problem
stringlengths 14
7.96k
| solution
stringlengths 3
10k
| answer
stringlengths 1
91
| problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
value | question_type
stringclasses 1
value | problem_type
stringclasses 8
values | problem_raw
stringlengths 14
7.96k
| solution_raw
stringlengths 3
10k
| metadata
dict | uuid
stringlengths 36
36
| id
int64 22.6k
612k
|
|---|---|---|---|---|---|---|---|---|---|---|---|
Find the largest integer less than 2012 all of whose divisors have at most two 1 's in their binary representations.
|
1536 Call a number good if all of its positive divisors have at most two 1's in their binary representations. Then, if $p$ is an odd prime divisor of a good number, $p$ must be of the form $2^{k}+1$. The only such primes less than 2012 are $3,5,17$, and 257 , so the only possible prime divisors of $n$ are $2,3,5,17$, and 257 .
Next, note that since $\left(2^{i}+1\right)\left(2^{j}+1\right)=2^{i+j}+2^{i}+2^{j}+1$, if either $i$ or $j$ is greater than 1 , then there will be at least 3 's in the binary representation of $\left(2^{i}+1\right)\left(2^{j}+1\right)$, so $\left(2^{i}+1\right)\left(2^{j}+1\right)$ cannot divide a good number. On the other hand, if $i=j=1$, then $\left(2^{1}+1\right)\left(2^{1}+1\right)=9=2^{3}+1$, so 9 is a good number and can divide a good number. Finally, note that since multiplication by 2 in binary just appends additional 0 s, so if $n$ is a good number, then $2 n$ is also a good number.
It therefore follows that any good number less than 2012 must be of the form $c \cdot 2^{k}$, where $c$ belongs to $\{1,3,5,9,17,257\}$ (and moreover, all such numbers are good). It is then straightforward to check that the largest such number is $1536=3 \cdot 2^{9}$.
|
1536
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the largest integer less than 2012 all of whose divisors have at most two 1 's in their binary representations.
|
1536 Call a number good if all of its positive divisors have at most two 1's in their binary representations. Then, if $p$ is an odd prime divisor of a good number, $p$ must be of the form $2^{k}+1$. The only such primes less than 2012 are $3,5,17$, and 257 , so the only possible prime divisors of $n$ are $2,3,5,17$, and 257 .
Next, note that since $\left(2^{i}+1\right)\left(2^{j}+1\right)=2^{i+j}+2^{i}+2^{j}+1$, if either $i$ or $j$ is greater than 1 , then there will be at least 3 's in the binary representation of $\left(2^{i}+1\right)\left(2^{j}+1\right)$, so $\left(2^{i}+1\right)\left(2^{j}+1\right)$ cannot divide a good number. On the other hand, if $i=j=1$, then $\left(2^{1}+1\right)\left(2^{1}+1\right)=9=2^{3}+1$, so 9 is a good number and can divide a good number. Finally, note that since multiplication by 2 in binary just appends additional 0 s, so if $n$ is a good number, then $2 n$ is also a good number.
It therefore follows that any good number less than 2012 must be of the form $c \cdot 2^{k}$, where $c$ belongs to $\{1,3,5,9,17,257\}$ (and moreover, all such numbers are good). It is then straightforward to check that the largest such number is $1536=3 \cdot 2^{9}$.
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-team-solutions.jsonl",
"problem_match": "\n3. [6]",
"solution_match": "\nAnswer: "
}
|
b7cdccb5-57ac-5842-b326-51523362d38e
| 609,052
|
Let $\pi$ be a permutation of the numbers from 2 through 2012. Find the largest possible value of $\log _{2} \pi(2) \cdot \log _{3} \pi(3) \cdots \log _{2012} \pi(2012)$.
|
1 Note that
$$
\begin{aligned}
\prod_{i=2}^{2012} \log _{i} \pi(i) & =\prod_{i=2}^{2012} \frac{\log \pi(i)}{\log i} \\
& =\frac{\prod_{i=2}^{2012} \log \pi(i)}{\prod_{i=2}^{2012} \log i} \\
& =1
\end{aligned}
$$
where the last equality holds since $\pi$ is a permutation of the numbers 2 through 2012 .
|
1
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $\pi$ be a permutation of the numbers from 2 through 2012. Find the largest possible value of $\log _{2} \pi(2) \cdot \log _{3} \pi(3) \cdots \log _{2012} \pi(2012)$.
|
1 Note that
$$
\begin{aligned}
\prod_{i=2}^{2012} \log _{i} \pi(i) & =\prod_{i=2}^{2012} \frac{\log \pi(i)}{\log i} \\
& =\frac{\prod_{i=2}^{2012} \log \pi(i)}{\prod_{i=2}^{2012} \log i} \\
& =1
\end{aligned}
$$
where the last equality holds since $\pi$ is a permutation of the numbers 2 through 2012 .
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-team-solutions.jsonl",
"problem_match": "\n4. [3]",
"solution_match": "\nAnswer: "
}
|
04b63981-7320-569f-961f-96f555d2bfbe
| 609,053
|
Let $\pi$ be a randomly chosen permutation of the numbers from 1 through 2012 . Find the probability that $\pi(\pi(2012))=2012$.
|
$\frac{1}{1006}$ There are two possibilities: either $\pi(2012)=2012$ or $\pi(2012)=i$ and $\pi(i)=2012$ for $i \neq 2012$. The first case occurs with probability $2011!/ 2012!=1 / 2012$, since any permutation on the remaining 2011 elements is possible. Similarly, for any fixed $i$, the second case occurs with probability $2010!/ 2012!=1 /(2011 \cdot 2012)$, since any permutation on the remaining 2010 elements is possible. Since there are 2011 possible values for $i$, and since our two possibilities are disjoint, the overall probability that $\pi(\pi(2012))=2012$ equals
$$
\frac{1}{2012}+(2011) \frac{1}{2011 \cdot 2012}=\frac{1}{1006}
$$
|
\frac{1}{1006}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $\pi$ be a randomly chosen permutation of the numbers from 1 through 2012 . Find the probability that $\pi(\pi(2012))=2012$.
|
$\frac{1}{1006}$ There are two possibilities: either $\pi(2012)=2012$ or $\pi(2012)=i$ and $\pi(i)=2012$ for $i \neq 2012$. The first case occurs with probability $2011!/ 2012!=1 / 2012$, since any permutation on the remaining 2011 elements is possible. Similarly, for any fixed $i$, the second case occurs with probability $2010!/ 2012!=1 /(2011 \cdot 2012)$, since any permutation on the remaining 2010 elements is possible. Since there are 2011 possible values for $i$, and since our two possibilities are disjoint, the overall probability that $\pi(\pi(2012))=2012$ equals
$$
\frac{1}{2012}+(2011) \frac{1}{2011 \cdot 2012}=\frac{1}{1006}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-team-solutions.jsonl",
"problem_match": "\n5. [4]",
"solution_match": "\nAnswer: "
}
|
da647abb-616a-54b5-986c-5d58e33752ff
| 609,054
|
Let $\pi$ be a permutation of the numbers from 1 through 2012. What is the maximum possible number of integers $n$ with $1 \leq n \leq 2011$ such that $\pi(n)$ divides $\pi(n+1)$ ?
|
1006 Since any proper divisor of $n$ must be less than or equal to $n / 2$, none of the numbers greater than 1006 can divide any other number less than or equal to 2012. Since there are at most 1006 values of $n$ for which $\pi(n) \leq 1006$, this means that there can be at most 1006 values of $n$ for which $\pi(n)$ divides $\pi(n+1)$.
On the other hand, there exists a permutation for which $\pi(n)$ divides $\pi(n+1)$ for exactly 1006 values of $n$, namely the permutation:
$$
\left(1,2,2^{2}, 2^{3}, \ldots, 2^{10}, 3,2 \cdot 3,2^{2} \cdot 3,2^{3} \cdot 3, \ldots, 2^{9} \cdot 3,5, \ldots\right)
$$
Formally, for each odd number $\ell \leq 2012$, we construct the sequence $\ell, 2 \ell, 4 \ell, \ldots, 2^{k} \ell$, where $k$ is the largest integer such that $2^{k} \ell \leq 2012$. We then concatenate all of these sequences to form a permutation of the numbers 1 through $\ell$ (note that no number occurs in more than one sequence). It follows that if $\pi(n) \leq 1006$, then $\pi(n+1)$ will equal $2 \pi(n)$, and therefore $\pi(n)$ will divide $\pi(n+1)$ for all 1006 values of $n$ satisfying $1 \leq \pi(n) \leq 1006$.
|
1006
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $\pi$ be a permutation of the numbers from 1 through 2012. What is the maximum possible number of integers $n$ with $1 \leq n \leq 2011$ such that $\pi(n)$ divides $\pi(n+1)$ ?
|
1006 Since any proper divisor of $n$ must be less than or equal to $n / 2$, none of the numbers greater than 1006 can divide any other number less than or equal to 2012. Since there are at most 1006 values of $n$ for which $\pi(n) \leq 1006$, this means that there can be at most 1006 values of $n$ for which $\pi(n)$ divides $\pi(n+1)$.
On the other hand, there exists a permutation for which $\pi(n)$ divides $\pi(n+1)$ for exactly 1006 values of $n$, namely the permutation:
$$
\left(1,2,2^{2}, 2^{3}, \ldots, 2^{10}, 3,2 \cdot 3,2^{2} \cdot 3,2^{3} \cdot 3, \ldots, 2^{9} \cdot 3,5, \ldots\right)
$$
Formally, for each odd number $\ell \leq 2012$, we construct the sequence $\ell, 2 \ell, 4 \ell, \ldots, 2^{k} \ell$, where $k$ is the largest integer such that $2^{k} \ell \leq 2012$. We then concatenate all of these sequences to form a permutation of the numbers 1 through $\ell$ (note that no number occurs in more than one sequence). It follows that if $\pi(n) \leq 1006$, then $\pi(n+1)$ will equal $2 \pi(n)$, and therefore $\pi(n)$ will divide $\pi(n+1)$ for all 1006 values of $n$ satisfying $1 \leq \pi(n) \leq 1006$.
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-team-solutions.jsonl",
"problem_match": "\n6. [6]",
"solution_match": "\nAnswer: "
}
|
e77e8990-c8e7-5ad7-a516-99de9cf537ce
| 609,055
|
Let $A_{1} A_{2} \ldots A_{100}$ be the vertices of a regular 100-gon. Let $\pi$ be a randomly chosen permutation of the numbers from 1 through 100. The segments $A_{\pi(1)} A_{\pi(2)}, A_{\pi(2)} A_{\pi(3)}, \ldots, A_{\pi(99)} A_{\pi(100)}, A_{\pi(100)} A_{\pi(1)}$ are drawn. Find the expected number of pairs of line segments that intersect at a point in the interior of the 100-gon.
|
| $\frac{4850}{3}$ |
| :---: |
| By linearity of expectation, the expected number of total intersections is equal to | the sum of the probabilities that any given intersection will occur.
Let us compute the probability $p_{i, j}$ that $A_{\pi(i)} A_{\pi(i+1)}$ intersects $A_{\pi(j)} A_{\pi(j+1)}$ (where $1 \leq i, j \leq 100$, $i \neq j$, and indices are taken modulo 100). Note first that if $j=i+1$, then these two segments share vertex $\pi(i+1)$ and therefore will not intersect in the interior of the 100-gon; similarly, if $i=j+1$, these two segments will also not intersect. On the other hand, if $\pi(i), \pi(i+1), \pi(j)$, and $\pi(j+1)$ are all distinct, then there is a $1 / 3$ chance that $A_{\pi(i)} A_{\pi(i+1)}$ intersects $A_{\pi(j)} A_{\pi(j+1)}$; in any set of four
points that form a convex quadrilateral, exactly one of the three ways of pairing the points into two pairs (two pairs of opposite sides and the two diagonals) forms two segments that intersect inside the quadrilateral (namely, the two diagonals).
Now, there are 100 ways to choose a value for $i$, and 97 ways to choose a value for $j$ which is not $i$, $i+1$, or $i-1$, there are 9700 ordered pairs $(i, j)$ where $p_{i, j}=1 / 3$. Since each pair is counted twice (once as $(i, j)$ and once as $(j, i)$ ), there are $9700 / 2=4850$ distinct possible intersections, each of which occurs with probability $1 / 3$, so the expected number of intersections is equal to $4850 / 3$.
|
\frac{4850}{3}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $A_{1} A_{2} \ldots A_{100}$ be the vertices of a regular 100-gon. Let $\pi$ be a randomly chosen permutation of the numbers from 1 through 100. The segments $A_{\pi(1)} A_{\pi(2)}, A_{\pi(2)} A_{\pi(3)}, \ldots, A_{\pi(99)} A_{\pi(100)}, A_{\pi(100)} A_{\pi(1)}$ are drawn. Find the expected number of pairs of line segments that intersect at a point in the interior of the 100-gon.
|
| $\frac{4850}{3}$ |
| :---: |
| By linearity of expectation, the expected number of total intersections is equal to | the sum of the probabilities that any given intersection will occur.
Let us compute the probability $p_{i, j}$ that $A_{\pi(i)} A_{\pi(i+1)}$ intersects $A_{\pi(j)} A_{\pi(j+1)}$ (where $1 \leq i, j \leq 100$, $i \neq j$, and indices are taken modulo 100). Note first that if $j=i+1$, then these two segments share vertex $\pi(i+1)$ and therefore will not intersect in the interior of the 100-gon; similarly, if $i=j+1$, these two segments will also not intersect. On the other hand, if $\pi(i), \pi(i+1), \pi(j)$, and $\pi(j+1)$ are all distinct, then there is a $1 / 3$ chance that $A_{\pi(i)} A_{\pi(i+1)}$ intersects $A_{\pi(j)} A_{\pi(j+1)}$; in any set of four
points that form a convex quadrilateral, exactly one of the three ways of pairing the points into two pairs (two pairs of opposite sides and the two diagonals) forms two segments that intersect inside the quadrilateral (namely, the two diagonals).
Now, there are 100 ways to choose a value for $i$, and 97 ways to choose a value for $j$ which is not $i$, $i+1$, or $i-1$, there are 9700 ordered pairs $(i, j)$ where $p_{i, j}=1 / 3$. Since each pair is counted twice (once as $(i, j)$ and once as $(j, i)$ ), there are $9700 / 2=4850$ distinct possible intersections, each of which occurs with probability $1 / 3$, so the expected number of intersections is equal to $4850 / 3$.
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-team-solutions.jsonl",
"problem_match": "\n7. [8]",
"solution_match": "\nAnswer: "
}
|
c9b07c5b-ac34-54e0-8db1-f10c5cabec91
| 609,056
|
$A B C$ is a triangle with $A B=15, B C=14$, and $C A=13$. The altitude from $A$ to $B C$ is extended to meet the circumcircle of $A B C$ at $D$. Find $A D$.
|
$\boxed{\frac{63}{4}}$ Let the altitude from $A$ to $BC$ meet $BC$ at $E$. The altitude $AE$ has length 12; one way to see this is that it splits the triangle $A B C$ into a $9-12-15$ right triangle and a $5-12-13$ right triangle; from this, we also know that $B E=9$ and $C E=5$.
Now, by Power of a Point, $A E \cdot D E=B E \cdot C E$, so $D E=(B E \cdot C E) / A E=(9 \cdot 5) /(12)=15 / 4$. It then follows that $A D=A E+D E=63 / 4$.
|
\frac{63}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
$A B C$ is a triangle with $A B=15, B C=14$, and $C A=13$. The altitude from $A$ to $B C$ is extended to meet the circumcircle of $A B C$ at $D$. Find $A D$.
|
$\boxed{\frac{63}{4}}$ Let the altitude from $A$ to $BC$ meet $BC$ at $E$. The altitude $AE$ has length 12; one way to see this is that it splits the triangle $A B C$ into a $9-12-15$ right triangle and a $5-12-13$ right triangle; from this, we also know that $B E=9$ and $C E=5$.
Now, by Power of a Point, $A E \cdot D E=B E \cdot C E$, so $D E=(B E \cdot C E) / A E=(9 \cdot 5) /(12)=15 / 4$. It then follows that $A D=A E+D E=63 / 4$.
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-team-solutions.jsonl",
"problem_match": "\n8. [4]",
"solution_match": "\nAnswer: "
}
|
c5853f17-7391-5bbb-8f4a-c7148ff35fdd
| 609,057
|
If $4^{4^{4}}=\sqrt[128]{2^{2^{2^{n}}}}$, find $n$.
|
4 We rewrite the left hand side as
$$
\left(2^{2}\right)^{4^{4}}=2^{2 \cdot 4^{4}}=2^{2^{9}},
$$
and the right hand side as
$$
\left(2^{2^{2^{n}}}\right)^{\frac{1}{128}}=2^{2^{\left(2^{n}-7\right)}} .
$$
Equating, we find $2^{n}-7=9$, yielding $n=4$.
|
4
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
If $4^{4^{4}}=\sqrt[128]{2^{2^{2^{n}}}}$, find $n$.
|
4 We rewrite the left hand side as
$$
\left(2^{2}\right)^{4^{4}}=2^{2 \cdot 4^{4}}=2^{2^{9}},
$$
and the right hand side as
$$
\left(2^{2^{2^{n}}}\right)^{\frac{1}{128}}=2^{2^{\left(2^{n}-7\right)}} .
$$
Equating, we find $2^{n}-7=9$, yielding $n=4$.
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-thm-solutions.jsonl",
"problem_match": "\n1. [3]",
"solution_match": "\nAnswer: "
}
|
854d53b3-5f55-5d55-96d9-bb762ca34af1
| 609,060
|
If $x^{x}=2012^{2012^{2013}}$, find $x$.
|
$2012^{2012}$ We have
$$
2012^{2012^{2013}}=2012^{2012 \cdot 2012^{2012}}=\left(2012^{2012}\right)^{2012^{2012}} .
$$
Thus, $x=2012^{2012}$.
|
2012^{2012}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
If $x^{x}=2012^{2012^{2013}}$, find $x$.
|
$2012^{2012}$ We have
$$
2012^{2012^{2013}}=2012^{2012 \cdot 2012^{2012}}=\left(2012^{2012}\right)^{2012^{2012}} .
$$
Thus, $x=2012^{2012}$.
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-thm-solutions.jsonl",
"problem_match": "\n2. [4]",
"solution_match": "\nAnswer: "
}
|
c42bc5b1-1204-502a-9aff-08f5f2d7ce48
| 609,061
|
Find the smallest positive integer $n$ such that $\underbrace{2^{2 \cdot 2}}_{n}>3^{3^{3^{3}}}$. (The notation $\underbrace{2^{2 \cdot 2^{2}}}_{n}$ is used to denote a power tower with $n 2$ 's. For example, $\underbrace{2^{2^{2}}}_{n}$ with $n=4$ would equal $2^{2^{2^{2}}}$.)
|
6 Clearly, $n \geq 5$. When we take $n=5$, we have
$$
2^{2^{2^{2^{2}}}}=2^{2^{16}}<3^{3^{27}}=3^{3^{3^{3}}} .
$$
On the other hand, when $n=6$, we have
$$
2^{2^{2^{2^{2^{2}}}}}=2^{2^{65536}}=4^{2^{65535}}>4^{4^{27}}>3^{3^{2^{27}}}=3^{3^{3^{3}}} .
$$
Our answer is thus $n=6$.
|
6
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the smallest positive integer $n$ such that $\underbrace{2^{2 \cdot 2}}_{n}>3^{3^{3^{3}}}$. (The notation $\underbrace{2^{2 \cdot 2^{2}}}_{n}$ is used to denote a power tower with $n 2$ 's. For example, $\underbrace{2^{2^{2}}}_{n}$ with $n=4$ would equal $2^{2^{2^{2}}}$.)
|
6 Clearly, $n \geq 5$. When we take $n=5$, we have
$$
2^{2^{2^{2^{2}}}}=2^{2^{16}}<3^{3^{27}}=3^{3^{3^{3}}} .
$$
On the other hand, when $n=6$, we have
$$
2^{2^{2^{2^{2^{2}}}}}=2^{2^{65536}}=4^{2^{65535}}>4^{4^{27}}>3^{3^{2^{27}}}=3^{3^{3^{3}}} .
$$
Our answer is thus $n=6$.
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-thm-solutions.jsonl",
"problem_match": "\n3. [4]",
"solution_match": "\nAnswer: "
}
|
7fb7270f-6277-5842-86dd-08d331a246eb
| 609,062
|
Given any positive integer, we can write the integer in base 12 and add together the digits of its base 12 representation. We perform this operation on the number $7^{6^{5^{3^{2^{1}}}}}$ repeatedly until a single base 12 digit remains. Find this digit.
|
4 For a positive integer $n$, let $s(n)$ be the sum of digits when $n$ is expressed in base 12 . We claim that $s(n) \equiv n(\bmod 11)$ for all positive integers $n$. Indeed, if $n=d_{k} 12^{k}+d_{k-1} 12^{k-1}+\cdots+d_{0}$ with each $d_{i}$ an integer between 0 and 11 , inclusive, because $12 \equiv 1(\bmod 11)$, reducing modulo 11 gives exactly $s(n)$. Thus, our answer is congruent to $N=7^{6^{5^{4^{3^{1}}}}}$ modulo 11 , and furthermore must be a one-digit integer in base 12; these two conditions uniquely determine the answer.
By Fermat's Little Theorem, $7^{10} \equiv 1(\bmod 11)$, and also observe that $6^{5^{4^{3^{2^{1}}}}} \equiv 6(\bmod 10)$ because $6 \cong 0(\bmod 2)$ and $6 \cong 1(\bmod 5)$. Thus, $N \equiv 7^{6} \equiv 343^{2} \equiv 2^{2} \equiv 4(\bmod 11)$, which is our answer. (Additionally, we note that this process of writing the number in base twelve and summing the digits must eventually terminate because the value decreases after each step.)
|
4
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Given any positive integer, we can write the integer in base 12 and add together the digits of its base 12 representation. We perform this operation on the number $7^{6^{5^{3^{2^{1}}}}}$ repeatedly until a single base 12 digit remains. Find this digit.
|
4 For a positive integer $n$, let $s(n)$ be the sum of digits when $n$ is expressed in base 12 . We claim that $s(n) \equiv n(\bmod 11)$ for all positive integers $n$. Indeed, if $n=d_{k} 12^{k}+d_{k-1} 12^{k-1}+\cdots+d_{0}$ with each $d_{i}$ an integer between 0 and 11 , inclusive, because $12 \equiv 1(\bmod 11)$, reducing modulo 11 gives exactly $s(n)$. Thus, our answer is congruent to $N=7^{6^{5^{4^{3^{1}}}}}$ modulo 11 , and furthermore must be a one-digit integer in base 12; these two conditions uniquely determine the answer.
By Fermat's Little Theorem, $7^{10} \equiv 1(\bmod 11)$, and also observe that $6^{5^{4^{3^{2^{1}}}}} \equiv 6(\bmod 10)$ because $6 \cong 0(\bmod 2)$ and $6 \cong 1(\bmod 5)$. Thus, $N \equiv 7^{6} \equiv 343^{2} \equiv 2^{2} \equiv 4(\bmod 11)$, which is our answer. (Additionally, we note that this process of writing the number in base twelve and summing the digits must eventually terminate because the value decreases after each step.)
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-thm-solutions.jsonl",
"problem_match": "\n5. [7]",
"solution_match": "\nAnswer: "
}
|
12d6d0d6-4163-584c-814a-d54025d85fcc
| 609,064
|
A rectangular piece of paper with vertices $A B C D$ is being cut by a pair of scissors. The pair of scissors starts at vertex $A$, and then cuts along the angle bisector of $D A B$ until it reaches another edge of the paper. One of the two resulting pieces of paper has 4 times the area of the other piece. What is the ratio of the longer side of the original paper to the shorter side?
|
$\sqrt{\frac{5}{2}}$ Without loss of generality, let $A B>A D$, and let $x=A D, y=A B$. Let the cut along the angle bisector of $\angle D A B$ meet $C D$ at $E$. Note that $A D E$ is a 45-45-90 triangle, so $D E=A D=x$, and $E C=y-x$. Now, $[A D E]=\frac{x^{2}}{2}$, and $[A E C B]=x\left(y-\frac{x}{2}\right)=4[A D E]$. Equating and dividing both sides by $x$, we find that $2 x=y-\frac{x}{2}$, so $y / x=\frac{5}{2}$.
|
\sqrt{\frac{5}{2}}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A rectangular piece of paper with vertices $A B C D$ is being cut by a pair of scissors. The pair of scissors starts at vertex $A$, and then cuts along the angle bisector of $D A B$ until it reaches another edge of the paper. One of the two resulting pieces of paper has 4 times the area of the other piece. What is the ratio of the longer side of the original paper to the shorter side?
|
$\sqrt{\frac{5}{2}}$ Without loss of generality, let $A B>A D$, and let $x=A D, y=A B$. Let the cut along the angle bisector of $\angle D A B$ meet $C D$ at $E$. Note that $A D E$ is a 45-45-90 triangle, so $D E=A D=x$, and $E C=y-x$. Now, $[A D E]=\frac{x^{2}}{2}$, and $[A E C B]=x\left(y-\frac{x}{2}\right)=4[A D E]$. Equating and dividing both sides by $x$, we find that $2 x=y-\frac{x}{2}$, so $y / x=\frac{5}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-thm-solutions.jsonl",
"problem_match": "\n6. [3]",
"solution_match": "\nAnswer: "
}
|
dfaf47c4-e7cc-5e7f-a613-49bbde12433a
| 609,065
|
The game of rock-scissors is played just like rock-paper-scissors, except that neither player is allowed to play paper. You play against a poorly-designed computer program that plays rock with $50 \%$ probability and scissors with $50 \%$ probability. If you play optimally against the computer, find the probability that after 8 games you have won at least 4 .
|
$\frac{163}{256}$ Since rock will always win against scissors, the optimum strategy is for you to always play rock; then, you win a game if and only if the computer plays scissors. Let $p_{n}$ be the probability that the computer plays scissors $n$ times; we want $p_{0}+p_{1}+p_{2}+p_{3}+p_{4}$. Note that by symmetry, $p_{n}=p_{8-n}$ for $n=0,1, \ldots, 8$, and because $p_{0}+p_{1}+\cdots+p_{8}=1, p_{0}+\cdots+p_{3}=p_{5}+\cdots+p_{8}=\left(1-p_{4}\right) / 2$. Our answer will thus be $\left(1+p_{4}\right) / 2$.
If the computer is to play scissors exactly 4 times, there are $\binom{8}{4}$ ways in which it can do so, compared to $2^{8}$ possible combinations of eight plays. Thus, $p_{4}=\binom{8}{4} / 2^{8}=35 / 128$. Our answer is thus $\frac{1+\frac{35}{128}}{2}=$ $\frac{163}{256}$.
|
\frac{163}{256}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
The game of rock-scissors is played just like rock-paper-scissors, except that neither player is allowed to play paper. You play against a poorly-designed computer program that plays rock with $50 \%$ probability and scissors with $50 \%$ probability. If you play optimally against the computer, find the probability that after 8 games you have won at least 4 .
|
$\frac{163}{256}$ Since rock will always win against scissors, the optimum strategy is for you to always play rock; then, you win a game if and only if the computer plays scissors. Let $p_{n}$ be the probability that the computer plays scissors $n$ times; we want $p_{0}+p_{1}+p_{2}+p_{3}+p_{4}$. Note that by symmetry, $p_{n}=p_{8-n}$ for $n=0,1, \ldots, 8$, and because $p_{0}+p_{1}+\cdots+p_{8}=1, p_{0}+\cdots+p_{3}=p_{5}+\cdots+p_{8}=\left(1-p_{4}\right) / 2$. Our answer will thus be $\left(1+p_{4}\right) / 2$.
If the computer is to play scissors exactly 4 times, there are $\binom{8}{4}$ ways in which it can do so, compared to $2^{8}$ possible combinations of eight plays. Thus, $p_{4}=\binom{8}{4} / 2^{8}=35 / 128$. Our answer is thus $\frac{1+\frac{35}{128}}{2}=$ $\frac{163}{256}$.
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-thm-solutions.jsonl",
"problem_match": "\n7. [4]",
"solution_match": "\nAnswer: "
}
|
936201b7-e571-5f61-8d33-f0f59ef2a5c1
| 609,066
|
In the game of rock-paper-scissors-lizard-Spock, rock defeats scissors and lizard, paper defeats rock and Spock, scissors defeats paper and lizard, lizard defeats paper and Spock, and Spock defeats rock and scissors, as shown in the below diagram. As before, if two players choose the same move, then there is a draw. If three people each play a game of rock-paper-scissors-lizard-Spock at the same time by choosing one of the five moves at random, what is the probability that one player beats the other two?
|
$\frac{12}{25}$ Let the three players be $A, B, C$. Our answer will simply be the sum of the probability that $A$ beats both $B$ and $C$, the probability that $B$ beats both $C$ and $A$, and the probability that $C$ beats $A$ and $B$, because these events are all mutually exclusive. By symmetry, these three probabilities are the same, so we only need to compute the probability that $A$ beats both $B$ and $C$. Given $A$ 's play, the probability that $B$ 's play loses to that of $A$ is $2 / 5$, and similarly for $C$. Thus, our answer is $3 \cdot\left(\frac{2}{5}\right) \cdot\left(\frac{2}{5}\right)=\frac{12}{25}$.
|
\frac{12}{25}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In the game of rock-paper-scissors-lizard-Spock, rock defeats scissors and lizard, paper defeats rock and Spock, scissors defeats paper and lizard, lizard defeats paper and Spock, and Spock defeats rock and scissors, as shown in the below diagram. As before, if two players choose the same move, then there is a draw. If three people each play a game of rock-paper-scissors-lizard-Spock at the same time by choosing one of the five moves at random, what is the probability that one player beats the other two?
|
$\frac{12}{25}$ Let the three players be $A, B, C$. Our answer will simply be the sum of the probability that $A$ beats both $B$ and $C$, the probability that $B$ beats both $C$ and $A$, and the probability that $C$ beats $A$ and $B$, because these events are all mutually exclusive. By symmetry, these three probabilities are the same, so we only need to compute the probability that $A$ beats both $B$ and $C$. Given $A$ 's play, the probability that $B$ 's play loses to that of $A$ is $2 / 5$, and similarly for $C$. Thus, our answer is $3 \cdot\left(\frac{2}{5}\right) \cdot\left(\frac{2}{5}\right)=\frac{12}{25}$.
