problem
stringlengths 14
10.4k
| solution
stringlengths 1
24.1k
| answer
stringlengths 1
250
⌀ | problem_is_valid
stringclasses 4
values | solution_is_valid
stringclasses 3
values | question_type
stringclasses 4
values | problem_type
stringclasses 8
values | problem_raw
stringlengths 14
10.4k
| solution_raw
stringlengths 1
24.1k
| metadata
dict |
|---|---|---|---|---|---|---|---|---|---|
Let $f(n)=\sum_{k=2}^{\infty} \frac{1}{k^{n} \cdot k!}$. Calculate $\sum_{n=2}^{\infty} f(n)$.
|
$3-e$
$$
\begin{aligned}
\sum_{n=2}^{\infty} f(n) & =\sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{k^{n} \cdot k!} \\
& =\sum_{k=2}^{\infty} \frac{1}{k!} \sum_{n=2}^{\infty} \frac{1}{k^{n}} \\
& =\sum_{k=2}^{\infty} \frac{1}{k!} \cdot \frac{1}{k(k-1)} \\
& =\sum_{k=2}^{\infty} \frac{1}{(k-1)!} \cdot \frac{1}{k^{2}(k-1)}
\end{aligned}
$$
$$
\begin{aligned}
& =\sum_{k=2}^{\infty} \frac{1}{(k-1)!}\left(\frac{1}{k-1}-\frac{1}{k^{2}}-\frac{1}{k}\right) \\
& =\sum_{k=2}^{\infty}\left(\frac{1}{(k-1)(k-1)!}-\frac{1}{k \cdot k!}-\frac{1}{k!}\right) \\
& =\sum_{k=2}^{\infty}\left(\frac{1}{(k-1)(k-1)!}-\frac{1}{k \cdot k!}\right)-\sum_{k=2}^{\infty} \frac{1}{k!} \\
& =\frac{1}{1 \cdot 1!}-\left(e-\frac{1}{0!}-\frac{1}{1!}\right) \\
& =3-e
\end{aligned}
$$
|
3-e
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Let $f(n)=\sum_{k=2}^{\infty} \frac{1}{k^{n} \cdot k!}$. Calculate $\sum_{n=2}^{\infty} f(n)$.
|
$3-e$
$$
\begin{aligned}
\sum_{n=2}^{\infty} f(n) & =\sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{k^{n} \cdot k!} \\
& =\sum_{k=2}^{\infty} \frac{1}{k!} \sum_{n=2}^{\infty} \frac{1}{k^{n}} \\
& =\sum_{k=2}^{\infty} \frac{1}{k!} \cdot \frac{1}{k(k-1)} \\
& =\sum_{k=2}^{\infty} \frac{1}{(k-1)!} \cdot \frac{1}{k^{2}(k-1)}
\end{aligned}
$$
$$
\begin{aligned}
& =\sum_{k=2}^{\infty} \frac{1}{(k-1)!}\left(\frac{1}{k-1}-\frac{1}{k^{2}}-\frac{1}{k}\right) \\
& =\sum_{k=2}^{\infty}\left(\frac{1}{(k-1)(k-1)!}-\frac{1}{k \cdot k!}-\frac{1}{k!}\right) \\
& =\sum_{k=2}^{\infty}\left(\frac{1}{(k-1)(k-1)!}-\frac{1}{k \cdot k!}\right)-\sum_{k=2}^{\infty} \frac{1}{k!} \\
& =\frac{1}{1 \cdot 1!}-\left(e-\frac{1}{0!}-\frac{1}{1!}\right) \\
& =3-e
\end{aligned}
$$
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [6]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-calc-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $x(t)$ be a solution to the differential equation
$$
\left(x+x^{\prime}\right)^{2}+x \cdot x^{\prime \prime}=\cos t
$$
with $x(0)=x^{\prime}(0)=\sqrt{\frac{2}{5}}$. Compute $x\left(\frac{\pi}{4}\right)$.
|
$\frac{\sqrt[4]{450}}{5}$ Rewrite the equation as $x^{2}+2 x x^{\prime}+\left(x x^{\prime}\right)^{\prime}=\cos t$. Let $y=x^{2}$, so $y^{\prime}=2 x x^{\prime}$ and the equation becomes $y+y^{\prime}+\frac{1}{2} y^{\prime \prime}=\cos t$. The term $\cos t$ suggests that the particular solution should be in the form $A \sin t+B \cos t$. By substitution and coefficient comparison, we get $A=\frac{4}{5}$ and $B=\frac{2}{5}$. Since the function $y(t)=\frac{4}{5} \sin t+\frac{2}{5} \cos t$ already satisfies the initial conditions $y(0)=x(0)^{2}=\frac{2}{5}$ and $y^{\prime}(0)=2 x(0) x^{\prime}(0)=\frac{4}{5}$, the function $y$ also solves the initial value problem. Note that since $x$ is positive at $t=0$ and $y=x^{2}$ never reaches zero before $t$ reaches $\frac{\pi}{4}$, the value of $x\left(\frac{\pi}{4}\right)$ must be positive. Therefore, $x\left(\frac{\pi}{4}\right)=+\sqrt{y\left(\frac{\pi}{4}\right)}=\sqrt{\frac{6}{5} \cdot \frac{\sqrt{2}}{2}}=\frac{\sqrt[4]{450}}{5}$.
|
\frac{\sqrt[4]{450}}{5}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Let $x(t)$ be a solution to the differential equation
$$
\left(x+x^{\prime}\right)^{2}+x \cdot x^{\prime \prime}=\cos t
$$
with $x(0)=x^{\prime}(0)=\sqrt{\frac{2}{5}}$. Compute $x\left(\frac{\pi}{4}\right)$.
|
$\frac{\sqrt[4]{450}}{5}$ Rewrite the equation as $x^{2}+2 x x^{\prime}+\left(x x^{\prime}\right)^{\prime}=\cos t$. Let $y=x^{2}$, so $y^{\prime}=2 x x^{\prime}$ and the equation becomes $y+y^{\prime}+\frac{1}{2} y^{\prime \prime}=\cos t$. The term $\cos t$ suggests that the particular solution should be in the form $A \sin t+B \cos t$. By substitution and coefficient comparison, we get $A=\frac{4}{5}$ and $B=\frac{2}{5}$. Since the function $y(t)=\frac{4}{5} \sin t+\frac{2}{5} \cos t$ already satisfies the initial conditions $y(0)=x(0)^{2}=\frac{2}{5}$ and $y^{\prime}(0)=2 x(0) x^{\prime}(0)=\frac{4}{5}$, the function $y$ also solves the initial value problem. Note that since $x$ is positive at $t=0$ and $y=x^{2}$ never reaches zero before $t$ reaches $\frac{\pi}{4}$, the value of $x\left(\frac{\pi}{4}\right)$ must be positive. Therefore, $x\left(\frac{\pi}{4}\right)=+\sqrt{y\left(\frac{\pi}{4}\right)}=\sqrt{\frac{6}{5} \cdot \frac{\sqrt{2}}{2}}=\frac{\sqrt[4]{450}}{5}$.
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [7]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-calc-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $f(n)=\sum_{k=1}^{n} \frac{1}{k}$. Then there exists constants $\gamma, c$, and $d$ such that
$$
f(n)=\ln (n)+\gamma+\frac{c}{n}+\frac{d}{n^{2}}+O\left(\frac{1}{n^{3}}\right)
$$
where the $O\left(\frac{1}{n^{3}}\right)$ means terms of order $\frac{1}{n^{3}}$ or lower. Compute the ordered pair $(c, d)$.
|
$\left(\frac{1}{2},-\frac{1}{12}\right)$ From the given formula, we pull out the term $\frac{k}{n^{3}}$ from $O\left(\frac{1}{n^{4}}\right)$, making $f(n)=$ $\log (n)+\gamma+\frac{c}{n}+\frac{d}{n^{2}}+\frac{k}{n^{3}}+O\left(\frac{1}{n^{4}}\right)$. Therefore,
$f(n+1)-f(n)=\log \left(\frac{n+1}{n}\right)-c\left(\frac{1}{n}-\frac{1}{n+1}\right)-d\left(\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right)-k\left(\frac{1}{n^{3}}-\frac{1}{(n+1)^{3}}\right)+O\left(\frac{1}{n^{4}}\right)$.
For the left hand side, $f(n+1)-f(n)=\frac{1}{n+1}$. By substituting $x=\frac{1}{n}$, the formula above becomes
$$
\frac{x}{x+1}=\log (1+x)-c x^{2} \cdot \frac{1}{x+1}-d x^{3} \cdot \frac{x+2}{(x+1)^{2}}-k x^{4} \cdot \frac{x^{2}+3 x+3}{(x+1)^{3}}+O\left(x^{4}\right)
$$
Because $x$ is on the order of $\frac{1}{n}, \frac{1}{(x+1)^{3}}$ is on the order of a constant. Therefore, all the terms in the expansion of $k x^{4} \cdot \frac{x^{2}+3 x+3}{(x+1)^{3}}$ are of order $x^{4}$ or higher, so we can collapse it into $O\left(x^{4}\right)$. Using the Taylor expansions, we get
$$
x\left(1-x+x^{2}\right)+O\left(x^{4}\right)=\left(x-\frac{1}{2} x^{2}+\frac{1}{3} x^{3}\right)-c x^{2}(1-x)-d x^{3}(2)+O\left(x^{4}\right) .
$$
Coefficient comparison gives $c=\frac{1}{2}$ and $d=-\frac{1}{12}$.
|
\left(\frac{1}{2},-\frac{1}{12}\right)
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Let $f(n)=\sum_{k=1}^{n} \frac{1}{k}$. Then there exists constants $\gamma, c$, and $d$ such that
$$
f(n)=\ln (n)+\gamma+\frac{c}{n}+\frac{d}{n^{2}}+O\left(\frac{1}{n^{3}}\right)
$$
where the $O\left(\frac{1}{n^{3}}\right)$ means terms of order $\frac{1}{n^{3}}$ or lower. Compute the ordered pair $(c, d)$.
|
$\left(\frac{1}{2},-\frac{1}{12}\right)$ From the given formula, we pull out the term $\frac{k}{n^{3}}$ from $O\left(\frac{1}{n^{4}}\right)$, making $f(n)=$ $\log (n)+\gamma+\frac{c}{n}+\frac{d}{n^{2}}+\frac{k}{n^{3}}+O\left(\frac{1}{n^{4}}\right)$. Therefore,
$f(n+1)-f(n)=\log \left(\frac{n+1}{n}\right)-c\left(\frac{1}{n}-\frac{1}{n+1}\right)-d\left(\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right)-k\left(\frac{1}{n^{3}}-\frac{1}{(n+1)^{3}}\right)+O\left(\frac{1}{n^{4}}\right)$.
For the left hand side, $f(n+1)-f(n)=\frac{1}{n+1}$. By substituting $x=\frac{1}{n}$, the formula above becomes
$$
\frac{x}{x+1}=\log (1+x)-c x^{2} \cdot \frac{1}{x+1}-d x^{3} \cdot \frac{x+2}{(x+1)^{2}}-k x^{4} \cdot \frac{x^{2}+3 x+3}{(x+1)^{3}}+O\left(x^{4}\right)
$$
Because $x$ is on the order of $\frac{1}{n}, \frac{1}{(x+1)^{3}}$ is on the order of a constant. Therefore, all the terms in the expansion of $k x^{4} \cdot \frac{x^{2}+3 x+3}{(x+1)^{3}}$ are of order $x^{4}$ or higher, so we can collapse it into $O\left(x^{4}\right)$. Using the Taylor expansions, we get
$$
x\left(1-x+x^{2}\right)+O\left(x^{4}\right)=\left(x-\frac{1}{2} x^{2}+\frac{1}{3} x^{3}\right)-c x^{2}(1-x)-d x^{3}(2)+O\left(x^{4}\right) .
$$
Coefficient comparison gives $c=\frac{1}{2}$ and $d=-\frac{1}{12}$.
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. [8]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-calc-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $S=\{1,2,3,4,5,6,7,8,9,10\}$. How many (potentially empty) subsets $T$ of $S$ are there such that, for all $x$, if $x$ is in $T$ and $2 x$ is in $S$ then $2 x$ is also in $T$ ?
|
180 We partition the elements of $S$ into the following subsets: $\{1,2,4,8\},\{3,6\},\{5,10\}$, $\{7\},\{9\}$. Consider the first subset, $\{1,2,4,8\}$. Say 2 is an element of $T$. Because $2 \cdot 2=4$ is in $S, 4$ must also be in $T$. Furthermore, since $4 \cdot 2=8$ is in $S, 8$ must also be in $T$. So if $T$ contains 2 , it must also contain 4 and 8 . Similarly, if $T$ contains 1 , it must also contain 2,4 , and 8 . So $T$ can contain the following subsets of the subset $\{1,2,4,8\}$ : the empty set, $\{8\},\{4,8\},\{2,4,8\}$, or $\{1,2,4,8\}$. This gives 5 possibilities for the first subset. In general, we see that if $T$ contains an element $q$ of one of these subsets, it must also contain the elements in that subset that are larger than $q$, because we created the subsets for this to be true. So there are 3 possibilities for $\{3,6\}, 3$ for $\{5,10\}, 2$ for $\{7\}$, and 2 for $\{9\}$. This gives a total of $5 \cdot 3 \cdot 3 \cdot 2 \cdot 2=180$ possible subsets $T$.
|
180
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $S=\{1,2,3,4,5,6,7,8,9,10\}$. How many (potentially empty) subsets $T$ of $S$ are there such that, for all $x$, if $x$ is in $T$ and $2 x$ is in $S$ then $2 x$ is also in $T$ ?
|
180 We partition the elements of $S$ into the following subsets: $\{1,2,4,8\},\{3,6\},\{5,10\}$, $\{7\},\{9\}$. Consider the first subset, $\{1,2,4,8\}$. Say 2 is an element of $T$. Because $2 \cdot 2=4$ is in $S, 4$ must also be in $T$. Furthermore, since $4 \cdot 2=8$ is in $S, 8$ must also be in $T$. So if $T$ contains 2 , it must also contain 4 and 8 . Similarly, if $T$ contains 1 , it must also contain 2,4 , and 8 . So $T$ can contain the following subsets of the subset $\{1,2,4,8\}$ : the empty set, $\{8\},\{4,8\},\{2,4,8\}$, or $\{1,2,4,8\}$. This gives 5 possibilities for the first subset. In general, we see that if $T$ contains an element $q$ of one of these subsets, it must also contain the elements in that subset that are larger than $q$, because we created the subsets for this to be true. So there are 3 possibilities for $\{3,6\}, 3$ for $\{5,10\}, 2$ for $\{7\}$, and 2 for $\{9\}$. This gives a total of $5 \cdot 3 \cdot 3 \cdot 2 \cdot 2=180$ possible subsets $T$.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n1. [2]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
How many positive integers less than or equal to 240 can be expressed as a sum of distinct factorials? Consider 0! and 1! to be distinct.
|
39 Note that $1=0!, 2=0!+1!, 3=0!+2$ !, and $4=0!+1!+2$ !. These are the only numbers less than 6 that can be written as the sum of factorials. The only other factorials less than 240 are $3!=6,4!=24$, and $5!=120$. So a positive integer less than or equal to 240 can only contain 3 !, 4 !, 5 !, and/or one of $1,2,3$, or 4 in its sum. If it contains any factorial larger than 5 !, it will be larger than 240 . So a sum less than or equal to 240 will will either include 3 ! or not ( 2 ways), 4 ! or not ( 2 ways), 5 ! or not ( 2 ways), and add an additional $0,1,2,3$ or 4 ( 5 ways). This gives $2 \cdot 2 \cdot 2 \cdot 5=40$ integers less than 240 . However, we want only positive integers, so we must not count 0 . So there are 39 such positive integers.
|
39
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many positive integers less than or equal to 240 can be expressed as a sum of distinct factorials? Consider 0! and 1! to be distinct.
|
39 Note that $1=0!, 2=0!+1!, 3=0!+2$ !, and $4=0!+1!+2$ !. These are the only numbers less than 6 that can be written as the sum of factorials. The only other factorials less than 240 are $3!=6,4!=24$, and $5!=120$. So a positive integer less than or equal to 240 can only contain 3 !, 4 !, 5 !, and/or one of $1,2,3$, or 4 in its sum. If it contains any factorial larger than 5 !, it will be larger than 240 . So a sum less than or equal to 240 will will either include 3 ! or not ( 2 ways), 4 ! or not ( 2 ways), 5 ! or not ( 2 ways), and add an additional $0,1,2,3$ or 4 ( 5 ways). This gives $2 \cdot 2 \cdot 2 \cdot 5=40$ integers less than 240 . However, we want only positive integers, so we must not count 0 . So there are 39 such positive integers.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [3]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
How many ways are there to choose 2010 functions $f_{1}, \ldots, f_{2010}$ from $\{0,1\}$ to $\{0,1\}$ such that $f_{2010} \circ f_{2009} \circ \cdots \circ f_{1}$ is constant? Note: a function $g$ is constant if $g(a)=g(b)$ for all $a, b$ in the domain of $g$.
|
$4^{2010}-2^{2010}$ If all 2010 functions are bijective ${ }^{1}$, then the composition $f_{2010} \circ f_{2009} \circ \cdots \circ f_{1}$ will be bijective also, and therefore not constant. If, however, one of $f_{1}, \ldots, f_{2010}$ is not bijective, say $f_{k}$, then $f_{k}(0)=f_{k}(1)=q$, so $f_{2010} \circ f_{2009} \circ \cdots \circ f_{k+1} \circ f_{k} \circ \cdots f_{1}(0)=f_{2010} \circ f_{2009} \circ \cdots \circ f_{k+1}(q)=$ $f_{2010} \circ f_{2009} \circ \cdots \circ f_{k+1} \circ f_{k} \circ \cdots f_{1}(1)$. So the composition will be constant unless all $f_{i}$ are bijective. Since there are 4 possible functions ${ }^{2}$ from $\{0,1\}$ to $\{0,1\}$ and 2 of them are bijective, we subtract the cases where all the functions are bijective from the total to get $4^{2010}-2^{2010}$.
|
4^{2010}-2^{2010}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many ways are there to choose 2010 functions $f_{1}, \ldots, f_{2010}$ from $\{0,1\}$ to $\{0,1\}$ such that $f_{2010} \circ f_{2009} \circ \cdots \circ f_{1}$ is constant? Note: a function $g$ is constant if $g(a)=g(b)$ for all $a, b$ in the domain of $g$.
|
$4^{2010}-2^{2010}$ If all 2010 functions are bijective ${ }^{1}$, then the composition $f_{2010} \circ f_{2009} \circ \cdots \circ f_{1}$ will be bijective also, and therefore not constant. If, however, one of $f_{1}, \ldots, f_{2010}$ is not bijective, say $f_{k}$, then $f_{k}(0)=f_{k}(1)=q$, so $f_{2010} \circ f_{2009} \circ \cdots \circ f_{k+1} \circ f_{k} \circ \cdots f_{1}(0)=f_{2010} \circ f_{2009} \circ \cdots \circ f_{k+1}(q)=$ $f_{2010} \circ f_{2009} \circ \cdots \circ f_{k+1} \circ f_{k} \circ \cdots f_{1}(1)$. So the composition will be constant unless all $f_{i}$ are bijective. Since there are 4 possible functions ${ }^{2}$ from $\{0,1\}$ to $\{0,1\}$ and 2 of them are bijective, we subtract the cases where all the functions are bijective from the total to get $4^{2010}-2^{2010}$.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. [4]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Manya has a stack of $85=1+4+16+64$ blocks comprised of 4 layers (the $k$ th layer from the top has $4^{k-1}$ blocks; see the diagram below). Each block rests on 4 smaller blocks, each with dimensions half those of the larger block. Laura removes blocks one at a time from this stack, removing only blocks that currently have no blocks on top of them. Find the number of ways Laura can remove precisely 5 blocks from Manya's stack (the order in which they are removed matters).
|
3384 Each time Laura removes a block, 4 additional blocks are exposed, increasing the total number of exposed blocks by 3 . She removes 5 blocks, for a total of $1 \cdot 4 \cdot 7 \cdot 10 \cdot 13$ ways. However, the stack originally only has 4 layers, so we must subtract the cases where removing a block on the bottom layer does not expose any new blocks. There are $1 \cdot 4 \cdot 4 \cdot 4 \cdot 4=256$ of these (the last factor of 4 is from the 4 blocks that we counted as being exposed, but were not actually). So our final answer is $1 \cdot 4 \cdot 7 \cdot 10 \cdot 13-1 \cdot 4 \cdot 4 \cdot 4 \cdot 4=3384$.
|
3384
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Manya has a stack of $85=1+4+16+64$ blocks comprised of 4 layers (the $k$ th layer from the top has $4^{k-1}$ blocks; see the diagram below). Each block rests on 4 smaller blocks, each with dimensions half those of the larger block. Laura removes blocks one at a time from this stack, removing only blocks that currently have no blocks on top of them. Find the number of ways Laura can remove precisely 5 blocks from Manya's stack (the order in which they are removed matters).
|
3384 Each time Laura removes a block, 4 additional blocks are exposed, increasing the total number of exposed blocks by 3 . She removes 5 blocks, for a total of $1 \cdot 4 \cdot 7 \cdot 10 \cdot 13$ ways. However, the stack originally only has 4 layers, so we must subtract the cases where removing a block on the bottom layer does not expose any new blocks. There are $1 \cdot 4 \cdot 4 \cdot 4 \cdot 4=256$ of these (the last factor of 4 is from the 4 blocks that we counted as being exposed, but were not actually). So our final answer is $1 \cdot 4 \cdot 7 \cdot 10 \cdot 13-1 \cdot 4 \cdot 4 \cdot 4 \cdot 4=3384$.
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [4]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
John needs to pay 2010 dollars for his dinner. He has an unlimited supply of 2, 5, and 10 dollar notes. In how many ways can he pay?
|
20503
Let the number of 2,5 , and 10 dollar notes John can use be $x, y$, and $z$ respectively. We wish to find the number of nonnegative integer solutions to $2 x+5 y+10 z=2010$. Consider this equation $\bmod 2$. Because $2 x, 10 z$, and 2010 are even, $5 y$ must also be even, so $y$ must be even. Now consider the equation mod 5 . Because $5 y, 10 z$, and 2010 are divisible by $5,2 x$ must also be divisible by 5 , so $x$ must be divisible by 5 . So both $2 x$ and $5 y$ are divisible by 10 . So the equation is equivalent to $10 x^{\prime}+10 y^{\prime}+10 z=2010$, or $x^{\prime}+y^{\prime}+z=201$, with $x^{\prime}, y^{\prime}$, and $z$ nonnegative integers. There is a well-known bijection between solutions of this equation and picking 2 of 203 balls in a row on the table (explained in further detail below), so there are $\binom{203}{2}=20503$ ways.
The bijection between solutions of $x^{\prime}+y^{\prime}+z=201$ and arrangements of 203 balls in a row is as follows. Given a solution of the equation, we put $x^{\prime}$ white balls in a row, then a black ball, then $y^{\prime}$ white balls, then a black ball, then $z$ white balls. This is like having 203 balls in a row on a table and picking two of them to be black. To go from an arrangement of balls to a solution of the equation, we just read off $x^{\prime}, y^{\prime}$, and $z$ from the number of white balls in a row. There are $\binom{203}{2}$ ways to choose 2 of 203 balls to be black, so there are $\binom{203}{2}$ solutions to $x^{\prime}+y^{\prime}+z=201$.
|
20503
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
John needs to pay 2010 dollars for his dinner. He has an unlimited supply of 2, 5, and 10 dollar notes. In how many ways can he pay?
|
20503
Let the number of 2,5 , and 10 dollar notes John can use be $x, y$, and $z$ respectively. We wish to find the number of nonnegative integer solutions to $2 x+5 y+10 z=2010$. Consider this equation $\bmod 2$. Because $2 x, 10 z$, and 2010 are even, $5 y$ must also be even, so $y$ must be even. Now consider the equation mod 5 . Because $5 y, 10 z$, and 2010 are divisible by $5,2 x$ must also be divisible by 5 , so $x$ must be divisible by 5 . So both $2 x$ and $5 y$ are divisible by 10 . So the equation is equivalent to $10 x^{\prime}+10 y^{\prime}+10 z=2010$, or $x^{\prime}+y^{\prime}+z=201$, with $x^{\prime}, y^{\prime}$, and $z$ nonnegative integers. There is a well-known bijection between solutions of this equation and picking 2 of 203 balls in a row on the table (explained in further detail below), so there are $\binom{203}{2}=20503$ ways.
The bijection between solutions of $x^{\prime}+y^{\prime}+z=201$ and arrangements of 203 balls in a row is as follows. Given a solution of the equation, we put $x^{\prime}$ white balls in a row, then a black ball, then $y^{\prime}$ white balls, then a black ball, then $z$ white balls. This is like having 203 balls in a row on a table and picking two of them to be black. To go from an arrangement of balls to a solution of the equation, we just read off $x^{\prime}, y^{\prime}$, and $z$ from the number of white balls in a row. There are $\binom{203}{2}$ ways to choose 2 of 203 balls to be black, so there are $\binom{203}{2}$ solutions to $x^{\prime}+y^{\prime}+z=201$.
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n5. [5]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
An ant starts out at $(0,0)$. Each second, if it is currently at the square $(x, y)$, it can move to $(x-1, y-1),(x-1, y+1),(x+1, y-1)$, or $(x+1, y+1)$. In how many ways can it end up at $(2010,2010)$ after 4020 seconds?
|
$\binom{4020}{1005}^{2}$ Note that each of the coordinates either increases or decreases the $x$ and $y$ coordinates by 1. In order to reach 2010 after 4020 steps, each of the coordinates must be increased 3015 times and decreased 1005 times. A permutation of 3015 plusses and 1005 minuses for each of $x$ and $y$ uniquely corresponds to a path the ant could take to $(2010,2010)$, because we can take ordered pairs from the two lists and match them up to a valid step the ant can take. So the number of ways the ant can end up at $(2010,2010)$ after 4020 seconds is equal to the number of ways to arrange plusses and minuses for both $x$ and $y$, or $\left(\binom{4020}{1005}\right)^{2}$.
|
\binom{4020}{1005}^{2}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
An ant starts out at $(0,0)$. Each second, if it is currently at the square $(x, y)$, it can move to $(x-1, y-1),(x-1, y+1),(x+1, y-1)$, or $(x+1, y+1)$. In how many ways can it end up at $(2010,2010)$ after 4020 seconds?
|
$\binom{4020}{1005}^{2}$ Note that each of the coordinates either increases or decreases the $x$ and $y$ coordinates by 1. In order to reach 2010 after 4020 steps, each of the coordinates must be increased 3015 times and decreased 1005 times. A permutation of 3015 plusses and 1005 minuses for each of $x$ and $y$ uniquely corresponds to a path the ant could take to $(2010,2010)$, because we can take ordered pairs from the two lists and match them up to a valid step the ant can take. So the number of ways the ant can end up at $(2010,2010)$ after 4020 seconds is equal to the number of ways to arrange plusses and minuses for both $x$ and $y$, or $\left(\binom{4020}{1005}\right)^{2}$.
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n6. [5]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
For each integer $x$ with $1 \leq x \leq 10$, a point is randomly placed at either $(x, 1)$ or $(x,-1)$ with equal probability. What is the expected area of the convex hull of these points? Note: the convex hull of a finite set is the smallest convex polygon containing it.
|
$\frac{1793}{\frac{1728}{128}}$ Let $n=10$. Given a random variable $X$, let $\mathbb{E}(X)$ denote its expected value. If all points are collinear, then the convex hull has area zero. This happens with probability $\frac{2}{2^{n}}$ (either all points are at $y=1$ or all points are at $y=-1$ ). Otherwise, the points form a trapezoid with height 2 (the trapezoid is possibly degenerate, but this won't matter for our calculation). Let $x_{1, l}$ be the $x$-coordinate of the left-most point at $y=1$ and $x_{1, r}$ be the $x$-coordinate of the right-most point at $y=1$. Define $x_{-1, l}$ and $x_{-1, r}$ similarly for $y=-1$. Then the area of the trapezoid is
$$
2 \cdot \frac{\left(x_{1, r}-x_{1, l}\right)+\left(x_{-1, r}-x_{-1, l}\right)}{2}=x_{1, r}+x_{-1, r}-x_{1, l}-x_{-1, l} .
$$
The expected area of the convex hull (assuming the points are not all collinear) is then, by linearity of expectation,
$$
\mathbb{E}\left(x_{1, r}+x_{-1, r}-x_{1, l}-x_{-1, l}\right)=\mathbb{E}\left(x_{1, r}\right)+\mathbb{E}\left(x_{-1, r}\right)-\mathbb{E}\left(x_{1, l}\right)-\mathbb{E}\left(x_{-1, l}\right) .
$$
We need only compute the expected values given in the above equation. Note that $x_{1, r}$ is equal to $k$ with probability $\frac{2^{k-1}}{2^{n}-2}$, except that it is equal to $n$ with probability $\frac{2^{n-1}-1}{2^{n}-2}$ (the denominator is $2^{n}-2$ instead of $2^{n}$ because we need to exclude the case where all points are collinear). Therefore, the expected value of $x_{1, r}$ is equal to
$$
\begin{aligned}
& \frac{1}{2^{n}-2}\left(\left(\sum_{k=1}^{n} k \cdot 2^{k-1}\right)-n \cdot 1\right) \\
& \quad=\frac{1}{2^{n}-2}\left(\left(1+2+\cdots+2^{n-1}\right)+\left(2+4+\cdots+2^{n-1}\right)+\cdots+2^{n-1}-n\right) \\
& =\frac{1}{2^{n}-2}\left(\left(2^{n}-1\right)+\left(2^{n}-2\right)+\cdots+\left(2^{n}-2^{n-1}\right)-n\right) \\
& =\frac{1}{2^{n}-2}\left(n \cdot 2^{n}-\left(2^{n}-1\right)-n\right) \\
& \quad=(n-1) \frac{2^{n}-1}{2^{n}-2}
\end{aligned}
$$
Similarly, the expected value of $x_{-1, r}$ is also $(n-1) \frac{2^{n}-1}{2^{n}-2}$. By symmetry, the expected value of both $x_{1, l}$ and $x_{-1, l}$ is $n+1-(n-1) \frac{\frac{2}{n}^{n}-1}{2^{n}-2}$. This says that if the points are not all collinear then the expected area is $2 \cdot\left((n-1) \frac{2^{n}-1}{2^{n-1}-1}-(n+1)\right)$. So, the expected area is
$$
\begin{aligned}
\frac{2}{2^{n}} \cdot 0+(1- & \left.\frac{2}{2^{n}}\right) \cdot 2 \cdot\left((n-1) \frac{2^{n}-1}{2^{n-1}-1}-(n+1)\right) \\
& =2 \cdot \frac{2^{n-1}-1}{2^{n-1}} \cdot\left((n-1) \frac{2^{n}-1}{2^{n-1}-1}-(n+1)\right) \\
& =2 \cdot \frac{(n-1)\left(2^{n}-1\right)-(n+1)\left(2^{n-1}-1\right)}{2^{n-1}} \\
& =2 \cdot \frac{((2 n-2)-(n+1)) 2^{n-1}+2}{2^{n-1}} \\
& =2 n-6+\frac{1}{2^{n-3}}
\end{aligned}
$$
Plugging in $n=10$, we get $14+\frac{1}{128}=\frac{1793}{128}$.
|
\frac{1793}{128}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
For each integer $x$ with $1 \leq x \leq 10$, a point is randomly placed at either $(x, 1)$ or $(x,-1)$ with equal probability. What is the expected area of the convex hull of these points? Note: the convex hull of a finite set is the smallest convex polygon containing it.
|
$\frac{1793}{\frac{1728}{128}}$ Let $n=10$. Given a random variable $X$, let $\mathbb{E}(X)$ denote its expected value. If all points are collinear, then the convex hull has area zero. This happens with probability $\frac{2}{2^{n}}$ (either all points are at $y=1$ or all points are at $y=-1$ ). Otherwise, the points form a trapezoid with height 2 (the trapezoid is possibly degenerate, but this won't matter for our calculation). Let $x_{1, l}$ be the $x$-coordinate of the left-most point at $y=1$ and $x_{1, r}$ be the $x$-coordinate of the right-most point at $y=1$. Define $x_{-1, l}$ and $x_{-1, r}$ similarly for $y=-1$. Then the area of the trapezoid is
$$
2 \cdot \frac{\left(x_{1, r}-x_{1, l}\right)+\left(x_{-1, r}-x_{-1, l}\right)}{2}=x_{1, r}+x_{-1, r}-x_{1, l}-x_{-1, l} .
$$
The expected area of the convex hull (assuming the points are not all collinear) is then, by linearity of expectation,
$$
\mathbb{E}\left(x_{1, r}+x_{-1, r}-x_{1, l}-x_{-1, l}\right)=\mathbb{E}\left(x_{1, r}\right)+\mathbb{E}\left(x_{-1, r}\right)-\mathbb{E}\left(x_{1, l}\right)-\mathbb{E}\left(x_{-1, l}\right) .
$$
We need only compute the expected values given in the above equation. Note that $x_{1, r}$ is equal to $k$ with probability $\frac{2^{k-1}}{2^{n}-2}$, except that it is equal to $n$ with probability $\frac{2^{n-1}-1}{2^{n}-2}$ (the denominator is $2^{n}-2$ instead of $2^{n}$ because we need to exclude the case where all points are collinear). Therefore, the expected value of $x_{1, r}$ is equal to
$$
\begin{aligned}
& \frac{1}{2^{n}-2}\left(\left(\sum_{k=1}^{n} k \cdot 2^{k-1}\right)-n \cdot 1\right) \\
& \quad=\frac{1}{2^{n}-2}\left(\left(1+2+\cdots+2^{n-1}\right)+\left(2+4+\cdots+2^{n-1}\right)+\cdots+2^{n-1}-n\right) \\
& =\frac{1}{2^{n}-2}\left(\left(2^{n}-1\right)+\left(2^{n}-2\right)+\cdots+\left(2^{n}-2^{n-1}\right)-n\right) \\
& =\frac{1}{2^{n}-2}\left(n \cdot 2^{n}-\left(2^{n}-1\right)-n\right) \\
& \quad=(n-1) \frac{2^{n}-1}{2^{n}-2}
\end{aligned}
$$
Similarly, the expected value of $x_{-1, r}$ is also $(n-1) \frac{2^{n}-1}{2^{n}-2}$. By symmetry, the expected value of both $x_{1, l}$ and $x_{-1, l}$ is $n+1-(n-1) \frac{\frac{2}{n}^{n}-1}{2^{n}-2}$. This says that if the points are not all collinear then the expected area is $2 \cdot\left((n-1) \frac{2^{n}-1}{2^{n-1}-1}-(n+1)\right)$. So, the expected area is
$$
\begin{aligned}
\frac{2}{2^{n}} \cdot 0+(1- & \left.\frac{2}{2^{n}}\right) \cdot 2 \cdot\left((n-1) \frac{2^{n}-1}{2^{n-1}-1}-(n+1)\right) \\
& =2 \cdot \frac{2^{n-1}-1}{2^{n-1}} \cdot\left((n-1) \frac{2^{n}-1}{2^{n-1}-1}-(n+1)\right) \\
& =2 \cdot \frac{(n-1)\left(2^{n}-1\right)-(n+1)\left(2^{n-1}-1\right)}{2^{n-1}} \\
& =2 \cdot \frac{((2 n-2)-(n+1)) 2^{n-1}+2}{2^{n-1}} \\
& =2 n-6+\frac{1}{2^{n-3}}
\end{aligned}
$$
Plugging in $n=10$, we get $14+\frac{1}{128}=\frac{1793}{128}$.
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [6]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
How many functions $f$ from $\{-1005, \ldots, 1005\}$ to $\{-2010, \ldots, 2010\}$ are there such that the following two conditions are satisfied?
- If $a<b$ then $f(a)<f(b)$.
- There is no $n$ in $\{-1005, \ldots, 1005\}$ such that $|f(n)|=|n|$.
|
$\cdots$ Note: the intended answer was $\binom{4019}{2011}$, but the original answer was incorrect. The correct answer is:
1173346782666677300072441773814388000553179587006710786401225043842699552460942166630860 5302966355504513409792805200762540756742811158611534813828022157596601875355477425764387 2333935841666957750009216404095352456877594554817419353494267665830087436353494075828446 0070506487793628698617665091500712606599653369601270652785265395252421526230453391663029 1476263072382369363170971857101590310272130771639046414860423440232291348986940615141526 0247281998288175423628757177754777309519630334406956881890655029018130367627043067425502 2334151384481231298380228052789795136259575164777156839054346649261636296328387580363485 2904329986459861362633348204891967272842242778625137520975558407856496002297523759366027
1506637984075036473724713869804364399766664507880042495122618597629613572449327653716600 6715747717529280910646607622693561789482959920478796128008380531607300324374576791477561 5881495035032334387221203759898494171708240222856256961757026746724252966598328065735933 6668742613422094179386207330487537984173936781232801614775355365060827617078032786368164 8860839124954588222610166915992867657815394480973063139752195206598739798365623873142903 28539769699667459275254643229234106717245366005816917271187760792
This obviously cannot be computed by hand, but there is a polynomial-time dynamic programming algorithm that will compute it.
|
not found
|
Yes
|
Problem not solved
|
math-word-problem
|
Combinatorics
|
How many functions $f$ from $\{-1005, \ldots, 1005\}$ to $\{-2010, \ldots, 2010\}$ are there such that the following two conditions are satisfied?
- If $a<b$ then $f(a)<f(b)$.
- There is no $n$ in $\{-1005, \ldots, 1005\}$ such that $|f(n)|=|n|$.
|
$\cdots$ Note: the intended answer was $\binom{4019}{2011}$, but the original answer was incorrect. The correct answer is:
1173346782666677300072441773814388000553179587006710786401225043842699552460942166630860 5302966355504513409792805200762540756742811158611534813828022157596601875355477425764387 2333935841666957750009216404095352456877594554817419353494267665830087436353494075828446 0070506487793628698617665091500712606599653369601270652785265395252421526230453391663029 1476263072382369363170971857101590310272130771639046414860423440232291348986940615141526 0247281998288175423628757177754777309519630334406956881890655029018130367627043067425502 2334151384481231298380228052789795136259575164777156839054346649261636296328387580363485 2904329986459861362633348204891967272842242778625137520975558407856496002297523759366027
1506637984075036473724713869804364399766664507880042495122618597629613572449327653716600 6715747717529280910646607622693561789482959920478796128008380531607300324374576791477561 5881495035032334387221203759898494171708240222856256961757026746724252966598328065735933 6668742613422094179386207330487537984173936781232801614775355365060827617078032786368164 8860839124954588222610166915992867657815394480973063139752195206598739798365623873142903 28539769699667459275254643229234106717245366005816917271187760792
This obviously cannot be computed by hand, but there is a polynomial-time dynamic programming algorithm that will compute it.
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [6]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Rosencrantz and Guildenstern are playing a game where they repeatedly flip coins. Rosencrantz wins if 1 heads followed by 2009 tails appears. Guildenstern wins if 2010 heads come in a row. They will flip coins until someone wins. What is the probability that Rosencrantz wins?
|
$\frac{2^{2009}-1}{3 \cdot 2^{2008}-1}$ We can assume the first throw is heads (because neither player can win starting from a string of only tails). Let $x$ be the probability that Rosencrantz wins. Let $y$ be the probability that Rosencrantz wins after HT.
Whenever there is a string of less than 2009 tails followed by a heads, the heads basically means the two are starting from the beginning, where Rosencrantz has probability $x$ of winning.
We also know that $x=y\left(1-\frac{1}{2^{2009}}\right)$. This is because from the initial heads there is a $\left(1-\frac{1}{2^{2009}}\right)$ chance Rosencrantz doesn't lose, and in this case the last two flips are HT, in which case Rosencrantz has probability $y$ of winning.
If the first two throws are HT, there is a $\frac{1}{2^{2008}}$ chance Rosencrantz wins; otherwise, there is eventually a heads, and so we are back in the case of starting from a heads, which corresponds to $x$. Therefore, $y=\frac{1}{2^{2008}}+x\left(1-\frac{1}{2^{2008}}\right)$. Putting this together with the previous equation, we get:
$$
\begin{aligned}
x & =\left(\frac{1}{2^{2008}}+x\left(1-\frac{1}{2^{2008}}\right)\right)\left(1-\frac{1}{2^{2009}}\right) \\
\Longrightarrow \quad x & =\left(\frac{1+2^{2008} x-x}{2^{2008}}\right)\left(\frac{2^{2009}-1}{\left.2^{2009}\right)}\right. \\
\Longrightarrow \quad 2^{4017} x & =x\left(2^{4017}-2^{2009}-2^{2008}+1\right)+2^{2009}-1 \\
\Longrightarrow \quad x & =\frac{2^{2009}-1}{2^{2009}+2^{2008}-1},
\end{aligned}
$$
so the answer is $\frac{2^{2009}-1}{2^{2009}+2^{2008}-1}=\frac{2^{2009}-1}{3 \cdot 2^{2008}-1}$.
|
\frac{2^{2009}-1}{3 \cdot 2^{2008}-1}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Rosencrantz and Guildenstern are playing a game where they repeatedly flip coins. Rosencrantz wins if 1 heads followed by 2009 tails appears. Guildenstern wins if 2010 heads come in a row. They will flip coins until someone wins. What is the probability that Rosencrantz wins?
|
$\frac{2^{2009}-1}{3 \cdot 2^{2008}-1}$ We can assume the first throw is heads (because neither player can win starting from a string of only tails). Let $x$ be the probability that Rosencrantz wins. Let $y$ be the probability that Rosencrantz wins after HT.
Whenever there is a string of less than 2009 tails followed by a heads, the heads basically means the two are starting from the beginning, where Rosencrantz has probability $x$ of winning.
We also know that $x=y\left(1-\frac{1}{2^{2009}}\right)$. This is because from the initial heads there is a $\left(1-\frac{1}{2^{2009}}\right)$ chance Rosencrantz doesn't lose, and in this case the last two flips are HT, in which case Rosencrantz has probability $y$ of winning.
If the first two throws are HT, there is a $\frac{1}{2^{2008}}$ chance Rosencrantz wins; otherwise, there is eventually a heads, and so we are back in the case of starting from a heads, which corresponds to $x$. Therefore, $y=\frac{1}{2^{2008}}+x\left(1-\frac{1}{2^{2008}}\right)$. Putting this together with the previous equation, we get:
$$
\begin{aligned}
x & =\left(\frac{1}{2^{2008}}+x\left(1-\frac{1}{2^{2008}}\right)\right)\left(1-\frac{1}{2^{2009}}\right) \\
\Longrightarrow \quad x & =\left(\frac{1+2^{2008} x-x}{2^{2008}}\right)\left(\frac{2^{2009}-1}{\left.2^{2009}\right)}\right. \\
\Longrightarrow \quad 2^{4017} x & =x\left(2^{4017}-2^{2009}-2^{2008}+1\right)+2^{2009}-1 \\
\Longrightarrow \quad x & =\frac{2^{2009}-1}{2^{2009}+2^{2008}-1},
\end{aligned}
$$
so the answer is $\frac{2^{2009}-1}{2^{2009}+2^{2008}-1}=\frac{2^{2009}-1}{3 \cdot 2^{2008}-1}$.
