $y+x$ are positive integers of the same parity and $\operatorname{gcd}(x, y)=1$, we have $y-x=1$ and $y+x=19$ as the only possibility, yielding $y=10, x=9$ and $u v=4$. Finally we have $n=a b=x^{2} u v=324$.
|
{
"exam": "BalticWay",
"problem_label": "17",
"problem_match": "\n17.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n17.",
"tier": "T3",
"year": "1997"
}
|
a) Prove the existence of two infinite sets $A$ and $B$, not necessarily disjoint, of non-negative integers such that each non-negative integer $n$ is uniquely representable in the form $n=a+b$ with $a \in A, b \in B$.
b) Prove that for each such pair $(A, B)$, either $A$ or $B$ contains only multiples of some integer $k>1$.
|
a) Let $A$ be the set of non-negative integers whose only non-zero decimal digits are in even positions counted from the right, and $B$ the set of nonnegative integers whose only non-zero decimal digits are in odd positions counted from the right. It is obvious that $A$ and $B$ have the required property.
b) Since the only possible representation of 0 is $0+0$, we have $0 \in A \cap B$. The only possible representations of 1 are $1+0$ and $0+1$. Hence 1 must belong to at least one of the sets $A$ and $B$. Let $1 \in A$, and let $k$ be the smallest positive integer such that $k \notin A$. Then $k>1$. If any number $b$ with $0<b<k$ belonged to $B$, it would have the two representations $b+0$ and $0+b$. Hence no such number belongs to $B$. Also, in $k=a+b$ with $a \in A$ and $b \in B$ the number $b$ cannot be 0 since then $a=k$, contradicting the assumption that $k \notin A$. Hence $b=k$, and $k \in B$.
Consider the decomposition of $A$ into the union $A_{1} \cup A_{2} \cup \cdots$ of its maximal subsets $A_{1}, A_{2}, \ldots$ of consecutive numbers, where each element of $A_{1}$ is less than each element of $A_{2}$ etc. In particular, $A_{1}=\{0,1, \ldots, k-1\}$. By our assumption the set of all non-negative integers is the union of nonintersecting sets $A_{n}+b=\left\{a+b \mid a \in A_{n}\right\}$ with $n \in \mathbb{N}$ and $b \in B$, each of these consisting of some number of consecutive integers. We will show that each subset $A_{n}$ has exactly $k$ elements. Indeed, suppose $m$ is the smallest index for which the number $l$ of elements in $A_{m}$ is different from $k$, then $l<k$ since $A_{m}+0$ and $A_{m}+k$ do not overlap. Denoting by $c$ the smallest element of $A_{m}$, we have $c+k-1 \notin A$, so $c+k-1=a+b$ with $a \in A$
and $0 \neq b \in B$. Hence, $b \geqslant k$ and $a<c$. Suppose $a \in A_{n}$, then $n<m$ and the subset $A_{n}$ has $k$ elements. But then $A_{n}+b$ overlaps with either $A_{m}+0$ or $A_{m}+k$, a contradiction.
Hence, the set of non-negative integers is the union of non-intersecting sets $A_{n}+b$ with $n \in \mathbb{N}$ and $b \in B$, each of which consists of $k$ consecutive integers. The smallest element of each of these subsets is a multiple of $k$. Since each integer $b \in B$ is the smallest element of $A_{1}+b$, it follows that each $b \in B$ is a multiple of $k$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
a) Prove the existence of two infinite sets $A$ and $B$, not necessarily disjoint, of non-negative integers such that each non-negative integer $n$ is uniquely representable in the form $n=a+b$ with $a \in A, b \in B$.
b) Prove that for each such pair $(A, B)$, either $A$ or $B$ contains only multiples of some integer $k>1$.
|
a) Let $A$ be the set of non-negative integers whose only non-zero decimal digits are in even positions counted from the right, and $B$ the set of nonnegative integers whose only non-zero decimal digits are in odd positions counted from the right. It is obvious that $A$ and $B$ have the required property.
b) Since the only possible representation of 0 is $0+0$, we have $0 \in A \cap B$. The only possible representations of 1 are $1+0$ and $0+1$. Hence 1 must belong to at least one of the sets $A$ and $B$. Let $1 \in A$, and let $k$ be the smallest positive integer such that $k \notin A$. Then $k>1$. If any number $b$ with $0<b<k$ belonged to $B$, it would have the two representations $b+0$ and $0+b$. Hence no such number belongs to $B$. Also, in $k=a+b$ with $a \in A$ and $b \in B$ the number $b$ cannot be 0 since then $a=k$, contradicting the assumption that $k \notin A$. Hence $b=k$, and $k \in B$.
Consider the decomposition of $A$ into the union $A_{1} \cup A_{2} \cup \cdots$ of its maximal subsets $A_{1}, A_{2}, \ldots$ of consecutive numbers, where each element of $A_{1}$ is less than each element of $A_{2}$ etc. In particular, $A_{1}=\{0,1, \ldots, k-1\}$. By our assumption the set of all non-negative integers is the union of nonintersecting sets $A_{n}+b=\left\{a+b \mid a \in A_{n}\right\}$ with $n \in \mathbb{N}$ and $b \in B$, each of these consisting of some number of consecutive integers. We will show that each subset $A_{n}$ has exactly $k$ elements. Indeed, suppose $m$ is the smallest index for which the number $l$ of elements in $A_{m}$ is different from $k$, then $l<k$ since $A_{m}+0$ and $A_{m}+k$ do not overlap. Denoting by $c$ the smallest element of $A_{m}$, we have $c+k-1 \notin A$, so $c+k-1=a+b$ with $a \in A$
and $0 \neq b \in B$. Hence, $b \geqslant k$ and $a<c$. Suppose $a \in A_{n}$, then $n<m$ and the subset $A_{n}$ has $k$ elements. But then $A_{n}+b$ overlaps with either $A_{m}+0$ or $A_{m}+k$, a contradiction.
Hence, the set of non-negative integers is the union of non-intersecting sets $A_{n}+b$ with $n \in \mathbb{N}$ and $b \in B$, each of which consists of $k$ consecutive integers. The smallest element of each of these subsets is a multiple of $k$. Since each integer $b \in B$ is the smallest element of $A_{1}+b$, it follows that each $b \in B$ is a multiple of $k$.
|
{
"exam": "BalticWay",
"problem_label": "18",
"problem_match": "\n18.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n18.",
"tier": "T3",
"year": "1997"
}
|
In a forest each of $n$ animals $(n \geqslant 3)$ lives in its own cave, and there is exactly one separate path between any two of these caves. Before the election for King of the Forest some of the animals make an election campaign. Each campaign-making animal visits each of the other caves exactly once,
uses only the paths for moving from cave to cave, never turns from one path to another between the caves and returns to its own cave in the end of its campaign. It is also known that no path between two caves is used by more than one campaign-making animal.
a) Prove that for any prime $n$, the maximum possible number of campaign-making animals is $\frac{n-1}{2}$;
b) Find the maximum number of campaign-making animals for $n=9$.
|
Answer: b) 4 .
a) As each campaign-making animal uses exactly $n$ paths and the total number of paths is $\frac{n(n-1)}{2}$, the number of campaign-making animals cannot exceed $\frac{n-1}{2}$. Labeling the caves by integers $0,1,2, \ldots, n-1$, we can construct $\frac{n-1}{2}$ non-intersecting campaign routes as follows:
$$
\begin{aligned}
& 0 \rightarrow 1 \rightarrow 2 \rightarrow 3 \rightarrow \ldots \rightarrow n \rightarrow 0 \\
& 0 \rightarrow 2 \rightarrow 4 \rightarrow 6 \rightarrow \ldots \rightarrow n-1 \rightarrow 0 \\
& 0 \rightarrow 3 \rightarrow 6 \rightarrow 9 \rightarrow \ldots \rightarrow n-2 \rightarrow 0 \\
& \ldots \ldots \ldots+\cdots \\
& 0 \rightarrow \frac{n-1}{2} \rightarrow n-1 \rightarrow \ldots \rightarrow \frac{n+1}{2} \rightarrow 0
\end{aligned}
$$
(As each of these cyclic routes passes through any cave, the $\frac{n-1}{2}$ campaign-making animals can be chosen arbitrarily).
b) As noted above, the number of campaign-making animals cannot exceed $\frac{9-1}{2}=4$. The 4 non-intersecting campaign routes can be constructed as follows:
$$
\begin{aligned}
& 0 \rightarrow 1 \rightarrow 2 \rightarrow 8 \rightarrow 3 \rightarrow 7 \rightarrow 4 \rightarrow 6 \rightarrow 5 \rightarrow 0 \\
& 0 \rightarrow 2 \rightarrow 3 \rightarrow 1 \rightarrow 4 \rightarrow 8 \rightarrow 5 \rightarrow 7 \rightarrow 6 \rightarrow 0 \\
& 0 \rightarrow 3 \rightarrow 4 \rightarrow 2 \rightarrow 5 \rightarrow 1 \rightarrow 6 \rightarrow 8 \rightarrow 7 \rightarrow 0 \\
& 0 \rightarrow 4 \rightarrow 5 \rightarrow 3 \rightarrow 6 \rightarrow 2 \rightarrow 7 \rightarrow 1 \rightarrow 8 \rightarrow 0
\end{aligned}
$$
Remark. In fact it can be proved that the maximal number of nonintersecting Hamiltonian cycles in a complete graph on $n$ vertices (that is what the problem actually asks for) is equal to $\left\lfloor\frac{n-1}{2}\right\rfloor$ for any integer $n$. The proof uses a construction similar to the one shown in part b) of the above solution.
|
4
|
Yes
|
Yes
|
proof
|
Combinatorics
|
In a forest each of $n$ animals $(n \geqslant 3)$ lives in its own cave, and there is exactly one separate path between any two of these caves. Before the election for King of the Forest some of the animals make an election campaign. Each campaign-making animal visits each of the other caves exactly once,
uses only the paths for moving from cave to cave, never turns from one path to another between the caves and returns to its own cave in the end of its campaign. It is also known that no path between two caves is used by more than one campaign-making animal.
a) Prove that for any prime $n$, the maximum possible number of campaign-making animals is $\frac{n-1}{2}$;
b) Find the maximum number of campaign-making animals for $n=9$.
|
Answer: b) 4 .
a) As each campaign-making animal uses exactly $n$ paths and the total number of paths is $\frac{n(n-1)}{2}$, the number of campaign-making animals cannot exceed $\frac{n-1}{2}$. Labeling the caves by integers $0,1,2, \ldots, n-1$, we can construct $\frac{n-1}{2}$ non-intersecting campaign routes as follows:
$$
\begin{aligned}
& 0 \rightarrow 1 \rightarrow 2 \rightarrow 3 \rightarrow \ldots \rightarrow n \rightarrow 0 \\
& 0 \rightarrow 2 \rightarrow 4 \rightarrow 6 \rightarrow \ldots \rightarrow n-1 \rightarrow 0 \\
& 0 \rightarrow 3 \rightarrow 6 \rightarrow 9 \rightarrow \ldots \rightarrow n-2 \rightarrow 0 \\
& \ldots \ldots \ldots+\cdots \\
& 0 \rightarrow \frac{n-1}{2} \rightarrow n-1 \rightarrow \ldots \rightarrow \frac{n+1}{2} \rightarrow 0
\end{aligned}
$$
(As each of these cyclic routes passes through any cave, the $\frac{n-1}{2}$ campaign-making animals can be chosen arbitrarily).
b) As noted above, the number of campaign-making animals cannot exceed $\frac{9-1}{2}=4$. The 4 non-intersecting campaign routes can be constructed as follows:
$$
\begin{aligned}
& 0 \rightarrow 1 \rightarrow 2 \rightarrow 8 \rightarrow 3 \rightarrow 7 \rightarrow 4 \rightarrow 6 \rightarrow 5 \rightarrow 0 \\
& 0 \rightarrow 2 \rightarrow 3 \rightarrow 1 \rightarrow 4 \rightarrow 8 \rightarrow 5 \rightarrow 7 \rightarrow 6 \rightarrow 0 \\
& 0 \rightarrow 3 \rightarrow 4 \rightarrow 2 \rightarrow 5 \rightarrow 1 \rightarrow 6 \rightarrow 8 \rightarrow 7 \rightarrow 0 \\
& 0 \rightarrow 4 \rightarrow 5 \rightarrow 3 \rightarrow 6 \rightarrow 2 \rightarrow 7 \rightarrow 1 \rightarrow 8 \rightarrow 0
\end{aligned}
$$
Remark. In fact it can be proved that the maximal number of nonintersecting Hamiltonian cycles in a complete graph on $n$ vertices (that is what the problem actually asks for) is equal to $\left\lfloor\frac{n-1}{2}\right\rfloor$ for any integer $n$. The proof uses a construction similar to the one shown in part b) of the above solution.
|
{
"exam": "BalticWay",
"problem_label": "19",
"problem_match": "\n19.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n19.",
"tier": "T3",
"year": "1997"
}
|
Twelve cards lie in a row. The cards are of three kinds: with both sides white, both sides black, or with a white and a black side. Initially, nine of the twelve cards have a black side up. The cards 1-6 are turned, and subsequently four of the twelve cards have a black side up. Now cards $4-9$ are turned, and six cards have a black side up. Finally, the cards 1-3 and 10-12 are turned, after which five cards have a black side up. How many cards of each kind are there?
## Solutions
|
Answer: there are 9 cards with one black and one white side and 3 cards with both sides white.
Divide the cards into four types according to the table below.
| Type | Initially up | Initially down |
| :---: | :---: | :---: |
| $A$ | black | white |
| $B$ | white | black |
| $C$ | white | white |
| $D$ | black | black |
When the cards 1-6 were turned, the number of cards with a black side up decreased by 5 . Hence among the cards 1-6 there are five of type $A$ and one of type $C$ or $D$. The result of all three moves is that cards $7-12$ have been turned over, hence among these cards there must be four of type $A$, and the combination of the other two must be one of the following:
(a) one of type $A$ and one of type $B$;
(b) one of type $C$ and one of type $D$;
(c) both of type $C$;
(d) both of type $D$.
Hence the unknown card among the cards 1-6 cannot be of type $D$, since this would make too many cards having a black side up initially. For the same reason, the alternatives (a), (b) and (d) are impossible. Hence there were nine cards of type $A$ and three cards of type $C$.