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-thm-solutions.jsonl",
"problem_match": "\n8. [4]",
"solution_match": "\nAnswer: "
}
|
67432a2b-a0ba-507e-818f-110a6f6c2aab
| 609,067
|
64 people are in a single elimination rock-paper-scissors tournament, which consists of a 6-round knockout bracket. Each person has a different rock-paper-scissors skill level, and in any game, the person with the higher skill level will always win. For how many players $P$ is it possible that $P$ wins the first four rounds that he plays?
(A 6-round knockout bracket is a tournament which works as follows:
(a) In the first round, all 64 competitors are paired into 32 groups, and the two people in each group play each other. The winners advance to the second round, and the losers are eliminated.
(b) In the second round, the remaining 32 players are paired into 16 groups. Again, the winner of each group proceeds to the next round, while the loser is eliminated.
(c) Each round proceeds in a similar way, eliminating half of the remaining players. After the sixth round, only one player will not have been eliminated. That player is declared the champion.)
|
49 Note that a sub-bracket, that is, a subset of games of the tournament that themselves constitute a bracket, is always won by the person with the highest skill level. Therefore, a person wins her first four rounds if and only if she has the highest skill level among the people in her 16 -person sub-bracket. This is possible for all but the people with the $16-1=15$ lowest skill levels, so our answer is $64-15=49$.
|
49
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
64 people are in a single elimination rock-paper-scissors tournament, which consists of a 6-round knockout bracket. Each person has a different rock-paper-scissors skill level, and in any game, the person with the higher skill level will always win. For how many players $P$ is it possible that $P$ wins the first four rounds that he plays?
(A 6-round knockout bracket is a tournament which works as follows:
(a) In the first round, all 64 competitors are paired into 32 groups, and the two people in each group play each other. The winners advance to the second round, and the losers are eliminated.
(b) In the second round, the remaining 32 players are paired into 16 groups. Again, the winner of each group proceeds to the next round, while the loser is eliminated.
(c) Each round proceeds in a similar way, eliminating half of the remaining players. After the sixth round, only one player will not have been eliminated. That player is declared the champion.)
|
49 Note that a sub-bracket, that is, a subset of games of the tournament that themselves constitute a bracket, is always won by the person with the highest skill level. Therefore, a person wins her first four rounds if and only if she has the highest skill level among the people in her 16 -person sub-bracket. This is possible for all but the people with the $16-1=15$ lowest skill levels, so our answer is $64-15=49$.
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-thm-solutions.jsonl",
"problem_match": "\n9. [6]",
"solution_match": "\nAnswer: "
}
|
b294f0d1-5343-57c9-92af-ddc76f66926b
| 609,068
|
In a game of rock-paper-scissors with $n$ people, the following rules are used to determine a champion:
(a) In a round, each person who has not been eliminated randomly chooses one of rock, paper, or scissors to play.
(b) If at least one person plays rock, at least one person plays paper, and at least one person plays scissors, then the round is declared a tie and no one is eliminated. If everyone makes the same move, then the round is also declared a tie.
(c) If exactly two moves are represented, then everyone who made the losing move is eliminated from playing in all further rounds (for example, in a game with 8 people, if 5 people play rock and 3 people play scissors, then the 3 who played scissors are eliminated).
(d) The rounds continue until only one person has not been eliminated. That person is declared the champion and the game ends.
If a game begins with 4 people, what is the expected value of the number of rounds required for a champion to be determined?
|
$\sqrt[\frac{45}{14}]{ }$ For each positive integer $n$, let $E_{n}$ denote the expected number of rounds required to determine a winner among $n$ people. Clearly, $E_{1}=0$. When $n=2$, on the first move, there is a $\frac{1}{3}$ probability that there is a tie, and a $\frac{2}{3}$ probability that a winner is determined. In the first case, the expected number of additional rounds needed is exactly $E_{2}$; in the second, it is $E_{1}$. Therefore, we get the relation
$$
E_{2}=\frac{1}{3}\left(E_{2}+1\right)+\frac{2}{3}\left(E_{1}+1\right),
$$
from which it follows that $E_{2}=\frac{3}{2}$.
Next, if $n=3$, with probability $\frac{1}{9}$ there is only one distinct play among the three players, and with probability $\frac{6}{27}=\frac{2}{9}$ all three players make different plays. In both of these cases, no players are eliminated. In all remaining situations, which occur with total probability $\frac{2}{3}$, two players make one play and the third makes a distinct play; with probability $\frac{1}{3}$ two players are eliminated and with probability $\frac{1}{3}$ one player is eliminated. This gives the relation
$$
E_{3}=\frac{1}{3}\left(E_{3}+1\right)+\frac{1}{3}\left(E_{2}+1\right)+\frac{1}{3}\left(E_{1}+1\right),
$$
from which we find that $E_{3}=\frac{9}{4}$.
Finally, suppose $n=4$. With probability $\frac{1}{27}$, all four players make the same play, and with probability $\frac{3 \cdot 6 \cdot 2}{81}=\frac{4}{9}$, two players make one play, and the other two players make the other two plays; in both cases no players are eliminated, with total probability $\frac{1}{27}+\frac{4}{9}=\frac{13}{27}$ over the two cases. With probability $\frac{6 \cdot 4}{81}=\frac{8}{27}$, three players make one play and the fourth makes another; thus, there is a probability of $\frac{4}{27}$ for exactly one player being eliminated and a probability of $\frac{4}{27}$ of three players being eliminated.
Then, there is a remaining probability of $\frac{6 \cdot 3}{81}=\frac{2}{9}$, two players make one play and the other two players make another. Similar analysis from before yields
$$
E_{4}=\frac{13}{27}\left(E_{4}+1\right)+\frac{4}{27}\left(E_{3}+1\right)+\frac{2}{9}\left(E_{2}+1\right)+\frac{4}{27}\left(E_{1}+1\right)
$$
so it follows that $E_{4}=\frac{45}{14}$.
|
\frac{45}{14}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In a game of rock-paper-scissors with $n$ people, the following rules are used to determine a champion:
(a) In a round, each person who has not been eliminated randomly chooses one of rock, paper, or scissors to play.
(b) If at least one person plays rock, at least one person plays paper, and at least one person plays scissors, then the round is declared a tie and no one is eliminated. If everyone makes the same move, then the round is also declared a tie.
(c) If exactly two moves are represented, then everyone who made the losing move is eliminated from playing in all further rounds (for example, in a game with 8 people, if 5 people play rock and 3 people play scissors, then the 3 who played scissors are eliminated).
(d) The rounds continue until only one person has not been eliminated. That person is declared the champion and the game ends.
If a game begins with 4 people, what is the expected value of the number of rounds required for a champion to be determined?
|
$\sqrt[\frac{45}{14}]{ }$ For each positive integer $n$, let $E_{n}$ denote the expected number of rounds required to determine a winner among $n$ people. Clearly, $E_{1}=0$. When $n=2$, on the first move, there is a $\frac{1}{3}$ probability that there is a tie, and a $\frac{2}{3}$ probability that a winner is determined. In the first case, the expected number of additional rounds needed is exactly $E_{2}$; in the second, it is $E_{1}$. Therefore, we get the relation
$$
E_{2}=\frac{1}{3}\left(E_{2}+1\right)+\frac{2}{3}\left(E_{1}+1\right),
$$
from which it follows that $E_{2}=\frac{3}{2}$.
Next, if $n=3$, with probability $\frac{1}{9}$ there is only one distinct play among the three players, and with probability $\frac{6}{27}=\frac{2}{9}$ all three players make different plays. In both of these cases, no players are eliminated. In all remaining situations, which occur with total probability $\frac{2}{3}$, two players make one play and the third makes a distinct play; with probability $\frac{1}{3}$ two players are eliminated and with probability $\frac{1}{3}$ one player is eliminated. This gives the relation
$$
E_{3}=\frac{1}{3}\left(E_{3}+1\right)+\frac{1}{3}\left(E_{2}+1\right)+\frac{1}{3}\left(E_{1}+1\right),
$$
from which we find that $E_{3}=\frac{9}{4}$.
Finally, suppose $n=4$. With probability $\frac{1}{27}$, all four players make the same play, and with probability $\frac{3 \cdot 6 \cdot 2}{81}=\frac{4}{9}$, two players make one play, and the other two players make the other two plays; in both cases no players are eliminated, with total probability $\frac{1}{27}+\frac{4}{9}=\frac{13}{27}$ over the two cases. With probability $\frac{6 \cdot 4}{81}=\frac{8}{27}$, three players make one play and the fourth makes another; thus, there is a probability of $\frac{4}{27}$ for exactly one player being eliminated and a probability of $\frac{4}{27}$ of three players being eliminated.
Then, there is a remaining probability of $\frac{6 \cdot 3}{81}=\frac{2}{9}$, two players make one play and the other two players make another. Similar analysis from before yields
$$
E_{4}=\frac{13}{27}\left(E_{4}+1\right)+\frac{4}{27}\left(E_{3}+1\right)+\frac{2}{9}\left(E_{2}+1\right)+\frac{4}{27}\left(E_{1}+1\right)
$$
so it follows that $E_{4}=\frac{45}{14}$.
|
{
"resource_path": "HarvardMIT/segmented/en-161-2012-nov-thm-solutions.jsonl",
"problem_match": "\n10. [8]",
"solution_match": "\nAnswer: "
}
|
cd4464e5-4f74-5ca7-b607-4becf4334126
| 609,069
|
Let $x$ and $y$ be real numbers with $x>y$ such that $x^{2} y^{2}+x^{2}+y^{2}+2 x y=40$ and $x y+x+y=8$. Find the value of $x$.
|
$\quad 3+\sqrt{7}$ We have $(x y)^{2}+(x+y)^{2}=40$ and $x y+(x+y)=8$. Squaring the second equation and subtracting the first gives $x y(x+y)=12$ so $x y, x+y$ are the roots of the quadratic $a^{2}-8 a+12=0$. It follows that $\{x y, x+y\}=\{2,6\}$. If $x+y=2$ and $x y=6$, then $x, y$ are the roots of the quadratic $b^{2}-2 b+6=0$, which are non-real, so in fact $x+y=6$ and $x y=2$, and $x, y$ are the roots of the quadratic $b^{2}-6 b+2=0$. Because $x>y$, we take the larger root, which is $\frac{6+\sqrt{28}}{2}=3+\sqrt{7}$.
|
3+\sqrt{7}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $x$ and $y$ be real numbers with $x>y$ such that $x^{2} y^{2}+x^{2}+y^{2}+2 x y=40$ and $x y+x+y=8$. Find the value of $x$.
|
$\quad 3+\sqrt{7}$ We have $(x y)^{2}+(x+y)^{2}=40$ and $x y+(x+y)=8$. Squaring the second equation and subtracting the first gives $x y(x+y)=12$ so $x y, x+y$ are the roots of the quadratic $a^{2}-8 a+12=0$. It follows that $\{x y, x+y\}=\{2,6\}$. If $x+y=2$ and $x y=6$, then $x, y$ are the roots of the quadratic $b^{2}-2 b+6=0$, which are non-real, so in fact $x+y=6$ and $x y=2$, and $x, y$ are the roots of the quadratic $b^{2}-6 b+2=0$. Because $x>y$, we take the larger root, which is $\frac{6+\sqrt{28}}{2}=3+\sqrt{7}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-alg-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nAnswer: "
}
|
5bc51951-a1b7-5aed-88b2-1e37b5e5dde5
| 609,070
|
Let $\left\{a_{n}\right\}_{n \geq 1}$ be an arithmetic sequence and $\left\{g_{n}\right\}_{n \geq 1}$ be a geometric sequence such that the first four terms of $\left\{a_{n}+g_{n}\right\}$ are $0,0,1$, and 0 , in that order. What is the 10 th term of $\left\{a_{n}+g_{n}\right\}$ ?
|
$\quad-54$ Let the terms of the geometric sequence be $a, r a, r^{2} a, r^{3} a$. Then, the terms of the arithmetic sequence are $-a,-r a,-r^{2} a+1,-r^{3} a$. However, if the first two terms of this sequence are $-a,-r a$, the next two terms must also be $(-2 r+1) a,(-3 r+2) a$. It is clear that $a \neq 0$ because $a_{3}+g_{3} \neq 0$, so $-r^{3}=-3 r+2 \Rightarrow r=1$ or -2 . However, we see from the arithmetic sequence that $r=1$ is impossible, so $r=-2$. Finally, by considering $a_{3}$, we see that $-4 a+1=5 a$, so $a=1 / 9$. We also see that $a_{n}=(3 n-4) a$ and $g_{n}=(-2)^{n-1} a$, so our answer is $a_{10}+g_{10}=(26-512) a=-486 a=-54$.
|
-54
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $\left\{a_{n}\right\}_{n \geq 1}$ be an arithmetic sequence and $\left\{g_{n}\right\}_{n \geq 1}$ be a geometric sequence such that the first four terms of $\left\{a_{n}+g_{n}\right\}$ are $0,0,1$, and 0 , in that order. What is the 10 th term of $\left\{a_{n}+g_{n}\right\}$ ?
|
$\quad-54$ Let the terms of the geometric sequence be $a, r a, r^{2} a, r^{3} a$. Then, the terms of the arithmetic sequence are $-a,-r a,-r^{2} a+1,-r^{3} a$. However, if the first two terms of this sequence are $-a,-r a$, the next two terms must also be $(-2 r+1) a,(-3 r+2) a$. It is clear that $a \neq 0$ because $a_{3}+g_{3} \neq 0$, so $-r^{3}=-3 r+2 \Rightarrow r=1$ or -2 . However, we see from the arithmetic sequence that $r=1$ is impossible, so $r=-2$. Finally, by considering $a_{3}$, we see that $-4 a+1=5 a$, so $a=1 / 9$. We also see that $a_{n}=(3 n-4) a$ and $g_{n}=(-2)^{n-1} a$, so our answer is $a_{10}+g_{10}=(26-512) a=-486 a=-54$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-alg-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nAnswer: "
}
|
3a43afe7-0035-538c-be5d-052682749f74
| 609,071
|
Let $S$ be the set of integers of the form $2^{x}+2^{y}+2^{z}$, where $x, y, z$ are pairwise distinct non-negative integers. Determine the 100 th smallest element of $S$.
|
$577 S$ is the set of positive integers with exactly three ones in its binary representation. The number of such integers with at most $d$ total bits is $\binom{d}{3}$, and noting that $\binom{9}{3}=84$ and $\binom{10}{3}=120$, we want the 16 th smallest integer of the form $2^{9}+2^{x}+2^{y}$, where $y<x<9$. Ignoring the $2^{9}$ term, there are $\binom{d^{\prime}}{2}$ positive integers of the form $2^{x}+2^{y}$ with at most $d^{\prime}$ total bits. Because $\binom{6}{2}=15$, our answer is $2^{9}+2^{6}+2^{0}=577$. (By a bit, we mean a digit in base 2.)
|
577
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $S$ be the set of integers of the form $2^{x}+2^{y}+2^{z}$, where $x, y, z$ are pairwise distinct non-negative integers. Determine the 100 th smallest element of $S$.
|
$577 S$ is the set of positive integers with exactly three ones in its binary representation. The number of such integers with at most $d$ total bits is $\binom{d}{3}$, and noting that $\binom{9}{3}=84$ and $\binom{10}{3}=120$, we want the 16 th smallest integer of the form $2^{9}+2^{x}+2^{y}$, where $y<x<9$. Ignoring the $2^{9}$ term, there are $\binom{d^{\prime}}{2}$ positive integers of the form $2^{x}+2^{y}$ with at most $d^{\prime}$ total bits. Because $\binom{6}{2}=15$, our answer is $2^{9}+2^{6}+2^{0}=577$. (By a bit, we mean a digit in base 2.)
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-alg-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nAnswer: "
}
|
da45ee3c-d14d-5fb2-83f3-ec000b470235
| 609,072
|
Determine all real values of $A$ for which there exist distinct complex numbers $x_{1}, x_{2}$ such that the following three equations hold:
$$
\begin{aligned}
x_{1}\left(x_{1}+1\right) & =A \\
x_{2}\left(x_{2}+1\right) & =A \\
x_{1}^{4}+3 x_{1}^{3}+5 x_{1} & =x_{2}^{4}+3 x_{2}^{3}+5 x_{2} .
\end{aligned}
$$
|
$\quad-7$ Applying polynomial division,
$$
\begin{aligned}
x_{1}^{4}+3 x_{1}^{3}+5 x_{1} & =\left(x_{1}^{2}+x_{1}-A\right)\left(x_{1}^{2}+2 x_{1}+(A-2)\right)+(A+7) x_{1}+A(A-2) \\
& =(A+7) x_{1}+A(A-2)
\end{aligned}
$$
Thus, in order for the last equation to hold, we need $(A+7) x_{1}=(A+7) x_{2}$, from which it follows that $A=-7$. These steps are reversible, so $A=-7$ indeed satisfies the needed condition.
|
-7
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Determine all real values of $A$ for which there exist distinct complex numbers $x_{1}, x_{2}$ such that the following three equations hold:
$$
\begin{aligned}
x_{1}\left(x_{1}+1\right) & =A \\
x_{2}\left(x_{2}+1\right) & =A \\
x_{1}^{4}+3 x_{1}^{3}+5 x_{1} & =x_{2}^{4}+3 x_{2}^{3}+5 x_{2} .
\end{aligned}
$$
|
$\quad-7$ Applying polynomial division,
$$
\begin{aligned}
x_{1}^{4}+3 x_{1}^{3}+5 x_{1} & =\left(x_{1}^{2}+x_{1}-A\right)\left(x_{1}^{2}+2 x_{1}+(A-2)\right)+(A+7) x_{1}+A(A-2) \\
& =(A+7) x_{1}+A(A-2)
\end{aligned}
$$
Thus, in order for the last equation to hold, we need $(A+7) x_{1}=(A+7) x_{2}$, from which it follows that $A=-7$. These steps are reversible, so $A=-7$ indeed satisfies the needed condition.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-alg-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nAnswer: "
}
|
2116f144-7fda-5167-bafe-f6bbe6840b8a
| 609,073
|
Let $a$ and $b$ be real numbers, and let $r, s$, and $t$ be the roots of $f(x)=x^{3}+a x^{2}+b x-1$. Also, $g(x)=x^{3}+m x^{2}+n x+p$ has roots $r^{2}, s^{2}$, and $t^{2}$. If $g(-1)=-5$, find the maximum possible value of $b$.
|
$\sqrt[{1+\sqrt{5}}]{ }$ By Vieta's Formulae, $m=-\left(r^{2}+s^{2}+t^{2}\right)=-a^{2}+2 b, n=r^{2} s^{2}+s^{2} t^{2}+t^{2} r^{2}=$ $b^{2}+2 a$, and $p=-1$. Therefore, $g(-1)=-1-a^{2}+2 b-b^{2}-2 a-1=-5 \Leftrightarrow(a+1)^{2}+(b-1)^{2}=5$. This is an equation of a circle, so $b$ reaches its maximum when $a+1=0 \Rightarrow a=-1$. When $a=-1$, $b=1 \pm \sqrt{5}$, so the maximum is $1+\sqrt{5}$.
|
1+\sqrt{5}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a$ and $b$ be real numbers, and let $r, s$, and $t$ be the roots of $f(x)=x^{3}+a x^{2}+b x-1$. Also, $g(x)=x^{3}+m x^{2}+n x+p$ has roots $r^{2}, s^{2}$, and $t^{2}$. If $g(-1)=-5$, find the maximum possible value of $b$.
|
$\sqrt[{1+\sqrt{5}}]{ }$ By Vieta's Formulae, $m=-\left(r^{2}+s^{2}+t^{2}\right)=-a^{2}+2 b, n=r^{2} s^{2}+s^{2} t^{2}+t^{2} r^{2}=$ $b^{2}+2 a$, and $p=-1$. Therefore, $g(-1)=-1-a^{2}+2 b-b^{2}-2 a-1=-5 \Leftrightarrow(a+1)^{2}+(b-1)^{2}=5$. This is an equation of a circle, so $b$ reaches its maximum when $a+1=0 \Rightarrow a=-1$. When $a=-1$, $b=1 \pm \sqrt{5}$, so the maximum is $1+\sqrt{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-alg-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nAnswer: "
}
|
7df6dc02-e941-56bd-90d9-853a7f58101f
| 609,074
|
Find the number of integers $n$ such that
$$
1+\left\lfloor\frac{100 n}{101}\right\rfloor=\left\lceil\frac{99 n}{100}\right\rceil
$$
|
10100 Consider $f(n)=\left\lceil\frac{99 n}{100}\right\rceil-\left\lfloor\frac{100 n}{101}\right\rfloor$. Note that $f(n+10100)=\left\lceil\frac{99 n}{100}+99 \cdot 101\right\rceil-$ $\left\lfloor\frac{100 n}{101}+100^{2}\right\rfloor=f(n)+99 \cdot 101-100^{2}=f(n)-1$. Thus, for each residue class $r$ modulo 10100, there is exactly one value of $n$ for which $f(n)=1$ and $n \equiv r(\bmod 10100)$. It follows immediately that the answer is 10100 .
|
10100
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the number of integers $n$ such that
$$
1+\left\lfloor\frac{100 n}{101}\right\rfloor=\left\lceil\frac{99 n}{100}\right\rceil
$$
|
10100 Consider $f(n)=\left\lceil\frac{99 n}{100}\right\rceil-\left\lfloor\frac{100 n}{101}\right\rfloor$. Note that $f(n+10100)=\left\lceil\frac{99 n}{100}+99 \cdot 101\right\rceil-$ $\left\lfloor\frac{100 n}{101}+100^{2}\right\rfloor=f(n)+99 \cdot 101-100^{2}=f(n)-1$. Thus, for each residue class $r$ modulo 10100, there is exactly one value of $n$ for which $f(n)=1$ and $n \equiv r(\bmod 10100)$. It follows immediately that the answer is 10100 .
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-alg-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nAnswer: "
}
|
88b426a4-72d7-595b-89d8-f621e6e85b82
| 609,075
|
Compute
$$
\sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty} \frac{a_{1}+a_{2}+\cdots+a_{7}}{3^{a_{1}+a_{2}+\cdots+a_{7}}}
$$
|
$15309 / 256$ Note that, since this is symmetric in $a_{1}$ through $a_{7}$,
$$
\begin{aligned}
\sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty} \frac{a_{1}+a_{2}+\cdots+a_{7}}{3^{a_{1}+a_{2}+\cdots+a_{7}}} & =7 \sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty} \frac{a_{1}}{3^{a_{1}+a_{2}+\cdots+a_{7}}} \\
& =7\left(\sum_{a_{1}=0}^{\infty} \frac{a_{1}}{3^{a_{1}}}\right)^{\left(\sum_{a=0}^{\infty} \frac{1}{3^{a}}\right)^{6}}
\end{aligned}
$$
If $S=\sum \frac{a}{3^{a}}$, then $3 S-S=\sum \frac{1}{3^{a}}=3 / 2$, so $S=3 / 4$. It follows that the answer equals $7 \cdot \frac{3}{4} \cdot\left(\frac{3}{2}\right)^{6}=\frac{15309}{256}$. Alternatively, let $f(z)=\sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty} z^{a_{1}+a_{2}+\cdots+a_{7}}$. Note that we can rewrite $f(z)=$ $\left(\sum_{a=0}^{\infty} z^{a}\right)^{7}=\frac{1}{(1-z)^{7}}$. Furthermore, note that $z f^{\prime}(z)=\sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty}\left(a_{1}+a_{2}+\cdots+\right.$ $\left.a_{7}\right) z^{a_{1}+a_{2}+\cdots+a_{7}}$, so the sum in question is simply $\frac{f^{\prime}(1 / 3)}{3}$. Since $f^{\prime}(x)=\frac{7}{(1-z)^{8}}$, it follows that the sum is equal to $\frac{7 \cdot 3^{7}}{2^{8}}=\frac{15309}{256}$.
|
\frac{15309}{256}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Compute
$$
\sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty} \frac{a_{1}+a_{2}+\cdots+a_{7}}{3^{a_{1}+a_{2}+\cdots+a_{7}}}
$$
|
$15309 / 256$ Note that, since this is symmetric in $a_{1}$ through $a_{7}$,
$$
\begin{aligned}
\sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty} \frac{a_{1}+a_{2}+\cdots+a_{7}}{3^{a_{1}+a_{2}+\cdots+a_{7}}} & =7 \sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty} \frac{a_{1}}{3^{a_{1}+a_{2}+\cdots+a_{7}}} \\
& =7\left(\sum_{a_{1}=0}^{\infty} \frac{a_{1}}{3^{a_{1}}}\right)^{\left(\sum_{a=0}^{\infty} \frac{1}{3^{a}}\right)^{6}}
\end{aligned}
$$
If $S=\sum \frac{a}{3^{a}}$, then $3 S-S=\sum \frac{1}{3^{a}}=3 / 2$, so $S=3 / 4$. It follows that the answer equals $7 \cdot \frac{3}{4} \cdot\left(\frac{3}{2}\right)^{6}=\frac{15309}{256}$. Alternatively, let $f(z)=\sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty} z^{a_{1}+a_{2}+\cdots+a_{7}}$. Note that we can rewrite $f(z)=$ $\left(\sum_{a=0}^{\infty} z^{a}\right)^{7}=\frac{1}{(1-z)^{7}}$. Furthermore, note that $z f^{\prime}(z)=\sum_{a_{1}=0}^{\infty} \sum_{a_{2}=0}^{\infty} \cdots \sum_{a_{7}=0}^{\infty}\left(a_{1}+a_{2}+\cdots+\right.$ $\left.a_{7}\right) z^{a_{1}+a_{2}+\cdots+a_{7}}$, so the sum in question is simply $\frac{f^{\prime}(1 / 3)}{3}$. Since $f^{\prime}(x)=\frac{7}{(1-z)^{8}}$, it follows that the sum is equal to $\frac{7 \cdot 3^{7}}{2^{8}}=\frac{15309}{256}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-alg-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nAnswer: "
}
|
2f2fcfdd-a6e6-5323-a8b5-f9f5123b3b16
| 609,076
|
Let $z$ be a non-real complex number with $z^{23}=1$. Compute
$$
\sum_{k=0}^{22} \frac{1}{1+z^{k}+z^{2 k}}
$$
|
$46 / 3$ First solution: Note that
$$
\sum_{k=0}^{22} \frac{1}{1+z^{k}+z^{2 k}}=\frac{1}{3}+\sum_{k=1}^{22} \frac{1-z^{k}}{1-z^{3 k}}=\frac{1}{3}+\sum_{k=1}^{22} \frac{1-\left(z^{24}\right)^{k}}{1-z^{3 k}}=\frac{1}{3}+\sum_{k=1}^{22} \sum_{\ell=0}^{7} z^{3 k \ell}
$$
3 and 23 are prime, so every non-zero residue modulo 23 appears in an exponent in the last sum exactly 7 times, and the summand 1 appears 22 times. Because the sum of the 23 rd roots of unity is zero, our answer is $\frac{1}{3}+(22-7)=\frac{46}{3}$.
Second solution: For an alternate approach, we first prove the following identity for an arbitrary complex number $a$ :
$$
\sum_{k=0}^{22} \frac{1}{a-z^{k}}=\frac{23 a^{22}}{a^{23}-1}
$$
To see this, let $f(x)=x^{23}-1=(x-1)(x-z)\left(x-z^{2}\right) \ldots\left(x-z^{22}\right)$. Note that the sum in question is merely $\frac{f^{\prime}(a)}{f(a)}$, from which the identity follows.
Now, returning to our orignal sum, let $\omega \neq 1$ satisfy $\omega^{3}=1$. Then
$$
\begin{aligned}
\sum_{k=0}^{22} \frac{1}{1+z^{k}+z^{2 k}} & =\frac{1}{\omega^{2}-\omega} \sum_{k=0}^{22} \frac{1}{\omega-z^{k}}-\frac{1}{\omega^{2}-z^{k}} \\
& =\frac{1}{\omega^{2}-\omega}\left(\sum_{k=0}^{22} \frac{1}{\omega-z^{k}}-\sum_{k=0}^{22} \frac{1}{\omega^{2}-z^{k}}\right) \\
& =\frac{1}{\omega^{2}-\omega}\left(\frac{23 \omega^{22}}{\omega^{23}-1}-\frac{23 \omega^{44}}{\omega^{46}-1}\right) \\
& =\frac{23}{\omega^{2}-\omega}\left(\frac{\omega}{\omega^{2}-1}-\frac{\omega^{2}}{\omega-1}\right) \\
& =\frac{23}{\omega^{2}-\omega} \frac{\left(\omega^{2}-\omega\right)-\left(\omega-\omega^{2}\right)}{2-\omega-\omega^{2}} \\
& =\frac{46}{3}
\end{aligned}
$$
|
\frac{46}{3}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $z$ be a non-real complex number with $z^{23}=1$. Compute
$$
\sum_{k=0}^{22} \frac{1}{1+z^{k}+z^{2 k}}
$$
|
$46 / 3$ First solution: Note that
$$
\sum_{k=0}^{22} \frac{1}{1+z^{k}+z^{2 k}}=\frac{1}{3}+\sum_{k=1}^{22} \frac{1-z^{k}}{1-z^{3 k}}=\frac{1}{3}+\sum_{k=1}^{22} \frac{1-\left(z^{24}\right)^{k}}{1-z^{3 k}}=\frac{1}{3}+\sum_{k=1}^{22} \sum_{\ell=0}^{7} z^{3 k \ell}
$$
3 and 23 are prime, so every non-zero residue modulo 23 appears in an exponent in the last sum exactly 7 times, and the summand 1 appears 22 times. Because the sum of the 23 rd roots of unity is zero, our answer is $\frac{1}{3}+(22-7)=\frac{46}{3}$.