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [7]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
In a $16 \times 16$ table of integers, each row and column contains at most 4 distinct integers. What is the maximum number of distinct integers that there can be in the whole table?
|
49 First, we show that 50 is too big. Assume for sake of contradiction that a labeling with at least 50 distinct integers exists. By the Pigeonhole Principle, there must be at least one row, say the first row, with at least 4 distinct integers in it; in this case, that is exactly 4 , since that is the maximum number of distinct integers in one row. Then, in the remaining 15 rows there must be at least 46 distinct integers (these 46 will also be distinct from the 4 in the first row). Using Pigeonhole again, there will be another row, say the second row, with 4 distinct integers in it. Call the set of integers in the first and second rows $S$. Because the 4 distinct integers in the second row are distinct from the 4 in the first row, there are 8 distinct values in the first two rows, so $|S|=8$. Now consider the subcolumns containing the cells in rows 3 to 16 . In each subcolumn, there are at most 2 values not in $S$, because there are already two distinct values in that column from the cells in the first two rows. So, the maximum number of distinct values in the table is $16 \cdot 2+8=40$, a contradiction. So a valid labeling must have fewer than 50 distinct integers. Below, we show by example that 49 is attainable.
| 1 | 17 | 33 | - | - | - | - | - | - | - | - | - | - | - | - | - |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| - | 2 | 18 | 34 | - | - | - | - | - | - | - | - | - | - | - | - |
| - | - | 3 | 19 | 35 | - | - | - | - | - | - | - | - | - | - | - |
| - | - | - | 4 | 20 | 36 | - | - | - | - | - | - | - | - | - | - |
| - | - | - | - | 5 | 21 | 37 | - | - | - | - | - | - | - | - | - |
| - | - | - | - | - | 6 | 22 | 38 | - | - | - | - | - | - | - | - |
| - | - | - | - | - | - | 7 | 23 | 39 | - | - | - | - | - | - | - |
| - | - | - | - | - | - | - | 8 | 24 | 40 | - | - | - | - | - | - |
| - | - | - | - | - | - | - | - | 9 | 25 | 41 | - | - | - | - | - |
| - | - | - | - | - | - | - | - | - | 10 | 26 | 42 | - | - | - | - |
| - | - | - | - | - | - | - | - | - | - | 11 | 27 | 43 | - | - | - |
| - | - | - | - | - | - | - | - | - | - | - | 12 | 28 | 44 | - | - |
| - | - | - | - | - | - | - | - | - | - | - | - | 13 | 29 | 45 | - |
| - | - | - | - | - | - | - | - | - | - | - | - | - | 14 | 30 | 46 |
| 47 | - | - | - | - | - | - | - | - | - | - | - | - | - | 15 | 31 |
| 32 | 48 | - | - | - | - | - | - | - | - | - | - | - | - | - | 16 |
Cells that do not contain a number are colored with color 49 .
|
49
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In a $16 \times 16$ table of integers, each row and column contains at most 4 distinct integers. What is the maximum number of distinct integers that there can be in the whole table?
|
49 First, we show that 50 is too big. Assume for sake of contradiction that a labeling with at least 50 distinct integers exists. By the Pigeonhole Principle, there must be at least one row, say the first row, with at least 4 distinct integers in it; in this case, that is exactly 4 , since that is the maximum number of distinct integers in one row. Then, in the remaining 15 rows there must be at least 46 distinct integers (these 46 will also be distinct from the 4 in the first row). Using Pigeonhole again, there will be another row, say the second row, with 4 distinct integers in it. Call the set of integers in the first and second rows $S$. Because the 4 distinct integers in the second row are distinct from the 4 in the first row, there are 8 distinct values in the first two rows, so $|S|=8$. Now consider the subcolumns containing the cells in rows 3 to 16 . In each subcolumn, there are at most 2 values not in $S$, because there are already two distinct values in that column from the cells in the first two rows. So, the maximum number of distinct values in the table is $16 \cdot 2+8=40$, a contradiction. So a valid labeling must have fewer than 50 distinct integers. Below, we show by example that 49 is attainable.
| 1 | 17 | 33 | - | - | - | - | - | - | - | - | - | - | - | - | - |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| - | 2 | 18 | 34 | - | - | - | - | - | - | - | - | - | - | - | - |
| - | - | 3 | 19 | 35 | - | - | - | - | - | - | - | - | - | - | - |
| - | - | - | 4 | 20 | 36 | - | - | - | - | - | - | - | - | - | - |
| - | - | - | - | 5 | 21 | 37 | - | - | - | - | - | - | - | - | - |
| - | - | - | - | - | 6 | 22 | 38 | - | - | - | - | - | - | - | - |
| - | - | - | - | - | - | 7 | 23 | 39 | - | - | - | - | - | - | - |
| - | - | - | - | - | - | - | 8 | 24 | 40 | - | - | - | - | - | - |
| - | - | - | - | - | - | - | - | 9 | 25 | 41 | - | - | - | - | - |
| - | - | - | - | - | - | - | - | - | 10 | 26 | 42 | - | - | - | - |
| - | - | - | - | - | - | - | - | - | - | 11 | 27 | 43 | - | - | - |
| - | - | - | - | - | - | - | - | - | - | - | 12 | 28 | 44 | - | - |
| - | - | - | - | - | - | - | - | - | - | - | - | 13 | 29 | 45 | - |
| - | - | - | - | - | - | - | - | - | - | - | - | - | 14 | 30 | 46 |
| 47 | - | - | - | - | - | - | - | - | - | - | - | - | - | 15 | 31 |
| 32 | 48 | - | - | - | - | - | - | - | - | - | - | - | - | - | 16 |
Cells that do not contain a number are colored with color 49 .
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. [8]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-comb-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Suppose that $x$ and $y$ are positive reals such that
$$
x-y^{2}=3, \quad x^{2}+y^{4}=13
$$
Find $x$.
|
$\frac{3+\sqrt{17}}{2}$ Squaring both sides of $x-y^{2}=3$ gives $x^{2}+y^{4}-2 x y^{2}=9$. Subtract this equation from twice the second given to get $x^{2}+2 x y^{2}+y^{4}=17 \Longrightarrow x+y^{2}= \pm 17$. Combining this equation with the first given, we see that $x=\frac{3 \pm \sqrt{17}}{2}$. Since $x$ is a positive real, $x$ must be $\frac{3+\sqrt{17}}{2}$.
|
\frac{3+\sqrt{17}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose that $x$ and $y$ are positive reals such that
$$
x-y^{2}=3, \quad x^{2}+y^{4}=13
$$
Find $x$.
|
$\frac{3+\sqrt{17}}{2}$ Squaring both sides of $x-y^{2}=3$ gives $x^{2}+y^{4}-2 x y^{2}=9$. Subtract this equation from twice the second given to get $x^{2}+2 x y^{2}+y^{4}=17 \Longrightarrow x+y^{2}= \pm 17$. Combining this equation with the first given, we see that $x=\frac{3 \pm \sqrt{17}}{2}$. Since $x$ is a positive real, $x$ must be $\frac{3+\sqrt{17}}{2}$.
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [3]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $S=\{1,2,3,4,5,6,7,8,9,10\}$. How many (potentially empty) subsets $T$ of $S$ are there such that, for all $x$, if $x$ is in $T$ and $2 x$ is in $S$ then $2 x$ is also in $T$ ?
|
180 We partition the elements of $S$ into the following subsets: $\{1,2,4,8\},\{3,6\},\{5,10\}$, $\{7\},\{9\}$. Consider the first subset, $\{1,2,4,8\}$. Say 2 is an element of $T$. Because $2 \cdot 2=4$ is in $S, 4$ must also be in $T$. Furthermore, since $4 \cdot 2=8$ is in $S, 8$ must also be in $T$. So if $T$ contains 2 , it must also contain 4 and 8 . Similarly, if $T$ contains 1 , it must also contain 2,4 , and 8 . So $T$ can contain the following subsets of the subset $\{1,2,4,8\}$ : the empty set, $\{8\},\{4,8\},\{2,4,8\}$, or $\{1,2,4,8\}$. This gives 5 possibilities for the first subset. In general, we see that if $T$ contains an element $q$ of one of these subsets, it must also contain the elements in that subset that are larger than $q$, because we created the subsets for this to be true. So there are 3 possibilities for $\{3,6\}, 3$ for $\{5,10\}, 2$ for $\{7\}$, and 2 for $\{9\}$. This gives a total of $5 \cdot 3 \cdot 3 \cdot 2 \cdot 2=180$ possible subsets $T$.
|
180
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $S=\{1,2,3,4,5,6,7,8,9,10\}$. How many (potentially empty) subsets $T$ of $S$ are there such that, for all $x$, if $x$ is in $T$ and $2 x$ is in $S$ then $2 x$ is also in $T$ ?
|
180 We partition the elements of $S$ into the following subsets: $\{1,2,4,8\},\{3,6\},\{5,10\}$, $\{7\},\{9\}$. Consider the first subset, $\{1,2,4,8\}$. Say 2 is an element of $T$. Because $2 \cdot 2=4$ is in $S, 4$ must also be in $T$. Furthermore, since $4 \cdot 2=8$ is in $S, 8$ must also be in $T$. So if $T$ contains 2 , it must also contain 4 and 8 . Similarly, if $T$ contains 1 , it must also contain 2,4 , and 8 . So $T$ can contain the following subsets of the subset $\{1,2,4,8\}$ : the empty set, $\{8\},\{4,8\},\{2,4,8\}$, or $\{1,2,4,8\}$. This gives 5 possibilities for the first subset. In general, we see that if $T$ contains an element $q$ of one of these subsets, it must also contain the elements in that subset that are larger than $q$, because we created the subsets for this to be true. So there are 3 possibilities for $\{3,6\}, 3$ for $\{5,10\}, 2$ for $\{7\}$, and 2 for $\{9\}$. This gives a total of $5 \cdot 3 \cdot 3 \cdot 2 \cdot 2=180$ possible subsets $T$.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [3]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convex pentagon that has an area of $\frac{7}{10}$ of the area of the original rectangle. Find the ratio of the longer side of the rectangle to the shorter side of the rectangle.
|
$\sqrt{5}$
Given a polygon $P_{1} P_{2} \cdots P_{k}$, let $\left[P_{1} P_{2} \cdots P_{k}\right]$ denote its area. Let $A B C D$ be the rectangle. Suppose we fold $B$ across $\overline{A C}$, and let $E$ be the intersection of $\overline{A D}$ and $\overline{B^{\prime} C}$. Then we end up with the pentagon $A C D E B^{\prime}$, depicted above. Let's suppose, without loss of generality, that $A B C D$ has area 1. Then $\triangle A E C$ must have area $\frac{3}{10}$, since
$$
\begin{aligned}
{[A B C D] } & =[A B C]+[A C D] \\
& =\left[A B^{\prime} C\right]+[A C D] \\
& =\left[A B^{\prime} E\right]+2[A E C]+[E D C] \\
& =\left[A C D E B^{\prime}\right]+[A E C] \\
& =\frac{7}{10}[A B C D]+[A E C],
\end{aligned}
$$
That is, $[A E C]=\frac{3}{10}[A B C D]=\frac{3}{10}$.
Since $\triangle E C D$ is congruent to $\triangle E A B^{\prime}$, both triangles have area $\frac{1}{5}$. Note that $\triangle A B^{\prime} C, \triangle A B C$, and $\triangle C D A$ are all congruent, and all have area $\frac{1}{2}$. Since $\triangle A E C$ and $\triangle E D C$ share altitude $\overline{D C}$, $\frac{D E}{E A}=\frac{[D E C]}{[A E C]}=\frac{2}{3}$. Because $\triangle C A E$ is isosceles, $C E=E A$. Let $A E=3 x$. The $C E=3 x, D E=2 x$, and $C D=x \sqrt{9-4}=x \sqrt{5}$. Then $\frac{A D}{D C}=\frac{A E+E D}{D C}=\frac{3+2}{\sqrt{5}}=\sqrt{5}$.
|
\sqrt{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convex pentagon that has an area of $\frac{7}{10}$ of the area of the original rectangle. Find the ratio of the longer side of the rectangle to the shorter side of the rectangle.
|
$\sqrt{5}$
Given a polygon $P_{1} P_{2} \cdots P_{k}$, let $\left[P_{1} P_{2} \cdots P_{k}\right]$ denote its area. Let $A B C D$ be the rectangle. Suppose we fold $B$ across $\overline{A C}$, and let $E$ be the intersection of $\overline{A D}$ and $\overline{B^{\prime} C}$. Then we end up with the pentagon $A C D E B^{\prime}$, depicted above. Let's suppose, without loss of generality, that $A B C D$ has area 1. Then $\triangle A E C$ must have area $\frac{3}{10}$, since
$$
\begin{aligned}
{[A B C D] } & =[A B C]+[A C D] \\
& =\left[A B^{\prime} C\right]+[A C D] \\
& =\left[A B^{\prime} E\right]+2[A E C]+[E D C] \\
& =\left[A C D E B^{\prime}\right]+[A E C] \\
& =\frac{7}{10}[A B C D]+[A E C],
\end{aligned}
$$
That is, $[A E C]=\frac{3}{10}[A B C D]=\frac{3}{10}$.
Since $\triangle E C D$ is congruent to $\triangle E A B^{\prime}$, both triangles have area $\frac{1}{5}$. Note that $\triangle A B^{\prime} C, \triangle A B C$, and $\triangle C D A$ are all congruent, and all have area $\frac{1}{2}$. Since $\triangle A E C$ and $\triangle E D C$ share altitude $\overline{D C}$, $\frac{D E}{E A}=\frac{[D E C]}{[A E C]}=\frac{2}{3}$. Because $\triangle C A E$ is isosceles, $C E=E A$. Let $A E=3 x$. The $C E=3 x, D E=2 x$, and $C D=x \sqrt{9-4}=x \sqrt{5}$. Then $\frac{A D}{D C}=\frac{A E+E D}{D C}=\frac{3+2}{\sqrt{5}}=\sqrt{5}$.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. [4]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $S_{0}=0$ and let $S_{k}$ equal $a_{1}+2 a_{2}+\ldots+k a_{k}$ for $k \geq 1$. Define $a_{i}$ to be 1 if $S_{i-1}<i$ and -1 if $S_{i-1} \geq i$. What is the largest $k \leq 2010$ such that $S_{k}=0$ ?
|
1092 Suppose that $S_{N}=0$ for some $N \geq 0$. Then $a_{N+1}=1$ because $N+1 \geq S_{N}$. The following table lists the values of $a_{k}$ and $S_{k}$ for a few $k \geq N$ :
| $k$ | $a_{k}$ | $S_{k}$ |
| :--- | ---: | :--- |
| $N$ | | 0 |
| $N+1$ | 1 | $N+1$ |
| $N+2$ | 1 | $2 N+3$ |
| $N+3$ | -1 | $N$ |
| $N+4$ | 1 | $2 N+4$ |
| $N+5$ | -1 | $N-1$ |
| $N+6$ | 1 | $2 N+5$ |
| $N+7$ | -1 | $N-2$ |
We see inductively that, for every $i \geq 1$,
$$
S_{N+2 i}=2 N+2+i
$$
and
$$
S_{N+1+2 i}=N+1-i
$$
thus $S_{3 N+3}=0$ is the next $k$ for which $S_{k}=0$. The values of $k$ for which $S_{k}=0$ satisfy the recurrence relation $p_{n+1}=3 p_{n}+3$, and we compute that the first terms of the sequence are $0,3,12,39,120,363,1092$; hence 1092 is our answer.
|
1092
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $S_{0}=0$ and let $S_{k}$ equal $a_{1}+2 a_{2}+\ldots+k a_{k}$ for $k \geq 1$. Define $a_{i}$ to be 1 if $S_{i-1}<i$ and -1 if $S_{i-1} \geq i$. What is the largest $k \leq 2010$ such that $S_{k}=0$ ?
|
1092 Suppose that $S_{N}=0$ for some $N \geq 0$. Then $a_{N+1}=1$ because $N+1 \geq S_{N}$. The following table lists the values of $a_{k}$ and $S_{k}$ for a few $k \geq N$ :
| $k$ | $a_{k}$ | $S_{k}$ |
| :--- | ---: | :--- |
| $N$ | | 0 |
| $N+1$ | 1 | $N+1$ |
| $N+2$ | 1 | $2 N+3$ |
| $N+3$ | -1 | $N$ |
| $N+4$ | 1 | $2 N+4$ |
| $N+5$ | -1 | $N-1$ |
| $N+6$ | 1 | $2 N+5$ |
| $N+7$ | -1 | $N-2$ |
We see inductively that, for every $i \geq 1$,
$$
S_{N+2 i}=2 N+2+i
$$
and
$$
S_{N+1+2 i}=N+1-i
$$
thus $S_{3 N+3}=0$ is the next $k$ for which $S_{k}=0$. The values of $k$ for which $S_{k}=0$ satisfy the recurrence relation $p_{n+1}=3 p_{n}+3$, and we compute that the first terms of the sequence are $0,3,12,39,120,363,1092$; hence 1092 is our answer.
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [4]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Manya has a stack of $85=1+4+16+64$ blocks comprised of 4 layers (the $k$ th layer from the top has $4^{k-1}$ blocks; see the diagram below). Each block rests on 4 smaller blocks, each with dimensions half those of the larger block. Laura removes blocks one at a time from this stack, removing only blocks that currently have no blocks on top of them. Find the number of ways Laura can remove precisely 5 blocks from Manya's stack (the order in which they are removed matters).
|
3384 Each time Laura removes a block, 4 additional blocks are exposed, increasing the total number of exposed blocks by 3 . She removes 5 blocks, for a total of $1 \cdot 4 \cdot 7 \cdot 10 \cdot 13$ ways. However, the stack originally only has 4 layers, so we must subtract the cases where removing a block on the bottom layer does not expose any new blocks. There are $1 \cdot 4 \cdot 4 \cdot 4 \cdot 4=256$ of these (the last factor of 4 is from the 4 blocks that we counted as being exposed, but were not actually). So our final answer is $1 \cdot 4 \cdot 7 \cdot 10 \cdot 13-1 \cdot 4 \cdot 4 \cdot 4 \cdot 4=3384$.
|
3384
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Manya has a stack of $85=1+4+16+64$ blocks comprised of 4 layers (the $k$ th layer from the top has $4^{k-1}$ blocks; see the diagram below). Each block rests on 4 smaller blocks, each with dimensions half those of the larger block. Laura removes blocks one at a time from this stack, removing only blocks that currently have no blocks on top of them. Find the number of ways Laura can remove precisely 5 blocks from Manya's stack (the order in which they are removed matters).
|
3384 Each time Laura removes a block, 4 additional blocks are exposed, increasing the total number of exposed blocks by 3 . She removes 5 blocks, for a total of $1 \cdot 4 \cdot 7 \cdot 10 \cdot 13$ ways. However, the stack originally only has 4 layers, so we must subtract the cases where removing a block on the bottom layer does not expose any new blocks. There are $1 \cdot 4 \cdot 4 \cdot 4 \cdot 4=256$ of these (the last factor of 4 is from the 4 blocks that we counted as being exposed, but were not actually). So our final answer is $1 \cdot 4 \cdot 7 \cdot 10 \cdot 13-1 \cdot 4 \cdot 4 \cdot 4 \cdot 4=3384$.
|
{
"exam": "HMMT",
"problem_label": "5",
"problem_match": "\n5. [4]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
John needs to pay 2010 dollars for his dinner. He has an unlimited supply of 2, 5, and 10 dollar notes. In how many ways can he pay?
|
20503
Let the number of 2,5 , and 10 dollar notes John can use be $x, y$, and $z$ respectively. We wish to find the number of nonnegative integer solutions to $2 x+5 y+10 z=2010$. Consider this equation $\bmod 2$. Because $2 x, 10 z$, and 2010 are even, $5 y$ must also be even, so $y$ must be even. Now consider the equation $\bmod 5$. Because $5 y, 10 z$, and 2010 are divisible by $5,2 x$ must also be divisible by 5 , so $x$ must be divisible by 5 . So both $2 x$ and $5 y$ are divisible by 10 . So the equation is equivalent to $10 x^{\prime}+10 y^{\prime}+10 z=2010$, or $x^{\prime}+y^{\prime}+z=201$, with $x^{\prime}, y^{\prime}$, and $z$ nonnegative integers. There is a well-known bijection between solutions of this equation and picking 2 of 203 balls in a row on the table (explained in further detail below), so there are $\binom{203}{2}=20503$ ways.
The bijection between solutions of $x^{\prime}+y^{\prime}+z=201$ and arrangements of 203 balls in a row is as follows. Given a solution of the equation, we put $x^{\prime}$ white balls in a row, then a black ball, then $y^{\prime}$ white balls, then a black ball, then $z$ white balls. This is like having 203 balls in a row on a table and picking two of them to be black. To go from an arrangement of balls to a solution of the equation, we just read off $x^{\prime}, y^{\prime}$, and $z$ from the number of white balls in a row. There are $\binom{203}{2}$ ways to choose 2 of 203 balls to be black, so there are $\binom{203}{2}$ solutions to $x^{\prime}+y^{\prime}+z=201$.
|
20503
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
John needs to pay 2010 dollars for his dinner. He has an unlimited supply of 2, 5, and 10 dollar notes. In how many ways can he pay?
|
20503
Let the number of 2,5 , and 10 dollar notes John can use be $x, y$, and $z$ respectively. We wish to find the number of nonnegative integer solutions to $2 x+5 y+10 z=2010$. Consider this equation $\bmod 2$. Because $2 x, 10 z$, and 2010 are even, $5 y$ must also be even, so $y$ must be even. Now consider the equation $\bmod 5$. Because $5 y, 10 z$, and 2010 are divisible by $5,2 x$ must also be divisible by 5 , so $x$ must be divisible by 5 . So both $2 x$ and $5 y$ are divisible by 10 . So the equation is equivalent to $10 x^{\prime}+10 y^{\prime}+10 z=2010$, or $x^{\prime}+y^{\prime}+z=201$, with $x^{\prime}, y^{\prime}$, and $z$ nonnegative integers. There is a well-known bijection between solutions of this equation and picking 2 of 203 balls in a row on the table (explained in further detail below), so there are $\binom{203}{2}=20503$ ways.
The bijection between solutions of $x^{\prime}+y^{\prime}+z=201$ and arrangements of 203 balls in a row is as follows. Given a solution of the equation, we put $x^{\prime}$ white balls in a row, then a black ball, then $y^{\prime}$ white balls, then a black ball, then $z$ white balls. This is like having 203 balls in a row on a table and picking two of them to be black. To go from an arrangement of balls to a solution of the equation, we just read off $x^{\prime}, y^{\prime}$, and $z$ from the number of white balls in a row. There are $\binom{203}{2}$ ways to choose 2 of 203 balls to be black, so there are $\binom{203}{2}$ solutions to $x^{\prime}+y^{\prime}+z=201$.
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n6. [5]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl",
"solution_match": "\n## Answer: ",
"tier": "T4",
"year": "2010"
}
|
Suppose that a polynomial of the form $p(x)=x^{2010} \pm x^{2009} \pm \cdots \pm x \pm 1$ has no real roots. What is the maximum possible number of coefficients of -1 in $p$ ?
|
1005 Let $p(x)$ be a polynomial with the maximum number of minus signs.
$p(x)$ cannot have more than 1005 minus signs, otherwise $p(1)<0$ and $p(2) \geq 2^{2010}-2^{2009}-\ldots-2-1=$ 1 , which implies, by the Intermediate Value Theorem, that $p$ must have a root greater than 1.
Let $p(x)=\frac{x^{2011}+1}{x+1}=x^{2010}-x^{2009}+x^{2008}-\ldots-x+1 .-1$ is the only real root of $x^{2011}+1=0$ but $p(-1)=2011$; therefore $p$ has no real roots. Since $p$ has 1005 minus signs, it is the desired polynomial.
|
1005
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose that a polynomial of the form $p(x)=x^{2010} \pm x^{2009} \pm \cdots \pm x \pm 1$ has no real roots. What is the maximum possible number of coefficients of -1 in $p$ ?
|
1005 Let $p(x)$ be a polynomial with the maximum number of minus signs.
$p(x)$ cannot have more than 1005 minus signs, otherwise $p(1)<0$ and $p(2) \geq 2^{2010}-2^{2009}-\ldots-2-1=$ 1 , which implies, by the Intermediate Value Theorem, that $p$ must have a root greater than 1.
Let $p(x)=\frac{x^{2011}+1}{x+1}=x^{2010}-x^{2009}+x^{2008}-\ldots-x+1 .-1$ is the only real root of $x^{2011}+1=0$ but $p(-1)=2011$; therefore $p$ has no real roots. Since $p$ has 1005 minus signs, it is the desired polynomial.
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [6]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
A sphere is the set of points at a fixed positive distance $r$ from its center. Let $\mathcal{S}$ be a set of 2010dimensional spheres. Suppose that the number of points lying on every element of $\mathcal{S}$ is a finite number $n$. Find the maximum possible value of $n$.
|
2 The answer is 2 for any number of dimensions. We prove this by induction on the dimension.
Note that 1-dimensional spheres are pairs of points, and 2-dimensional spheres are circles.
Base case, $d=2$ : The intersection of two circles is either a circle (if the original circles are identical, and in the same place), a pair of points, a single point (if the circles are tangent), or the empty set. Thus, in dimension 2 , the largest finite number of intersection points is 2 , because the number of pairwise intersection points is 0,1 , or 2 for distinct circles.
We now prove that the intersection of two $k$-dimensional spheres is either the empty set, a $(k-1)$ dimensional sphere, a $k$-dimensional sphere (which only occurs if the original spheres are identical and coincident). Consider two spheres in $k$-dimensional space, and impose a coordinate system such that the centers of the two spheres lie on one coordinate axis. Then the equations for the two spheres become identical in all but one coordinate:
$$
\begin{aligned}
& \left(x_{1}-a_{1}\right)^{2}+x_{2}^{2}+\ldots+x_{k}^{2}=r_{1}^{2} \\
& \left(x_{1}-a_{2}\right)^{2}+x_{2}^{2}+\ldots+x_{k}^{2}=r_{2}^{2}
\end{aligned}
$$
If $a_{1}=a_{2}$, the spheres are concentric, and so they are either nonintersecting or coincident, intersecting in a $k$-dimensional sphere. If $a_{1} \neq a_{2}$, then subtracting the equations and solving for $x_{1}$ yields $x_{1}=\frac{r_{1}^{2}-a_{1}^{2}-r_{2}^{2}+a_{2}^{2}}{2\left(a_{2}-a_{2}\right)}$. Plugging this in to either equation above yields a single equation equation that describes a $(k-1)$-dimensional sphere.
Assume we are in dimension $d$, and suppose for induction that for all $k$ less than $d$, any two distinct $k$-dimensional spheres intersecting in a finite number of points intersect in at most two points. Suppose we have a collection of $d$-dimensional spheres $s_{1}, s_{2}, \ldots, s_{m}$. Without loss of generality, suppose the $s_{i}$ are distinct. Let $t_{i}$ be the intersection of $s_{i}$ and $s_{i+1}$ for $1 \leq i<m$. If any $t_{i}$ are the empty set, then the intersection of the $t_{i}$ is empty. None of the $t_{i}$ is a $d$-dimensional sphere because the $s_{i}$ are distinct. Thus each of $t_{1}, t_{2}, \ldots, t_{m-1}$ is a $(d-1)$-dimensional sphere, and the intersection of all of them is the same as the intersection of the $d$-dimensional spheres. We can then apply the inductive hypothesis to find that $t_{1}, \ldots, t_{m-1}$ intersect in at most two points. Thus, by induction, a set of spheres in any dimension which intersect at only finitely many points intersect at at most two points.
We now exhibit a set of $2^{2009}$ 2010-dimensional spheres, and prove that their intersection contains exactly two points. Take the spheres with radii $\sqrt{2013}$ and centers $(0, \pm 1, \pm 1, \ldots, \pm 1)$, where the sign of each coordinate is independent from the sign of every other coordinate. Because of our choice of radius, all these spheres pass through the points $( \pm 2,0,0, \ldots 0)$. Then the intersection is the set of points $\left(x_{1}, x_{2}, \ldots, x_{2010}\right)$ which satisfy the equations $x_{1}^{2}+\left(x_{2} \pm 1\right)^{2}+\cdots+\left(x_{2010} \pm 1\right)^{2}=2013$. The only solutions to these equations are the points $( \pm 2,0,0, \ldots, 0)$ (since $\left(x_{i}+1\right)^{2}$ must be the same as $\left(x_{i}-1\right)^{2}$ for all $i>1$, because we may hold all but one of the $\pm$ choices constant, and change the remaining one).
|
2
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A sphere is the set of points at a fixed positive distance $r$ from its center. Let $\mathcal{S}$ be a set of 2010dimensional spheres. Suppose that the number of points lying on every element of $\mathcal{S}$ is a finite number $n$. Find the maximum possible value of $n$.
|
2 The answer is 2 for any number of dimensions. We prove this by induction on the dimension.
Note that 1-dimensional spheres are pairs of points, and 2-dimensional spheres are circles.
Base case, $d=2$ : The intersection of two circles is either a circle (if the original circles are identical, and in the same place), a pair of points, a single point (if the circles are tangent), or the empty set. Thus, in dimension 2 , the largest finite number of intersection points is 2 , because the number of pairwise intersection points is 0,1 , or 2 for distinct circles.
We now prove that the intersection of two $k$-dimensional spheres is either the empty set, a $(k-1)$ dimensional sphere, a $k$-dimensional sphere (which only occurs if the original spheres are identical and coincident). Consider two spheres in $k$-dimensional space, and impose a coordinate system such that the centers of the two spheres lie on one coordinate axis. Then the equations for the two spheres become identical in all but one coordinate:
$$
\begin{aligned}
& \left(x_{1}-a_{1}\right)^{2}+x_{2}^{2}+\ldots+x_{k}^{2}=r_{1}^{2} \\
& \left(x_{1}-a_{2}\right)^{2}+x_{2}^{2}+\ldots+x_{k}^{2}=r_{2}^{2}
\end{aligned}
$$
If $a_{1}=a_{2}$, the spheres are concentric, and so they are either nonintersecting or coincident, intersecting in a $k$-dimensional sphere. If $a_{1} \neq a_{2}$, then subtracting the equations and solving for $x_{1}$ yields $x_{1}=\frac{r_{1}^{2}-a_{1}^{2}-r_{2}^{2}+a_{2}^{2}}{2\left(a_{2}-a_{2}\right)}$. Plugging this in to either equation above yields a single equation equation that describes a $(k-1)$-dimensional sphere.
Assume we are in dimension $d$, and suppose for induction that for all $k$ less than $d$, any two distinct $k$-dimensional spheres intersecting in a finite number of points intersect in at most two points. Suppose we have a collection of $d$-dimensional spheres $s_{1}, s_{2}, \ldots, s_{m}$. Without loss of generality, suppose the $s_{i}$ are distinct. Let $t_{i}$ be the intersection of $s_{i}$ and $s_{i+1}$ for $1 \leq i<m$. If any $t_{i}$ are the empty set, then the intersection of the $t_{i}$ is empty. None of the $t_{i}$ is a $d$-dimensional sphere because the $s_{i}$ are distinct. Thus each of $t_{1}, t_{2}, \ldots, t_{m-1}$ is a $(d-1)$-dimensional sphere, and the intersection of all of them is the same as the intersection of the $d$-dimensional spheres. We can then apply the inductive hypothesis to find that $t_{1}, \ldots, t_{m-1}$ intersect in at most two points. Thus, by induction, a set of spheres in any dimension which intersect at only finitely many points intersect at at most two points.
We now exhibit a set of $2^{2009}$ 2010-dimensional spheres, and prove that their intersection contains exactly two points. Take the spheres with radii $\sqrt{2013}$ and centers $(0, \pm 1, \pm 1, \ldots, \pm 1)$, where the sign of each coordinate is independent from the sign of every other coordinate. Because of our choice of radius, all these spheres pass through the points $( \pm 2,0,0, \ldots 0)$. Then the intersection is the set of points $\left(x_{1}, x_{2}, \ldots, x_{2010}\right)$ which satisfy the equations $x_{1}^{2}+\left(x_{2} \pm 1\right)^{2}+\cdots+\left(x_{2010} \pm 1\right)^{2}=2013$. The only solutions to these equations are the points $( \pm 2,0,0, \ldots, 0)$ (since $\left(x_{i}+1\right)^{2}$ must be the same as $\left(x_{i}-1\right)^{2}$ for all $i>1$, because we may hold all but one of the $\pm$ choices constant, and change the remaining one).
|
{
"exam": "HMMT",
"problem_label": "5",
"problem_match": "\n8. [6]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Three unit circles $\omega_{1}, \omega_{2}$, and $\omega_{3}$ in the plane have the property that each circle passes through the centers of the other two. A square $S$ surrounds the three circles in such a way that each of its four sides is tangent to at least one of $\omega_{1}, \omega_{2}$ and $\omega_{3}$. Find the side length of the square $S$.
|
$\frac{\sqrt{6}+\sqrt{2}+8}{4}$
By the Pigeonhole Principle, two of the sides must be tangent to the same circle, say $\omega_{1}$. Since $S$ surrounds the circles, these two sides must be adjacent, so we can let $A$ denote the common vertex of the two sides tangent to $\omega_{1}$. Let $B, C$, and $D$ be the other vertices of $S$ in clockwise order, and let $P, Q$, and $R$ be the centers of $\omega_{1}, \omega_{2}$, and $\omega_{3}$ respectively, and suppose WLOG that they are also in clockwise order. Then $A C$ passes through the center of $\omega_{1}$, and by symmetry ( since $A B=A D$ ) it
must also pass through the other intersection point of $\omega_{2}$ and $\omega_{3}$. That is, $A C$ is the radical axis of $\omega_{2}$ and $\omega_{3}$.
Now, let $M$ and $N$ be the feet of the perpendiculars from $P$ and $R$, respectively, to side $A D$. Let $E$ and $F$ be the feet of the perpendiculars from $P$ to $A B$ and from $R$ to $D C$, respectively. Then $P E A M$ and $N R F D$ are rectangles, and $P E$ and $R F$ are radii of $\omega_{1}$ and $\omega_{2}$ respectively. Thus $A M=E P=1$ and $N D=R F=1$. Finally, we have
$$
\begin{aligned}
M N & =P R \cdot \cos \left(180^{\circ}-\angle E P R\right) \\
& =\cos \left(180^{\circ}-E P Q-R P Q\right) \\
& =-\cos \left(\left(270^{\circ}-60^{\circ}\right) / 2+60^{\circ}\right) \\
& =-\cos \left(165^{\circ}\right) \\
& =\cos \left(15^{\circ}\right) \\
& =\frac{\sqrt{6}+\sqrt{2}}{4} .
\end{aligned}
$$
Thus $A D=A M+M N+N D=1+\frac{\sqrt{6}+\sqrt{2}}{4}+1=\frac{\sqrt{6}+\sqrt{2}+8}{4}$ as claimed.
|
\frac{\sqrt{6}+\sqrt{2}+8}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Three unit circles $\omega_{1}, \omega_{2}$, and $\omega_{3}$ in the plane have the property that each circle passes through the centers of the other two. A square $S$ surrounds the three circles in such a way that each of its four sides is tangent to at least one of $\omega_{1}, \omega_{2}$ and $\omega_{3}$. Find the side length of the square $S$.
|
$\frac{\sqrt{6}+\sqrt{2}+8}{4}$
By the Pigeonhole Principle, two of the sides must be tangent to the same circle, say $\omega_{1}$. Since $S$ surrounds the circles, these two sides must be adjacent, so we can let $A$ denote the common vertex of the two sides tangent to $\omega_{1}$. Let $B, C$, and $D$ be the other vertices of $S$ in clockwise order, and let $P, Q$, and $R$ be the centers of $\omega_{1}, \omega_{2}$, and $\omega_{3}$ respectively, and suppose WLOG that they are also in clockwise order. Then $A C$ passes through the center of $\omega_{1}$, and by symmetry ( since $A B=A D$ ) it
must also pass through the other intersection point of $\omega_{2}$ and $\omega_{3}$. That is, $A C$ is the radical axis of $\omega_{2}$ and $\omega_{3}$.
Now, let $M$ and $N$ be the feet of the perpendiculars from $P$ and $R$, respectively, to side $A D$. Let $E$ and $F$ be the feet of the perpendiculars from $P$ to $A B$ and from $R$ to $D C$, respectively. Then $P E A M$ and $N R F D$ are rectangles, and $P E$ and $R F$ are radii of $\omega_{1}$ and $\omega_{2}$ respectively. Thus $A M=E P=1$ and $N D=R F=1$. Finally, we have
$$
\begin{aligned}
M N & =P R \cdot \cos \left(180^{\circ}-\angle E P R\right) \\
& =\cos \left(180^{\circ}-E P Q-R P Q\right) \\
& =-\cos \left(\left(270^{\circ}-60^{\circ}\right) / 2+60^{\circ}\right) \\
& =-\cos \left(165^{\circ}\right) \\
& =\cos \left(15^{\circ}\right) \\
& =\frac{\sqrt{6}+\sqrt{2}}{4} .
\end{aligned}
$$
Thus $A D=A M+M N+N D=1+\frac{\sqrt{6}+\sqrt{2}}{4}+1=\frac{\sqrt{6}+\sqrt{2}+8}{4}$ as claimed.
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n9. [7]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $a, b, c, x, y$, and $z$ be complex numbers such that
$$
a=\frac{b+c}{x-2}, \quad b=\frac{c+a}{y-2}, \quad c=\frac{a+b}{z-2} .
$$
If $x y+y z+z x=67$ and $x+y+z=2010$, find the value of $x y z$.
|
-5892 Manipulate the equations to get a common denominator: $a=\frac{b+c}{x-2} \Longrightarrow x-2=$ $\frac{b+c}{a} \Longrightarrow x-1=\frac{a+b+c}{a} \Longrightarrow \frac{1}{x-1}=\frac{a}{a+b+c}$; similarly, $\frac{1}{y-1}=\frac{b}{a+b+c}$ and $\frac{1}{z-1}=\frac{c}{a+b+c}$. Thus
$$
\begin{aligned}
\frac{1}{x-1}+\frac{1}{y-1}+\frac{1}{z-1} & =1 \\
(y-1)(z-1)+(x-1)(z-1)+(x-1)(y-1) & =(x-1)(y-1)(z-1) \\
x y+y z+z x-2(x+y+z)+3 & =x y z-(x y+y z+z x)+(x+y+z)-1 \\
x y z-2(x y+y z+z x)+3(x+y+z)-4 & =0 \\
x y z-2(67)+3(2010)-4 & =0 \\
x y z & =-5892
\end{aligned}
$$
|
-5892
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a, b, c, x, y$, and $z$ be complex numbers such that
$$
a=\frac{b+c}{x-2}, \quad b=\frac{c+a}{y-2}, \quad c=\frac{a+b}{z-2} .
$$
If $x y+y z+z x=67$ and $x+y+z=2010$, find the value of $x y z$.
|
-5892 Manipulate the equations to get a common denominator: $a=\frac{b+c}{x-2} \Longrightarrow x-2=$ $\frac{b+c}{a} \Longrightarrow x-1=\frac{a+b+c}{a} \Longrightarrow \frac{1}{x-1}=\frac{a}{a+b+c}$; similarly, $\frac{1}{y-1}=\frac{b}{a+b+c}$ and $\frac{1}{z-1}=\frac{c}{a+b+c}$. Thus
$$
\begin{aligned}
\frac{1}{x-1}+\frac{1}{y-1}+\frac{1}{z-1} & =1 \\
(y-1)(z-1)+(x-1)(z-1)+(x-1)(y-1) & =(x-1)(y-1)(z-1) \\
x y+y z+z x-2(x+y+z)+3 & =x y z-(x y+y z+z x)+(x+y+z)-1 \\
x y z-2(x y+y z+z x)+3(x+y+z)-4 & =0 \\
x y z-2(67)+3(2010)-4 & =0 \\
x y z & =-5892
\end{aligned}
$$
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. [8]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Below is pictured a regular seven-pointed star. Find the measure of angle $a$ in radians.
|
| $\frac{3 \pi}{7}$ |
| :---: | The measure of the interior angle of a point of the star is $\frac{\pi}{7}$ because it is an inscribed angle on the circumcircle which intercepts a seventh of the circle ${ }^{1}$
Consider the triangle shown above in bold. Because the sum of the angles in any triangle is $\pi$,
$$
2 \varphi+3\left(\frac{\pi}{7}\right)=\pi=2 \varphi+a
$$
Canceling the $2 \varphi$ on the right-hand side and on the left-hand side, we obtain
$$
a=\frac{3 \pi}{7} .
$$
|
\frac{3 \pi}{7}
|
Incomplete
|
Yes
|
math-word-problem
|
Geometry
|
Below is pictured a regular seven-pointed star. Find the measure of angle $a$ in radians.
|
| $\frac{3 \pi}{7}$ |
| :---: | The measure of the interior angle of a point of the star is $\frac{\pi}{7}$ because it is an inscribed angle on the circumcircle which intercepts a seventh of the circle ${ }^{1}$
Consider the triangle shown above in bold. Because the sum of the angles in any triangle is $\pi$,
$$
2 \varphi+3\left(\frac{\pi}{7}\right)=\pi=2 \varphi+a
$$
Canceling the $2 \varphi$ on the right-hand side and on the left-hand side, we obtain
$$
a=\frac{3 \pi}{7} .
$$
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [3]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
The rank of a rational number $q$ is the unique $k$ for which $q=\frac{1}{a_{1}}+\cdots+\frac{1}{a_{k}}$, where each $a_{i}$ is the smallest positive integer such that $q \geq \frac{1}{a_{1}}+\cdots+\frac{1}{a_{i}}$. Let $q$ be the largest rational number less than $\frac{1}{4}$ with rank 3 , and suppose the expression for $q$ is $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}$. Find the ordered triple $\left(a_{1}, a_{2}, a_{3}\right)$.
|
$(5,21,421)$ Suppose that $A$ and $B$ were rational numbers of rank 3 less than $\frac{1}{4}$, and let $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ be positive integers so that $A=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}$ and $B=\frac{1}{b_{1}}+\frac{1}{b_{2}}+\frac{1}{b_{3}}$ are the expressions for $A$ and $B$ as stated in the problem. If $b_{1}<a_{1}$ then $A<\frac{1}{a_{1}-1} \leq \frac{1}{b_{1}}<B$. In other words, of all the rationals less than $\frac{1}{4}$ with rank 3 , those that have $a_{1}=5$ are greater than those that have $a_{1}=6,7,8, \ldots$ Therefore we can "build" $q$ greedily, adding the largest unit fraction that keeps $q$ less than $\frac{1}{4}$ :
$\frac{1}{5}$ is the largest unit fraction less than $\frac{1}{4}$, hence $a_{1}=5$;
$\frac{1}{27}$ is the largest unit fraction less than $\frac{1}{4}-\frac{1}{5}$, hence $a_{2}=21$;
$\frac{1}{421}$ is the largest unit fraction less than $\frac{1}{4}-\frac{1}{5}-\frac{1}{21}$, hence $a_{3}=421$.
|
(5,21,421)
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
The rank of a rational number $q$ is the unique $k$ for which $q=\frac{1}{a_{1}}+\cdots+\frac{1}{a_{k}}$, where each $a_{i}$ is the smallest positive integer such that $q \geq \frac{1}{a_{1}}+\cdots+\frac{1}{a_{i}}$. Let $q$ be the largest rational number less than $\frac{1}{4}$ with rank 3 , and suppose the expression for $q$ is $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}$. Find the ordered triple $\left(a_{1}, a_{2}, a_{3}\right)$.
|
$(5,21,421)$ Suppose that $A$ and $B$ were rational numbers of rank 3 less than $\frac{1}{4}$, and let $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ be positive integers so that $A=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}$ and $B=\frac{1}{b_{1}}+\frac{1}{b_{2}}+\frac{1}{b_{3}}$ are the expressions for $A$ and $B$ as stated in the problem. If $b_{1}<a_{1}$ then $A<\frac{1}{a_{1}-1} \leq \frac{1}{b_{1}}<B$. In other words, of all the rationals less than $\frac{1}{4}$ with rank 3 , those that have $a_{1}=5$ are greater than those that have $a_{1}=6,7,8, \ldots$ Therefore we can "build" $q$ greedily, adding the largest unit fraction that keeps $q$ less than $\frac{1}{4}$ :
$\frac{1}{5}$ is the largest unit fraction less than $\frac{1}{4}$, hence $a_{1}=5$;
$\frac{1}{27}$ is the largest unit fraction less than $\frac{1}{4}-\frac{1}{5}$, hence $a_{2}=21$;
$\frac{1}{421}$ is the largest unit fraction less than $\frac{1}{4}-\frac{1}{5}-\frac{1}{21}$, hence $a_{3}=421$.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [3]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
How many positive integers less than or equal to 240 can be expressed as a sum of distinct factorials? Consider 0 ! and 1 ! to be distinct.
|
39 Note that $1=0!, 2=0!+1!, 3=0!+2$ !, and $4=0!+1!+2$ !. These are the only numbers less than 6 that can be written as the sum of factorials. The only other factorials less than 240 are $3!=6,4!=24$, and $5!=120$. So a positive integer less than or equal to 240 can only contain $3!, 4!, 5$ !, and/or one of $1,2,3$, or 4 in its sum. If it contains any factorial larger than 5 !, it will be larger than 240 . So a sum less than or equal to 240 will will either include 3 ! or not ( 2 ways), 4 ! or not ( 2 ways), 5 ! or not ( 2 ways), and add an additional $0,1,2,3$ or 4 ( 5 ways). This gives $2 \cdot 2 \cdot 2 \cdot 5=40$ integers less than 240 . However, we want only positive integers, so we must not count 0 . So there are 39 such positive integers.
|
39
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many positive integers less than or equal to 240 can be expressed as a sum of distinct factorials? Consider 0 ! and 1 ! to be distinct.
|
39 Note that $1=0!, 2=0!+1!, 3=0!+2$ !, and $4=0!+1!+2$ !. These are the only numbers less than 6 that can be written as the sum of factorials. The only other factorials less than 240 are $3!=6,4!=24$, and $5!=120$. So a positive integer less than or equal to 240 can only contain $3!, 4!, 5$ !, and/or one of $1,2,3$, or 4 in its sum. If it contains any factorial larger than 5 !, it will be larger than 240 . So a sum less than or equal to 240 will will either include 3 ! or not ( 2 ways), 4 ! or not ( 2 ways), 5 ! or not ( 2 ways), and add an additional $0,1,2,3$ or 4 ( 5 ways). This gives $2 \cdot 2 \cdot 2 \cdot 5=40$ integers less than 240 . However, we want only positive integers, so we must not count 0 . So there are 39 such positive integers.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. [4]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
For $0 \leq y \leq 2$, let $D_{y}$ be the half-disk of diameter 2 with one vertex at $(0, y)$, the other vertex on the positive $x$-axis, and the curved boundary further from the origin than the straight boundary. Find the area of the union of $D_{y}$ for all $0 \leq y \leq 2$.
|
$\pi$
From the picture above, we see that the union of the half-disks will be a quarter-circle with radius 2 , and therefore area $\pi$. To prove that this is the case, we first prove that the boundary of every half-disk intersects the quarter-circle with radius 2 , and then that the half-disk is internally tangent to the quarter-circle at that point. This is sufficient because it is clear from the diagram that we need not worry about covering the interior of the quarter-circle.