Alternative solution. Denote by $a_{1}, a_{2}, \ldots, a_{12}$ the sides of each card that are initially visible, and by $b_{1}, b_{2}, \ldots, b_{12}$ the initially invisible sides each of these is either white or black. The conditions of the problem imply the following:
(a) there are 9 black and 3 white sides among $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$, $a_{7}, a_{8}, a_{9}, a_{10}, a_{11}, a_{12}$
(b) there are 4 black and 8 white sides among $b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, a_{7}$, $a_{8}, a_{9}, a_{10}, a_{11}, a_{12}$;
(c) there are 6 black and 6 white sides among $b_{1}, b_{2}, b_{3}, a_{4}, a_{5}, a_{6}, b_{7}$, $b_{8}, b_{9}, a_{10}, a_{11}, a_{12}$;
(d) there are 5 black and 7 white sides among $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, b_{7}$, $b_{8}, b_{9}, b_{10}, b_{11}, b_{12}$.
Cases (b) and (d) together enumerate each of the sides $a_{i}$ and $b_{i}$ exactly once - hence there are 9 black and 15 white sides altogether. Therefore, all existing black sides are enumerated in (a), implying that we have 9 cards with one black and one white side, and the remaining 3 cards have both sides white.
|
9 \text{ cards with one black and one white side and 3 cards with both sides white}
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
Twelve cards lie in a row. The cards are of three kinds: with both sides white, both sides black, or with a white and a black side. Initially, nine of the twelve cards have a black side up. The cards 1-6 are turned, and subsequently four of the twelve cards have a black side up. Now cards $4-9$ are turned, and six cards have a black side up. Finally, the cards 1-3 and 10-12 are turned, after which five cards have a black side up. How many cards of each kind are there?
## Solutions
|
Answer: there are 9 cards with one black and one white side and 3 cards with both sides white.
Divide the cards into four types according to the table below.
| Type | Initially up | Initially down |
| :---: | :---: | :---: |
| $A$ | black | white |
| $B$ | white | black |
| $C$ | white | white |
| $D$ | black | black |
When the cards 1-6 were turned, the number of cards with a black side up decreased by 5 . Hence among the cards 1-6 there are five of type $A$ and one of type $C$ or $D$. The result of all three moves is that cards $7-12$ have been turned over, hence among these cards there must be four of type $A$, and the combination of the other two must be one of the following:
(a) one of type $A$ and one of type $B$;
(b) one of type $C$ and one of type $D$;
(c) both of type $C$;
(d) both of type $D$.
Hence the unknown card among the cards 1-6 cannot be of type $D$, since this would make too many cards having a black side up initially. For the same reason, the alternatives (a), (b) and (d) are impossible. Hence there were nine cards of type $A$ and three cards of type $C$.
Alternative solution. Denote by $a_{1}, a_{2}, \ldots, a_{12}$ the sides of each card that are initially visible, and by $b_{1}, b_{2}, \ldots, b_{12}$ the initially invisible sides each of these is either white or black. The conditions of the problem imply the following:
(a) there are 9 black and 3 white sides among $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$, $a_{7}, a_{8}, a_{9}, a_{10}, a_{11}, a_{12}$
(b) there are 4 black and 8 white sides among $b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, a_{7}$, $a_{8}, a_{9}, a_{10}, a_{11}, a_{12}$;
(c) there are 6 black and 6 white sides among $b_{1}, b_{2}, b_{3}, a_{4}, a_{5}, a_{6}, b_{7}$, $b_{8}, b_{9}, a_{10}, a_{11}, a_{12}$;
(d) there are 5 black and 7 white sides among $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, b_{7}$, $b_{8}, b_{9}, b_{10}, b_{11}, b_{12}$.
Cases (b) and (d) together enumerate each of the sides $a_{i}$ and $b_{i}$ exactly once - hence there are 9 black and 15 white sides altogether. Therefore, all existing black sides are enumerated in (a), implying that we have 9 cards with one black and one white side, and the remaining 3 cards have both sides white.
|
{
"exam": "BalticWay",
"problem_label": "20",
"problem_match": "\n20.",
"resource_path": "BalticWay/segmented/en-bw97sol.jsonl",
"solution_match": "\n20.",
"tier": "T3",
"year": "1997"
}
|
Let $\mathbb{Z}^{+}$be the set of all positive integers. Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$ satisfying the following conditions for all $x, y \in \mathbb{Z}^{+}$:
$$
\begin{aligned}
f(x, x) & =x, \\
f(x, y) & =f(y, x), \\
(x+y) f(x, y) & =y f(x, x+y) .
\end{aligned}
$$
|
Answer: $f(x, y)=\operatorname{lcm}(x, y)$ is the only such function.
We first show that there is at most one such function $f$. Let $z \geqslant 2$ be an integer. Knowing the values $f(x, y)$ for all $x, y$ with $0<x, y<z$, we compute $f(x, z)$ for $0<x<z$ using the third equation (with $y=z-x$ ); then from the first two equations we get the values $f(z, y)$ for $0<y \leqslant z$. Hence, if $f$ exists then it is unique.
Experimenting a little, we can guess that $f(x, y)$ is the least common multiple of $x$ and $y$. It remains to verify that the least-common-multiple function satisfies the given equations. The first two are clear, and for the third one:
$$
\begin{aligned}
(x+y) \cdot \operatorname{lcm}(x, y) & =(x+y) \cdot \frac{x y}{\operatorname{gcd}(x, y)}=y \cdot \frac{x(x+y)}{\operatorname{gcd}(x, x+y)}= \\
& =y \cdot \operatorname{lcm}(x, x+y) .
\end{aligned}
$$
|
f(x, y)=\operatorname{lcm}(x, y)
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $\mathbb{Z}^{+}$be the set of all positive integers. Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$ satisfying the following conditions for all $x, y \in \mathbb{Z}^{+}$:
$$
\begin{aligned}
f(x, x) & =x, \\
f(x, y) & =f(y, x), \\
(x+y) f(x, y) & =y f(x, x+y) .
\end{aligned}
$$
|
Answer: $f(x, y)=\operatorname{lcm}(x, y)$ is the only such function.
We first show that there is at most one such function $f$. Let $z \geqslant 2$ be an integer. Knowing the values $f(x, y)$ for all $x, y$ with $0<x, y<z$, we compute $f(x, z)$ for $0<x<z$ using the third equation (with $y=z-x$ ); then from the first two equations we get the values $f(z, y)$ for $0<y \leqslant z$. Hence, if $f$ exists then it is unique.
Experimenting a little, we can guess that $f(x, y)$ is the least common multiple of $x$ and $y$. It remains to verify that the least-common-multiple function satisfies the given equations. The first two are clear, and for the third one:
$$
\begin{aligned}
(x+y) \cdot \operatorname{lcm}(x, y) & =(x+y) \cdot \frac{x y}{\operatorname{gcd}(x, y)}=y \cdot \frac{x(x+y)}{\operatorname{gcd}(x, x+y)}= \\
& =y \cdot \operatorname{lcm}(x, x+y) .
\end{aligned}
$$
|
{
"exam": "BalticWay",
"problem_label": "1",
"problem_match": "\n1.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n1.",
"tier": "T3",
"year": "1998"
}
|
A triple of positive integers $(a, b, c)$ is called quasi-Pythagorean if there exists a triangle with lengths of the sides $a, b, c$ and the angle opposite to the side $c$ equal to $120^{\circ}$. Prove that if $(a, b, c)$ is a quasi-Pythagorean triple then $c$ has a prime divisor greater than 5 .
|
By the cosine law, a triple of positive integers $(a, b, c)$ is quasi-Pythagorean if and only if
$$
c^{2}=a^{2}+a b+b^{2}
$$
If a triple $(a, b, c)$ with a common divisor $d>1$ satisfies (1), then so does the reduced triple $\left(\frac{a}{d}, \frac{b}{d}, \frac{c}{d}\right)$. Hence it suffices to prove that in every irreducible quasi-Pythagorean triple the greatest term $c$ has a prime divisor greater than 5. Actually, we will show that in that case every prime divisor of $c$ is greater than 5 .
Let $(a, b, c)$ be an irreducible triple satisfying (1). Note that then $a, b$ and $c$ are pairwise coprime. We have to show that $c$ is not divisible by 2,3 or 5 .
If $c$ were even, then $a$ and $b$ (coprime to $c$ ) should be odd, and (1) would not hold.
Suppose now that $c$ is divisible by 3 , and rewrite (1) as
$$
4 c^{2}=(a+2 b)^{2}+3 a^{2}
$$
Then $a+2 b$ must be divisible by 3 . Since $a$ is coprime to $c$, the number $3 a^{2}$ is not divisible by 9 . This yields a contradiction since the remaining terms in (2) are divisible by 9 .
Finally, suppose $c$ is divisible by 5 (and hence $a$ is not). Again we get a contradiction with (2) since the square of every integer is congruent to 0 ,
1 or -1 modulo 5 ; so $4 c^{2}-3 a^{2} \equiv \pm 2(\bmod 5)$ and it cannot be equal to $(a+2 b)^{2}$. This completes the proof.
Remark. A yet stronger claim is true: If $a$ and $b$ are coprime, then every prime divisor $p>3$ of $a^{2}+a b+b^{2}$ is of the form $p=6 k+1$. (Hence every prime divisor of $c$ in an irreducible quasi-Pythagorean triple $(a, b, c)$ has such a form.)
This stronger claim can be proved by observing that $p$ does not divide $a$ and the number $g=(a+2 b) a^{(p-3) / 2}$ is an integer whose square satisfies
$$
\begin{aligned}
g^{2} & =(a+2 b)^{2} a^{p-3}=\left(4\left(a^{2}+a b+b^{2}\right)-3 a^{2}\right) a^{p-3} \equiv-3 a^{p-1} \equiv \\
& \equiv-3(\bmod p)
\end{aligned}
$$
Hence -3 is a quadratic residue modulo $p$. This is known to be true only for primes of the form $6 k+1$; proofs can be found in many books on number theory, e.g. [1].
Reference. [1] K. Ireland, M. Rosen, A Classical Introduction to Modern Number Theory, Second Edition, Springer-Verlag, New York 1990.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
A triple of positive integers $(a, b, c)$ is called quasi-Pythagorean if there exists a triangle with lengths of the sides $a, b, c$ and the angle opposite to the side $c$ equal to $120^{\circ}$. Prove that if $(a, b, c)$ is a quasi-Pythagorean triple then $c$ has a prime divisor greater than 5 .
|
By the cosine law, a triple of positive integers $(a, b, c)$ is quasi-Pythagorean if and only if
$$
c^{2}=a^{2}+a b+b^{2}
$$
If a triple $(a, b, c)$ with a common divisor $d>1$ satisfies (1), then so does the reduced triple $\left(\frac{a}{d}, \frac{b}{d}, \frac{c}{d}\right)$. Hence it suffices to prove that in every irreducible quasi-Pythagorean triple the greatest term $c$ has a prime divisor greater than 5. Actually, we will show that in that case every prime divisor of $c$ is greater than 5 .
Let $(a, b, c)$ be an irreducible triple satisfying (1). Note that then $a, b$ and $c$ are pairwise coprime. We have to show that $c$ is not divisible by 2,3 or 5 .
If $c$ were even, then $a$ and $b$ (coprime to $c$ ) should be odd, and (1) would not hold.
Suppose now that $c$ is divisible by 3 , and rewrite (1) as
$$
4 c^{2}=(a+2 b)^{2}+3 a^{2}
$$
Then $a+2 b$ must be divisible by 3 . Since $a$ is coprime to $c$, the number $3 a^{2}$ is not divisible by 9 . This yields a contradiction since the remaining terms in (2) are divisible by 9 .
Finally, suppose $c$ is divisible by 5 (and hence $a$ is not). Again we get a contradiction with (2) since the square of every integer is congruent to 0 ,
1 or -1 modulo 5 ; so $4 c^{2}-3 a^{2} \equiv \pm 2(\bmod 5)$ and it cannot be equal to $(a+2 b)^{2}$. This completes the proof.
Remark. A yet stronger claim is true: If $a$ and $b$ are coprime, then every prime divisor $p>3$ of $a^{2}+a b+b^{2}$ is of the form $p=6 k+1$. (Hence every prime divisor of $c$ in an irreducible quasi-Pythagorean triple $(a, b, c)$ has such a form.)
This stronger claim can be proved by observing that $p$ does not divide $a$ and the number $g=(a+2 b) a^{(p-3) / 2}$ is an integer whose square satisfies
$$
\begin{aligned}
g^{2} & =(a+2 b)^{2} a^{p-3}=\left(4\left(a^{2}+a b+b^{2}\right)-3 a^{2}\right) a^{p-3} \equiv-3 a^{p-1} \equiv \\
& \equiv-3(\bmod p)
\end{aligned}
$$
Hence -3 is a quadratic residue modulo $p$. This is known to be true only for primes of the form $6 k+1$; proofs can be found in many books on number theory, e.g. [1].
Reference. [1] K. Ireland, M. Rosen, A Classical Introduction to Modern Number Theory, Second Edition, Springer-Verlag, New York 1990.
|
{
"exam": "BalticWay",
"problem_label": "2",
"problem_match": "\n2.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n2.",
"tier": "T3",
"year": "1998"
}
|
Find all pairs of positive integers $x, y$ which satisfy the equation
$$
2 x^{2}+5 y^{2}=11(x y-11) \text {. }
$$
|
Answer: $x=14, y=27$.
Rewriting the equation as $2 x^{2}-x y+5 y^{2}-10 x y=-121$ and factoring we get:
$$
(2 x-y) \cdot(5 y-x)=121 .
$$
Both factors must be of the same sign. If they were both negative, we would have $2 x<y<\frac{x}{5}$, a contradiction. Hence the last equation represents the number 121 as the product of two positive integers: $a=2 x-y$ and $b=5 y-x$, and $(a, b)$ must be one of the pairs $(1,121),(11,11)$ or $(121,1)$. Examining these three possibilities we find that only the first one yields integer values of $x$ and $y$, namely, $(x, y)=(14,27)$. Hence this pair is the unique solution of the original equation.
|
(x, y)=(14,27)
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all pairs of positive integers $x, y$ which satisfy the equation
$$
2 x^{2}+5 y^{2}=11(x y-11) \text {. }
$$
|
Answer: $x=14, y=27$.