Second solution: For an alternate approach, we first prove the following identity for an arbitrary complex number $a$ :
$$
\sum_{k=0}^{22} \frac{1}{a-z^{k}}=\frac{23 a^{22}}{a^{23}-1}
$$
To see this, let $f(x)=x^{23}-1=(x-1)(x-z)\left(x-z^{2}\right) \ldots\left(x-z^{22}\right)$. Note that the sum in question is merely $\frac{f^{\prime}(a)}{f(a)}$, from which the identity follows.
Now, returning to our orignal sum, let $\omega \neq 1$ satisfy $\omega^{3}=1$. Then
$$
\begin{aligned}
\sum_{k=0}^{22} \frac{1}{1+z^{k}+z^{2 k}} & =\frac{1}{\omega^{2}-\omega} \sum_{k=0}^{22} \frac{1}{\omega-z^{k}}-\frac{1}{\omega^{2}-z^{k}} \\
& =\frac{1}{\omega^{2}-\omega}\left(\sum_{k=0}^{22} \frac{1}{\omega-z^{k}}-\sum_{k=0}^{22} \frac{1}{\omega^{2}-z^{k}}\right) \\
& =\frac{1}{\omega^{2}-\omega}\left(\frac{23 \omega^{22}}{\omega^{23}-1}-\frac{23 \omega^{44}}{\omega^{46}-1}\right) \\
& =\frac{23}{\omega^{2}-\omega}\left(\frac{\omega}{\omega^{2}-1}-\frac{\omega^{2}}{\omega-1}\right) \\
& =\frac{23}{\omega^{2}-\omega} \frac{\left(\omega^{2}-\omega\right)-\left(\omega-\omega^{2}\right)}{2-\omega-\omega^{2}} \\
& =\frac{46}{3}
\end{aligned}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-alg-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nAnswer: "
}
|
40b88bb9-0f45-5fb8-959b-9144b9737d43
| 609,078
|
Let $N$ be a positive integer whose decimal representation contains 11235 as a contiguous substring, and let $k$ be a positive integer such that $10^{k}>N$. Find the minimum possible value of
$$
\frac{10^{k}-1}{\operatorname{gcd}\left(N, 10^{k}-1\right)}
$$
|
89 Set $m=\frac{10^{k}-1}{\operatorname{gcd}\left(N, 10^{k}-1\right)}$. Then, in lowest terms, $\frac{N}{10^{k}-1}=\frac{a}{m}$ for some integer $a$. On the other hand, the decimal expansion of $\frac{N}{10^{k}-1}$ simply consists of the decimal expansion of $N$, possibly with some padded zeros, repeating. Since $N$ contains 11235 as a contiguous substring, the decimal representation of $\frac{a}{m}$ must as well.
Conversely, if $m$ is relatively prime to 10 and if there exists an $a$ such that the decimal representation of $\frac{a}{m}$ contains the substring 11235 , we claim that $m$ is an attainable value for $\frac{10^{k}-1}{\operatorname{gcd}\left(N, 10^{k}-1\right)}$. To see this, note that since $m$ is relatively prime to 10 , there exists a value of $k$ such that $m$ divides $10^{k}-1$ (for example, $k=\phi(m)$ ). Letting $m s=10^{k}-1$ and $N=a s$, it follows that $\frac{a}{m}=\frac{a s}{m s}=\frac{N}{10^{k}-1}$. Since the decimal expansion of this fraction contains the substring 11235 , it follows that $N$ must also, and therefore $m$ is an attainable value.
We are therefore looking for a fraction $\frac{a}{m}$ which contains the substring 11235 in its decimal expansion. Since $1,1,2,3$, and 5 are the first five Fibonacci numbers, it makes sense to look at the value of the infinite series
$$
\sum_{i=1}^{\infty} \frac{F_{i}}{10^{i}}
$$
A simple generating function argument shows that $\sum_{i=1}^{\infty} F_{i} x^{i}=\frac{x}{1-x-x^{2}}$, so substituting $x=1 / 10$ leads us to the fraction 10/89 (which indeed begins $0.11235 \ldots$ ).
How do we know no smaller values of $m$ are possible? Well, if $a^{\prime} / m^{\prime}$ contains the substring 11235 somewhere in its infinitely repeating decimal expansion, then note that there is an $i$ such that the decimal expansion of the fractional part of $10^{i}\left(a^{\prime} / m^{\prime}\right)$ begins with $0.11235 \ldots$ We can therefore, without loss of generality, assume that the decimal representation of $a^{\prime} / m^{\prime}$ begins $0.11235 \ldots$ But since the decimal representation of $10 / 89$ begins $0.11235 \ldots$, it follows that
$$
\left|\frac{10}{89}-\frac{a^{\prime}}{m^{\prime}}\right| \leq 10^{-5}
$$
On the other hand, this absolute difference, if non-zero, is at least $\frac{1}{89 m^{\prime}}$. If $m^{\prime}<89$, this is at least $\frac{1}{89^{2}}>10^{-5}$, and therefore no smaller values of $m^{\prime}$ are possible.
|
89
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $N$ be a positive integer whose decimal representation contains 11235 as a contiguous substring, and let $k$ be a positive integer such that $10^{k}>N$. Find the minimum possible value of
$$
\frac{10^{k}-1}{\operatorname{gcd}\left(N, 10^{k}-1\right)}
$$
|
89 Set $m=\frac{10^{k}-1}{\operatorname{gcd}\left(N, 10^{k}-1\right)}$. Then, in lowest terms, $\frac{N}{10^{k}-1}=\frac{a}{m}$ for some integer $a$. On the other hand, the decimal expansion of $\frac{N}{10^{k}-1}$ simply consists of the decimal expansion of $N$, possibly with some padded zeros, repeating. Since $N$ contains 11235 as a contiguous substring, the decimal representation of $\frac{a}{m}$ must as well.
Conversely, if $m$ is relatively prime to 10 and if there exists an $a$ such that the decimal representation of $\frac{a}{m}$ contains the substring 11235 , we claim that $m$ is an attainable value for $\frac{10^{k}-1}{\operatorname{gcd}\left(N, 10^{k}-1\right)}$. To see this, note that since $m$ is relatively prime to 10 , there exists a value of $k$ such that $m$ divides $10^{k}-1$ (for example, $k=\phi(m)$ ). Letting $m s=10^{k}-1$ and $N=a s$, it follows that $\frac{a}{m}=\frac{a s}{m s}=\frac{N}{10^{k}-1}$. Since the decimal expansion of this fraction contains the substring 11235 , it follows that $N$ must also, and therefore $m$ is an attainable value.
We are therefore looking for a fraction $\frac{a}{m}$ which contains the substring 11235 in its decimal expansion. Since $1,1,2,3$, and 5 are the first five Fibonacci numbers, it makes sense to look at the value of the infinite series
$$
\sum_{i=1}^{\infty} \frac{F_{i}}{10^{i}}
$$
A simple generating function argument shows that $\sum_{i=1}^{\infty} F_{i} x^{i}=\frac{x}{1-x-x^{2}}$, so substituting $x=1 / 10$ leads us to the fraction 10/89 (which indeed begins $0.11235 \ldots$ ).
How do we know no smaller values of $m$ are possible? Well, if $a^{\prime} / m^{\prime}$ contains the substring 11235 somewhere in its infinitely repeating decimal expansion, then note that there is an $i$ such that the decimal expansion of the fractional part of $10^{i}\left(a^{\prime} / m^{\prime}\right)$ begins with $0.11235 \ldots$ We can therefore, without loss of generality, assume that the decimal representation of $a^{\prime} / m^{\prime}$ begins $0.11235 \ldots$ But since the decimal representation of $10 / 89$ begins $0.11235 \ldots$, it follows that
$$
\left|\frac{10}{89}-\frac{a^{\prime}}{m^{\prime}}\right| \leq 10^{-5}
$$
On the other hand, this absolute difference, if non-zero, is at least $\frac{1}{89 m^{\prime}}$. If $m^{\prime}<89$, this is at least $\frac{1}{89^{2}}>10^{-5}$, and therefore no smaller values of $m^{\prime}$ are possible.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-alg-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nAnswer: "
}
|
d9a8ed62-9a22-5ed4-8ae2-623da92c5a1b
| 609,079
|
A standard 52 -card deck contains cards of 4 suits and 13 numbers, with exactly one card for each pairing of suit and number. If Maya draws two cards with replacement from this deck, what is the probability that the two cards have the same suit or have the same number, but not both?
|
$\quad \frac{15}{52}$ After drawing the first card, there are 12 other cards from the same suit and 3 other cards with the same number, so the probability is $\frac{12+3}{52}$.
|
\frac{15}{52}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A standard 52 -card deck contains cards of 4 suits and 13 numbers, with exactly one card for each pairing of suit and number. If Maya draws two cards with replacement from this deck, what is the probability that the two cards have the same suit or have the same number, but not both?
|
$\quad \frac{15}{52}$ After drawing the first card, there are 12 other cards from the same suit and 3 other cards with the same number, so the probability is $\frac{12+3}{52}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-comb-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nAnswer: "
}
|
6d1883db-faa9-570c-a41b-001e36c511e3
| 609,080
|
On a game show, Merble will be presented with a series of 2013 marbles, each of which is either red or blue on the outside. Each time he sees a marble, he can either keep it or pass, but cannot return to a previous marble; he receives 3 points for keeping a red marble, loses 2 points for keeping a blue marble, and gains 0 points for passing. All distributions of colors are equally likely and Merble can only see the color of his current marble. If his goal is to end with exactly one point and he plays optimally, what is the probability that he fails?
|
$\frac{1}{2^{2012}}$ First, we note that if all the marbles are red or all are blue, then it is impossible for Merble to win; we claim that he can guarantee himself a win in every other case. In particular, his strategy should be to keep the first red and first blue marble that he encounters, and to ignore all the others. Consequently, the probability that he cannot win is $\frac{2}{2^{2013}}=\frac{1}{2^{2012}}$.
|
\frac{1}{2^{2012}}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
On a game show, Merble will be presented with a series of 2013 marbles, each of which is either red or blue on the outside. Each time he sees a marble, he can either keep it or pass, but cannot return to a previous marble; he receives 3 points for keeping a red marble, loses 2 points for keeping a blue marble, and gains 0 points for passing. All distributions of colors are equally likely and Merble can only see the color of his current marble. If his goal is to end with exactly one point and he plays optimally, what is the probability that he fails?
|
$\frac{1}{2^{2012}}$ First, we note that if all the marbles are red or all are blue, then it is impossible for Merble to win; we claim that he can guarantee himself a win in every other case. In particular, his strategy should be to keep the first red and first blue marble that he encounters, and to ignore all the others. Consequently, the probability that he cannot win is $\frac{2}{2^{2013}}=\frac{1}{2^{2012}}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-comb-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nAnswer: "
}
|
73bb02b1-912f-5c4a-93da-74c6dc8ab894
| 609,082
|
How many orderings $\left(a_{1}, \ldots, a_{8}\right)$ of $(1,2, \ldots, 8)$ exist such that $a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+a_{7}-a_{8}=0$ ?
|
4608 We can divide the numbers up based on whether they have a + or - before them. Both the numbers following +'s and -'s must add up to 18 . Without loss of generality, we can assume the +'s contain the number 1 (and add a factor of 2 at the end to account for this). The possible 4 -element sets containing a 1 which add to 18 are $\{1,2,7,8\},\{1,3,6,8\},\{1,4,5,8\},\{1,4,6,7\}$. Additionally, there are 4 ! ways to order the numbers following a + and 4 ! ways to order the numbers following a - . Thus the total number of possibilities is $4 \times 2 \times 4!\times 4!=4608$.
|
4608
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many orderings $\left(a_{1}, \ldots, a_{8}\right)$ of $(1,2, \ldots, 8)$ exist such that $a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+a_{7}-a_{8}=0$ ?
|
4608 We can divide the numbers up based on whether they have a + or - before them. Both the numbers following +'s and -'s must add up to 18 . Without loss of generality, we can assume the +'s contain the number 1 (and add a factor of 2 at the end to account for this). The possible 4 -element sets containing a 1 which add to 18 are $\{1,2,7,8\},\{1,3,6,8\},\{1,4,5,8\},\{1,4,6,7\}$. Additionally, there are 4 ! ways to order the numbers following a + and 4 ! ways to order the numbers following a - . Thus the total number of possibilities is $4 \times 2 \times 4!\times 4!=4608$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-comb-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nAnswer: "
}
|
9cf2e7ab-97a3-555b-8fde-9891f487a7a3
| 609,083
|
At a certain chocolate company, each bar is 1 unit long. To make the bars more interesting, the company has decided to combine dark and white chocolate pieces. The process starts with two bars, one completely dark and one completely white. At each step of the process, a new number $p$ is chosen uniformly at random between 0 and 1 . Each of the two bars is cut $p$ units from the left, and the pieces on the left are switched: each is grafted onto the opposite bar where the other piece of length $p$ was previously attached. For example, the bars might look like this after the first step:
Each step after the first operates on the bars resulting from the previous step. After a total of 100 steps, what is the probability that on each bar, the chocolate $1 / 3$ units from the left is the same type of chocolate as that $2 / 3$ units from the left?
|
$\frac{1}{2}\left[\left(\frac{1}{3}\right)^{100}+1\right]$ If the values of $p$ chosen are $p_{1}, \ldots, p_{100}$, then note that the color of a bar changes at each value of $p_{i}$. Consequently, we want to find the probability that exactly an even number of $p_{i}$ are in $\left(\frac{1}{3}, \frac{2}{3}\right)$. Summing, this is equal to
$$
\binom{100}{0}\left(\frac{1}{3}\right)^{0}\left(\frac{2}{3}\right)^{100}+\binom{100}{2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{98}+\ldots\binom{100}{100}\left(\frac{1}{3}\right)^{100}\left(\frac{2}{3}\right)^{0}
$$
To compute, we note that this is equal to
$$
\frac{1}{2}\left[\left(\frac{2}{3}+\frac{1}{3}\right)^{100}+\left(\frac{2}{3}-\frac{1}{3}\right)^{100}\right]
$$
after expanding using the binomial theorem, since any terms with odd exponents are cancelled out between the two terms.
|
\frac{1}{2}\left[\left(\frac{1}{3}\right)^{100}+1\right]
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
At a certain chocolate company, each bar is 1 unit long. To make the bars more interesting, the company has decided to combine dark and white chocolate pieces. The process starts with two bars, one completely dark and one completely white. At each step of the process, a new number $p$ is chosen uniformly at random between 0 and 1 . Each of the two bars is cut $p$ units from the left, and the pieces on the left are switched: each is grafted onto the opposite bar where the other piece of length $p$ was previously attached. For example, the bars might look like this after the first step:
Each step after the first operates on the bars resulting from the previous step. After a total of 100 steps, what is the probability that on each bar, the chocolate $1 / 3$ units from the left is the same type of chocolate as that $2 / 3$ units from the left?
|
$\frac{1}{2}\left[\left(\frac{1}{3}\right)^{100}+1\right]$ If the values of $p$ chosen are $p_{1}, \ldots, p_{100}$, then note that the color of a bar changes at each value of $p_{i}$. Consequently, we want to find the probability that exactly an even number of $p_{i}$ are in $\left(\frac{1}{3}, \frac{2}{3}\right)$. Summing, this is equal to
$$
\binom{100}{0}\left(\frac{1}{3}\right)^{0}\left(\frac{2}{3}\right)^{100}+\binom{100}{2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{98}+\ldots\binom{100}{100}\left(\frac{1}{3}\right)^{100}\left(\frac{2}{3}\right)^{0}
$$
To compute, we note that this is equal to
$$
\frac{1}{2}\left[\left(\frac{2}{3}+\frac{1}{3}\right)^{100}+\left(\frac{2}{3}-\frac{1}{3}\right)^{100}\right]
$$
after expanding using the binomial theorem, since any terms with odd exponents are cancelled out between the two terms.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-comb-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nAnswer: "
}
|
ceab6a66-8f8c-56fd-a390-3801e78b93f8
| 609,084
|
Values $a_{1}, \ldots, a_{2013}$ are chosen independently and at random from the set $\{1, \ldots, 2013\}$. What is expected number of distinct values in the set $\left\{a_{1}, \ldots, a_{2013}\right\}$ ?
|
$\frac{2013^{2013}-2012^{2013}}{2013^{2012}}$ For each $n \in\{1,2, \ldots, 2013\}$, let $X_{n}=1$ if $n$ appears in $\left\{a_{1}, a_{2}, \ldots, a_{2013}\right\}$ and 0 otherwise. Defined this way, $\mathrm{E}\left[X_{n}\right]$ is the probability that $n$ appears in $\left\{a_{1}, a_{2}, \ldots, a_{2013}\right\}$. Since each $a_{i}(1 \leq i \leq 2013)$ is not $n$ with probability 2012/2013, the probability that $n$ is none of the $a_{i}$ 's is $\left(\frac{2012}{2013}\right)^{2013}$, so $\mathrm{E}\left[X_{n}\right]$, the probability that $n$ is one of the $a_{i}$ 's, is $1-\left(\frac{2012}{2013}\right)^{2013}$. The expected number of distinct values in $\left\{a_{1}, \ldots, a_{2013}\right\}$ is the expected number of $n \in\{1,2, \ldots, 2013\}$ such that $X_{n}=1$, that is, the expected value of $X_{1}+X_{2}+\cdots+X_{2013}$. By linearity of expectation, $\mathrm{E}\left[X_{1}+X_{2}+\cdots+X_{2013}\right]=\mathrm{E}\left[X_{1}\right]+\mathrm{E}\left[X_{2}\right]+\cdots+\mathrm{E}\left[X_{n}\right]=2013\left(1-\left(\frac{2012}{2013}\right)^{2013}\right)=\frac{2013^{2013}-2012^{2013}}{2013^{2012}}$.
|
\frac{2013^{2013}-2012^{2013}}{2013^{2012}}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Values $a_{1}, \ldots, a_{2013}$ are chosen independently and at random from the set $\{1, \ldots, 2013\}$. What is expected number of distinct values in the set $\left\{a_{1}, \ldots, a_{2013}\right\}$ ?
|
$\frac{2013^{2013}-2012^{2013}}{2013^{2012}}$ For each $n \in\{1,2, \ldots, 2013\}$, let $X_{n}=1$ if $n$ appears in $\left\{a_{1}, a_{2}, \ldots, a_{2013}\right\}$ and 0 otherwise. Defined this way, $\mathrm{E}\left[X_{n}\right]$ is the probability that $n$ appears in $\left\{a_{1}, a_{2}, \ldots, a_{2013}\right\}$. Since each $a_{i}(1 \leq i \leq 2013)$ is not $n$ with probability 2012/2013, the probability that $n$ is none of the $a_{i}$ 's is $\left(\frac{2012}{2013}\right)^{2013}$, so $\mathrm{E}\left[X_{n}\right]$, the probability that $n$ is one of the $a_{i}$ 's, is $1-\left(\frac{2012}{2013}\right)^{2013}$. The expected number of distinct values in $\left\{a_{1}, \ldots, a_{2013}\right\}$ is the expected number of $n \in\{1,2, \ldots, 2013\}$ such that $X_{n}=1$, that is, the expected value of $X_{1}+X_{2}+\cdots+X_{2013}$. By linearity of expectation, $\mathrm{E}\left[X_{1}+X_{2}+\cdots+X_{2013}\right]=\mathrm{E}\left[X_{1}\right]+\mathrm{E}\left[X_{2}\right]+\cdots+\mathrm{E}\left[X_{n}\right]=2013\left(1-\left(\frac{2012}{2013}\right)^{2013}\right)=\frac{2013^{2013}-2012^{2013}}{2013^{2012}}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-comb-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nAnswer: "
}
|
13d399af-cc7a-52ec-ba44-d5a7196ecb46
| 609,085
|
A single-elimination ping-pong tournament has $2^{2013}$ players, seeded in order of ability. If the player with seed $x$ plays the player with seed $y$, then it is possible for $x$ to win if and only if $x \leq y+3$. For how many players $P$ it is possible for $P$ to win? (In each round of a single elimination tournament, the remaining players are randomly paired up; each player plays against the other player in his pair, with the winner from each pair progressing to the next round and the loser eliminated. This is repeated until there is only one player remaining.)
|
6038 We calculate the highest seed $n$ that can win. Below, we say that a player $x$ vicariously defeats a player $y$ if $x$ defeats $y$ directly or indirectly through some chain (i.e. $x$ defeats $x_{1}$, who defeated $x_{2}, \ldots$, who defeated $x_{n}$, who defeated $y$ for some players $x_{1}, \ldots, x_{n}$ ).
We first consider the highest seeds that are capable of making the semifinals. The eventual winner must be able to beat two of these players and thus must be able to beat the second best player in the semifinals. The seed of the player who vicariously beats the 1 -seed is maximized if 1 loses to 4 in the first round, 4 to 7 in the second round, etc. Therefore $3 \cdot 2011+1=6034$ is the maximum value of the highest seed in the semifinals. If 1 , and 2 are in different quarters of the draw, then by a similar argument 6035 is the largest possible value of the second best player in the semis, and thus 6038 is the highest that can win. If 1 and 2 are in the same quarter, then in one round the highest remaining seed will not be able to go up by 3 , when the player who has vicariously beaten 1 plays the player who vicariously beat 2 , so $3 \cdot 2011-1=6032$ is the highest player the semifinalist from that quarter could be. But then the eventual winner still must be seeded at most 6 above this player, and thus 6038 is still the upper bound.
Therefore 6038 is the worst seed that could possibly win, and can do so if $6034,6035,6036,6038$ all make the semis, which is possible (it is not difficult to construct such a tournament). Then, note that any player $x$ with a lower seed can also win for some tournament - in particular, it suffices to take
the tournament where it is possible for player 6038 to win and switch the positions of 6038 and $x$. Consequently, there are 6038 players for whom it is possible to win under some tournament.
|
6038
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A single-elimination ping-pong tournament has $2^{2013}$ players, seeded in order of ability. If the player with seed $x$ plays the player with seed $y$, then it is possible for $x$ to win if and only if $x \leq y+3$. For how many players $P$ it is possible for $P$ to win? (In each round of a single elimination tournament, the remaining players are randomly paired up; each player plays against the other player in his pair, with the winner from each pair progressing to the next round and the loser eliminated. This is repeated until there is only one player remaining.)
|
6038 We calculate the highest seed $n$ that can win. Below, we say that a player $x$ vicariously defeats a player $y$ if $x$ defeats $y$ directly or indirectly through some chain (i.e. $x$ defeats $x_{1}$, who defeated $x_{2}, \ldots$, who defeated $x_{n}$, who defeated $y$ for some players $x_{1}, \ldots, x_{n}$ ).
We first consider the highest seeds that are capable of making the semifinals. The eventual winner must be able to beat two of these players and thus must be able to beat the second best player in the semifinals. The seed of the player who vicariously beats the 1 -seed is maximized if 1 loses to 4 in the first round, 4 to 7 in the second round, etc. Therefore $3 \cdot 2011+1=6034$ is the maximum value of the highest seed in the semifinals. If 1 , and 2 are in different quarters of the draw, then by a similar argument 6035 is the largest possible value of the second best player in the semis, and thus 6038 is the highest that can win. If 1 and 2 are in the same quarter, then in one round the highest remaining seed will not be able to go up by 3 , when the player who has vicariously beaten 1 plays the player who vicariously beat 2 , so $3 \cdot 2011-1=6032$ is the highest player the semifinalist from that quarter could be. But then the eventual winner still must be seeded at most 6 above this player, and thus 6038 is still the upper bound.
Therefore 6038 is the worst seed that could possibly win, and can do so if $6034,6035,6036,6038$ all make the semis, which is possible (it is not difficult to construct such a tournament). Then, note that any player $x$ with a lower seed can also win for some tournament - in particular, it suffices to take
the tournament where it is possible for player 6038 to win and switch the positions of 6038 and $x$. Consequently, there are 6038 players for whom it is possible to win under some tournament.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-comb-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nAnswer: "
}
|
620daf0b-f027-50a6-aa9c-0a58eb375d40
| 609,086
|
It is known that exactly one of the three (distinguishable) musketeers stole the truffles. Each musketeer makes one statement, in which he either claims that one of the three is guilty, or claims that one of the three is innocent. It is possible for two or more of the musketeers to make the same statement. After hearing their claims, and knowing that exactly one musketeer lied, the inspector is able to deduce who stole the truffles. How many ordered triplets of statements could have been made?
|
99 We divide into cases, based on the number of distinct people that statements are made about.
- The statements are made about 3 distinct people. Then, since exactly one person is guilty, and because exactly one of the three lied, there are either zero statements of guilt or two statements of guilt possible; in either case, it is impossible by symmetry to determine who is guilty or innocent.
- The statements are made about 2 distinct people or 1 distinct person. Then, either at least two of the statements are the same, or all are different.
- If two statements are the same, then those two statements must be true because only one musketeer lied. Consequently, the lone statement must be false. If all the statements are about the same person, there there must be 2 guilty claims and 1 innocence claim (otherwise we would not know which of the other two people was guilty). Then, there are 3 choices for who the statement is about and 3 choices for who makes the innocence claim, for a $3 \cdot 3=9$ triplets of statements. Meanwhile, if the statements are about two different people, then this is doable unless both of the distinct statements imply guilt for the person concerned (i.e. where there are two guilty accusations against one person and one claim of innocence against another). Consequently, there are 3 sets of statements that can be made, $3 \cdot 2=6$ ways to determine who they are made about, and 3 ways to determine who makes which statement, for a total of $3 \cdot 6 \cdot 3=54$ triplets in this case.
- If all the statements are different, then they must be about two different people. Here, there must be one person, who we will call A, who has both a claim of innocence and an accusation of guilt against him. The last statement must concern another person, B. If the statement accuses B of being guilty, then we can deduce that he is the guilty one. On the other hand, if the statement claims that B is innocent, either of the other two musketeers could be guilty. Consequently, there are $3 \cdot 2=6$ ways to choose A and B , and $3!=6$ ways to choose who makes which statement, for a total of $6 \cdot 6=36$ triplets of statements.
In total, we have $9+54+36=99$ possible triplets of statements.
|
99
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
It is known that exactly one of the three (distinguishable) musketeers stole the truffles. Each musketeer makes one statement, in which he either claims that one of the three is guilty, or claims that one of the three is innocent. It is possible for two or more of the musketeers to make the same statement. After hearing their claims, and knowing that exactly one musketeer lied, the inspector is able to deduce who stole the truffles. How many ordered triplets of statements could have been made?
|
99 We divide into cases, based on the number of distinct people that statements are made about.
- The statements are made about 3 distinct people. Then, since exactly one person is guilty, and because exactly one of the three lied, there are either zero statements of guilt or two statements of guilt possible; in either case, it is impossible by symmetry to determine who is guilty or innocent.
- The statements are made about 2 distinct people or 1 distinct person. Then, either at least two of the statements are the same, or all are different.
- If two statements are the same, then those two statements must be true because only one musketeer lied. Consequently, the lone statement must be false. If all the statements are about the same person, there there must be 2 guilty claims and 1 innocence claim (otherwise we would not know which of the other two people was guilty). Then, there are 3 choices for who the statement is about and 3 choices for who makes the innocence claim, for a $3 \cdot 3=9$ triplets of statements. Meanwhile, if the statements are about two different people, then this is doable unless both of the distinct statements imply guilt for the person concerned (i.e. where there are two guilty accusations against one person and one claim of innocence against another). Consequently, there are 3 sets of statements that can be made, $3 \cdot 2=6$ ways to determine who they are made about, and 3 ways to determine who makes which statement, for a total of $3 \cdot 6 \cdot 3=54$ triplets in this case.
- If all the statements are different, then they must be about two different people. Here, there must be one person, who we will call A, who has both a claim of innocence and an accusation of guilt against him. The last statement must concern another person, B. If the statement accuses B of being guilty, then we can deduce that he is the guilty one. On the other hand, if the statement claims that B is innocent, either of the other two musketeers could be guilty. Consequently, there are $3 \cdot 2=6$ ways to choose A and B , and $3!=6$ ways to choose who makes which statement, for a total of $6 \cdot 6=36$ triplets of statements.
In total, we have $9+54+36=99$ possible triplets of statements.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-comb-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nAnswer: "
}
|
25068e4f-a69e-54af-8198-fcef4ce161ce
| 609,087
|
Given a permutation $\sigma$ of $\{1,2, \ldots, 2013\}$, let $f(\sigma)$ to be the number of fixed points of $\sigma$ - that is, the number of $k \in\{1,2, \ldots, 2013\}$ such that $\sigma(k)=k$. If $S$ is the set of all possible permutations $\sigma$, compute
$$
\sum_{\sigma \in S} f(\sigma)^{4}
$$
(Here, a permutation $\sigma$ is a bijective mapping from $\{1,2, \ldots, 2013\}$ to $\{1,2, \ldots, 2013\}$.)
|
$15(2013!)$ First, note that
$$
\sum_{\sigma \in S} f(\sigma)^{4}=\sum_{\sigma \in S} \sum_{1 \leq a_{1}, a_{2}, a_{3}, a_{4} \leq 2013} g\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)
$$
where $g\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)=1$ if all $a_{i}$ are fixed points of $\sigma$ and 0 otherwise. (The $a_{i}$ 's need not be distinct.) Switching the order of summation, we find that the desired sum is
$$
\sum_{1 \leq a_{1}, a_{2}, a_{3}, a_{4} \leq 2013} \sum_{\sigma \in S} g\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)
$$
Note that the inner sum is equal to the number of permutations on $\{1,2, \ldots, 2013\}$ that fix $a_{1}, a_{2}, a_{3}$, and $a_{4}$. This depends on the number of distinct values the $a_{i}$ s take. If they take on exactly $k$ distinct values, then the inner sum will evaluate to $(2013-k)$ !, because $\sigma$ can be any permutation of the remaining $2013-k$ elements. (For example, if $a_{1}=a_{2}$ but $a_{1}, a_{3}$, and $a_{4}$ are distinct, the inner sum is 2010 ! because $\sigma$ can be any permutation that fixes $a_{1}, a_{3}$, and $a_{4}$.)