Let $O$ be the origin. For a given half-disk $D_{y}$, label the vertex on the $y$-axis $A$ and the vertex on the $x$-axis $B$. Let $M$ be the midpoint of line segment $\overline{A B}$. Draw segment $O M$, and extend it until it intersects the curved boundary of $D_{y}$. Label the intersection point $C$. This construction is shown in the diagram below.
We first prove that $C$ lies on the quarter-circle, centered at the origin, with radius 2 . Since $M$ is the midpoint of $\overline{A B}$, and $A$ is on the $y$-axis, $M$ is horizontally halfway between $B$ and the $y$-axis. Since
$O$ and $B$ are on the $x$-axis (which is perpendicular to the $y$-axis), segments $\overline{O M}$ and $\overline{M B}$ have the same length. Since $M$ is the midpoint of $\overline{A B}$, and $A B=2, O M=1$. Since $D_{y}$ is a half-disk with radius 1 , all points on its curved boundary are 1 away from its center, $M$. Then $C$ is 2 away from the origin, and the quarter-circle consists of all points which are 2 away from the origin. Thus, $C$ is an intersection of the half-disk $D_{y}$ with the positive quarter-circle of radius 2 .
It remains to show that the half-disk $D_{y}$ is internally tangent to the quarter-circle. Since $\overline{O C}$ is a radius of the quarter-circle, it is perpendicular to the tangent of the quarter-circle at $C$. Since $\overline{M C}$ is a radius of the half-disk, it is perpendicular to the tangent of the half-disk at $C$. Then the tangents lines of the half-disk and the quarter-circle coincide, and the half-disk is tangent to the quarter-circle. It is obvious from the diagram that the half-disk lies at least partially inside of the quarter-circle, the half-disk $D_{y}$ is internally tangent to the quarter-circle.
Then the union of the half-disks is be a quarter-circle with radius 2 , and has area $\pi$.
|
\pi
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
For $0 \leq y \leq 2$, let $D_{y}$ be the half-disk of diameter 2 with one vertex at $(0, y)$, the other vertex on the positive $x$-axis, and the curved boundary further from the origin than the straight boundary. Find the area of the union of $D_{y}$ for all $0 \leq y \leq 2$.
|
$\pi$
From the picture above, we see that the union of the half-disks will be a quarter-circle with radius 2 , and therefore area $\pi$. To prove that this is the case, we first prove that the boundary of every half-disk intersects the quarter-circle with radius 2 , and then that the half-disk is internally tangent to the quarter-circle at that point. This is sufficient because it is clear from the diagram that we need not worry about covering the interior of the quarter-circle.
Let $O$ be the origin. For a given half-disk $D_{y}$, label the vertex on the $y$-axis $A$ and the vertex on the $x$-axis $B$. Let $M$ be the midpoint of line segment $\overline{A B}$. Draw segment $O M$, and extend it until it intersects the curved boundary of $D_{y}$. Label the intersection point $C$. This construction is shown in the diagram below.
We first prove that $C$ lies on the quarter-circle, centered at the origin, with radius 2 . Since $M$ is the midpoint of $\overline{A B}$, and $A$ is on the $y$-axis, $M$ is horizontally halfway between $B$ and the $y$-axis. Since
$O$ and $B$ are on the $x$-axis (which is perpendicular to the $y$-axis), segments $\overline{O M}$ and $\overline{M B}$ have the same length. Since $M$ is the midpoint of $\overline{A B}$, and $A B=2, O M=1$. Since $D_{y}$ is a half-disk with radius 1 , all points on its curved boundary are 1 away from its center, $M$. Then $C$ is 2 away from the origin, and the quarter-circle consists of all points which are 2 away from the origin. Thus, $C$ is an intersection of the half-disk $D_{y}$ with the positive quarter-circle of radius 2 .
It remains to show that the half-disk $D_{y}$ is internally tangent to the quarter-circle. Since $\overline{O C}$ is a radius of the quarter-circle, it is perpendicular to the tangent of the quarter-circle at $C$. Since $\overline{M C}$ is a radius of the half-disk, it is perpendicular to the tangent of the half-disk at $C$. Then the tangents lines of the half-disk and the quarter-circle coincide, and the half-disk is tangent to the quarter-circle. It is obvious from the diagram that the half-disk lies at least partially inside of the quarter-circle, the half-disk $D_{y}$ is internally tangent to the quarter-circle.
Then the union of the half-disks is be a quarter-circle with radius 2 , and has area $\pi$.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n4. [4]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Suppose that there exist nonzero complex numbers $a, b, c$, and $d$ such that $k$ is a root of both the equations $a x^{3}+b x^{2}+c x+d=0$ and $b x^{3}+c x^{2}+d x+a=0$. Find all possible values of $k$ (including complex values).
|
$1,-1, i,-i$ Let $k$ be a root of both polynomials. Multiplying the first polynomial by $k$ and subtracting the second, we have $a k^{4}-a=0$, which means that $k$ is either $1,-1, i$, or $-i$. If $a=b=c=d=1$, then $-1, i$, and $-i$ are roots of both polynomials. If $a=b=c=1$ and $d=-3$, then 1 is a root of both polynomials. So $k$ can be $1,-1, i$, and $-i$.
|
1, -1, i, -i
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose that there exist nonzero complex numbers $a, b, c$, and $d$ such that $k$ is a root of both the equations $a x^{3}+b x^{2}+c x+d=0$ and $b x^{3}+c x^{2}+d x+a=0$. Find all possible values of $k$ (including complex values).
|
$1,-1, i,-i$ Let $k$ be a root of both polynomials. Multiplying the first polynomial by $k$ and subtracting the second, we have $a k^{4}-a=0$, which means that $k$ is either $1,-1, i$, or $-i$. If $a=b=c=d=1$, then $-1, i$, and $-i$ are roots of both polynomials. If $a=b=c=1$ and $d=-3$, then 1 is a root of both polynomials. So $k$ can be $1,-1, i$, and $-i$.
|
{
"exam": "HMMT",
"problem_label": "5",
"problem_match": "\n5. [5]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $A B C D$ be an isosceles trapezoid such that $A B=10, B C=15, C D=28$, and $D A=15$. There is a point $E$ such that $\triangle A E D$ and $\triangle A E B$ have the same area and such that $E C$ is minimal. Find $E C$.
|
| $\frac{216}{\sqrt{145}}$ |
| :---: |
The locus of points $E$ such that $[A E D]=[A E B]$ forms a line, since area is a linear function of the coordinates of $E$; setting the areas equal gives a linear equation in the coordinates $E^{2}$. Note that $A$ and $M$, the midpoint of $\overline{D B}$, are on this line; $A$ because both areas are 0 , and $M$ because the triangles share an altitude, and bases $\overline{D M}$ and $\overline{M B}$ are equal in length. Then $\overline{A M}$ is the set of points satisfying the area condition. The point $E$, then, is such that $\triangle A E C$ is a right angle (to make the distance minimal) and $E$ lies on $\overline{A M}$.
Let $X$ be the point of intersection of $\overline{A M}$ and $\overline{C D}$. Then $\triangle A M B \sim \triangle X M D$, and since $M D=B M$, they are in fact congruent. Thus $D X=A B=10$, and $X C=18$. Similarly, $B X=15$, so $A B X D$ is a parallelogram. Let $Y$ be the foot of the perpendicular from $A$ to $\overline{D C}$, so that $D Y=\frac{D C-A B}{2}=9$. Then
[^1]$A Y=\sqrt{A D^{2}-D Y^{2}}=\sqrt{225-81}=12$. Then $Y X=D X-D Y=1$ and $A X=\sqrt{A Y^{2}+Y X^{2}}=$ $\sqrt{144+1}=\sqrt{145}$. Since both $\triangle A X Y$ and $\triangle C X E$ have a right angle, and $\angle E X C$ and $\angle Y X A$ are congruent because they are vertical angles, $\triangle A X Y \sim \triangle C X E$. Then $\frac{C E}{A Y}=\frac{C X}{A X}$, so $C E=12 \cdot \frac{18}{\sqrt{145}}=$ $\frac{216}{\sqrt{145}}$.
|
\frac{216}{\sqrt{145}}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C D$ be an isosceles trapezoid such that $A B=10, B C=15, C D=28$, and $D A=15$. There is a point $E$ such that $\triangle A E D$ and $\triangle A E B$ have the same area and such that $E C$ is minimal. Find $E C$.
|
| $\frac{216}{\sqrt{145}}$ |
| :---: |
The locus of points $E$ such that $[A E D]=[A E B]$ forms a line, since area is a linear function of the coordinates of $E$; setting the areas equal gives a linear equation in the coordinates $E^{2}$. Note that $A$ and $M$, the midpoint of $\overline{D B}$, are on this line; $A$ because both areas are 0 , and $M$ because the triangles share an altitude, and bases $\overline{D M}$ and $\overline{M B}$ are equal in length. Then $\overline{A M}$ is the set of points satisfying the area condition. The point $E$, then, is such that $\triangle A E C$ is a right angle (to make the distance minimal) and $E$ lies on $\overline{A M}$.
Let $X$ be the point of intersection of $\overline{A M}$ and $\overline{C D}$. Then $\triangle A M B \sim \triangle X M D$, and since $M D=B M$, they are in fact congruent. Thus $D X=A B=10$, and $X C=18$. Similarly, $B X=15$, so $A B X D$ is a parallelogram. Let $Y$ be the foot of the perpendicular from $A$ to $\overline{D C}$, so that $D Y=\frac{D C-A B}{2}=9$. Then
[^1]$A Y=\sqrt{A D^{2}-D Y^{2}}=\sqrt{225-81}=12$. Then $Y X=D X-D Y=1$ and $A X=\sqrt{A Y^{2}+Y X^{2}}=$ $\sqrt{144+1}=\sqrt{145}$. Since both $\triangle A X Y$ and $\triangle C X E$ have a right angle, and $\angle E X C$ and $\angle Y X A$ are congruent because they are vertical angles, $\triangle A X Y \sim \triangle C X E$. Then $\frac{C E}{A Y}=\frac{C X}{A X}$, so $C E=12 \cdot \frac{18}{\sqrt{145}}=$ $\frac{216}{\sqrt{145}}$.
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n6. [5]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Suppose that $x$ and $y$ are complex numbers such that $x+y=1$ and that $x^{20}+y^{20}=20$. Find the sum of all possible values of $x^{2}+y^{2}$.
|
-90 We have $x^{2}+y^{2}+2 x y=1$. Define $a=2 x y$ and $b=x^{2}+y^{2}$ for convenience. Then $a+b=1$ and $b-a=x^{2}+y^{2}-2 x y=(x-y)^{2}=2 b-1$ so that $x, y=\frac{\sqrt{2 b-1} \pm 1}{2}$. Then
\[
$$
\begin{aligned}
x^{20}+y^{20} & =\left(\frac{\sqrt{2 b-1}+1}{2}\right)^{20}+\left(\frac{\sqrt{2 b-1}-1}{2}\right)^{20} \\
& =\frac{1}{2^{20}}\left[(\sqrt{2 b-1}+1)^{20}+(\sqrt{2 b-1}-1)^{20}\right] \\
& =\frac{2}{2^{20}}\left[(\sqrt{2 b-1})^{20}+\binom{20}{2}(\sqrt{2 b-1})^{18}+\binom{20}{4}(\sqrt{2 b-1})^{16}+\ldots\right] \\
& =\frac{2}{2^{20}}\left[(2 b-1)^{10}+\binom{20}{2}(2 b-1)^{9}+\binom{20}{4}(2 b-1)^{8}+\ldots\right] \\
& =20
\end{aligned}
$$
\]
We want to find the sum of distinct roots of the above polynomial in $b$; we first prove that the original polynomial is square-free. The conditions $x+y=1$ and $x^{20}+y^{20}=20$ imply that $x^{20}+(1-x)^{20}-20=0$; let $p(x)=x^{20}+(1-x)^{20}-20 . p$ is square-free if and only if $G C D\left(p, p^{\prime}\right)=c$ for some constant $c$ :
$$
\begin{aligned}
G C D\left(p, p^{\prime}\right) & =G C D\left(x^{20}+(1-x)^{20}-20,20\left(x^{19}-(1-x)^{19}\right)\right) \\
& =G C D\left(x^{20}-x(1-x)^{19}+(1-x)^{19}-20,20\left(x^{19}-(1-x)^{19}\right)\right) \\
& =G C D\left((1-x)^{19}-20, x^{19}-(1-x)^{19}\right) \\
& =G C D\left((1-x)^{19}-20, x^{19}-20\right)
\end{aligned}
$$
The roots of $x^{19}-20$ are $\sqrt[19]{20^{k}} \exp \left(\frac{2 \pi i k}{19}\right)$ for some $k=0,1, \ldots, 18$; the roots of $(1-x)^{19}-20$ are $1-\sqrt[19]{20^{k}} \exp \left(\frac{2 \pi i k}{19}\right)$ for some $k=0,1, \ldots, 18$. If $x^{19}-20$ and $(1-x)^{19}-20$ share a common root, then there exist integers $m$, $n$ such that $\sqrt[19]{20^{m}} \exp \left(\frac{2 \pi i m}{19}\right)=1-\sqrt[19]{20^{n}} \exp \left(\frac{2 \pi i n}{19}\right)$; since the imaginary parts of both sides must be the same, we have $m=n$ and $\sqrt[19]{20^{m}} \exp \left(\frac{2 \pi i m}{19}\right)=\frac{1}{2} \Longrightarrow 20^{m}=\frac{1}{2^{19}}$, a contradiction. Thus we have proved that the polynomial in $x$ has no double roots. Since for each $b$ there exists a unique pair $(x, y)$ (up to permutations) that satisfies $x^{2}+y^{2}=b$ and $(x+y)^{2}=1$, the polynomial in $b$ has no double roots.
Let the coefficient of $b^{n}$ in the above equation be $\left[b^{n}\right]$. By Vieta's Formulas, the sum of all possible values of $b=x^{2}+y^{2}$ is equal to $-\frac{\left[b^{9}\right]}{\left[b^{10}\right]} . \quad\left[b^{10}\right]=\frac{2}{2^{20}}\left(2^{10}\right)$ and $\left[b^{9}\right]=\frac{2}{2^{20}}\left(-\binom{10}{1} 2^{9}+\binom{20}{2} 2^{9}\right)$; thus $-\frac{\left[b^{9}\right]}{\left[b^{10}\right]}=-\frac{\binom{10}{1} 2^{9}-\binom{20}{2} 2^{9}}{2^{10}}=-90$.
|
-90
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose that $x$ and $y$ are complex numbers such that $x+y=1$ and that $x^{20}+y^{20}=20$. Find the sum of all possible values of $x^{2}+y^{2}$.
|
-90 We have $x^{2}+y^{2}+2 x y=1$. Define $a=2 x y$ and $b=x^{2}+y^{2}$ for convenience. Then $a+b=1$ and $b-a=x^{2}+y^{2}-2 x y=(x-y)^{2}=2 b-1$ so that $x, y=\frac{\sqrt{2 b-1} \pm 1}{2}$. Then
\[
$$
\begin{aligned}
x^{20}+y^{20} & =\left(\frac{\sqrt{2 b-1}+1}{2}\right)^{20}+\left(\frac{\sqrt{2 b-1}-1}{2}\right)^{20} \\
& =\frac{1}{2^{20}}\left[(\sqrt{2 b-1}+1)^{20}+(\sqrt{2 b-1}-1)^{20}\right] \\
& =\frac{2}{2^{20}}\left[(\sqrt{2 b-1})^{20}+\binom{20}{2}(\sqrt{2 b-1})^{18}+\binom{20}{4}(\sqrt{2 b-1})^{16}+\ldots\right] \\
& =\frac{2}{2^{20}}\left[(2 b-1)^{10}+\binom{20}{2}(2 b-1)^{9}+\binom{20}{4}(2 b-1)^{8}+\ldots\right] \\
& =20
\end{aligned}
$$
\]
We want to find the sum of distinct roots of the above polynomial in $b$; we first prove that the original polynomial is square-free. The conditions $x+y=1$ and $x^{20}+y^{20}=20$ imply that $x^{20}+(1-x)^{20}-20=0$; let $p(x)=x^{20}+(1-x)^{20}-20 . p$ is square-free if and only if $G C D\left(p, p^{\prime}\right)=c$ for some constant $c$ :
$$
\begin{aligned}
G C D\left(p, p^{\prime}\right) & =G C D\left(x^{20}+(1-x)^{20}-20,20\left(x^{19}-(1-x)^{19}\right)\right) \\
& =G C D\left(x^{20}-x(1-x)^{19}+(1-x)^{19}-20,20\left(x^{19}-(1-x)^{19}\right)\right) \\
& =G C D\left((1-x)^{19}-20, x^{19}-(1-x)^{19}\right) \\
& =G C D\left((1-x)^{19}-20, x^{19}-20\right)
\end{aligned}
$$
The roots of $x^{19}-20$ are $\sqrt[19]{20^{k}} \exp \left(\frac{2 \pi i k}{19}\right)$ for some $k=0,1, \ldots, 18$; the roots of $(1-x)^{19}-20$ are $1-\sqrt[19]{20^{k}} \exp \left(\frac{2 \pi i k}{19}\right)$ for some $k=0,1, \ldots, 18$. If $x^{19}-20$ and $(1-x)^{19}-20$ share a common root, then there exist integers $m$, $n$ such that $\sqrt[19]{20^{m}} \exp \left(\frac{2 \pi i m}{19}\right)=1-\sqrt[19]{20^{n}} \exp \left(\frac{2 \pi i n}{19}\right)$; since the imaginary parts of both sides must be the same, we have $m=n$ and $\sqrt[19]{20^{m}} \exp \left(\frac{2 \pi i m}{19}\right)=\frac{1}{2} \Longrightarrow 20^{m}=\frac{1}{2^{19}}$, a contradiction. Thus we have proved that the polynomial in $x$ has no double roots. Since for each $b$ there exists a unique pair $(x, y)$ (up to permutations) that satisfies $x^{2}+y^{2}=b$ and $(x+y)^{2}=1$, the polynomial in $b$ has no double roots.
Let the coefficient of $b^{n}$ in the above equation be $\left[b^{n}\right]$. By Vieta's Formulas, the sum of all possible values of $b=x^{2}+y^{2}$ is equal to $-\frac{\left[b^{9}\right]}{\left[b^{10}\right]} . \quad\left[b^{10}\right]=\frac{2}{2^{20}}\left(2^{10}\right)$ and $\left[b^{9}\right]=\frac{2}{2^{20}}\left(-\binom{10}{1} 2^{9}+\binom{20}{2} 2^{9}\right)$; thus $-\frac{\left[b^{9}\right]}{\left[b^{10}\right]}=-\frac{\binom{10}{1} 2^{9}-\binom{20}{2} 2^{9}}{2^{10}}=-90$.
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [5]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
An ant starts out at $(0,0)$. Each second, if it is currently at the square $(x, y)$, it can move to $(x-1, y-1),(x-1, y+1),(x+1, y-1)$, or $(x+1, y+1)$. In how many ways can it end up at $(2010,2010)$ after 4020 seconds?
|
\( \binom{4020}{1005}^2 \). Note that each of the coordinates either increases or decreases the x and y- coordinates by 1. In order to reach 2010 after 4020 steps, each of the coordinates must be increased 3015 times and decreased 1005 times. A permutation of 3015 plusses and 1005 minuses for each of $x$ and $y$ uniquely corresponds to a path the ant could take to $(2010,2010)$, because we can take ordered pairs from the two lists and match them up to a valid step the ant can take. So the number of ways the ant can end up at $(2010,2010)$ after 4020 seconds is equal to the number of ways to arrange plusses and minuses for both $x$ and $y$, or $\left(\binom{4020}{1005}\right)^{2}$.
|
\binom{4020}{1005}^2
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
An ant starts out at $(0,0)$. Each second, if it is currently at the square $(x, y)$, it can move to $(x-1, y-1),(x-1, y+1),(x+1, y-1)$, or $(x+1, y+1)$. In how many ways can it end up at $(2010,2010)$ after 4020 seconds?
|
\( \binom{4020}{1005}^2 \). Note that each of the coordinates either increases or decreases the x and y- coordinates by 1. In order to reach 2010 after 4020 steps, each of the coordinates must be increased 3015 times and decreased 1005 times. A permutation of 3015 plusses and 1005 minuses for each of $x$ and $y$ uniquely corresponds to a path the ant could take to $(2010,2010)$, because we can take ordered pairs from the two lists and match them up to a valid step the ant can take. So the number of ways the ant can end up at $(2010,2010)$ after 4020 seconds is equal to the number of ways to arrange plusses and minuses for both $x$ and $y$, or $\left(\binom{4020}{1005}\right)^{2}$.
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [6]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
You are standing in an infinitely long hallway with sides given by the lines $x=0$ and $x=6$. You start at $(3,0)$ and want to get to $(3,6)$. Furthermore, at each instant you want your distance to $(3,6)$ to either decrease or stay the same. What is the area of the set of points that you could pass through on your journey from $(3,0)$ to $(3,6)$ ?
|
$9 \sqrt{3}+\frac{21 \pi}{2}$
If you draw concentric circles around the destination point, the condition is equivalent to the restriction that you must always go inwards towards the destination. In the diagram above, the regions through which you might pass are shaded.
We find the areas of regions $\mathrm{A}, \mathrm{B}$, and C separately, and add them up (doubling the area of region A, because there are two of them).
The hypotenuse of triangle A is of length 6 , and the base is of length 3 , so it is a $\frac{\pi}{6}-\frac{\pi}{3}-\frac{\pi}{2}$ triangle (30-60-90 triangle) with area $\frac{9 \sqrt{3}}{2}$. Then the total area of the regions labeled A is $9 \sqrt{3}$.
Since the angle of triangle A nearest the center of the circle (the destination point) is $\frac{\pi}{3}$, sector B has central angle $\frac{\pi}{3}$. Then the area of sector B is $\frac{1}{2} r^{2} \theta=\frac{1}{2} \cdot 36 \cdot \frac{\pi}{3}=6 \pi$.
Region C is a half-disc of radius 3 , so its area is $\frac{9 \pi}{2}$.
Thus, the total area is $9 \sqrt{3}+\frac{21 \pi}{2}$.
|
9 \sqrt{3}+\frac{21 \pi}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
You are standing in an infinitely long hallway with sides given by the lines $x=0$ and $x=6$. You start at $(3,0)$ and want to get to $(3,6)$. Furthermore, at each instant you want your distance to $(3,6)$ to either decrease or stay the same. What is the area of the set of points that you could pass through on your journey from $(3,0)$ to $(3,6)$ ?
|
$9 \sqrt{3}+\frac{21 \pi}{2}$
If you draw concentric circles around the destination point, the condition is equivalent to the restriction that you must always go inwards towards the destination. In the diagram above, the regions through which you might pass are shaded.
We find the areas of regions $\mathrm{A}, \mathrm{B}$, and C separately, and add them up (doubling the area of region A, because there are two of them).
The hypotenuse of triangle A is of length 6 , and the base is of length 3 , so it is a $\frac{\pi}{6}-\frac{\pi}{3}-\frac{\pi}{2}$ triangle (30-60-90 triangle) with area $\frac{9 \sqrt{3}}{2}$. Then the total area of the regions labeled A is $9 \sqrt{3}$.
Since the angle of triangle A nearest the center of the circle (the destination point) is $\frac{\pi}{3}$, sector B has central angle $\frac{\pi}{3}$. Then the area of sector B is $\frac{1}{2} r^{2} \theta=\frac{1}{2} \cdot 36 \cdot \frac{\pi}{3}=6 \pi$.
Region C is a half-disc of radius 3 , so its area is $\frac{9 \pi}{2}$.
Thus, the total area is $9 \sqrt{3}+\frac{21 \pi}{2}$.
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n9. [7]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
In a $16 \times 16$ table of integers, each row and column contains at most 4 distinct integers. What is the maximum number of distinct integers that there can be in the whole table?
|
49 First, we show that 50 is too big. Assume for sake of contradiction that a labeling with at least 50 distinct integers exists. By the Pigeonhole Principle, there must be at least one row, say the first row, with at least 4 distinct integers in it; in this case, that is exactly 4 , since that is the maximum number of distinct integers in one row. Then, in the remaining 15 rows there must be at
least 46 distinct integers (these 46 will also be distinct from the 4 in the first row). Using Pigeonhole again, there will be another row, say the second row, with 4 distinct integers in it. Call the set of integers in the first and second rows $S$. Because the 4 distinct integers in the second row are distinct from the 4 in the first row, there are 8 distinct values in the first two rows, so $|S|=8$. Now consider the subcolumns containing the cells in rows 3 to 16 . In each subcolumn, there are at most 2 values not in $S$, because there are already two distinct values in that column from the cells in the first two rows. So, the maximum number of distinct values in the table is $16 \cdot 2+8=40$, a contradiction. So a valid labeling must have fewer than 50 distinct integers. Below, we show by example that 49 is attainable.
| 1 | 17 | 33 | - | - | - | - | - | - | - | - | - | - | - | - | - |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| - | 2 | 18 | 34 | - | - | - | - | - | - | - | - | - | - | - | - |
| - | - | 3 | 19 | 35 | - | - | - | - | - | - | - | - | - | - | - |
| - | - | - | 4 | 20 | 36 | - | - | - | - | - | - | - | - | - | - |
| - | - | - | - | 5 | 21 | 37 | - | - | - | - | - | - | - | - | - |
| - | - | - | - | - | 6 | 22 | 38 | - | - | - | - | - | - | - | - |
| - | - | - | - | - | - | 7 | 23 | 39 | - | - | - | - | - | - | - |
| - | - | - | - | - | - | - | 8 | 24 | 40 | - | - | - | - | - | - |
| - | - | - | - | - | - | - | - | 9 | 25 | 41 | - | - | - | - | - |
| - | - | - | - | - | - | - | - | - | 10 | 26 | 42 | - | - | - | - |
| - | - | - | - | - | - | - | - | - | - | 11 | 27 | 43 | - | - | - |
| - | - | - | - | - | - | - | - | - | - | - | 12 | 28 | 44 | - | - |
| - | - | - | - | - | - | - | - | - | - | - | - | 13 | 29 | 45 | - |
| - | - | - | - | - | - | - | - | - | - | - | - | - | 14 | 30 | 46 |
| 47 | - | - | - | - | - | - | - | - | - | - | - | - | - | 15 | 31 |
| 32 | 48 | - | - | - | - | - | - | - | - | - | - | - | - | - | 16 |
Cells that do not contain a number are colored with color 49 .
[^0]: 1/http://en.wikipedia.org/wiki/Inscribed_angle_theorem
[^1]: ${ }^{2}$ The notation $\left[A_{1} A_{2} \ldots A_{n}\right]$ means the area of polygon $A_{1} A_{2} \ldots A_{n}$.
|
49
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In a $16 \times 16$ table of integers, each row and column contains at most 4 distinct integers. What is the maximum number of distinct integers that there can be in the whole table?
|
49 First, we show that 50 is too big. Assume for sake of contradiction that a labeling with at least 50 distinct integers exists. By the Pigeonhole Principle, there must be at least one row, say the first row, with at least 4 distinct integers in it; in this case, that is exactly 4 , since that is the maximum number of distinct integers in one row. Then, in the remaining 15 rows there must be at
least 46 distinct integers (these 46 will also be distinct from the 4 in the first row). Using Pigeonhole again, there will be another row, say the second row, with 4 distinct integers in it. Call the set of integers in the first and second rows $S$. Because the 4 distinct integers in the second row are distinct from the 4 in the first row, there are 8 distinct values in the first two rows, so $|S|=8$. Now consider the subcolumns containing the cells in rows 3 to 16 . In each subcolumn, there are at most 2 values not in $S$, because there are already two distinct values in that column from the cells in the first two rows. So, the maximum number of distinct values in the table is $16 \cdot 2+8=40$, a contradiction. So a valid labeling must have fewer than 50 distinct integers. Below, we show by example that 49 is attainable.
| 1 | 17 | 33 | - | - | - | - | - | - | - | - | - | - | - | - | - |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| - | 2 | 18 | 34 | - | - | - | - | - | - | - | - | - | - | - | - |
| - | - | 3 | 19 | 35 | - | - | - | - | - | - | - | - | - | - | - |
| - | - | - | 4 | 20 | 36 | - | - | - | - | - | - | - | - | - | - |
| - | - | - | - | 5 | 21 | 37 | - | - | - | - | - | - | - | - | - |
| - | - | - | - | - | 6 | 22 | 38 | - | - | - | - | - | - | - | - |
| - | - | - | - | - | - | 7 | 23 | 39 | - | - | - | - | - | - | - |
| - | - | - | - | - | - | - | 8 | 24 | 40 | - | - | - | - | - | - |
| - | - | - | - | - | - | - | - | 9 | 25 | 41 | - | - | - | - | - |
| - | - | - | - | - | - | - | - | - | 10 | 26 | 42 | - | - | - | - |
| - | - | - | - | - | - | - | - | - | - | 11 | 27 | 43 | - | - | - |
| - | - | - | - | - | - | - | - | - | - | - | 12 | 28 | 44 | - | - |
| - | - | - | - | - | - | - | - | - | - | - | - | 13 | 29 | 45 | - |
| - | - | - | - | - | - | - | - | - | - | - | - | - | 14 | 30 | 46 |
| 47 | - | - | - | - | - | - | - | - | - | - | - | - | - | 15 | 31 |
| 32 | 48 | - | - | - | - | - | - | - | - | - | - | - | - | - | 16 |
Cells that do not contain a number are colored with color 49 .
[^0]: 1/http://en.wikipedia.org/wiki/Inscribed_angle_theorem
[^1]: ${ }^{2}$ The notation $\left[A_{1} A_{2} \ldots A_{n}\right]$ means the area of polygon $A_{1} A_{2} \ldots A_{n}$.
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. [8]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-gen2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Below is pictured a regular seven-pointed star. Find the measure of angle $a$ in radians.
|
| $3 \pi$ |
| :---: |
| 7 | The measure of the interior angle of a point of the star is $\frac{\pi}{7}$ because it is an inscribed angle on the circumcircle which intercepts a seventh of the circle ${ }^{1}$
Consider the triangle shown above in bold. Because the sum of the angles in any triangle is $\pi$,
$$
2 \varphi+3\left(\frac{\pi}{7}\right)=\pi=2 \varphi+a
$$
Canceling the $2 \varphi$ on the right-hand side and on the left-hand side, we obtain
$$
a=\frac{3 \pi}{7} .
$$
|
\frac{3 \pi}{7}
|
Incomplete
|
Yes
|
math-word-problem
|
Geometry
|
Below is pictured a regular seven-pointed star. Find the measure of angle $a$ in radians.
|
| $3 \pi$ |
| :---: |
| 7 | The measure of the interior angle of a point of the star is $\frac{\pi}{7}$ because it is an inscribed angle on the circumcircle which intercepts a seventh of the circle ${ }^{1}$
Consider the triangle shown above in bold. Because the sum of the angles in any triangle is $\pi$,
$$
2 \varphi+3\left(\frac{\pi}{7}\right)=\pi=2 \varphi+a
$$
Canceling the $2 \varphi$ on the right-hand side and on the left-hand side, we obtain
$$
a=\frac{3 \pi}{7} .
$$
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [3]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-geo-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convex pentagon that has an area of $\frac{7}{10}$ of the area of the original rectangle. Find the ratio of the longer side of the rectangle to the shorter side of the rectangle.
|
$\sqrt{5}$
[^0]
Given a polygon $P_{1} P_{2} \cdots P_{k}$, let $\left[P_{1} P_{2} \cdots P_{k}\right]$ denote its area. Let $A B C D$ be the rectangle. Suppose we fold $B$ across $\overline{A C}$, and let $E$ be the intersection of $\overline{A D}$ and $\overline{B^{\prime} C}$. Then we end up with the pentagon $A C D E B^{\prime}$, depicted above. Let's suppose, without loss of generality, that $A B C D$ has area 1. Then $\triangle A E C$ must have area $\frac{3}{10}$, since
$$
\begin{aligned}
{[A B C D] } & =[A B C]+[A C D] \\
& =\left[A B^{\prime} C\right]+[A C D] \\
& =\left[A B^{\prime} E\right]+2[A E C]+[E D C] \\
& =\left[A C D E B^{\prime}\right]+[A E C] \\
& =\frac{7}{10}[A B C D]+[A E C]
\end{aligned}
$$
That is, $[A E C]=\frac{3}{10}[A B C D]=\frac{3}{10}$.
Since $\triangle E C D$ is congruent to $\triangle E A B^{\prime}$, both triangles have area $\frac{1}{5}$. Note that $\triangle A B^{\prime} C, \triangle A B C$, and $\triangle C D A$ are all congruent, and all have area $\frac{1}{2}$. Since $\triangle A E C$ and $\triangle E D C$ share altitude $\overline{D C}$, $\frac{D E}{E A}=\frac{[D E C]}{[A E C]}=\frac{2}{3}$. Because $\triangle C A E$ is isosceles, $C E=E A$. Let $A E=3 x$. The $C E=3 x, D E=2 x$, and $C D=x \sqrt{9-4}=x \sqrt{5}$. Then $\frac{A D}{D C}=\frac{A E+E D}{D C}=\frac{3+2}{\sqrt{5}}=\sqrt{5}$.
|
\sqrt{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convex pentagon that has an area of $\frac{7}{10}$ of the area of the original rectangle. Find the ratio of the longer side of the rectangle to the shorter side of the rectangle.
|
$\sqrt{5}$
[^0]
Given a polygon $P_{1} P_{2} \cdots P_{k}$, let $\left[P_{1} P_{2} \cdots P_{k}\right]$ denote its area. Let $A B C D$ be the rectangle. Suppose we fold $B$ across $\overline{A C}$, and let $E$ be the intersection of $\overline{A D}$ and $\overline{B^{\prime} C}$. Then we end up with the pentagon $A C D E B^{\prime}$, depicted above. Let's suppose, without loss of generality, that $A B C D$ has area 1. Then $\triangle A E C$ must have area $\frac{3}{10}$, since
$$
\begin{aligned}
{[A B C D] } & =[A B C]+[A C D] \\
& =\left[A B^{\prime} C\right]+[A C D] \\
& =\left[A B^{\prime} E\right]+2[A E C]+[E D C] \\
& =\left[A C D E B^{\prime}\right]+[A E C] \\
& =\frac{7}{10}[A B C D]+[A E C]
\end{aligned}
$$
That is, $[A E C]=\frac{3}{10}[A B C D]=\frac{3}{10}$.
Since $\triangle E C D$ is congruent to $\triangle E A B^{\prime}$, both triangles have area $\frac{1}{5}$. Note that $\triangle A B^{\prime} C, \triangle A B C$, and $\triangle C D A$ are all congruent, and all have area $\frac{1}{2}$. Since $\triangle A E C$ and $\triangle E D C$ share altitude $\overline{D C}$, $\frac{D E}{E A}=\frac{[D E C]}{[A E C]}=\frac{2}{3}$. Because $\triangle C A E$ is isosceles, $C E=E A$. Let $A E=3 x$. The $C E=3 x, D E=2 x$, and $C D=x \sqrt{9-4}=x \sqrt{5}$. Then $\frac{A D}{D C}=\frac{A E+E D}{D C}=\frac{3+2}{\sqrt{5}}=\sqrt{5}$.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [3]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-geo-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
For $0 \leq y \leq 2$, let $D_{y}$ be the half-disk of diameter 2 with one vertex at $(0, y)$, the other vertex on the positive $x$-axis, and the curved boundary further from the origin than the straight boundary. Find the area of the union of $D_{y}$ for all $0 \leq y \leq 2$.
|
$\pi$
From the picture above, we see that the union of the half-disks will be a quarter-circle with radius 2 , and therefore area $\pi$. To prove that this is the case, we first prove that the boundary of every half-disk intersects the quarter-circle with radius 2 , and then that the half-disk is internally tangent to the quarter-circle at that point. This is sufficient because it is clear from the diagram that we need not worry about covering the interior of the quarter-circle.
Let $O$ be the origin. For a given half-disk $D_{y}$, label the vertex on the $y$-axis $A$ and the vertex on the $x$-axis $B$. Let $M$ be the midpoint of line segment $\overline{A B}$. Draw segment $O M$, and extend it until it intersects the curved boundary of $D_{y}$. Label the intersection point $C$. This construction is shown in the diagram below.
We first prove that $C$ lies on the quarter-circle, centered at the origin, with radius 2 . Since $M$ is the midpoint of $\overline{A B}$, and $A$ is on the $y$-axis, $M$ is horizontally halfway between $B$ and the $y$-axis. Since $O$ and $B$ are on the $x$-axis (which is perpendicular to the $y$-axis), segments $\overline{O M}$ and $\overline{M B}$ have the same length. Since $M$ is the midpoint of $\overline{A B}$, and $A B=2, O M=1$. Since $D_{y}$ is a half-disk with radius 1 , all points on its curved boundary are 1 away from its center, $M$. Then $C$ is 2 away from the origin, and the quarter-circle consists of all points which are 2 away from the origin. Thus, $C$ is an intersection of the half-disk $D_{y}$ with the positive quarter-circle of radius 2 .
It remains to show that the half-disk $D_{y}$ is internally tangent to the quarter-circle. Since $\overline{O C}$ is a radius of the quarter-circle, it is perpendicular to the tangent of the quarter-circle at $C$. Since $\overline{M C}$ is a radius of the half-disk, it is perpendicular to the tangent of the half-disk at $C$. Then the tangents lines of the half-disk and the quarter-circle coincide, and the half-disk is tangent to the quarter-circle. It is obvious from the diagram that the half-disk lies at least partially inside of the quarter-circle, the half-disk $D_{y}$ is internally tangent to the quarter-circle.
Then the union of the half-disks is be a quarter-circle with radius 2 , and has area $\pi$.
|
\pi
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
For $0 \leq y \leq 2$, let $D_{y}$ be the half-disk of diameter 2 with one vertex at $(0, y)$, the other vertex on the positive $x$-axis, and the curved boundary further from the origin than the straight boundary. Find the area of the union of $D_{y}$ for all $0 \leq y \leq 2$.
|
$\pi$
From the picture above, we see that the union of the half-disks will be a quarter-circle with radius 2 , and therefore area $\pi$. To prove that this is the case, we first prove that the boundary of every half-disk intersects the quarter-circle with radius 2 , and then that the half-disk is internally tangent to the quarter-circle at that point. This is sufficient because it is clear from the diagram that we need not worry about covering the interior of the quarter-circle.
Let $O$ be the origin. For a given half-disk $D_{y}$, label the vertex on the $y$-axis $A$ and the vertex on the $x$-axis $B$. Let $M$ be the midpoint of line segment $\overline{A B}$. Draw segment $O M$, and extend it until it intersects the curved boundary of $D_{y}$. Label the intersection point $C$. This construction is shown in the diagram below.
We first prove that $C$ lies on the quarter-circle, centered at the origin, with radius 2 . Since $M$ is the midpoint of $\overline{A B}$, and $A$ is on the $y$-axis, $M$ is horizontally halfway between $B$ and the $y$-axis. Since $O$ and $B$ are on the $x$-axis (which is perpendicular to the $y$-axis), segments $\overline{O M}$ and $\overline{M B}$ have the same length. Since $M$ is the midpoint of $\overline{A B}$, and $A B=2, O M=1$. Since $D_{y}$ is a half-disk with radius 1 , all points on its curved boundary are 1 away from its center, $M$. Then $C$ is 2 away from the origin, and the quarter-circle consists of all points which are 2 away from the origin. Thus, $C$ is an intersection of the half-disk $D_{y}$ with the positive quarter-circle of radius 2 .
It remains to show that the half-disk $D_{y}$ is internally tangent to the quarter-circle. Since $\overline{O C}$ is a radius of the quarter-circle, it is perpendicular to the tangent of the quarter-circle at $C$. Since $\overline{M C}$ is a radius of the half-disk, it is perpendicular to the tangent of the half-disk at $C$. Then the tangents lines of the half-disk and the quarter-circle coincide, and the half-disk is tangent to the quarter-circle. It is obvious from the diagram that the half-disk lies at least partially inside of the quarter-circle, the half-disk $D_{y}$ is internally tangent to the quarter-circle.