Rewriting the equation as $2 x^{2}-x y+5 y^{2}-10 x y=-121$ and factoring we get:
$$
(2 x-y) \cdot(5 y-x)=121 .
$$
Both factors must be of the same sign. If they were both negative, we would have $2 x<y<\frac{x}{5}$, a contradiction. Hence the last equation represents the number 121 as the product of two positive integers: $a=2 x-y$ and $b=5 y-x$, and $(a, b)$ must be one of the pairs $(1,121),(11,11)$ or $(121,1)$. Examining these three possibilities we find that only the first one yields integer values of $x$ and $y$, namely, $(x, y)=(14,27)$. Hence this pair is the unique solution of the original equation.
|
{
"exam": "BalticWay",
"problem_label": "3",
"problem_match": "\n3.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n3.",
"tier": "T3",
"year": "1998"
}
|
Let $P$ be a polynomial with integer coefficients. Suppose that for $n=1,2,3, \ldots, 1998$ the number $P(n)$ is a three-digit positive integer. Prove that the polynomial $P$ has no integer roots.
|
Let $m$ be an arbitrary integer and define $n \in\{1,2, \ldots, 1998\}$ to be such that $m \equiv n(\bmod 1998)$. Then $P(m) \equiv P(n)(\bmod 1998)$. Since $P(n)$ as a three-digit number cannot be divisible by 1998 , then $P(m)$ cannot be equal to 0 . Hence $P$ has no integer roots.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $P$ be a polynomial with integer coefficients. Suppose that for $n=1,2,3, \ldots, 1998$ the number $P(n)$ is a three-digit positive integer. Prove that the polynomial $P$ has no integer roots.
|
Let $m$ be an arbitrary integer and define $n \in\{1,2, \ldots, 1998\}$ to be such that $m \equiv n(\bmod 1998)$. Then $P(m) \equiv P(n)(\bmod 1998)$. Since $P(n)$ as a three-digit number cannot be divisible by 1998 , then $P(m)$ cannot be equal to 0 . Hence $P$ has no integer roots.
|
{
"exam": "BalticWay",
"problem_label": "4",
"problem_match": "\n4.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n4.",
"tier": "T3",
"year": "1998"
}
|
Let $a$ be an odd digit and $b$ an even digit. Prove that for every positive integer $n$ there exists a positive integer, divisible by $2^{n}$, whose decimal representation contains no digits other than $a$ and $b$.
|
If $b=0$, then $N=10^{n} a$ meets the demands. For the sequel, suppose $b \neq 0$.
Let $n$ be fixed. We prove that if $1 \leqslant k \leqslant n$, then we can find a positive integer $m_{k}<5^{k}$ such that the last $k$ digits of $m_{k} 2^{n}$ are all $a$ or $b$. Clearly, for $k=1$ we can find $m_{1}$ with $1 \leqslant m_{1} \leqslant 4$ such that $m_{1} 2^{n}$ ends with the digit $b$. (This corresponds to solving the congruence $m_{1} 2^{n-1} \equiv \frac{b}{2}$ modulo 5.) If $n=1$, we are done. Hence let $n \geqslant 2$.
Assume that for a certain $k$ with $1 \leqslant k<n$ we have found the integer $m_{k}$. Let $c$ be the $(k+1)$-st digit from the right of $m_{k} 2^{n}$ (i.e., the coefficient of $10^{k}$ in its decimal representation). Consider the number $5^{k} 2^{n}$ : it ends with precisely $k$ zeros, and the last non-zero digit is even; call it $d$. For any $r$, the corresponding digit of the number $m_{k} 2^{n}+r 5^{k} 2^{n}$ will be $c+r d$ modulo 10. By a suitable choice of $r \leqslant 4$ we can make this digit be either $a$ or $b$, according to whether $c$ is odd or even. (As before, this corresponds to solving one of the congruences $r \cdot \frac{d}{2} \equiv \frac{a-c}{2}$ or $r \cdot \frac{d}{2} \equiv \frac{b-c}{2}$ modulo 5.) Now, let $m_{k+1}=m_{k}+r 5^{k}$. The last $k+1$ digits of $m_{k+1} 2^{n}$ are all $a$ or $b$. As $m_{k+1}<5^{k}+4 \cdot 5^{k}=5^{k+1}$, we see that $m_{k+1}$ has the required properties. This process can be continued until we obtain a number $m_{n}$ such that the last $n$ digits of $N=m_{n} 2^{n}$ are $a$ or $b$. Since $m_{n}<5^{n}$, the number $N$ has at most $n$ digits, all of which are $a$ or $b$.
Alternative solution. The case $b=0$ is handled as in the first solution. Assume that $b \neq 0$. We prove the statement by induction on $n$, postulating, in addition, that $N$ (the integer we are looking for) must be an $n$-digit number.
For $n=1$ we take the one-digit number $b$. Assume the claim is true for a certain $n \geqslant 1$, with $N \equiv 0\left(\bmod 2^{n}\right)$ having exactly $n$ digits, all $a$ or $b$; thus $N<10^{n}$. Define
$$
N^{*}= \begin{cases}10^{n} b+N & \text { if } N \equiv 0\left(\bmod 2^{n+1}\right) \\ 10^{n} a+N & \text { if } N \equiv 2^{n}\left(\bmod 2^{n+1}\right) .\end{cases}
$$
Clearly, $N^{*}$ is an $(n+1)$-digit number, satisfying
$$
N^{*} \equiv \begin{cases}0+0\left(\bmod 2^{n+1}\right) & \text { in the first case } \\ 2^{n}+2^{n}\left(\bmod 2^{n+1}\right) & \text { in the second case. }\end{cases}
$$
In both cases $N^{*}$ is divisible by $2^{n+1}$, and we have the induction claim. The result follows.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a$ be an odd digit and $b$ an even digit. Prove that for every positive integer $n$ there exists a positive integer, divisible by $2^{n}$, whose decimal representation contains no digits other than $a$ and $b$.
|
If $b=0$, then $N=10^{n} a$ meets the demands. For the sequel, suppose $b \neq 0$.
Let $n$ be fixed. We prove that if $1 \leqslant k \leqslant n$, then we can find a positive integer $m_{k}<5^{k}$ such that the last $k$ digits of $m_{k} 2^{n}$ are all $a$ or $b$. Clearly, for $k=1$ we can find $m_{1}$ with $1 \leqslant m_{1} \leqslant 4$ such that $m_{1} 2^{n}$ ends with the digit $b$. (This corresponds to solving the congruence $m_{1} 2^{n-1} \equiv \frac{b}{2}$ modulo 5.) If $n=1$, we are done. Hence let $n \geqslant 2$.
Assume that for a certain $k$ with $1 \leqslant k<n$ we have found the integer $m_{k}$. Let $c$ be the $(k+1)$-st digit from the right of $m_{k} 2^{n}$ (i.e., the coefficient of $10^{k}$ in its decimal representation). Consider the number $5^{k} 2^{n}$ : it ends with precisely $k$ zeros, and the last non-zero digit is even; call it $d$. For any $r$, the corresponding digit of the number $m_{k} 2^{n}+r 5^{k} 2^{n}$ will be $c+r d$ modulo 10. By a suitable choice of $r \leqslant 4$ we can make this digit be either $a$ or $b$, according to whether $c$ is odd or even. (As before, this corresponds to solving one of the congruences $r \cdot \frac{d}{2} \equiv \frac{a-c}{2}$ or $r \cdot \frac{d}{2} \equiv \frac{b-c}{2}$ modulo 5.) Now, let $m_{k+1}=m_{k}+r 5^{k}$. The last $k+1$ digits of $m_{k+1} 2^{n}$ are all $a$ or $b$. As $m_{k+1}<5^{k}+4 \cdot 5^{k}=5^{k+1}$, we see that $m_{k+1}$ has the required properties. This process can be continued until we obtain a number $m_{n}$ such that the last $n$ digits of $N=m_{n} 2^{n}$ are $a$ or $b$. Since $m_{n}<5^{n}$, the number $N$ has at most $n$ digits, all of which are $a$ or $b$.
Alternative solution. The case $b=0$ is handled as in the first solution. Assume that $b \neq 0$. We prove the statement by induction on $n$, postulating, in addition, that $N$ (the integer we are looking for) must be an $n$-digit number.
For $n=1$ we take the one-digit number $b$. Assume the claim is true for a certain $n \geqslant 1$, with $N \equiv 0\left(\bmod 2^{n}\right)$ having exactly $n$ digits, all $a$ or $b$; thus $N<10^{n}$. Define
$$
N^{*}= \begin{cases}10^{n} b+N & \text { if } N \equiv 0\left(\bmod 2^{n+1}\right) \\ 10^{n} a+N & \text { if } N \equiv 2^{n}\left(\bmod 2^{n+1}\right) .\end{cases}
$$
Clearly, $N^{*}$ is an $(n+1)$-digit number, satisfying
$$
N^{*} \equiv \begin{cases}0+0\left(\bmod 2^{n+1}\right) & \text { in the first case } \\ 2^{n}+2^{n}\left(\bmod 2^{n+1}\right) & \text { in the second case. }\end{cases}
$$
In both cases $N^{*}$ is divisible by $2^{n+1}$, and we have the induction claim. The result follows.
|
{
"exam": "BalticWay",
"problem_label": "5",
"problem_match": "\n5.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n5.",
"tier": "T3",
"year": "1998"
}
|
Let $P$ be a polynomial of degree 6 and let $a, b$ be real numbers such that $0<a<b$. Suppose that $P(a)=P(-a), P(b)=P(-b)$ and $P^{\prime}(0)=0$. Prove that $P(x)=P(-x)$ for all real $x$.
|
The polynomial $Q(x)=P(x)-P(-x)$, of degree at most 5 , has roots at $-b,-a, 0, a$ and $b$; these are five distinct numbers. Moreover, $Q^{\prime}(0)=0$, showing that $Q$ has a multiple root at 0 . Thus $Q$ must be the constant 0 , i.e. $P(x)=P(-x)$ for all $x$.
|
P(x)=P(-x)
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $P$ be a polynomial of degree 6 and let $a, b$ be real numbers such that $0<a<b$. Suppose that $P(a)=P(-a), P(b)=P(-b)$ and $P^{\prime}(0)=0$. Prove that $P(x)=P(-x)$ for all real $x$.
|
The polynomial $Q(x)=P(x)-P(-x)$, of degree at most 5 , has roots at $-b,-a, 0, a$ and $b$; these are five distinct numbers. Moreover, $Q^{\prime}(0)=0$, showing that $Q$ has a multiple root at 0 . Thus $Q$ must be the constant 0 , i.e. $P(x)=P(-x)$ for all $x$.
|
{
"exam": "BalticWay",
"problem_label": "6",
"problem_match": "\n6.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n6.",
"tier": "T3",
"year": "1998"
}
|
Let $\mathbb{R}$ be the set of all real numbers. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying for all $x, y \in \mathbb{R}$ the equation
$$
f(x)+f(y)=f(f(x) f(y)) .
$$
|
Answer: $f(x) \equiv 0$ is the only such function.
Choose an arbitrary real number $x_{0}$ and denote $f\left(x_{0}\right)=c$. Setting $x=y=x_{0}$ in the equation we obtain $f\left(c^{2}\right)=2 c$. For $x=y=c^{2}$ the equation now gives $f\left(4 c^{2}\right)=4 c$. On the other hand, substituting $x=x_{0}$ and $y=4 c^{2}$ we obtain $f\left(4 c^{2}\right)=5 c$. Hence $4 c=5 c$, implying $c=0$. As $x_{0}$ was chosen arbitrarily, we have $f(x)=0$ for all real numbers $x$.
Obviously, the function $f(x) \equiv 0$ satisfies the equation. So it is the only solution.
|
f(x) \equiv 0
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $\mathbb{R}$ be the set of all real numbers. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying for all $x, y \in \mathbb{R}$ the equation
$$
f(x)+f(y)=f(f(x) f(y)) .
$$
|
Answer: $f(x) \equiv 0$ is the only such function.
Choose an arbitrary real number $x_{0}$ and denote $f\left(x_{0}\right)=c$. Setting $x=y=x_{0}$ in the equation we obtain $f\left(c^{2}\right)=2 c$. For $x=y=c^{2}$ the equation now gives $f\left(4 c^{2}\right)=4 c$. On the other hand, substituting $x=x_{0}$ and $y=4 c^{2}$ we obtain $f\left(4 c^{2}\right)=5 c$. Hence $4 c=5 c$, implying $c=0$. As $x_{0}$ was chosen arbitrarily, we have $f(x)=0$ for all real numbers $x$.
Obviously, the function $f(x) \equiv 0$ satisfies the equation. So it is the only solution.
|
{
"exam": "BalticWay",
"problem_label": "7",
"problem_match": "\n7.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n7.",
"tier": "T3",
"year": "1998"
}
|
Let $P_{k}(x)=1+x+x^{2}+\cdots+x^{k-1}$. Show that
$$
\sum_{k=1}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) P_{k}(x)=2^{n-1} P_{n}\left(\frac{1+x}{2}\right)
$$
for every real number $x$ and every positive integer $n$.
|
Let $A$ and $B$ be the left- and right-hand side of the claimed formula, respectively. Since
$$
(1-x) P_{k}(x)=1-x^{k},
$$
we get
$$
(1-x) \cdot A=\sum_{k=1}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)\left(1-x^{k}\right)=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)\left(1-x^{k}\right)=2^{n}-(1+x)^{n}
$$
and
$$
\begin{aligned}
(1-x) \cdot B & =2\left(1-\frac{1+x}{2}\right) \cdot 2^{n-1} P_{n}\left(\frac{1+x}{2}\right)= \\
& =2^{n}\left(1-\left(\frac{1+x}{2}\right)^{n}\right)=2^{n}-(1+x)^{n} .
\end{aligned}
$$
Thus $A=B$ for all real numbers $x \neq 1$. Since both $A$ and $B$ are polynomials, they coincide also for $x=1$.
Remark. The desired equality can be also proved without multiplication by $(1-x)$, just via regrouping the terms of the expanded $P_{k}$ 's and some more manipulation; this approach is more cumbersome.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $P_{k}(x)=1+x+x^{2}+\cdots+x^{k-1}$. Show that
$$
\sum_{k=1}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) P_{k}(x)=2^{n-1} P_{n}\left(\frac{1+x}{2}\right)
$$
for every real number $x$ and every positive integer $n$.
|
Let $A$ and $B$ be the left- and right-hand side of the claimed formula, respectively. Since
$$
(1-x) P_{k}(x)=1-x^{k},
$$
we get
$$
(1-x) \cdot A=\sum_{k=1}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)\left(1-x^{k}\right)=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)\left(1-x^{k}\right)=2^{n}-(1+x)^{n}
$$
and
$$
\begin{aligned}
(1-x) \cdot B & =2\left(1-\frac{1+x}{2}\right) \cdot 2^{n-1} P_{n}\left(\frac{1+x}{2}\right)= \\
& =2^{n}\left(1-\left(\frac{1+x}{2}\right)^{n}\right)=2^{n}-(1+x)^{n} .
\end{aligned}
$$
Thus $A=B$ for all real numbers $x \neq 1$. Since both $A$ and $B$ are polynomials, they coincide also for $x=1$.