Now, suppose we are given which of the $a_{i}$ are equal (for example, we could be given $a_{1}=a_{2}$ but $a_{1}, a_{3}, a_{4}$ mutually distinct, as per the above example). Assuming there are $k$ distinct values among the $a_{i}$, there are $2013(2013-1) \cdots(2013-k+1)$ ways to choose the $a_{i}$. At this point, there are $(2013-k)$ ! ways to choose $\sigma$ on the remaining $(2013-k)$ values such that it fixes the $a_{i}$, for a total of 2013 ! choices for $\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)$ such that $g\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)=1$ and the $a_{i}$ satisfy the correct equality relations.
Thus the answer is 2013 ! times the number of ways to choose equivalence classes on the $a_{i}$, so the problem reduces to finding the number of ways to partition 4 elements into nonempty sets. This process can be accelerated by doing casework based on the number of sets:
(a) One set must contain all four elements, only one possibility. (i.e. all the $a_{i}$ s are equal)
(b) Either one set contains 3 elements and the other contains the fourth ( 4 possibilities) or one set contains 2 elements and the other contains the other two ( 3 possibilities). (i.e. there are two distinct values of $a_{i}$ )
(c) One set contains two elements, the other two each contain one. There are $\binom{4}{2}=6$ ways to choose the two elements in the set with two elements, and this uniquely determines the partition. (i.e. there are three distinct values of $a_{i}$ )
(d) All sets contain one element, in which case there is only one possibility. (i.e. all the $a_{i}$ are distinct)
Thus the number of ways to construct such a partition is $1+4+3+6+1=15$, and our answer is $15 \cdot 2013$ !.
|
15(2013!)
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Given a permutation $\sigma$ of $\{1,2, \ldots, 2013\}$, let $f(\sigma)$ to be the number of fixed points of $\sigma$ - that is, the number of $k \in\{1,2, \ldots, 2013\}$ such that $\sigma(k)=k$. If $S$ is the set of all possible permutations $\sigma$, compute
$$
\sum_{\sigma \in S} f(\sigma)^{4}
$$
(Here, a permutation $\sigma$ is a bijective mapping from $\{1,2, \ldots, 2013\}$ to $\{1,2, \ldots, 2013\}$.)
|
$15(2013!)$ First, note that
$$
\sum_{\sigma \in S} f(\sigma)^{4}=\sum_{\sigma \in S} \sum_{1 \leq a_{1}, a_{2}, a_{3}, a_{4} \leq 2013} g\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)
$$
where $g\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)=1$ if all $a_{i}$ are fixed points of $\sigma$ and 0 otherwise. (The $a_{i}$ 's need not be distinct.) Switching the order of summation, we find that the desired sum is
$$
\sum_{1 \leq a_{1}, a_{2}, a_{3}, a_{4} \leq 2013} \sum_{\sigma \in S} g\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)
$$
Note that the inner sum is equal to the number of permutations on $\{1,2, \ldots, 2013\}$ that fix $a_{1}, a_{2}, a_{3}$, and $a_{4}$. This depends on the number of distinct values the $a_{i}$ s take. If they take on exactly $k$ distinct values, then the inner sum will evaluate to $(2013-k)$ !, because $\sigma$ can be any permutation of the remaining $2013-k$ elements. (For example, if $a_{1}=a_{2}$ but $a_{1}, a_{3}$, and $a_{4}$ are distinct, the inner sum is 2010 ! because $\sigma$ can be any permutation that fixes $a_{1}, a_{3}$, and $a_{4}$.)
Now, suppose we are given which of the $a_{i}$ are equal (for example, we could be given $a_{1}=a_{2}$ but $a_{1}, a_{3}, a_{4}$ mutually distinct, as per the above example). Assuming there are $k$ distinct values among the $a_{i}$, there are $2013(2013-1) \cdots(2013-k+1)$ ways to choose the $a_{i}$. At this point, there are $(2013-k)$ ! ways to choose $\sigma$ on the remaining $(2013-k)$ values such that it fixes the $a_{i}$, for a total of 2013 ! choices for $\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)$ such that $g\left(\sigma, a_{1}, a_{2}, a_{3}, a_{4}\right)=1$ and the $a_{i}$ satisfy the correct equality relations.
Thus the answer is 2013 ! times the number of ways to choose equivalence classes on the $a_{i}$, so the problem reduces to finding the number of ways to partition 4 elements into nonempty sets. This process can be accelerated by doing casework based on the number of sets:
(a) One set must contain all four elements, only one possibility. (i.e. all the $a_{i}$ s are equal)
(b) Either one set contains 3 elements and the other contains the fourth ( 4 possibilities) or one set contains 2 elements and the other contains the other two ( 3 possibilities). (i.e. there are two distinct values of $a_{i}$ )
(c) One set contains two elements, the other two each contain one. There are $\binom{4}{2}=6$ ways to choose the two elements in the set with two elements, and this uniquely determines the partition. (i.e. there are three distinct values of $a_{i}$ )
(d) All sets contain one element, in which case there is only one possibility. (i.e. all the $a_{i}$ are distinct)
Thus the number of ways to construct such a partition is $1+4+3+6+1=15$, and our answer is $15 \cdot 2013$ !.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-comb-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nAnswer: "
}
|
bd688a61-a1a3-5646-b636-8bdce746e364
| 609,088
|
Jarris the triangle is playing in the $(x, y)$ plane. Let his maximum $y$ coordinate be $k$. Given that he has side lengths 6,8 , and 10 and that no part of him is below the $x$-axis, find the minimum possible value of $k$.
|
$\quad \frac{24}{5}$ By playing around, we find that Jarris should have his hypotenuse flat on the $x$ axis. The desired minimum value of $k$ is then the length of the altitude to the hypotenuse. Thus, by computing the area of the triangle in two ways, $\frac{1}{2} \cdot 10 \cdot k=\frac{1}{2} \cdot 6 \cdot 8$ and so $k=\frac{24}{5}$.
|
\frac{24}{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Jarris the triangle is playing in the $(x, y)$ plane. Let his maximum $y$ coordinate be $k$. Given that he has side lengths 6,8 , and 10 and that no part of him is below the $x$-axis, find the minimum possible value of $k$.
|
$\quad \frac{24}{5}$ By playing around, we find that Jarris should have his hypotenuse flat on the $x$ axis. The desired minimum value of $k$ is then the length of the altitude to the hypotenuse. Thus, by computing the area of the triangle in two ways, $\frac{1}{2} \cdot 10 \cdot k=\frac{1}{2} \cdot 6 \cdot 8$ and so $k=\frac{24}{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-geo-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nAnswer: "
}
|
60b37fae-48b3-5132-a1a6-d24e013b8353
| 609,090
|
Let $A B C D$ be an isosceles trapezoid such that $A D=B C, A B=3$, and $C D=8$. Let $E$ be a point in the plane such that $B C=E C$ and $A E \perp E C$. Compute $A E$.
|
$2 \sqrt{6}$ Let $r=B C=E C=A D . \triangle A C E$ has right angle at $E$, so by the Pythagorean Theorem,
$$
A E^{2}=A C^{2}-C E^{2}=A C^{2}-r^{2}
$$
Let the height of $\triangle A C D$ at $A$ intersect $D C$ at $F$. Once again, by the Pythagorean Theorem,
$$
A C^{2}=F C^{2}+A F^{2}=\left(\frac{8-3}{2}+3\right)^{2}+A D^{2}-D F^{2}=\left(\frac{11}{2}\right)^{2}+r^{2}-\left(\frac{5}{2}\right)^{2}
$$
Plugging into the first equation,
$$
A E^{2}=\left(\frac{11}{2}\right)^{2}+r^{2}-\left(\frac{5}{2}\right)^{2}-r^{2}
$$
so $A E=2 \sqrt{6}$.
|
2 \sqrt{6}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C D$ be an isosceles trapezoid such that $A D=B C, A B=3$, and $C D=8$. Let $E$ be a point in the plane such that $B C=E C$ and $A E \perp E C$. Compute $A E$.
|
$2 \sqrt{6}$ Let $r=B C=E C=A D . \triangle A C E$ has right angle at $E$, so by the Pythagorean Theorem,
$$
A E^{2}=A C^{2}-C E^{2}=A C^{2}-r^{2}
$$
Let the height of $\triangle A C D$ at $A$ intersect $D C$ at $F$. Once again, by the Pythagorean Theorem,
$$
A C^{2}=F C^{2}+A F^{2}=\left(\frac{8-3}{2}+3\right)^{2}+A D^{2}-D F^{2}=\left(\frac{11}{2}\right)^{2}+r^{2}-\left(\frac{5}{2}\right)^{2}
$$
Plugging into the first equation,
$$
A E^{2}=\left(\frac{11}{2}\right)^{2}+r^{2}-\left(\frac{5}{2}\right)^{2}-r^{2}
$$
so $A E=2 \sqrt{6}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-geo-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nAnswer: "
}
|
1eb5cdac-e884-51d4-b422-82f15e996bee
| 609,091
|
Let $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ be a convex hexagon such that $A_{i} A_{i+2} \| A_{i+3} A_{i+5}$ for $i=1,2,3$ (we take $A_{i+6}=A_{i}$ for each $i$. Segment $A_{i} A_{i+2}$ intersects segment $A_{i+1} A_{i+3}$ at $B_{i}$, for $1 \leq i \leq 6$, as shown. Furthermore, suppose that $\triangle A_{1} A_{3} A_{5} \cong \triangle A_{4} A_{6} A_{2}$. Given that $\left[A_{1} B_{5} B_{6}\right]=1,\left[A_{2} B_{6} B_{1}\right]=$ 4, and $\left[A_{3} B_{1} B_{2}\right]=9$ (by $[X Y Z]$ we mean the area of $\triangle X Y Z$ ), determine the area of hexagon $B_{1} B_{2} B_{3} B_{4} B_{5} B_{6}$.
|
22 Because $B_{6} A_{3} B_{3} A_{6}$ and $B_{1} A_{4} B_{4} A_{1}$ are parallelograms, $B_{6} A_{3}=A_{6} B_{3}$ and $A_{1} B_{1}=$ $A_{4} B_{4}$. By the congruence of the large triangles $A_{1} A_{3} A_{5}$ and $A_{2} A_{4} A_{6}, A_{1} A_{3}=A_{4} A_{6}$. Thus, $B_{6} A_{3}+$ $A_{1} B_{1}-A_{1} A_{3}=A_{6} B_{3}+A_{4} B_{4}-A_{4} A_{6}$, so $B_{6} B_{1}=B_{3} B_{4}$. Similarly, opposite sides of hexagon $B_{1} B_{2} B_{3} B_{4} B_{5} B_{6}$ are equal, and implying that the triangles opposite each other on the outside of this hexagon are congruent.
Furthermore, by definition $B_{5} B_{6}\left\|A_{3} A_{5}, B_{3} B_{4}\right\| A_{1} A_{3}, B_{6} B_{1} \| A_{4} A_{6}$ and $B_{1} B_{2} \| A_{1} A_{5}$. Let the area of triangle $A_{1} A_{3} A_{5}$ and triangle $A_{2} A_{4} A_{6}$ be $k^{2}$. Then, by similar triangles,
$$
\begin{aligned}
& \sqrt{\frac{1}{k^{2}}}=\frac{A_{1} B_{6}}{A_{1} A_{3}} \\
& \sqrt{\frac{4}{k^{2}}}=\frac{B_{6} B_{1}}{A_{4} A_{6}}=\frac{B_{1} B_{6}}{A_{1} A_{3}} \\
& \sqrt{\frac{9}{k^{2}}}=\frac{A_{3} B_{1}}{A_{1} A_{3}}
\end{aligned}
$$
Summing yields $6 / k=1$, so $k^{2}=36$. To finish, the area of $B_{1} B_{2} B_{3} B_{4} B_{5} B_{6}$ is equivalent to the area of the triangle $A_{1} A_{3} A_{5}$ minus the areas of the smaller triangles provided in the hypothesis. Thus, our answer is $36-1-4-9=22$.
|
22
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ be a convex hexagon such that $A_{i} A_{i+2} \| A_{i+3} A_{i+5}$ for $i=1,2,3$ (we take $A_{i+6}=A_{i}$ for each $i$. Segment $A_{i} A_{i+2}$ intersects segment $A_{i+1} A_{i+3}$ at $B_{i}$, for $1 \leq i \leq 6$, as shown. Furthermore, suppose that $\triangle A_{1} A_{3} A_{5} \cong \triangle A_{4} A_{6} A_{2}$. Given that $\left[A_{1} B_{5} B_{6}\right]=1,\left[A_{2} B_{6} B_{1}\right]=$ 4, and $\left[A_{3} B_{1} B_{2}\right]=9$ (by $[X Y Z]$ we mean the area of $\triangle X Y Z$ ), determine the area of hexagon $B_{1} B_{2} B_{3} B_{4} B_{5} B_{6}$.
|
22 Because $B_{6} A_{3} B_{3} A_{6}$ and $B_{1} A_{4} B_{4} A_{1}$ are parallelograms, $B_{6} A_{3}=A_{6} B_{3}$ and $A_{1} B_{1}=$ $A_{4} B_{4}$. By the congruence of the large triangles $A_{1} A_{3} A_{5}$ and $A_{2} A_{4} A_{6}, A_{1} A_{3}=A_{4} A_{6}$. Thus, $B_{6} A_{3}+$ $A_{1} B_{1}-A_{1} A_{3}=A_{6} B_{3}+A_{4} B_{4}-A_{4} A_{6}$, so $B_{6} B_{1}=B_{3} B_{4}$. Similarly, opposite sides of hexagon $B_{1} B_{2} B_{3} B_{4} B_{5} B_{6}$ are equal, and implying that the triangles opposite each other on the outside of this hexagon are congruent.
Furthermore, by definition $B_{5} B_{6}\left\|A_{3} A_{5}, B_{3} B_{4}\right\| A_{1} A_{3}, B_{6} B_{1} \| A_{4} A_{6}$ and $B_{1} B_{2} \| A_{1} A_{5}$. Let the area of triangle $A_{1} A_{3} A_{5}$ and triangle $A_{2} A_{4} A_{6}$ be $k^{2}$. Then, by similar triangles,
$$
\begin{aligned}
& \sqrt{\frac{1}{k^{2}}}=\frac{A_{1} B_{6}}{A_{1} A_{3}} \\
& \sqrt{\frac{4}{k^{2}}}=\frac{B_{6} B_{1}}{A_{4} A_{6}}=\frac{B_{1} B_{6}}{A_{1} A_{3}} \\
& \sqrt{\frac{9}{k^{2}}}=\frac{A_{3} B_{1}}{A_{1} A_{3}}
\end{aligned}
$$
Summing yields $6 / k=1$, so $k^{2}=36$. To finish, the area of $B_{1} B_{2} B_{3} B_{4} B_{5} B_{6}$ is equivalent to the area of the triangle $A_{1} A_{3} A_{5}$ minus the areas of the smaller triangles provided in the hypothesis. Thus, our answer is $36-1-4-9=22$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-geo-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nAnswer: "
}
|
c8ccd192-0997-5a5c-83ee-59739b376340
| 609,092
|
In triangle $A B C, \angle A=45^{\circ}$ and $M$ is the midpoint of $\overline{B C}$. $\overline{A M}$ intersects the circumcircle of $A B C$ for the second time at $D$, and $A M=2 M D$. Find $\cos \angle A O D$, where $O$ is the circumcenter of $A B C$.
|
$-\frac{1}{8} \angle B A C=45^{\circ}$, so $\angle B O C=90^{\circ}$. If the radius of the circumcircle is $r, B C=\sqrt{2} r$, and $B M=C M=\frac{\sqrt{2}}{2} r$. By power of a point, $B M \cdot C M=A M \cdot D M$, so $A M=r$ and $D M=\frac{1}{2} r$, and $A D=\frac{3}{2} r$. Using the law of cosines on triangle $A O D$ gives $\cos \angle A O D=-\frac{1}{8}$.
|
-\frac{1}{8}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
In triangle $A B C, \angle A=45^{\circ}$ and $M$ is the midpoint of $\overline{B C}$. $\overline{A M}$ intersects the circumcircle of $A B C$ for the second time at $D$, and $A M=2 M D$. Find $\cos \angle A O D$, where $O$ is the circumcenter of $A B C$.
|
$-\frac{1}{8} \angle B A C=45^{\circ}$, so $\angle B O C=90^{\circ}$. If the radius of the circumcircle is $r, B C=\sqrt{2} r$, and $B M=C M=\frac{\sqrt{2}}{2} r$. By power of a point, $B M \cdot C M=A M \cdot D M$, so $A M=r$ and $D M=\frac{1}{2} r$, and $A D=\frac{3}{2} r$. Using the law of cosines on triangle $A O D$ gives $\cos \angle A O D=-\frac{1}{8}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-geo-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nAnswer: "
}
|
f6a91e31-edfd-54a2-bfee-03f800b20c68
| 609,094
|
Let $A B C$ be an obtuse triangle with circumcenter $O$ such that $\angle A B C=15^{\circ}$ and $\angle B A C>90^{\circ}$. Suppose that $A O$ meets $B C$ at $D$, and that $O D^{2}+O C \cdot D C=O C^{2}$. Find $\angle C$.
|
35 Let the radius of the circumcircle of $\triangle A B C$ be $r$.
$$
\begin{gathered}
O D^{2}+O C \cdot C D=O C^{2} \\
O C \cdot C D=O C^{2}-O D^{2} \\
O C \cdot C D=(O C+O D)(O C-O D) \\
O C \cdot C D=(r+O D)(r-O D)
\end{gathered}
$$
By the power of the point at D ,
$$
\begin{gathered}
O C \cdot C D=B D \cdot D C \\
r=B D
\end{gathered}
$$
Then, $\triangle O B D$ and $\triangle O A B$ and $\triangle A O C$ are isosceles triangles. Let $\angle D O B=\alpha . \angle B A O=90-\frac{\alpha}{2}$. In $\triangle A B D, 15+90-\frac{\alpha}{2}=\alpha$. This means that $\alpha=70$. Furthermore, $\angle A C B$ intercepts minor arc $A B$, thus $\angle A C B=\frac{\angle A O B}{2}=\frac{70}{2}=35$
|
35
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be an obtuse triangle with circumcenter $O$ such that $\angle A B C=15^{\circ}$ and $\angle B A C>90^{\circ}$. Suppose that $A O$ meets $B C$ at $D$, and that $O D^{2}+O C \cdot D C=O C^{2}$. Find $\angle C$.
|
35 Let the radius of the circumcircle of $\triangle A B C$ be $r$.
$$
\begin{gathered}
O D^{2}+O C \cdot C D=O C^{2} \\
O C \cdot C D=O C^{2}-O D^{2} \\
O C \cdot C D=(O C+O D)(O C-O D) \\
O C \cdot C D=(r+O D)(r-O D)
\end{gathered}
$$
By the power of the point at D ,
$$
\begin{gathered}
O C \cdot C D=B D \cdot D C \\
r=B D
\end{gathered}
$$
Then, $\triangle O B D$ and $\triangle O A B$ and $\triangle A O C$ are isosceles triangles. Let $\angle D O B=\alpha . \angle B A O=90-\frac{\alpha}{2}$. In $\triangle A B D, 15+90-\frac{\alpha}{2}=\alpha$. This means that $\alpha=70$. Furthermore, $\angle A C B$ intercepts minor arc $A B$, thus $\angle A C B=\frac{\angle A O B}{2}=\frac{70}{2}=35$
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-geo-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nAnswer: "
}
|
10391c4d-a874-508a-b172-be165f3d79ae
| 609,096
|
Let $A B C D$ be a convex quadrilateral. Extend line $C D$ past $D$ to meet line $A B$ at $P$ and extend line $C B$ past $B$ to meet line $A D$ at $Q$. Suppose that line $A C$ bisects $\angle B A D$. If $A D=\frac{7}{4}, A P=\frac{21}{2}$, and $A B=\frac{14}{11}$, compute $A Q$.
|
$\frac{42}{13}$ We prove the more general statement $\frac{1}{A B}+\frac{1}{A P}=\frac{1}{A D}+\frac{1}{A Q}$, from which the answer easily follows.
Denote $\angle B A C=\angle C A D=\gamma, \angle B C A=\alpha, \angle A C D=\beta$. Then we have that by the law of sines, $\frac{A C}{A B}+\frac{A C}{A P}=\frac{\sin (\gamma+\alpha)}{\sin (\alpha)}+\frac{\sin (\gamma-\beta)}{\sin (\beta)}=\frac{\sin (\gamma-\alpha)}{\sin (\alpha)}+\frac{\sin (\gamma+\beta)}{\sin (\beta)}=\frac{A C}{A D}+\frac{A C}{A Q}$ where we have simply used the sine addition formula for the middle step.
Dividing the whole equation by $A C$ gives the desired formula, from which we compute $A Q=\left(\frac{11}{14}+\right.$ $\left.\frac{2}{21}-\frac{4}{7}\right)^{-1}=\frac{42}{13}$.
|
\frac{42}{13}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C D$ be a convex quadrilateral. Extend line $C D$ past $D$ to meet line $A B$ at $P$ and extend line $C B$ past $B$ to meet line $A D$ at $Q$. Suppose that line $A C$ bisects $\angle B A D$. If $A D=\frac{7}{4}, A P=\frac{21}{2}$, and $A B=\frac{14}{11}$, compute $A Q$.
|
$\frac{42}{13}$ We prove the more general statement $\frac{1}{A B}+\frac{1}{A P}=\frac{1}{A D}+\frac{1}{A Q}$, from which the answer easily follows.
Denote $\angle B A C=\angle C A D=\gamma, \angle B C A=\alpha, \angle A C D=\beta$. Then we have that by the law of sines, $\frac{A C}{A B}+\frac{A C}{A P}=\frac{\sin (\gamma+\alpha)}{\sin (\alpha)}+\frac{\sin (\gamma-\beta)}{\sin (\beta)}=\frac{\sin (\gamma-\alpha)}{\sin (\alpha)}+\frac{\sin (\gamma+\beta)}{\sin (\beta)}=\frac{A C}{A D}+\frac{A C}{A Q}$ where we have simply used the sine addition formula for the middle step.
Dividing the whole equation by $A C$ gives the desired formula, from which we compute $A Q=\left(\frac{11}{14}+\right.$ $\left.\frac{2}{21}-\frac{4}{7}\right)^{-1}=\frac{42}{13}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-geo-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nAnswer: "
}
|
6b1b88ea-8e7c-5b20-b137-c2cafc2064f6
| 609,097
|
Pentagon $A B C D E$ is given with the following conditions:
(a) $\angle C B D+\angle D A E=\angle B A D=45^{\circ}, \angle B C D+\angle D E A=300^{\circ}$
(b) $\frac{B A}{D A}=\frac{2 \sqrt{2}}{3}, C D=\frac{7 \sqrt{5}}{3}$, and $D E=\frac{15 \sqrt{2}}{4}$
(c) $A D^{2} \cdot B C=A B \cdot A E \cdot B D$
Compute $B D$.
|
$\sqrt{\sqrt{39}}$ As a preliminary, we may compute that by the law of cosines, the ratio $\frac{A D}{B D}=\frac{3}{\sqrt{5}}$. Now, construct the point $P$ in triangle $A B D$ such that $\triangle A P B \sim \triangle A E D$. Observe that $\frac{A P}{A D}=$ $\frac{A E \cdot A B}{A D \cdot A D}=\frac{B C}{B D}$ (where we have used first the similarity and then condition 3). Furthermore, $\angle C B D=$ $\angle D A B-\angle D A E=\angle D A B-\angle P A B=\angle P A D$ so by SAS, we have that $\triangle C B D \sim \triangle P A D$.
Therefore, by the similar triangles, we may compute $P B=D E \cdot \frac{A B}{A D}=5$ and $P D=C D \cdot \frac{A D}{B D}=7$. Furthermore, $\angle B P D=360-\angle B P A-\angle D P A=360-\angle B C D-\angle D E A=60$ and therefore, by the law of cosines, we have that $B D=\sqrt{39}$.
|
\sqrt{39}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Pentagon $A B C D E$ is given with the following conditions:
(a) $\angle C B D+\angle D A E=\angle B A D=45^{\circ}, \angle B C D+\angle D E A=300^{\circ}$
(b) $\frac{B A}{D A}=\frac{2 \sqrt{2}}{3}, C D=\frac{7 \sqrt{5}}{3}$, and $D E=\frac{15 \sqrt{2}}{4}$
(c) $A D^{2} \cdot B C=A B \cdot A E \cdot B D$
Compute $B D$.
|
$\sqrt{\sqrt{39}}$ As a preliminary, we may compute that by the law of cosines, the ratio $\frac{A D}{B D}=\frac{3}{\sqrt{5}}$. Now, construct the point $P$ in triangle $A B D$ such that $\triangle A P B \sim \triangle A E D$. Observe that $\frac{A P}{A D}=$ $\frac{A E \cdot A B}{A D \cdot A D}=\frac{B C}{B D}$ (where we have used first the similarity and then condition 3). Furthermore, $\angle C B D=$ $\angle D A B-\angle D A E=\angle D A B-\angle P A B=\angle P A D$ so by SAS, we have that $\triangle C B D \sim \triangle P A D$.
Therefore, by the similar triangles, we may compute $P B=D E \cdot \frac{A B}{A D}=5$ and $P D=C D \cdot \frac{A D}{B D}=7$. Furthermore, $\angle B P D=360-\angle B P A-\angle D P A=360-\angle B C D-\angle D E A=60$ and therefore, by the law of cosines, we have that $B D=\sqrt{39}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-geo-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nAnswer: "
}
|
f530980e-011a-55fd-b1c4-6a321d2fc20e
| 609,098
|
Triangle $A B C$ is inscribed in a circle $\omega$. Let the bisector of angle $A$ meet $\omega$ at $D$ and $B C$ at $E$. Let the reflections of $A$ across $D$ and $C$ be $D^{\prime}$ and $C^{\prime}$, respectively. Suppose that $\angle A=60^{\circ}, A B=3$, and $A E=4$. If the tangent to $\omega$ at $A$ meets line $B C$ at $P$, and the circumcircle of $A P D^{\prime}$ meets line $B C$ at $F$ (other than $P$ ), compute $F C^{\prime}$.
|
$2 \sqrt{13-6 \sqrt{3}}$ First observe that by angle chasing, $\angle P A E=180-\frac{1}{2} \angle B A C-\angle A B C=$ $\angle A E P$, so by the cyclic quadrilateral $A P D^{\prime} F, \angle E F D^{\prime}=\angle P A E=\angle P E A=\angle D^{\prime} E F$. Thus, $E D^{\prime} F$ is isosceles.
Define $B^{\prime}$ to be the reflection of $A$ about $B$, and observe that $B^{\prime} C^{\prime} \| E F$ and $B^{\prime} D^{\prime} C^{\prime}$ is isosceles. It follows that $B^{\prime} E F C^{\prime}$ is an isosceles trapezoid, so $F C^{\prime}=B^{\prime} E$, which by the law of cosines, is equal to $\sqrt{A B^{\prime 2}+A E^{2}-2 A B \cdot A E \cos 30}=2 \sqrt{13-6 \sqrt{3}}$.
|
2 \sqrt{13-6 \sqrt{3}}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Triangle $A B C$ is inscribed in a circle $\omega$. Let the bisector of angle $A$ meet $\omega$ at $D$ and $B C$ at $E$. Let the reflections of $A$ across $D$ and $C$ be $D^{\prime}$ and $C^{\prime}$, respectively. Suppose that $\angle A=60^{\circ}, A B=3$, and $A E=4$. If the tangent to $\omega$ at $A$ meets line $B C$ at $P$, and the circumcircle of $A P D^{\prime}$ meets line $B C$ at $F$ (other than $P$ ), compute $F C^{\prime}$.
|
$2 \sqrt{13-6 \sqrt{3}}$ First observe that by angle chasing, $\angle P A E=180-\frac{1}{2} \angle B A C-\angle A B C=$ $\angle A E P$, so by the cyclic quadrilateral $A P D^{\prime} F, \angle E F D^{\prime}=\angle P A E=\angle P E A=\angle D^{\prime} E F$. Thus, $E D^{\prime} F$ is isosceles.