Then the union of the half-disks is be a quarter-circle with radius 2 , and has area $\pi$.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. [4]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-geo-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $A B C D$ be an isosceles trapezoid such that $A B=10, B C=15, C D=28$, and $D A=15$. There is a point $E$ such that $\triangle A E D$ and $\triangle A E B$ have the same area and such that $E C$ is minimal. Find $E C$.
|
$\frac{216}{\sqrt{145}}$
Geometry Subject Test
The locus of points $E$ such that $[A E D]=[A E B]$ forms a line, since area is a linear function of the coordinates of $E$; setting the areas equal gives a linear equation in the coordinates $E^{2}$. Note that $A$ and $M$, the midpoint of $\overline{D B}$, are on this line; $A$ because both areas are 0 , and $M$ because the triangles share an altitude, and bases $\overline{D M}$ and $\overline{M B}$ are equal in length. Then $\overline{A M}$ is the set of points satisfying the area condition. The point $E$, then, is such that $\triangle A E C$ is a right angle (to make the distance minimal) and $E$ lies on $\overline{A M}$.
Let $X$ be the point of intersection of $\overline{A M}$ and $\overline{C D}$. Then $\triangle A M B \sim \triangle X M D$, and since $M D=B M$, they are in fact congruent. Thus $D X=A B=10$, and $X C=18$. Similarly, $B X=15$, so $A B X D$ is a parallelogram. Let $Y$ be the foot of the perpendicular from $A$ to $\overline{D C}$, so that $D Y=\frac{D C-A B}{2}=9$. Then $A Y=\sqrt{A D^{2}-D Y^{2}}=\sqrt{225-81}=12$. Then $Y X=D X-D Y=1$ and $A X=\sqrt{A Y^{2}+Y X^{2}}=$ $\sqrt{144+1}=\sqrt{145}$. Since both $\triangle A X Y$ and $\triangle C X E$ have a right angle, and $\angle E X C$ and $\angle Y X A$ are congruent because they are vertical angles, $\triangle A X Y \sim \triangle C X E$. Then $\frac{C E}{A Y}=\frac{C X}{A X}$, so $C E=12 \cdot \frac{18}{\sqrt{145}}=$ $\frac{216}{\sqrt{145}}$.
|
\frac{216}{\sqrt{145}}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C D$ be an isosceles trapezoid such that $A B=10, B C=15, C D=28$, and $D A=15$. There is a point $E$ such that $\triangle A E D$ and $\triangle A E B$ have the same area and such that $E C$ is minimal. Find $E C$.
|
$\frac{216}{\sqrt{145}}$
Geometry Subject Test
The locus of points $E$ such that $[A E D]=[A E B]$ forms a line, since area is a linear function of the coordinates of $E$; setting the areas equal gives a linear equation in the coordinates $E^{2}$. Note that $A$ and $M$, the midpoint of $\overline{D B}$, are on this line; $A$ because both areas are 0 , and $M$ because the triangles share an altitude, and bases $\overline{D M}$ and $\overline{M B}$ are equal in length. Then $\overline{A M}$ is the set of points satisfying the area condition. The point $E$, then, is such that $\triangle A E C$ is a right angle (to make the distance minimal) and $E$ lies on $\overline{A M}$.
Let $X$ be the point of intersection of $\overline{A M}$ and $\overline{C D}$. Then $\triangle A M B \sim \triangle X M D$, and since $M D=B M$, they are in fact congruent. Thus $D X=A B=10$, and $X C=18$. Similarly, $B X=15$, so $A B X D$ is a parallelogram. Let $Y$ be the foot of the perpendicular from $A$ to $\overline{D C}$, so that $D Y=\frac{D C-A B}{2}=9$. Then $A Y=\sqrt{A D^{2}-D Y^{2}}=\sqrt{225-81}=12$. Then $Y X=D X-D Y=1$ and $A X=\sqrt{A Y^{2}+Y X^{2}}=$ $\sqrt{144+1}=\sqrt{145}$. Since both $\triangle A X Y$ and $\triangle C X E$ have a right angle, and $\angle E X C$ and $\angle Y X A$ are congruent because they are vertical angles, $\triangle A X Y \sim \triangle C X E$. Then $\frac{C E}{A Y}=\frac{C X}{A X}$, so $C E=12 \cdot \frac{18}{\sqrt{145}}=$ $\frac{216}{\sqrt{145}}$.
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [4]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-geo-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
A sphere is the set of points at a fixed positive distance $r$ from its center. Let $\mathcal{S}$ be a set of 2010dimensional spheres. Suppose that the number of points lying on every element of $\mathcal{S}$ is a finite number $n$. Find the maximum possible value of $n$.
|
2 The answer is 2 for any number of dimensions. We prove this by induction on the dimension.
Note that 1-dimensional spheres are pairs of points, and 2-dimensional spheres are circles.
Base case, $d=2$ : The intersection of two circles is either a circle (if the original circles are identical, and in the same place), a pair of points, a single point (if the circles are tangent), or the empty set. Thus, in dimension 2, the largest finite number of intersection points is 2 , because the number of pairwise intersection points is 0,1 , or 2 for distinct circles.
We now prove that the intersection of two $k$-dimensional spheres is either the empty set, a $(k-1)$ dimensional sphere, a $k$-dimensional sphere (which only occurs if the original spheres are identical and coincident). Consider two spheres in $k$-dimensional space, and impose a coordinate system such that the centers of the two spheres lie on one coordinate axis. Then the equations for the two spheres become identical in all but one coordinate:
$$
\begin{aligned}
& \left(x_{1}-a_{1}\right)^{2}+x_{2}^{2}+\ldots+x_{k}^{2}=r_{1}^{2} \\
& \left(x_{1}-a_{2}\right)^{2}+x_{2}^{2}+\ldots+x_{k}^{2}=r_{2}^{2}
\end{aligned}
$$
If $a_{1}=a_{2}$, the spheres are concentric, and so they are either nonintersecting or coincident, intersecting in a $k$-dimensional sphere. If $a_{1} \neq a_{2}$, then subtracting the equations and solving for $x_{1}$ yields $x_{1}=\frac{r_{1}^{2}-a_{1}^{2}-r_{2}^{2}+a_{2}^{2}}{2\left(a_{2}-a_{2}\right)}$. Plugging this in to either equation above yields a single equation equation that describes a $(k-1)$-dimensional sphere.
Assume we are in dimension $d$, and suppose for induction that for all $k$ less than $d$, any two distinct $k$-dimensional spheres intersecting in a finite number of points intersect in at most two points. Suppose we have a collection of $d$-dimensional spheres $s_{1}, s_{2}, \ldots, s_{m}$. Without loss of generality, suppose the $s_{i}$ are distinct. Let $t_{i}$ be the intersection of $s_{i}$ and $s_{i+1}$ for $1 \leq i<m$. If any $t_{i}$ are the empty set, then the intersection of the $t_{i}$ is empty. None of the $t_{i}$ is a $d$-dimensional sphere because the $s_{i}$ are distinct. Thus each of $t_{1}, t_{2}, \ldots, t_{m-1}$ is a $(d-1)$-dimensional sphere, and the intersection of all of them is the same as the intersection of the $d$-dimensional spheres. We can then apply the inductive hypothesis to find that $t_{1}, \ldots, t_{m-1}$ intersect in at most two points. Thus, by induction, a set of spheres in any dimension which intersect at only finitely many points intersect at at most two points.
We now exhibit a set of $2^{2009} 2010$-dimensional spheres, and prove that their intersection contains exactly two points. Take the spheres with radii $\sqrt{2013}$ and centers $(0, \pm 1, \pm 1, \ldots, \pm 1)$, where the sign of each coordinate is independent from the sign of every other coordinate. Because of our choice of radius, all these spheres pass through the points $( \pm 2,0,0, \ldots 0)$. Then the intersection is the set of
[^1]points $\left(x_{1}, x_{2}, \ldots, x_{2010}\right)$ which satisfy the equations $x_{1}^{2}+\left(x_{2} \pm 1\right)^{2}+\cdots+\left(x_{2010} \pm 1\right)^{2}=2013$. The only solutions to these equations are the points $( \pm 2,0,0, \ldots, 0)$ (since $\left(x_{i}+1\right)^{2}$ must be the same as $\left(x_{i}-1\right)^{2}$ for all $i>1$, because we may hold all but one of the $\pm$ choices constant, and change the remaining one).
|
2
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A sphere is the set of points at a fixed positive distance $r$ from its center. Let $\mathcal{S}$ be a set of 2010dimensional spheres. Suppose that the number of points lying on every element of $\mathcal{S}$ is a finite number $n$. Find the maximum possible value of $n$.
|
2 The answer is 2 for any number of dimensions. We prove this by induction on the dimension.
Note that 1-dimensional spheres are pairs of points, and 2-dimensional spheres are circles.
Base case, $d=2$ : The intersection of two circles is either a circle (if the original circles are identical, and in the same place), a pair of points, a single point (if the circles are tangent), or the empty set. Thus, in dimension 2, the largest finite number of intersection points is 2 , because the number of pairwise intersection points is 0,1 , or 2 for distinct circles.
We now prove that the intersection of two $k$-dimensional spheres is either the empty set, a $(k-1)$ dimensional sphere, a $k$-dimensional sphere (which only occurs if the original spheres are identical and coincident). Consider two spheres in $k$-dimensional space, and impose a coordinate system such that the centers of the two spheres lie on one coordinate axis. Then the equations for the two spheres become identical in all but one coordinate:
$$
\begin{aligned}
& \left(x_{1}-a_{1}\right)^{2}+x_{2}^{2}+\ldots+x_{k}^{2}=r_{1}^{2} \\
& \left(x_{1}-a_{2}\right)^{2}+x_{2}^{2}+\ldots+x_{k}^{2}=r_{2}^{2}
\end{aligned}
$$
If $a_{1}=a_{2}$, the spheres are concentric, and so they are either nonintersecting or coincident, intersecting in a $k$-dimensional sphere. If $a_{1} \neq a_{2}$, then subtracting the equations and solving for $x_{1}$ yields $x_{1}=\frac{r_{1}^{2}-a_{1}^{2}-r_{2}^{2}+a_{2}^{2}}{2\left(a_{2}-a_{2}\right)}$. Plugging this in to either equation above yields a single equation equation that describes a $(k-1)$-dimensional sphere.
Assume we are in dimension $d$, and suppose for induction that for all $k$ less than $d$, any two distinct $k$-dimensional spheres intersecting in a finite number of points intersect in at most two points. Suppose we have a collection of $d$-dimensional spheres $s_{1}, s_{2}, \ldots, s_{m}$. Without loss of generality, suppose the $s_{i}$ are distinct. Let $t_{i}$ be the intersection of $s_{i}$ and $s_{i+1}$ for $1 \leq i<m$. If any $t_{i}$ are the empty set, then the intersection of the $t_{i}$ is empty. None of the $t_{i}$ is a $d$-dimensional sphere because the $s_{i}$ are distinct. Thus each of $t_{1}, t_{2}, \ldots, t_{m-1}$ is a $(d-1)$-dimensional sphere, and the intersection of all of them is the same as the intersection of the $d$-dimensional spheres. We can then apply the inductive hypothesis to find that $t_{1}, \ldots, t_{m-1}$ intersect in at most two points. Thus, by induction, a set of spheres in any dimension which intersect at only finitely many points intersect at at most two points.
We now exhibit a set of $2^{2009} 2010$-dimensional spheres, and prove that their intersection contains exactly two points. Take the spheres with radii $\sqrt{2013}$ and centers $(0, \pm 1, \pm 1, \ldots, \pm 1)$, where the sign of each coordinate is independent from the sign of every other coordinate. Because of our choice of radius, all these spheres pass through the points $( \pm 2,0,0, \ldots 0)$. Then the intersection is the set of
[^1]points $\left(x_{1}, x_{2}, \ldots, x_{2010}\right)$ which satisfy the equations $x_{1}^{2}+\left(x_{2} \pm 1\right)^{2}+\cdots+\left(x_{2010} \pm 1\right)^{2}=2013$. The only solutions to these equations are the points $( \pm 2,0,0, \ldots, 0)$ (since $\left(x_{i}+1\right)^{2}$ must be the same as $\left(x_{i}-1\right)^{2}$ for all $i>1$, because we may hold all but one of the $\pm$ choices constant, and change the remaining one).
|
{
"exam": "HMMT",
"problem_label": "5",
"problem_match": "\n5. [4]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-geo-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Three unit circles $\omega_{1}, \omega_{2}$, and $\omega_{3}$ in the plane have the property that each circle passes through the centers of the other two. A square $S$ surrounds the three circles in such a way that each of its four sides is tangent to at least one of $\omega_{1}, \omega_{2}$ and $\omega_{3}$. Find the side length of the square $S$.
|
$\frac{\sqrt{6}+\sqrt{2}+8}{4}$
By the Pigeonhole Principle, two of the sides must be tangent to the same circle, say $\omega_{1}$. Since $S$ surrounds the circles, these two sides must be adjacent, so we can let $A$ denote the common vertex of the two sides tangent to $\omega_{1}$. Let $B, C$, and $D$ be the other vertices of $S$ in clockwise order, and let $P, Q$, and $R$ be the centers of $\omega_{1}, \omega_{2}$, and $\omega_{3}$ respectively, and suppose WLOG that they are also in clockwise order. Then $A C$ passes through the center of $\omega_{1}$, and by symmetry (since $A B=A D$ ) it must also pass through the other intersection point of $\omega_{2}$ and $\omega_{3}$. That is, $A C$ is the radical axis of $\omega_{2}$ and $\omega_{3}$.
Now, let $M$ and $N$ be the feet of the perpendiculars from $P$ and $R$, respectively, to side $A D$. Let $E$ and $F$ be the feet of the perpendiculars from $P$ to $A B$ and from $R$ to $D C$, respectively. Then $P E A M$ and $N R F D$ are rectangles, and $P E$ and $R F$ are radii of $\omega_{1}$ and $\omega_{2}$ respectively. Thus $A M=E P=1$ and $N D=R F=1$. Finally, we have
$$
\begin{aligned}
M N & =P R \cdot \cos \left(180^{\circ}-\angle E P R\right) \\
& =\cos \left(180^{\circ}-E P Q-R P Q\right) \\
& =-\cos \left(\left(270^{\circ}-60^{\circ}\right) / 2+60^{\circ}\right) \\
& =-\cos \left(165^{\circ}\right) \\
& =\cos \left(15^{\circ}\right) \\
& =\frac{\sqrt{6}+\sqrt{2}}{4}
\end{aligned}
$$
Thus $A D=A M+M N+N D=1+\frac{\sqrt{6}+\sqrt{2}}{4}+1=\frac{\sqrt{6}+\sqrt{2}+8}{4}$ as claimed.
|
\frac{\sqrt{6}+\sqrt{2}+8}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Three unit circles $\omega_{1}, \omega_{2}$, and $\omega_{3}$ in the plane have the property that each circle passes through the centers of the other two. A square $S$ surrounds the three circles in such a way that each of its four sides is tangent to at least one of $\omega_{1}, \omega_{2}$ and $\omega_{3}$. Find the side length of the square $S$.
|
$\frac{\sqrt{6}+\sqrt{2}+8}{4}$
By the Pigeonhole Principle, two of the sides must be tangent to the same circle, say $\omega_{1}$. Since $S$ surrounds the circles, these two sides must be adjacent, so we can let $A$ denote the common vertex of the two sides tangent to $\omega_{1}$. Let $B, C$, and $D$ be the other vertices of $S$ in clockwise order, and let $P, Q$, and $R$ be the centers of $\omega_{1}, \omega_{2}$, and $\omega_{3}$ respectively, and suppose WLOG that they are also in clockwise order. Then $A C$ passes through the center of $\omega_{1}$, and by symmetry (since $A B=A D$ ) it must also pass through the other intersection point of $\omega_{2}$ and $\omega_{3}$. That is, $A C$ is the radical axis of $\omega_{2}$ and $\omega_{3}$.
Now, let $M$ and $N$ be the feet of the perpendiculars from $P$ and $R$, respectively, to side $A D$. Let $E$ and $F$ be the feet of the perpendiculars from $P$ to $A B$ and from $R$ to $D C$, respectively. Then $P E A M$ and $N R F D$ are rectangles, and $P E$ and $R F$ are radii of $\omega_{1}$ and $\omega_{2}$ respectively. Thus $A M=E P=1$ and $N D=R F=1$. Finally, we have
$$
\begin{aligned}
M N & =P R \cdot \cos \left(180^{\circ}-\angle E P R\right) \\
& =\cos \left(180^{\circ}-E P Q-R P Q\right) \\
& =-\cos \left(\left(270^{\circ}-60^{\circ}\right) / 2+60^{\circ}\right) \\
& =-\cos \left(165^{\circ}\right) \\
& =\cos \left(15^{\circ}\right) \\
& =\frac{\sqrt{6}+\sqrt{2}}{4}
\end{aligned}
$$
Thus $A D=A M+M N+N D=1+\frac{\sqrt{6}+\sqrt{2}}{4}+1=\frac{\sqrt{6}+\sqrt{2}+8}{4}$ as claimed.
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n6. [5]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-geo-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
You are standing in an infinitely long hallway with sides given by the lines $x=0$ and $x=6$. You start at $(3,0)$ and want to get to $(3,6)$. Furthermore, at each instant you want your distance to $(3,6)$ to either decrease or stay the same. What is the area of the set of points that you could pass through on your journey from $(3,0)$ to $(3,6)$ ?
|
$9 \sqrt{3}+\frac{21 \pi}{2}$
If you draw concentric circles around the destination point, the condition is equivalent to the restriction that you must always go inwards towards the destination. In the diagram above, the regions through which you might pass are shaded.
We find the areas of regions A, B, and C separately, and add them up (doubling the area of region A, because there are two of them).
The hypotenuse of triangle A is of length 6 , and the base is of length 3 , so it is a $\frac{\pi}{6}-\frac{\pi}{3}-\frac{\pi}{2}$ triangle (30-60-90 triangle) with area $\frac{9 \sqrt{3}}{2}$. Then the total area of the regions labeled A is $9 \sqrt{3}$.
Since the angle of triangle A nearest the center of the circle (the destination point) is $\frac{\pi}{3}$, sector B has central angle $\frac{\pi}{3}$. Then the area of sector B is $\frac{1}{2} r^{2} \theta=\frac{1}{2} \cdot 36 \cdot \frac{\pi}{3}=6 \pi$.
Region C is a half-disc of radius 3 , so its area is $\frac{9 \pi}{2}$.
Thus, the total area is $9 \sqrt{3}+\frac{21 \pi}{2}$.
|
9 \sqrt{3}+\frac{21 \pi}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
You are standing in an infinitely long hallway with sides given by the lines $x=0$ and $x=6$. You start at $(3,0)$ and want to get to $(3,6)$. Furthermore, at each instant you want your distance to $(3,6)$ to either decrease or stay the same. What is the area of the set of points that you could pass through on your journey from $(3,0)$ to $(3,6)$ ?
|
$9 \sqrt{3}+\frac{21 \pi}{2}$
If you draw concentric circles around the destination point, the condition is equivalent to the restriction that you must always go inwards towards the destination. In the diagram above, the regions through which you might pass are shaded.
We find the areas of regions A, B, and C separately, and add them up (doubling the area of region A, because there are two of them).
The hypotenuse of triangle A is of length 6 , and the base is of length 3 , so it is a $\frac{\pi}{6}-\frac{\pi}{3}-\frac{\pi}{2}$ triangle (30-60-90 triangle) with area $\frac{9 \sqrt{3}}{2}$. Then the total area of the regions labeled A is $9 \sqrt{3}$.
Since the angle of triangle A nearest the center of the circle (the destination point) is $\frac{\pi}{3}$, sector B has central angle $\frac{\pi}{3}$. Then the area of sector B is $\frac{1}{2} r^{2} \theta=\frac{1}{2} \cdot 36 \cdot \frac{\pi}{3}=6 \pi$.
Region C is a half-disc of radius 3 , so its area is $\frac{9 \pi}{2}$.
Thus, the total area is $9 \sqrt{3}+\frac{21 \pi}{2}$.
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [6]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-geo-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $O$ be the point $(0,0)$. Let $A, B, C$ be three points in the plane such that $A O=15, B O=15$, and $C O=7$, and such that the area of triangle $A B C$ is maximal. What is the length of the shortest side of $A B C$ ?
|
20 We claim that $O$ should be the orthocenter of the triangle $A B C$. If $O$ is not on an altitude of $\triangle A B C$, suppose (without loss of generality) that $\overline{A O}$ is not perpendicular to $\overline{B C}$. We can
rotate $A$ around $O$, leaving $B$ and $C$ fixed, to make $\overline{A O}$ perpendicular to $\overline{B C}$, which strictly increases the area. Therefore, if $[A B C]$ is maximal then $\triangle A B C$ is an isosceles triangle with orthocenter $O$ and base $\overline{A B}$.
Let $F$ be the foot of the perpendicular from $C$ to $\overline{A B}$. Since $\angle F O A$ and $\angle C O E$ are vertical, $\angle F A O=$ $\angle O C E$. Then $\triangle F A O$ is similar to $\triangle F C B$, so we have $\frac{A F}{O F}=\frac{C F}{B F}=\frac{O F+7}{A F}$, so $A F^{2}=O F^{2}+7 \cdot O F$. Since $A F^{2}=225-O F^{2}, 2 \cdot O F^{2}+7 \cdot O F-225=0$, so $O F=9$. Then $A F=12$, so $A B=24$ and $B C=20$. Thus, the length of the shortest side of $\triangle A B C$ is 20 .
|
20
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $O$ be the point $(0,0)$. Let $A, B, C$ be three points in the plane such that $A O=15, B O=15$, and $C O=7$, and such that the area of triangle $A B C$ is maximal. What is the length of the shortest side of $A B C$ ?
|
20 We claim that $O$ should be the orthocenter of the triangle $A B C$. If $O$ is not on an altitude of $\triangle A B C$, suppose (without loss of generality) that $\overline{A O}$ is not perpendicular to $\overline{B C}$. We can
rotate $A$ around $O$, leaving $B$ and $C$ fixed, to make $\overline{A O}$ perpendicular to $\overline{B C}$, which strictly increases the area. Therefore, if $[A B C]$ is maximal then $\triangle A B C$ is an isosceles triangle with orthocenter $O$ and base $\overline{A B}$.
Let $F$ be the foot of the perpendicular from $C$ to $\overline{A B}$. Since $\angle F O A$ and $\angle C O E$ are vertical, $\angle F A O=$ $\angle O C E$. Then $\triangle F A O$ is similar to $\triangle F C B$, so we have $\frac{A F}{O F}=\frac{C F}{B F}=\frac{O F+7}{A F}$, so $A F^{2}=O F^{2}+7 \cdot O F$. Since $A F^{2}=225-O F^{2}, 2 \cdot O F^{2}+7 \cdot O F-225=0$, so $O F=9$. Then $A F=12$, so $A B=24$ and $B C=20$. Thus, the length of the shortest side of $\triangle A B C$ is 20 .
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [6]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-geo-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $A B C D$ be a quadrilateral with an inscribed circle centered at $I$. Let $C I$ intersect $A B$ at $E$. If $\angle I D E=35^{\circ}, \angle A B C=70^{\circ}$, and $\angle B C D=60^{\circ}$, then what are all possible measures of $\angle C D A$ ?
|
$70^{\circ}$ and $160^{\circ}$
Arbitrarily defining $B$ and $C$ determines $I$ and $E$ up to reflections across $B C$. D lies on both the circle determined by $\angle E D I=35^{\circ}$ and the line through $C$ tangent to the circle (and on the opposite side of $B)$; since the intersection of a line and a circle has at most two points, there are only two cases for $A B C D$. The diagram below on the left shows the construction made in this solution, containing both cases. The diagram below on the right shows only the degenerate case.
Reflect $B$ across $E C$ to $B^{\prime}$; then $C B=C B^{\prime}$. Since $B A$ and $B C$ are tangent to the circle centered at $I, I B$ is the angle bisector of $\angle A B C$. Then $\angle I B E=\angle I B^{\prime} E=35^{\circ}$. If $B^{\prime}=D$, then $\angle A D C=$ $\angle E B^{\prime} C=70^{\circ}$. Otherwise, since $\angle I B^{\prime} E=35^{\circ}=\angle I D E$ (given), $E B^{\prime} D I$ is a cyclic quadrilateral. Then $\angle I E D=\angle I B^{\prime} D=35^{\circ}$ and $\angle B C I=\angle E C D=30^{\circ}$, so $\triangle C E D \sim \triangle C B I$.
Since $\angle C I D$ is exterior to $\triangle D I E, \angle C I D=\angle I D E+\angle D E I=70^{\circ}$. Then $\triangle C D I \sim \triangle C E B$. Because $E B^{\prime} D I$ is cyclic, $\angle I D C=\angle I E B^{\prime}=\angle I E B=180^{\circ}-70^{\circ}-30^{\circ}=80^{\circ}$. Then $\angle A D C=2 \angle I D C=160^{\circ}$.
Thus, the two possible measures are $70^{\circ}$ and $160^{\circ}$.
|
70^{\circ} \text{ and } 160^{\circ}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C D$ be a quadrilateral with an inscribed circle centered at $I$. Let $C I$ intersect $A B$ at $E$. If $\angle I D E=35^{\circ}, \angle A B C=70^{\circ}$, and $\angle B C D=60^{\circ}$, then what are all possible measures of $\angle C D A$ ?
|
$70^{\circ}$ and $160^{\circ}$
Arbitrarily defining $B$ and $C$ determines $I$ and $E$ up to reflections across $B C$. D lies on both the circle determined by $\angle E D I=35^{\circ}$ and the line through $C$ tangent to the circle (and on the opposite side of $B)$; since the intersection of a line and a circle has at most two points, there are only two cases for $A B C D$. The diagram below on the left shows the construction made in this solution, containing both cases. The diagram below on the right shows only the degenerate case.
Reflect $B$ across $E C$ to $B^{\prime}$; then $C B=C B^{\prime}$. Since $B A$ and $B C$ are tangent to the circle centered at $I, I B$ is the angle bisector of $\angle A B C$. Then $\angle I B E=\angle I B^{\prime} E=35^{\circ}$. If $B^{\prime}=D$, then $\angle A D C=$ $\angle E B^{\prime} C=70^{\circ}$. Otherwise, since $\angle I B^{\prime} E=35^{\circ}=\angle I D E$ (given), $E B^{\prime} D I$ is a cyclic quadrilateral. Then $\angle I E D=\angle I B^{\prime} D=35^{\circ}$ and $\angle B C I=\angle E C D=30^{\circ}$, so $\triangle C E D \sim \triangle C B I$.
Since $\angle C I D$ is exterior to $\triangle D I E, \angle C I D=\angle I D E+\angle D E I=70^{\circ}$. Then $\triangle C D I \sim \triangle C E B$. Because $E B^{\prime} D I$ is cyclic, $\angle I D C=\angle I E B^{\prime}=\angle I E B=180^{\circ}-70^{\circ}-30^{\circ}=80^{\circ}$. Then $\angle A D C=2 \angle I D C=160^{\circ}$.
Thus, the two possible measures are $70^{\circ}$ and $160^{\circ}$.
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [7]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-geo-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$. Segment $P Q$ is tangent to $\omega_{1}$ at $P$ and to $\omega_{2}$ at $Q$, and $A$ is closer to $P Q$ than $B$. Point $X$ is on $\omega_{1}$ such that $P X \| Q B$, and point $Y$ is on $\omega_{2}$ such that $Q Y \| P B$. Given that $\angle A P Q=30^{\circ}$ and $\angle P Q A=15^{\circ}$, find the ratio $A X / A Y$.
|
$2-\sqrt{3}$
Let $C$ be the fourth vertex of parallelogram $A P C Q$. The midpoint $M$ of $\overline{P Q}$ is the intersection of the diagonals of this parallelogram. Because $M$ has equal power ${ }^{3}$ with respect to the two circles $\omega_{1}$ and $\omega_{2}$, it lies on $\overleftrightarrow{A B}$, the circles' radical axis $4^{4}$. Therefore, $C$ lies on $\overleftrightarrow{A B}$ as well.
Using a series of parallel lines and inscribed arcs, we have:
$$
\angle A P C=\angle A P Q+\angle C P Q=\angle A P Q+\angle P Q A=\angle A B P+\angle Q B A=\angle P B Q=\angle X P B
$$
where the last equality follows from the fact that $P X \| Q B$.
[^2]We also know that $\angle B X P=180^{\circ}-\angle P A B=\angle C A P$, so triangles $B X P$ and $C A P$ are similar. By the spiral similarity theorem triangles $B P C$ and $X P A$ are similar, too.
By analogous reasoning, triangles $B Q C$ and $Y Q A$ are similar. Then we have:
$$
\frac{A X}{A Y}=\frac{A X / B C}{A Y / B C}=\frac{A P / C P}{A Q / C Q}=\frac{A P^{2}}{A Q^{2}}
$$
where the last inequality holds because $A P C Q$ is a parallelogram. Using the Law of Sines, the last expression equals $\frac{\sin ^{2} 15^{\circ}}{\sin ^{2} 30^{\circ}}=2-\sqrt{3}$.
[^3]
[^0]: ${ }^{1}$ http://en.wikipedia.org/wiki/Inscribed_angle_theorem
[^1]: ${ }^{2}$ The notation $\left[A_{1} A_{2} \ldots A_{n}\right]$ means the area of polygon $A_{1} A_{2} \ldots A_{n}$.
[^2]: $\sqrt[3]{ }$ http://en.wikipedia.org/wiki/Power_of_a_point
4 http://en.wikipedia.org/wiki/Radical_axis
[^3]: ${ }^{5}$ This theorem states that if $\triangle P A B$ and $\triangle P X Y$ are similar and oriented the same way, then $\triangle P A X$ and $\triangle P B Y$ are similar too. It is true because the first similarity implies that $A P / B P=X P / Y P$ and $\angle A P B=\angle X P Y$, which proves the second similarity.
|
2-\sqrt{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$. Segment $P Q$ is tangent to $\omega_{1}$ at $P$ and to $\omega_{2}$ at $Q$, and $A$ is closer to $P Q$ than $B$. Point $X$ is on $\omega_{1}$ such that $P X \| Q B$, and point $Y$ is on $\omega_{2}$ such that $Q Y \| P B$. Given that $\angle A P Q=30^{\circ}$ and $\angle P Q A=15^{\circ}$, find the ratio $A X / A Y$.
|
$2-\sqrt{3}$
Let $C$ be the fourth vertex of parallelogram $A P C Q$. The midpoint $M$ of $\overline{P Q}$ is the intersection of the diagonals of this parallelogram. Because $M$ has equal power ${ }^{3}$ with respect to the two circles $\omega_{1}$ and $\omega_{2}$, it lies on $\overleftrightarrow{A B}$, the circles' radical axis $4^{4}$. Therefore, $C$ lies on $\overleftrightarrow{A B}$ as well.
Using a series of parallel lines and inscribed arcs, we have:
$$
\angle A P C=\angle A P Q+\angle C P Q=\angle A P Q+\angle P Q A=\angle A B P+\angle Q B A=\angle P B Q=\angle X P B
$$
where the last equality follows from the fact that $P X \| Q B$.
[^2]We also know that $\angle B X P=180^{\circ}-\angle P A B=\angle C A P$, so triangles $B X P$ and $C A P$ are similar. By the spiral similarity theorem triangles $B P C$ and $X P A$ are similar, too.
By analogous reasoning, triangles $B Q C$ and $Y Q A$ are similar. Then we have:
$$
\frac{A X}{A Y}=\frac{A X / B C}{A Y / B C}=\frac{A P / C P}{A Q / C Q}=\frac{A P^{2}}{A Q^{2}}
$$
where the last inequality holds because $A P C Q$ is a parallelogram. Using the Law of Sines, the last expression equals $\frac{\sin ^{2} 15^{\circ}}{\sin ^{2} 30^{\circ}}=2-\sqrt{3}$.
[^3]
[^0]: ${ }^{1}$ http://en.wikipedia.org/wiki/Inscribed_angle_theorem
[^1]: ${ }^{2}$ The notation $\left[A_{1} A_{2} \ldots A_{n}\right]$ means the area of polygon $A_{1} A_{2} \ldots A_{n}$.
[^2]: $\sqrt[3]{ }$ http://en.wikipedia.org/wiki/Power_of_a_point
4 http://en.wikipedia.org/wiki/Radical_axis
[^3]: ${ }^{5}$ This theorem states that if $\triangle P A B$ and $\triangle P X Y$ are similar and oriented the same way, then $\triangle P A X$ and $\triangle P B Y$ are similar too. It is true because the first similarity implies that $A P / B P=X P / Y P$ and $\angle A P B=\angle X P Y$, which proves the second similarity.
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. [8]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-geo-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
If $A=10^{9}-987654321$ and $B=\frac{123456789+1}{10}$, what is the value of $\sqrt{A B}$ ?
|
12345679 Both $A$ and $B$ equal 12345679, so $\sqrt{A B}=12345679$ as well.
|
12345679
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
If $A=10^{9}-987654321$ and $B=\frac{123456789+1}{10}$, what is the value of $\sqrt{A B}$ ?
|
12345679 Both $A$ and $B$ equal 12345679, so $\sqrt{A B}=12345679$ as well.
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [4]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Suppose that $x, y$, and $z$ are non-negative real numbers such that $x+y+z=1$. What is the maximum possible value of $x+y^{2}+z^{3}$ ?
|
1 Since $0 \leq y, z \leq 1$, we have $y^{2} \leq y$ and $z^{3} \leq z$. Therefore $x+y^{2}+z^{3} \leq x+y+z=1$. We can get $x+y^{2}+z^{3}=1$ by setting $(x, y, z)=(1,0,0)$.
|
1
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose that $x, y$, and $z$ are non-negative real numbers such that $x+y+z=1$. What is the maximum possible value of $x+y^{2}+z^{3}$ ?
|
1 Since $0 \leq y, z \leq 1$, we have $y^{2} \leq y$ and $z^{3} \leq z$. Therefore $x+y^{2}+z^{3} \leq x+y+z=1$. We can get $x+y^{2}+z^{3}=1$ by setting $(x, y, z)=(1,0,0)$.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [4]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
In a group of people, there are 13 who like apples, 9 who like blueberries, 15 who like cantaloupe, and 6 who like dates. (A person can like more than 1 kind of fruit.) Each person who likes blueberries also likes exactly one of apples and cantaloupe. Each person who likes cantaloupe also likes exactly one of blueberries and dates. Find the minimum possible number of people in the group.
|
22 Everyone who likes cantaloupe likes exactly one of blueberries and dates. However, there are 15 people who like cantaloupe, 9 who like blueberries, and 6 who like dates. Thus, everyone who likes blueberries or dates must also like cantaloupes (because if any of them didn't, we would end up with less than 15 people who like cantaloupe).
Since everyone who likes blueberries likes cantaloupes, none of them can like apples. However, the 6 people who like both cantaloupe and dates can also like apples. So, we could have a group where 7 people like apples alone, 9 like blueberries and cantaloupe, and 6 like apples, cantaloupe, and dates. This gives 22 people in the group, which is optimal.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
22
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In a group of people, there are 13 who like apples, 9 who like blueberries, 15 who like cantaloupe, and 6 who like dates. (A person can like more than 1 kind of fruit.) Each person who likes blueberries also likes exactly one of apples and cantaloupe. Each person who likes cantaloupe also likes exactly one of blueberries and dates. Find the minimum possible number of people in the group.
|
22 Everyone who likes cantaloupe likes exactly one of blueberries and dates. However, there are 15 people who like cantaloupe, 9 who like blueberries, and 6 who like dates. Thus, everyone who likes blueberries or dates must also like cantaloupes (because if any of them didn't, we would end up with less than 15 people who like cantaloupe).
Since everyone who likes blueberries likes cantaloupes, none of them can like apples. However, the 6 people who like both cantaloupe and dates can also like apples. So, we could have a group where 7 people like apples alone, 9 like blueberries and cantaloupe, and 6 like apples, cantaloupe, and dates. This gives 22 people in the group, which is optimal.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. [4]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
To set up for a Fourth of July party, David is making a string of red, white, and blue balloons. He places them according to the following rules:
- No red balloon is adjacent to another red balloon.
- White balloons appear in groups of exactly two, and groups of white balloons are separated by at least two non-white balloons.
- Blue balloons appear in groups of exactly three, and groups of blue balloons are separated by at least three non-blue balloons.
If David uses over 600 balloons, determine the smallest number of red balloons that he can use.
|
99 It is possible to achieve 99 red balloons with the arrangement
$$
\text { WWBBBWW } \underbrace{\text { RBBBWWRBBBWW . . RBBBWW }}_{99 \text { RBBBWW's }},
$$
which contains $99 \cdot 6+7=601$ balloons.
Now assume that one can construct a chain with 98 or fewer red balloons. Then there can be 99 blocks of non-red balloons, which in total must contain more than 502 balloons. The only valid combinations of white and blue balloons are WWBBB, BBBWW, and WWBBBWW (Any others contain the subsequence BBBWWBBB , which is invalid). The sequence ... WWR must be followed by BBBWW; otherwise two groups of white balloons would be too close. Similarly, the sequence
RWW . . . must be preceded by WWBBB. It follows that WWBBBWW can be used at most once in a valid sequence, meaning that there can be at most $98 \cdot 5+7=497$ non-red balloons. Contradiction. Therefore the minimum is 99 red balloons. (Better if the party's outdoors; then we'd have 99 red balloons floating in the summer sky. :-p)
|
99
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
To set up for a Fourth of July party, David is making a string of red, white, and blue balloons. He places them according to the following rules:
- No red balloon is adjacent to another red balloon.
- White balloons appear in groups of exactly two, and groups of white balloons are separated by at least two non-white balloons.
- Blue balloons appear in groups of exactly three, and groups of blue balloons are separated by at least three non-blue balloons.
If David uses over 600 balloons, determine the smallest number of red balloons that he can use.
|
99 It is possible to achieve 99 red balloons with the arrangement
$$
\text { WWBBBWW } \underbrace{\text { RBBBWWRBBBWW . . RBBBWW }}_{99 \text { RBBBWW's }},
$$
which contains $99 \cdot 6+7=601$ balloons.
Now assume that one can construct a chain with 98 or fewer red balloons. Then there can be 99 blocks of non-red balloons, which in total must contain more than 502 balloons. The only valid combinations of white and blue balloons are WWBBB, BBBWW, and WWBBBWW (Any others contain the subsequence BBBWWBBB , which is invalid). The sequence ... WWR must be followed by BBBWW; otherwise two groups of white balloons would be too close. Similarly, the sequence
RWW . . . must be preceded by WWBBB. It follows that WWBBBWW can be used at most once in a valid sequence, meaning that there can be at most $98 \cdot 5+7=497$ non-red balloons. Contradiction. Therefore the minimum is 99 red balloons. (Better if the party's outdoors; then we'd have 99 red balloons floating in the summer sky. :-p)
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [5]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
You have a length of string and 7 beads in the 7 colors of the rainbow. You place the beads on the string as follows - you randomly pick a bead that you haven't used yet, then randomly add it to either the left end or the right end of the string. What is the probability that, at the end, the colors of the beads are the colors of the rainbow in order? (The string cannot be flipped, so the red bead must appear on the left side and the violet bead on the right side.)
|
$\frac{1}{5040}$ The threading method does not depend on the colors of the beads, so at the end all configurations are equally likely. Since there are $7!=5040$ configurations in total, the probability of any particular configuration is $\frac{1}{5040}$.
|
\frac{1}{5040}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
You have a length of string and 7 beads in the 7 colors of the rainbow. You place the beads on the string as follows - you randomly pick a bead that you haven't used yet, then randomly add it to either the left end or the right end of the string. What is the probability that, at the end, the colors of the beads are the colors of the rainbow in order? (The string cannot be flipped, so the red bead must appear on the left side and the violet bead on the right side.)
|
$\frac{1}{5040}$ The threading method does not depend on the colors of the beads, so at the end all configurations are equally likely. Since there are $7!=5040$ configurations in total, the probability of any particular configuration is $\frac{1}{5040}$.
|
{
"exam": "HMMT",
"problem_label": "5",
"problem_match": "\n5. [5]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
How many different numbers are obtainable from five 5 s by first concatenating some of the 5 s, then multiplying them together? For example, we could do $5 \cdot 55 \cdot 55,555 \cdot 55$, or 55555 , but not $5 \cdot 5$ or 2525 .
|
7 If we do 55555 , then we're done.
Note that $5,55,555$, and 5555 all have completely distinguishable prime factorizations. This means that if we are given a product of them, we can obtain the individual terms. The number of 5555's is the exponent of 101 , the number of 555 's is the exponent of 37 , the number of 55 's is the exponent of 11 minus the exponent of 101 , and the number of 5 's is just whatever we need to get the proper exponent of 5 . Then the answer is the number of ways we can split the five 5 's into groups of at least one. This is the number of unordered partitions of 5 , which is 7 .
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
7
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many different numbers are obtainable from five 5 s by first concatenating some of the 5 s, then multiplying them together? For example, we could do $5 \cdot 55 \cdot 55,555 \cdot 55$, or 55555 , but not $5 \cdot 5$ or 2525 .
|
7 If we do 55555 , then we're done.
Note that $5,55,555$, and 5555 all have completely distinguishable prime factorizations. This means that if we are given a product of them, we can obtain the individual terms. The number of 5555's is the exponent of 101 , the number of 555 's is the exponent of 37 , the number of 55 's is the exponent of 11 minus the exponent of 101 , and the number of 5 's is just whatever we need to get the proper exponent of 5 . Then the answer is the number of ways we can split the five 5 's into groups of at least one. This is the number of unordered partitions of 5 , which is 7 .
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n6. [5]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
What are the last 8 digits of
$$
11 \times 101 \times 1001 \times 10001 \times 100001 \times 1000001 \times 111 ?
$$
|
19754321 Multiply terms in a clever order.
$$
\begin{aligned}
11 \cdot 101 \cdot 10001 & =11,111,111 \\
111 \cdot 1001 \cdot 1000001 & =111,111,111,111
\end{aligned}
$$
The last eight digits of $11,111,111 \cdot 111,111,111,111$ are 87654321 . We then just need to compute the last 8 digits of $87654321 \cdot 100001=87654321+\ldots 32100000$, which are 19754321 .
|
19754321
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
What are the last 8 digits of
$$
11 \times 101 \times 1001 \times 10001 \times 100001 \times 1000001 \times 111 ?
$$
|
19754321 Multiply terms in a clever order.
$$
\begin{aligned}
11 \cdot 101 \cdot 10001 & =11,111,111 \\
111 \cdot 1001 \cdot 1000001 & =111,111,111,111
\end{aligned}
$$
The last eight digits of $11,111,111 \cdot 111,111,111,111$ are 87654321 . We then just need to compute the last 8 digits of $87654321 \cdot 100001=87654321+\ldots 32100000$, which are 19754321 .
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [6]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Each square in the following hexomino has side length 1. Find the minimum area of any rectangle that contains the entire hexomino.
|
$\frac{21}{2}$ If a rectangle contains the entire hexomino, it must also contain its convex hull, which is an origin-symmetric hexagon. It is fairly clear that the smallest rectangle that contains such a hexagon must share one set of parallel sides with the hexagon. There are three such rectangles, and checking them all, we find that the one shown below is the smallest. It has area $\frac{21}{2}$.
|
\frac{21}{2}
|
Incomplete
|
Yes
|
math-word-problem
|
Geometry
|
Each square in the following hexomino has side length 1. Find the minimum area of any rectangle that contains the entire hexomino.
|
$\frac{21}{2}$ If a rectangle contains the entire hexomino, it must also contain its convex hull, which is an origin-symmetric hexagon. It is fairly clear that the smallest rectangle that contains such a hexagon must share one set of parallel sides with the hexagon. There are three such rectangles, and checking them all, we find that the one shown below is the smallest. It has area $\frac{21}{2}$.