Remark. The desired equality can be also proved without multiplication by $(1-x)$, just via regrouping the terms of the expanded $P_{k}$ 's and some more manipulation; this approach is more cumbersome.
|
{
"exam": "BalticWay",
"problem_label": "8",
"problem_match": "\n8.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n8.",
"tier": "T3",
"year": "1998"
}
|
Let the numbers $\alpha, \beta$ satisfy $0<\alpha<\beta<\pi / 2$ and let $\gamma$ and $\delta$ be the numbers defined by the conditions:
(i) $0<\gamma<\pi / 2$, and $\tan \gamma$ is the arithmetic mean of $\tan \alpha$ and $\tan \beta$;
(ii) $0<\delta<\pi / 2$, and $\frac{1}{\cos \delta}$ is the arithmetic mean of $\frac{1}{\cos \alpha}$ and $\frac{1}{\cos \beta}$.
Prove that $\gamma<\delta$.
|
Let $f(t)=\sqrt{1+t^{2}}$. Since $f^{\prime \prime}(t)=\left(1+t^{2}\right)^{-3 / 2}>0$, the function $f(t)$ is
strictly convex on $(0, \infty)$. Consequently,
$$
\begin{aligned}
\frac{1}{\cos \gamma} & =\sqrt{1+\tan ^{2} \gamma}=f(\tan \gamma)=f\left(\frac{\tan \alpha+\tan \beta}{2}\right)< \\
& <\frac{f(\tan \alpha)+f(\tan \beta)}{2}=\frac{1}{2}\left(\frac{1}{\cos \alpha}+\frac{1}{\cos \beta}\right)=\frac{1}{\cos \delta},
\end{aligned}
$$
and hence $\gamma<\delta$.
Remark. The use of calculus can be avoided. We only need the midpointconvexity of $f$, i.e., the inequality
$$
\sqrt{1+\frac{1}{4}(u+v)^{2}}<\frac{1}{2} \sqrt{1+u^{2}}+\frac{1}{2} \sqrt{1+v^{2}}
$$
for $u, v>0$ and $u \neq v$, which is equivalent (via squaring) to
$$
1+u v<\sqrt{\left(1+u^{2}\right)\left(1+v^{2}\right)} .
$$
The latter inequality reduces (again by squaring) to $2 u v<u^{2}+v^{2}$, holding trivially.
Alternative solution. Draw a unit segment $O P$ in the plane and take points $A$ and $B$ on the same side of line $O P$ so that $\angle P O A=\angle P O B=90^{\circ}$, $\angle O P A=\alpha$ and $\angle O P B=\beta$ (see Figure 1). Then we have $|O A|=\tan \alpha$, $|O B|=\tan \beta,|P A|=\frac{1}{\cos \alpha}$ and $|P B|=\frac{1}{\cos \beta}$.
Figure 1
Let $C$ be the midpoint of the segment $A B$. By hypothesis, we have $|O C|=\frac{\tan \alpha+\tan \beta}{2}=\tan \gamma$, hence $\angle O P C=\gamma$ and $|P C|=\frac{1}{\cos \gamma}$. Let $Q$ be the point symmetric to $P$ with respect to $C$. The quadrilateral $P A Q B$ is a parallelogram, and therefore $|A Q|=|P B|=\frac{1}{\cos \beta}$. Eventually,
$$
\frac{2}{\cos \delta}=\frac{1}{\cos \alpha}+\frac{1}{\cos \beta}=|P A|+|A Q|>|P Q|=2 \cdot|P C|=\frac{2}{\cos \gamma},
$$
and hence $\delta>\gamma$.
Another solution. Set $x=\frac{\alpha+\beta}{2}$ and $y=\frac{\alpha-\beta}{2}$, then $\alpha=x+y, \beta=x-y$ and
$$
\begin{aligned}
\cos \alpha \cos \beta & =\frac{1}{2}(\cos 2 x+\cos 2 y)= \\
& =\frac{1}{2}\left(1-2 \sin ^{2} x\right)+\frac{1}{2}\left(2 \cos ^{2} y-1\right)=\cos ^{2} y-\sin ^{2} x .
\end{aligned}
$$
By the conditions of the problem,
$$
\tan \gamma=\frac{1}{2}\left(\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}\right)=\frac{1}{2} \cdot \frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\frac{\sin x \cos x}{\cos \alpha \cos \beta}
$$
and
$$
\frac{1}{\cos \delta}=\frac{1}{2}\left(\frac{1}{\cos \alpha}+\frac{1}{\cos \beta}\right)=\frac{1}{2} \cdot \frac{\cos \alpha+\cos \beta}{\cos \alpha \cos \beta}=\frac{\cos x \cos y}{\cos \alpha \cos \beta} .
$$
Using (3) we hence obtain
$$
\begin{aligned}
\tan ^{2} \delta-\tan ^{2} \gamma & =\frac{1}{\cos ^{2} \delta}-1-\tan ^{2} \gamma=\frac{\cos ^{2} x \cos ^{2} y-\sin ^{2} x \cos ^{2} x}{\cos ^{2} \alpha \cos ^{2} \beta}-1= \\
& =\frac{\cos ^{2} x\left(\cos ^{2} y-\sin ^{2} x\right)}{\left(\cos ^{2} y-\sin ^{2} x\right)^{2}}-1=\frac{\cos ^{2} x}{\cos ^{2} y-\sin ^{2} x}-1= \\
& =\frac{\cos ^{2} x-\cos ^{2} y+\sin ^{2} x}{\cos ^{2} y-\sin ^{2} x}=\frac{\sin ^{2} y}{\cos \alpha \cos \beta}>0,
\end{aligned}
$$
showing that $\delta>\gamma$.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let the numbers $\alpha, \beta$ satisfy $0<\alpha<\beta<\pi / 2$ and let $\gamma$ and $\delta$ be the numbers defined by the conditions:
(i) $0<\gamma<\pi / 2$, and $\tan \gamma$ is the arithmetic mean of $\tan \alpha$ and $\tan \beta$;
(ii) $0<\delta<\pi / 2$, and $\frac{1}{\cos \delta}$ is the arithmetic mean of $\frac{1}{\cos \alpha}$ and $\frac{1}{\cos \beta}$.
Prove that $\gamma<\delta$.
|
Let $f(t)=\sqrt{1+t^{2}}$. Since $f^{\prime \prime}(t)=\left(1+t^{2}\right)^{-3 / 2}>0$, the function $f(t)$ is
strictly convex on $(0, \infty)$. Consequently,
$$
\begin{aligned}
\frac{1}{\cos \gamma} & =\sqrt{1+\tan ^{2} \gamma}=f(\tan \gamma)=f\left(\frac{\tan \alpha+\tan \beta}{2}\right)< \\
& <\frac{f(\tan \alpha)+f(\tan \beta)}{2}=\frac{1}{2}\left(\frac{1}{\cos \alpha}+\frac{1}{\cos \beta}\right)=\frac{1}{\cos \delta},
\end{aligned}
$$
and hence $\gamma<\delta$.
Remark. The use of calculus can be avoided. We only need the midpointconvexity of $f$, i.e., the inequality
$$
\sqrt{1+\frac{1}{4}(u+v)^{2}}<\frac{1}{2} \sqrt{1+u^{2}}+\frac{1}{2} \sqrt{1+v^{2}}
$$
for $u, v>0$ and $u \neq v$, which is equivalent (via squaring) to
$$
1+u v<\sqrt{\left(1+u^{2}\right)\left(1+v^{2}\right)} .
$$
The latter inequality reduces (again by squaring) to $2 u v<u^{2}+v^{2}$, holding trivially.
Alternative solution. Draw a unit segment $O P$ in the plane and take points $A$ and $B$ on the same side of line $O P$ so that $\angle P O A=\angle P O B=90^{\circ}$, $\angle O P A=\alpha$ and $\angle O P B=\beta$ (see Figure 1). Then we have $|O A|=\tan \alpha$, $|O B|=\tan \beta,|P A|=\frac{1}{\cos \alpha}$ and $|P B|=\frac{1}{\cos \beta}$.
Figure 1
Let $C$ be the midpoint of the segment $A B$. By hypothesis, we have $|O C|=\frac{\tan \alpha+\tan \beta}{2}=\tan \gamma$, hence $\angle O P C=\gamma$ and $|P C|=\frac{1}{\cos \gamma}$. Let $Q$ be the point symmetric to $P$ with respect to $C$. The quadrilateral $P A Q B$ is a parallelogram, and therefore $|A Q|=|P B|=\frac{1}{\cos \beta}$. Eventually,
$$
\frac{2}{\cos \delta}=\frac{1}{\cos \alpha}+\frac{1}{\cos \beta}=|P A|+|A Q|>|P Q|=2 \cdot|P C|=\frac{2}{\cos \gamma},
$$
and hence $\delta>\gamma$.
Another solution. Set $x=\frac{\alpha+\beta}{2}$ and $y=\frac{\alpha-\beta}{2}$, then $\alpha=x+y, \beta=x-y$ and
$$
\begin{aligned}
\cos \alpha \cos \beta & =\frac{1}{2}(\cos 2 x+\cos 2 y)= \\
& =\frac{1}{2}\left(1-2 \sin ^{2} x\right)+\frac{1}{2}\left(2 \cos ^{2} y-1\right)=\cos ^{2} y-\sin ^{2} x .
\end{aligned}
$$
By the conditions of the problem,
$$
\tan \gamma=\frac{1}{2}\left(\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}\right)=\frac{1}{2} \cdot \frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\frac{\sin x \cos x}{\cos \alpha \cos \beta}
$$
and
$$
\frac{1}{\cos \delta}=\frac{1}{2}\left(\frac{1}{\cos \alpha}+\frac{1}{\cos \beta}\right)=\frac{1}{2} \cdot \frac{\cos \alpha+\cos \beta}{\cos \alpha \cos \beta}=\frac{\cos x \cos y}{\cos \alpha \cos \beta} .
$$
Using (3) we hence obtain
$$
\begin{aligned}
\tan ^{2} \delta-\tan ^{2} \gamma & =\frac{1}{\cos ^{2} \delta}-1-\tan ^{2} \gamma=\frac{\cos ^{2} x \cos ^{2} y-\sin ^{2} x \cos ^{2} x}{\cos ^{2} \alpha \cos ^{2} \beta}-1= \\
& =\frac{\cos ^{2} x\left(\cos ^{2} y-\sin ^{2} x\right)}{\left(\cos ^{2} y-\sin ^{2} x\right)^{2}}-1=\frac{\cos ^{2} x}{\cos ^{2} y-\sin ^{2} x}-1= \\
& =\frac{\cos ^{2} x-\cos ^{2} y+\sin ^{2} x}{\cos ^{2} y-\sin ^{2} x}=\frac{\sin ^{2} y}{\cos \alpha \cos \beta}>0,
\end{aligned}
$$
showing that $\delta>\gamma$.
|
{
"exam": "BalticWay",
"problem_label": "9",
"problem_match": "\n9.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n9.",
"tier": "T3",
"year": "1998"
}
|
Let $n \geqslant 4$ be an even integer. A regular $n$-gon and a regular $(n-1)$-gon are inscribed into the unit circle. For each vertex of the $n$-gon consider the distance from this vertex to the nearest vertex of the $(n-1)$-gon, measured along the circumference. Let $S$ be the sum of these $n$ distances. Prove that $S$ depends only on $n$, and not on the relative position of the two polygons.
|
For simplicity, take the length of the circle to be $2 n(n-1)$ rather than $2 \pi$. The vertices of the $(n-1)$-gon $A_{0} A_{1} \ldots A_{n-2}$ divide it into $n-1$ arcs of length $2 n$. By the pigeonhole principle, some two of the vertices of the $n$-gon $B_{0} B_{1} \ldots B_{n-1}$ lie in the same arc. Assume w.l.o.g. that $B_{0}$ and $B_{1}$ lie in the arc $A_{0} A_{1}$, with $B_{0}$ closer to $A_{0}$ and $B_{1}$ closer to $A_{1}$, and that $\left|A_{0} B_{0}\right| \leqslant\left|B_{1} A_{1}\right|$.
Consider the circle as the segment $[0,2 n(n-1)]$ of the real line, with both of its endpoints identified with the vertex $A_{0}$ and the numbers $2 n, 4 n, 6 n, \ldots$ identified accordingly with the vertices $A_{1}, A_{2}, A_{3}, \ldots$
For $k=0,1, \ldots, n-1$, let $x_{k}$ be the "coordinate" of the vertex $B_{k}$ of the $n$-gon. Each arc $B_{k} B_{k+1}$ has length $2(n-1)$. By the choice of labelling, we have
$$
0 \leqslant x_{0}<x_{1}=x_{0}+2(n-1) \leqslant 2 n
$$
and, moreover, $x_{0}-0 \leqslant 2 n-x_{1}$. Hence $0 \leqslant x_{0} \leqslant 1$.
Clearly, $x_{k}=x_{0}+2 k(n-1)$ for $k=0,1, \ldots, n-1$. It is not hard to see that $(2 k-1) n \leqslant x_{k} \leqslant 2 k n$ if $1 \leqslant k \leqslant \frac{n}{2}$, and $(2 k-2) n \leqslant x_{k} \leqslant(2 k-1) n$ if $\frac{n}{2}<k \leqslant n-1$. These inequalities are verified immediately by inserting $x_{k}=x_{0}+2 k(n-1)$ and taking into account that $0 \leqslant x_{0} \leqslant 1$.
Summing up, we have:
1) if $1 \leqslant k \leqslant \frac{n}{2}$, then $B_{k}$ lies between $A_{k-1}$ and $A_{k}$, closer to $A_{k}$; recalling that $A_{k}$ has "coordinate" $2 k n$, we see that the distance in question is equal to $2 k n-x_{k}=2 k-x_{0}$;
2) if $\frac{n}{2}<k \leqslant n-1$, then $B_{k}$ lies between $A_{k-1}$ and $A_{k}$, closer to $A_{k-1}$; the distance in question is equal to $x_{k}-(2 k-2) n=x_{0}-2 k+2 n$;
3) for $B_{0}$, the distance in question is $x_{0}$.
The sum of these distances evaluates to
$$
x_{0}+\sum_{k=1}^{n / 2}\left(2 k-x_{0}\right)+\sum_{k=n / 2+1}^{n-1}\left(x_{0}-2 k+2 n\right)
$$
Note that here $x_{0}$ appears half of the times with a plus sign and half of the times with a minus sign. Thus, eventually, all terms $x_{0}$ cancel out, and the value of $S$ does not depend on anything but $n$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $n \geqslant 4$ be an even integer. A regular $n$-gon and a regular $(n-1)$-gon are inscribed into the unit circle. For each vertex of the $n$-gon consider the distance from this vertex to the nearest vertex of the $(n-1)$-gon, measured along the circumference. Let $S$ be the sum of these $n$ distances. Prove that $S$ depends only on $n$, and not on the relative position of the two polygons.
|
For simplicity, take the length of the circle to be $2 n(n-1)$ rather than $2 \pi$. The vertices of the $(n-1)$-gon $A_{0} A_{1} \ldots A_{n-2}$ divide it into $n-1$ arcs of length $2 n$. By the pigeonhole principle, some two of the vertices of the $n$-gon $B_{0} B_{1} \ldots B_{n-1}$ lie in the same arc. Assume w.l.o.g. that $B_{0}$ and $B_{1}$ lie in the arc $A_{0} A_{1}$, with $B_{0}$ closer to $A_{0}$ and $B_{1}$ closer to $A_{1}$, and that $\left|A_{0} B_{0}\right| \leqslant\left|B_{1} A_{1}\right|$.