Define $B^{\prime}$ to be the reflection of $A$ about $B$, and observe that $B^{\prime} C^{\prime} \| E F$ and $B^{\prime} D^{\prime} C^{\prime}$ is isosceles. It follows that $B^{\prime} E F C^{\prime}$ is an isosceles trapezoid, so $F C^{\prime}=B^{\prime} E$, which by the law of cosines, is equal to $\sqrt{A B^{\prime 2}+A E^{2}-2 A B \cdot A E \cos 30}=2 \sqrt{13-6 \sqrt{3}}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-geo-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nAnswer: "
}
|
9eefcadf-1b69-55d1-9064-fb6f0f066158
| 609,099
|
Arpon chooses a positive real number $k$. For each positive integer $n$, he places a marker at the point $(n, n k)$ in the $(x, y)$ plane. Suppose that two markers whose $x$ coordinates differ by 4 have distance 31. What is the distance between the markers at $(7,7 k)$ and $(19,19 k)$ ?
|
93 The difference of the $x$-coordinates of the markers is $12=3 \cdot 4$. Thus, by similar triangles (where we draw right triangles whose legs are parallel to the axes and whose hypotenuses lie along the line $y=k x$ ), the distance between the markers is $3 \cdot 31=93$.
|
93
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Arpon chooses a positive real number $k$. For each positive integer $n$, he places a marker at the point $(n, n k)$ in the $(x, y)$ plane. Suppose that two markers whose $x$ coordinates differ by 4 have distance 31. What is the distance between the markers at $(7,7 k)$ and $(19,19 k)$ ?
|
93 The difference of the $x$-coordinates of the markers is $12=3 \cdot 4$. Thus, by similar triangles (where we draw right triangles whose legs are parallel to the axes and whose hypotenuses lie along the line $y=k x$ ), the distance between the markers is $3 \cdot 31=93$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n1. [4]",
"solution_match": "\nAnswer: "
}
|
99b95cc2-e552-5387-b5df-8ab3218fd0c4
| 609,100
|
The real numbers $x, y, z$ satisfy $0 \leq x \leq y \leq z \leq 4$. If their squares form an arithmetic progression with common difference 2 , determine the minimum possible value of $|x-y|+|y-z|$.
|
$4-2 \sqrt{3}$ Clearly $|x-y|+|y-z|=z-x=\frac{z^{2}-x^{2}}{z+x}=\frac{4}{z+x}$, which is minimized when $z=4$ and $x=\sqrt{12}$. Thus, our answer is $4-\sqrt{12}=4-2 \sqrt{3}$.
|
4-2 \sqrt{3}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
The real numbers $x, y, z$ satisfy $0 \leq x \leq y \leq z \leq 4$. If their squares form an arithmetic progression with common difference 2 , determine the minimum possible value of $|x-y|+|y-z|$.
|
$4-2 \sqrt{3}$ Clearly $|x-y|+|y-z|=z-x=\frac{z^{2}-x^{2}}{z+x}=\frac{4}{z+x}$, which is minimized when $z=4$ and $x=\sqrt{12}$. Thus, our answer is $4-\sqrt{12}=4-2 \sqrt{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n2. [4]",
"solution_match": "\nAnswer: "
}
|
c7e34a60-d592-5873-ae99-2f3c98abf1c5
| 609,101
|
Find the rightmost non-zero digit of the expansion of (20)(13!).
|
6 We can rewrite this as $(10 * 2)(13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)=$ $\left(10^{3}\right)(2 * 13 * 12 * 11 * 9 * 8 * 7 * 6 * 4 * 3)$; multiplying together the units digits for the terms not equal to 10 reveals that the rightmost non-zero digit is 6 .
|
6
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the rightmost non-zero digit of the expansion of (20)(13!).
|
6 We can rewrite this as $(10 * 2)(13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)=$ $\left(10^{3}\right)(2 * 13 * 12 * 11 * 9 * 8 * 7 * 6 * 4 * 3)$; multiplying together the units digits for the terms not equal to 10 reveals that the rightmost non-zero digit is 6 .
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n3. [4]",
"solution_match": "\nAnswer: "
}
|
2403c395-d291-50f5-bc84-5a5b9dbc09e2
| 609,102
|
Spencer is making burritos, each of which consists of one wrap and one filling. He has enough filling for up to four beef burritos and three chicken burritos. However, he only has five wraps for the burritos; in how many orders can he make exactly five burritos?
|
25 Spencer's burrito-making can include either 3, 2, or 1 chicken burrito; consequently, he has $\binom{5}{3}+\binom{5}{2}+\binom{5}{1}=25$ orders in which he can make burritos.
|
25
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Spencer is making burritos, each of which consists of one wrap and one filling. He has enough filling for up to four beef burritos and three chicken burritos. However, he only has five wraps for the burritos; in how many orders can he make exactly five burritos?
|
25 Spencer's burrito-making can include either 3, 2, or 1 chicken burrito; consequently, he has $\binom{5}{3}+\binom{5}{2}+\binom{5}{1}=25$ orders in which he can make burritos.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n4. [4]",
"solution_match": "\nAnswer: "
}
|
84338cf5-bd1a-5543-90bb-124f6a8e5c90
| 54,028
|
Rahul has ten cards face-down, which consist of five distinct pairs of matching cards. During each move of his game, Rahul chooses one card to turn face-up, looks at it, and then chooses another to turn face-up and looks at it. If the two face-up cards match, the game ends. If not, Rahul flips both cards face-down and keeps repeating this process. Initially, Rahul doesn't know which cards are which. Assuming that he has perfect memory, find the smallest number of moves after which he can guarantee that the game has ended.
|
4 Label the 10 cards $a_{1}, a_{2}, \ldots, a_{5}, b_{1}, b_{2}, \ldots, b_{5}$ such that $a_{i}$ and $b_{i}$ match for $1 \leq i \leq 5$.
First, we'll show that Rahul cannot always end the game in less than 4 moves, in particular, when he turns up his fifth card (during the third move), it is possible that the card he flips over is not one which he has yet encountered; consequently, he will not guarantee being able to match it, so he cannot guarantee that the game can end in three moves.
However, Rahul can always end the game in 4 moves. To do this, he can turn over 6 distinct cards in his first 3 moves. If we consider the 5 sets of cards $\left\{a_{1}, b_{1}\right\},\left\{a_{2}, b_{2}\right\},\left\{a_{3}, b_{3}\right\},\left\{a_{4}, b_{4}\right\},\left\{a_{5}, b_{5}\right\}$, then by the pigeonhole principle, at least 2 of the 6 revealed cards must be from the same set. Rahul can then turn over those 2 cards on the fourth move, ending the game.
|
4
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Rahul has ten cards face-down, which consist of five distinct pairs of matching cards. During each move of his game, Rahul chooses one card to turn face-up, looks at it, and then chooses another to turn face-up and looks at it. If the two face-up cards match, the game ends. If not, Rahul flips both cards face-down and keeps repeating this process. Initially, Rahul doesn't know which cards are which. Assuming that he has perfect memory, find the smallest number of moves after which he can guarantee that the game has ended.
|
4 Label the 10 cards $a_{1}, a_{2}, \ldots, a_{5}, b_{1}, b_{2}, \ldots, b_{5}$ such that $a_{i}$ and $b_{i}$ match for $1 \leq i \leq 5$.
First, we'll show that Rahul cannot always end the game in less than 4 moves, in particular, when he turns up his fifth card (during the third move), it is possible that the card he flips over is not one which he has yet encountered; consequently, he will not guarantee being able to match it, so he cannot guarantee that the game can end in three moves.
However, Rahul can always end the game in 4 moves. To do this, he can turn over 6 distinct cards in his first 3 moves. If we consider the 5 sets of cards $\left\{a_{1}, b_{1}\right\},\left\{a_{2}, b_{2}\right\},\left\{a_{3}, b_{3}\right\},\left\{a_{4}, b_{4}\right\},\left\{a_{5}, b_{5}\right\}$, then by the pigeonhole principle, at least 2 of the 6 revealed cards must be from the same set. Rahul can then turn over those 2 cards on the fourth move, ending the game.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n5. [5]",
"solution_match": "\nAnswer: "
}
|
917291e4-00a8-5f31-a6ee-75e6ea916959
| 54,040
|
Let $R$ be the region in the Cartesian plane of points $(x, y)$ satisfying $x \geq 0, y \geq 0$, and $x+y+$ $\lfloor x\rfloor+\lfloor y\rfloor \leq 5$. Determine the area of $R$.
|
| $\frac{9}{2}$ |
| :---: |
| We claim that a point in the first quadrant satisfies the desired property if the point | is below the line $x+y=3$ and does not satisfy the desired property if it is above the line.
To see this, for a point inside the region, $x+y<3$ and $\lfloor x\rfloor+\lfloor y\rfloor \leq x+y<3$ However, $\lfloor x\rfloor+\lfloor y\rfloor$ must equal to an integer. Thus, $\lfloor x\rfloor+\lfloor y\rfloor \leq 2$. Adding these two equations, $x+y+\lfloor x\rfloor+\lfloor y\rfloor<5$, which satisfies the desired property. Conversely, for a point outside the region, $\lfloor x\rfloor+\lfloor y\rfloor+\{x\}+\{y\}=x+y>3$ However, $\{x\}+\{y\}<2$. Thus, $\lfloor x\rfloor+\lfloor y\rfloor>1$, so $\lfloor x\rfloor+\lfloor y\rfloor \geq 2$, implying that $x+y+\lfloor x\rfloor+\lfloor y\rfloor>5$.
To finish, $R$ is the region bounded by the x -axis, the y -axis, and the line $x+y=3$ is a right triangle whose legs have length 3 . Consequently, $R$ has area $\frac{9}{2}$.
|
\frac{9}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $R$ be the region in the Cartesian plane of points $(x, y)$ satisfying $x \geq 0, y \geq 0$, and $x+y+$ $\lfloor x\rfloor+\lfloor y\rfloor \leq 5$. Determine the area of $R$.
|
| $\frac{9}{2}$ |
| :---: |
| We claim that a point in the first quadrant satisfies the desired property if the point | is below the line $x+y=3$ and does not satisfy the desired property if it is above the line.
To see this, for a point inside the region, $x+y<3$ and $\lfloor x\rfloor+\lfloor y\rfloor \leq x+y<3$ However, $\lfloor x\rfloor+\lfloor y\rfloor$ must equal to an integer. Thus, $\lfloor x\rfloor+\lfloor y\rfloor \leq 2$. Adding these two equations, $x+y+\lfloor x\rfloor+\lfloor y\rfloor<5$, which satisfies the desired property. Conversely, for a point outside the region, $\lfloor x\rfloor+\lfloor y\rfloor+\{x\}+\{y\}=x+y>3$ However, $\{x\}+\{y\}<2$. Thus, $\lfloor x\rfloor+\lfloor y\rfloor>1$, so $\lfloor x\rfloor+\lfloor y\rfloor \geq 2$, implying that $x+y+\lfloor x\rfloor+\lfloor y\rfloor>5$.
To finish, $R$ is the region bounded by the x -axis, the y -axis, and the line $x+y=3$ is a right triangle whose legs have length 3 . Consequently, $R$ has area $\frac{9}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n6. [5]",
"solution_match": "\nAnswer: "
}
|
4326cd90-c4e8-5eb2-86e0-4475b14792a1
| 609,103
|
Find the number of positive divisors $d$ of $15!=15 \cdot 14 \cdots 2 \cdot 1$ such that $\operatorname{gcd}(d, 60)=5$.
|
36 Since $\operatorname{gcd}(d, 60)=5$, we know that $d=5^{i} d^{\prime}$ for some integer $i>0$ and some integer $d^{\prime}$ which is relatively prime to 60 . Consequently, $d^{\prime}$ is a divisor of $(15!) / 5$; eliminating common factors with 60 gives that $d^{\prime}$ is a factor of $\left(7^{2}\right)(11)(13)$, which has $(2+1)(1+1)(1+1)=12$ factors. Finally, $i$ can be 1,2 , or 3 , so there are a total of $3 \cdot 12=36$ possibilities.
|
36
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the number of positive divisors $d$ of $15!=15 \cdot 14 \cdots 2 \cdot 1$ such that $\operatorname{gcd}(d, 60)=5$.
|
36 Since $\operatorname{gcd}(d, 60)=5$, we know that $d=5^{i} d^{\prime}$ for some integer $i>0$ and some integer $d^{\prime}$ which is relatively prime to 60 . Consequently, $d^{\prime}$ is a divisor of $(15!) / 5$; eliminating common factors with 60 gives that $d^{\prime}$ is a factor of $\left(7^{2}\right)(11)(13)$, which has $(2+1)(1+1)(1+1)=12$ factors. Finally, $i$ can be 1,2 , or 3 , so there are a total of $3 \cdot 12=36$ possibilities.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n7. [5]",
"solution_match": "\nAnswer: "
}
|
84452689-9788-541b-a578-fa457e565364
| 609,104
|
In a game, there are three indistinguishable boxes; one box contains two red balls, one contains two blue balls, and the last contains one ball of each color. To play, Raj first predicts whether he will draw two balls of the same color or two of different colors. Then, he picks a box, draws a ball at random, looks at the color, and replaces the ball in the same box. Finally, he repeats this; however, the boxes are not shuffled between draws, so he can determine whether he wants to draw again from the same box. Raj wins if he predicts correctly; if he plays optimally, what is the probability that he will win?
|
| $\frac{5}{6}$ | Call the box with two red balls box 1, the box with one of each color box 2 , and the |
| :---: | :---: | box with two blue balls box 3. Without loss of generality, assume that the first ball that Bob draws is red. If Bob picked box 1 , then he would have picked a red ball with probability 1 , and if Bob picked box 2 , then he would have picked a red ball with probability $\frac{1}{2}$. Therefore, the probability that he picked box 1 is $\frac{1}{1+\frac{1}{2}}=\frac{2}{3}$, and the probability that he picked box 2 is $\frac{1}{3}$. We will now consider both possible predictions and find which one gives a better probability of winning, assuming optimal play.
If Bob predicts that he will draw two balls of the same color, then there are two possible plays: he draws from the same box, or he draws from a different box. If he draws from the same box, then in the $\frac{2}{3}$ chance that he originally picked box 1 , he will always win, and in the $\frac{1}{3}$ chance that he picked box 2 , he will win with probability $\frac{1}{2}$, for a total probability of $\frac{2}{3}+\frac{1}{3} \cdot \frac{1}{2}=\frac{5}{6}$. If he draws from a different box, then if he originally picked box 1 , he will win with probability $\frac{1}{4}$ and if he originally picked box 2 , he will win with probability $\frac{1}{2}$, for a total probability of $\frac{2}{3} \cdot \frac{1}{4}+\frac{1}{3} \cdot \frac{1}{2}=\frac{1}{3}$.
If Bob predicts that he will draw two balls of different colors, then we can consider the same two possible plays. Using similar calculations, if he draws from the same box, then he will win with probability $\frac{1}{6}$, and if he draws from a different box, then he will win with probability $\frac{2}{3}$. Looking at all cases, Bob's best play is to predict that he will draw two balls of the same color and then draw the second ball from the same box, with a winning probability of $\frac{5}{6}$.
|
\frac{5}{6}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In a game, there are three indistinguishable boxes; one box contains two red balls, one contains two blue balls, and the last contains one ball of each color. To play, Raj first predicts whether he will draw two balls of the same color or two of different colors. Then, he picks a box, draws a ball at random, looks at the color, and replaces the ball in the same box. Finally, he repeats this; however, the boxes are not shuffled between draws, so he can determine whether he wants to draw again from the same box. Raj wins if he predicts correctly; if he plays optimally, what is the probability that he will win?
|
| $\frac{5}{6}$ | Call the box with two red balls box 1, the box with one of each color box 2 , and the |
| :---: | :---: | box with two blue balls box 3. Without loss of generality, assume that the first ball that Bob draws is red. If Bob picked box 1 , then he would have picked a red ball with probability 1 , and if Bob picked box 2 , then he would have picked a red ball with probability $\frac{1}{2}$. Therefore, the probability that he picked box 1 is $\frac{1}{1+\frac{1}{2}}=\frac{2}{3}$, and the probability that he picked box 2 is $\frac{1}{3}$. We will now consider both possible predictions and find which one gives a better probability of winning, assuming optimal play.
If Bob predicts that he will draw two balls of the same color, then there are two possible plays: he draws from the same box, or he draws from a different box. If he draws from the same box, then in the $\frac{2}{3}$ chance that he originally picked box 1 , he will always win, and in the $\frac{1}{3}$ chance that he picked box 2 , he will win with probability $\frac{1}{2}$, for a total probability of $\frac{2}{3}+\frac{1}{3} \cdot \frac{1}{2}=\frac{5}{6}$. If he draws from a different box, then if he originally picked box 1 , he will win with probability $\frac{1}{4}$ and if he originally picked box 2 , he will win with probability $\frac{1}{2}$, for a total probability of $\frac{2}{3} \cdot \frac{1}{4}+\frac{1}{3} \cdot \frac{1}{2}=\frac{1}{3}$.
If Bob predicts that he will draw two balls of different colors, then we can consider the same two possible plays. Using similar calculations, if he draws from the same box, then he will win with probability $\frac{1}{6}$, and if he draws from a different box, then he will win with probability $\frac{2}{3}$. Looking at all cases, Bob's best play is to predict that he will draw two balls of the same color and then draw the second ball from the same box, with a winning probability of $\frac{5}{6}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n8. [5]",
"solution_match": "\nAnswer: "
}
|
8de4a657-0ec0-526f-93e0-ae367c1c7787
| 609,105
|
I have 8 unit cubes of different colors, which I want to glue together into a $2 \times 2 \times 2$ cube. How many distinct $2 \times 2 \times 2$ cubes can I make? Rotations of the same cube are not considered distinct, but reflections are.
|
1680 Our goal is to first pin down the cube, so it can't rotate. Without loss of generality, suppose one of the unit cubes is purple, and let the purple cube be in the top left front position. Now, look at the three positions that share a face with the purple cube. There are $\binom{7}{3}$ ways to pick the three cubes that fill those positions and two ways to position them that are rotationally distinct. Now, we've taken care of any possible rotations, so there are simply 4 ! ways to position the final four cubes. Thus, our answer is $\binom{7}{3} \cdot 2 \cdot 4!=1680$ ways.
|
1680
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
I have 8 unit cubes of different colors, which I want to glue together into a $2 \times 2 \times 2$ cube. How many distinct $2 \times 2 \times 2$ cubes can I make? Rotations of the same cube are not considered distinct, but reflections are.
|
1680 Our goal is to first pin down the cube, so it can't rotate. Without loss of generality, suppose one of the unit cubes is purple, and let the purple cube be in the top left front position. Now, look at the three positions that share a face with the purple cube. There are $\binom{7}{3}$ ways to pick the three cubes that fill those positions and two ways to position them that are rotationally distinct. Now, we've taken care of any possible rotations, so there are simply 4 ! ways to position the final four cubes. Thus, our answer is $\binom{7}{3} \cdot 2 \cdot 4!=1680$ ways.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n9. [6]",
"solution_match": "\nAnswer: "
}
|
67b1eb4d-c449-5861-a6c5-01dd0a6e0963
| 609,106
|
Wesyu is a farmer, and she's building a cao (a relative of the cow) pasture. She starts with a triangle $A_{0} A_{1} A_{2}$ where angle $A_{0}$ is $90^{\circ}$, angle $A_{1}$ is $60^{\circ}$, and $A_{0} A_{1}$ is 1 . She then extends the pasture. First, she extends $A_{2} A_{0}$ to $A_{3}$ such that $A_{3} A_{0}=\frac{1}{2} A_{2} A_{0}$ and the new pasture is triangle $A_{1} A_{2} A_{3}$. Next, she extends $A_{3} A_{1}$ to $A_{4}$ such that $A_{4} A_{1}=\frac{1}{6} A_{3} A_{1}$. She continues, each time extending $A_{n} A_{n-2}$ to $A_{n+1}$ such that $A_{n+1} A_{n-2}=\frac{1}{2^{n}-2} A_{n} A_{n-2}$. What is the smallest $K$ such that her pasture never exceeds an area of $K$ ?
|
$\sqrt{3}$ First, note that for any $i$, after performing the operation on triangle $A_{i} A_{i+1} A_{i+2}$, the resulting pasture is triangle $A_{i+1} A_{i+2} A_{i+3}$. Let $K_{i}$ be the area of triangle $A_{i} A_{i+1} A_{i+2}$. From
$A_{n+1} A_{n-2}=\frac{1}{2^{n}-2} A_{n} A_{n-2}$ and $A_{n} A_{n+1}=A_{n} A_{n-2}+A_{n-2} A_{n+1}$, we have $A_{n} A_{n+1}=\left(1+\frac{1}{2^{n}-2}\right) A_{n} A_{n-2}$. We also know that the area of a triangle is half the product of its base and height, so if we let the base of triangle $A_{n-2} A_{n-1} A_{n}$ be $A_{n} A_{n-2}$, its area is $K_{n-2}=\frac{1}{2} h A_{n} A_{n-2}$. The area of triangle $A_{n-1} A_{n} A_{n+1}$ is $K_{n-1}=\frac{1}{2} h A_{n} A_{n+1}$. The $h$ 's are equal because the distance from $A_{n-1}$ to the base does not change.
We now have $\frac{K_{n-1}}{K_{n-2}}=\frac{A_{n} A_{n+1}}{A_{n} A_{n-2}}=1+\frac{1}{2^{n}-2}=\frac{2^{n}-1}{2^{n}-2}$. Therefore, $\frac{K_{1}}{K_{0}}=\frac{3}{2}, \frac{K_{2}}{K_{0}}=\frac{K_{2}}{K_{1}} \frac{K_{1}}{K_{0}}=\frac{7}{6} \cdot \frac{3}{2}=\frac{7}{4}$, $\frac{K_{3}}{K_{0}}=\frac{K_{3}}{K_{2}} \frac{K_{2}}{K_{0}}=\frac{15}{14} \cdot \frac{7}{4}=\frac{15}{8}$. We see the pattern $\frac{K_{n}}{K_{0}}=\frac{2^{n+1}-1}{2^{n}}$, which can be easily proven by induction. As $n$ approaches infinity, $\frac{K_{n}}{K_{0}}$ grows arbitrarily close to 2 , so the smallest $K$ such that the pasture never exceeds an area of $K$ is $2 K_{0}=\sqrt{3}$.
|
\sqrt{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Wesyu is a farmer, and she's building a cao (a relative of the cow) pasture. She starts with a triangle $A_{0} A_{1} A_{2}$ where angle $A_{0}$ is $90^{\circ}$, angle $A_{1}$ is $60^{\circ}$, and $A_{0} A_{1}$ is 1 . She then extends the pasture. First, she extends $A_{2} A_{0}$ to $A_{3}$ such that $A_{3} A_{0}=\frac{1}{2} A_{2} A_{0}$ and the new pasture is triangle $A_{1} A_{2} A_{3}$. Next, she extends $A_{3} A_{1}$ to $A_{4}$ such that $A_{4} A_{1}=\frac{1}{6} A_{3} A_{1}$. She continues, each time extending $A_{n} A_{n-2}$ to $A_{n+1}$ such that $A_{n+1} A_{n-2}=\frac{1}{2^{n}-2} A_{n} A_{n-2}$. What is the smallest $K$ such that her pasture never exceeds an area of $K$ ?
|
$\sqrt{3}$ First, note that for any $i$, after performing the operation on triangle $A_{i} A_{i+1} A_{i+2}$, the resulting pasture is triangle $A_{i+1} A_{i+2} A_{i+3}$. Let $K_{i}$ be the area of triangle $A_{i} A_{i+1} A_{i+2}$. From
$A_{n+1} A_{n-2}=\frac{1}{2^{n}-2} A_{n} A_{n-2}$ and $A_{n} A_{n+1}=A_{n} A_{n-2}+A_{n-2} A_{n+1}$, we have $A_{n} A_{n+1}=\left(1+\frac{1}{2^{n}-2}\right) A_{n} A_{n-2}$. We also know that the area of a triangle is half the product of its base and height, so if we let the base of triangle $A_{n-2} A_{n-1} A_{n}$ be $A_{n} A_{n-2}$, its area is $K_{n-2}=\frac{1}{2} h A_{n} A_{n-2}$. The area of triangle $A_{n-1} A_{n} A_{n+1}$ is $K_{n-1}=\frac{1}{2} h A_{n} A_{n+1}$. The $h$ 's are equal because the distance from $A_{n-1}$ to the base does not change.
We now have $\frac{K_{n-1}}{K_{n-2}}=\frac{A_{n} A_{n+1}}{A_{n} A_{n-2}}=1+\frac{1}{2^{n}-2}=\frac{2^{n}-1}{2^{n}-2}$. Therefore, $\frac{K_{1}}{K_{0}}=\frac{3}{2}, \frac{K_{2}}{K_{0}}=\frac{K_{2}}{K_{1}} \frac{K_{1}}{K_{0}}=\frac{7}{6} \cdot \frac{3}{2}=\frac{7}{4}$, $\frac{K_{3}}{K_{0}}=\frac{K_{3}}{K_{2}} \frac{K_{2}}{K_{0}}=\frac{15}{14} \cdot \frac{7}{4}=\frac{15}{8}$. We see the pattern $\frac{K_{n}}{K_{0}}=\frac{2^{n+1}-1}{2^{n}}$, which can be easily proven by induction. As $n$ approaches infinity, $\frac{K_{n}}{K_{0}}$ grows arbitrarily close to 2 , so the smallest $K$ such that the pasture never exceeds an area of $K$ is $2 K_{0}=\sqrt{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n10. [6]",
"solution_match": "\nAnswer: "
}
|
8f3470dc-cad6-5382-bb3a-0594379589d5
| 609,107
|
Compute the prime factorization of 1007021035035021007001 . (You should write your answer in the form $p_{1}^{e_{1}} p_{2}^{e_{2}} \ldots p_{k}^{e_{k}}$, where $p_{1}, \ldots, p_{k}$ are distinct prime numbers and $e_{1}, \ldots, e_{k}$ are positive integers.)
|
$7^{7} \cdot 11^{7} \cdot 13^{7}$ The number in question is
$$
\sum_{i=0}^{7}\binom{7}{i} 1000^{i}=(1000+1)^{7}=1001^{7}=7^{7} \cdot 11^{7} \cdot 13^{7}
$$
|
7^{7} \cdot 11^{7} \cdot 13^{7}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Compute the prime factorization of 1007021035035021007001 . (You should write your answer in the form $p_{1}^{e_{1}} p_{2}^{e_{2}} \ldots p_{k}^{e_{k}}$, where $p_{1}, \ldots, p_{k}$ are distinct prime numbers and $e_{1}, \ldots, e_{k}$ are positive integers.)
|
$7^{7} \cdot 11^{7} \cdot 13^{7}$ The number in question is
$$
\sum_{i=0}^{7}\binom{7}{i} 1000^{i}=(1000+1)^{7}=1001^{7}=7^{7} \cdot 11^{7} \cdot 13^{7}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n11. [6]",
"solution_match": "\nAnswer: "
}
|
5fac4612-b9bb-5205-a3b0-f51de9fa515c
| 609,108
|
For how many integers $1 \leq k \leq 2013$ does the decimal representation of $k^{k}$ end with a 1 ?
|
202 We claim that this is only possible if $k$ has a units digit of 1 . Clearly, it is true in these cases. Additionally, $k^{k}$ cannot have a units digit of 1 when $k$ has a units digit of $2,4,5,6$, or 8 . If $k$ has a units digit of 3 or 7 , then $k^{k}$ has a units digit of 1 if and only if $4 \mid k$, a contradiction. Similarly, if $k$ has a units digit of 9 , then $k^{k}$ has a units digit of 1 if and only if $2 \mid k$, also a contradiction. Since there are 202 integers between 1 and 2013, inclusive, with a units digit of 1 , there are 202 such $k$ which fulfill our criterion.
|
202
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
For how many integers $1 \leq k \leq 2013$ does the decimal representation of $k^{k}$ end with a 1 ?
|
202 We claim that this is only possible if $k$ has a units digit of 1 . Clearly, it is true in these cases. Additionally, $k^{k}$ cannot have a units digit of 1 when $k$ has a units digit of $2,4,5,6$, or 8 . If $k$ has a units digit of 3 or 7 , then $k^{k}$ has a units digit of 1 if and only if $4 \mid k$, a contradiction. Similarly, if $k$ has a units digit of 9 , then $k^{k}$ has a units digit of 1 if and only if $2 \mid k$, also a contradiction. Since there are 202 integers between 1 and 2013, inclusive, with a units digit of 1 , there are 202 such $k$ which fulfill our criterion.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n12. [6]",
"solution_match": "\nAnswer: "
}
|
7031ed4a-6182-5c7d-9d19-55bf96626696
| 609,109
|
Find the smallest positive integer $n$ such that $\frac{5^{n+1}+2^{n+1}}{5^{n}+2^{n}}>4.99$.
|
7 Writing $5^{n+1}=5 \cdot 5^{n}$ and $2^{n+1}=2 \cdot 2^{n}$ and cross-multiplying yields $0.01 \cdot 5^{n}>2.99 \cdot 2^{n}$, and re-arranging yields $(2.5)^{n}>299$. A straightforward calculation shows that the smallest $n$ for which this is true is $n=7$.
|
7
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find the smallest positive integer $n$ such that $\frac{5^{n+1}+2^{n+1}}{5^{n}+2^{n}}>4.99$.
|
7 Writing $5^{n+1}=5 \cdot 5^{n}$ and $2^{n+1}=2 \cdot 2^{n}$ and cross-multiplying yields $0.01 \cdot 5^{n}>2.99 \cdot 2^{n}$, and re-arranging yields $(2.5)^{n}>299$. A straightforward calculation shows that the smallest $n$ for which this is true is $n=7$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n13. [8]",
"solution_match": "\nAnswer: "
}
|
f7ccc7ae-1a13-522f-a162-602b9199298b
| 609,110
|
Consider triangle $A B C$ with $\angle A=2 \angle B$. The angle bisectors from $A$ and $C$ intersect at $D$, and the angle bisector from $C$ intersects $\overline{A B}$ at $E$. If $\frac{D E}{D C}=\frac{1}{3}$, compute $\frac{A B}{A C}$.
|
$\quad \frac{7}{9} \quad$ Let $A E=x$ and $B E=y$. Using angle-bisector theorem on $\triangle A C E$ we have $x: D E=A C: D C$, so $A C=3 x$. Using some angle chasing, it is simple to see that $\angle A D E=\angle A E D$, so $A D=A E=x$. Then, note that $\triangle C D A \sim \triangle C E B$, so $y:(D C+D E)=x: D C$, so $y: x=1+\frac{1}{3}=\frac{4}{3}$, so $A B=x+\frac{4}{3} x=\frac{7}{3} x$. Thus the desired answer is $A B: A C=\frac{7}{3} x: 3 x=\frac{7}{9}$.
|
\frac{7}{9}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Consider triangle $A B C$ with $\angle A=2 \angle B$. The angle bisectors from $A$ and $C$ intersect at $D$, and the angle bisector from $C$ intersects $\overline{A B}$ at $E$. If $\frac{D E}{D C}=\frac{1}{3}$, compute $\frac{A B}{A C}$.
|
$\quad \frac{7}{9} \quad$ Let $A E=x$ and $B E=y$. Using angle-bisector theorem on $\triangle A C E$ we have $x: D E=A C: D C$, so $A C=3 x$. Using some angle chasing, it is simple to see that $\angle A D E=\angle A E D$, so $A D=A E=x$. Then, note that $\triangle C D A \sim \triangle C E B$, so $y:(D C+D E)=x: D C$, so $y: x=1+\frac{1}{3}=\frac{4}{3}$, so $A B=x+\frac{4}{3} x=\frac{7}{3} x$. Thus the desired answer is $A B: A C=\frac{7}{3} x: 3 x=\frac{7}{9}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n14. [8]",
"solution_match": "\nAnswer: "
}
|
8d7da142-dcd4-5477-8234-9a2f817ab97a
| 609,111
|
Tim and Allen are playing a match of tenus. In a match of tenus, the two players play a series of games, each of which is won by one of the two players. The match ends when one player has won exactly two more games than the other player, at which point the player who has won more games wins the match. In odd-numbered games, Tim wins with probability $3 / 4$, and in the even-numbered games, Allen wins with probability $3 / 4$. What is the expected number of games in a match?