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [6]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Indecisive Andy starts out at the midpoint of the 1-unit-long segment $\overline{H T}$. He flips 2010 coins. On each flip, if the coin is heads, he moves halfway towards endpoint $H$, and if the coin is tails, he moves halfway towards endpoint $T$. After his 2010 moves, what is the expected distance between Andy and the midpoint of $\overline{H T}$ ?
|
| $\frac{1}{4}$ |
| :---: |
| Let Andy's position be $x$ units from the H end after 2009 flips. If Any moves towards | the $H$ end, he ends up at $\frac{x}{2}$, a distance of $\frac{1-x}{2}$ from the midpoint. If Andy moves towards the $T$ end, he ends up at $\frac{1+x}{2}$, a distance of $\frac{x}{2}$ from the midpoint. His expected distance from the midpoint is then
$$
\frac{\frac{1-x}{2}+\frac{x}{2}}{2}=\frac{1}{4} .
$$
Since this does not depend on $x, \frac{1}{4}$ is the answer.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND
|
\frac{1}{4}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Indecisive Andy starts out at the midpoint of the 1-unit-long segment $\overline{H T}$. He flips 2010 coins. On each flip, if the coin is heads, he moves halfway towards endpoint $H$, and if the coin is tails, he moves halfway towards endpoint $T$. After his 2010 moves, what is the expected distance between Andy and the midpoint of $\overline{H T}$ ?
|
| $\frac{1}{4}$ |
| :---: |
| Let Andy's position be $x$ units from the H end after 2009 flips. If Any moves towards | the $H$ end, he ends up at $\frac{x}{2}$, a distance of $\frac{1-x}{2}$ from the midpoint. If Andy moves towards the $T$ end, he ends up at $\frac{1+x}{2}$, a distance of $\frac{x}{2}$ from the midpoint. His expected distance from the midpoint is then
$$
\frac{\frac{1-x}{2}+\frac{x}{2}}{2}=\frac{1}{4} .
$$
Since this does not depend on $x, \frac{1}{4}$ is the answer.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [6]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $A B C$ be a triangle with $A B=8, B C=15$, and $A C=17$. Point $X$ is chosen at random on line segment $A B$. Point $Y$ is chosen at random on line segment $B C$. Point $Z$ is chosen at random on line segment $C A$. What is the expected area of triangle $X Y Z$ ?
|
15 Let $\mathbb{E}(X)$ denote the expected value of $X$, and let $[S]$ denote the area of $S$. Then
$$
\begin{aligned}
\mathbb{E}([\triangle X Y Z]) & =\mathbb{E}([\triangle A B C]-[\triangle X Y B]-[\triangle Z Y C]-[\triangle X B Z]) \\
& =[\triangle A B C]-\mathbb{E}([\triangle X Y B])-\mathbb{E}([\triangle Z Y C])-[\triangle X B Z])
\end{aligned}
$$
where the last step follows from linearity of expectation ${ }^{1}$. But $[\triangle X Y B]=\frac{1}{2} \cdot B X \cdot B Y \cdot \sin (B)$. The $\frac{1}{2} \sin (B)$ term is constant, and $B X$ and $B Y$ are both independent with expected values $\frac{A B}{2}$ and $\frac{B C}{2}$, respectively. Thus $\mathbb{E}([\triangle X Y B])=\frac{1}{8} A B \cdot B C \cdot \sin (B)=\frac{1}{4}[\triangle A B C]$. Similarly, $\mathbb{E}([\triangle Z Y C])=$ $\mathbb{E}([\triangle Z B X])=\frac{1}{4}[\triangle A B C]$.
Then we have $\mathbb{E}([\triangle X Y Z])=\left(1-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}\right)[\triangle A B C]=\frac{1}{4}[\triangle A B C]=15$.
Note: We can also solve this problem (and the more general case of polygons) by noting that the area of $X Y Z$ is linear in the coordinates of $X, Y$, and $Z$, so the expected area of $X Y Z$ is the same as the area of $X^{\prime} Y^{\prime} Z^{\prime}$, where $X^{\prime}$ is the expected location of $X, Y^{\prime}$ is the expected location of $Y$, and $Z^{\prime}$ is the expected location of $Z$. In our case, this corresponds to the midpoints of the three sides $A B$, $B C$, and $C A$.
|
15
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a triangle with $A B=8, B C=15$, and $A C=17$. Point $X$ is chosen at random on line segment $A B$. Point $Y$ is chosen at random on line segment $B C$. Point $Z$ is chosen at random on line segment $C A$. What is the expected area of triangle $X Y Z$ ?
|
15 Let $\mathbb{E}(X)$ denote the expected value of $X$, and let $[S]$ denote the area of $S$. Then
$$
\begin{aligned}
\mathbb{E}([\triangle X Y Z]) & =\mathbb{E}([\triangle A B C]-[\triangle X Y B]-[\triangle Z Y C]-[\triangle X B Z]) \\
& =[\triangle A B C]-\mathbb{E}([\triangle X Y B])-\mathbb{E}([\triangle Z Y C])-[\triangle X B Z])
\end{aligned}
$$
where the last step follows from linearity of expectation ${ }^{1}$. But $[\triangle X Y B]=\frac{1}{2} \cdot B X \cdot B Y \cdot \sin (B)$. The $\frac{1}{2} \sin (B)$ term is constant, and $B X$ and $B Y$ are both independent with expected values $\frac{A B}{2}$ and $\frac{B C}{2}$, respectively. Thus $\mathbb{E}([\triangle X Y B])=\frac{1}{8} A B \cdot B C \cdot \sin (B)=\frac{1}{4}[\triangle A B C]$. Similarly, $\mathbb{E}([\triangle Z Y C])=$ $\mathbb{E}([\triangle Z B X])=\frac{1}{4}[\triangle A B C]$.
Then we have $\mathbb{E}([\triangle X Y Z])=\left(1-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}\right)[\triangle A B C]=\frac{1}{4}[\triangle A B C]=15$.
Note: We can also solve this problem (and the more general case of polygons) by noting that the area of $X Y Z$ is linear in the coordinates of $X, Y$, and $Z$, so the expected area of $X Y Z$ is the same as the area of $X^{\prime} Y^{\prime} Z^{\prime}$, where $X^{\prime}$ is the expected location of $X, Y^{\prime}$ is the expected location of $Y$, and $Z^{\prime}$ is the expected location of $Z$. In our case, this corresponds to the midpoints of the three sides $A B$, $B C$, and $C A$.
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. [7]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
From the point $(x, y)$, a legal move is a move to $\left(\frac{x}{3}+u, \frac{y}{3}+v\right)$, where $u$ and $v$ are real numbers such that $u^{2}+v^{2} \leq 1$. What is the area of the set of points that can be reached from $(0,0)$ in a finite number of legal moves?
|
| $\frac{9 \pi}{4}$ |
| :---: |
| We claim that the set of points is the disc with radius $\frac{3}{2}$ centered at the origin, which | clearly has area $\frac{9 \pi}{4}$.
First, we show that the set is contained in this disc. This is because if we are currently at a distance of $r$ from the origin, then we can't end up at a distance of greater than $\frac{r}{3}+1$ from the origin after a single move. Since $\frac{r}{3}+1<\frac{3}{2}$ if $r<\frac{3}{2}$, we will always end up in the disc of radius $\frac{3}{2}$ if we start in it. Since the origin is inside this disc, any finite number of moves will leave us inside this disc.
Next, we show that all points in this disc can be reached in a finite number of moves. Indeed, after one move we can get all points within a distance of 1 . After two moves, we can get all points within a distance of $\frac{4}{3}$. After three moves, we can get all points within a distance of $\frac{13}{9}$. In general, after $n$ moves we can get all points within a distance of $\frac{3}{2}-\frac{1}{2 \cdot 3^{k-1}}$. This means that for any distance $d<\frac{3}{2}$, we will eventually get all points within a distance of $d$, so all points in the disc of radius $\frac{3}{2}$ can be reached after some number of moves.
|
\frac{9 \pi}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
From the point $(x, y)$, a legal move is a move to $\left(\frac{x}{3}+u, \frac{y}{3}+v\right)$, where $u$ and $v$ are real numbers such that $u^{2}+v^{2} \leq 1$. What is the area of the set of points that can be reached from $(0,0)$ in a finite number of legal moves?
|
| $\frac{9 \pi}{4}$ |
| :---: |
| We claim that the set of points is the disc with radius $\frac{3}{2}$ centered at the origin, which | clearly has area $\frac{9 \pi}{4}$.
First, we show that the set is contained in this disc. This is because if we are currently at a distance of $r$ from the origin, then we can't end up at a distance of greater than $\frac{r}{3}+1$ from the origin after a single move. Since $\frac{r}{3}+1<\frac{3}{2}$ if $r<\frac{3}{2}$, we will always end up in the disc of radius $\frac{3}{2}$ if we start in it. Since the origin is inside this disc, any finite number of moves will leave us inside this disc.
Next, we show that all points in this disc can be reached in a finite number of moves. Indeed, after one move we can get all points within a distance of 1 . After two moves, we can get all points within a distance of $\frac{4}{3}$. After three moves, we can get all points within a distance of $\frac{13}{9}$. In general, after $n$ moves we can get all points within a distance of $\frac{3}{2}-\frac{1}{2 \cdot 3^{k-1}}$. This means that for any distance $d<\frac{3}{2}$, we will eventually get all points within a distance of $d$, so all points in the disc of radius $\frac{3}{2}$ can be reached after some number of moves.
|
{
"exam": "HMMT",
"problem_label": "11",
"problem_match": "\n11. [7]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
How many different collections of 9 letters are there? A letter can appear multiple times in a collection. Two collections are equal if each letter appears the same number of times in both collections.
|
$\left.\begin{array}{c}34 \\ 9\end{array}\right)$ We put these collections in bijections with binary strings of length 34 containing 9 zeroes and 25 ones. Take any such string - the 9 zeroes will correspond to the 9 letters in the collection. If there are $n$ ones before a zero, then that zero corresponds to the $(n+1)$ st letter of the alphabet. This scheme is an injective map from the binary strings to the collections, and it has an inverse, so the number of collections is $\binom{34}{9}$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
\binom{34}{9}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many different collections of 9 letters are there? A letter can appear multiple times in a collection. Two collections are equal if each letter appears the same number of times in both collections.
|
$\left.\begin{array}{c}34 \\ 9\end{array}\right)$ We put these collections in bijections with binary strings of length 34 containing 9 zeroes and 25 ones. Take any such string - the 9 zeroes will correspond to the 9 letters in the collection. If there are $n$ ones before a zero, then that zero corresponds to the $(n+1)$ st letter of the alphabet. This scheme is an injective map from the binary strings to the collections, and it has an inverse, so the number of collections is $\binom{34}{9}$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"exam": "HMMT",
"problem_label": "12",
"problem_match": "\n12. [7]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
A triangle in the $x y$-plane is such that when projected onto the $x$-axis, $y$-axis, and the line $y=x$, the results are line segments whose endpoints are $(1,0)$ and $(5,0),(0,8)$ and $(0,13)$, and $(5,5)$ and $(7.5,7.5)$, respectively. What is the triangle's area?
|
$\frac{17}{2}$ Sketch the lines $x=1, x=5, y=8, y=13, y=10-x$, and $y=15-x$. The triangle has to be contained in the hexagonal region contained in all these lines. If all the projections are correct, every other vertex of the hexagon must be a vertex of the triangle, which gives us two possibilities for the triangle. One of these triangles has vertices at $(2,8),(1,13)$, and $(5,10)$, and has an area of $\frac{17}{2}$. It is easy to check that the other triangle has the same area, so the answer is unique.
|
\frac{17}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A triangle in the $x y$-plane is such that when projected onto the $x$-axis, $y$-axis, and the line $y=x$, the results are line segments whose endpoints are $(1,0)$ and $(5,0),(0,8)$ and $(0,13)$, and $(5,5)$ and $(7.5,7.5)$, respectively. What is the triangle's area?
|
$\frac{17}{2}$ Sketch the lines $x=1, x=5, y=8, y=13, y=10-x$, and $y=15-x$. The triangle has to be contained in the hexagonal region contained in all these lines. If all the projections are correct, every other vertex of the hexagon must be a vertex of the triangle, which gives us two possibilities for the triangle. One of these triangles has vertices at $(2,8),(1,13)$, and $(5,10)$, and has an area of $\frac{17}{2}$. It is easy to check that the other triangle has the same area, so the answer is unique.
|
{
"exam": "HMMT",
"problem_label": "13",
"problem_match": "\n13. [8]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
In how many ways can you fill a $3 \times 3$ table with the numbers 1 through 9 (each used once) such that all pairs of adjacent numbers (sharing one side) are relatively prime?
|
2016 The numbers can be separated into four sets. Numbers in the set $A=\{1,5,7\}$ can be placed next to anything. The next two sets are $B=\{2,4,8\}$ and $C=\{3,9\}$. The number 6 , which forms the final set $D$, can only be placed next to elements of $A$. The elements of each group can be interchanged without violating the condition, so without loss of generality, we can pretend we have three 1 's, three 2's, two 3 's, and one 6 , as long as we multiply our answer by $3!3!2$ ! at the end. The available arrangements are, grouped by the position of the 6 , are:
When 6 is in contact with three numbers:
| 1 | 2 | 3 |
| :--- | :--- | :--- |
| 6 | 1 | 2 |
| 1 | 2 | 3 |
When 6 is in contact with two numbers:
| 6 | 1 | 2 | | | |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 2 | 3 | 6 1 2 <br> 1 1 3 <br> 2 3 1 | 2 | 3 |
The next two can be flipped diagonally to create different arrangements:
| 6 | 1 | 2 | | | |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 2 | 3 | 6 1 | 2 | |
| 1 | 2 | 3 | | | |
| 1 | 3 | 2 | 3 | 1 | 2 |
Those seven arrangements can be rotated 90, 180, and 270 degrees about the center to generate a total of 28 arrangements. $28 \cdot 3!3!2!=2016$.
|
2016
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In how many ways can you fill a $3 \times 3$ table with the numbers 1 through 9 (each used once) such that all pairs of adjacent numbers (sharing one side) are relatively prime?
|
2016 The numbers can be separated into four sets. Numbers in the set $A=\{1,5,7\}$ can be placed next to anything. The next two sets are $B=\{2,4,8\}$ and $C=\{3,9\}$. The number 6 , which forms the final set $D$, can only be placed next to elements of $A$. The elements of each group can be interchanged without violating the condition, so without loss of generality, we can pretend we have three 1 's, three 2's, two 3 's, and one 6 , as long as we multiply our answer by $3!3!2$ ! at the end. The available arrangements are, grouped by the position of the 6 , are:
When 6 is in contact with three numbers:
| 1 | 2 | 3 |
| :--- | :--- | :--- |
| 6 | 1 | 2 |
| 1 | 2 | 3 |
When 6 is in contact with two numbers:
| 6 | 1 | 2 | | | |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 2 | 3 | 6 1 2 <br> 1 1 3 <br> 2 3 1 | 2 | 3 |
The next two can be flipped diagonally to create different arrangements:
| 6 | 1 | 2 | | | |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 2 | 3 | 6 1 | 2 | |
| 1 | 2 | 3 | | | |
| 1 | 3 | 2 | 3 | 1 | 2 |
Those seven arrangements can be rotated 90, 180, and 270 degrees about the center to generate a total of 28 arrangements. $28 \cdot 3!3!2!=2016$.
|
{
"exam": "HMMT",
"problem_label": "14",
"problem_match": "\n14. [8]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Pick a random integer between 0 and 4095, inclusive. Write it in base 2 (without any leading zeroes). What is the expected number of consecutive digits that are not the same (that is, the expected number of occurrences of either 01 or 10 in the base 2 representation)?
|
$\frac{20481}{4096}$ Note that every number in the range can be written as a 12-digit binary string. For $i=1,2, \ldots 11$, let $R_{i}$ be a random variable which is 1 if the $i$ th and $(i+1)$ st digits differ in a randomly chosen number in the range. By linearity of expectation, $E\left(\sum_{i} R_{i}\right)=\sum E\left(R_{i}\right)$.
Since we choose every binary string of length 12 with equal probability, the sum of the expectations is $\frac{11}{2}$. However, this is not the expected number of 01 s and $10 s$ - we need to subtract the occasions where the leading digit is zero. There is a $\frac{1}{2}$ chance that the number starts with a 0 , in which case we must ignore the first digit change - unless the number was 0 , in which case there are no digit changes. Therefore, our answer is $\frac{11}{2}-\frac{1}{2}+\frac{1}{4096}=\frac{20481}{4096}$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
\frac{20481}{4096}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Pick a random integer between 0 and 4095, inclusive. Write it in base 2 (without any leading zeroes). What is the expected number of consecutive digits that are not the same (that is, the expected number of occurrences of either 01 or 10 in the base 2 representation)?
|
$\frac{20481}{4096}$ Note that every number in the range can be written as a 12-digit binary string. For $i=1,2, \ldots 11$, let $R_{i}$ be a random variable which is 1 if the $i$ th and $(i+1)$ st digits differ in a randomly chosen number in the range. By linearity of expectation, $E\left(\sum_{i} R_{i}\right)=\sum E\left(R_{i}\right)$.
Since we choose every binary string of length 12 with equal probability, the sum of the expectations is $\frac{11}{2}$. However, this is not the expected number of 01 s and $10 s$ - we need to subtract the occasions where the leading digit is zero. There is a $\frac{1}{2}$ chance that the number starts with a 0 , in which case we must ignore the first digit change - unless the number was 0 , in which case there are no digit changes. Therefore, our answer is $\frac{11}{2}-\frac{1}{2}+\frac{1}{4096}=\frac{20481}{4096}$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"exam": "HMMT",
"problem_label": "15",
"problem_match": "\n15. [8]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Jessica has three marbles colored red, green, and blue. She randomly selects a non-empty subset of them (such that each subset is equally likely) and puts them in a bag. You then draw three marbles from the bag with replacement. The colors you see are red, blue, red. What is the probability that the only marbles in the bag are red and blue?
|
| $\frac{27}{35}$ |
| :---: |
| There are two possible sets of marbles in the bag, $\{$ red,blue\} and \{red,blue,green\}. | Initially, both these sets are equally likely to be in the bag. However, the probability of red, blue, red being drawn from a set $S$ of marbles is proportional to $|S|^{-3}$, as long as red and blue are both in $S$. By Bayes's Rule, we must weight the probability of these two sets by $|S|^{-3}$. The answer is $\frac{(1 / 2)^{3}}{(1 / 2)^{3}+(1 / 3)^{3}}$.
|
\frac{27}{35}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Jessica has three marbles colored red, green, and blue. She randomly selects a non-empty subset of them (such that each subset is equally likely) and puts them in a bag. You then draw three marbles from the bag with replacement. The colors you see are red, blue, red. What is the probability that the only marbles in the bag are red and blue?
|
| $\frac{27}{35}$ |
| :---: |
| There are two possible sets of marbles in the bag, $\{$ red,blue\} and \{red,blue,green\}. | Initially, both these sets are equally likely to be in the bag. However, the probability of red, blue, red being drawn from a set $S$ of marbles is proportional to $|S|^{-3}$, as long as red and blue are both in $S$. By Bayes's Rule, we must weight the probability of these two sets by $|S|^{-3}$. The answer is $\frac{(1 / 2)^{3}}{(1 / 2)^{3}+(1 / 3)^{3}}$.
|
{
"exam": "HMMT",
"problem_label": "16",
"problem_match": "\n16. [9]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
An ant starts at the origin, facing in the positive $x$-direction. Each second, it moves 1 unit forward, then turns counterclockwise by $\sin ^{-1}\left(\frac{3}{5}\right)$ degrees. What is the least upper bound on the distance between the ant and the origin? (The least upper bound is the smallest real number $r$ that is at least as big as every distance that the ant ever is from the origin.)
|
$\sqrt{10}$ We claim that the points the ant visits lie on a circle of radius $\frac{\sqrt{10}}{2}$. We show this by saying that the ant stays a constant distance $\frac{\sqrt{10}}{2}$ from the point $\left(\frac{1}{2}, \frac{3}{2}\right)$.
Suppose the ant moves on a plane $P$. Consider a transformation of the plane $P^{\prime}$ such that after the first move, the ant is at the origin of $P^{\prime}$ and facing in the direction of the $x^{\prime}$ axis (on $P^{\prime}$ ). The transformation to get from $P$ to $P^{\prime}$ can be gotten by rotating $P$ about the origin counterclockwise through an angle $\sin ^{-1}\left(\frac{3}{5}\right)$ and then translating it 1 unit to the right. Observe that the point $\left(\frac{1}{2}, \frac{3}{2}\right)$ is fixed under this transformation, which can be shown through the expression $\left(\frac{1}{2}+\frac{3}{2} i\right)\left(\frac{4}{5}+\frac{3}{5} i\right)+1=\frac{1}{2}+\frac{3}{2} i$. It follows that at every point the ant stops, it will always be the same distance from $\left(\frac{1}{2}, \frac{3}{2}\right)$. Since it starts at $(0,0)$, this fixed distance is $\frac{\sqrt{10}}{2}$.
Since $\sin ^{-1}\left(\frac{3}{5}\right)$ is not a rational multiple of $\pi$, the points the ant stops at form a dense subset of the circle in question. As a result, the least upper bound on the distance between the ant and the origin is the diameter of the circle, which is $\sqrt{10}$.
|
\sqrt{10}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
An ant starts at the origin, facing in the positive $x$-direction. Each second, it moves 1 unit forward, then turns counterclockwise by $\sin ^{-1}\left(\frac{3}{5}\right)$ degrees. What is the least upper bound on the distance between the ant and the origin? (The least upper bound is the smallest real number $r$ that is at least as big as every distance that the ant ever is from the origin.)
|
$\sqrt{10}$ We claim that the points the ant visits lie on a circle of radius $\frac{\sqrt{10}}{2}$. We show this by saying that the ant stays a constant distance $\frac{\sqrt{10}}{2}$ from the point $\left(\frac{1}{2}, \frac{3}{2}\right)$.
Suppose the ant moves on a plane $P$. Consider a transformation of the plane $P^{\prime}$ such that after the first move, the ant is at the origin of $P^{\prime}$ and facing in the direction of the $x^{\prime}$ axis (on $P^{\prime}$ ). The transformation to get from $P$ to $P^{\prime}$ can be gotten by rotating $P$ about the origin counterclockwise through an angle $\sin ^{-1}\left(\frac{3}{5}\right)$ and then translating it 1 unit to the right. Observe that the point $\left(\frac{1}{2}, \frac{3}{2}\right)$ is fixed under this transformation, which can be shown through the expression $\left(\frac{1}{2}+\frac{3}{2} i\right)\left(\frac{4}{5}+\frac{3}{5} i\right)+1=\frac{1}{2}+\frac{3}{2} i$. It follows that at every point the ant stops, it will always be the same distance from $\left(\frac{1}{2}, \frac{3}{2}\right)$. Since it starts at $(0,0)$, this fixed distance is $\frac{\sqrt{10}}{2}$.
Since $\sin ^{-1}\left(\frac{3}{5}\right)$ is not a rational multiple of $\pi$, the points the ant stops at form a dense subset of the circle in question. As a result, the least upper bound on the distance between the ant and the origin is the diameter of the circle, which is $\sqrt{10}$.
|
{
"exam": "HMMT",
"problem_label": "17",
"problem_match": "\n17. [9]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Find two lines of symmetry of the graph of the function $y=x+\frac{1}{x}$. Express your answer as two equations of the form $y=a x+b$.
|
$y=(1+\sqrt{2}) x$ and $y=(1-\sqrt{2}) x$ The graph of the function $y=x+\frac{1}{x}$ is a hyperbola. We can see this more clearly by writing it out in the standard form $x^{2}-x y+1=0$ or $\left(\frac{y}{2}\right)^{2}-\left(x-\frac{1}{2} y\right)^{2}=1$. The hyperbola has asymptotes given by $x=0$ and $y=x$, so the lines of symmetry will be the (interior and exterior) angle bisectors of these two lines. This means that they will be $y=\tan \left(67.5^{\circ}\right) x$ and $y=-\cot \left(67.5^{\circ}\right) x$, which, using the tangent half-angle formula $\tan \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos (x)}{1-\cos (x)}}$, gives the two lines $y=(1+\sqrt{2}) x$ and $y=(1-\sqrt{2}) x$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
y=(1+\sqrt{2}) x \text{ and } y=(1-\sqrt{2}) x
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find two lines of symmetry of the graph of the function $y=x+\frac{1}{x}$. Express your answer as two equations of the form $y=a x+b$.
|
$y=(1+\sqrt{2}) x$ and $y=(1-\sqrt{2}) x$ The graph of the function $y=x+\frac{1}{x}$ is a hyperbola. We can see this more clearly by writing it out in the standard form $x^{2}-x y+1=0$ or $\left(\frac{y}{2}\right)^{2}-\left(x-\frac{1}{2} y\right)^{2}=1$. The hyperbola has asymptotes given by $x=0$ and $y=x$, so the lines of symmetry will be the (interior and exterior) angle bisectors of these two lines. This means that they will be $y=\tan \left(67.5^{\circ}\right) x$ and $y=-\cot \left(67.5^{\circ}\right) x$, which, using the tangent half-angle formula $\tan \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos (x)}{1-\cos (x)}}$, gives the two lines $y=(1+\sqrt{2}) x$ and $y=(1-\sqrt{2}) x$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"exam": "HMMT",
"problem_label": "18",
"problem_match": "\n18. [9]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
A 5-dimensional ant starts at one vertex of a 5 -dimensional hypercube of side length 1 . A move is when the ant travels from one vertex to another vertex at a distance of $\sqrt{2}$ away. How many ways can the ant make 5 moves and end up on the same vertex it started at?
|
6240 We let the cube lie in $\mathbb{R}^{5}$ with each corner with coordinates 1 or 0 . Assume the ant starts at $(0,0,0,0,0)$. Every move the ant adds or subtracts 1 to two of the places. Note that this means the ant can only land on a vertex with the sum of its coordinates an even number. Every move the ant has $\binom{5}{2}=10$ choices.
From any vertex there are 10 two-move sequences that put the ant at the same vertex it started at.
There are 6 two-move sequences to move from one vertex to a different, chosen vertex. If your chosen vertex differs from your current vertex by 2 of the 5 coordinates, your first move corrects for one of these two. There are 2 ways to choose which coordinate to correct for on the first move, and there are 3 ways to choose the second coordinate you change, yielding 6 sequences. If your chosen vertex differs from your current vertex by 4 of the 5 coordinates, each move corrects for two of these four. This yields $\binom{4}{2}=6$ sequences.
Finally, there are 60 three-move sequences that put the ant at the same vertex it started at. There are 10 ways to choose the first move, and there are 6 ways to make two moves to return to your original position.
The motion of the ant can be split into two cases.
Case 1: After the 3rd move the ant is on the vertex it started at. There are $(60)(10)=600$ different possible paths.
Case 2: After the third move the ant is on a vertex different from the one it started on. There are $\left(10^{3}-60\right)(6)=(940)(6)=5640$ different possible paths.
So there are 6240 total possible paths.
|
6240
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A 5-dimensional ant starts at one vertex of a 5 -dimensional hypercube of side length 1 . A move is when the ant travels from one vertex to another vertex at a distance of $\sqrt{2}$ away. How many ways can the ant make 5 moves and end up on the same vertex it started at?
|
6240 We let the cube lie in $\mathbb{R}^{5}$ with each corner with coordinates 1 or 0 . Assume the ant starts at $(0,0,0,0,0)$. Every move the ant adds or subtracts 1 to two of the places. Note that this means the ant can only land on a vertex with the sum of its coordinates an even number. Every move the ant has $\binom{5}{2}=10$ choices.
From any vertex there are 10 two-move sequences that put the ant at the same vertex it started at.
There are 6 two-move sequences to move from one vertex to a different, chosen vertex. If your chosen vertex differs from your current vertex by 2 of the 5 coordinates, your first move corrects for one of these two. There are 2 ways to choose which coordinate to correct for on the first move, and there are 3 ways to choose the second coordinate you change, yielding 6 sequences. If your chosen vertex differs from your current vertex by 4 of the 5 coordinates, each move corrects for two of these four. This yields $\binom{4}{2}=6$ sequences.
Finally, there are 60 three-move sequences that put the ant at the same vertex it started at. There are 10 ways to choose the first move, and there are 6 ways to make two moves to return to your original position.
The motion of the ant can be split into two cases.
Case 1: After the 3rd move the ant is on the vertex it started at. There are $(60)(10)=600$ different possible paths.
Case 2: After the third move the ant is on a vertex different from the one it started on. There are $\left(10^{3}-60\right)(6)=(940)(6)=5640$ different possible paths.
So there are 6240 total possible paths.
|
{
"exam": "HMMT",
"problem_label": "19",
"problem_match": "\n19. [10]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Find the volume of the set of points $(x, y, z)$ satisfying
$$
\begin{aligned}
x, y, z & \geq 0 \\
x+y & \leq 1 \\
y+z & \leq 1 \\
z+x & \leq 1
\end{aligned}
$$
|
$\frac{1}{4}$ Without loss of generality, assume that $x \geq y$ - half the volume of the solid is on this side of the plane $x=y$. For each value of $c$ from 0 to $\frac{1}{2}$, the region of the intersection of this half of the solid with the plane $y=c$ is a trapezoid. The trapezoid has height $1-2 c$ and average base $\frac{1}{2}$, so it has an area of $\frac{1}{2}-c$.
The total volume of this region is $\frac{1}{2}$ times the average area of the trapezoids, which is $\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}$. Double that to get the total volume, which is $\frac{1}{4}$.
|
\frac{1}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Find the volume of the set of points $(x, y, z)$ satisfying
$$
\begin{aligned}
x, y, z & \geq 0 \\
x+y & \leq 1 \\
y+z & \leq 1 \\
z+x & \leq 1
\end{aligned}
$$
|
$\frac{1}{4}$ Without loss of generality, assume that $x \geq y$ - half the volume of the solid is on this side of the plane $x=y$. For each value of $c$ from 0 to $\frac{1}{2}$, the region of the intersection of this half of the solid with the plane $y=c$ is a trapezoid. The trapezoid has height $1-2 c$ and average base $\frac{1}{2}$, so it has an area of $\frac{1}{2}-c$.
The total volume of this region is $\frac{1}{2}$ times the average area of the trapezoids, which is $\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}$. Double that to get the total volume, which is $\frac{1}{4}$.
|
{
"exam": "HMMT",
"problem_label": "20",
"problem_match": "\n20. [10]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $\triangle A B C$ be a scalene triangle. Let $h_{a}$ be the locus of points $P$ such that $|P B-P C|=|A B-A C|$. Let $h_{b}$ be the locus of points $P$ such that $|P C-P A|=|B C-B A|$. Let $h_{c}$ be the locus of points $P$ such that $|P A-P B|=|C A-C B|$. In how many points do all of $h_{a}, h_{b}$, and $h_{c}$ concur?
|
2 The idea is similar to the proof that the angle bisectors concur or that the perpendicular bisectors concur. Assume WLOG that $B C>A B>C A$. Note that $h_{a}$ and $h_{b}$ are both hyperbolas. Therefore, $h_{a}$ and $h_{b}$ intersect in four points (each branch of $h_{a}$ intersects exactly once with each branch of $h_{b}$ ). Note that the branches of $h_{a}$ correspond to the cases when $P B>P C$ and when $P B<P C$. Similarly, the branches of $h_{b}$ correspond to the cases when $P C>P A$ and $P C<P A$.
If either $P A<P B<P C$ or $P C<P B<P A$ (which each happens for exactly one point of intersection of $h_{a}$ and $h_{b}$ ), then $|P C-P A|=|P C-P B|+|P B-P A|=|A B-A C|+|B C-B A|=|B C-A C|$, and so $P$ also lies on $h_{c}$. So, exactly two of the four points of intersection of $h_{a}$ and $h_{b}$ lie on $h_{c}$, meaning that $h_{a}, h_{b}$, and $h_{c}$ concur in four points.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
2
|
Yes
|
Incomplete
|
math-word-problem
|
Geometry
|
Let $\triangle A B C$ be a scalene triangle. Let $h_{a}$ be the locus of points $P$ such that $|P B-P C|=|A B-A C|$. Let $h_{b}$ be the locus of points $P$ such that $|P C-P A|=|B C-B A|$. Let $h_{c}$ be the locus of points $P$ such that $|P A-P B|=|C A-C B|$. In how many points do all of $h_{a}, h_{b}$, and $h_{c}$ concur?
|
2 The idea is similar to the proof that the angle bisectors concur or that the perpendicular bisectors concur. Assume WLOG that $B C>A B>C A$. Note that $h_{a}$ and $h_{b}$ are both hyperbolas. Therefore, $h_{a}$ and $h_{b}$ intersect in four points (each branch of $h_{a}$ intersects exactly once with each branch of $h_{b}$ ). Note that the branches of $h_{a}$ correspond to the cases when $P B>P C$ and when $P B<P C$. Similarly, the branches of $h_{b}$ correspond to the cases when $P C>P A$ and $P C<P A$.
If either $P A<P B<P C$ or $P C<P B<P A$ (which each happens for exactly one point of intersection of $h_{a}$ and $h_{b}$ ), then $|P C-P A|=|P C-P B|+|P B-P A|=|A B-A C|+|B C-B A|=|B C-A C|$, and so $P$ also lies on $h_{c}$. So, exactly two of the four points of intersection of $h_{a}$ and $h_{b}$ lie on $h_{c}$, meaning that $h_{a}, h_{b}$, and $h_{c}$ concur in four points.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"exam": "HMMT",
"problem_label": "21",
"problem_match": "\n21. [10]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
You are the general of an army. You and the opposing general both have an equal number of troops to distribute among three battlefields. Whoever has more troops on a battlefield always wins (you win ties). An order is an ordered triple of non-negative real numbers $(x, y, z)$ such that $x+y+z=1$, and corresponds to sending a fraction $x$ of the troops to the first field, $y$ to the second, and $z$ to the third. Suppose that you give the order $\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{2}\right)$ and that the other general issues an order chosen uniformly at random from all possible orders. What is the probability that you win two out of the three battles?
|
| $\frac{5}{8}$ |
| :---: |
| Let $x$ | be the portion of soldiers the opposing general sends to the first battlefield, and $y$ the portion he sends to the second. Then $1-x-y$ is the portion he sends to the third. Then $x \geq 0$, $y \geq 0$, and $x+y \leq 1$. Furthermore, you win if one of the three conditions is satisfied: $x \leq \frac{1}{4}$ and $y \leq \frac{1}{4}, x \leq \frac{1}{4}$ and $1-x-y \leq \frac{1}{2}$, or $y \leq \frac{1}{4}$ and $1-x-y \leq \frac{1}{2}$. This is illustrated in the picture below.
Guts Round
This triangle is a linear projection of the region of feasible orders, so it preserves area and probability ratios. The probability that you win, then is given by the portion of the triangle that satisfies one of the three above constraints - in other words, the area of the shaded region divided by the area of the entire triangle. We can easily calculate this to be $\frac{\frac{5}{16}}{\frac{1}{2}}=\frac{5}{8}$.
|
\frac{5}{8}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
You are the general of an army. You and the opposing general both have an equal number of troops to distribute among three battlefields. Whoever has more troops on a battlefield always wins (you win ties). An order is an ordered triple of non-negative real numbers $(x, y, z)$ such that $x+y+z=1$, and corresponds to sending a fraction $x$ of the troops to the first field, $y$ to the second, and $z$ to the third. Suppose that you give the order $\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{2}\right)$ and that the other general issues an order chosen uniformly at random from all possible orders. What is the probability that you win two out of the three battles?
|
| $\frac{5}{8}$ |
| :---: |
| Let $x$ | be the portion of soldiers the opposing general sends to the first battlefield, and $y$ the portion he sends to the second. Then $1-x-y$ is the portion he sends to the third. Then $x \geq 0$, $y \geq 0$, and $x+y \leq 1$. Furthermore, you win if one of the three conditions is satisfied: $x \leq \frac{1}{4}$ and $y \leq \frac{1}{4}, x \leq \frac{1}{4}$ and $1-x-y \leq \frac{1}{2}$, or $y \leq \frac{1}{4}$ and $1-x-y \leq \frac{1}{2}$. This is illustrated in the picture below.
Guts Round
This triangle is a linear projection of the region of feasible orders, so it preserves area and probability ratios. The probability that you win, then is given by the portion of the triangle that satisfies one of the three above constraints - in other words, the area of the shaded region divided by the area of the entire triangle. We can easily calculate this to be $\frac{\frac{5}{16}}{\frac{1}{2}}=\frac{5}{8}$.
|
{
"exam": "HMMT",
"problem_label": "22",
"problem_match": "\n22. [12]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
In the country of Francisca, there are 2010 cities, some of which are connected by roads. Between any two cities, there is a unique path which runs along the roads and which does not pass through any city twice. What is the maximum possible number of cities in Francisca which have at least 3 roads running out of them?
|
1004 The restrictions on how roads connect cities directly imply that the graph of the cities of Francisca with the roads as edges is a tree. Therefore the sum of the degrees of all the vertices is $2009 \cdot 2=4018$. Suppose that $b$ vertices have degree $\geq 3$. The other $2010-b$ vertices must have a degree of at least 1 , so $3 b+(2010-b) \leq 4018$ or $2 b \leq 2008$. So $b$ is at most 1004 . We can achieve $b=1004$ with the following graph:
|
1004
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In the country of Francisca, there are 2010 cities, some of which are connected by roads. Between any two cities, there is a unique path which runs along the roads and which does not pass through any city twice. What is the maximum possible number of cities in Francisca which have at least 3 roads running out of them?
|
1004 The restrictions on how roads connect cities directly imply that the graph of the cities of Francisca with the roads as edges is a tree. Therefore the sum of the degrees of all the vertices is $2009 \cdot 2=4018$. Suppose that $b$ vertices have degree $\geq 3$. The other $2010-b$ vertices must have a degree of at least 1 , so $3 b+(2010-b) \leq 4018$ or $2 b \leq 2008$. So $b$ is at most 1004 . We can achieve $b=1004$ with the following graph:
|
{
"exam": "HMMT",
"problem_label": "23",
"problem_match": "\n23. [12]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Define a sequence of polynomials as follows: let $a_{1}=3 x^{2}-x$, let $a_{2}=3 x^{2}-7 x+3$, and for $n \geq 1$, let $a_{n+2}=\frac{5}{2} a_{n+1}-a_{n}$. As $n$ tends to infinity, what is the limit of the sum of the roots of $a_{n}$ ?
|
$\frac{13}{3}$ By using standard methods for solving linear recurrences ${ }^{2}$, we see that this recurrence has a characteristic polynomial of $x^{2}-\frac{5}{2} x+1=\left(x-\frac{1}{2}\right)(x-2)$, hence $a_{n}(x)=c(x) \cdot 2^{n}+d(x) \cdot 2^{-n}$ for some polynomials $c$ and $d$. Plugging in $n=1$ and $n=2$ gives
$$
2 c(x)+\frac{1}{2} d(x)=3 x^{2}-x
$$
and
$$
4 c(x)+\frac{1}{4} d(x)=3 x^{2}-7 x+3
$$
Subtracting the first equation from two times the second equation gives $6 c(x)=3 x^{2}-13 x+6$, so $c(x)=\frac{3 x^{2}-13 x+6}{6}$. As $n$ grows large, the $c(x) 2^{n}$ term dominates compared to the $d(x) 2^{-n}$ term, so the roots of $a_{n}(x)$ converge to the roots of $c(x)$. Thus the roots of $a_{n}(x)$ converge to the roots of $3 x^{2}-13 x+6$, which by Vieta's formula ${ }^{3}$ have a sum of $\frac{13}{3}$.
|
\frac{13}{3}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Define a sequence of polynomials as follows: let $a_{1}=3 x^{2}-x$, let $a_{2}=3 x^{2}-7 x+3$, and for $n \geq 1$, let $a_{n+2}=\frac{5}{2} a_{n+1}-a_{n}$. As $n$ tends to infinity, what is the limit of the sum of the roots of $a_{n}$ ?
|
$\frac{13}{3}$ By using standard methods for solving linear recurrences ${ }^{2}$, we see that this recurrence has a characteristic polynomial of $x^{2}-\frac{5}{2} x+1=\left(x-\frac{1}{2}\right)(x-2)$, hence $a_{n}(x)=c(x) \cdot 2^{n}+d(x) \cdot 2^{-n}$ for some polynomials $c$ and $d$. Plugging in $n=1$ and $n=2$ gives
$$
2 c(x)+\frac{1}{2} d(x)=3 x^{2}-x
$$
and
$$
4 c(x)+\frac{1}{4} d(x)=3 x^{2}-7 x+3
$$
Subtracting the first equation from two times the second equation gives $6 c(x)=3 x^{2}-13 x+6$, so $c(x)=\frac{3 x^{2}-13 x+6}{6}$. As $n$ grows large, the $c(x) 2^{n}$ term dominates compared to the $d(x) 2^{-n}$ term, so the roots of $a_{n}(x)$ converge to the roots of $c(x)$. Thus the roots of $a_{n}(x)$ converge to the roots of $3 x^{2}-13 x+6$, which by Vieta's formula ${ }^{3}$ have a sum of $\frac{13}{3}$.
|
{
"exam": "HMMT",
"problem_label": "24",
"problem_match": "\n24. [12]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ have the property that $f(\{1,2,3\})$ and $f(f(\{1,2,3\}))$ are disjoint?
|
94 Let $f(\{1,2,3\})$ be $A$. Then $A \cap f(A)=\emptyset$, so $A$ must be a subset of $\{4,5\}$. If $B=\{4,5\}$, there are $2^{3}-2$ ways to assign each element in $\{1,2,3\}$ to a value in $\{4,5\}$, and 9 ways to assign each element of $\{4,5\}$ to a value in $\{1,2,3\}$, for a total of 54 choices of $f$. If $A=\{4\}$, there is 1 possible value for each element of $\{1,2,3\}, 4$ ways to assign $\{4\}$ with a value from $\{1,2,3,5\}$, and 5 ways to assign a value to $\{5\}$. Similarly, if $A=\{5\}$, there are $4 \cdot 5=20$ choices for $f$. In total, there are $54+20 \cdot 2=94$ possible functions.
[^1]26. [15] Express the following in closed form, as a function of $x$ :
$\sin ^{2}(x)+\sin ^{2}(2 x) \cos ^{2}(x)+\sin ^{2}(4 x) \cos ^{2}(2 x) \cos ^{2}(x)+\cdots+\sin ^{2}\left(2^{2010} x\right) \cos ^{2}\left(2^{2009} x\right) \cdots \cos ^{2}(2 x) \cos ^{2}(x)$.
|
94
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ have the property that $f(\{1,2,3\})$ and $f(f(\{1,2,3\}))$ are disjoint?
|
94 Let $f(\{1,2,3\})$ be $A$. Then $A \cap f(A)=\emptyset$, so $A$ must be a subset of $\{4,5\}$. If $B=\{4,5\}$, there are $2^{3}-2$ ways to assign each element in $\{1,2,3\}$ to a value in $\{4,5\}$, and 9 ways to assign each element of $\{4,5\}$ to a value in $\{1,2,3\}$, for a total of 54 choices of $f$. If $A=\{4\}$, there is 1 possible value for each element of $\{1,2,3\}, 4$ ways to assign $\{4\}$ with a value from $\{1,2,3,5\}$, and 5 ways to assign a value to $\{5\}$. Similarly, if $A=\{5\}$, there are $4 \cdot 5=20$ choices for $f$. In total, there are $54+20 \cdot 2=94$ possible functions.