Consider the circle as the segment $[0,2 n(n-1)]$ of the real line, with both of its endpoints identified with the vertex $A_{0}$ and the numbers $2 n, 4 n, 6 n, \ldots$ identified accordingly with the vertices $A_{1}, A_{2}, A_{3}, \ldots$
For $k=0,1, \ldots, n-1$, let $x_{k}$ be the "coordinate" of the vertex $B_{k}$ of the $n$-gon. Each arc $B_{k} B_{k+1}$ has length $2(n-1)$. By the choice of labelling, we have
$$
0 \leqslant x_{0}<x_{1}=x_{0}+2(n-1) \leqslant 2 n
$$
and, moreover, $x_{0}-0 \leqslant 2 n-x_{1}$. Hence $0 \leqslant x_{0} \leqslant 1$.
Clearly, $x_{k}=x_{0}+2 k(n-1)$ for $k=0,1, \ldots, n-1$. It is not hard to see that $(2 k-1) n \leqslant x_{k} \leqslant 2 k n$ if $1 \leqslant k \leqslant \frac{n}{2}$, and $(2 k-2) n \leqslant x_{k} \leqslant(2 k-1) n$ if $\frac{n}{2}<k \leqslant n-1$. These inequalities are verified immediately by inserting $x_{k}=x_{0}+2 k(n-1)$ and taking into account that $0 \leqslant x_{0} \leqslant 1$.
Summing up, we have:
1) if $1 \leqslant k \leqslant \frac{n}{2}$, then $B_{k}$ lies between $A_{k-1}$ and $A_{k}$, closer to $A_{k}$; recalling that $A_{k}$ has "coordinate" $2 k n$, we see that the distance in question is equal to $2 k n-x_{k}=2 k-x_{0}$;
2) if $\frac{n}{2}<k \leqslant n-1$, then $B_{k}$ lies between $A_{k-1}$ and $A_{k}$, closer to $A_{k-1}$; the distance in question is equal to $x_{k}-(2 k-2) n=x_{0}-2 k+2 n$;
3) for $B_{0}$, the distance in question is $x_{0}$.
The sum of these distances evaluates to
$$
x_{0}+\sum_{k=1}^{n / 2}\left(2 k-x_{0}\right)+\sum_{k=n / 2+1}^{n-1}\left(x_{0}-2 k+2 n\right)
$$
Note that here $x_{0}$ appears half of the times with a plus sign and half of the times with a minus sign. Thus, eventually, all terms $x_{0}$ cancel out, and the value of $S$ does not depend on anything but $n$.
|
{
"exam": "BalticWay",
"problem_label": "10",
"problem_match": "\n10.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n10.",
"tier": "T3",
"year": "1998"
}
|
Let $a, b$ and $c$ be the lengths of the sides of a triangle with circumradius $R$. Prove that
$$
R \geqslant \frac{a^{2}+b^{2}}{2 \sqrt{2 a^{2}+2 b^{2}-c^{2}}} .
$$
When does equality hold?
|
Answer: equality holds if $a=b$ or the angle opposite to $c$ is equal to $90^{\circ}$. Denote the angles opposite to the sides $a, b, c$ by $A, B, C$, respectively. By the law of sines we have $a=2 R \sin A, b=2 R \sin B, c=2 R \sin C$. Hence, the given inequality is equivalent to each of the following:
$$
\begin{aligned}
& R \geqslant \frac{4 R^{2}\left(\sin ^{2} A+\sin ^{2} B\right)}{2 \sqrt{8 R^{2}\left(\sin ^{2} A+\sin ^{2} B\right)-4 R^{2} \sin ^{2} C}}, \\
& 2\left(\sin ^{2} A+\sin ^{2} B\right)-\sin ^{2} C \geqslant\left(\sin ^{2} A+\sin ^{2} B\right)^{2} \\
& \left(\sin ^{2} A+\sin ^{2} B\right)\left(2-\sin ^{2} A-\sin ^{2} B\right) \geqslant \sin ^{2} C \\
& \left(\sin ^{2} A+\sin ^{2} B\right)\left(\cos ^{2} A+\cos ^{2} B\right) \geqslant \sin ^{2} C
\end{aligned}
$$
The last inequality follows from the Cauchy-Schwarz inequality:
$$
\begin{aligned}
& \left(\sin ^{2} A+\sin ^{2} B\right)\left(\cos ^{2} B+\cos ^{2} A\right) \geqslant \\
& \quad \geqslant(\sin A \cdot \cos B+\sin B \cdot \cos A)^{2}=\sin ^{2} C .
\end{aligned}
$$
Equality requires that $\sin A=\lambda \cos B$ and $\sin B=\lambda \cos A$ for a certain real number $\lambda$, implying that $\lambda$ is positive and $A, B$ are acute angles. From these two equations we conclude that $\sin 2 A=\sin 2 B$. This means that either $2 A=2 B$ or $2 A+2 B=\pi$; in other words, $a=b$ or $C=90^{\circ}$. In each of these two cases the inequality indeed turns into equality.
Figure 2
Alternative solution. Let $A, B, C$ be the respective vertices of the triangle, $O$ be its circumcentre and $M$ be the midpoint of $A B$ (see Figure 2). The length $m_{c}=|C M|$ of the median drawn from $C$ is expressed by the well-
known formula
$$
4 m_{c}^{2}=2 a^{2}+2 b^{2}-c^{2} .
$$
Hence the inequality of the problem can be rewritten as $4 R m_{c} \geqslant a^{2}+b^{2}$, or $8 R m_{c} \geqslant 4 m_{c}^{2}+c^{2}$. The last inequality is equivalent to
$$
\left|m_{c}-R\right| \leqslant \sqrt{R^{2}-(c / 2)^{2}},
$$
or ||$M C|-| O C|| \leqslant|O M|$, which is the triangle inequality for triangle COM .
Equality holds if and only if the points $C, O, M$ are collinear. This happens if and only if $a=b$ or $\angle C=90^{\circ}$.
Remark. Yet another solution can be obtained by setting $R=\frac{a b c}{4 S}$ (where $S$ denotes the area of the triangle) and expressing $S$ by Heron's formula. After squaring both sides, cross-multiplying and cancelling a lot, the inequality reduces to $\left(a^{2}-b^{2}\right)^{2}\left(a^{2}+b^{2}-c^{2}\right)^{2} \geqslant 0$, with equality if $a=b$ or $a^{2}+b^{2}=c^{2}$.
Figure 3
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a, b$ and $c$ be the lengths of the sides of a triangle with circumradius $R$. Prove that
$$
R \geqslant \frac{a^{2}+b^{2}}{2 \sqrt{2 a^{2}+2 b^{2}-c^{2}}} .
$$
When does equality hold?
|
Answer: equality holds if $a=b$ or the angle opposite to $c$ is equal to $90^{\circ}$. Denote the angles opposite to the sides $a, b, c$ by $A, B, C$, respectively. By the law of sines we have $a=2 R \sin A, b=2 R \sin B, c=2 R \sin C$. Hence, the given inequality is equivalent to each of the following:
$$
\begin{aligned}
& R \geqslant \frac{4 R^{2}\left(\sin ^{2} A+\sin ^{2} B\right)}{2 \sqrt{8 R^{2}\left(\sin ^{2} A+\sin ^{2} B\right)-4 R^{2} \sin ^{2} C}}, \\
& 2\left(\sin ^{2} A+\sin ^{2} B\right)-\sin ^{2} C \geqslant\left(\sin ^{2} A+\sin ^{2} B\right)^{2} \\
& \left(\sin ^{2} A+\sin ^{2} B\right)\left(2-\sin ^{2} A-\sin ^{2} B\right) \geqslant \sin ^{2} C \\
& \left(\sin ^{2} A+\sin ^{2} B\right)\left(\cos ^{2} A+\cos ^{2} B\right) \geqslant \sin ^{2} C
\end{aligned}
$$
The last inequality follows from the Cauchy-Schwarz inequality:
$$
\begin{aligned}
& \left(\sin ^{2} A+\sin ^{2} B\right)\left(\cos ^{2} B+\cos ^{2} A\right) \geqslant \\
& \quad \geqslant(\sin A \cdot \cos B+\sin B \cdot \cos A)^{2}=\sin ^{2} C .
\end{aligned}
$$
Equality requires that $\sin A=\lambda \cos B$ and $\sin B=\lambda \cos A$ for a certain real number $\lambda$, implying that $\lambda$ is positive and $A, B$ are acute angles. From these two equations we conclude that $\sin 2 A=\sin 2 B$. This means that either $2 A=2 B$ or $2 A+2 B=\pi$; in other words, $a=b$ or $C=90^{\circ}$. In each of these two cases the inequality indeed turns into equality.
Figure 2
Alternative solution. Let $A, B, C$ be the respective vertices of the triangle, $O$ be its circumcentre and $M$ be the midpoint of $A B$ (see Figure 2). The length $m_{c}=|C M|$ of the median drawn from $C$ is expressed by the well-
known formula
$$
4 m_{c}^{2}=2 a^{2}+2 b^{2}-c^{2} .
$$
Hence the inequality of the problem can be rewritten as $4 R m_{c} \geqslant a^{2}+b^{2}$, or $8 R m_{c} \geqslant 4 m_{c}^{2}+c^{2}$. The last inequality is equivalent to
$$
\left|m_{c}-R\right| \leqslant \sqrt{R^{2}-(c / 2)^{2}},
$$
or ||$M C|-| O C|| \leqslant|O M|$, which is the triangle inequality for triangle COM .
Equality holds if and only if the points $C, O, M$ are collinear. This happens if and only if $a=b$ or $\angle C=90^{\circ}$.
Remark. Yet another solution can be obtained by setting $R=\frac{a b c}{4 S}$ (where $S$ denotes the area of the triangle) and expressing $S$ by Heron's formula. After squaring both sides, cross-multiplying and cancelling a lot, the inequality reduces to $\left(a^{2}-b^{2}\right)^{2}\left(a^{2}+b^{2}-c^{2}\right)^{2} \geqslant 0$, with equality if $a=b$ or $a^{2}+b^{2}=c^{2}$.
Figure 3
|
{
"exam": "BalticWay",
"problem_label": "11",
"problem_match": "\n11.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n11.",
"tier": "T3",
"year": "1998"
}
|
In a triangle $A B C, \angle B A C=90^{\circ}$. Point $D$ lies on the side $B C$ and satisfies $\angle B D A=2 \angle B A D$. Prove that
$$
\frac{1}{|A D|}=\frac{1}{2}\left(\frac{1}{|B D|}+\frac{1}{|C D|}\right)
$$
|
Let $O$ be the circumcentre of triangle $A B C$ (i.e., the midpoint of $B C$ ) and let $A D$ meet the circumcircle again at $E$ (see Figure 3). Then $\angle B O E=2 \angle B A E=\angle C D E$, showing that $|D E|=|O E|$. Triangles $A D C$ and $B D E$ are similar; hence $\frac{|A D|}{|B D|}=\frac{|C D|}{|D E|}, \frac{|A D|}{|C D|}=\frac{|B D|}{|D E|}$ and finally
$$
\frac{|A D|}{|B D|}+\frac{|A D|}{|C D|}=\frac{|C D|}{|D E|}+\frac{|B D|}{|D E|}=\frac{|B C|}{|D E|}=\frac{|B C|}{|O E|}=2
$$
which is equivalent to the equality we have to prove.
Alternative solution. Let $\angle B A D=\alpha$ and $\angle C A D=\beta$. By the conditions of the problem, $\alpha+\beta=90^{\circ}$ (hence $\sin \beta=\cos \alpha$ ), $\angle B D A=2 \alpha$ and $\angle C D A=2 \beta$. By the law of sines,
$$
\frac{|A D|}{|B D|}=\frac{\sin 3 \alpha}{\sin \alpha}=3-4 \sin ^{2} \alpha
$$
and
$$
\frac{|A D|}{|C D|}=\frac{\sin 3 \beta}{\sin \beta}=3-4 \sin ^{2} \beta=3-4 \cos ^{2} \alpha .
$$
Adding these two equalities we get the claimed one.
Figure 4
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In a triangle $A B C, \angle B A C=90^{\circ}$. Point $D$ lies on the side $B C$ and satisfies $\angle B D A=2 \angle B A D$. Prove that
$$
\frac{1}{|A D|}=\frac{1}{2}\left(\frac{1}{|B D|}+\frac{1}{|C D|}\right)
$$
|
Let $O$ be the circumcentre of triangle $A B C$ (i.e., the midpoint of $B C$ ) and let $A D$ meet the circumcircle again at $E$ (see Figure 3). Then $\angle B O E=2 \angle B A E=\angle C D E$, showing that $|D E|=|O E|$. Triangles $A D C$ and $B D E$ are similar; hence $\frac{|A D|}{|B D|}=\frac{|C D|}{|D E|}, \frac{|A D|}{|C D|}=\frac{|B D|}{|D E|}$ and finally
$$
\frac{|A D|}{|B D|}+\frac{|A D|}{|C D|}=\frac{|C D|}{|D E|}+\frac{|B D|}{|D E|}=\frac{|B C|}{|D E|}=\frac{|B C|}{|O E|}=2
$$
which is equivalent to the equality we have to prove.
Alternative solution. Let $\angle B A D=\alpha$ and $\angle C A D=\beta$. By the conditions of the problem, $\alpha+\beta=90^{\circ}$ (hence $\sin \beta=\cos \alpha$ ), $\angle B D A=2 \alpha$ and $\angle C D A=2 \beta$. By the law of sines,
$$
\frac{|A D|}{|B D|}=\frac{\sin 3 \alpha}{\sin \alpha}=3-4 \sin ^{2} \alpha
$$
and
$$
\frac{|A D|}{|C D|}=\frac{\sin 3 \beta}{\sin \beta}=3-4 \sin ^{2} \beta=3-4 \cos ^{2} \alpha .
$$
Adding these two equalities we get the claimed one.