|
$\quad \frac{16}{3}$ Let the answer be $E$. If Tim wins the first game and Allen wins the second game or vice versa, which occurs with probability $(3 / 4)^{2}+(1 / 4)^{2}=5 / 8$, the expected number of additional games is just $E$, so the expected total number of games is $E+2$. If, on the other hand, one of Tim and Allen wins both of the first two games, with probability $1-(5 / 8)=3 / 8$, there are exactly 2 games in the match. It follows that
$$
E=\frac{3}{8} \cdot 2+\frac{5}{8} \cdot(E+2)
$$
and solving gives $E=\frac{16}{3}$.
|
\frac{16}{3}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Tim and Allen are playing a match of tenus. In a match of tenus, the two players play a series of games, each of which is won by one of the two players. The match ends when one player has won exactly two more games than the other player, at which point the player who has won more games wins the match. In odd-numbered games, Tim wins with probability $3 / 4$, and in the even-numbered games, Allen wins with probability $3 / 4$. What is the expected number of games in a match?
|
$\quad \frac{16}{3}$ Let the answer be $E$. If Tim wins the first game and Allen wins the second game or vice versa, which occurs with probability $(3 / 4)^{2}+(1 / 4)^{2}=5 / 8$, the expected number of additional games is just $E$, so the expected total number of games is $E+2$. If, on the other hand, one of Tim and Allen wins both of the first two games, with probability $1-(5 / 8)=3 / 8$, there are exactly 2 games in the match. It follows that
$$
E=\frac{3}{8} \cdot 2+\frac{5}{8} \cdot(E+2)
$$
and solving gives $E=\frac{16}{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n15. [8]",
"solution_match": "\nAnswer: "
}
|
38bca082-a4a5-5254-b9a1-77242417ae2c
| 609,112
|
The walls of a room are in the shape of a triangle $A B C$ with $\angle A B C=90^{\circ}, \angle B A C=60^{\circ}$, and $A B=6$. Chong stands at the midpoint of $B C$ and rolls a ball toward $A B$. Suppose that the ball bounces off $A B$, then $A C$, then returns exactly to Chong. Find the length of the path of the ball.
|
$\quad 3 \sqrt{21}$ Let $C^{\prime}$ be the reflection of $C$ across $A B$ and $B^{\prime}$ be the reflection of $B$ across $A C^{\prime}$; note that $B^{\prime}, A, C$ are collinear by angle chasing. The image of the path under these reflections is just the line segment $M M^{\prime}$, where $M$ is the midpoint of $B C$ and $M^{\prime}$ is the midpoint of $B^{\prime} C^{\prime}$, so our answer is just the length of $M M^{\prime}$. Applying the Law of Cosines to triangle $M^{\prime} C^{\prime} M$, we have $M M^{\prime 2}=27+243-2 \cdot 3 \sqrt{3} \cdot 9 \sqrt{3} \cdot \frac{1}{2}=189$, so $M M^{\prime}=3 \sqrt{21}$.
|
3 \sqrt{21}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
The walls of a room are in the shape of a triangle $A B C$ with $\angle A B C=90^{\circ}, \angle B A C=60^{\circ}$, and $A B=6$. Chong stands at the midpoint of $B C$ and rolls a ball toward $A B$. Suppose that the ball bounces off $A B$, then $A C$, then returns exactly to Chong. Find the length of the path of the ball.
|
$\quad 3 \sqrt{21}$ Let $C^{\prime}$ be the reflection of $C$ across $A B$ and $B^{\prime}$ be the reflection of $B$ across $A C^{\prime}$; note that $B^{\prime}, A, C$ are collinear by angle chasing. The image of the path under these reflections is just the line segment $M M^{\prime}$, where $M$ is the midpoint of $B C$ and $M^{\prime}$ is the midpoint of $B^{\prime} C^{\prime}$, so our answer is just the length of $M M^{\prime}$. Applying the Law of Cosines to triangle $M^{\prime} C^{\prime} M$, we have $M M^{\prime 2}=27+243-2 \cdot 3 \sqrt{3} \cdot 9 \sqrt{3} \cdot \frac{1}{2}=189$, so $M M^{\prime}=3 \sqrt{21}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n16. [8]",
"solution_match": "\nAnswer: "
}
|
cfbe03e3-7f72-5f02-a51b-26a8c61e4e95
| 609,113
|
The lines $y=x, y=2 x$, and $y=3 x$ are the three medians of a triangle with perimeter 1. Find the length of the longest side of the triangle.
|
| $\frac{\sqrt{58}}{2+\sqrt{34}+\sqrt{58}}$ |
| :---: | The three medians of a triangle contain its vertices, so the three vertices of the triangle are $(a, a),(b, 2 b)$ and $(c, 3 c)$ for some $a, b$, and $c$. Then, the midpoint of $(a, a)$ and $(b, 2 b)$, which is $\left(\frac{a+b}{2}, \frac{a+2 b}{2}\right)$, must lie along the line $y=3 x$. Therefore,
$$
\begin{aligned}
\frac{a+2 b}{2} & =3 \cdot \frac{a+b}{2} \\
a+2 b & =3 a+3 b \\
-2 a & =b
\end{aligned}
$$
Similarly, the midpoint of $(b, 2 b)$ and $(c, 3 c)$, which is $\left(\frac{b+c}{2}, \frac{2 b+3 c}{2}\right)$, must lie along the line $y=x$. Therefore,
$$
\begin{aligned}
\frac{2 b+3 c}{2} & =\frac{b+c}{2} \\
2 b+3 c & =b+c \\
b & =-2 c \\
c & =-\frac{1}{2} b=a
\end{aligned}
$$
From this, three points can be represented as $(a, a),(-2 a,-4 a)$, and $(a, 3 a)$. Using the distance formula, the three side lengths of the triangle are $2|a|, \sqrt{34}|a|$, and $\sqrt{58}|a|$. Since the perimeter of the triangle is 1 , we find that $|a|=\frac{1}{2+\sqrt{34}+\sqrt{58}}$ and therefore the longest side length is $\frac{\sqrt{58}}{2+\sqrt{34}+\sqrt{58}}$.
|
\frac{\sqrt{58}}{2+\sqrt{34}+\sqrt{58}}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
The lines $y=x, y=2 x$, and $y=3 x$ are the three medians of a triangle with perimeter 1. Find the length of the longest side of the triangle.
|
| $\frac{\sqrt{58}}{2+\sqrt{34}+\sqrt{58}}$ |
| :---: | The three medians of a triangle contain its vertices, so the three vertices of the triangle are $(a, a),(b, 2 b)$ and $(c, 3 c)$ for some $a, b$, and $c$. Then, the midpoint of $(a, a)$ and $(b, 2 b)$, which is $\left(\frac{a+b}{2}, \frac{a+2 b}{2}\right)$, must lie along the line $y=3 x$. Therefore,
$$
\begin{aligned}
\frac{a+2 b}{2} & =3 \cdot \frac{a+b}{2} \\
a+2 b & =3 a+3 b \\
-2 a & =b
\end{aligned}
$$
Similarly, the midpoint of $(b, 2 b)$ and $(c, 3 c)$, which is $\left(\frac{b+c}{2}, \frac{2 b+3 c}{2}\right)$, must lie along the line $y=x$. Therefore,
$$
\begin{aligned}
\frac{2 b+3 c}{2} & =\frac{b+c}{2} \\
2 b+3 c & =b+c \\
b & =-2 c \\
c & =-\frac{1}{2} b=a
\end{aligned}
$$
From this, three points can be represented as $(a, a),(-2 a,-4 a)$, and $(a, 3 a)$. Using the distance formula, the three side lengths of the triangle are $2|a|, \sqrt{34}|a|$, and $\sqrt{58}|a|$. Since the perimeter of the triangle is 1 , we find that $|a|=\frac{1}{2+\sqrt{34}+\sqrt{58}}$ and therefore the longest side length is $\frac{\sqrt{58}}{2+\sqrt{34}+\sqrt{58}}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n17. [11]",
"solution_match": "\nAnswer: "
}
|
956c0548-ff1b-5ba1-b6c5-fe205be2da20
| 609,114
|
Define the sequence of positive integers $\left\{a_{n}\right\}$ as follows. Let $a_{1}=1, a_{2}=3$, and for each $n>2$, let $a_{n}$ be the result of expressing $a_{n-1}$ in base $n-1$, then reading the resulting numeral in base $n$, then adding 2 (in base $n$ ). For example, $a_{2}=3_{10}=11_{2}$, so $a_{3}=11_{3}+2_{3}=6_{10}$. Express $a_{2013}$ in base ten.
|
23097 We claim that for nonnegative integers $m$ and for $0 \leq n<3 \cdot 2^{m}, a_{3 \cdot 2^{m}+n}=$ $\left(3 \cdot 2^{m}+n\right)(m+2)+2 n$. We will prove this by induction; the base case for $a_{3}=6$ (when $m=0$, $n=0$ ) is given in the problem statement. Now, suppose that this is true for some pair $m$ and $n$. We will divide this into two cases:
- Case 1: $n<3 \cdot 2^{m}-1$. Then, we want to prove that this is true for $m$ and $n+1$. In particular, writing $a_{3 \cdot 2^{m}+n}$ in base $3 \cdot 2^{m}+n$ results in the digits $m+2$ and $2 n$. Consequently, reading it in base $3 \cdot 2^{m}+n+1$ gives $a_{3 \cdot 2^{m}+n+1}=2+\left(3 \cdot 2^{m}+n+1\right)(m+2)+2 n=\left(2 \cdot 2^{m}+n+1\right)(m+2)+2(n+1)$, as desired.
- Case 2: $n=3 \cdot 2^{m}-1$. Then, we want to prove that this is true for $m+1$ and 0 . Similarly to the previous case, we get that $a_{3 \cdot 2^{m}+n+1}=a_{3 \cdot 2^{m+1}}=2+\left(3 \cdot 2^{m}+n+1\right)(m+2)+2 n=$ $2+\left(3 \cdot 2^{m+1}\right)(m+2)+2\left(3 \cdot 2^{m}-1\right)=\left(3 \cdot 2^{m+1}+0\right)((m+1)+2)+2(0)$, as desired.
In both cases, we have proved our claim.
|
23097
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Define the sequence of positive integers $\left\{a_{n}\right\}$ as follows. Let $a_{1}=1, a_{2}=3$, and for each $n>2$, let $a_{n}$ be the result of expressing $a_{n-1}$ in base $n-1$, then reading the resulting numeral in base $n$, then adding 2 (in base $n$ ). For example, $a_{2}=3_{10}=11_{2}$, so $a_{3}=11_{3}+2_{3}=6_{10}$. Express $a_{2013}$ in base ten.
|
23097 We claim that for nonnegative integers $m$ and for $0 \leq n<3 \cdot 2^{m}, a_{3 \cdot 2^{m}+n}=$ $\left(3 \cdot 2^{m}+n\right)(m+2)+2 n$. We will prove this by induction; the base case for $a_{3}=6$ (when $m=0$, $n=0$ ) is given in the problem statement. Now, suppose that this is true for some pair $m$ and $n$. We will divide this into two cases:
- Case 1: $n<3 \cdot 2^{m}-1$. Then, we want to prove that this is true for $m$ and $n+1$. In particular, writing $a_{3 \cdot 2^{m}+n}$ in base $3 \cdot 2^{m}+n$ results in the digits $m+2$ and $2 n$. Consequently, reading it in base $3 \cdot 2^{m}+n+1$ gives $a_{3 \cdot 2^{m}+n+1}=2+\left(3 \cdot 2^{m}+n+1\right)(m+2)+2 n=\left(2 \cdot 2^{m}+n+1\right)(m+2)+2(n+1)$, as desired.
- Case 2: $n=3 \cdot 2^{m}-1$. Then, we want to prove that this is true for $m+1$ and 0 . Similarly to the previous case, we get that $a_{3 \cdot 2^{m}+n+1}=a_{3 \cdot 2^{m+1}}=2+\left(3 \cdot 2^{m}+n+1\right)(m+2)+2 n=$ $2+\left(3 \cdot 2^{m+1}\right)(m+2)+2\left(3 \cdot 2^{m}-1\right)=\left(3 \cdot 2^{m+1}+0\right)((m+1)+2)+2(0)$, as desired.
In both cases, we have proved our claim.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n18. [11]",
"solution_match": "\nAnswer: "
}
|
eaf102dc-4b01-556b-bb77-4a4f3feb5c32
| 609,115
|
An isosceles trapezoid $A B C D$ with bases $A B$ and $C D$ has $A B=13, C D=17$, and height 3 . Let $E$ be the intersection of $A C$ and $B D$. Circles $\Omega$ and $\omega$ are circumscribed about triangles $A B E$ and $C D E$. Compute the sum of the radii of $\Omega$ and $\omega$.
|
$\quad 39$ Let $\Omega$ have center $O$ and radius $R$ and let $\omega$ have center $P$ and radius $M$. Let $Q$ be the intersection of $A B$ and $O E$. Note that $O E$ is the perpendicular bisector of $A B$ because the trapezoid is isosceles. Also, we see $O E$ is the circumradius of $\Omega$. On the other hand, we know by similarity of $\triangle A E B$ and $\triangle C E D$ that $Q E=\frac{13}{13+17} \cdot 3=\frac{13}{30} \cdot 3$. And, because $B Q=13 / 2$ and is perpendicular to $O Q$, we can apply the Pythagorean theorem to $\triangle O Q B$ to see $O Q=\sqrt{R^{2}-\left(\frac{13}{2}\right)^{2}}$. Since $O E=O Q+Q E, R=\frac{13}{30} \cdot 3+\sqrt{R^{2}-\left(\frac{13}{2}\right)^{2}}$. Solving this equation for $R$ yields $R=\frac{13}{30} \cdot 39$. Since by similarity $M=\frac{17}{13} R$, we know $R+M=\frac{30}{13} R$, so $R+M=39$.
|
39
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
An isosceles trapezoid $A B C D$ with bases $A B$ and $C D$ has $A B=13, C D=17$, and height 3 . Let $E$ be the intersection of $A C$ and $B D$. Circles $\Omega$ and $\omega$ are circumscribed about triangles $A B E$ and $C D E$. Compute the sum of the radii of $\Omega$ and $\omega$.
|
$\quad 39$ Let $\Omega$ have center $O$ and radius $R$ and let $\omega$ have center $P$ and radius $M$. Let $Q$ be the intersection of $A B$ and $O E$. Note that $O E$ is the perpendicular bisector of $A B$ because the trapezoid is isosceles. Also, we see $O E$ is the circumradius of $\Omega$. On the other hand, we know by similarity of $\triangle A E B$ and $\triangle C E D$ that $Q E=\frac{13}{13+17} \cdot 3=\frac{13}{30} \cdot 3$. And, because $B Q=13 / 2$ and is perpendicular to $O Q$, we can apply the Pythagorean theorem to $\triangle O Q B$ to see $O Q=\sqrt{R^{2}-\left(\frac{13}{2}\right)^{2}}$. Since $O E=O Q+Q E, R=\frac{13}{30} \cdot 3+\sqrt{R^{2}-\left(\frac{13}{2}\right)^{2}}$. Solving this equation for $R$ yields $R=\frac{13}{30} \cdot 39$. Since by similarity $M=\frac{17}{13} R$, we know $R+M=\frac{30}{13} R$, so $R+M=39$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n19. [11]",
"solution_match": "\nAnswer: "
}
|
bf28cea9-8647-572e-bd39-9f95e190c0d2
| 609,116
|
The polynomial $f(x)=x^{3}-3 x^{2}-4 x+4$ has three real roots $r_{1}, r_{2}$, and $r_{3}$. Let $g(x)=$ $x^{3}+a x^{2}+b x+c$ be the polynomial which has roots $s_{1}, s_{2}$, and $s_{3}$, where $s_{1}=r_{1}+r_{2} z+r_{3} z^{2}$, $s_{2}=r_{1} z+r_{2} z^{2}+r_{3}, s_{3}=r_{1} z^{2}+r_{2}+r_{3} z$, and $z=\frac{-1+i \sqrt{3}}{2}$. Find the real part of the sum of the coefficients of $g(x)$.
|
-26 Note that $z=e^{\frac{2 \pi}{3} i}=\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}$, so that $z^{3}=1$ and $z^{2}+z+1=0$. Also, $s_{2}=s_{1} z$ and $s_{3}=s_{1} z^{2}$.
Then, the sum of the coefficients of $g(x)$ is $g(1)=\left(1-s_{1}\right)\left(1-s_{2}\right)\left(1-s_{3}\right)=\left(1-s_{1}\right)\left(1-s_{1} z\right)\left(1-s_{1} z^{2}\right)=$ $1-\left(1+z+z^{2}\right) s_{1}+\left(z+z^{2}+z^{3}\right) s_{1}^{2}-z^{3} s_{1}^{3}=1-s_{1}^{3}$.
Meanwhile, $s_{1}^{3}=\left(r_{1}+r_{2} z+r_{3} z^{2}\right)^{3}=r_{1}^{3}+r_{2}^{3}+r_{3}^{3}+3 r_{1}^{2} r_{2} z+3 r_{1}^{2} r_{3} z^{2}+3 r_{2}^{2} r_{3} z+3 r_{2}^{2} r_{1} z^{2}+3 r_{3}^{2} r_{1} z+$ $3 r_{3}^{2} r_{2} z^{2}+6 r_{1} r_{2} r_{3}$.
Since the real parts of both $z$ and $z^{2}$ are $-\frac{1}{2}$, and since all of $r_{1}, r_{2}$, and $r_{3}$ are real, the real part of $s_{1}^{3}$ is $r_{1}^{3}+r_{2}^{3}+r_{3}^{3}-\frac{3}{2}\left(r_{1}^{2} r_{2}+\cdots+r_{3}^{2} r_{2}\right)+6 r_{1} r_{2} r_{3}=\left(r_{1}+r_{2}+r_{3}\right)^{3}-\frac{9}{2}\left(r_{1}+r_{2}+r_{3}\right)\left(r_{1} r_{2}+r_{2} r_{3}+r_{3} r_{1}\right)+\frac{27}{2} r_{1} r_{2} r_{3}=$ $3^{3}-\frac{9}{2} \cdot 3 \cdot-4+\frac{27}{2} \cdot-4=27$.
Therefore, the answer is $1-27=-26$.
|
-26
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
The polynomial $f(x)=x^{3}-3 x^{2}-4 x+4$ has three real roots $r_{1}, r_{2}$, and $r_{3}$. Let $g(x)=$ $x^{3}+a x^{2}+b x+c$ be the polynomial which has roots $s_{1}, s_{2}$, and $s_{3}$, where $s_{1}=r_{1}+r_{2} z+r_{3} z^{2}$, $s_{2}=r_{1} z+r_{2} z^{2}+r_{3}, s_{3}=r_{1} z^{2}+r_{2}+r_{3} z$, and $z=\frac{-1+i \sqrt{3}}{2}$. Find the real part of the sum of the coefficients of $g(x)$.
|
-26 Note that $z=e^{\frac{2 \pi}{3} i}=\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}$, so that $z^{3}=1$ and $z^{2}+z+1=0$. Also, $s_{2}=s_{1} z$ and $s_{3}=s_{1} z^{2}$.
Then, the sum of the coefficients of $g(x)$ is $g(1)=\left(1-s_{1}\right)\left(1-s_{2}\right)\left(1-s_{3}\right)=\left(1-s_{1}\right)\left(1-s_{1} z\right)\left(1-s_{1} z^{2}\right)=$ $1-\left(1+z+z^{2}\right) s_{1}+\left(z+z^{2}+z^{3}\right) s_{1}^{2}-z^{3} s_{1}^{3}=1-s_{1}^{3}$.
Meanwhile, $s_{1}^{3}=\left(r_{1}+r_{2} z+r_{3} z^{2}\right)^{3}=r_{1}^{3}+r_{2}^{3}+r_{3}^{3}+3 r_{1}^{2} r_{2} z+3 r_{1}^{2} r_{3} z^{2}+3 r_{2}^{2} r_{3} z+3 r_{2}^{2} r_{1} z^{2}+3 r_{3}^{2} r_{1} z+$ $3 r_{3}^{2} r_{2} z^{2}+6 r_{1} r_{2} r_{3}$.
Since the real parts of both $z$ and $z^{2}$ are $-\frac{1}{2}$, and since all of $r_{1}, r_{2}$, and $r_{3}$ are real, the real part of $s_{1}^{3}$ is $r_{1}^{3}+r_{2}^{3}+r_{3}^{3}-\frac{3}{2}\left(r_{1}^{2} r_{2}+\cdots+r_{3}^{2} r_{2}\right)+6 r_{1} r_{2} r_{3}=\left(r_{1}+r_{2}+r_{3}\right)^{3}-\frac{9}{2}\left(r_{1}+r_{2}+r_{3}\right)\left(r_{1} r_{2}+r_{2} r_{3}+r_{3} r_{1}\right)+\frac{27}{2} r_{1} r_{2} r_{3}=$ $3^{3}-\frac{9}{2} \cdot 3 \cdot-4+\frac{27}{2} \cdot-4=27$.
Therefore, the answer is $1-27=-26$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n20. [11]",
"solution_match": "\nAnswer: "
}
|
6d14024c-a5d3-5197-950c-8853f1ee07a9
| 609,117
|
Find the number of positive integers $j \leq 3^{2013}$ such that
$$
j=\sum_{k=0}^{m}\left((-1)^{k} \cdot 3^{a_{k}}\right)
$$
for some strictly increasing sequence of nonnegative integers $\left\{a_{k}\right\}$. For example, we may write $3=3^{1}$ and $55=3^{0}-3^{3}+3^{4}$, but 4 cannot be written in this form.
|
$2^{2013}$ Clearly $m$ must be even, or the sum would be negative. Furthermore, if $a_{m} \leq 2013$, the sum cannot exceed $3^{2013}$ since $j=3^{a_{m}}+\sum_{k=0}^{m-1}\left((-1)^{k} \cdot 3^{a_{k}}\right) \leq 3^{a_{m}}$. Likewise, if $a_{m}>2013$, then the sum necessarily exceeds $3^{2013}$, which is not hard to see by applying the Triangle Inequality and summing a geometric series. Hence, the elements of $\left\{a_{k}\right\}$ can be any subset of $\{0,1, \ldots, 2013\}$ with an odd number of elements. Since the number of even-sized subsets is equal to the number of odd-sized elements, there are $\frac{2^{2014}}{2}=2^{2013}$ such subsets.
Now, it suffices to show that given such an $\left\{a_{k}\right\}$, the value of $j$ can only be obtained in this way. Suppose for the the sake of contradiction that there exist two such sequences $\left\{a_{k}\right\}_{0 \leq k \leq m_{a}}$ and $\left\{b_{k}\right\}_{0 \leq k \leq m_{b}}$ which produce the same value of $j$ for $j$ positive or negative, where we choose $\left\{a_{k}\right\},\left\{b_{k}\right\}$ such that $\min \left(m_{a}, m_{b}\right)$ is as small as possible. Then, we note that since $3^{a_{0}}+3^{a_{1}}+\ldots+3^{\left(a_{\left.m_{a}-1\right)}\right.} \leq 3^{0}+3^{1}+$ $\ldots+3^{\left(a_{m_{a}-1}\right)}<2\left(3^{\left(a_{m_{a}}-1\right)}\right)$, we have that $\sum_{k=0}^{m_{a}}\left((-1)^{k} \cdot 3^{a_{k}}\right)>3^{\left(a_{\left.m_{a}-1\right)}\right)}$. Similarly, we get that $3^{\left(a_{m_{b}}-1\right)} \geq \sum_{k=0}^{m_{b}}\left((-1)^{k} \cdot 3^{a_{k}}\right)>3^{\left(m_{b}-1\right)}$; for the two to be equal, we must have $m_{a}=m_{b}$. However, this means that the sequences obtained by removing $a_{m_{a}}$ and $a_{m_{b}}$ from $\left\{a_{k}\right\}\left\{b_{k}\right\}$ have smaller maximum value but still produce the same alternating sum, contradicting our original assumption.
|
2^{2013}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the number of positive integers $j \leq 3^{2013}$ such that
$$
j=\sum_{k=0}^{m}\left((-1)^{k} \cdot 3^{a_{k}}\right)
$$
for some strictly increasing sequence of nonnegative integers $\left\{a_{k}\right\}$. For example, we may write $3=3^{1}$ and $55=3^{0}-3^{3}+3^{4}$, but 4 cannot be written in this form.
|
$2^{2013}$ Clearly $m$ must be even, or the sum would be negative. Furthermore, if $a_{m} \leq 2013$, the sum cannot exceed $3^{2013}$ since $j=3^{a_{m}}+\sum_{k=0}^{m-1}\left((-1)^{k} \cdot 3^{a_{k}}\right) \leq 3^{a_{m}}$. Likewise, if $a_{m}>2013$, then the sum necessarily exceeds $3^{2013}$, which is not hard to see by applying the Triangle Inequality and summing a geometric series. Hence, the elements of $\left\{a_{k}\right\}$ can be any subset of $\{0,1, \ldots, 2013\}$ with an odd number of elements. Since the number of even-sized subsets is equal to the number of odd-sized elements, there are $\frac{2^{2014}}{2}=2^{2013}$ such subsets.
Now, it suffices to show that given such an $\left\{a_{k}\right\}$, the value of $j$ can only be obtained in this way. Suppose for the the sake of contradiction that there exist two such sequences $\left\{a_{k}\right\}_{0 \leq k \leq m_{a}}$ and $\left\{b_{k}\right\}_{0 \leq k \leq m_{b}}$ which produce the same value of $j$ for $j$ positive or negative, where we choose $\left\{a_{k}\right\},\left\{b_{k}\right\}$ such that $\min \left(m_{a}, m_{b}\right)$ is as small as possible. Then, we note that since $3^{a_{0}}+3^{a_{1}}+\ldots+3^{\left(a_{\left.m_{a}-1\right)}\right.} \leq 3^{0}+3^{1}+$ $\ldots+3^{\left(a_{m_{a}-1}\right)}<2\left(3^{\left(a_{m_{a}}-1\right)}\right)$, we have that $\sum_{k=0}^{m_{a}}\left((-1)^{k} \cdot 3^{a_{k}}\right)>3^{\left(a_{\left.m_{a}-1\right)}\right)}$. Similarly, we get that $3^{\left(a_{m_{b}}-1\right)} \geq \sum_{k=0}^{m_{b}}\left((-1)^{k} \cdot 3^{a_{k}}\right)>3^{\left(m_{b}-1\right)}$; for the two to be equal, we must have $m_{a}=m_{b}$. However, this means that the sequences obtained by removing $a_{m_{a}}$ and $a_{m_{b}}$ from $\left\{a_{k}\right\}\left\{b_{k}\right\}$ have smaller maximum value but still produce the same alternating sum, contradicting our original assumption.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n21. [14]",
"solution_match": "\nAnswer: "
}
|
67ef8dc9-d83c-5288-b9b0-740afb784ead
| 609,118
|
Sherry and Val are playing a game. Sherry has a deck containing 2011 red cards and 2012 black cards, shuffled randomly. Sherry flips these cards over one at a time, and before she flips each card over, Val guesses whether it is red or black. If Val guesses correctly, she wins 1 dollar; otherwise, she loses 1 dollar. In addition, Val must guess red exactly 2011 times. If Val plays optimally, what is her expected profit from this game?
|
$\frac{1}{4023}$ We will prove by induction on $r+b$ that the expected profit for guessing if there are $r$ red cards, $b$ black cards, and where $g$ guesses must be red, is equal to $(b-r)+\frac{2(r-b)}{(r+b)} g$. It is not difficult to check that this holds in the cases $(r, b, g)=(1,0,0),(0,1,0),(1,0,1),(0,1,1)$. Then, suppose that this is true as long as the number of cards is strictly less than $r+b$; we will prove that it also holds true when there are $r$ red and $b$ blue cards.
Let $f(r, b, g)$ be her expected profit under these conditions. If she guesses red, her expected profit is
$$
\frac{r}{r+b}(1+f(r-1, b, g-1))+\frac{b}{r+b}(-1+f(r, b-1, g-1))=(b-r)+\frac{2(r-b)}{(r+b)} g
$$
Similarly, if she guesses black, her expected profit is
$$
\frac{r}{r+b}(-1+f(r-1, b, g))+\frac{b}{r+b}(1+f(r, b-1, g))=(b-r)+\frac{2(r-b)}{(r+b)} g .
$$
Plugging in the our starting values gives an expected profit of $\frac{1}{4023}$.
|
\frac{1}{4023}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Sherry and Val are playing a game. Sherry has a deck containing 2011 red cards and 2012 black cards, shuffled randomly. Sherry flips these cards over one at a time, and before she flips each card over, Val guesses whether it is red or black. If Val guesses correctly, she wins 1 dollar; otherwise, she loses 1 dollar. In addition, Val must guess red exactly 2011 times. If Val plays optimally, what is her expected profit from this game?
|
$\frac{1}{4023}$ We will prove by induction on $r+b$ that the expected profit for guessing if there are $r$ red cards, $b$ black cards, and where $g$ guesses must be red, is equal to $(b-r)+\frac{2(r-b)}{(r+b)} g$. It is not difficult to check that this holds in the cases $(r, b, g)=(1,0,0),(0,1,0),(1,0,1),(0,1,1)$. Then, suppose that this is true as long as the number of cards is strictly less than $r+b$; we will prove that it also holds true when there are $r$ red and $b$ blue cards.