[^1]26. [15] Express the following in closed form, as a function of $x$ :
$\sin ^{2}(x)+\sin ^{2}(2 x) \cos ^{2}(x)+\sin ^{2}(4 x) \cos ^{2}(2 x) \cos ^{2}(x)+\cdots+\sin ^{2}\left(2^{2010} x\right) \cos ^{2}\left(2^{2009} x\right) \cdots \cos ^{2}(2 x) \cos ^{2}(x)$.
|
{
"exam": "HMMT",
"problem_label": "25",
"problem_match": "\n25. [15]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ have the property that $f(\{1,2,3\})$ and $f(f(\{1,2,3\}))$ are disjoint?
|
$1-\frac{\sin ^{2}\left(2^{2011} x\right)}{4^{2011} \sin ^{2}(x)}$ Note that
\[
$$
\begin{aligned}
& \sin ^{2}(x)+\sin ^{2}(2 x) \cos ^{2}(x)+\cdots+\sin ^{2}\left(2^{2010} x\right) \cos ^{2}\left(2^{2009} x\right) \cdots \cos ^{2}(2 x) \cos ^{2}(x) \\
& \quad=\left(1-\cos ^{2}(x)\right)+\left(1-\cos ^{2}(2 x)\right) \cos ^{2}(x)+\cdots+\left(1-\cos ^{2}\left(2^{2010} x\right)\right) \cos ^{2}\left(2^{2009} x\right) \cdots \cos ^{2}(x)
\end{aligned}
$$
\]
which telescopes to $1-\cos ^{2}(x) \cos ^{2}(2 x) \cos ^{2}(4 x) \cdots \cos ^{2}\left(2^{2010} x\right)$. To evaluate $\cos ^{2}(x) \cos ^{2}(2 x) \cdots \cos ^{2}\left(2^{2010} x\right)$, multiply and divide by $\sin ^{2}(x)$. We then get
$$
1-\frac{\sin ^{2}(x) \cos ^{2}(x) \cos ^{2}(2 x) \cdots \cos ^{2}\left(2^{2010} x\right)}{\sin ^{2}(x)}
$$
Using the double-angle formula for sin, we get that $\sin ^{2}(y) \cos ^{2}(y)=\frac{\sin ^{2}(2 y)}{4}$. Applying this 2011 times makes the above expression
$$
1-\frac{\sin ^{2}\left(2^{2011} x\right)}{4^{2011} \sin ^{2}(x)}
$$
which is in closed form.
|
not found
|
Yes
|
Problem not solved
|
math-word-problem
|
Combinatorics
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ have the property that $f(\{1,2,3\})$ and $f(f(\{1,2,3\}))$ are disjoint?
|
$1-\frac{\sin ^{2}\left(2^{2011} x\right)}{4^{2011} \sin ^{2}(x)}$ Note that
\[
$$
\begin{aligned}
& \sin ^{2}(x)+\sin ^{2}(2 x) \cos ^{2}(x)+\cdots+\sin ^{2}\left(2^{2010} x\right) \cos ^{2}\left(2^{2009} x\right) \cdots \cos ^{2}(2 x) \cos ^{2}(x) \\
& \quad=\left(1-\cos ^{2}(x)\right)+\left(1-\cos ^{2}(2 x)\right) \cos ^{2}(x)+\cdots+\left(1-\cos ^{2}\left(2^{2010} x\right)\right) \cos ^{2}\left(2^{2009} x\right) \cdots \cos ^{2}(x)
\end{aligned}
$$
\]
which telescopes to $1-\cos ^{2}(x) \cos ^{2}(2 x) \cos ^{2}(4 x) \cdots \cos ^{2}\left(2^{2010} x\right)$. To evaluate $\cos ^{2}(x) \cos ^{2}(2 x) \cdots \cos ^{2}\left(2^{2010} x\right)$, multiply and divide by $\sin ^{2}(x)$. We then get
$$
1-\frac{\sin ^{2}(x) \cos ^{2}(x) \cos ^{2}(2 x) \cdots \cos ^{2}\left(2^{2010} x\right)}{\sin ^{2}(x)}
$$
Using the double-angle formula for sin, we get that $\sin ^{2}(y) \cos ^{2}(y)=\frac{\sin ^{2}(2 y)}{4}$. Applying this 2011 times makes the above expression
$$
1-\frac{\sin ^{2}\left(2^{2011} x\right)}{4^{2011} \sin ^{2}(x)}
$$
which is in closed form.
|
{
"exam": "HMMT",
"problem_label": "25",
"problem_match": "\n25. [15]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Suppose that there are real numbers $a, b, c \geq 1$ and that there are positive reals $x, y, z$ such that
$$
\begin{aligned}
a^{x}+b^{y}+c^{z} & =4 \\
x a^{x}+y b^{y}+z c^{z} & =6 \\
x^{2} a^{x}+y^{2} b^{y}+z^{2} c^{z} & =9
\end{aligned}
$$
What is the maximum possible value of $c$ ?
|
$\sqrt[3]{4}$ The Cauchy-Schwarz inequality states that given 2 sequences of $n$ real numbers $x_{1}, x_{2}, \ldots, x_{n}$ and $y_{1}, y_{2}, \ldots, y_{n}$, then $\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\right)\left(y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}\right) \geq\left(x_{1} y_{1}+x_{2} y_{2}+\ldots+x_{n} y_{n}\right)^{2}$ with equality holding if and only if $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}=\ldots=\frac{x_{n}}{y_{n}}$. Applying this to $\left\{a^{x / 2}, b^{y / 2}, c^{z / 2}\right\}$ and $\left\{x a^{x / 2}, y b^{y / 2}, z c^{z / 2}\right\}$ yields $\left(a^{x}+b^{y}+c^{z}\right)\left(x^{2} a^{x}+y^{2} b^{y}+z^{2} c^{z}\right) \geq\left(x a^{x}+y b^{y}+z b^{z}\right)^{2}$ with equality holding if and only if $x=y=z$.
However, equality does hold (both sides evaluate to 36 ), so $x=y=z$. The second equation then becomes $x\left(a^{x}+b^{x}+c^{x}\right)=6$, which implies $x=\frac{3}{2}$. Then we have $a^{3 / 2}+b^{3 / 2}+c^{3 / 2}=4$. To maximize $c$, we minimize $a$ and $b$ by setting $a=b=1$. Then $c^{3 / 2}=2$ or $c=\sqrt[3]{4}$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
\sqrt[3]{4}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose that there are real numbers $a, b, c \geq 1$ and that there are positive reals $x, y, z$ such that
$$
\begin{aligned}
a^{x}+b^{y}+c^{z} & =4 \\
x a^{x}+y b^{y}+z c^{z} & =6 \\
x^{2} a^{x}+y^{2} b^{y}+z^{2} c^{z} & =9
\end{aligned}
$$
What is the maximum possible value of $c$ ?
|
$\sqrt[3]{4}$ The Cauchy-Schwarz inequality states that given 2 sequences of $n$ real numbers $x_{1}, x_{2}, \ldots, x_{n}$ and $y_{1}, y_{2}, \ldots, y_{n}$, then $\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\right)\left(y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}\right) \geq\left(x_{1} y_{1}+x_{2} y_{2}+\ldots+x_{n} y_{n}\right)^{2}$ with equality holding if and only if $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}=\ldots=\frac{x_{n}}{y_{n}}$. Applying this to $\left\{a^{x / 2}, b^{y / 2}, c^{z / 2}\right\}$ and $\left\{x a^{x / 2}, y b^{y / 2}, z c^{z / 2}\right\}$ yields $\left(a^{x}+b^{y}+c^{z}\right)\left(x^{2} a^{x}+y^{2} b^{y}+z^{2} c^{z}\right) \geq\left(x a^{x}+y b^{y}+z b^{z}\right)^{2}$ with equality holding if and only if $x=y=z$.
However, equality does hold (both sides evaluate to 36 ), so $x=y=z$. The second equation then becomes $x\left(a^{x}+b^{x}+c^{x}\right)=6$, which implies $x=\frac{3}{2}$. Then we have $a^{3 / 2}+b^{3 / 2}+c^{3 / 2}=4$. To maximize $c$, we minimize $a$ and $b$ by setting $a=b=1$. Then $c^{3 / 2}=2$ or $c=\sqrt[3]{4}$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"exam": "HMMT",
"problem_label": "27",
"problem_match": "\n27. [15]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Danielle Bellatrix Robinson is organizing a poker tournament with 9 people. The tournament will have 4 rounds, and in each round the 9 players are split into 3 groups of 3 . During the tournament, each player plays every other player exactly once. How many different ways can Danielle divide the 9 people into three groups in each round to satisfy these requirements?
|
20160 We first split the 9 people up arbitrarily into groups of 3 . There are $\frac{\binom{9}{3}\binom{6}{3}\binom{3}{3}}{3!}=280$ ways of doing this. Without loss of generality, label the people 1 through 9 so that the first round groups are $\{1,2,3\},\{4,5,6\}$, and $\{7,8,9\}$. We will use this numbering to count the number of ways DBR can divide the 9 players in rounds 2,3 , and 4 .
In round 2 , because players 1,2 , and 3 are together in the first round, they must be in separate groups, and likewise for $\{4,5,6\}$ and $\{7,8,9\}$. Disregarding ordering of the three groups in a single round, we will first place 1,2 , and 3 into their groups, then count the number of ways to place $\{4,5,6\}$ and $\{7,8,9\}$ in the groups with them. We do this by placing one member from each of $\{4,5,6\}$ and $\{7,8,9\}$ into each group. There are $(3!)^{2}$ ways to do this. Now, because of symmetry, we can use the round 2 grouping $\{1,4,7\},\{2,5,8\},\{3,6,9\}$ to list out the remaining possibilities for round 3 and 4 groupings.
Casework shows that there are 2 ways to group the players in the remaining two rounds. We multiply $280 \cdot(3!)^{2} \cdot 2$ to get 20160.
|
20160
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Danielle Bellatrix Robinson is organizing a poker tournament with 9 people. The tournament will have 4 rounds, and in each round the 9 players are split into 3 groups of 3 . During the tournament, each player plays every other player exactly once. How many different ways can Danielle divide the 9 people into three groups in each round to satisfy these requirements?
|
20160 We first split the 9 people up arbitrarily into groups of 3 . There are $\frac{\binom{9}{3}\binom{6}{3}\binom{3}{3}}{3!}=280$ ways of doing this. Without loss of generality, label the people 1 through 9 so that the first round groups are $\{1,2,3\},\{4,5,6\}$, and $\{7,8,9\}$. We will use this numbering to count the number of ways DBR can divide the 9 players in rounds 2,3 , and 4 .
In round 2 , because players 1,2 , and 3 are together in the first round, they must be in separate groups, and likewise for $\{4,5,6\}$ and $\{7,8,9\}$. Disregarding ordering of the three groups in a single round, we will first place 1,2 , and 3 into their groups, then count the number of ways to place $\{4,5,6\}$ and $\{7,8,9\}$ in the groups with them. We do this by placing one member from each of $\{4,5,6\}$ and $\{7,8,9\}$ into each group. There are $(3!)^{2}$ ways to do this. Now, because of symmetry, we can use the round 2 grouping $\{1,4,7\},\{2,5,8\},\{3,6,9\}$ to list out the remaining possibilities for round 3 and 4 groupings.
Casework shows that there are 2 ways to group the players in the remaining two rounds. We multiply $280 \cdot(3!)^{2} \cdot 2$ to get 20160.
|
{
"exam": "HMMT",
"problem_label": "28",
"problem_match": "\n28. [18]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Compute the remainder when
$$
\sum_{k=1}^{30303} k^{k}
$$
is divided by 101 .
|
29 The main idea is the following lemma:
Lemma. For any non-negative integer $n$ and prime $p, \sum_{k=n+1}^{n+p^{2}-p} k^{k} \equiv 1(\bmod p)$.
Proof. Note that $a^{b}$ depends only on the value of $a(\bmod p)$ and the value of $b(\bmod p-1)$. Since $p$ and $p-1$ are relatively prime, the Chinese Remainder Theorem ${ }^{4}$ implies that any $p^{2}-p$ consecutive integers will take on each possible pair of a residue $\bmod p$ and a residue $\bmod p-1$. In other words, if we let $(a, b)=(k(\bmod p), k(\bmod p-1))$, then as $k$ ranges through $p^{2}-p$ consecutive integers, $(a, b)$ will range through all $p^{2}-p$ ordered pairs of residues $\bmod p$ and residues mod $p-1$. This implies that
$$
\sum_{k=n+1}^{n+p^{2}-p} k^{k} \equiv \sum_{b=1}^{p-1} \sum_{a=1}^{p} a^{b} .
$$
It is well-known that $\sum_{a=1}^{p} a^{b}=\left\{\begin{array}{rl}-1 & p-1 \mid b \\ 0 & p-1 \nmid b\end{array}\right.$. We will sketch a proof here. When $p-1 \mid b$, the result follows from Fermat's Little Theorem ${ }^{5}$. When $p-1 \nmid b$, it suffices to consider the case when $b \mid p-1$, since the $b$ th powers $\bmod p$ are the same as the $\operatorname{gcd}(b, p-1)$ th powers $\bmod p$, and there are an equal number of every non-zero $b$ th power. But in this case, the $b$ th powers are just the solutions to $x^{\frac{p-1}{b}}-1$, which add up to zero by Vieta's formulas.
Now, using the formula for $\sum a^{b}$, we get that
$$
\sum_{b=1}^{p-1} \sum_{a=1}^{p} a^{b} \equiv-1 \quad(\bmod p)
$$
which completes the lemma.
We now apply the lemma with $p=101$ and $n=3,10103$, and 20103 to get that $\sum_{k=1}^{30303} k^{k} \equiv$ $\left(\sum_{k=1}^{3} k^{k}\right)-3$. But $\sum_{k=1}^{3} k^{k}=1^{1}+2^{2}+3^{3}=1+4+27=32$, so the answer is $32-3=29$.
|
29
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Compute the remainder when
$$
\sum_{k=1}^{30303} k^{k}
$$
is divided by 101 .
|
29 The main idea is the following lemma:
Lemma. For any non-negative integer $n$ and prime $p, \sum_{k=n+1}^{n+p^{2}-p} k^{k} \equiv 1(\bmod p)$.
Proof. Note that $a^{b}$ depends only on the value of $a(\bmod p)$ and the value of $b(\bmod p-1)$. Since $p$ and $p-1$ are relatively prime, the Chinese Remainder Theorem ${ }^{4}$ implies that any $p^{2}-p$ consecutive integers will take on each possible pair of a residue $\bmod p$ and a residue $\bmod p-1$. In other words, if we let $(a, b)=(k(\bmod p), k(\bmod p-1))$, then as $k$ ranges through $p^{2}-p$ consecutive integers, $(a, b)$ will range through all $p^{2}-p$ ordered pairs of residues $\bmod p$ and residues mod $p-1$. This implies that
$$
\sum_{k=n+1}^{n+p^{2}-p} k^{k} \equiv \sum_{b=1}^{p-1} \sum_{a=1}^{p} a^{b} .
$$
It is well-known that $\sum_{a=1}^{p} a^{b}=\left\{\begin{array}{rl}-1 & p-1 \mid b \\ 0 & p-1 \nmid b\end{array}\right.$. We will sketch a proof here. When $p-1 \mid b$, the result follows from Fermat's Little Theorem ${ }^{5}$. When $p-1 \nmid b$, it suffices to consider the case when $b \mid p-1$, since the $b$ th powers $\bmod p$ are the same as the $\operatorname{gcd}(b, p-1)$ th powers $\bmod p$, and there are an equal number of every non-zero $b$ th power. But in this case, the $b$ th powers are just the solutions to $x^{\frac{p-1}{b}}-1$, which add up to zero by Vieta's formulas.
Now, using the formula for $\sum a^{b}$, we get that
$$
\sum_{b=1}^{p-1} \sum_{a=1}^{p} a^{b} \equiv-1 \quad(\bmod p)
$$
which completes the lemma.
We now apply the lemma with $p=101$ and $n=3,10103$, and 20103 to get that $\sum_{k=1}^{30303} k^{k} \equiv$ $\left(\sum_{k=1}^{3} k^{k}\right)-3$. But $\sum_{k=1}^{3} k^{k}=1^{1}+2^{2}+3^{3}=1+4+27=32$, so the answer is $32-3=29$.
|
{
"exam": "HMMT",
"problem_label": "29",
"problem_match": "\n29. [18]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
A monomial term $x_{i_{1}} x_{i_{2}} \ldots x_{i_{k}}$ in the variables $x_{1}, x_{2}, \ldots x_{8}$ is square-free if $i_{1}, i_{2}, \ldots i_{k}$ are distinct. (A constant term such as 1 is considered square-free.) What is the sum of the coefficients of the squarefree terms in the following product?
$$
\prod_{1 \leq i<j \leq 8}\left(1+x_{i} x_{j}\right)
$$
|
764 Let $a_{n}$ be the sum of the coefficients of the square-terms in the product $\prod_{1 \leq i<j \leq n}(1+$ $x_{i} x_{j}$ ). Square-free terms in this product come in two types: either they include $x_{n}$, or they do not. The sum of the coefficients of the terms that include $x_{n}$ is $(n-1) a_{n-2}$, since we can choose any of the $n-1$ other variables to be paired with $x_{n}$, and then choose any square-free term from the product taken over the other $n-2$ variables. The sum of the coefficients of the terms that do not include $x_{n}$ are $a_{n-1}$, because they all come from the product over the other $n-1$ variables. Therefore, $a_{n}=a_{n-1}+(n-1) a_{n-2}$.
We use this recursion to find $a_{8}$. As base cases, $a_{0}$ and $a_{1}$ are both 1 . Then $a_{2}=2, a_{3}=4, a_{4}=10$, $a_{5}=26, a_{6}=76, a_{7}=232$, and finally, $a_{8}=764$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
764
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A monomial term $x_{i_{1}} x_{i_{2}} \ldots x_{i_{k}}$ in the variables $x_{1}, x_{2}, \ldots x_{8}$ is square-free if $i_{1}, i_{2}, \ldots i_{k}$ are distinct. (A constant term such as 1 is considered square-free.) What is the sum of the coefficients of the squarefree terms in the following product?
$$
\prod_{1 \leq i<j \leq 8}\left(1+x_{i} x_{j}\right)
$$
|
764 Let $a_{n}$ be the sum of the coefficients of the square-terms in the product $\prod_{1 \leq i<j \leq n}(1+$ $x_{i} x_{j}$ ). Square-free terms in this product come in two types: either they include $x_{n}$, or they do not. The sum of the coefficients of the terms that include $x_{n}$ is $(n-1) a_{n-2}$, since we can choose any of the $n-1$ other variables to be paired with $x_{n}$, and then choose any square-free term from the product taken over the other $n-2$ variables. The sum of the coefficients of the terms that do not include $x_{n}$ are $a_{n-1}$, because they all come from the product over the other $n-1$ variables. Therefore, $a_{n}=a_{n-1}+(n-1) a_{n-2}$.
We use this recursion to find $a_{8}$. As base cases, $a_{0}$ and $a_{1}$ are both 1 . Then $a_{2}=2, a_{3}=4, a_{4}=10$, $a_{5}=26, a_{6}=76, a_{7}=232$, and finally, $a_{8}=764$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"exam": "HMMT",
"problem_label": "30",
"problem_match": "\n30. [18]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
In the Democratic Republic of Irun, 5 people are voting in an election among 5 candidates. If each person votes for a single candidate at random, what is the expected number of candidates that will be voted for?
|
| $\frac{2101}{625}$ |
| :---: |
| The probability that a chosen candidate will receive no votes at all is $\left(\frac{4}{5}\right)^{5}$, or the | probability that every person will vote for someone other than that one candidate. Then the probability that a chosen candidate will receive at least one vote is $1-\left(\frac{4}{5}\right)^{5}=\frac{2101}{3125}$. By linearity of expectations, the final answer is this probability times the number of candidates, or $5 \cdot \frac{2101}{3125}=\frac{2101}{625}$.
|
\frac{2101}{625}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In the Democratic Republic of Irun, 5 people are voting in an election among 5 candidates. If each person votes for a single candidate at random, what is the expected number of candidates that will be voted for?
|
| $\frac{2101}{625}$ |
| :---: |
| The probability that a chosen candidate will receive no votes at all is $\left(\frac{4}{5}\right)^{5}$, or the | probability that every person will vote for someone other than that one candidate. Then the probability that a chosen candidate will receive at least one vote is $1-\left(\frac{4}{5}\right)^{5}=\frac{2101}{3125}$. By linearity of expectations, the final answer is this probability times the number of candidates, or $5 \cdot \frac{2101}{3125}=\frac{2101}{625}$.
|
{
"exam": "HMMT",
"problem_label": "31",
"problem_match": "\n31. [21]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
There are 101 people participating in a Secret Santa gift exchange. As usual each person is randomly assigned another person for whom (s)he has to get a gift, such that each person gives and receives exactly one gift and no one gives a gift to themself. What is the probability that the first person neither gives gifts to or receives gifts from the second or third person? Express your answer as a decimal rounded to five decimal places.
|
0.96039 Let $D_{k}$ denote the number of derangements of $\{1,2, \ldots, k\}$. (A derangement is a permutation in which no element appears in its original position.)
Call the first three people $A, B$, and $C$. Let $X \rightarrow Y$ denote that $X$ gives a gift to $Y$ and let $X \nrightarrow Y$ denote that $X$ gives a gift to anyone other than $Y$. We are fine unless we have $A \rightarrow B, B \rightarrow A$, $A \rightarrow C$, or $C \rightarrow A$. We will compute the number of ways for various things to occur, then combine it into what we want.
There are $D_{n-1}$ ways to have $A \rightarrow B \nrightarrow A$. This is because if $A \rightarrow B$, we can treat $A$ and $B$ as a single vertex, and since $B \nrightarrow A$, we have a derangement. Similarly, there are $D_{n-1}$ ways to have $B \rightarrow A \nrightarrow B$. Thirdly, there are $D_{n-2}$ ways to have $A \rightarrow B \rightarrow A$, since $D_{n-2}$ says how many ways the other $n-2$ people can exchange their gifts. So there are $2 D_{n-1}+D_{n-2}$ ways to have a conflict between $A$ and $B$.
Similarly, there are $2 D_{n-1}+D_{n-2}$ ways to have a conflict between $A$ and $C$.
Using similar arguments, we can show that there are $D_{n-2}$ ways for $B \rightarrow A \rightarrow C \nrightarrow B$ to occur and $D_{n-3}$ ways for $B \rightarrow A \rightarrow C \rightarrow B$ to occur. We get the same results when $B$ and $C$ are reversed. This gives $2 D_{n-2}+2 D_{n-3}$ ways for a conflict to occur within all three people.
By the Principle of Inclusion-Exclusion, the total number of ways to have a conflict is
(\# conflicts between $A$ and $B)+(\#$ conflicts between $A$ and $C)-(\#$ conflicts between $A, B$, and $C)$,
which evaluates to $4 D_{n-1}-2 D_{n-3}$.
Approximating $D_{n}$ as $\frac{n!}{e}$ (the actual formula is this quantity rounded to the nearest integer, so this is a great approximation), we find that the probability of no conflicts is
$1-\frac{4 D_{n-1}-2 D_{n-3}}{D_{n}} \approx 1-4\left(\frac{(n-1)!/ e}{n!/ e}\right)-2\left(\frac{(n-3)!/ e}{n!/ e}\right)=\frac{n(n-1)(n-2)-4(n-1)(n-2)-2}{n(n-1)(n-2)}$.
Substitute $m=n-1$ (this makes $m=100$, so the expression is easier to evaluate) to get a probability of
$$
\begin{gathered}
\frac{m^{3}-m-4 m^{2}+4 m-2}{m^{3}-m}=\frac{m^{3}-4 m^{2}+3 m-2}{m^{3}-m}=\frac{1,000,000-40,000+300-2}{100 \cdot 9999}=\frac{960298}{100 \cdot 9999} \\
=0.960208 \cdot(1.000100010001 \ldots)=0.960208+0.0000960208+\ldots=0.9603940 \ldots
\end{gathered}
$$
To five decimal places, the desired probability is 0.96039 .
|
0.96039
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
There are 101 people participating in a Secret Santa gift exchange. As usual each person is randomly assigned another person for whom (s)he has to get a gift, such that each person gives and receives exactly one gift and no one gives a gift to themself. What is the probability that the first person neither gives gifts to or receives gifts from the second or third person? Express your answer as a decimal rounded to five decimal places.
|
0.96039 Let $D_{k}$ denote the number of derangements of $\{1,2, \ldots, k\}$. (A derangement is a permutation in which no element appears in its original position.)
Call the first three people $A, B$, and $C$. Let $X \rightarrow Y$ denote that $X$ gives a gift to $Y$ and let $X \nrightarrow Y$ denote that $X$ gives a gift to anyone other than $Y$. We are fine unless we have $A \rightarrow B, B \rightarrow A$, $A \rightarrow C$, or $C \rightarrow A$. We will compute the number of ways for various things to occur, then combine it into what we want.
There are $D_{n-1}$ ways to have $A \rightarrow B \nrightarrow A$. This is because if $A \rightarrow B$, we can treat $A$ and $B$ as a single vertex, and since $B \nrightarrow A$, we have a derangement. Similarly, there are $D_{n-1}$ ways to have $B \rightarrow A \nrightarrow B$. Thirdly, there are $D_{n-2}$ ways to have $A \rightarrow B \rightarrow A$, since $D_{n-2}$ says how many ways the other $n-2$ people can exchange their gifts. So there are $2 D_{n-1}+D_{n-2}$ ways to have a conflict between $A$ and $B$.
Similarly, there are $2 D_{n-1}+D_{n-2}$ ways to have a conflict between $A$ and $C$.
Using similar arguments, we can show that there are $D_{n-2}$ ways for $B \rightarrow A \rightarrow C \nrightarrow B$ to occur and $D_{n-3}$ ways for $B \rightarrow A \rightarrow C \rightarrow B$ to occur. We get the same results when $B$ and $C$ are reversed. This gives $2 D_{n-2}+2 D_{n-3}$ ways for a conflict to occur within all three people.
By the Principle of Inclusion-Exclusion, the total number of ways to have a conflict is
(\# conflicts between $A$ and $B)+(\#$ conflicts between $A$ and $C)-(\#$ conflicts between $A, B$, and $C)$,
which evaluates to $4 D_{n-1}-2 D_{n-3}$.
Approximating $D_{n}$ as $\frac{n!}{e}$ (the actual formula is this quantity rounded to the nearest integer, so this is a great approximation), we find that the probability of no conflicts is
$1-\frac{4 D_{n-1}-2 D_{n-3}}{D_{n}} \approx 1-4\left(\frac{(n-1)!/ e}{n!/ e}\right)-2\left(\frac{(n-3)!/ e}{n!/ e}\right)=\frac{n(n-1)(n-2)-4(n-1)(n-2)-2}{n(n-1)(n-2)}$.
Substitute $m=n-1$ (this makes $m=100$, so the expression is easier to evaluate) to get a probability of
$$
\begin{gathered}
\frac{m^{3}-m-4 m^{2}+4 m-2}{m^{3}-m}=\frac{m^{3}-4 m^{2}+3 m-2}{m^{3}-m}=\frac{1,000,000-40,000+300-2}{100 \cdot 9999}=\frac{960298}{100 \cdot 9999} \\
=0.960208 \cdot(1.000100010001 \ldots)=0.960208+0.0000960208+\ldots=0.9603940 \ldots
\end{gathered}
$$
To five decimal places, the desired probability is 0.96039 .
|
{
"exam": "HMMT",
"problem_label": "32",
"problem_match": "\n32. [21]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $a_{1}=3$, and for $n>1$, let $a_{n}$ be the largest real number such that
$$
4\left(a_{n-1}^{2}+a_{n}^{2}\right)=10 a_{n-1} a_{n}-9
$$
What is the largest positive integer less than $a_{8}$ ?
|
335 Let $t_{n}$ be the larger real such that $a_{n}=t_{n}+\frac{1}{t_{n}}$. Then $t_{1}=\frac{3+\sqrt{5}}{2}$. We claim that $t_{n}=2 t_{n-1}$. Writing the recurrence as a quadratic polynomial in $a_{n}$, we have:
$$
4 a_{n}^{2}-10 a_{n-1} a_{n}+4 a_{n-1}^{2}+9=0
$$
Using the quadratic formula, we see that $a_{n}=\frac{5}{4} a_{n-1}+\frac{3}{4} \sqrt{a_{n-1}^{2}-4}$. (We ignore the negative square root, since $a_{n}$ is the largest real number satisfying the polynomial.) Substituting $t_{n-1}+\frac{1}{t_{n-1}}$ for $a_{n-1}$, we see that $\sqrt{a_{n-1}^{2}-4}=\sqrt{t_{n-1}^{2}-2+\frac{1}{t_{n-1}^{2}}}$, so we have:
$$
a_{n}=\frac{5}{4}\left(t_{n-1}+\frac{1}{t_{n-1}}\right)+\frac{3}{4} \sqrt{\left(t_{n-1}-\frac{1}{t_{n-1}}\right)^{2}}=2 t_{n-1}+\frac{1}{2 t_{n-1}}
$$
so $t_{n}=2 t_{n-1}$, as claimed. Then $a_{8}=\frac{128(3+\sqrt{5})}{2}+\frac{2}{128(3+\sqrt{5})}$. The second term is vanishingly small, so $\left\lfloor a_{8}\right\rfloor=\lfloor 64(3+\sqrt{5})\rfloor$. We approximate $\sqrt{5}$ to two decimal places as 2.24 , making this expression $\lfloor 335.36\rfloor=335$. Since our value of $\sqrt{5}$ is correct to within 0.005 , the decimal is correct to within 0.32 , which means the final answer is exact.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
335
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a_{1}=3$, and for $n>1$, let $a_{n}$ be the largest real number such that
$$
4\left(a_{n-1}^{2}+a_{n}^{2}\right)=10 a_{n-1} a_{n}-9
$$
What is the largest positive integer less than $a_{8}$ ?
|
335 Let $t_{n}$ be the larger real such that $a_{n}=t_{n}+\frac{1}{t_{n}}$. Then $t_{1}=\frac{3+\sqrt{5}}{2}$. We claim that $t_{n}=2 t_{n-1}$. Writing the recurrence as a quadratic polynomial in $a_{n}$, we have:
$$
4 a_{n}^{2}-10 a_{n-1} a_{n}+4 a_{n-1}^{2}+9=0
$$
Using the quadratic formula, we see that $a_{n}=\frac{5}{4} a_{n-1}+\frac{3}{4} \sqrt{a_{n-1}^{2}-4}$. (We ignore the negative square root, since $a_{n}$ is the largest real number satisfying the polynomial.) Substituting $t_{n-1}+\frac{1}{t_{n-1}}$ for $a_{n-1}$, we see that $\sqrt{a_{n-1}^{2}-4}=\sqrt{t_{n-1}^{2}-2+\frac{1}{t_{n-1}^{2}}}$, so we have:
$$
a_{n}=\frac{5}{4}\left(t_{n-1}+\frac{1}{t_{n-1}}\right)+\frac{3}{4} \sqrt{\left(t_{n-1}-\frac{1}{t_{n-1}}\right)^{2}}=2 t_{n-1}+\frac{1}{2 t_{n-1}}
$$
so $t_{n}=2 t_{n-1}$, as claimed. Then $a_{8}=\frac{128(3+\sqrt{5})}{2}+\frac{2}{128(3+\sqrt{5})}$. The second term is vanishingly small, so $\left\lfloor a_{8}\right\rfloor=\lfloor 64(3+\sqrt{5})\rfloor$. We approximate $\sqrt{5}$ to two decimal places as 2.24 , making this expression $\lfloor 335.36\rfloor=335$. Since our value of $\sqrt{5}$ is correct to within 0.005 , the decimal is correct to within 0.32 , which means the final answer is exact.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"exam": "HMMT",
"problem_label": "33",
"problem_match": "\n33. [21]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
3000 people each go into one of three rooms randomly. What is the most likely value for the maximum number of people in any of the rooms? Your score for this problem will be 0 if you write down a number less than or equal to 1000 . Otherwise, it will be $25-27 \frac{|A-C|}{\min (A, C)-1000}$.
|
1019 To get a rough approximation, we can use the fact that a sum of identical random variables converges to a Gaussian distribution ${ }^{6}$ in this case with a mean of 1000 and a variance of $3000 \cdot \frac{2}{9}=667$. Since $\sqrt{667} \approx 26,1026$ is a good guess, as Gaussians tend to differ from their mean by approximately their variance.
The actual answer was computed with the following python program:
```
facts = [0]*3001
facts[0]=1
for a in range(1,3001):
facts[a]=a*facts[a-1]
def binom(n,k):
return facts[n]/(facts[k]*facts[n-k])
```
[^3]```
maxes = [0]*3001
M = 1075
for a in range(0,3001):
for b in range(0,3001-a):
c = 3000-a-b
m = max (a,max (b,c))
if m < M:
maxes[m] += facts[3000]/(facts[a]*facts[b]*facts[c])
print [a,b]
best = 1000
for a in range(1000,1050):
print maxes[a],a
if maxes[best] <= maxes[a]:
best = a
print maxes[best]
print best
```
We can use arguments involving the Chernoff bound ${ }^{7}$ to show that the answer is necessarily less than 1075. Alternately, if we wanted to be really careful, we could just set $M=3001$, but then we'd have to wait a while for the script to finish.
|
1019
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
3000 people each go into one of three rooms randomly. What is the most likely value for the maximum number of people in any of the rooms? Your score for this problem will be 0 if you write down a number less than or equal to 1000 . Otherwise, it will be $25-27 \frac{|A-C|}{\min (A, C)-1000}$.
|
1019 To get a rough approximation, we can use the fact that a sum of identical random variables converges to a Gaussian distribution ${ }^{6}$ in this case with a mean of 1000 and a variance of $3000 \cdot \frac{2}{9}=667$. Since $\sqrt{667} \approx 26,1026$ is a good guess, as Gaussians tend to differ from their mean by approximately their variance.
The actual answer was computed with the following python program:
```
facts = [0]*3001
facts[0]=1
for a in range(1,3001):
facts[a]=a*facts[a-1]
def binom(n,k):
return facts[n]/(facts[k]*facts[n-k])
```
[^3]```
maxes = [0]*3001
M = 1075
for a in range(0,3001):
for b in range(0,3001-a):
c = 3000-a-b
m = max (a,max (b,c))
if m < M:
maxes[m] += facts[3000]/(facts[a]*facts[b]*facts[c])
print [a,b]
best = 1000
for a in range(1000,1050):
print maxes[a],a
if maxes[best] <= maxes[a]:
best = a
print maxes[best]
print best
```
We can use arguments involving the Chernoff bound ${ }^{7}$ to show that the answer is necessarily less than 1075. Alternately, if we wanted to be really careful, we could just set $M=3001$, but then we'd have to wait a while for the script to finish.
|
{
"exam": "HMMT",
"problem_label": "34",
"problem_match": "\n34. [25]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Call an positive integer almost-square if it can be written as $a \cdot b$, where $a$ and $b$ are integers and $a \leq b \leq \frac{4}{3} a$. How many almost-square positive integers are less than or equal to 1000000 ? Your score will be equal to $25-65 \frac{|A-C|}{\min (A, C)}$.
|
130348 To get a good estimate for the number of almost-square integers, note that any number of the form $a \cdot b$, with $b \leq \frac{4}{3} a$, will be by definition almost-square. Let's assume that it's relatively unlikely that a number is almost-square in more than one way. Then the number of almostsquare numbers less than $n$ will be approximately
$$
\sum_{a=1}^{\sqrt{n}} \sum_{b=a}^{\frac{4}{3} a} 1=\frac{1}{3} \sum_{a=1}^{\sqrt{n}} a=\frac{1}{6} \sqrt{n}(\sqrt{n}+1)
$$
which is about $\frac{n}{6}$. So, $\frac{n}{6}$ will be a fairly good estimate for the number of almost-square numbers less than $n$, making 160000 a reasonable guess.
We can do better, though. For example, we summed $\frac{a}{3}$ all the way up to $\sqrt{n}$, but we are really overcounting here because when $a$ is close to $\sqrt{n}, a \cdot b$ will be less than $n$ only when $b \leq \frac{n}{a}$, as opposed to $b \leq \frac{4 a}{3}$. So we should really be taking the sum
[^4]\[
$$
\begin{aligned}
& \sum_{a=1}^{\sqrt{\frac{3 n}{4}}} \sum_{b=a}^{\frac{4 a}{3}} 1+\sum_{a=\sqrt{\frac{3 n}{4}}}^{\sqrt{n}} \sum_{b=a}^{\frac{n}{a}} 1 \\
& \quad=\sum_{a=1}^{\sqrt{\frac{3 n}{4}}} \frac{a}{3}+\sum_{a=\sqrt{\frac{3 n}{4}}}^{\sqrt{n}}\left(\frac{n}{a}-a\right) \\
& \quad \approx \frac{1}{6} \frac{3 n}{4}+n\left(\log (\sqrt{n})-\log \left(\sqrt{\frac{3 n}{4}}\right)\right)-\left(\frac{n}{2}-\frac{3 n}{8}\right) \\
& \quad=\frac{n}{8}+n \frac{\log (4)-\log (3)}{2}-\frac{n}{8} \\
& \quad=n \frac{\log (4)-\log (3)}{2}
\end{aligned}
$$
\]
In the process of taking the sum, we saw that we had something between $\frac{n}{8}$ and $\frac{n}{6}$, so we could also guess something between 166000 and 125000 , which would give us about 145000 , an even better answer. If we actually calculate $\frac{\log (4)-\log (3)}{2}$, we see that it's about 0.14384 , so 143840 would be the best guess if we were to use this strategy. In reality, we would want to round down a bit in both cases, since we are overcounting (because numbers could be square-free in multiple ways), so we should probably answer something like 140000 .
A final refinement to our calculation (and perhaps easier than the previous one), is to assume that the products $a \cdot b$ that we consider are randomly distributed between 1 and $n$, and to compute the expected number of distinct numbers we end up with. This is the same type of problem as number 31 on this contest, and we compute that if we randomly distribute $k$ numbers between 1 and $n$ then we expect to end up with $n\left(1-\left(1-\frac{1}{n}\right)^{k}\right)$ distinct numbers. When $k=n \frac{\log (4)-\log (3)}{2}$, we get that this equals
$$
\begin{aligned}
n\left(1-\left(\left(1-\frac{1}{n}\right)^{n}\right)^{\frac{\log (4)-\log (3)}{2}}\right) & =n\left(1-\sqrt{e^{\log (3)-\log (4)}}\right) \\
& =n\left(1-\sqrt{\frac{3}{4}}\right) \\
& =n\left(1-\frac{\sqrt{3}}{2}\right) \\
& \approx 0.134 n
\end{aligned}
$$
Giving us an answer of 134000 , which is very close to the correct answer.
The actual answer was found by computer, using the following $\mathrm{C}++$ program:
```
#include <stdio.h>
using namespace std;
bool isAlmostSquare(int n){
for(int k=1;k*k<=n;k++)
if(n%k==0 && 3*(n/k) <= 4*k) return true;
return false;
}
```
```
int main(){
int c = 0;
for(int n=1;n<=1000000;n++)
if(isAlmostSquare(n)) c++;
printf("%d\n",c);
return 0;
}
```
|
130348
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Call an positive integer almost-square if it can be written as $a \cdot b$, where $a$ and $b$ are integers and $a \leq b \leq \frac{4}{3} a$. How many almost-square positive integers are less than or equal to 1000000 ? Your score will be equal to $25-65 \frac{|A-C|}{\min (A, C)}$.
|
130348 To get a good estimate for the number of almost-square integers, note that any number of the form $a \cdot b$, with $b \leq \frac{4}{3} a$, will be by definition almost-square. Let's assume that it's relatively unlikely that a number is almost-square in more than one way. Then the number of almostsquare numbers less than $n$ will be approximately
$$
\sum_{a=1}^{\sqrt{n}} \sum_{b=a}^{\frac{4}{3} a} 1=\frac{1}{3} \sum_{a=1}^{\sqrt{n}} a=\frac{1}{6} \sqrt{n}(\sqrt{n}+1)
$$
which is about $\frac{n}{6}$. So, $\frac{n}{6}$ will be a fairly good estimate for the number of almost-square numbers less than $n$, making 160000 a reasonable guess.
We can do better, though. For example, we summed $\frac{a}{3}$ all the way up to $\sqrt{n}$, but we are really overcounting here because when $a$ is close to $\sqrt{n}, a \cdot b$ will be less than $n$ only when $b \leq \frac{n}{a}$, as opposed to $b \leq \frac{4 a}{3}$. So we should really be taking the sum
[^4]\[
$$
\begin{aligned}
& \sum_{a=1}^{\sqrt{\frac{3 n}{4}}} \sum_{b=a}^{\frac{4 a}{3}} 1+\sum_{a=\sqrt{\frac{3 n}{4}}}^{\sqrt{n}} \sum_{b=a}^{\frac{n}{a}} 1 \\
& \quad=\sum_{a=1}^{\sqrt{\frac{3 n}{4}}} \frac{a}{3}+\sum_{a=\sqrt{\frac{3 n}{4}}}^{\sqrt{n}}\left(\frac{n}{a}-a\right) \\
& \quad \approx \frac{1}{6} \frac{3 n}{4}+n\left(\log (\sqrt{n})-\log \left(\sqrt{\frac{3 n}{4}}\right)\right)-\left(\frac{n}{2}-\frac{3 n}{8}\right) \\
& \quad=\frac{n}{8}+n \frac{\log (4)-\log (3)}{2}-\frac{n}{8} \\
& \quad=n \frac{\log (4)-\log (3)}{2}
\end{aligned}
$$
\]
In the process of taking the sum, we saw that we had something between $\frac{n}{8}$ and $\frac{n}{6}$, so we could also guess something between 166000 and 125000 , which would give us about 145000 , an even better answer. If we actually calculate $\frac{\log (4)-\log (3)}{2}$, we see that it's about 0.14384 , so 143840 would be the best guess if we were to use this strategy. In reality, we would want to round down a bit in both cases, since we are overcounting (because numbers could be square-free in multiple ways), so we should probably answer something like 140000 .
A final refinement to our calculation (and perhaps easier than the previous one), is to assume that the products $a \cdot b$ that we consider are randomly distributed between 1 and $n$, and to compute the expected number of distinct numbers we end up with. This is the same type of problem as number 31 on this contest, and we compute that if we randomly distribute $k$ numbers between 1 and $n$ then we expect to end up with $n\left(1-\left(1-\frac{1}{n}\right)^{k}\right)$ distinct numbers. When $k=n \frac{\log (4)-\log (3)}{2}$, we get that this equals
$$
\begin{aligned}
n\left(1-\left(\left(1-\frac{1}{n}\right)^{n}\right)^{\frac{\log (4)-\log (3)}{2}}\right) & =n\left(1-\sqrt{e^{\log (3)-\log (4)}}\right) \\
& =n\left(1-\sqrt{\frac{3}{4}}\right) \\
& =n\left(1-\frac{\sqrt{3}}{2}\right) \\
& \approx 0.134 n
\end{aligned}
$$
Giving us an answer of 134000 , which is very close to the correct answer.