Figure 4
|
{
"exam": "BalticWay",
"problem_label": "12",
"problem_match": "\n12.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n12.",
"tier": "T3",
"year": "1998"
}
|
In a convex pentagon $A B C D E$, the sides $A E$ and $B C$ are parallel and $\angle A D E=\angle B D C$. The diagonals $A C$ and $B E$ intersect at $P$. Prove that $\angle E A D=\angle B D P$ and $\angle C B D=\angle A D P$.
|
Let $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ be the circumcircles of triangles $A E D$ and $B C D$, respectively. Let $D P$ meet $\mathcal{C}_{2}$ for the second time at $F$ (see Figure 4). Since $\angle A D E=\angle B D C$, the ratio of the lengths of the segments $E A$ and $B C$ is equal to the ratio of the radii of $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$. Thus the homothety with centre $P$ that takes $A E$ to $C B$, also transforms $\mathcal{C}_{1}$ onto $\mathcal{C}_{2}$. The same homothety transforms the arc $D E$ of $\mathcal{C}_{1}$ onto the $\operatorname{arc} F B$ of $\mathcal{C}_{2}$. Therefore $\angle E A D=\angle B D F=\angle B D P$. The second equality is proved similarly.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In a convex pentagon $A B C D E$, the sides $A E$ and $B C$ are parallel and $\angle A D E=\angle B D C$. The diagonals $A C$ and $B E$ intersect at $P$. Prove that $\angle E A D=\angle B D P$ and $\angle C B D=\angle A D P$.
|
Let $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ be the circumcircles of triangles $A E D$ and $B C D$, respectively. Let $D P$ meet $\mathcal{C}_{2}$ for the second time at $F$ (see Figure 4). Since $\angle A D E=\angle B D C$, the ratio of the lengths of the segments $E A$ and $B C$ is equal to the ratio of the radii of $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$. Thus the homothety with centre $P$ that takes $A E$ to $C B$, also transforms $\mathcal{C}_{1}$ onto $\mathcal{C}_{2}$. The same homothety transforms the arc $D E$ of $\mathcal{C}_{1}$ onto the $\operatorname{arc} F B$ of $\mathcal{C}_{2}$. Therefore $\angle E A D=\angle B D F=\angle B D P$. The second equality is proved similarly.
|
{
"exam": "BalticWay",
"problem_label": "13",
"problem_match": "\n13.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n13.",
"tier": "T3",
"year": "1998"
}
|
Given a triangle $A B C$ with $|A B|<|A C|$. The line passing through $B$ and parallel to $A C$ meets the external bisector of angle $B A C$ at $D$. The line passing through $C$ and parallel to $A B$ meets this bisector at $E$. Point $F$ lies on the side $A C$ and satisfies the equality $|F C|=|A B|$. Prove that $|D F|=|F E|$.
|
Since the lines $B D$ and $A C$ are parallel and since $A D$ is the external bisector of $\angle B A C$, we have $\angle B A D=\angle B D A$; denote their common size by $\alpha$ (see Figure 5). Also $\angle C A E=\angle C E A=\alpha$, implying $|A B|=|B D|$ and $|A C|=|C E|$. Let $B^{\prime}, C^{\prime}, F^{\prime}$ be the feet of the perpendiculars from
Figure 5
the points $B, C, F$ to line $D E$. From $|F C|=|A B|$ we obtain
$$
\left|B^{\prime} F^{\prime}\right|=(|A B|+|A F|) \cos \alpha=|A C| \cos \alpha=\left|A C^{\prime}\right|=\left|C^{\prime} E\right|
$$
and
$$
\left|D B^{\prime}\right|=|B D| \cos \alpha=|F C| \cos \alpha=\left|F^{\prime} C^{\prime}\right|,
$$
Thus $\left|D F^{\prime}\right|=\left|F^{\prime} E\right|$, whence $|D F|=|F E|$.
Figure 6
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Given a triangle $A B C$ with $|A B|<|A C|$. The line passing through $B$ and parallel to $A C$ meets the external bisector of angle $B A C$ at $D$. The line passing through $C$ and parallel to $A B$ meets this bisector at $E$. Point $F$ lies on the side $A C$ and satisfies the equality $|F C|=|A B|$. Prove that $|D F|=|F E|$.
|
Since the lines $B D$ and $A C$ are parallel and since $A D$ is the external bisector of $\angle B A C$, we have $\angle B A D=\angle B D A$; denote their common size by $\alpha$ (see Figure 5). Also $\angle C A E=\angle C E A=\alpha$, implying $|A B|=|B D|$ and $|A C|=|C E|$. Let $B^{\prime}, C^{\prime}, F^{\prime}$ be the feet of the perpendiculars from
Figure 5
the points $B, C, F$ to line $D E$. From $|F C|=|A B|$ we obtain
$$
\left|B^{\prime} F^{\prime}\right|=(|A B|+|A F|) \cos \alpha=|A C| \cos \alpha=\left|A C^{\prime}\right|=\left|C^{\prime} E\right|
$$
and
$$
\left|D B^{\prime}\right|=|B D| \cos \alpha=|F C| \cos \alpha=\left|F^{\prime} C^{\prime}\right|,
$$
Thus $\left|D F^{\prime}\right|=\left|F^{\prime} E\right|$, whence $|D F|=|F E|$.
Figure 6
|
{
"exam": "BalticWay",
"problem_label": "14",
"problem_match": "\n14.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n14.",
"tier": "T3",
"year": "1998"
}
|
Given an acute triangle $A B C$. Point $D$ is the foot of the perpendicular from $A$ to $B C$. Point $E$ lies on the segment $A D$ and satisfies the equation
$$
\frac{|A E|}{|E D|}=\frac{|C D|}{|D B|}
$$
Point $F$ is the foot of the perpendicular from $D$ to $B E$. Prove that $\angle A F C=90^{\circ}$.
|
Complete the rectangle $A D C P$ (see Figure 6). In view of
$$
\frac{|A E|}{|E D|}=\frac{|C D|}{|D B|}=\frac{|A P|}{|D B|}
$$
the points $B, E, P$ are collinear. Therefore $\angle D F P=90^{\circ}$, and so $F$ lies on the circumcircle of the rectangle $A D C P$ with diameter $A C$; hence $\angle A F C=90^{\circ}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Given an acute triangle $A B C$. Point $D$ is the foot of the perpendicular from $A$ to $B C$. Point $E$ lies on the segment $A D$ and satisfies the equation
$$
\frac{|A E|}{|E D|}=\frac{|C D|}{|D B|}
$$
Point $F$ is the foot of the perpendicular from $D$ to $B E$. Prove that $\angle A F C=90^{\circ}$.
|
Complete the rectangle $A D C P$ (see Figure 6). In view of
$$
\frac{|A E|}{|E D|}=\frac{|C D|}{|D B|}=\frac{|A P|}{|D B|}
$$
the points $B, E, P$ are collinear. Therefore $\angle D F P=90^{\circ}$, and so $F$ lies on the circumcircle of the rectangle $A D C P$ with diameter $A C$; hence $\angle A F C=90^{\circ}$.
|
{
"exam": "BalticWay",
"problem_label": "15",
"problem_match": "\n15.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n15.",
"tier": "T3",
"year": "1998"
}
|
Is it possible to cover a $13 \times 13$ chessboard with forty-two tiles of size $4 \times 1$ so that only the central square of the chessboard remains uncovered? (It is assumed that each tile covers exactly four squares of the chessboard, and the tiles do not overlap.)
|
Answer: no.
Label the horizontal rows by integers from 1 to 13. Assume that the tiling is possible, and let $a_{i}$ be the number of vertical tiles with their outer squares in rows $i$ and $i+3$. Then $b_{i}=a_{i}+a_{i-1}+a_{i-2}+a_{i-3}$ is the number of vertical tiles intersecting row $i$ (here we assume $a_{j}=0$ if $j \leqslant 0$ ). Since there are 13 squares in each row, and each horizontal tile covers four (i.e. an even number) of these, then $b_{i}$ must be odd for all $1 \leqslant i \leqslant 13$ except for $b_{7}$, which must be even.
We now get that $a_{1}=b_{1}$ is odd, $a_{2}$ is even (since $b_{2}=a_{2}+a_{1}$ is odd), and similarly $a_{3}$ and $a_{4}$ are even. Since $b_{5}=a_{5}+a_{4}+a_{3}+a_{2}$ is odd, then $a_{5}$ must be odd. Continuing this way we find that $a_{6}$ is even, $a_{7}$ is odd (since $b_{7}$ is even), $a_{8}$ is odd, $a_{9}$ is odd and $a_{10}$ is even. Obviously $a_{i}=0$ for $i>10$, as no tile is allowed to extend beyond the edge of the board. But then $b_{13}=a_{10}$ must be both even and odd, a contradiction.
Alternative solution. Colour the squares of the board black and white in the following pattern. In the first (top) row, let the two leftmost squares be black, the next two be white, the next two black, the next two white, and so on (at the right end there remains a single black square). In the second row, let the colouring be reciprocal to that of the first row (two white squares, two black squares, and so on). If the rows are labelled by 1 through 13 , let all the odd-indexed rows be coloured as the first row, and all the even-indexed ones as the second row (see Figure 7).
Note that there are more black squares than white squares in the board. Each $4 \times 1$ tile, no matter how placed, covers two black squares and two white squares. Thus if a tiling leaves a single square uncovered, this square must be black. But the central square of the board is white. Hence such a tiling is impossible.
Another solution. Colour the squares in four colours as follows: colour all squares in the 1-st column green, all squares in the 2-nd column black, all squares in the 3 -rd column white, all squares in the 4 -th column red, all squares in the 5 -th column green, all squares in the 6 -th column black etc., leaving only the central square uncoloured (see Figure 8). Altogether we have $3 \cdot 13=39$ black squares and $3 \cdot 13-1=38$ white squares. Since
each $4 \times 1$ tile covers either one square of each colour or all four squares of the same colour, then the difference of the numbers of black and white squares must be divisible by 4 . Since $39-38=1$ is not divisible by 4 , the required tiling does not exist.
Figure 7
G BWR G B WR G BWR G
Figure 8
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Is it possible to cover a $13 \times 13$ chessboard with forty-two tiles of size $4 \times 1$ so that only the central square of the chessboard remains uncovered? (It is assumed that each tile covers exactly four squares of the chessboard, and the tiles do not overlap.)
|
Answer: no.
Label the horizontal rows by integers from 1 to 13. Assume that the tiling is possible, and let $a_{i}$ be the number of vertical tiles with their outer squares in rows $i$ and $i+3$. Then $b_{i}=a_{i}+a_{i-1}+a_{i-2}+a_{i-3}$ is the number of vertical tiles intersecting row $i$ (here we assume $a_{j}=0$ if $j \leqslant 0$ ). Since there are 13 squares in each row, and each horizontal tile covers four (i.e. an even number) of these, then $b_{i}$ must be odd for all $1 \leqslant i \leqslant 13$ except for $b_{7}$, which must be even.
We now get that $a_{1}=b_{1}$ is odd, $a_{2}$ is even (since $b_{2}=a_{2}+a_{1}$ is odd), and similarly $a_{3}$ and $a_{4}$ are even. Since $b_{5}=a_{5}+a_{4}+a_{3}+a_{2}$ is odd, then $a_{5}$ must be odd. Continuing this way we find that $a_{6}$ is even, $a_{7}$ is odd (since $b_{7}$ is even), $a_{8}$ is odd, $a_{9}$ is odd and $a_{10}$ is even. Obviously $a_{i}=0$ for $i>10$, as no tile is allowed to extend beyond the edge of the board. But then $b_{13}=a_{10}$ must be both even and odd, a contradiction.
Alternative solution. Colour the squares of the board black and white in the following pattern. In the first (top) row, let the two leftmost squares be black, the next two be white, the next two black, the next two white, and so on (at the right end there remains a single black square). In the second row, let the colouring be reciprocal to that of the first row (two white squares, two black squares, and so on). If the rows are labelled by 1 through 13 , let all the odd-indexed rows be coloured as the first row, and all the even-indexed ones as the second row (see Figure 7).
Note that there are more black squares than white squares in the board. Each $4 \times 1$ tile, no matter how placed, covers two black squares and two white squares. Thus if a tiling leaves a single square uncovered, this square must be black. But the central square of the board is white. Hence such a tiling is impossible.
Another solution. Colour the squares in four colours as follows: colour all squares in the 1-st column green, all squares in the 2-nd column black, all squares in the 3 -rd column white, all squares in the 4 -th column red, all squares in the 5 -th column green, all squares in the 6 -th column black etc., leaving only the central square uncoloured (see Figure 8). Altogether we have $3 \cdot 13=39$ black squares and $3 \cdot 13-1=38$ white squares. Since
each $4 \times 1$ tile covers either one square of each colour or all four squares of the same colour, then the difference of the numbers of black and white squares must be divisible by 4 . Since $39-38=1$ is not divisible by 4 , the required tiling does not exist.
Figure 7
G BWR G B WR G BWR G
Figure 8
|
{
"exam": "BalticWay",
"problem_label": "16",
"problem_match": "\n16.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n16.",
"tier": "T3",
"year": "1998"
}
|
Let $n$ and $k$ be positive integers. There are $n k$ objects (of the same size) and $k$ boxes, each of which can hold $n$ objects. Each object is coloured in one of $k$ different colours. Show that the objects can be packed in the boxes so that each box holds objects of at most two colours.
|
If $k=1$, it is obvious how to do the packing. Now assume $k>1$. There are not more than $n$ objects of a certain colour - say, pink - and also not fewer than $n$ objects of some other colour - say, grey. Pack all pink objects into one box; if there is space left, fill the box up with grey objects. Then remove that box together with its contents; the problem gets reduced to an analogous one with $k-1$ boxes and $k-1$ colours. Assuming inductively that the task can be done in that case, we see that it can also be done for $k$ boxes and colours. The general result follows by induction.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n$ and $k$ be positive integers. There are $n k$ objects (of the same size) and $k$ boxes, each of which can hold $n$ objects. Each object is coloured in one of $k$ different colours. Show that the objects can be packed in the boxes so that each box holds objects of at most two colours.
|
If $k=1$, it is obvious how to do the packing. Now assume $k>1$. There are not more than $n$ objects of a certain colour - say, pink - and also not fewer than $n$ objects of some other colour - say, grey. Pack all pink objects into one box; if there is space left, fill the box up with grey objects. Then remove that box together with its contents; the problem gets reduced to an analogous one with $k-1$ boxes and $k-1$ colours. Assuming inductively that the task can be done in that case, we see that it can also be done for $k$ boxes and colours. The general result follows by induction.
|
{
"exam": "BalticWay",
"problem_label": "17",
"problem_match": "\n17.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n17.",
"tier": "T3",
"year": "1998"
}
|
Determine all positive integers $n$ for which there exists a set $S$ with the following properties:
(i) $S$ consists of $n$ positive integers, all smaller than $2^{n-1}$;
(ii) for any two distinct subsets $A$ and $B$ of $S$, the sum of the elements of $A$ is different from the sum of the elements of $B$.
|
Answer: all integers $n \geqslant 4$.