Let $f(r, b, g)$ be her expected profit under these conditions. If she guesses red, her expected profit is
$$
\frac{r}{r+b}(1+f(r-1, b, g-1))+\frac{b}{r+b}(-1+f(r, b-1, g-1))=(b-r)+\frac{2(r-b)}{(r+b)} g
$$
Similarly, if she guesses black, her expected profit is
$$
\frac{r}{r+b}(-1+f(r-1, b, g))+\frac{b}{r+b}(1+f(r, b-1, g))=(b-r)+\frac{2(r-b)}{(r+b)} g .
$$
Plugging in the our starting values gives an expected profit of $\frac{1}{4023}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n22. [14]",
"solution_match": "\nAnswer: "
}
|
a5b7378d-f21f-56e5-9503-107801407a54
| 609,119
|
Let $A B C D$ be a parallelogram with $A B=8, A D=11$, and $\angle B A D=60^{\circ}$. Let $X$ be on segment $C D$ with $C X / X D=1 / 3$ and $Y$ be on segment $A D$ with $A Y / Y D=1 / 2$. Let $Z$ be on segment $A B$ such that $A X, B Y$, and $D Z$ are concurrent. Determine the area of triangle $X Y Z$.
|
$\frac{19 \sqrt{3}}{2}$ Let $A X$ and $B D$ meet at $P$. We have $D P / P B=D X / A B=3 / 4$. Now, applying
Ceva's Theorem in triangle $A B D$, we see that
$$
\frac{A Z}{Z B}=\frac{D P}{P B} \cdot \frac{A Y}{Y D}=\frac{3}{4} \cdot \frac{1}{2}=\frac{3}{8}
$$
Now,
$$
\frac{[A Y Z]}{[A B C D]}=\frac{[A Y Z]}{2[A B D]}=\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{11}=\frac{1}{22},
$$
and similarly
$$
\frac{[D Y X]}{[A B C D]}=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4}=\frac{1}{4} .
$$
Also,
$$
\frac{[X C B Z]}{[A B C D]}=\frac{1}{2}\left(\frac{1}{4}+\frac{8}{11}\right)=\frac{43}{88}
$$
The area of $X Y Z$ is the rest of the fraction of the area of $A B C D$ not covered by the three above polygons, which by a straightforward calculation $19 / 88$ the area of $A B C D$, so our answer is
$$
8 \cdot 11 \cdot \sin 60^{\circ} \cdot \frac{19}{88}=\frac{19 \sqrt{3}}{2}
$$
|
\frac{19 \sqrt{3}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C D$ be a parallelogram with $A B=8, A D=11$, and $\angle B A D=60^{\circ}$. Let $X$ be on segment $C D$ with $C X / X D=1 / 3$ and $Y$ be on segment $A D$ with $A Y / Y D=1 / 2$. Let $Z$ be on segment $A B$ such that $A X, B Y$, and $D Z$ are concurrent. Determine the area of triangle $X Y Z$.
|
$\frac{19 \sqrt{3}}{2}$ Let $A X$ and $B D$ meet at $P$. We have $D P / P B=D X / A B=3 / 4$. Now, applying
Ceva's Theorem in triangle $A B D$, we see that
$$
\frac{A Z}{Z B}=\frac{D P}{P B} \cdot \frac{A Y}{Y D}=\frac{3}{4} \cdot \frac{1}{2}=\frac{3}{8}
$$
Now,
$$
\frac{[A Y Z]}{[A B C D]}=\frac{[A Y Z]}{2[A B D]}=\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{11}=\frac{1}{22},
$$
and similarly
$$
\frac{[D Y X]}{[A B C D]}=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4}=\frac{1}{4} .
$$
Also,
$$
\frac{[X C B Z]}{[A B C D]}=\frac{1}{2}\left(\frac{1}{4}+\frac{8}{11}\right)=\frac{43}{88}
$$
The area of $X Y Z$ is the rest of the fraction of the area of $A B C D$ not covered by the three above polygons, which by a straightforward calculation $19 / 88$ the area of $A B C D$, so our answer is
$$
8 \cdot 11 \cdot \sin 60^{\circ} \cdot \frac{19}{88}=\frac{19 \sqrt{3}}{2}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n23. [14]",
"solution_match": "\nAnswer: "
}
|
04ce83e9-038a-5be5-8271-6bc7dedd78ec
| 609,120
|
Given a point $p$ and a line segment $l$, let $d(p, l)$ be the distance between them. Let $A, B$, and $C$ be points in the plane such that $A B=6, B C=8, A C=10$. What is the area of the region in the $(x, y)$-plane formed by the ordered pairs $(x, y)$ such that there exists a point $P$ inside triangle $A B C$ with $d(P, A B)+x=d(P, B C)+y=d(P, A C)$ ?
|
$\quad \frac{288}{5}$. Place $A B C$ in the coordinate plane so that $A=(0,6), B=(0,0), C=(8,0)$. Consider a point $P=(a, b)$ inside triangle $A B C$. Clearly, $d(P, A B)=a, d(P, B C)=b$. Now, we see
that the area of triangle $A B C$ is $\frac{6.8}{2}=24$, but may also be computed by summing the areas of triangles $P A B, P B C, P C A$. The area of triangle $P A B$ is $\frac{6 \cdot a}{2}=3 a$, and similarly the area of triangle $P B C$ is $4 b$. Thus, it follows easily that $d(P, C A)=\frac{24-3 a-4 b}{5}$. Now, we have
$$
(x, y)=\left(\frac{24}{5}-\frac{8}{5} a-\frac{4}{b} b, \frac{24}{5}-\frac{3}{5} a-\frac{9}{5} b\right)
$$
The desired region is the set of $(x, y)$ obtained by those $(a, b)$ subject to the constraints $a \geq 0, b \geq$ $0,6 a+8 b \leq 48$.
Consequently, our region is the triangle whose vertices are obtained by evaluating $(x, y)$ at the vertices $(a, b)$ of the triangle. To see this, let $f(a, b)$ output the corresponding $(x, y)$ according to the above. Then, we can write every point $P$ in $A B C$ as $P=m(0,0)+n(0,6)+p(8,0)$ for some $m+n+p=1$. Then, $f(P)=m f(0,0)+n f(0,6)+p f(8,0)=m\left(\frac{24}{5}, \frac{24}{5}\right)+n(-8,0)+p(0,-6)$, so $f(P)$ ranges over the triangle with those three vertices.
Therefore, we need the area of the triangle with vertices $\left(\frac{24}{5}, \frac{24}{5}\right),(0,-6),(-8,0)$, which is easily computed (for example, using determinants) to be $\frac{288}{5}$.
|
\frac{288}{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Given a point $p$ and a line segment $l$, let $d(p, l)$ be the distance between them. Let $A, B$, and $C$ be points in the plane such that $A B=6, B C=8, A C=10$. What is the area of the region in the $(x, y)$-plane formed by the ordered pairs $(x, y)$ such that there exists a point $P$ inside triangle $A B C$ with $d(P, A B)+x=d(P, B C)+y=d(P, A C)$ ?
|
$\quad \frac{288}{5}$. Place $A B C$ in the coordinate plane so that $A=(0,6), B=(0,0), C=(8,0)$. Consider a point $P=(a, b)$ inside triangle $A B C$. Clearly, $d(P, A B)=a, d(P, B C)=b$. Now, we see
that the area of triangle $A B C$ is $\frac{6.8}{2}=24$, but may also be computed by summing the areas of triangles $P A B, P B C, P C A$. The area of triangle $P A B$ is $\frac{6 \cdot a}{2}=3 a$, and similarly the area of triangle $P B C$ is $4 b$. Thus, it follows easily that $d(P, C A)=\frac{24-3 a-4 b}{5}$. Now, we have
$$
(x, y)=\left(\frac{24}{5}-\frac{8}{5} a-\frac{4}{b} b, \frac{24}{5}-\frac{3}{5} a-\frac{9}{5} b\right)
$$
The desired region is the set of $(x, y)$ obtained by those $(a, b)$ subject to the constraints $a \geq 0, b \geq$ $0,6 a+8 b \leq 48$.
Consequently, our region is the triangle whose vertices are obtained by evaluating $(x, y)$ at the vertices $(a, b)$ of the triangle. To see this, let $f(a, b)$ output the corresponding $(x, y)$ according to the above. Then, we can write every point $P$ in $A B C$ as $P=m(0,0)+n(0,6)+p(8,0)$ for some $m+n+p=1$. Then, $f(P)=m f(0,0)+n f(0,6)+p f(8,0)=m\left(\frac{24}{5}, \frac{24}{5}\right)+n(-8,0)+p(0,-6)$, so $f(P)$ ranges over the triangle with those three vertices.
Therefore, we need the area of the triangle with vertices $\left(\frac{24}{5}, \frac{24}{5}\right),(0,-6),(-8,0)$, which is easily computed (for example, using determinants) to be $\frac{288}{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n24. [14]",
"solution_match": "\nAnswer: "
}
|
1eeb00de-a635-5488-86e0-71b37fc84476
| 609,121
|
The sequence $\left(z_{n}\right)$ of complex numbers satisfies the following properties:
- $z_{1}$ and $z_{2}$ are not real.
- $z_{n+2}=z_{n+1}^{2} z_{n}$ for all integers $n \geq 1$.
- $\frac{z_{n+3}}{z_{n}^{2}}$ is real for all integers $n \geq 1$.
- $\left|\frac{z_{3}}{z_{4}}\right|=\left|\frac{z_{4}}{z_{5}}\right|=2$.
Find the product of all possible values of $z_{1}$.
|
65536 All complex numbers can be expressed as $r(\cos \theta+i \sin \theta)=r e^{i \theta}$. Let $z_{n}$ be $r_{n} e^{i \theta_{n}}$.
$\frac{z_{n+3}}{z_{n}^{2}}=\frac{z_{n+2}^{2} z_{n+1}}{z_{n}^{2}}=\frac{z_{n+1}^{5} z_{n}^{2}}{z_{n}^{2}}=z_{n+1}^{5}$ is real for all $n \geq 1$, so $\theta_{n}=\frac{\pi k_{n}}{5}$ for all $n \geq 2$, where $k_{n}$ is an integer. $\theta_{1}+2 \theta_{2}=\theta_{3}$, so we may write $\theta_{1}=\frac{\pi k_{1}}{5}$ with $k_{1}$ an integer.
$\frac{r_{3}}{r_{4}}=\frac{r_{4}}{r_{5}} \Rightarrow r_{5}=\frac{r_{4}^{2}}{r_{3}}=r_{4}^{2} r_{3}$, so $r_{3}=1 . \frac{r_{3}}{r_{4}}=2 \Rightarrow r_{4}=\frac{1}{2}, r_{4}=r_{3}^{2} r_{2} \Rightarrow r_{2}=\frac{1}{2}$, and $r_{3}=r_{2}^{2} r_{1} \Rightarrow r_{1}=4$.
Therefore, the possible values of $z_{1}$ are the nonreal roots of the equation $x^{10}-4^{10}=0$, and the product of the eight possible values is $\frac{4^{10}}{4^{2}}=4^{8}=65536$. For these values of $z_{1}$, it is not difficult to construct a sequence which works, by choosing $z_{2}$ nonreal so that $\left|z_{2}\right|=\frac{1}{2}$.
|
65536
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
The sequence $\left(z_{n}\right)$ of complex numbers satisfies the following properties:
- $z_{1}$ and $z_{2}$ are not real.
- $z_{n+2}=z_{n+1}^{2} z_{n}$ for all integers $n \geq 1$.
- $\frac{z_{n+3}}{z_{n}^{2}}$ is real for all integers $n \geq 1$.
- $\left|\frac{z_{3}}{z_{4}}\right|=\left|\frac{z_{4}}{z_{5}}\right|=2$.
Find the product of all possible values of $z_{1}$.
|
65536 All complex numbers can be expressed as $r(\cos \theta+i \sin \theta)=r e^{i \theta}$. Let $z_{n}$ be $r_{n} e^{i \theta_{n}}$.
$\frac{z_{n+3}}{z_{n}^{2}}=\frac{z_{n+2}^{2} z_{n+1}}{z_{n}^{2}}=\frac{z_{n+1}^{5} z_{n}^{2}}{z_{n}^{2}}=z_{n+1}^{5}$ is real for all $n \geq 1$, so $\theta_{n}=\frac{\pi k_{n}}{5}$ for all $n \geq 2$, where $k_{n}$ is an integer. $\theta_{1}+2 \theta_{2}=\theta_{3}$, so we may write $\theta_{1}=\frac{\pi k_{1}}{5}$ with $k_{1}$ an integer.
$\frac{r_{3}}{r_{4}}=\frac{r_{4}}{r_{5}} \Rightarrow r_{5}=\frac{r_{4}^{2}}{r_{3}}=r_{4}^{2} r_{3}$, so $r_{3}=1 . \frac{r_{3}}{r_{4}}=2 \Rightarrow r_{4}=\frac{1}{2}, r_{4}=r_{3}^{2} r_{2} \Rightarrow r_{2}=\frac{1}{2}$, and $r_{3}=r_{2}^{2} r_{1} \Rightarrow r_{1}=4$.
Therefore, the possible values of $z_{1}$ are the nonreal roots of the equation $x^{10}-4^{10}=0$, and the product of the eight possible values is $\frac{4^{10}}{4^{2}}=4^{8}=65536$. For these values of $z_{1}$, it is not difficult to construct a sequence which works, by choosing $z_{2}$ nonreal so that $\left|z_{2}\right|=\frac{1}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n25. [17]",
"solution_match": "\nAnswer: "
}
|
f42c32b6-70cf-5bec-97c4-64ae2d58166b
| 609,122
|
Let $W$ be the hypercube $\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid 0 \leq x_{1}, x_{2}, x_{3}, x_{4} \leq 1\right\}$. The intersection of $W$ and a hyperplane parallel to $x_{1}+x_{2}+x_{3}+x_{4}=0$ is a non-degenerate 3 -dimensional polyhedron. What is the maximum number of faces of this polyhedron?
|
8 The number of faces in the polyhedron is equal to the number of distinct cells (3dimensional faces) of the hypercube whose interior the hyperplane intersects. However, it is possible to arrange the hyperplane such that it intersects all 8 cells. Namely, $x_{1}+x_{2}+x_{3}+x_{4}=\frac{3}{2}$ intersects all 8 cells because it passes through $\left(0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$ (which is on the cell $\left.x_{1}=0\right),\left(1, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}\right)$ (which is on the cell $x_{1}=1$ ), and the points of intersection with the other 6 cells can be found by permuting these quadruples.
|
8
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $W$ be the hypercube $\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid 0 \leq x_{1}, x_{2}, x_{3}, x_{4} \leq 1\right\}$. The intersection of $W$ and a hyperplane parallel to $x_{1}+x_{2}+x_{3}+x_{4}=0$ is a non-degenerate 3 -dimensional polyhedron. What is the maximum number of faces of this polyhedron?
|
8 The number of faces in the polyhedron is equal to the number of distinct cells (3dimensional faces) of the hypercube whose interior the hyperplane intersects. However, it is possible to arrange the hyperplane such that it intersects all 8 cells. Namely, $x_{1}+x_{2}+x_{3}+x_{4}=\frac{3}{2}$ intersects all 8 cells because it passes through $\left(0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$ (which is on the cell $\left.x_{1}=0\right),\left(1, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}\right)$ (which is on the cell $x_{1}=1$ ), and the points of intersection with the other 6 cells can be found by permuting these quadruples.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n27. [17]",
"solution_match": "\nAnswer: "
}
|
5684fccf-d20e-5ff7-a0c0-5650cf7ebe3c
| 609,124
|
Let $z_{0}+z_{1}+z_{2}+\cdots$ be an infinite complex geometric series such that $z_{0}=1$ and $z_{2013}=\frac{1}{2013^{2013}}$. Find the sum of all possible sums of this series.
|
$\frac{2013^{2014}}{2013^{2013}-1}$ Clearly, the possible common ratios are the 2013 roots $r_{1}, r_{2}, \ldots, r_{2013}$ of the equation $r^{2013}=\frac{1}{2013^{2013}}$. We want the sum of the values of $x_{n}=\frac{1}{1-r_{n}}$, so we consider the polynomial whose roots are $x_{1}, x_{2}, \ldots, x_{2013}$. It is easy to see that $\left(1-\frac{1}{x_{n}}\right)^{2013}=\frac{1}{2013^{2013}}$, so it follows that the $x_{n}$ are the roots of the polynomial equation $\frac{1}{2013^{2013}} x^{2013}-(x-1)^{2013}=0$. The leading coefficient of this polynomial is $\frac{1}{2013^{2013}}-1$, and it follows easily from the Binomial Theorem that the next coefficient is 2013, so our answer is, by Vieta's Formulae,
$$
-\frac{2013}{\frac{1}{2013^{2013}}-1}=\frac{2013^{2014}}{2013^{2013}-1}
$$
|
\frac{2013^{2014}}{2013^{2013}-1}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $z_{0}+z_{1}+z_{2}+\cdots$ be an infinite complex geometric series such that $z_{0}=1$ and $z_{2013}=\frac{1}{2013^{2013}}$. Find the sum of all possible sums of this series.
|
$\frac{2013^{2014}}{2013^{2013}-1}$ Clearly, the possible common ratios are the 2013 roots $r_{1}, r_{2}, \ldots, r_{2013}$ of the equation $r^{2013}=\frac{1}{2013^{2013}}$. We want the sum of the values of $x_{n}=\frac{1}{1-r_{n}}$, so we consider the polynomial whose roots are $x_{1}, x_{2}, \ldots, x_{2013}$. It is easy to see that $\left(1-\frac{1}{x_{n}}\right)^{2013}=\frac{1}{2013^{2013}}$, so it follows that the $x_{n}$ are the roots of the polynomial equation $\frac{1}{2013^{2013}} x^{2013}-(x-1)^{2013}=0$. The leading coefficient of this polynomial is $\frac{1}{2013^{2013}}-1$, and it follows easily from the Binomial Theorem that the next coefficient is 2013, so our answer is, by Vieta's Formulae,
$$
-\frac{2013}{\frac{1}{2013^{2013}}-1}=\frac{2013^{2014}}{2013^{2013}-1}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n28. [17]",
"solution_match": "\nAnswer: "
}
|
b33addd9-7fbe-5a2b-8f67-31ad6c94b809
| 609,125
|
How many positive integers $k$ are there such that
$$
\frac{k}{2013}(a+b)=\operatorname{lcm}(a, b)
$$
has a solution in positive integers $(a, b)$ ?
|
1006 First, we can let $h=\operatorname{gcd}(a, b)$ so that $(a, b)=(h A, h B)$ where $\operatorname{gcd}(A, B)=1$. Making these subtitutions yields $\frac{k}{2013}(h A+h B)=h A B$, so $k=\frac{2013 A B}{A+B}$. Because $A$ and $B$ are relatively prime, $A+B$ shares no common factors with neither $A$ nor $B$, so in order to have $k$ be an integer, $A+B$ must divide 2013, and since $A$ and $B$ are positive, $A+B>1$.
We first show that for different possible values of $A+B$, the values of $k$ generated are distinct. In particular, we need to show that $\frac{2013 A B}{A+B} \neq \frac{2013 A^{\prime} B^{\prime}}{A^{\prime}+B^{\prime}}$ whenever $A+B \neq A^{\prime}+B^{\prime}$. Assume that such an equality exists, and cross-multiplying yields $\stackrel{A+B}{A^{\prime}+B^{\prime}} B^{\prime}\left(A^{\prime}+B^{\prime}\right)=A^{\prime} B^{\prime}(A+B)$. Since $A B$ is relatively prime to $A+B$, we must have $A+B$ divide $A^{\prime}+B^{\prime}$. With a similar argument, we can show that $A^{\prime}+B^{\prime}$ must divide $A+B$, so $A+B=A^{\prime}+B^{\prime}$.
Now, we need to show that for the same denominator $A+B$, the values of $k$ generated are also distinct for some relatively prime non-ordered pair $(A, B)$. Let $n=A+B=C+D$. Assume that $\frac{2013 A B}{n}=\frac{2013 C D}{n}$, or equivalently, $A(n-A)=C(n-C)$. After some rearrangement, we have $(C+A)(C-A)=n(C-A)$ This implies that either $C=A$ or $C=n-A=B$. But in either case, $(C, D)$ is some permutation of $(A, B)$.
Our answer can therefore be obtained by summing up the totients of the factors of 2013 (excluding 1) and dividing by 2 since $(A, B)$ and $(B, A)$ correspond to the same $k$ value, so our answer is $\frac{2013-1}{2}=$ 1006.
Remark: It can be proven that the sum of the totients of all the factors of any positive integer $N$ equals $N$, but in this case, the sum of the totients can be computed by hand.
|
1006
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many positive integers $k$ are there such that
$$
\frac{k}{2013}(a+b)=\operatorname{lcm}(a, b)
$$
has a solution in positive integers $(a, b)$ ?
|
1006 First, we can let $h=\operatorname{gcd}(a, b)$ so that $(a, b)=(h A, h B)$ where $\operatorname{gcd}(A, B)=1$. Making these subtitutions yields $\frac{k}{2013}(h A+h B)=h A B$, so $k=\frac{2013 A B}{A+B}$. Because $A$ and $B$ are relatively prime, $A+B$ shares no common factors with neither $A$ nor $B$, so in order to have $k$ be an integer, $A+B$ must divide 2013, and since $A$ and $B$ are positive, $A+B>1$.
We first show that for different possible values of $A+B$, the values of $k$ generated are distinct. In particular, we need to show that $\frac{2013 A B}{A+B} \neq \frac{2013 A^{\prime} B^{\prime}}{A^{\prime}+B^{\prime}}$ whenever $A+B \neq A^{\prime}+B^{\prime}$. Assume that such an equality exists, and cross-multiplying yields $\stackrel{A+B}{A^{\prime}+B^{\prime}} B^{\prime}\left(A^{\prime}+B^{\prime}\right)=A^{\prime} B^{\prime}(A+B)$. Since $A B$ is relatively prime to $A+B$, we must have $A+B$ divide $A^{\prime}+B^{\prime}$. With a similar argument, we can show that $A^{\prime}+B^{\prime}$ must divide $A+B$, so $A+B=A^{\prime}+B^{\prime}$.
Now, we need to show that for the same denominator $A+B$, the values of $k$ generated are also distinct for some relatively prime non-ordered pair $(A, B)$. Let $n=A+B=C+D$. Assume that $\frac{2013 A B}{n}=\frac{2013 C D}{n}$, or equivalently, $A(n-A)=C(n-C)$. After some rearrangement, we have $(C+A)(C-A)=n(C-A)$ This implies that either $C=A$ or $C=n-A=B$. But in either case, $(C, D)$ is some permutation of $(A, B)$.
Our answer can therefore be obtained by summing up the totients of the factors of 2013 (excluding 1) and dividing by 2 since $(A, B)$ and $(B, A)$ correspond to the same $k$ value, so our answer is $\frac{2013-1}{2}=$ 1006.
Remark: It can be proven that the sum of the totients of all the factors of any positive integer $N$ equals $N$, but in this case, the sum of the totients can be computed by hand.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n30. [20]",
"solution_match": "\nAnswer: "
}
|
46675bb7-56d8-574e-aee2-0c88d7b54cd8
| 609,127
|
Let $A B C D$ be a quadrilateral inscribed in a unit circle with center $O$. Suppose that $\angle A O B=$ $\angle C O D=135^{\circ}, B C=1$. Let $B^{\prime}$ and $C^{\prime}$ be the reflections of $A$ across $B O$ and $C O$ respectively. Let $H_{1}$ and $H_{2}$ be the orthocenters of $A B^{\prime} C^{\prime}$ and $B C D$, respectively. If $M$ is the midpoint of $O H_{1}$, and $O^{\prime}$ is the reflection of $O$ about the midpoint of $M H_{2}$, compute $O O^{\prime}$.
|
$\frac{1}{4}(8-\sqrt{6}-3 \sqrt{2})$ Put the diagram on the complex plane with $O$ at the origin and $A$ at
- 1. Let $B$ have coordinate $b$ and $C$ have coordinate $c$. We obtain easily that $B^{\prime}$ is $b^{2}, C^{\prime}$ is $c^{2}$, and $D$ is $b c$. Therefore, $H_{1}$ is $1+b^{2}+c^{2}$ and $H_{2}$ is $b+c+b c$ (we have used the fact that for triangles on the unit circle, their orthocenter is the sum of the vertices). Finally, we have that $M$ is $\frac{1}{2}\left(1+b^{2}+c^{2}\right)$, so the reflection of $O$ about the midpoint of $M H_{2}$ is $\frac{1}{2}\left(1+b^{2}+c^{2}+2 b+2 c+2 b c\right)=\frac{1}{2}(b+c+1)^{2}$, so we just seek $\frac{1}{2}|b+c+1|^{2}$. But we know that $b=\operatorname{cis} 135^{\circ}$ and $c=\operatorname{cis} 195^{\circ}$, so we obtain that this value is $\frac{1}{4}(8-\sqrt{6}-3 \sqrt{2})$.
- 2. [20] For an even integer positive integer $n$ Kevin has a tape of length $4 n$ with marks at $-2 n,-2 n+$ $1, \ldots, 2 n-1,2 n$. He then randomly picks $n$ points in the set $-n,-n+1,-n+2, \ldots, n-1, n$, and places a stone on each of these points. We call a stone 'stuck' if it is on $2 n$ or $-2 n$, or either all the points to the right, or all the points to the left, all contain stones. Then, every minute, Kevin shifts the unstuck stones in the following manner:
- He picks an unstuck stone uniformly at random and then flips a fair coin.
- If the coin came up heads, he then moves that stone and every stone in the largest contiguous set containing that stone one point to the left. If the coin came up tails, he moves every stone in that set one point right instead.
- He repeats until all the stones are stuck.
Let $p_{k}$ be the probability that at the end of the process there are exactly $k$ stones in the right half. Evaluate
$$
\frac{p_{n-1}-p_{n-2}+p_{n-3}-\ldots+p_{3}-p_{2}+p_{1}}{p_{n-1}+p_{n-2}+p_{n-3}+\ldots+p_{3}+p_{2}+p_{1}}
$$
in terms of $n$.
|
\frac{1}{4}(8-\sqrt{6}-3 \sqrt{2})
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C D$ be a quadrilateral inscribed in a unit circle with center $O$. Suppose that $\angle A O B=$ $\angle C O D=135^{\circ}, B C=1$. Let $B^{\prime}$ and $C^{\prime}$ be the reflections of $A$ across $B O$ and $C O$ respectively. Let $H_{1}$ and $H_{2}$ be the orthocenters of $A B^{\prime} C^{\prime}$ and $B C D$, respectively. If $M$ is the midpoint of $O H_{1}$, and $O^{\prime}$ is the reflection of $O$ about the midpoint of $M H_{2}$, compute $O O^{\prime}$.
|
$\frac{1}{4}(8-\sqrt{6}-3 \sqrt{2})$ Put the diagram on the complex plane with $O$ at the origin and $A$ at
- 1. Let $B$ have coordinate $b$ and $C$ have coordinate $c$. We obtain easily that $B^{\prime}$ is $b^{2}, C^{\prime}$ is $c^{2}$, and $D$ is $b c$. Therefore, $H_{1}$ is $1+b^{2}+c^{2}$ and $H_{2}$ is $b+c+b c$ (we have used the fact that for triangles on the unit circle, their orthocenter is the sum of the vertices). Finally, we have that $M$ is $\frac{1}{2}\left(1+b^{2}+c^{2}\right)$, so the reflection of $O$ about the midpoint of $M H_{2}$ is $\frac{1}{2}\left(1+b^{2}+c^{2}+2 b+2 c+2 b c\right)=\frac{1}{2}(b+c+1)^{2}$, so we just seek $\frac{1}{2}|b+c+1|^{2}$. But we know that $b=\operatorname{cis} 135^{\circ}$ and $c=\operatorname{cis} 195^{\circ}$, so we obtain that this value is $\frac{1}{4}(8-\sqrt{6}-3 \sqrt{2})$.
- 2. [20] For an even integer positive integer $n$ Kevin has a tape of length $4 n$ with marks at $-2 n,-2 n+$ $1, \ldots, 2 n-1,2 n$. He then randomly picks $n$ points in the set $-n,-n+1,-n+2, \ldots, n-1, n$, and places a stone on each of these points. We call a stone 'stuck' if it is on $2 n$ or $-2 n$, or either all the points to the right, or all the points to the left, all contain stones. Then, every minute, Kevin shifts the unstuck stones in the following manner:
- He picks an unstuck stone uniformly at random and then flips a fair coin.
- If the coin came up heads, he then moves that stone and every stone in the largest contiguous set containing that stone one point to the left. If the coin came up tails, he moves every stone in that set one point right instead.
- He repeats until all the stones are stuck.
Let $p_{k}$ be the probability that at the end of the process there are exactly $k$ stones in the right half. Evaluate
$$
\frac{p_{n-1}-p_{n-2}+p_{n-3}-\ldots+p_{3}-p_{2}+p_{1}}{p_{n-1}+p_{n-2}+p_{n-3}+\ldots+p_{3}+p_{2}+p_{1}}
$$
in terms of $n$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n31. [20]",
"solution_match": "\nAnswer: "
}
|
e86f5b0d-fbd1-56a3-b513-e23cd402c7bc
| 609,128
|
Compute the value of $1^{25}+2^{24}+3^{23}+\ldots+24^{2}+25^{1}$. If your answer is $A$ and the correct answer is $C$, then your score on this problem will be $\left\lfloor 25 \min \left(\left(\frac{A}{C}\right)^{2},\left(\frac{C}{A}\right)^{2}\right)\right\rfloor$.
|
66071772829247409 The sum is extremely unimodal, so we want to approximate it using its largest term. Taking logs of each term, we see that the max occurs when $(26-n) \log n$ peaks, and taking derivatives gives
$x+x \log x=26$
From here it's easy to see that the answer is around 10 , and slightly less (it's actually about 8.3 , but in any case it's hard to find powers of anything except 10). Thus the largest term will be something like $10^{16}$, which is already an order of magnitude within the desired answer $6.6 \times 10^{16}$.