The actual answer was found by computer, using the following $\mathrm{C}++$ program:
```
#include <stdio.h>
using namespace std;
bool isAlmostSquare(int n){
for(int k=1;k*k<=n;k++)
if(n%k==0 && 3*(n/k) <= 4*k) return true;
return false;
}
```
```
int main(){
int c = 0;
for(int n=1;n<=1000000;n++)
if(isAlmostSquare(n)) c++;
printf("%d\n",c);
return 0;
}
```
|
{
"exam": "HMMT",
"problem_label": "35",
"problem_match": "\n35. [25]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Consider an infinite grid of unit squares. An $n$-omino is a subset of $n$ squares that is connected. Below are depicted examples of 8 -ominoes. Two $n$-ominoes are considered equivalent if one can be obtained from the other by translations and rotations. What is the number of distinct 15 -ominoes? Your score will be equal to $25-13|\ln (A)-\ln (C)|$.
|
3426576 We claim that there are approximately $\frac{3^{n-1}}{4} n$-ominoes. First, we define an order on the squares in an $n$-omino, as follows: we order the squares from left to right, and within a column, we order the squares from top to bottom.
We construct an $n$-omino by starting with a single square and attaching squares, one by one, as close to order we just defined as possible. After the first square is placed, the next square can be attached to its bottom or right side. If the second square is right of the first, then the third square can be attached to the top, right, or bottom edges of the second. The third square cannot be attached to the first square, because the first column must be completed before the second is begun. If the second square is below the first, the third square can be attached to the bottom or right sides of the second square, or the right side of the first way. In either case, there are 3 choices for the third square.
Suppose that $m$ squares have been added, and that $k$ squares are in the right-most column right now. If $k=1$, then the next square can be attached to the top, right, or bottom side of this square. If $k>1$, then the next square can be added to the bottom or right side of the last square in this column. The next square can also be added to the right of the top square in this column, so there are still 3 choices for the next square. Therefore, there are $3^{n-1}$ ways to build up the entire $n$-omino.
Almost all $n$-omines have no rotational symmetry, so they can be built in 4 different ways by this method. This correction gives us our estimate of $\frac{3^{n-1}}{4}$ for the number of $n$-ominoes. This reasoning is not exactly correct, because some $n$-ominoes are highly non-convex, and do not admit an in-order traversal like the one described above. If some columns in the $n$-ominoes have gaps, they are not enumerated by this construction, so $\frac{3^{n-1}}{4}$ is an underestimate. We estimate that there are 1195742 15 -ominoes. The actual value of 3426576 , which can be found in the Sloane Encyclopedia of Integer Sequences, differes from the estimate by less than a factor of 3 .
[^0]: ${ }^{1}$ See http://www-math.mit.edu/~spielman/AdvComplexity/handout.ps for an introduction to linearity of expectation and other important tools in probability
[^1]: ${ }^{2}$ See the sections "Linear homogeneous recurrence relations with constant coefficients" and "Theorem" at http://en. wikipedia.org/wiki/Recurrence_relation
${ }^{3}$ See http://mathworld.wolfram.com/VietasFormulas.html
[^2]: ${ }^{4}$ See http://en.wikipedia.org/wiki/Chinese_remainder_theorem
${ }^{5}$ See http://en.wikipedia.org/wiki/Fermat's_little_theorem
[^3]: ${ }^{6}$ See http://en.wikipedia.org/wiki/Central_limit_theorem
[^4]: ${ }^{7}$ See http://en.wikipedia.org/wiki/Chernoff_bound
|
3426576
|
Incomplete
|
Yes
|
math-word-problem
|
Combinatorics
|
Consider an infinite grid of unit squares. An $n$-omino is a subset of $n$ squares that is connected. Below are depicted examples of 8 -ominoes. Two $n$-ominoes are considered equivalent if one can be obtained from the other by translations and rotations. What is the number of distinct 15 -ominoes? Your score will be equal to $25-13|\ln (A)-\ln (C)|$.
|
3426576 We claim that there are approximately $\frac{3^{n-1}}{4} n$-ominoes. First, we define an order on the squares in an $n$-omino, as follows: we order the squares from left to right, and within a column, we order the squares from top to bottom.
We construct an $n$-omino by starting with a single square and attaching squares, one by one, as close to order we just defined as possible. After the first square is placed, the next square can be attached to its bottom or right side. If the second square is right of the first, then the third square can be attached to the top, right, or bottom edges of the second. The third square cannot be attached to the first square, because the first column must be completed before the second is begun. If the second square is below the first, the third square can be attached to the bottom or right sides of the second square, or the right side of the first way. In either case, there are 3 choices for the third square.
Suppose that $m$ squares have been added, and that $k$ squares are in the right-most column right now. If $k=1$, then the next square can be attached to the top, right, or bottom side of this square. If $k>1$, then the next square can be added to the bottom or right side of the last square in this column. The next square can also be added to the right of the top square in this column, so there are still 3 choices for the next square. Therefore, there are $3^{n-1}$ ways to build up the entire $n$-omino.
Almost all $n$-omines have no rotational symmetry, so they can be built in 4 different ways by this method. This correction gives us our estimate of $\frac{3^{n-1}}{4}$ for the number of $n$-ominoes. This reasoning is not exactly correct, because some $n$-ominoes are highly non-convex, and do not admit an in-order traversal like the one described above. If some columns in the $n$-ominoes have gaps, they are not enumerated by this construction, so $\frac{3^{n-1}}{4}$ is an underestimate. We estimate that there are 1195742 15 -ominoes. The actual value of 3426576 , which can be found in the Sloane Encyclopedia of Integer Sequences, differes from the estimate by less than a factor of 3 .
[^0]: ${ }^{1}$ See http://www-math.mit.edu/~spielman/AdvComplexity/handout.ps for an introduction to linearity of expectation and other important tools in probability
[^1]: ${ }^{2}$ See the sections "Linear homogeneous recurrence relations with constant coefficients" and "Theorem" at http://en. wikipedia.org/wiki/Recurrence_relation
${ }^{3}$ See http://mathworld.wolfram.com/VietasFormulas.html
[^2]: ${ }^{4}$ See http://en.wikipedia.org/wiki/Chinese_remainder_theorem
${ }^{5}$ See http://en.wikipedia.org/wiki/Fermat's_little_theorem
[^3]: ${ }^{6}$ See http://en.wikipedia.org/wiki/Central_limit_theorem
[^4]: ${ }^{7}$ See http://en.wikipedia.org/wiki/Chernoff_bound
|
{
"exam": "HMMT",
"problem_label": "36",
"problem_match": "\n36. [25]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
You are trying to sink a submarine. Every second, you launch a missile at a point of your choosing on the $x$-axis. If the submarine is at that point at that time, you sink it. A firing sequence is a sequence of real numbers that specify where you will fire at each second. For example, the firing sequence $2,3,5,6, \ldots$ means that you will fire at 2 after one second, 3 after two seconds, 5 after three seconds, 6 after four seconds, and so on.
(a) [5] Suppose that the submarine starts at the origin and travels along the positive $x$-axis with an (unknown) positive integer velocity. Show that there is a firing sequence that is guaranteed to hit the submarine eventually.
|
The firing sequence $1,4,9, \ldots, n^{2}, \ldots$ works. If the velocity of the submarine is $v$, then after $v$ seconds it will be at $x=v^{2}$, the same location where the mine explodes at time $v$.
(b) [10] Suppose now that the submarine starts at an unknown integer point on the non-negative $x$-axis and again travels with an unknown positive integer velocity. Show that there is still a firing sequence that is guaranteed to hit the submarine eventually.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
You are trying to sink a submarine. Every second, you launch a missile at a point of your choosing on the $x$-axis. If the submarine is at that point at that time, you sink it. A firing sequence is a sequence of real numbers that specify where you will fire at each second. For example, the firing sequence $2,3,5,6, \ldots$ means that you will fire at 2 after one second, 3 after two seconds, 5 after three seconds, 6 after four seconds, and so on.
(a) [5] Suppose that the submarine starts at the origin and travels along the positive $x$-axis with an (unknown) positive integer velocity. Show that there is a firing sequence that is guaranteed to hit the submarine eventually.
|
The firing sequence $1,4,9, \ldots, n^{2}, \ldots$ works. If the velocity of the submarine is $v$, then after $v$ seconds it will be at $x=v^{2}$, the same location where the mine explodes at time $v$.
(b) [10] Suppose now that the submarine starts at an unknown integer point on the non-negative $x$-axis and again travels with an unknown positive integer velocity. Show that there is still a firing sequence that is guaranteed to hit the submarine eventually.
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. ",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2010"
}
|
You are trying to sink a submarine. Every second, you launch a missile at a point of your choosing on the $x$-axis. If the submarine is at that point at that time, you sink it. A firing sequence is a sequence of real numbers that specify where you will fire at each second. For example, the firing sequence $2,3,5,6, \ldots$ means that you will fire at 2 after one second, 3 after two seconds, 5 after three seconds, 6 after four seconds, and so on.
(a) [5] Suppose that the submarine starts at the origin and travels along the positive $x$-axis with an (unknown) positive integer velocity. Show that there is a firing sequence that is guaranteed to hit the submarine eventually.
|
Represent the submarine's motion by an ordered pair ( $a, b$ ), where $a$ is the starting point of the submarine and $b$ is its velocity. We want to find a way to map each positive integer to a possible ordered pair so that every ordered pair is covered. This way, if we fire at $b_{n} n+a_{n}$ at time $n$, where $\left(a_{n}, b_{n}\right)$ is the point that $n$ maps to, then we will eventually hit the submarine. (Keep in mind that $b_{n} n+a_{n}$ would be the location of the submarine at time $n$.) There are many such ways to map the positive integers to possible points; here is one way:
$$
\begin{aligned}
& 1 \rightarrow(1,1), 2 \rightarrow(2,1), 3 \rightarrow(1,2), 4 \rightarrow(3,1), 5 \rightarrow(2,2), 6 \rightarrow(1,3), 7 \rightarrow(4,1), 8 \rightarrow(3,2), \\
& 9 \rightarrow(2,3), 10 \rightarrow(1,4), 11 \rightarrow(5,1), 12 \rightarrow(4,2), 13 \rightarrow(3,3), 14 \rightarrow(2,4), 15 \rightarrow(1,5), \ldots
\end{aligned}
$$
(The path of points trace out diagonal lines that sweep every lattice point in the coordinate plane.) Since we cover every point, we will eventually hit the submarine.
Remark: The mapping shown above is known as a bijection between the positive integers and ordered pairs of integers $(a, b)$ where $b>0$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
You are trying to sink a submarine. Every second, you launch a missile at a point of your choosing on the $x$-axis. If the submarine is at that point at that time, you sink it. A firing sequence is a sequence of real numbers that specify where you will fire at each second. For example, the firing sequence $2,3,5,6, \ldots$ means that you will fire at 2 after one second, 3 after two seconds, 5 after three seconds, 6 after four seconds, and so on.
(a) [5] Suppose that the submarine starts at the origin and travels along the positive $x$-axis with an (unknown) positive integer velocity. Show that there is a firing sequence that is guaranteed to hit the submarine eventually.
|
Represent the submarine's motion by an ordered pair ( $a, b$ ), where $a$ is the starting point of the submarine and $b$ is its velocity. We want to find a way to map each positive integer to a possible ordered pair so that every ordered pair is covered. This way, if we fire at $b_{n} n+a_{n}$ at time $n$, where $\left(a_{n}, b_{n}\right)$ is the point that $n$ maps to, then we will eventually hit the submarine. (Keep in mind that $b_{n} n+a_{n}$ would be the location of the submarine at time $n$.) There are many such ways to map the positive integers to possible points; here is one way:
$$
\begin{aligned}
& 1 \rightarrow(1,1), 2 \rightarrow(2,1), 3 \rightarrow(1,2), 4 \rightarrow(3,1), 5 \rightarrow(2,2), 6 \rightarrow(1,3), 7 \rightarrow(4,1), 8 \rightarrow(3,2), \\
& 9 \rightarrow(2,3), 10 \rightarrow(1,4), 11 \rightarrow(5,1), 12 \rightarrow(4,2), 13 \rightarrow(3,3), 14 \rightarrow(2,4), 15 \rightarrow(1,5), \ldots
\end{aligned}
$$
(The path of points trace out diagonal lines that sweep every lattice point in the coordinate plane.) Since we cover every point, we will eventually hit the submarine.
Remark: The mapping shown above is known as a bijection between the positive integers and ordered pairs of integers $(a, b)$ where $b>0$.
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. ",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2010"
}
|
Consider the following two-player game. Player 1 starts with a number, $N$. He then subtracts a proper divisor of $N$ from $N$ and gives the result to player 2 (a proper divisor of $N$ is a positive divisor of $N$ that is not equal to 1 or $N)$. Player 2 does the same thing with the number she gets from player 1 , and gives the result back to player 1. The two players continue until a player is given a prime number or 1 , at which point that player loses. For which values of $N$ does player 1 have a winning strategy?
|
All even numbers except for odd powers of 2 . First we show that if you are stuck with an odd number, then you are guaranteed to lose. Suppose you have an odd number $a b$, where $a$ and $b$ are odd numbers, and you choose to subtract $a$. You pass your opponent the number $a(b-1)$. This cannot be a power of 2 (otherwise $a$ is a power of 2 and hence $a=1$, which is not allowed), so your opponent can find an odd proper divisor of $a(b-1)$, and you will have a smaller odd number. Eventually you will get to an odd prime and lose.
Now consider even numbers that aren't powers of 2. As with before, you can find an odd proper divisor of $N$ and pass your opponent an odd number, so you are guaranteed to win.
Finally consider powers of 2 . If you have the number $N=2^{k}$, it would be unwise to choose a proper divisor other than $2^{k-1}$; otherwise you would give your opponent an even number that isn't a power of 2 . Therefore if $k$ is odd, you will end up with 2 and lose. If $k$ is even, though, your opponent will end up with 2 and you will win.
Therefore player 1 has a winning strategy for all even numbers except for odd powers of 2.
|
All even numbers except for odd powers of 2
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Consider the following two-player game. Player 1 starts with a number, $N$. He then subtracts a proper divisor of $N$ from $N$ and gives the result to player 2 (a proper divisor of $N$ is a positive divisor of $N$ that is not equal to 1 or $N)$. Player 2 does the same thing with the number she gets from player 1 , and gives the result back to player 1. The two players continue until a player is given a prime number or 1 , at which point that player loses. For which values of $N$ does player 1 have a winning strategy?
|
All even numbers except for odd powers of 2 . First we show that if you are stuck with an odd number, then you are guaranteed to lose. Suppose you have an odd number $a b$, where $a$ and $b$ are odd numbers, and you choose to subtract $a$. You pass your opponent the number $a(b-1)$. This cannot be a power of 2 (otherwise $a$ is a power of 2 and hence $a=1$, which is not allowed), so your opponent can find an odd proper divisor of $a(b-1)$, and you will have a smaller odd number. Eventually you will get to an odd prime and lose.
Now consider even numbers that aren't powers of 2. As with before, you can find an odd proper divisor of $N$ and pass your opponent an odd number, so you are guaranteed to win.
Finally consider powers of 2 . If you have the number $N=2^{k}$, it would be unwise to choose a proper divisor other than $2^{k-1}$; otherwise you would give your opponent an even number that isn't a power of 2 . Therefore if $k$ is odd, you will end up with 2 and lose. If $k$ is even, though, your opponent will end up with 2 and you will win.
Therefore player 1 has a winning strategy for all even numbers except for odd powers of 2.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [15]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Call a positive integer in base $10 k$-good if we can split it into two integers y and z , such that y is all digits on the left and z is all digits on the right, and such that $y=k \cdot z$. For example, 2010 is 2 -good because we can split it into 20 and 10 and $20=2 \cdot 10.20010$ is also 2 -good, because we can split it into 20 and 010 . In addition, it is 20 -good, because we can split it into 200 and 10 .
Show that there exists a 48 -good perfect square.
|
We wish to find integers $a, z$ such that $48 z \cdot 10^{a}+z=z\left(48 \cdot 10^{a}+1\right)$ a perfect square, where $z<10^{a}$. This would prove that there exists a 48 -good perfect square because we are pulling off the last $a$ digits of the number and get two integers $48 z$ and $z$. To make $z$ small by keeping the product a perfect square, we'd like $48 \cdot 10^{a}+1$ to be divisible by some reasonably large square. Take $a=42=\varphi(49)$. By Euler's theorem, $10^{42} \equiv 1(\bmod 49)$, so $48 \cdot 10^{a}+1$ is a multiple of 49 . Then we can take $z=\frac{48 \cdot 10^{a}+1}{49}$. (Clearly $z<10^{a}$, so we're fine.) Then we have $z\left(48 \cdot 10^{a}+1\right)=\left(\frac{48 \cdot 10^{42}+1}{7}\right)^{2}$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Call a positive integer in base $10 k$-good if we can split it into two integers y and z , such that y is all digits on the left and z is all digits on the right, and such that $y=k \cdot z$. For example, 2010 is 2 -good because we can split it into 20 and 10 and $20=2 \cdot 10.20010$ is also 2 -good, because we can split it into 20 and 010 . In addition, it is 20 -good, because we can split it into 200 and 10 .
Show that there exists a 48 -good perfect square.
|
We wish to find integers $a, z$ such that $48 z \cdot 10^{a}+z=z\left(48 \cdot 10^{a}+1\right)$ a perfect square, where $z<10^{a}$. This would prove that there exists a 48 -good perfect square because we are pulling off the last $a$ digits of the number and get two integers $48 z$ and $z$. To make $z$ small by keeping the product a perfect square, we'd like $48 \cdot 10^{a}+1$ to be divisible by some reasonably large square. Take $a=42=\varphi(49)$. By Euler's theorem, $10^{42} \equiv 1(\bmod 49)$, so $48 \cdot 10^{a}+1$ is a multiple of 49 . Then we can take $z=\frac{48 \cdot 10^{a}+1}{49}$. (Clearly $z<10^{a}$, so we're fine.) Then we have $z\left(48 \cdot 10^{a}+1\right)=\left(\frac{48 \cdot 10^{42}+1}{7}\right)^{2}$.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. [15]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2010"
}
|
Let
$$
\begin{gathered}
e^{x}+e^{y}=A \\
x e^{x}+y e^{y}=B \\
x^{2} e^{x}+y^{2} e^{y}=C \\
x^{3} e^{x}+y^{3} e^{y}=D \\
x^{4} e^{x}+y^{4} e^{y}=E .
\end{gathered}
$$
Prove that if $A, B, C$, and $D$ are all rational, then so is $E$.
|
We can express $x+y$ in two ways:
$$
\begin{aligned}
& x+y=\frac{A D-B C}{A C-B^{2}} \\
& x+y=\frac{A E-C^{2}}{A D-B C}
\end{aligned}
$$
(We have to be careful if $A C-B^{2}$ or $A D-B C$ is zero. We'll deal with that case later.) It is easy to check that these equations hold by substituting the expressions for $A, B, C, D$, and $E$. Setting these two expressions for $x+y$ equal to each other, we get
$$
\frac{A D-B C}{A C-B^{2}}=\frac{A E-C^{2}}{A D-B C}
$$
which we can easily solve for $E$ as a rational function of $A, B, C$, and $D$. Therefore if $A, B, C$, and $D$ are all rational, then $E$ will be rational as well.
Now, we have to check what happens if $A C-B^{2}=0$ or $A D-B C=0$. If $A C-B^{2}=0$, then writing down the expressions for $A, B$, and $C$ gives us that $(x-y)^{2} e^{x+y}=0$, meaning that $x=y$. If $x=y$, and $x \neq 0, A$ and $D$ are also non-zero, and $\frac{B}{A}=\frac{E}{D}=x$. Since $\frac{B}{A}$ is rational and $D$ is rational, this implies that $E$ is rational. If $x=y=0$, then $E=0$ and so is certainly rational.
We finally must check what happens if $A D-B C=0$. Since $A D-B C=(x+y)\left(A C-B^{2}\right)$, either $A C-B^{2}=0$ (a case we have already dealt with), or $x+y=0$. But if $x+y=0$ then $A E-C^{2}=0$, which implies that $E=\frac{C^{2}}{A}$ (we know that $A \neq 0$ because $e^{x}$ and $e^{y}$ are both positive). Since $A$ and $C$ are rational, this implies that $E$ is also rational.
So, we have shown $E$ to be rational in all cases, as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let
$$
\begin{gathered}
e^{x}+e^{y}=A \\
x e^{x}+y e^{y}=B \\
x^{2} e^{x}+y^{2} e^{y}=C \\
x^{3} e^{x}+y^{3} e^{y}=D \\
x^{4} e^{x}+y^{4} e^{y}=E .
\end{gathered}
$$
Prove that if $A, B, C$, and $D$ are all rational, then so is $E$.
|
We can express $x+y$ in two ways:
$$
\begin{aligned}
& x+y=\frac{A D-B C}{A C-B^{2}} \\
& x+y=\frac{A E-C^{2}}{A D-B C}
\end{aligned}
$$
(We have to be careful if $A C-B^{2}$ or $A D-B C$ is zero. We'll deal with that case later.) It is easy to check that these equations hold by substituting the expressions for $A, B, C, D$, and $E$. Setting these two expressions for $x+y$ equal to each other, we get
$$
\frac{A D-B C}{A C-B^{2}}=\frac{A E-C^{2}}{A D-B C}
$$
which we can easily solve for $E$ as a rational function of $A, B, C$, and $D$. Therefore if $A, B, C$, and $D$ are all rational, then $E$ will be rational as well.
Now, we have to check what happens if $A C-B^{2}=0$ or $A D-B C=0$. If $A C-B^{2}=0$, then writing down the expressions for $A, B$, and $C$ gives us that $(x-y)^{2} e^{x+y}=0$, meaning that $x=y$. If $x=y$, and $x \neq 0, A$ and $D$ are also non-zero, and $\frac{B}{A}=\frac{E}{D}=x$. Since $\frac{B}{A}$ is rational and $D$ is rational, this implies that $E$ is rational. If $x=y=0$, then $E=0$ and so is certainly rational.
We finally must check what happens if $A D-B C=0$. Since $A D-B C=(x+y)\left(A C-B^{2}\right)$, either $A C-B^{2}=0$ (a case we have already dealt with), or $x+y=0$. But if $x+y=0$ then $A E-C^{2}=0$, which implies that $E=\frac{C^{2}}{A}$ (we know that $A \neq 0$ because $e^{x}$ and $e^{y}$ are both positive). Since $A$ and $C$ are rational, this implies that $E$ is also rational.
So, we have shown $E$ to be rational in all cases, as desired.
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [20]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2010"
}
|
Show that, for every positive integer $n$, there exists a monic polynomial of degree $n$ with integer coefficients such that the coefficients are decreasing and the roots of the polynomial are all integers.
|
We claim we can find values $a$ and $b$ such that $p(x)=(x-a)(x+b)^{n}$ is a polynomial of degree $n+1$ that satisfies these constraints. We show that its coefficients are decreasing by finding a general formula for the coefficient of $x^{k}$.
The coefficient of $x^{k}$ is $b^{k}\binom{n}{k}-a b^{k-1}\binom{n}{k-1}$, which can be seen by expanding out $(x+b)^{n}$ and then multiplying by $(x-a)$. Then we must prove that
$$
b^{k+1}\binom{n}{k+1}-a b^{k}\binom{n}{k}<b^{k}\binom{n}{k}-a b^{k-1}\binom{n}{k-1}
$$
or
$$
a b^{k-1}\left(b\binom{n}{k}-\binom{n}{k-1}\right)>b^{k}\left(b\binom{n}{k+1}-\binom{n}{k}\right) .
$$
Choose $b>\max \left(\frac{\binom{n}{k}}{\binom{n}{k-1}}\right)$ in order to make sure the right-hand term in each product on each side of the inequality sign is positive (we'll be dividing by it, so this makes things much easier), and choose $a>\max \left(\frac{b\left(b\binom{n}{k+1}-\binom{n}{k}\right)}{b\binom{n}{k}-\binom{n}{k-1}}\right)$ to make sure the inequality always holds. Since there are only finite values that $k$ can take on given a fixed $n$ (namely, integers between 0 and $n$ inclusive), we can always find values of $a$ and $b$ that satisfy these constraints.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Show that, for every positive integer $n$, there exists a monic polynomial of degree $n$ with integer coefficients such that the coefficients are decreasing and the roots of the polynomial are all integers.
|
We claim we can find values $a$ and $b$ such that $p(x)=(x-a)(x+b)^{n}$ is a polynomial of degree $n+1$ that satisfies these constraints. We show that its coefficients are decreasing by finding a general formula for the coefficient of $x^{k}$.
The coefficient of $x^{k}$ is $b^{k}\binom{n}{k}-a b^{k-1}\binom{n}{k-1}$, which can be seen by expanding out $(x+b)^{n}$ and then multiplying by $(x-a)$. Then we must prove that
$$
b^{k+1}\binom{n}{k+1}-a b^{k}\binom{n}{k}<b^{k}\binom{n}{k}-a b^{k-1}\binom{n}{k-1}
$$
or
$$
a b^{k-1}\left(b\binom{n}{k}-\binom{n}{k-1}\right)>b^{k}\left(b\binom{n}{k+1}-\binom{n}{k}\right) .
$$
Choose $b>\max \left(\frac{\binom{n}{k}}{\binom{n}{k-1}}\right)$ in order to make sure the right-hand term in each product on each side of the inequality sign is positive (we'll be dividing by it, so this makes things much easier), and choose $a>\max \left(\frac{b\left(b\binom{n}{k+1}-\binom{n}{k}\right)}{b\binom{n}{k}-\binom{n}{k-1}}\right)$ to make sure the inequality always holds. Since there are only finite values that $k$ can take on given a fixed $n$ (namely, integers between 0 and $n$ inclusive), we can always find values of $a$ and $b$ that satisfy these constraints.
|
{
"exam": "HMMT",
"problem_label": "5",
"problem_match": "\n5. [20]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2010"
}
|
Let $S$ be a convex set in the plane with a finite area $a$. Prove that either $a=0$ or $S$ is bounded. Note: a set is bounded if it is contained in a circle of finite radius. Note: a set is convex if, whenever two points $A$ and $B$ are in the set, the line segment between them is also in the set.
|
If all points in $S$ lie on a straight line, then $a=0$.
Otherwise we may pick three points $A, B$, and $C$ that are not collinear. Let $\omega$ be the incircle of $\triangle A B C$, with $I$ its center and $r$ its radius. Since $S$ is convex, $S$ must contain $\omega$.
Suppose $S$ also contains a point $X$ at a distance $d$ from $I$, with $d>R$. We will show that $d \leq \sqrt{r^{2}+\frac{a^{2}}{r^{2}}}$, which implies that the $S$ is bounded since all points are contained within the circle centered at $I$ of radius $\sqrt{r^{2}+\frac{a^{2}}{r^{2}}}$.
Let $Y$ and $Z$ be on $\omega$ such that $\overline{X Y}$ and $\overline{X Z}$ are tangents to $\omega$. Because $S$ is convex, it must contain kite $I Y X Z$, whose area we can compute in terms of $d$ and $r$.
Let $M$ be the midpoint of $\overline{Y Z}$. Since $\triangle I Y X \sim \triangle I M Y$, we know that $\frac{I M}{I Y}=\frac{I Y}{I X}$, that is, $I M=$ $\frac{(I Y)^{2}}{I X}=\frac{r^{2}}{d}$. Then $M Y=\sqrt{r^{2}-\frac{r^{4}}{d^{2}}}=r \sqrt{1-\left(\frac{r}{d}\right)^{2}}=\frac{1}{2} Y Z$.
The area of $I Y X Z$ is $\frac{1}{2}(Y Z)(I X)=r d \sqrt{1-\left(\frac{r}{d}\right)^{2}}=r \sqrt{d^{2}-r^{2}}$. This must be less than or equal to $a$, the area of $S$. This yields $a^{2} \geq r^{2} d^{2}-r^{4}$ or $d^{2} \leq r^{2}+\frac{a^{2}}{r^{2}}$. It follows that $d \leq \sqrt{r^{2}+\frac{a^{2}}{r^{2}}}$, as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $S$ be a convex set in the plane with a finite area $a$. Prove that either $a=0$ or $S$ is bounded. Note: a set is bounded if it is contained in a circle of finite radius. Note: a set is convex if, whenever two points $A$ and $B$ are in the set, the line segment between them is also in the set.
|
If all points in $S$ lie on a straight line, then $a=0$.
Otherwise we may pick three points $A, B$, and $C$ that are not collinear. Let $\omega$ be the incircle of $\triangle A B C$, with $I$ its center and $r$ its radius. Since $S$ is convex, $S$ must contain $\omega$.
Suppose $S$ also contains a point $X$ at a distance $d$ from $I$, with $d>R$. We will show that $d \leq \sqrt{r^{2}+\frac{a^{2}}{r^{2}}}$, which implies that the $S$ is bounded since all points are contained within the circle centered at $I$ of radius $\sqrt{r^{2}+\frac{a^{2}}{r^{2}}}$.
Let $Y$ and $Z$ be on $\omega$ such that $\overline{X Y}$ and $\overline{X Z}$ are tangents to $\omega$. Because $S$ is convex, it must contain kite $I Y X Z$, whose area we can compute in terms of $d$ and $r$.
Let $M$ be the midpoint of $\overline{Y Z}$. Since $\triangle I Y X \sim \triangle I M Y$, we know that $\frac{I M}{I Y}=\frac{I Y}{I X}$, that is, $I M=$ $\frac{(I Y)^{2}}{I X}=\frac{r^{2}}{d}$. Then $M Y=\sqrt{r^{2}-\frac{r^{4}}{d^{2}}}=r \sqrt{1-\left(\frac{r}{d}\right)^{2}}=\frac{1}{2} Y Z$.
The area of $I Y X Z$ is $\frac{1}{2}(Y Z)(I X)=r d \sqrt{1-\left(\frac{r}{d}\right)^{2}}=r \sqrt{d^{2}-r^{2}}$. This must be less than or equal to $a$, the area of $S$. This yields $a^{2} \geq r^{2} d^{2}-r^{4}$ or $d^{2} \leq r^{2}+\frac{a^{2}}{r^{2}}$. It follows that $d \leq \sqrt{r^{2}+\frac{a^{2}}{r^{2}}}$, as desired.
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n6. [20]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2010"
}
|
Point $P$ lies inside a convex pentagon $A F Q D C$ such that $F P D Q$ is a parallelogram. Given that $\angle F A Q=\angle P A C=10^{\circ}$, and $\angle P F A=\angle P D C=15^{\circ}$. What is $\angle A Q C ?$
|
$\frac{\pi}{12}$ Let $C^{\prime}$ be the point such that there is a spiral similarity between $\triangle A F P$ and $\triangle A Q C^{\prime}$. In other words, one triangle can be formed from the other by dilating and rotating about one of the
triangle's vertices (in this case, $A$ ). We will show that $C^{\prime}$ is $C$, so our answer will be $\angle A Q C=$ $\angle A Q C^{\prime}=\angle A F P=15^{\circ}$. By the spiral similarity theorem, $\triangle A F Q \sim \triangle A P C^{\prime}$ (this is intuitive by looking at a diagram), so $\angle P A C=\angle F A Q=10^{\circ}$, so to show that $C^{\prime}$ is $C$, it is sufficient to show that $\angle P D C^{\prime}=15^{\circ}$.
Let $X$ be the fourth point of the parallelogram $F P C^{\prime} X$ (see the above diagram). The angle between lines $\overline{F P}$ and $\overline{F A}$ is $15^{\circ}$. Since $\overline{X C^{\prime}} \| \overline{F P}$, the angle between $\overline{F A}$ and $\overline{X C^{\prime}}$ is $15^{\circ}$ as well. In addition, the angle between $\overline{Q A}$ and $\overline{Q C^{\prime}}$ is $\angle A Q C^{\prime}=15^{\circ}$, so $\angle X C^{\prime} Q=\angle F A Q$. Further, because $F P C^{\prime} X$ is a parallelogram, $\frac{Q C^{\prime}}{X C^{\prime}}=\frac{Q C^{\prime}}{F P}$. By similar triangles $\triangle A F P$ and $\triangle A Q C^{\prime}, \frac{Q C^{\prime}}{F P}=\frac{Q A}{F A}$. By SAS similarity, there is a spiral similarity between $\triangle X C^{\prime} Q$ and $\triangle F A Q$, so $\angle F Q X=\angle A Q C^{\prime}=15^{\circ}$.
Note that the segments $\overline{F P}, \overline{X C^{\prime}}$, and $\overline{Q D}$ are all parallel and equal in length. Therefore, $\triangle F Q X \cong$ $\triangle P D C^{\prime}$ are congruent, and $\angle P D C^{\prime}=15^{\circ}$ as well. So $C^{\prime}$ is $C$, and $\angle A Q C=15^{\circ}$.
|
15^{\circ}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Point $P$ lies inside a convex pentagon $A F Q D C$ such that $F P D Q$ is a parallelogram. Given that $\angle F A Q=\angle P A C=10^{\circ}$, and $\angle P F A=\angle P D C=15^{\circ}$. What is $\angle A Q C ?$
|
$\frac{\pi}{12}$ Let $C^{\prime}$ be the point such that there is a spiral similarity between $\triangle A F P$ and $\triangle A Q C^{\prime}$. In other words, one triangle can be formed from the other by dilating and rotating about one of the
triangle's vertices (in this case, $A$ ). We will show that $C^{\prime}$ is $C$, so our answer will be $\angle A Q C=$ $\angle A Q C^{\prime}=\angle A F P=15^{\circ}$. By the spiral similarity theorem, $\triangle A F Q \sim \triangle A P C^{\prime}$ (this is intuitive by looking at a diagram), so $\angle P A C=\angle F A Q=10^{\circ}$, so to show that $C^{\prime}$ is $C$, it is sufficient to show that $\angle P D C^{\prime}=15^{\circ}$.
Let $X$ be the fourth point of the parallelogram $F P C^{\prime} X$ (see the above diagram). The angle between lines $\overline{F P}$ and $\overline{F A}$ is $15^{\circ}$. Since $\overline{X C^{\prime}} \| \overline{F P}$, the angle between $\overline{F A}$ and $\overline{X C^{\prime}}$ is $15^{\circ}$ as well. In addition, the angle between $\overline{Q A}$ and $\overline{Q C^{\prime}}$ is $\angle A Q C^{\prime}=15^{\circ}$, so $\angle X C^{\prime} Q=\angle F A Q$. Further, because $F P C^{\prime} X$ is a parallelogram, $\frac{Q C^{\prime}}{X C^{\prime}}=\frac{Q C^{\prime}}{F P}$. By similar triangles $\triangle A F P$ and $\triangle A Q C^{\prime}, \frac{Q C^{\prime}}{F P}=\frac{Q A}{F A}$. By SAS similarity, there is a spiral similarity between $\triangle X C^{\prime} Q$ and $\triangle F A Q$, so $\angle F Q X=\angle A Q C^{\prime}=15^{\circ}$.
Note that the segments $\overline{F P}, \overline{X C^{\prime}}$, and $\overline{Q D}$ are all parallel and equal in length. Therefore, $\triangle F Q X \cong$ $\triangle P D C^{\prime}$ are congruent, and $\angle P D C^{\prime}=15^{\circ}$ as well. So $C^{\prime}$ is $C$, and $\angle A Q C=15^{\circ}$.
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [25]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
A knight moves on a two-dimensional grid. From any square, it can move 2 units in one axisparallel direction, then move 1 unit in an orthogonal direction, the way a regular knight moves in a game of chess. The knight starts at the origin. As it moves, it keeps track of a number $t$, which is initially 0 . When the knight lands at the point $(a, b)$, the number is changed from $x$ to $a x+b$.
Show that, for any integers $a$ and $b$, it is possible for the knight to land at the points $(1, a)$ and $(-1, a)$ with $t$ equal to $b$.
|
For convenience, we will refer to $(a, b)$ as $[a x+b]$, the function it represents. This will make it easier to follow the trajectory of $t$ over a given sequence of moves.
Suppose we start at $[x+1]$ with $t=a$. Taking the path $[x+1] \rightarrow[-x] \rightarrow[x-1] \rightarrow[-x] \rightarrow[x+1]$ will yield $t=a+2$. So we can go from $t=a$ to $t=a+2$ at $[x+1]$.
We can also move until we get to $[-3]$, then go $[-3] \rightarrow[x-1]$ to end up with $t=-4$ at $[x-1]$. But going $[x-1] \rightarrow[3 x] \rightarrow[x-1]$ means we can go from $t=a$ to $t=3 a-1$ at $x-1$. Since we can start with $t=-4$, this means we can therefore get arbitrarily small even and odd numbers at $[x-1]$, hence also $[3 x]$, hence also at $[x+1]$.
This implies we can get any value of $t$ we want at $[x+1]$, so we can also get any value of $t$ we want at $[-x],[x-1],[-x-2],[x-3]$, etc., as well as $[-x+2],[x+3],[-x+4],[x+5]$, etc. We can do a similar thing starting at $[-x+1]$ to get from $t=a$ to $t=a+2$, and use the $[-x-1] \rightarrow[-3 x] \rightarrow[-x-1]$ loop to get arbitrarily small integers of both parities. So we can get any value of $t$ we want at all points of the form $[ \pm x+k]$ for any integer $k$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
A knight moves on a two-dimensional grid. From any square, it can move 2 units in one axisparallel direction, then move 1 unit in an orthogonal direction, the way a regular knight moves in a game of chess. The knight starts at the origin. As it moves, it keeps track of a number $t$, which is initially 0 . When the knight lands at the point $(a, b)$, the number is changed from $x$ to $a x+b$.
Show that, for any integers $a$ and $b$, it is possible for the knight to land at the points $(1, a)$ and $(-1, a)$ with $t$ equal to $b$.
|
For convenience, we will refer to $(a, b)$ as $[a x+b]$, the function it represents. This will make it easier to follow the trajectory of $t$ over a given sequence of moves.
Suppose we start at $[x+1]$ with $t=a$. Taking the path $[x+1] \rightarrow[-x] \rightarrow[x-1] \rightarrow[-x] \rightarrow[x+1]$ will yield $t=a+2$. So we can go from $t=a$ to $t=a+2$ at $[x+1]$.
We can also move until we get to $[-3]$, then go $[-3] \rightarrow[x-1]$ to end up with $t=-4$ at $[x-1]$. But going $[x-1] \rightarrow[3 x] \rightarrow[x-1]$ means we can go from $t=a$ to $t=3 a-1$ at $x-1$. Since we can start with $t=-4$, this means we can therefore get arbitrarily small even and odd numbers at $[x-1]$, hence also $[3 x]$, hence also at $[x+1]$.
This implies we can get any value of $t$ we want at $[x+1]$, so we can also get any value of $t$ we want at $[-x],[x-1],[-x-2],[x-3]$, etc., as well as $[-x+2],[x+3],[-x+4],[x+5]$, etc. We can do a similar thing starting at $[-x+1]$ to get from $t=a$ to $t=a+2$, and use the $[-x-1] \rightarrow[-3 x] \rightarrow[-x-1]$ loop to get arbitrarily small integers of both parities. So we can get any value of $t$ we want at all points of the form $[ \pm x+k]$ for any integer $k$.
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [30]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2010"
}
|
Let $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}$ be a polynomial with complex coefficients such that $a_{i} \neq 0$ for all $i$. Prove that $|r| \leq 2 \max _{i=0}^{n-1}\left|\frac{a_{i-1}}{a_{i}}\right|$ for all roots $r$ of all such polynomials $p$. Here we let $|z|$ denote the absolute value of the complex number $z$.
|
If $r$ is a root, then $-a_{n} r^{n}=a_{n-1} r^{n-1}+\ldots+a_{0}$. By the Triangle Inequality, $\left|-a_{n} r^{n}\right| \leq$ $\left|a_{n-1} r^{n-1}\right|+\ldots+\left|a_{0}\right|$. Rearranging this inequality yields $\left|a_{n} r^{n}\right|-\left|a_{n-1} r^{n-1}\right|-\ldots-\left|a_{0}\right| \leq 0$.
Now suppose $|r|=k \max \left|\frac{a_{i-1}}{a_{i}}\right|$. Applying this over values of $i$ ranging from $m+1$ to $n$ (assuming $m+1 \leq n$ ), we get $\left|a_{m} r^{m}\right| \leq \frac{\left|a_{n} r^{n}\right|}{k^{n-m}}$. This, along with the above equation, yields:
$$
\left|a_{n} r^{n}\right| \cdot\left(1-\frac{1}{k}-\frac{1}{k^{2}}-\frac{1}{k^{3}}-\ldots-\frac{1}{k^{n}}\right)=0
$$
This is only true when $a_{n}=0, r=0$, or $\left(1-\frac{1}{k}-\frac{1}{k^{2}}-\ldots\right)=0$. The first option is impossible by the constraints in the problem. The second option implies $k=0$. The third option implies that $k<2$; otherwise ( $1-\frac{1}{k}-\frac{1}{k^{2}}-\ldots-\frac{1}{k^{n}}$ ) would always remain positive. Either way, $|r| \leq 2 \max _{i=0}^{n-1}\left|\frac{a_{i-1}}{a_{i}}\right|$.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}$ be a polynomial with complex coefficients such that $a_{i} \neq 0$ for all $i$. Prove that $|r| \leq 2 \max _{i=0}^{n-1}\left|\frac{a_{i-1}}{a_{i}}\right|$ for all roots $r$ of all such polynomials $p$. Here we let $|z|$ denote the absolute value of the complex number $z$.
|
If $r$ is a root, then $-a_{n} r^{n}=a_{n-1} r^{n-1}+\ldots+a_{0}$. By the Triangle Inequality, $\left|-a_{n} r^{n}\right| \leq$ $\left|a_{n-1} r^{n-1}\right|+\ldots+\left|a_{0}\right|$. Rearranging this inequality yields $\left|a_{n} r^{n}\right|-\left|a_{n-1} r^{n-1}\right|-\ldots-\left|a_{0}\right| \leq 0$.
Now suppose $|r|=k \max \left|\frac{a_{i-1}}{a_{i}}\right|$. Applying this over values of $i$ ranging from $m+1$ to $n$ (assuming $m+1 \leq n$ ), we get $\left|a_{m} r^{m}\right| \leq \frac{\left|a_{n} r^{n}\right|}{k^{n-m}}$. This, along with the above equation, yields:
$$
\left|a_{n} r^{n}\right| \cdot\left(1-\frac{1}{k}-\frac{1}{k^{2}}-\frac{1}{k^{3}}-\ldots-\frac{1}{k^{n}}\right)=0
$$
This is only true when $a_{n}=0, r=0$, or $\left(1-\frac{1}{k}-\frac{1}{k^{2}}-\ldots\right)=0$. The first option is impossible by the constraints in the problem. The second option implies $k=0$. The third option implies that $k<2$; otherwise ( $1-\frac{1}{k}-\frac{1}{k^{2}}-\ldots-\frac{1}{k^{n}}$ ) would always remain positive. Either way, $|r| \leq 2 \max _{i=0}^{n-1}\left|\frac{a_{i-1}}{a_{i}}\right|$.
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [30]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team1-solutions.jsonl",
"solution_match": "\nSolution: ",
"tier": "T4",
"year": "2010"
}
|
Call an $2 n$-digit base-10 number special if we can split its digits into two sets of size $n$ such that the sum of the numbers in the two sets is the same. Let $p_{n}$ be the probability that a randomly-chosen $2 n$-digit number is special. (We allow leading zeros in $2 n$-digit numbers).