Direct search shows that there is no such set $S$ for $n=1,2,3$. For $n=4$ we can take $S=\{3,5,6,7\}$. If, for a certain $n \geqslant 4$ we have a set $S=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ as needed, then the set $S^{*}=\left\{1,2 a_{1}, 2 a_{2}, \ldots, 2 a_{n}\right\}$ satisfies the requirements for $n+1$. Hence a set with the required properties exists if and only if $n \geqslant 4$.
|
n \geqslant 4
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Determine all positive integers $n$ for which there exists a set $S$ with the following properties:
(i) $S$ consists of $n$ positive integers, all smaller than $2^{n-1}$;
(ii) for any two distinct subsets $A$ and $B$ of $S$, the sum of the elements of $A$ is different from the sum of the elements of $B$.
|
Answer: all integers $n \geqslant 4$.
Direct search shows that there is no such set $S$ for $n=1,2,3$. For $n=4$ we can take $S=\{3,5,6,7\}$. If, for a certain $n \geqslant 4$ we have a set $S=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ as needed, then the set $S^{*}=\left\{1,2 a_{1}, 2 a_{2}, \ldots, 2 a_{n}\right\}$ satisfies the requirements for $n+1$. Hence a set with the required properties exists if and only if $n \geqslant 4$.
|
{
"exam": "BalticWay",
"problem_label": "18",
"problem_match": "\n18.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n18.",
"tier": "T3",
"year": "1998"
}
|
Consider a ping-pong match between two teams, each consisting of 1000 players. Each player played against each player of the other team exactly once (there are no draws in ping-pong). Prove that there exist ten players, all from the same team, such that every member of the other team has lost his game against at least one of those ten players.
|
We start with the following observation: In a match between two teams (not necessarily of equal sizes), there exists in one of the teams a player who won his games with at least half of the members of the other team.
Indeed: suppose there is no such player. If the teams consist of $m$ and $n$ members then the players of the first team jointly won less than $m \cdot \frac{n}{2}$ games, and the players of the second team jointly won less than $m \cdot \frac{n}{2}$ games - this is a contradiction since the total number of games played is $m n$, and in each game there must have been a winner.
Returning to the original problem (with two equal teams of size 1000), choose a player who won his games with at least half of the members of the other team - such a player exists, according to the observation above, and we shall call his team "first" and the other team "second" in the sequel. Mark this player with a white hat and remove from further consideration all those players of the second team who lost their games to him. Applying the same observation to the first team (complete) and the second team truncated as explained above, we again find a player (in the first or in the second team) who won with at least half of the other team members. Mark him with a white hat, too, and remove the players who lost to him from further consideration.
We repeat this procedure until there are no players left in one of the teams; say, in team $Y$. This means that the white-hatted players of team $X$ constitute a group with the required property (every member of team $Y$ has lost his game to at least one player from that group). Each time when a player of team $X$ was receiving a white hat, the size of team $Y$ was reduced at least by half; and since initially the size was a number less than $2^{10}$, this could not happen more than ten times.
Hence the white-hatted group from team $X$ consists of not more than ten players. If there are fewer than ten, round the group up to ten with any players.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Consider a ping-pong match between two teams, each consisting of 1000 players. Each player played against each player of the other team exactly once (there are no draws in ping-pong). Prove that there exist ten players, all from the same team, such that every member of the other team has lost his game against at least one of those ten players.
|
We start with the following observation: In a match between two teams (not necessarily of equal sizes), there exists in one of the teams a player who won his games with at least half of the members of the other team.
Indeed: suppose there is no such player. If the teams consist of $m$ and $n$ members then the players of the first team jointly won less than $m \cdot \frac{n}{2}$ games, and the players of the second team jointly won less than $m \cdot \frac{n}{2}$ games - this is a contradiction since the total number of games played is $m n$, and in each game there must have been a winner.
Returning to the original problem (with two equal teams of size 1000), choose a player who won his games with at least half of the members of the other team - such a player exists, according to the observation above, and we shall call his team "first" and the other team "second" in the sequel. Mark this player with a white hat and remove from further consideration all those players of the second team who lost their games to him. Applying the same observation to the first team (complete) and the second team truncated as explained above, we again find a player (in the first or in the second team) who won with at least half of the other team members. Mark him with a white hat, too, and remove the players who lost to him from further consideration.
We repeat this procedure until there are no players left in one of the teams; say, in team $Y$. This means that the white-hatted players of team $X$ constitute a group with the required property (every member of team $Y$ has lost his game to at least one player from that group). Each time when a player of team $X$ was receiving a white hat, the size of team $Y$ was reduced at least by half; and since initially the size was a number less than $2^{10}$, this could not happen more than ten times.
Hence the white-hatted group from team $X$ consists of not more than ten players. If there are fewer than ten, round the group up to ten with any players.
|
{
"exam": "BalticWay",
"problem_label": "19",
"problem_match": "\n19.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n19.",
"tier": "T3",
"year": "1998"
}
|
We say that an integer $m$ covers the number 1998 if $1,9,9,8$ appear in this order as digits of $m$. (For instance, 1998 is covered by 215993698 but not by 213326798 .) Let $k(n)$ be the number of positive integers that cover 1998 and have exactly $n$ digits $(n \geqslant 5)$, all different from 0 . What is the remainder of $k(n)$ in division by 8 ?
## Solutions
|
Answer: 1.
Let $1 \leqslant g<h<i<j \leqslant n$ be fixed integers. Consider all $n$-digit numbers $a=\overline{a_{1} a_{2} \ldots a_{n}}$ with all digits non-zero, such that $a_{g}=1, a_{h}=9, a_{i}=9$, $a_{j}=8$ and this quadruple 1998 is the leftmost one in $a$; that is,
$$
\begin{cases}a_{l} \neq 1 & \text { if } l<g ; \\ a_{l} \neq 9 & \text { if } g<l<h ; \\ a_{l} \neq 9 & \text { if } h<l<i \\ a_{l} \neq 8 & \text { if } i<l<j\end{cases}
$$
There are $k_{g h i j}(n)=8^{g-1} \cdot 8^{h-g-1} \cdot 8^{i-h-1} \cdot 8^{j-i-1} \cdot 9^{n-j}$ such numbers $a$. Obviously, $k_{g h i j}(n) \equiv 1(\bmod 8)$ for $g=1, h=2, i=3, j=4$, and $k_{g h i j}(n) \equiv 0(\bmod 8)$ in all other cases. Since $k(n)$ is obtained by summing up the values of $k_{g h i j}(n)$ over all possible choicecs of $g, h, i, j$, the remainder we are looking for is 1 .
|
1
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
We say that an integer $m$ covers the number 1998 if $1,9,9,8$ appear in this order as digits of $m$. (For instance, 1998 is covered by 215993698 but not by 213326798 .) Let $k(n)$ be the number of positive integers that cover 1998 and have exactly $n$ digits $(n \geqslant 5)$, all different from 0 . What is the remainder of $k(n)$ in division by 8 ?
## Solutions
|
Answer: 1.
Let $1 \leqslant g<h<i<j \leqslant n$ be fixed integers. Consider all $n$-digit numbers $a=\overline{a_{1} a_{2} \ldots a_{n}}$ with all digits non-zero, such that $a_{g}=1, a_{h}=9, a_{i}=9$, $a_{j}=8$ and this quadruple 1998 is the leftmost one in $a$; that is,
$$
\begin{cases}a_{l} \neq 1 & \text { if } l<g ; \\ a_{l} \neq 9 & \text { if } g<l<h ; \\ a_{l} \neq 9 & \text { if } h<l<i \\ a_{l} \neq 8 & \text { if } i<l<j\end{cases}
$$
There are $k_{g h i j}(n)=8^{g-1} \cdot 8^{h-g-1} \cdot 8^{i-h-1} \cdot 8^{j-i-1} \cdot 9^{n-j}$ such numbers $a$. Obviously, $k_{g h i j}(n) \equiv 1(\bmod 8)$ for $g=1, h=2, i=3, j=4$, and $k_{g h i j}(n) \equiv 0(\bmod 8)$ in all other cases. Since $k(n)$ is obtained by summing up the values of $k_{g h i j}(n)$ over all possible choicecs of $g, h, i, j$, the remainder we are looking for is 1 .
|
{
"exam": "BalticWay",
"problem_label": "20",
"problem_match": "\n20.",
"resource_path": "BalticWay/segmented/en-bw98sol.jsonl",
"solution_match": "\n20.",
"tier": "T3",
"year": "1998"
}
|
Determine all real numbers $a, b, c, d$ that satisfy the following system of equations.
$$
\left\{\begin{array}{r}
a b c+a b+b c+c a+a+b+c=1 \\
b c d+b c+c d+d b+b+c+d=9 \\
c d a+c d+d a+a c+c+d+a=9 \\
d a b+d a+a b+b d+d+a+b=9
\end{array}\right.
$$
|
Answer: $a=b=c=\sqrt[3]{2}-1, d=5 \sqrt[3]{2}-1$.
Substituting $A=a+1, B=b+1, C=c+1, D=d+1$, we obtain
$$
\begin{aligned}
& A B C=2 \\
& B C D=10 \\
& C D A=10 \\
& D A B=10
\end{aligned}
$$
Multiplying (1), (2), (3) gives $C^{3}(A B D)^{2}=200$, which together with (4) implies $C^{3}=2$. Similarly we find $A^{3}=B^{3}=2$ and $D^{3}=250$. Therefore the only solution is $a=b=c=\sqrt[3]{2}-1, d=5 \sqrt[3]{2}-1$.
|
a=b=c=\sqrt[3]{2}-1, d=5 \sqrt[3]{2}-1
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Determine all real numbers $a, b, c, d$ that satisfy the following system of equations.
$$
\left\{\begin{array}{r}
a b c+a b+b c+c a+a+b+c=1 \\
b c d+b c+c d+d b+b+c+d=9 \\
c d a+c d+d a+a c+c+d+a=9 \\
d a b+d a+a b+b d+d+a+b=9
\end{array}\right.
$$
|
Answer: $a=b=c=\sqrt[3]{2}-1, d=5 \sqrt[3]{2}-1$.
Substituting $A=a+1, B=b+1, C=c+1, D=d+1$, we obtain
$$
\begin{aligned}
& A B C=2 \\
& B C D=10 \\
& C D A=10 \\
& D A B=10
\end{aligned}
$$
Multiplying (1), (2), (3) gives $C^{3}(A B D)^{2}=200$, which together with (4) implies $C^{3}=2$. Similarly we find $A^{3}=B^{3}=2$ and $D^{3}=250$. Therefore the only solution is $a=b=c=\sqrt[3]{2}-1, d=5 \sqrt[3]{2}-1$.
|
{
"exam": "BalticWay",
"problem_label": "1",
"problem_match": "\n1.",
"resource_path": "BalticWay/segmented/en-bw99sol.jsonl",
"solution_match": "\n1.",
"tier": "T3",
"year": "1999"
}
|
Determine all positive integers $n$ with the property that the third root of $n$ is obtained by removing the last three decimal digits of $n$.
|
Answer: 32768 is the only such integer.
If $n=m^{3}$ is a solution, then $m$ satisfies $1000 m \leqslant m^{3}<1000(m+1)$. From the first inequality, we get $m^{2} \geqslant 1000$, or $m \geqslant 32$. By the second inequality, we then have
$$
m^{2}<1000 \cdot \frac{m+1}{m} \leqslant 1000 \cdot \frac{33}{32}=1000+\frac{1000}{32} \leqslant 1032,
$$
or $m \leqslant 32$. Hence, $m=32$ and $n=m^{3}=32768$ is the only solution.
|
32768
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Determine all positive integers $n$ with the property that the third root of $n$ is obtained by removing the last three decimal digits of $n$.
|
Answer: 32768 is the only such integer.
If $n=m^{3}$ is a solution, then $m$ satisfies $1000 m \leqslant m^{3}<1000(m+1)$. From the first inequality, we get $m^{2} \geqslant 1000$, or $m \geqslant 32$. By the second inequality, we then have
$$
m^{2}<1000 \cdot \frac{m+1}{m} \leqslant 1000 \cdot \frac{33}{32}=1000+\frac{1000}{32} \leqslant 1032,
$$
or $m \leqslant 32$. Hence, $m=32$ and $n=m^{3}=32768$ is the only solution.
|
{
"exam": "BalticWay",
"problem_label": "2",
"problem_match": "\n2.",
"resource_path": "BalticWay/segmented/en-bw99sol.jsonl",
"solution_match": "\n2.",
"tier": "T3",
"year": "1999"
}
|
Determine all positive integers $n \geqslant 3$ such that the inequality
$$
a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n-1} a_{n}+a_{n} a_{1} \leqslant 0
$$
holds for all real numbers $a_{1}, a_{2}, \ldots, a_{n}$ which satisfy $a_{1}+\cdots+a_{n}=0$.
|
Answer: $n=3$ and $n=4$.
For $n=3$ we have
$$
\begin{aligned}
& a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}=\frac{\left(a_{1}+a_{2}+a_{3}\right)^{2}-\left(a_{1}^{2}+a_{2}^{3}+a_{3}^{2}\right)}{2} \leqslant \\
& \quad \leqslant \frac{\left(a_{1}+a_{2}+a_{3}\right)^{2}}{2}=0 .
\end{aligned}
$$
For $n=4$, applying the AM-GM inequality we have
$$
\begin{aligned}
& a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}+a_{4} a_{1}=\left(a_{1}+a_{3}\right)\left(a_{2}+a_{4}\right) \leqslant \\
& \\
& \leqslant \frac{\left(a_{1}+a_{2}+a_{3}+a_{4}\right)^{2}}{4}=0 .
\end{aligned}
$$
For $n \geqslant 5$ take $a_{1}=-1, a_{2}=-2, a_{3}=a_{4}=\cdots=a_{n-2}=0, a_{n-1}=2$, $a_{n}=1$. This gives
$$
a_{1} a_{2}+a_{2} a_{3}+\ldots+a_{n-1} a_{n}+a_{n} a_{1}=2+2-1=3>0 .
$$
|
n=3 \text{ and } n=4
|
Yes
|
Yes
|
math-word-problem
|
Inequalities
|
Determine all positive integers $n \geqslant 3$ such that the inequality
$$
a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n-1} a_{n}+a_{n} a_{1} \leqslant 0
$$
holds for all real numbers $a_{1}, a_{2}, \ldots, a_{n}$ which satisfy $a_{1}+\cdots+a_{n}=0$.
|
Answer: $n=3$ and $n=4$.
For $n=3$ we have
$$
\begin{aligned}
& a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}=\frac{\left(a_{1}+a_{2}+a_{3}\right)^{2}-\left(a_{1}^{2}+a_{2}^{3}+a_{3}^{2}\right)}{2} \leqslant \\
& \quad \leqslant \frac{\left(a_{1}+a_{2}+a_{3}\right)^{2}}{2}=0 .