(a) [20] The sequence $p_{n}$ converges to a constant $c$. Find $c$.
|
$\frac{1}{2}$ We first claim that if a $2 n$-digit number $x$ has at least eight 0 's and at least eight 1's and the sum of its digits is even, then $x$ is special.
Let $A$ be a set of eight 0 's and eight 1's and let $B$ be the set of all the other digits. We split $b$ arbitrily into two sets $Y$ and $Z$ of equal size. If $\left|\sum_{y \in Y} y-\sum_{z \in Z} z\right|>8$, then we swap the biggest element of the set with the bigger sum with the smallest element of the other set. This transposition always decreases the absolute value of the sum: in the worst case, a 9 from the bigger set is swapped for a 0 from the smaller set, which changes the difference by at most 18 . Therefore, after a finite number of steps, we will have $\left|\sum_{y \in Y} y-\sum_{z \in Z} z\right| \leq 8$.
Note that this absolute value is even, since the sum of all the digits is even. Without loss of generality, suppose that $\sum_{y \in Y} y-\sum_{z \in Z} z$ is $2 k$, where $0 \leq k \leq 4$. If we add $k 0$ 's and $8-k$ 's to $Y$, and we add the other elements of $A$ to $Z$, then the two sets will balance, so $x$ is special.
(b) [45] Let $q_{n}=p_{n}-c$. There exists a unique positive constant $r$ such that $\frac{q_{n}}{r^{n}}$ converges to a constant $d$. Find $r$ and $d$.
|
not found
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Call an $2 n$-digit base-10 number special if we can split its digits into two sets of size $n$ such that the sum of the numbers in the two sets is the same. Let $p_{n}$ be the probability that a randomly-chosen $2 n$-digit number is special. (We allow leading zeros in $2 n$-digit numbers).
(a) [20] The sequence $p_{n}$ converges to a constant $c$. Find $c$.
|
$\frac{1}{2}$ We first claim that if a $2 n$-digit number $x$ has at least eight 0 's and at least eight 1's and the sum of its digits is even, then $x$ is special.
Let $A$ be a set of eight 0 's and eight 1's and let $B$ be the set of all the other digits. We split $b$ arbitrily into two sets $Y$ and $Z$ of equal size. If $\left|\sum_{y \in Y} y-\sum_{z \in Z} z\right|>8$, then we swap the biggest element of the set with the bigger sum with the smallest element of the other set. This transposition always decreases the absolute value of the sum: in the worst case, a 9 from the bigger set is swapped for a 0 from the smaller set, which changes the difference by at most 18 . Therefore, after a finite number of steps, we will have $\left|\sum_{y \in Y} y-\sum_{z \in Z} z\right| \leq 8$.
Note that this absolute value is even, since the sum of all the digits is even. Without loss of generality, suppose that $\sum_{y \in Y} y-\sum_{z \in Z} z$ is $2 k$, where $0 \leq k \leq 4$. If we add $k 0$ 's and $8-k$ 's to $Y$, and we add the other elements of $A$ to $Z$, then the two sets will balance, so $x$ is special.
(b) [45] Let $q_{n}=p_{n}-c$. There exists a unique positive constant $r$ such that $\frac{q_{n}}{r^{n}}$ converges to a constant $d$. Find $r$ and $d$.
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. ",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Call an $2 n$-digit base-10 number special if we can split its digits into two sets of size $n$ such that the sum of the numbers in the two sets is the same. Let $p_{n}$ be the probability that a randomly-chosen $2 n$-digit number is special. (We allow leading zeros in $2 n$-digit numbers).
(a) [20] The sequence $p_{n}$ converges to a constant $c$. Find $c$.
|
$r=\frac{1}{4}, d=-1$ To get the next asymptotic term after the constant term of $\frac{1}{2}$, we need to consider what happens when the digit sum is even; we want to find the probability that such a number isn't balanced. We claim that the configuration that contributes the vast majority of unbalanced numbers is when all numbers are even and the sum is $2 \bmod 4$, or such a configuration with all numbers increased by 1. Clearly this gives $q_{n}$ being asymptotic to $-\frac{1}{2} \cdot 2 \cdot\left(\frac{1}{2}\right)^{2 n}=-\left(\frac{1}{4}\right)^{n}$, so $r=\frac{1}{4}$ and $d=-1$.
To prove the claim, first note that the asymptotic probability that there are at most 4 digits that occur more than 10 times is asymptotically much smaller than $\left(\frac{1}{2}\right)^{n}$, so we can assume that there exist 5 digits that each occur at least 10 times. If any of those digits are consecutive, then the digit sum being even implies that the number is balanced (by an argument similar to part (a)).
So, we can assume that none of the numbers are consecutive. We would like to say that this implies that the numbers are either $0,2,4,6,8$ or $1,3,5,7,9$. However, we can't quite say this yet, as we need to rule out possibilities like $0,2,4,7,9$. In this case, though, we can just pair 0 and 7 up with 2 and 4 ; by using the same argument as in part (a), except using 0 and 7 both (to get a sum of 7 ) and 2 and 4 both (to get a sum of 6 ) to balance out the two sets at the end.
In general, if there is ever a gap of size 3 , consider the number right after it and the 3 numbers before it (so we have $k-4, k-2, k, k+3$ for some $k$ ), and pair them up such that one pair has a sum that's exactly one more than the other (i.e. pair $k-4$ with $k+3$ and $k-2$ with $k$ ). Since we again have pairs of numbers whose sums differ by 1 , we can use the technique from part (a) of balancing out the sets at the end.
So, we can assume there is no gap of size 3 , which together with the condition that no two numbers are adjacent implies that the 5 digits are either $0,2,4,6,8$ or $1,3,5,7,9$. For the remainder of the solution, we will deal with the $0,2,4,6,8$ case, since it is symmetric with the other case under the transformation $x \mapsto 9-x$.
If we can distribute the odd digits into two sets $S_{1}$ and $S_{2}$ such that (i) the differense in sums of $S_{1}$ and $S_{2}$ is small; and (ii) the difference in sums of $S_{1}$ and $S_{2}$, plus the sum of the even digits, is divisible by 4 , then the same argument as in part (a) implies that the number is good.
In fact, if there are any odd digits, then we can use them at the beginning to fix the parity mod 4 (by adding them all in such that the sums of the two sets remain close, and then potentially switching one with an even digit). Therefore, if there are any odd digits then the number is good.
Also, even if there are no odd digits, if the sum of the digits is divisible by 4 then the number is good.
So, we have shown that almost all non-good numbers come from having all numbers being even with a digit sum that is $2 \bmod 4$, or the analogous case under the mapping $x \mapsto 9-x$. This formalizes the claim we made in the first paragraph, so $r=\frac{1}{4}$ and $d=-1$, as claimed.
|
r=\frac{1}{4}, d=-1
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Call an $2 n$-digit base-10 number special if we can split its digits into two sets of size $n$ such that the sum of the numbers in the two sets is the same. Let $p_{n}$ be the probability that a randomly-chosen $2 n$-digit number is special. (We allow leading zeros in $2 n$-digit numbers).
(a) [20] The sequence $p_{n}$ converges to a constant $c$. Find $c$.
|
$r=\frac{1}{4}, d=-1$ To get the next asymptotic term after the constant term of $\frac{1}{2}$, we need to consider what happens when the digit sum is even; we want to find the probability that such a number isn't balanced. We claim that the configuration that contributes the vast majority of unbalanced numbers is when all numbers are even and the sum is $2 \bmod 4$, or such a configuration with all numbers increased by 1. Clearly this gives $q_{n}$ being asymptotic to $-\frac{1}{2} \cdot 2 \cdot\left(\frac{1}{2}\right)^{2 n}=-\left(\frac{1}{4}\right)^{n}$, so $r=\frac{1}{4}$ and $d=-1$.
To prove the claim, first note that the asymptotic probability that there are at most 4 digits that occur more than 10 times is asymptotically much smaller than $\left(\frac{1}{2}\right)^{n}$, so we can assume that there exist 5 digits that each occur at least 10 times. If any of those digits are consecutive, then the digit sum being even implies that the number is balanced (by an argument similar to part (a)).
So, we can assume that none of the numbers are consecutive. We would like to say that this implies that the numbers are either $0,2,4,6,8$ or $1,3,5,7,9$. However, we can't quite say this yet, as we need to rule out possibilities like $0,2,4,7,9$. In this case, though, we can just pair 0 and 7 up with 2 and 4 ; by using the same argument as in part (a), except using 0 and 7 both (to get a sum of 7 ) and 2 and 4 both (to get a sum of 6 ) to balance out the two sets at the end.
In general, if there is ever a gap of size 3 , consider the number right after it and the 3 numbers before it (so we have $k-4, k-2, k, k+3$ for some $k$ ), and pair them up such that one pair has a sum that's exactly one more than the other (i.e. pair $k-4$ with $k+3$ and $k-2$ with $k$ ). Since we again have pairs of numbers whose sums differ by 1 , we can use the technique from part (a) of balancing out the sets at the end.
So, we can assume there is no gap of size 3 , which together with the condition that no two numbers are adjacent implies that the 5 digits are either $0,2,4,6,8$ or $1,3,5,7,9$. For the remainder of the solution, we will deal with the $0,2,4,6,8$ case, since it is symmetric with the other case under the transformation $x \mapsto 9-x$.
If we can distribute the odd digits into two sets $S_{1}$ and $S_{2}$ such that (i) the differense in sums of $S_{1}$ and $S_{2}$ is small; and (ii) the difference in sums of $S_{1}$ and $S_{2}$, plus the sum of the even digits, is divisible by 4 , then the same argument as in part (a) implies that the number is good.
In fact, if there are any odd digits, then we can use them at the beginning to fix the parity mod 4 (by adding them all in such that the sums of the two sets remain close, and then potentially switching one with an even digit). Therefore, if there are any odd digits then the number is good.
Also, even if there are no odd digits, if the sum of the digits is divisible by 4 then the number is good.
So, we have shown that almost all non-good numbers come from having all numbers being even with a digit sum that is $2 \bmod 4$, or the analogous case under the mapping $x \mapsto 9-x$. This formalizes the claim we made in the first paragraph, so $r=\frac{1}{4}$ and $d=-1$, as claimed.
|
{
"exam": "HMMT",
"problem_label": "10",
"problem_match": "\n10. ",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team1-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
How many ways are there to place pawns on an $8 \times 8$ chessboard, so that there is at most 1 pawn in each horizontal row? Express your answer in the form $p_{1}^{e_{1}} \cdot p_{2}^{e_{2}} \cdots$, where the $p_{i}$ are distinct primes and the $e_{i}$ are positive integers.
|
$3^{16}$ If there is at most 1 pawn in each row, then each row of the chessboard may have either 0 or 1 pawn somewhere in the row. There is 1 case if there are no pawns in the row. There are 8 possible cases if there is 1 pawn in the row, one case for each square in the row. Hence for each row, there are 9 possible pawn arrangements. There are 8 rows, thus we have $9^{8}=3^{16}$.
|
3^{16}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many ways are there to place pawns on an $8 \times 8$ chessboard, so that there is at most 1 pawn in each horizontal row? Express your answer in the form $p_{1}^{e_{1}} \cdot p_{2}^{e_{2}} \cdots$, where the $p_{i}$ are distinct primes and the $e_{i}$ are positive integers.
|
$3^{16}$ If there is at most 1 pawn in each row, then each row of the chessboard may have either 0 or 1 pawn somewhere in the row. There is 1 case if there are no pawns in the row. There are 8 possible cases if there is 1 pawn in the row, one case for each square in the row. Hence for each row, there are 9 possible pawn arrangements. There are 8 rows, thus we have $9^{8}=3^{16}$.
|
{
"exam": "HMMT",
"problem_label": "1",
"problem_match": "\n1. [10]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
In the following figure, a regular hexagon of side length 1 is attached to a semicircle of diameter 1. What is the longest distance between any two points in the figure?
|
$\frac{1+\sqrt{13}}{2}$ Inspection shows that one point must be on the semicircle and the other must be on the side of the hexagon directly opposite the edge with the semicircle, the bottom edge of the hexagon in the above diagram. Let $O$ be the center of the semicircle and let $M$ be the midpoint of the bottom edge.
We will determine the longest distance between points in the figure by comparing the lengths of all the segments with one endpoint on the bottom edge and the other endpoint on the semicircle. Fix a point $A$ on the bottom edge of the hexagon. Suppose that $B$ is chosen on the semicircle such that $A B$ is as long as possible. Let $C$ be the circle centered at $A$ with radius $A B$. If $C$ is not tangent to the semicircle, then part of the semicircle is outside $C$, so we could pick a $B^{\prime}$ on the semicircle such that $A B^{\prime}$ is longer than $A B$. So $C$ must be tangent to the semicircle, and $A B$ must pass through $O$.
Then $O B$ is always $\frac{1}{2}$, no matter which $A$ we choose on the bottom edge. All that remains is maximizing $A O$. This length is the hypotenuse of a right triangle with the fixed height $M O$, so it is maximized when $A M$ is as large as possible - when $A$ is an endpoint of the bottom edge. Note that $M O=2 \cdot \frac{\sqrt{3}}{2}$, and that $A M$ can be at most $\frac{1}{2}$, so $A O$ can be at most $\frac{\sqrt{13}}{2}$. So the maximum distance between two points in the diagram is $A O+O B=\frac{1+\sqrt{13}}{2}$.
|
\frac{1+\sqrt{13}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
In the following figure, a regular hexagon of side length 1 is attached to a semicircle of diameter 1. What is the longest distance between any two points in the figure?
|
$\frac{1+\sqrt{13}}{2}$ Inspection shows that one point must be on the semicircle and the other must be on the side of the hexagon directly opposite the edge with the semicircle, the bottom edge of the hexagon in the above diagram. Let $O$ be the center of the semicircle and let $M$ be the midpoint of the bottom edge.
We will determine the longest distance between points in the figure by comparing the lengths of all the segments with one endpoint on the bottom edge and the other endpoint on the semicircle. Fix a point $A$ on the bottom edge of the hexagon. Suppose that $B$ is chosen on the semicircle such that $A B$ is as long as possible. Let $C$ be the circle centered at $A$ with radius $A B$. If $C$ is not tangent to the semicircle, then part of the semicircle is outside $C$, so we could pick a $B^{\prime}$ on the semicircle such that $A B^{\prime}$ is longer than $A B$. So $C$ must be tangent to the semicircle, and $A B$ must pass through $O$.
Then $O B$ is always $\frac{1}{2}$, no matter which $A$ we choose on the bottom edge. All that remains is maximizing $A O$. This length is the hypotenuse of a right triangle with the fixed height $M O$, so it is maximized when $A M$ is as large as possible - when $A$ is an endpoint of the bottom edge. Note that $M O=2 \cdot \frac{\sqrt{3}}{2}$, and that $A M$ can be at most $\frac{1}{2}$, so $A O$ can be at most $\frac{\sqrt{13}}{2}$. So the maximum distance between two points in the diagram is $A O+O B=\frac{1+\sqrt{13}}{2}$.
|
{
"exam": "HMMT",
"problem_label": "2",
"problem_match": "\n2. [10]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}$, where each $a_{i}$ is either 1 or -1 . Let $r$ be a root of $p$. If $|r|>\frac{15}{8}$, what is the minimum possible value of $n$ ?
|
4 We claim that $n=4$ is the answer. First, we show that $n>3$. Suppose that $n \leq 3$. Let $r$ be the root of the polynomial with $|r| \geq \frac{15}{8}$. Then, by the Triangle Inequality, we have:
$$
\begin{gathered}
\left|a_{n} r^{n}\right|=\left|a_{n-1} r^{n-1}+a_{n-2} r^{n-2}+\ldots+a_{0}\right| \leq\left|a_{n-1} r^{n-1}\right|+\left|a_{n-2} r^{n-2}\right|+\ldots+\left|a_{0}\right| \\
|r|^{n} \leq\left|r^{n-1}\right|+\left|r^{n-2}\right|+\ldots+|1|=\frac{|r|^{n}-1}{|r|-1} \\
|r|^{n+1}-2|r|^{n}+1 \leq 0 \Rightarrow 1 \leq|r|^{n}(2-|r|)
\end{gathered}
$$
The right-hand side is increasing in $n$, for $|r|>1$, so it is bounded by $|r|^{3}(2-|r|)$. This expression is decreasing in $r$ for $r \geq \frac{3}{2}$. When $|r|=\frac{15}{8}$, then the right-hand side is less than 1 , which violates the inequalities. Therefore $n>3$. Now, we claim that there is a 4 th degree polynomial with a root $r$ with $|r| \geq \frac{15}{8}$. Let $p(x)=x^{4}-x^{3}-x^{2}-x-1$. Then $p\left(\frac{15}{8}\right)<0$ and $p(2)>2$. By the Intermediate Value Theorem, $p(x)$ has such a root $r$.
|
4
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}$, where each $a_{i}$ is either 1 or -1 . Let $r$ be a root of $p$. If $|r|>\frac{15}{8}$, what is the minimum possible value of $n$ ?
|
4 We claim that $n=4$ is the answer. First, we show that $n>3$. Suppose that $n \leq 3$. Let $r$ be the root of the polynomial with $|r| \geq \frac{15}{8}$. Then, by the Triangle Inequality, we have:
$$
\begin{gathered}
\left|a_{n} r^{n}\right|=\left|a_{n-1} r^{n-1}+a_{n-2} r^{n-2}+\ldots+a_{0}\right| \leq\left|a_{n-1} r^{n-1}\right|+\left|a_{n-2} r^{n-2}\right|+\ldots+\left|a_{0}\right| \\
|r|^{n} \leq\left|r^{n-1}\right|+\left|r^{n-2}\right|+\ldots+|1|=\frac{|r|^{n}-1}{|r|-1} \\
|r|^{n+1}-2|r|^{n}+1 \leq 0 \Rightarrow 1 \leq|r|^{n}(2-|r|)
\end{gathered}
$$
The right-hand side is increasing in $n$, for $|r|>1$, so it is bounded by $|r|^{3}(2-|r|)$. This expression is decreasing in $r$ for $r \geq \frac{3}{2}$. When $|r|=\frac{15}{8}$, then the right-hand side is less than 1 , which violates the inequalities. Therefore $n>3$. Now, we claim that there is a 4 th degree polynomial with a root $r$ with $|r| \geq \frac{15}{8}$. Let $p(x)=x^{4}-x^{3}-x^{2}-x-1$. Then $p\left(\frac{15}{8}\right)<0$ and $p(2)>2$. By the Intermediate Value Theorem, $p(x)$ has such a root $r$.
|
{
"exam": "HMMT",
"problem_label": "3",
"problem_match": "\n3. [15]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Find all 4-digit integers of the form $a a b b$ (when written in base 10) that are perfect squares.
|
7744 Let $x$ be an integer such that $x^{2}$ is of the desired form. Then $1100 a+11 b=x^{2}$. Then $x^{2}$ is divisible by 11 , which means $x$ is divisible by 11 . Then for some integer, $y, x=11 y$. Then $1100 a+11 b=11^{2} y^{2} \Rightarrow 100 a+b=11 y^{2}$. This means that $100 a+b \equiv 0(\bmod 11) \Rightarrow a+b \equiv 0(\bmod 11)$. Because $a$ and $b$ must be nonzero digits, we have $2 \leq a, b \leq 9$, so we can write $b=11-a$.
Replacing $b$ in the equation derived above, we obtain $99 a+11=11 y^{2} \Rightarrow 9 a+1=y^{2}$. We check the possible values of $a$ from 2 to 9 , and only $a=7$ yields a perfect square. When $a=7, b=4$, so the only perfect square of for $a a b b$ is 7744 .
|
7744
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find all 4-digit integers of the form $a a b b$ (when written in base 10) that are perfect squares.
|
7744 Let $x$ be an integer such that $x^{2}$ is of the desired form. Then $1100 a+11 b=x^{2}$. Then $x^{2}$ is divisible by 11 , which means $x$ is divisible by 11 . Then for some integer, $y, x=11 y$. Then $1100 a+11 b=11^{2} y^{2} \Rightarrow 100 a+b=11 y^{2}$. This means that $100 a+b \equiv 0(\bmod 11) \Rightarrow a+b \equiv 0(\bmod 11)$. Because $a$ and $b$ must be nonzero digits, we have $2 \leq a, b \leq 9$, so we can write $b=11-a$.
Replacing $b$ in the equation derived above, we obtain $99 a+11=11 y^{2} \Rightarrow 9 a+1=y^{2}$. We check the possible values of $a$ from 2 to 9 , and only $a=7$ yields a perfect square. When $a=7, b=4$, so the only perfect square of for $a a b b$ is 7744 .
|
{
"exam": "HMMT",
"problem_label": "4",
"problem_match": "\n4. [20]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Compute
$$
\sum_{n=1}^{98} \frac{2}{\sqrt{n}+\sqrt{n+2}}+\frac{1}{\sqrt{n+1}+\sqrt{n+2}}
$$
|
$\quad 3 \sqrt{11}-2 \sqrt{2}+19$ Rationalizing the denominator of both terms in the summation yields $\sqrt{n+2}-\sqrt{n}+\sqrt{n+2}-\sqrt{n+1}=2 \sqrt{n+2}-(\sqrt{n}+\sqrt{n+1})$. Then the sum $\sum_{n=1}^{98} 2 \sqrt{n+2}-(\sqrt{n}+\sqrt{n+1})$ telescopes. All terms cancel except for $-(\sqrt{1}+\sqrt{2})-\sqrt{2}+2 \sqrt{99}+2 \sqrt{100}-\sqrt{99}=3 \sqrt{11}-2 \sqrt{2}+19$.
|
3 \sqrt{11}-2 \sqrt{2}+19
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Compute
$$
\sum_{n=1}^{98} \frac{2}{\sqrt{n}+\sqrt{n+2}}+\frac{1}{\sqrt{n+1}+\sqrt{n+2}}
$$
|
$\quad 3 \sqrt{11}-2 \sqrt{2}+19$ Rationalizing the denominator of both terms in the summation yields $\sqrt{n+2}-\sqrt{n}+\sqrt{n+2}-\sqrt{n+1}=2 \sqrt{n+2}-(\sqrt{n}+\sqrt{n+1})$. Then the sum $\sum_{n=1}^{98} 2 \sqrt{n+2}-(\sqrt{n}+\sqrt{n+1})$ telescopes. All terms cancel except for $-(\sqrt{1}+\sqrt{2})-\sqrt{2}+2 \sqrt{99}+2 \sqrt{100}-\sqrt{99}=3 \sqrt{11}-2 \sqrt{2}+19$.
|
{
"exam": "HMMT",
"problem_label": "5",
"problem_match": "\n5. [25]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Into how many regions can a circle be cut by 10 parabolas?
|
201 We will consider the general case of $n$ parabolas, for which the answer is $2 n^{2}+1$.
We will start with some rough intuition, then fill in the details afterwards. The intuition is that, if we make the parabolas steep enough, we can basically treat them as two parallel lines. Furthermore, the number of regions is given in terms of the number of intersections of the parabolas that occur within the circle, since every time two parabolas cross a new region is created. Since two pairs of parallel lines intersect in 4 points, and pairs of parabolas also intersect in 4 points, as long as we can always make all 4 points of intersection lie inside the circle, the parallel lines case is the best we can do.
In other words, the answer is the same as the answer if we were trying to add ten pairs of parallel lines. We can compute the answer for pairs of parallel lines as follows - when we add the $k$ th set of parallel lines, there are already $2 k-2$ lines that the two new lines can intersect, meaning that each of the lines adds $2 k-1$ new regions ${ }^{1}$. This means that we add $4 k-2$ regions when adding the $k$ th set of lines, making the answer $1+2+6+10+14+\cdots+(4 n-2)=1+2(1+3+5+7+\cdots+(2 n-1))=1+2 \cdot n^{2}=2 n^{2}+1$.
Now that we have sketched out the solution, we will fill in the details more rigorously. First, if there are $n$ parabolas inside the circle, and the they intersect in $K$ points total, then we claim that the number of regions the circle is divided into will be at most $K+n+r+1$, where $r$ is the number of parabolas that intersect the circle itself in exactly four points.
We will prove this by induction. In the base case of $n=0$, we are just saying that the circle itself consists of exactly one region.
To prove the inductive step, suppose that we have $n$ parabolas with $K$ points of intersection. We want to show that if we add an additional parabola, and this parabola intersects the other parabolas in $p$ points, then this new parabola adds either $p+1$ or $p+2$ regions to the circle, and that we get $p+2$ regions if and only if the parabola intersects the circle in exactly four points.
We will do this by considering how many regions the parabola cuts through, following its path from when it initially enters the circle to when it exits the circle for the last time. When it initially enters the circle, it cuts through one region, thereby increasing the number of regions by on ${ }^{2}$. Then, for each other parabola that this parabola crosses, we cut through one additional region. It is also possible for the parabola to leave and then re-enter the circle, which happens if and only if the parabola intersects
[^0]the circle in four points, and also adds one additional region. Therefore, the number of regions is either $p+1$ or $p+2$, and it is $p+2$ if and only if the parabola intersects the circle in four points. This completes the induction and proves the claim.
So, we are left with trying to maximize $K+n+r+1$. Since a pair of parabolas intersects in at most 4 points, and there are $\binom{n}{2}$ pairs of parabolas, we have $K \leq 4\binom{n}{2}=2 n^{2}-2 n$. Also, $r \leq n$, so $K+n+r+1 \leq 2 n^{2}+1$. On the other hand, as explained in the paragraphs giving the intuition, we can attain $2 n^{2}+1$ by making the parabolas sufficiently steep that they act like pairs of parallel lines. Therefore, the answer is $2 n^{2}+1$, as claimed.
|
201
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Into how many regions can a circle be cut by 10 parabolas?
|
201 We will consider the general case of $n$ parabolas, for which the answer is $2 n^{2}+1$.
We will start with some rough intuition, then fill in the details afterwards. The intuition is that, if we make the parabolas steep enough, we can basically treat them as two parallel lines. Furthermore, the number of regions is given in terms of the number of intersections of the parabolas that occur within the circle, since every time two parabolas cross a new region is created. Since two pairs of parallel lines intersect in 4 points, and pairs of parabolas also intersect in 4 points, as long as we can always make all 4 points of intersection lie inside the circle, the parallel lines case is the best we can do.
In other words, the answer is the same as the answer if we were trying to add ten pairs of parallel lines. We can compute the answer for pairs of parallel lines as follows - when we add the $k$ th set of parallel lines, there are already $2 k-2$ lines that the two new lines can intersect, meaning that each of the lines adds $2 k-1$ new regions ${ }^{1}$. This means that we add $4 k-2$ regions when adding the $k$ th set of lines, making the answer $1+2+6+10+14+\cdots+(4 n-2)=1+2(1+3+5+7+\cdots+(2 n-1))=1+2 \cdot n^{2}=2 n^{2}+1$.
Now that we have sketched out the solution, we will fill in the details more rigorously. First, if there are $n$ parabolas inside the circle, and the they intersect in $K$ points total, then we claim that the number of regions the circle is divided into will be at most $K+n+r+1$, where $r$ is the number of parabolas that intersect the circle itself in exactly four points.
We will prove this by induction. In the base case of $n=0$, we are just saying that the circle itself consists of exactly one region.
To prove the inductive step, suppose that we have $n$ parabolas with $K$ points of intersection. We want to show that if we add an additional parabola, and this parabola intersects the other parabolas in $p$ points, then this new parabola adds either $p+1$ or $p+2$ regions to the circle, and that we get $p+2$ regions if and only if the parabola intersects the circle in exactly four points.
We will do this by considering how many regions the parabola cuts through, following its path from when it initially enters the circle to when it exits the circle for the last time. When it initially enters the circle, it cuts through one region, thereby increasing the number of regions by on ${ }^{2}$. Then, for each other parabola that this parabola crosses, we cut through one additional region. It is also possible for the parabola to leave and then re-enter the circle, which happens if and only if the parabola intersects
[^0]the circle in four points, and also adds one additional region. Therefore, the number of regions is either $p+1$ or $p+2$, and it is $p+2$ if and only if the parabola intersects the circle in four points. This completes the induction and proves the claim.
So, we are left with trying to maximize $K+n+r+1$. Since a pair of parabolas intersects in at most 4 points, and there are $\binom{n}{2}$ pairs of parabolas, we have $K \leq 4\binom{n}{2}=2 n^{2}-2 n$. Also, $r \leq n$, so $K+n+r+1 \leq 2 n^{2}+1$. On the other hand, as explained in the paragraphs giving the intuition, we can attain $2 n^{2}+1$ by making the parabolas sufficiently steep that they act like pairs of parallel lines. Therefore, the answer is $2 n^{2}+1$, as claimed.
|
{
"exam": "HMMT",
"problem_label": "6",
"problem_match": "\n6. [25]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Evaluate
$$
\sum_{k=1}^{2010} \cos ^{2}(k)
$$
|
$1005+\frac{\sin (4021)-\sin (1)}{4 \sin (1)}$ We use the identity $\cos ^{2}(k)=\frac{1+\cos (2 x)}{2}$. Then our expression evalutes to $1005+\frac{(\cos (2)+\ldots+\cos (4020))}{2}$.
To evaluate $\cos (2)+\cdots+\cos (4020)$, let $y=\cos (2)+\cdots+\cos (4020) \Rightarrow y(\sin (1))=\cos (2) \sin (1)+$ $\ldots+\cos (4020) \sin (1)$. Observe that for any $x, \cos (x) \sin (1)=\frac{\sin (x+1)-\sin (x-1)}{2}$. Then $y(\sin (1))=$ $\frac{\sin (3)-\sin (1)}{2}+\frac{\sin (5)-\sin (3)}{2}+\ldots+\frac{\sin 4021-\sin (4019)}{2}$. This is clearly a telescoping sum; we get $y(\sin (1))=$ $\frac{\sin (4021)-\sin (1)}{2}$. Then we have the desired $y=\frac{\sin (4021)-\sin (1)}{2 \sin (1)}$. Then our original expression evaluates to $1005+\frac{\sin (4021)-\sin (1)}{4 \sin (1)}$.
|
1005+\frac{\sin (4021)-\sin (1)}{4 \sin (1)}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Evaluate
$$
\sum_{k=1}^{2010} \cos ^{2}(k)
$$
|
$1005+\frac{\sin (4021)-\sin (1)}{4 \sin (1)}$ We use the identity $\cos ^{2}(k)=\frac{1+\cos (2 x)}{2}$. Then our expression evalutes to $1005+\frac{(\cos (2)+\ldots+\cos (4020))}{2}$.
To evaluate $\cos (2)+\cdots+\cos (4020)$, let $y=\cos (2)+\cdots+\cos (4020) \Rightarrow y(\sin (1))=\cos (2) \sin (1)+$ $\ldots+\cos (4020) \sin (1)$. Observe that for any $x, \cos (x) \sin (1)=\frac{\sin (x+1)-\sin (x-1)}{2}$. Then $y(\sin (1))=$ $\frac{\sin (3)-\sin (1)}{2}+\frac{\sin (5)-\sin (3)}{2}+\ldots+\frac{\sin 4021-\sin (4019)}{2}$. This is clearly a telescoping sum; we get $y(\sin (1))=$ $\frac{\sin (4021)-\sin (1)}{2}$. Then we have the desired $y=\frac{\sin (4021)-\sin (1)}{2 \sin (1)}$. Then our original expression evaluates to $1005+\frac{\sin (4021)-\sin (1)}{4 \sin (1)}$.
|
{
"exam": "HMMT",
"problem_label": "7",
"problem_match": "\n7. [30]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Consider the following two-player game. Player 1 starts with a number, $N$. He then subtracts a proper divisor of $N$ from $N$ and gives the result to player 2 (a proper divisor of $N$ is a positive divisor of $N$ that is not equal to 1 or $N)$. Player 2 does the same thing with the number she gets from player 1 , and gives the result back to player 1. The two players continue until a player is given a prime number, at which point that player loses. For how many values of $N$ between 2 and 100 inclusive does player 1 have a winning strategy?
|
47 We claim that player 1 has a winning strategy if and only if $N$ is even and not an odd power of 2 .
First we show that if you are stuck with an odd number, then you are guaranteed to lose. Suppose you have an odd number $a b$, where $a$ and $b$ are odd numbers, and you choose to subtract $a$. You pass your opponent the number $a(b-1)$. This cannot be a power of 2 (otherwise $a$ is a power of 2 and hence $a=1$, which is not allowed), so your opponent can find an odd proper divisor of $a(b-1)$ (such as $a$ ), and you will have a smaller odd number. Eventually you will get to an odd prime and lose.
Now consider even numbers that aren't powers of 2. As with before, you can find an odd proper divisor of $N$ and pass your opponent an odd number, so you are guaranteed to win.
Finally consider powers of 2 . If you have the number $N=2^{k}$, it would be unwise to choose a proper divisor other than $2^{k-1}$; otherwise you would give your opponent an even number that isn't a power of 2 . Therefore if $k$ is odd, you will end up with 2 and lose. If $k$ is even, though, your opponent will end up with 2 and you will win.
Therefore player 1 has a winning strategy for all even numbers except for odd powers of 2 .
|
47
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Consider the following two-player game. Player 1 starts with a number, $N$. He then subtracts a proper divisor of $N$ from $N$ and gives the result to player 2 (a proper divisor of $N$ is a positive divisor of $N$ that is not equal to 1 or $N)$. Player 2 does the same thing with the number she gets from player 1 , and gives the result back to player 1. The two players continue until a player is given a prime number, at which point that player loses. For how many values of $N$ between 2 and 100 inclusive does player 1 have a winning strategy?
|
47 We claim that player 1 has a winning strategy if and only if $N$ is even and not an odd power of 2 .
First we show that if you are stuck with an odd number, then you are guaranteed to lose. Suppose you have an odd number $a b$, where $a$ and $b$ are odd numbers, and you choose to subtract $a$. You pass your opponent the number $a(b-1)$. This cannot be a power of 2 (otherwise $a$ is a power of 2 and hence $a=1$, which is not allowed), so your opponent can find an odd proper divisor of $a(b-1)$ (such as $a$ ), and you will have a smaller odd number. Eventually you will get to an odd prime and lose.
Now consider even numbers that aren't powers of 2. As with before, you can find an odd proper divisor of $N$ and pass your opponent an odd number, so you are guaranteed to win.
Finally consider powers of 2 . If you have the number $N=2^{k}$, it would be unwise to choose a proper divisor other than $2^{k-1}$; otherwise you would give your opponent an even number that isn't a power of 2 . Therefore if $k$ is odd, you will end up with 2 and lose. If $k$ is even, though, your opponent will end up with 2 and you will win.
Therefore player 1 has a winning strategy for all even numbers except for odd powers of 2 .
|
{
"exam": "HMMT",
"problem_label": "8",
"problem_match": "\n8. [30]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Let $S$ be the set of ordered pairs of integers $(x, y)$ with $1 \leq x \leq 5$ and $1 \leq y \leq 3$. How many subsets $R$ of $S$ have the property that all the points of $R$ lie on the graph of a single cubic? A cubic is a polynomial of the form $y=a x^{3}+b x^{2}+c x+d$, where $a, b, c$, and $d$ are real numbers (meaning that $a$ is allowed to be 0 ).
|
796 We observe that $R$ must contain at most 1 point from each column of $S$, because no function can contain more than 1 point with the same $x$-coordinate. Therefore, $|R| \leq 5(|R|$ denotes the number of elements of $R$ ). Note that 4 points determine a cubic, so if $R$ is any subset of points in distinct columns and $|R| \leq 4$, then $R$ has the desired property. There are $4^{5}$ ways to choose at most 1
point from each column and $3^{5}$ ways to choose exactly 1 point from each column. There are therefore $4^{5}-3^{5}=781$ subsets $R$ of $S$ such that $|R| \leq 4$ and all points of $R$ lie in distinct columns. As noted, these sets all automatically have the desired property.
Now we consider all sets $R$ of size 5 . As before, each point in $R$ must come from a different column. Let us shift our origin to $(3,2)$, and let $p$ be the polynomial containing all 5 points of $R$. Then $R=\{(-2, p(-2)),(-1, p(-1)),(0, p(0)),(1, p(1)),(2, p(2))\}$.
By the method of finite differences $3^{3}$ or alternately by Lagrange Interpolation there is a unique polynomial $p$ of degree less than 5 going through 5 specified points, and this polynomial is of degree less than 4 if and only if $p(-2)-4 p(-1)+6 p(0)-4 p(1)+p(2)=0$.
Then $p(-2)+p(2)+6 p(0)=4(p(-1)+p(1))$, where $p(-2)+p(2) \in\{-2-1,0,1,2\}, p(-1)+p(1) \in$ $\{-2,-1,0,1,2\}$, and $p(0) \in\{-1,0,1\}$. We know that $6 p(0)$ and $4(p(-1)+p(1))$ are necessarily even, thus we must have $p(-2)+p(2) \in\{-2,0,2\}$ in order for the equation to be satisfied.
Let $(a, b, c)=(p(-2)+p(2), 6 p(0), 4(p(-1)+p(1)))$. The possible values of $(a, b, c)$ that are solutions to $a+b=c$ are then $\{(-2,-6,-8),(-2,6,4),(0,0,0),(2,-6,-4),(2,6,8)\}$.
If $(a, b, c)=(-2,-6,-8)$, then we need $p(-2)+p(2)=-2, p(0)=-1, p(-1)+p(1)=-2$. There is only 1 possible solution to each of these equations: $(p(-2), p(2))=(-1,-1)$ for the first one, $p(0)=-1$ for the second, and $(p(1))=(-1,-1)$ for the third. Hence there is 1 possible subset $R$ for the case $(a, b, c)=(-2,-6,-8)$.
If $(a, b, c)=(-2,6,4)$, then there is again 1 possible solution to $p(-2)+p(2)=1$. There are two solutions to $p(-1)+p(1)=1:(p(-1), p(1))=(0,1),(1,0)$. Also, $p(0)$ can only be 1 , so there are 2 possible subsets for this case.
If $(a, b, c)=(0,0,0)$, then there there are 3 possible solutions to $p(-2)+p(2)=0:(p(-2), p(2))=$ $(-1,1),(0,0),(1,-1)$. Similarly, there are 3 possible solutions to $p(-1)+p(1)=0$. Also, $p(0)$ can only be 0 , so there are 9 possible subsets for this case.
If $(a, b, c)=(2,-6,-4)$, then there is 1 possible solution to $p(-2)+p(2)=2:(p(-2), p(2))=(1,1)$. There are 2 possible solutions to $p(-1)+p(1)=-1:(p(-1), p(1))=(0,-1),(-1,0)$. Also, $p(0)$ can only be -1 , so there are 2 possible subsets for this case.
If $(a, b, c)=(2,6,8)$, then there is 1 possible solution to $p(-2)+p(2)=2$, as shown above. There is 1 solution to $p(-1)+p(1)=2:(p(-1), p(1))=(1,1)$. Also, $p(0)$ can only be 1 , so there is 1 possible subset for this case.
Then there are $1+2+9+2+1=15$ total possible subsets of size 5 that can be fit to a polynomial of degree less than 4 . Hence there are $781+15=796$ possible subsets total.
|
796
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $S$ be the set of ordered pairs of integers $(x, y)$ with $1 \leq x \leq 5$ and $1 \leq y \leq 3$. How many subsets $R$ of $S$ have the property that all the points of $R$ lie on the graph of a single cubic? A cubic is a polynomial of the form $y=a x^{3}+b x^{2}+c x+d$, where $a, b, c$, and $d$ are real numbers (meaning that $a$ is allowed to be 0 ).
|
796 We observe that $R$ must contain at most 1 point from each column of $S$, because no function can contain more than 1 point with the same $x$-coordinate. Therefore, $|R| \leq 5(|R|$ denotes the number of elements of $R$ ). Note that 4 points determine a cubic, so if $R$ is any subset of points in distinct columns and $|R| \leq 4$, then $R$ has the desired property. There are $4^{5}$ ways to choose at most 1
point from each column and $3^{5}$ ways to choose exactly 1 point from each column. There are therefore $4^{5}-3^{5}=781$ subsets $R$ of $S$ such that $|R| \leq 4$ and all points of $R$ lie in distinct columns. As noted, these sets all automatically have the desired property.
Now we consider all sets $R$ of size 5 . As before, each point in $R$ must come from a different column. Let us shift our origin to $(3,2)$, and let $p$ be the polynomial containing all 5 points of $R$. Then $R=\{(-2, p(-2)),(-1, p(-1)),(0, p(0)),(1, p(1)),(2, p(2))\}$.
By the method of finite differences $3^{3}$ or alternately by Lagrange Interpolation there is a unique polynomial $p$ of degree less than 5 going through 5 specified points, and this polynomial is of degree less than 4 if and only if $p(-2)-4 p(-1)+6 p(0)-4 p(1)+p(2)=0$.
Then $p(-2)+p(2)+6 p(0)=4(p(-1)+p(1))$, where $p(-2)+p(2) \in\{-2-1,0,1,2\}, p(-1)+p(1) \in$ $\{-2,-1,0,1,2\}$, and $p(0) \in\{-1,0,1\}$. We know that $6 p(0)$ and $4(p(-1)+p(1))$ are necessarily even, thus we must have $p(-2)+p(2) \in\{-2,0,2\}$ in order for the equation to be satisfied.
Let $(a, b, c)=(p(-2)+p(2), 6 p(0), 4(p(-1)+p(1)))$. The possible values of $(a, b, c)$ that are solutions to $a+b=c$ are then $\{(-2,-6,-8),(-2,6,4),(0,0,0),(2,-6,-4),(2,6,8)\}$.
If $(a, b, c)=(-2,-6,-8)$, then we need $p(-2)+p(2)=-2, p(0)=-1, p(-1)+p(1)=-2$. There is only 1 possible solution to each of these equations: $(p(-2), p(2))=(-1,-1)$ for the first one, $p(0)=-1$ for the second, and $(p(1))=(-1,-1)$ for the third. Hence there is 1 possible subset $R$ for the case $(a, b, c)=(-2,-6,-8)$.
If $(a, b, c)=(-2,6,4)$, then there is again 1 possible solution to $p(-2)+p(2)=1$. There are two solutions to $p(-1)+p(1)=1:(p(-1), p(1))=(0,1),(1,0)$. Also, $p(0)$ can only be 1 , so there are 2 possible subsets for this case.
If $(a, b, c)=(0,0,0)$, then there there are 3 possible solutions to $p(-2)+p(2)=0:(p(-2), p(2))=$ $(-1,1),(0,0),(1,-1)$. Similarly, there are 3 possible solutions to $p(-1)+p(1)=0$. Also, $p(0)$ can only be 0 , so there are 9 possible subsets for this case.
If $(a, b, c)=(2,-6,-4)$, then there is 1 possible solution to $p(-2)+p(2)=2:(p(-2), p(2))=(1,1)$. There are 2 possible solutions to $p(-1)+p(1)=-1:(p(-1), p(1))=(0,-1),(-1,0)$. Also, $p(0)$ can only be -1 , so there are 2 possible subsets for this case.
If $(a, b, c)=(2,6,8)$, then there is 1 possible solution to $p(-2)+p(2)=2$, as shown above. There is 1 solution to $p(-1)+p(1)=2:(p(-1), p(1))=(1,1)$. Also, $p(0)$ can only be 1 , so there is 1 possible subset for this case.
Then there are $1+2+9+2+1=15$ total possible subsets of size 5 that can be fit to a polynomial of degree less than 4 . Hence there are $781+15=796$ possible subsets total.
|
{
"exam": "HMMT",
"problem_label": "9",
"problem_match": "\n9. [35]",
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"solution_match": "\nAnswer: ",
"tier": "T4",
"year": "2010"
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.