\end{aligned}
$$
For $n=4$, applying the AM-GM inequality we have
$$
\begin{aligned}
& a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}+a_{4} a_{1}=\left(a_{1}+a_{3}\right)\left(a_{2}+a_{4}\right) \leqslant \\
& \\
& \leqslant \frac{\left(a_{1}+a_{2}+a_{3}+a_{4}\right)^{2}}{4}=0 .
\end{aligned}
$$
For $n \geqslant 5$ take $a_{1}=-1, a_{2}=-2, a_{3}=a_{4}=\cdots=a_{n-2}=0, a_{n-1}=2$, $a_{n}=1$. This gives
$$
a_{1} a_{2}+a_{2} a_{3}+\ldots+a_{n-1} a_{n}+a_{n} a_{1}=2+2-1=3>0 .
$$
|
{
"exam": "BalticWay",
"problem_label": "3",
"problem_match": "\n3.",
"resource_path": "BalticWay/segmented/en-bw99sol.jsonl",
"solution_match": "\n3.",
"tier": "T3",
"year": "1999"
}
|
For all positive real numbers $x$ and $y$ let
$$
f(x, y)=\min \left(x, \frac{y}{x^{2}+y^{2}}\right) .
$$
Show that there exist $x_{0}$ and $y_{0}$ such that $f(x, y) \leqslant f\left(x_{0}, y_{0}\right)$ for all positive $x$ and $y$, and find $f\left(x_{0}, y_{0}\right)$.
|
Answer: the maximum value is $f\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}$.
We shall make use of the inequality $x^{2}+y^{2} \geqslant 2 x y$. If $x \leqslant \frac{y}{x^{2}+y^{2}}$, then
$$
x \leqslant \frac{y}{x^{2}+y^{2}} \leqslant \frac{y}{2 x y}=\frac{1}{2 x},
$$
implying $x \leqslant \frac{1}{\sqrt{2}}$, and the equality holds if and only if $x=y=\frac{1}{\sqrt{2}}$.
If $x>\frac{1}{\sqrt{2}}$, then
$$
\frac{y}{x^{2}+y^{2}} \leqslant \frac{y}{2 x y}=\frac{1}{2 x}<\frac{1}{\sqrt{2}} .
$$
Hence always at least one of $x$ and $\frac{y}{x^{2}+y^{2}}$ does not exceed $\frac{1}{\sqrt{2}}$. Consequently $f(x, y) \leqslant \frac{1}{\sqrt{2}}$, with an equality if and only if $x=y=\frac{1}{\sqrt{2}}$.
|
\frac{1}{\sqrt{2}}
|
Yes
|
Yes
|
proof
|
Algebra
|
For all positive real numbers $x$ and $y$ let
$$
f(x, y)=\min \left(x, \frac{y}{x^{2}+y^{2}}\right) .
$$
Show that there exist $x_{0}$ and $y_{0}$ such that $f(x, y) \leqslant f\left(x_{0}, y_{0}\right)$ for all positive $x$ and $y$, and find $f\left(x_{0}, y_{0}\right)$.
|
Answer: the maximum value is $f\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}$.
We shall make use of the inequality $x^{2}+y^{2} \geqslant 2 x y$. If $x \leqslant \frac{y}{x^{2}+y^{2}}$, then
$$
x \leqslant \frac{y}{x^{2}+y^{2}} \leqslant \frac{y}{2 x y}=\frac{1}{2 x},
$$
implying $x \leqslant \frac{1}{\sqrt{2}}$, and the equality holds if and only if $x=y=\frac{1}{\sqrt{2}}$.
If $x>\frac{1}{\sqrt{2}}$, then
$$
\frac{y}{x^{2}+y^{2}} \leqslant \frac{y}{2 x y}=\frac{1}{2 x}<\frac{1}{\sqrt{2}} .
$$
Hence always at least one of $x$ and $\frac{y}{x^{2}+y^{2}}$ does not exceed $\frac{1}{\sqrt{2}}$. Consequently $f(x, y) \leqslant \frac{1}{\sqrt{2}}$, with an equality if and only if $x=y=\frac{1}{\sqrt{2}}$.
|
{
"exam": "BalticWay",
"problem_label": "4",
"problem_match": "\n4.",
"resource_path": "BalticWay/segmented/en-bw99sol.jsonl",
"solution_match": "\n4.",
"tier": "T3",
"year": "1999"
}
|
The point $(a, b)$ lies on the circle $x^{2}+y^{2}=1$. The tangent to the circle at this point meets the parabola $y=x^{2}+1$ at exactly one point. Find all such points $(a, b)$.
|
Answer: $(-1,0),(1,0),(0,1),\left(-\frac{2 \sqrt{6}}{5},-\frac{1}{5}\right),\left(\frac{2 \sqrt{6}}{5},-\frac{1}{5}\right)$.
Since any non-vertical line intersecting the parabola $y=x^{2}+1$ has exactly two intersection points with it, the line mentioned in the problem must be either vertical or a common tangent to the circle and the parabola. The only vertical lines with the required property are the lines $x=1$ and $x=-1$, which meet the circle in the points $(1,0)$ and $(-1,0)$, respectively.
Now, consider a line $y=k x+l$. It touches the circle if and only if the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=1 \\
y=k x+l
\end{array}\right.
$$
has a unique solution, or equivalently the equation $x^{2}+(k x+l)^{2}=1$ has unique solution, i.e. if and only if
$$
D_{1}=4 k^{2} l^{2}-4\left(1+k^{2}\right)\left(l^{2}-1\right)=4\left(k^{2}-l^{2}+1\right)=0 \text {, }
$$
or $l^{2}-k^{2}=1$. The line is tangent to the parabola if and only if the system
$$
\left\{\begin{array}{l}
y=x^{2}+1 \\
y=k x+l
\end{array}\right.
$$
has a unique solution, or equivalently the equation $x^{2}=k x+l-1$ has unique solution, i.e. if and only if
$$
D_{2}=k^{2}-4(1-l)=k^{2}+4 l-4=0 \text {. }
$$
From the system of equations
$$
\left\{\begin{array}{l}
l^{2}-k^{2}=1 \\
k^{2}+4 l-4=0
\end{array}\right.
$$
we have $l^{2}+4 l-5=0$, which has two solutions $l=1$ and $l=-5$. Hence the last system of equations has the solutions $k=0, l=1$ and $k= \pm 2 \sqrt{6}$, $l=-5$. From (5) we now have $(0,1)$ and $\left( \pm \frac{2 \sqrt{6}}{5},-\frac{1}{5}\right)$ as the possible points of tangency on the circle.
|
(-1,0),(1,0),(0,1),\left(-\frac{2 \sqrt{6}}{5},-\frac{1}{5}\right),\left(\frac{2 \sqrt{6}}{5},-\frac{1}{5}\right)
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
The point $(a, b)$ lies on the circle $x^{2}+y^{2}=1$. The tangent to the circle at this point meets the parabola $y=x^{2}+1$ at exactly one point. Find all such points $(a, b)$.
|
Answer: $(-1,0),(1,0),(0,1),\left(-\frac{2 \sqrt{6}}{5},-\frac{1}{5}\right),\left(\frac{2 \sqrt{6}}{5},-\frac{1}{5}\right)$.
Since any non-vertical line intersecting the parabola $y=x^{2}+1$ has exactly two intersection points with it, the line mentioned in the problem must be either vertical or a common tangent to the circle and the parabola. The only vertical lines with the required property are the lines $x=1$ and $x=-1$, which meet the circle in the points $(1,0)$ and $(-1,0)$, respectively.
Now, consider a line $y=k x+l$. It touches the circle if and only if the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=1 \\
y=k x+l
\end{array}\right.
$$
has a unique solution, or equivalently the equation $x^{2}+(k x+l)^{2}=1$ has unique solution, i.e. if and only if
$$
D_{1}=4 k^{2} l^{2}-4\left(1+k^{2}\right)\left(l^{2}-1\right)=4\left(k^{2}-l^{2}+1\right)=0 \text {, }
$$
or $l^{2}-k^{2}=1$. The line is tangent to the parabola if and only if the system
$$
\left\{\begin{array}{l}
y=x^{2}+1 \\
y=k x+l
\end{array}\right.
$$
has a unique solution, or equivalently the equation $x^{2}=k x+l-1$ has unique solution, i.e. if and only if
$$
D_{2}=k^{2}-4(1-l)=k^{2}+4 l-4=0 \text {. }
$$
From the system of equations
$$
\left\{\begin{array}{l}
l^{2}-k^{2}=1 \\
k^{2}+4 l-4=0
\end{array}\right.
$$
we have $l^{2}+4 l-5=0$, which has two solutions $l=1$ and $l=-5$. Hence the last system of equations has the solutions $k=0, l=1$ and $k= \pm 2 \sqrt{6}$, $l=-5$. From (5) we now have $(0,1)$ and $\left( \pm \frac{2 \sqrt{6}}{5},-\frac{1}{5}\right)$ as the possible points of tangency on the circle.
|
{
"exam": "BalticWay",
"problem_label": "5",
"problem_match": "\n5.",
"resource_path": "BalticWay/segmented/en-bw99sol.jsonl",
"solution_match": "\n5.",
"tier": "T3",
"year": "1999"
}
|
What is the least number of moves it takes a knight to get from one corner of an $n \times n$ chessboard, where $n \geqslant 4$, to the diagonally opposite corner?
|
Answer: $2 \cdot\left\lfloor\frac{n+1}{3}\right\rfloor$.
Label the squares by pairs of integers $(x, y), x, y=1, \ldots, n$, and consider a sequence of moves that takes the knight from square $(1,1)$ to square $(n, n)$.
The total increment of $x+y$ is $2(n-1)$, and the maximal increment in each move is 3 . Furthermore, the parity of $x+y$ shifts in each move, and $1+1$ and $n+n$ are both even. Hence, the number of moves is even and larger than or equal to $\frac{2 \cdot(n-1)}{3}$. If $N=2 m$ is the least integer that satisfies these conditions, then $m$ is the least integer that satisfies $m \geqslant \frac{n-1}{3}$, i.e. $m=\left\lfloor\frac{n+1}{3}\right\rfloor$.
$n=4$
$n=5$
$n=6$
Figure 1
For $n=4, n=5$ and $n=6$ the sequences of moves are easily found that take the knight from square $(1,1)$ to square $(n, n)$ in 2,4 and 4 moves,
respectively (see Figure 1). In particular, the knight may get from square $(k, k)$ to square $(k+3, k+3)$ in 2 moves. Hence, by simple induction, for any $n$ the knight can get from square $(1,1)$ to square $(n, n)$ in a number of moves equal to twice the integer part of $\frac{n+1}{3}$, which is the minimal possible number of moves.
|
2 \cdot\left\lfloor\frac{n+1}{3}\right\rfloor
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
What is the least number of moves it takes a knight to get from one corner of an $n \times n$ chessboard, where $n \geqslant 4$, to the diagonally opposite corner?
|
Answer: $2 \cdot\left\lfloor\frac{n+1}{3}\right\rfloor$.
Label the squares by pairs of integers $(x, y), x, y=1, \ldots, n$, and consider a sequence of moves that takes the knight from square $(1,1)$ to square $(n, n)$.
The total increment of $x+y$ is $2(n-1)$, and the maximal increment in each move is 3 . Furthermore, the parity of $x+y$ shifts in each move, and $1+1$ and $n+n$ are both even. Hence, the number of moves is even and larger than or equal to $\frac{2 \cdot(n-1)}{3}$. If $N=2 m$ is the least integer that satisfies these conditions, then $m$ is the least integer that satisfies $m \geqslant \frac{n-1}{3}$, i.e. $m=\left\lfloor\frac{n+1}{3}\right\rfloor$.
$n=4$
$n=5$
$n=6$
Figure 1
For $n=4, n=5$ and $n=6$ the sequences of moves are easily found that take the knight from square $(1,1)$ to square $(n, n)$ in 2,4 and 4 moves,
respectively (see Figure 1). In particular, the knight may get from square $(k, k)$ to square $(k+3, k+3)$ in 2 moves. Hence, by simple induction, for any $n$ the knight can get from square $(1,1)$ to square $(n, n)$ in a number of moves equal to twice the integer part of $\frac{n+1}{3}$, which is the minimal possible number of moves.
|
{
"exam": "BalticWay",
"problem_label": "6",
"problem_match": "\n6.",
"resource_path": "BalticWay/segmented/en-bw99sol.jsonl",
"solution_match": "\n6.",
"tier": "T3",
"year": "1999"
}
|
Two squares on an $8 \times 8$ chessboard are called adjacent if they have a common edge or common corner. Is it possible for a king to begin in some
square and visit all squares exactly once in such a way that all moves except the first are made into squares adjacent to an even number of squares already visited?
|
Answer: No, it is not possible.
Consider the set $S$ of all (non-ordered) pairs of adjacent squares. Call an element of $S$ treated if the king has visited both its squares. After the first move there is one treated pair. Each subsequent move creates a further even number of treated pairs. So after each move the total number of treated pairs is odd. If the king could complete his tour then the total number of pairs of adjacent squares (i.e. the number of elements of $S$ ) would have to be odd. But the number of elements of $S$ is even as can be seen by the following argument. Rotation by 180 degrees around the centre of the board induces a bijection of $S$ onto itself. This bijection leaves precisely two pairs fixed, namely the pairs of squares sharing only a common corner at the middle of the board. It follows that the number of elements of $S$ is even.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Two squares on an $8 \times 8$ chessboard are called adjacent if they have a common edge or common corner. Is it possible for a king to begin in some
square and visit all squares exactly once in such a way that all moves except the first are made into squares adjacent to an even number of squares already visited?
|
Answer: No, it is not possible.
Consider the set $S$ of all (non-ordered) pairs of adjacent squares. Call an element of $S$ treated if the king has visited both its squares. After the first move there is one treated pair. Each subsequent move creates a further even number of treated pairs. So after each move the total number of treated pairs is odd. If the king could complete his tour then the total number of pairs of adjacent squares (i.e. the number of elements of $S$ ) would have to be odd. But the number of elements of $S$ is even as can be seen by the following argument. Rotation by 180 degrees around the centre of the board induces a bijection of $S$ onto itself. This bijection leaves precisely two pairs fixed, namely the pairs of squares sharing only a common corner at the middle of the board. It follows that the number of elements of $S$ is even.
|
{
"exam": "BalticWay",
"problem_label": "7",
"problem_match": "\n7.",
"resource_path": "BalticWay/segmented/en-bw99sol.jsonl",
"solution_match": "\n7.",
"tier": "T3",
"year": "1999"
}
|
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