To do better we'd really need to understand the behavior of the function $x^{26-x}$, but what approximately happens is that only the four or five largest terms in the sum are of any substantial size; thus it is reasonable here to pick some constant from 4 to 20 to multiply our guess $10^{16}$; any guess between $4.0 \times 10^{16}$ and $2.0 \times 10^{17}$ is reasonable.
|
66071772829247409
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Compute the value of $1^{25}+2^{24}+3^{23}+\ldots+24^{2}+25^{1}$. If your answer is $A$ and the correct answer is $C$, then your score on this problem will be $\left\lfloor 25 \min \left(\left(\frac{A}{C}\right)^{2},\left(\frac{C}{A}\right)^{2}\right)\right\rfloor$.
|
66071772829247409 The sum is extremely unimodal, so we want to approximate it using its largest term. Taking logs of each term, we see that the max occurs when $(26-n) \log n$ peaks, and taking derivatives gives
$x+x \log x=26$
From here it's easy to see that the answer is around 10 , and slightly less (it's actually about 8.3 , but in any case it's hard to find powers of anything except 10). Thus the largest term will be something like $10^{16}$, which is already an order of magnitude within the desired answer $6.6 \times 10^{16}$.
To do better we'd really need to understand the behavior of the function $x^{26-x}$, but what approximately happens is that only the four or five largest terms in the sum are of any substantial size; thus it is reasonable here to pick some constant from 4 to 20 to multiply our guess $10^{16}$; any guess between $4.0 \times 10^{16}$ and $2.0 \times 10^{17}$ is reasonable.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n33. [25]",
"solution_match": "\nAnswer: "
}
|
ebf4534e-bf7c-52ba-a31c-714b68130c3f
| 609,129
|
For how many unordered sets $\{a, b, c, d\}$ of positive integers, none of which exceed 168 , do there exist integers $w, x, y, z$ such that $(-1)^{w} a+(-1)^{x} b+(-1)^{y} c+(-1)^{z} d=168$ ? If your answer is $A$ and the correct answer is $C$, then your score on this problem will be $\left\lfloor 25 e^{-3 \frac{|C-A|}{C}}\right\rfloor$.
|
761474 As an approximation, we assume $a, b, c, d$ are ordered to begin with (so we have to divide by 24 later) and add to 168 with a unique choice of signs; then, it suffices to count $e+f+g+h=168$ with each $e, f, g, h$ in $[-168,168]$ and then divide by 24 (we drop the condition that none of them can be zero because it shouldn't affect the answer that much).
One way to do this is generating functions. We want the coefficient of $t^{168}$ in the generating function $\left(t^{-168}+t^{-167}+\ldots+t^{167}+t^{168}\right)^{4}=\left(t^{169}-t^{-168}\right)^{4} /(t-1)^{4}$
Clearing the negative powers, it suffices to find the coefficient of $t^{840}$ in
$\left(t^{337}-1\right)^{4} /(t-1)^{4}=\left(1-4 t^{337}+6 t^{674}-\ldots\right) \frac{1}{(t-1)^{4}}$.
To do this we expand the bottom as a power series in $t$ :
$\frac{1}{(t-1)^{4}}=\sum_{n \geq 0}\binom{n+3}{3} t^{n}$
It remains to calculate $\binom{840+3}{3}-4 \cdot\binom{840-337+3}{3}+6 \cdot\binom{840-674+3}{3}$. This is almost exactly equal to $\frac{1}{6}\left(843^{3}-4 \cdot 506^{3}+6 \cdot 169^{3}\right) \approx 1.83 \times 10^{7}$.
Dividing by 24 , we arrive at an estimation 762500 . Even if we use a bad approximation $\frac{1}{6 \cdot 24}\left(850^{3}-4\right.$. $500^{3}+6 \cdot 150^{3}$ ) we get approximately 933000 , which is fairly close to the answer.
|
761474
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
For how many unordered sets $\{a, b, c, d\}$ of positive integers, none of which exceed 168 , do there exist integers $w, x, y, z$ such that $(-1)^{w} a+(-1)^{x} b+(-1)^{y} c+(-1)^{z} d=168$ ? If your answer is $A$ and the correct answer is $C$, then your score on this problem will be $\left\lfloor 25 e^{-3 \frac{|C-A|}{C}}\right\rfloor$.
|
761474 As an approximation, we assume $a, b, c, d$ are ordered to begin with (so we have to divide by 24 later) and add to 168 with a unique choice of signs; then, it suffices to count $e+f+g+h=168$ with each $e, f, g, h$ in $[-168,168]$ and then divide by 24 (we drop the condition that none of them can be zero because it shouldn't affect the answer that much).
One way to do this is generating functions. We want the coefficient of $t^{168}$ in the generating function $\left(t^{-168}+t^{-167}+\ldots+t^{167}+t^{168}\right)^{4}=\left(t^{169}-t^{-168}\right)^{4} /(t-1)^{4}$
Clearing the negative powers, it suffices to find the coefficient of $t^{840}$ in
$\left(t^{337}-1\right)^{4} /(t-1)^{4}=\left(1-4 t^{337}+6 t^{674}-\ldots\right) \frac{1}{(t-1)^{4}}$.
To do this we expand the bottom as a power series in $t$ :
$\frac{1}{(t-1)^{4}}=\sum_{n \geq 0}\binom{n+3}{3} t^{n}$
It remains to calculate $\binom{840+3}{3}-4 \cdot\binom{840-337+3}{3}+6 \cdot\binom{840-674+3}{3}$. This is almost exactly equal to $\frac{1}{6}\left(843^{3}-4 \cdot 506^{3}+6 \cdot 169^{3}\right) \approx 1.83 \times 10^{7}$.
Dividing by 24 , we arrive at an estimation 762500 . Even if we use a bad approximation $\frac{1}{6 \cdot 24}\left(850^{3}-4\right.$. $500^{3}+6 \cdot 150^{3}$ ) we get approximately 933000 , which is fairly close to the answer.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n34. [25]",
"solution_match": "\nAnswer: "
}
|
7f712a38-77e9-5c36-9004-1e46b3854240
| 609,130
|
Let $P$ be the number to partition 2013 into an ordered tuple of prime numbers? What is $\log _{2}(P)$ ? If your answer is $A$ and the correct answer is $C$, then your score on this problem will be $\left\lfloor\frac{125}{2}\left(\min \left(\frac{C}{A}, \frac{A}{C}\right)-\frac{3}{5}\right)\right\rfloor$ or zero, whichever is larger.
|
614.519... We use the following facts and heuristics.
(1) The ordered partitions of $n$ into any positive integers (not just primes) is $2^{n-1}$. This can be guessed by checking small cases and finding a pattern, and is not difficult to prove.
(2) The partitions of $\frac{2013}{n}$ into any positive integers equals the partitions of 2013 into integers from the set $\{n, 2 n, 3 n, \cdots\}$.
(3) The small numbers matter more when considering partitions.
(4) The set of primes $\{2,3,5,7, \cdots\}$ is close in size (near the small numbers) to $\{3,6,9, \cdots\}$ or $\{2,4,6, \cdots\}$.
(5) The prime numbers get very sparse compared to the above two sets in the larger numbers.
Thus, using these heuristics, the number of partitions of 2013 into primes is approximately $2^{\frac{2013}{3}-1}$ or $2^{\frac{2013}{2}-1}$, which, taking logarithms, give 670 and 1005.5 , respectively. By (5), we should estimate something that is slightly less than these numbers.
|
614.519
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $P$ be the number to partition 2013 into an ordered tuple of prime numbers? What is $\log _{2}(P)$ ? If your answer is $A$ and the correct answer is $C$, then your score on this problem will be $\left\lfloor\frac{125}{2}\left(\min \left(\frac{C}{A}, \frac{A}{C}\right)-\frac{3}{5}\right)\right\rfloor$ or zero, whichever is larger.
|
614.519... We use the following facts and heuristics.
(1) The ordered partitions of $n$ into any positive integers (not just primes) is $2^{n-1}$. This can be guessed by checking small cases and finding a pattern, and is not difficult to prove.
(2) The partitions of $\frac{2013}{n}$ into any positive integers equals the partitions of 2013 into integers from the set $\{n, 2 n, 3 n, \cdots\}$.
(3) The small numbers matter more when considering partitions.
(4) The set of primes $\{2,3,5,7, \cdots\}$ is close in size (near the small numbers) to $\{3,6,9, \cdots\}$ or $\{2,4,6, \cdots\}$.
(5) The prime numbers get very sparse compared to the above two sets in the larger numbers.
Thus, using these heuristics, the number of partitions of 2013 into primes is approximately $2^{\frac{2013}{3}-1}$ or $2^{\frac{2013}{2}-1}$, which, taking logarithms, give 670 and 1005.5 , respectively. By (5), we should estimate something that is slightly less than these numbers.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-guts-solutions.jsonl",
"problem_match": "\n35. [25]",
"solution_match": "\nAnswer: "
}
|
c98708b5-58f2-52a6-aa62-a68196c351d9
| 609,131
|
Let $a$ and $b$ be real numbers such that $\frac{a b}{a^{2}+b^{2}}=\frac{1}{4}$. Find all possible values of $\frac{\left|a^{2}-b^{2}\right|}{a^{2}+b^{2}}$.
|
$\frac{\sqrt{3}}{2}$ The hypothesis statement is equivalent to
$$
\begin{gathered}
a^{2}+b^{2}=4 a b \\
1:(a+b)^{2}=6 a b \\
2:(a-b)^{2}=2 a b
\end{gathered}
$$
Multiplying equations 1 and 2,
$$
\begin{aligned}
& \left(a^{2}-b^{2}\right)^{2}=12(a b)^{2} \\
& \left|a^{2}-b^{2}\right|= \pm \sqrt{12} a b
\end{aligned}
$$
Since the left hand side and $a b$ are both positive,
$$
\begin{gathered}
\left|a^{2}-b^{2}\right|=\sqrt{12} a b \\
\frac{\left|a^{2}-b^{2}\right|}{a^{2}+b^{2}}=\frac{\sqrt{12} a b}{a^{2}+b^{2}}=\frac{\sqrt{12}}{4}=\frac{\sqrt{3}}{2}
\end{gathered}
$$
(It is clear that such $a$ and $b$ exist: for example, we can take $a=1$ and solve for $b$ by way of the quadratic formula.)
|
\frac{\sqrt{3}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a$ and $b$ be real numbers such that $\frac{a b}{a^{2}+b^{2}}=\frac{1}{4}$. Find all possible values of $\frac{\left|a^{2}-b^{2}\right|}{a^{2}+b^{2}}$.
|
$\frac{\sqrt{3}}{2}$ The hypothesis statement is equivalent to
$$
\begin{gathered}
a^{2}+b^{2}=4 a b \\
1:(a+b)^{2}=6 a b \\
2:(a-b)^{2}=2 a b
\end{gathered}
$$
Multiplying equations 1 and 2,
$$
\begin{aligned}
& \left(a^{2}-b^{2}\right)^{2}=12(a b)^{2} \\
& \left|a^{2}-b^{2}\right|= \pm \sqrt{12} a b
\end{aligned}
$$
Since the left hand side and $a b$ are both positive,
$$
\begin{gathered}
\left|a^{2}-b^{2}\right|=\sqrt{12} a b \\
\frac{\left|a^{2}-b^{2}\right|}{a^{2}+b^{2}}=\frac{\sqrt{12} a b}{a^{2}+b^{2}}=\frac{\sqrt{12}}{4}=\frac{\sqrt{3}}{2}
\end{gathered}
$$
(It is clear that such $a$ and $b$ exist: for example, we can take $a=1$ and solve for $b$ by way of the quadratic formula.)
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-team-solutions.jsonl",
"problem_match": "\n1. [10]",
"solution_match": "\nAnswer: "
}
|
de09857b-4f6a-5022-8170-bbe64512e0a4
| 609,133
|
A cafe has 3 tables and 5 individual counter seats. People enter in groups of size between 1 and 4 , inclusive, and groups never share a table. A group of more than 1 will always try to sit at a table, but will sit in counter seats if no tables are available. Conversely, a group of 1 will always try to sit at the counter first. One morning, $M$ groups consisting of a total of $N$ people enter and sit down. Then, a single person walks in, and realizes that all the tables and counter seats are occupied by some person or group. What is the minimum possible value of $M+N$ ?
|
16 We first show that $M+N \geq 16$. Consider the point right before the last table is occupied. We have two cases: first suppose there exists at least one open counter seat. Then, every table must contribute at least 3 to the value of $M+N$, because no groups of 1 will have taken a table with one of the counter seats open. By the end, the counter must contribute at least $5+2=7$ to $M+N$, as there must be at least two groups sitting at the counter. It follows that $M+N \geq 16$. For the second case, assume the counter is full right before the last table is taken. Then, everybody sitting at the counter must have entered as a singleton, since they entered when a table was still available. Consequently, the counter must contribute 10 to $M+N$, and each table contributes at least 2 , so once again $M+N \geq 16$.
Now, $M+N=16$ is achievable with eight groups of one, who first fill the counter seats, then the three tables. Thus, our answer is 16 .
|
16
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A cafe has 3 tables and 5 individual counter seats. People enter in groups of size between 1 and 4 , inclusive, and groups never share a table. A group of more than 1 will always try to sit at a table, but will sit in counter seats if no tables are available. Conversely, a group of 1 will always try to sit at the counter first. One morning, $M$ groups consisting of a total of $N$ people enter and sit down. Then, a single person walks in, and realizes that all the tables and counter seats are occupied by some person or group. What is the minimum possible value of $M+N$ ?
|
16 We first show that $M+N \geq 16$. Consider the point right before the last table is occupied. We have two cases: first suppose there exists at least one open counter seat. Then, every table must contribute at least 3 to the value of $M+N$, because no groups of 1 will have taken a table with one of the counter seats open. By the end, the counter must contribute at least $5+2=7$ to $M+N$, as there must be at least two groups sitting at the counter. It follows that $M+N \geq 16$. For the second case, assume the counter is full right before the last table is taken. Then, everybody sitting at the counter must have entered as a singleton, since they entered when a table was still available. Consequently, the counter must contribute 10 to $M+N$, and each table contributes at least 2 , so once again $M+N \geq 16$.
Now, $M+N=16$ is achievable with eight groups of one, who first fill the counter seats, then the three tables. Thus, our answer is 16 .
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-team-solutions.jsonl",
"problem_match": "\n2. [15]",
"solution_match": "\nAnswer: "
}
|
f7f42159-e663-5db2-ace6-1ee99ac9eecf
| 609,134
|
Let $A B C$ be a triangle with circumcenter $O$ such that $A C=7$. Suppose that the circumcircle of $A O C$ is tangent to $B C$ at $C$ and intersects the line $A B$ at $A$ and $F$. Let $F O$ intersect $B C$ at $E$. Compute BE.
|
$E B=\frac{7}{2} \quad O$ is the circumcenter of $\triangle A B C \Longrightarrow A O=C O \Longrightarrow O C A=\angle O A C$. Because $A C$ is an inscribed arc of circumcircle $\triangle A O C, \angle O C A=\angle O F A$. Furthermore $B C$ is tangent to circumcircle $\triangle A O C$, so $\angle O A C=\angle O C B$. However, again using the fact that $O$ is the circumcenter of $\triangle A B C, \angle O C B=\angle O B C$.
We now have that $C O$ bisects $\angle A C B$, so it follows that triangle $C A=C B$. Also, by AA similarity we have $E O B \sim E B F$. Thus, $E B^{2}=E O \cdot E F=E C^{2}$ by the similarity and power of a point, so $E B=B C / 2=A C / 2=7 / 2$.
|
\frac{7}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a triangle with circumcenter $O$ such that $A C=7$. Suppose that the circumcircle of $A O C$ is tangent to $B C$ at $C$ and intersects the line $A B$ at $A$ and $F$. Let $F O$ intersect $B C$ at $E$. Compute BE.
|
$E B=\frac{7}{2} \quad O$ is the circumcenter of $\triangle A B C \Longrightarrow A O=C O \Longrightarrow O C A=\angle O A C$. Because $A C$ is an inscribed arc of circumcircle $\triangle A O C, \angle O C A=\angle O F A$. Furthermore $B C$ is tangent to circumcircle $\triangle A O C$, so $\angle O A C=\angle O C B$. However, again using the fact that $O$ is the circumcenter of $\triangle A B C, \angle O C B=\angle O B C$.
We now have that $C O$ bisects $\angle A C B$, so it follows that triangle $C A=C B$. Also, by AA similarity we have $E O B \sim E B F$. Thus, $E B^{2}=E O \cdot E F=E C^{2}$ by the similarity and power of a point, so $E B=B C / 2=A C / 2=7 / 2$.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-team-solutions.jsonl",
"problem_match": "\n3. [15]",
"solution_match": "\nAnswer: "
}
|
89c66519-03b3-5758-a6bd-0c96e8b29fd7
| 609,135
|
Thaddeus is given a $2013 \times 2013$ array of integers each between 1 and 2013 , inclusive. He is allowed two operations:
- 1. Choose a row, and subtract 1 from each entry.
- 2. Chooses a column, and add 1 to each entry.
He would like to get an array where all integers are divisible by 2013. On how many arrays is this possible?
|
$2013^{4025}$ We claim that the set of grids on which it is possible to obtain an array of all zeroes (mod 2013) is indexed by ordered 4025-tuples of residues (mod 2013), corresponding to the starting entries in the first row and first column of the grid, giving the answer of $2013^{4025}$. To do this, we show that given after fixing all of the entries in the first row and column, there is a unique starting grid which can become an array of all zeroes after applying the appropriate operations.
Let $a_{i, j}$ be the entry in the $i$-th row and the $j$-th column. Suppose there is a sequence of operations giving all zeroes in the array; let $r_{i}$ be the number of times we operate on row $i$, and let $c_{j}$ be the number of times we operate on column $j$. It is enough to take all of these values to be residues modulo 2013. Clearly, $a_{i, j}+r_{i}+c_{j} \equiv 0(\bmod 2013)$ for each $i, j$. In particular, $r_{1}+c_{1} \equiv a_{1,1}$. Now, for each $i, j$, we have
$$
\begin{aligned}
a_{i, j} & \equiv-r_{i}-c_{j} \\
& \equiv\left(a_{i, 1}+c_{1}\right)+\left(a_{1, j}+r_{1}\right) \\
& \equiv a_{i, 1}+a_{1, j}-a_{1,1},
\end{aligned}
$$
which is fixed. Thus, there rest of the entries in the grid are forced.
Conversely, if we set $a_{i, j}$ to be the appropriate representative of the residue class of $a_{i, 1}+a_{1, j}-a_{1,1}$ modulo 2013, we may take $r_{i} \equiv-a_{i, 1}(\bmod 2013)$, and $c_{j} \equiv a_{1,1}-a_{1, j}(\bmod 2013)$ for each $i, j$. It is clear that $a_{i, j}+r_{i}+c_{j} \equiv 0(\bmod 2013)$ for each $i, j$, so we're done.
|
2013^{4025}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Thaddeus is given a $2013 \times 2013$ array of integers each between 1 and 2013 , inclusive. He is allowed two operations:
- 1. Choose a row, and subtract 1 from each entry.
- 2. Chooses a column, and add 1 to each entry.
He would like to get an array where all integers are divisible by 2013. On how many arrays is this possible?
|
$2013^{4025}$ We claim that the set of grids on which it is possible to obtain an array of all zeroes (mod 2013) is indexed by ordered 4025-tuples of residues (mod 2013), corresponding to the starting entries in the first row and first column of the grid, giving the answer of $2013^{4025}$. To do this, we show that given after fixing all of the entries in the first row and column, there is a unique starting grid which can become an array of all zeroes after applying the appropriate operations.
Let $a_{i, j}$ be the entry in the $i$-th row and the $j$-th column. Suppose there is a sequence of operations giving all zeroes in the array; let $r_{i}$ be the number of times we operate on row $i$, and let $c_{j}$ be the number of times we operate on column $j$. It is enough to take all of these values to be residues modulo 2013. Clearly, $a_{i, j}+r_{i}+c_{j} \equiv 0(\bmod 2013)$ for each $i, j$. In particular, $r_{1}+c_{1} \equiv a_{1,1}$. Now, for each $i, j$, we have
$$
\begin{aligned}
a_{i, j} & \equiv-r_{i}-c_{j} \\
& \equiv\left(a_{i, 1}+c_{1}\right)+\left(a_{1, j}+r_{1}\right) \\
& \equiv a_{i, 1}+a_{1, j}-a_{1,1},
\end{aligned}
$$
which is fixed. Thus, there rest of the entries in the grid are forced.
Conversely, if we set $a_{i, j}$ to be the appropriate representative of the residue class of $a_{i, 1}+a_{1, j}-a_{1,1}$ modulo 2013, we may take $r_{i} \equiv-a_{i, 1}(\bmod 2013)$, and $c_{j} \equiv a_{1,1}-a_{1, j}(\bmod 2013)$ for each $i, j$. It is clear that $a_{i, j}+r_{i}+c_{j} \equiv 0(\bmod 2013)$ for each $i, j$, so we're done.
|
{
"resource_path": "HarvardMIT/segmented/en-162-2013-feb-team-solutions.jsonl",
"problem_match": "\n5. [25]",
"solution_match": "\nAnswer: "
}
|
1e6a7210-ee8d-5df2-b86c-36ba7f2163ea
| 609,137
|
What is the smallest non-square positive integer that is the product of four prime numbers (not necessarily distinct)?
|
$\quad 24$ The smallest two integers that are the product of four primes are $2^{4}=16$ and $2^{3} \cdot 3=24$. Since 16 is a perfect square and 24 is not, the answer is 24 .
|
24
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
What is the smallest non-square positive integer that is the product of four prime numbers (not necessarily distinct)?
|
$\quad 24$ The smallest two integers that are the product of four primes are $2^{4}=16$ and $2^{3} \cdot 3=24$. Since 16 is a perfect square and 24 is not, the answer is 24 .
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-gen-solutions.jsonl",
"problem_match": "\n1. [2]",
"solution_match": "\nAnswer: "
}
|
ac41d9ab-bc86-57cb-8cd0-e5ecf3a499e8
| 75,160
|
Plot points $A, B, C$ at coordinates $(0,0),(0,1)$, and $(1,1)$ in the plane, respectively. Let $S$ denote the union of the two line segments $A B$ and $B C$. Let $X_{1}$ be the area swept out when Bobby rotates $S$ counterclockwise 45 degrees about point $A$. Let $X_{2}$ be the area swept out when Calvin rotates $S$ clockwise 45 degrees about point $A$. Find $\frac{X_{1}+X_{2}}{2}$.
|
$\sqrt{\frac{\pi}{4}}$ It's easy to see $X_{1}=X_{2}$. Simple cutting and pasting shows that $X_{1}$ equals the area of $\frac{1}{8}$ of a circle with radius $A C=\sqrt{2}$, so $\frac{X_{1}+X_{2}}{2}=X_{1}=\frac{1}{8} \pi(\sqrt{2})^{2}=\frac{\pi}{4}$.
|
\frac{\pi}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Plot points $A, B, C$ at coordinates $(0,0),(0,1)$, and $(1,1)$ in the plane, respectively. Let $S$ denote the union of the two line segments $A B$ and $B C$. Let $X_{1}$ be the area swept out when Bobby rotates $S$ counterclockwise 45 degrees about point $A$. Let $X_{2}$ be the area swept out when Calvin rotates $S$ clockwise 45 degrees about point $A$. Find $\frac{X_{1}+X_{2}}{2}$.
|
$\sqrt{\frac{\pi}{4}}$ It's easy to see $X_{1}=X_{2}$. Simple cutting and pasting shows that $X_{1}$ equals the area of $\frac{1}{8}$ of a circle with radius $A C=\sqrt{2}$, so $\frac{X_{1}+X_{2}}{2}=X_{1}=\frac{1}{8} \pi(\sqrt{2})^{2}=\frac{\pi}{4}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-gen-solutions.jsonl",
"problem_match": "\n2. [3]",
"solution_match": "\nAnswer: "
}
|
bbea4678-6334-5e03-9b12-4b9ec7e298d8
| 609,148
|
A 24-hour digital clock shows times $h: m: s$, where $h, m$, and $s$ are integers with $0 \leq h \leq 23$, $0 \leq m \leq 59$, and $0 \leq s \leq 59$. How many times $h: m: s$ satisfy $h+m=s$ ?
|
1164 We are solving $h+m=s$ in $0 \leq s \leq 59,0 \leq m \leq 59$, and $0 \leq h \leq 23$. If $s \geq 24$, each $h$ corresponds to exactly 1 solution, so we get $24(59-23)=24(36)$ in this case. If $s \leq 23$, we want the number of nonnegative integer solutions to $h+m \leq 23$, which by lattice point counting (or balls and urns) is $\binom{23+2}{2}=(23+2)(23+1) / 2=25 \cdot 12$. Thus our total is $12(72+25)=12(100-3)=1164$.
|
1164
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A 24-hour digital clock shows times $h: m: s$, where $h, m$, and $s$ are integers with $0 \leq h \leq 23$, $0 \leq m \leq 59$, and $0 \leq s \leq 59$. How many times $h: m: s$ satisfy $h+m=s$ ?
|
1164 We are solving $h+m=s$ in $0 \leq s \leq 59,0 \leq m \leq 59$, and $0 \leq h \leq 23$. If $s \geq 24$, each $h$ corresponds to exactly 1 solution, so we get $24(59-23)=24(36)$ in this case. If $s \leq 23$, we want the number of nonnegative integer solutions to $h+m \leq 23$, which by lattice point counting (or balls and urns) is $\binom{23+2}{2}=(23+2)(23+1) / 2=25 \cdot 12$. Thus our total is $12(72+25)=12(100-3)=1164$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-gen-solutions.jsonl",
"problem_match": "\n3. [4]",
"solution_match": "\nAnswer: "
}
|
a964c2de-f71c-5fa9-b01b-a5b9e803b06a
| 609,149
|
A 50 -card deck consists of 4 cards labeled " $i$ " for $i=1,2, \ldots, 12$ and 2 cards labeled " 13 ". If Bob randomly chooses 2 cards from the deck without replacement, what is the probability that his 2 cards have the same label?
|
$\frac{73}{1225}$ All pairs of distinct cards (where we distinguish cards even with the same label) are equally likely. There are $\binom{2}{2}+12\binom{4}{2}=73$ pairs of cards with the same label and $\binom{50}{2}=100 \cdot \frac{49}{4}=1225$ pairs of cards overall, so the desired probability is $\frac{73}{1225}$.
|
\frac{73}{1225}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A 50 -card deck consists of 4 cards labeled " $i$ " for $i=1,2, \ldots, 12$ and 2 cards labeled " 13 ". If Bob randomly chooses 2 cards from the deck without replacement, what is the probability that his 2 cards have the same label?
|
$\frac{73}{1225}$ All pairs of distinct cards (where we distinguish cards even with the same label) are equally likely. There are $\binom{2}{2}+12\binom{4}{2}=73$ pairs of cards with the same label and $\binom{50}{2}=100 \cdot \frac{49}{4}=1225$ pairs of cards overall, so the desired probability is $\frac{73}{1225}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-gen-solutions.jsonl",
"problem_match": "\n4. [4]",
"solution_match": "\nAnswer: "
}
|
117a942e-7400-5933-8edc-27a988be730f
| 609,150
|
Let $A B C$ be an isosceles triangle with $A B=A C$. Let $D$ and $E$ be the midpoints of segments $A B$ and $A C$, respectively. Suppose that there exists a point $F$ on ray $\overrightarrow{D E}$ outside of $A B C$ such that triangle $B F A$ is similar to triangle $A B C$. Compute $\frac{A B}{B C}$.
|
$\sqrt{2}$ Let $\alpha=\angle A B C=\angle A C B, A B=2 x$, and $B C=2 y$, so $A D=D B=A E=E C=x$ and $D E=y$. Since $\triangle B F A \sim \triangle A B C$ and $B A=A C$, we in fact have $\triangle B F A \cong \triangle A B C$, so $B F=B A=2 x, F A=2 y$, and $\angle D A F=\alpha$. But $D E \| B C$ yields $\angle A D F=\angle A B C=\alpha$ as well, whence $\triangle F A D \sim \triangle A B C$ gives $\frac{2 y}{x}=\frac{F A}{A D}=\frac{A B}{B C}=\frac{2 x}{2 y} \Longrightarrow \frac{A B}{B C}=\frac{x}{y}=\sqrt{2}$.
|
\sqrt{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be an isosceles triangle with $A B=A C$. Let $D$ and $E$ be the midpoints of segments $A B$ and $A C$, respectively. Suppose that there exists a point $F$ on ray $\overrightarrow{D E}$ outside of $A B C$ such that triangle $B F A$ is similar to triangle $A B C$. Compute $\frac{A B}{B C}$.
|
$\sqrt{2}$ Let $\alpha=\angle A B C=\angle A C B, A B=2 x$, and $B C=2 y$, so $A D=D B=A E=E C=x$ and $D E=y$. Since $\triangle B F A \sim \triangle A B C$ and $B A=A C$, we in fact have $\triangle B F A \cong \triangle A B C$, so $B F=B A=2 x, F A=2 y$, and $\angle D A F=\alpha$. But $D E \| B C$ yields $\angle A D F=\angle A B C=\alpha$ as well, whence $\triangle F A D \sim \triangle A B C$ gives $\frac{2 y}{x}=\frac{F A}{A D}=\frac{A B}{B C}=\frac{2 x}{2 y} \Longrightarrow \frac{A B}{B C}=\frac{x}{y}=\sqrt{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-gen-solutions.jsonl",
"problem_match": "\n5. [5]",
"solution_match": "\nAnswer: "
}
|
6b534ae3-e862-5805-9fb9-a0e4ecec1248
| 609,151
|
Find the number of positive integer divisors of 12 ! that leave a remainder of 1 when divided by 3 .
|
66 First we factor $12!=2^{10} 3^{5} 5^{2} 7^{1} 11^{1}$, and note that $2,5,11 \equiv-1(\bmod 3)$ while $7 \equiv 1$ $(\bmod 3)$. The desired divisors are precisely $2^{a} 5^{b} 7^{c} 11^{d}$ with $0 \leq a \leq 10,0 \leq b \leq 2,0 \leq c \leq 1,0 \leq d \leq 1$, and $a+b+d$ even. But then for any choice of $a, b$, exactly one $d \in\{0,1\}$ makes $a+b+d$ even, so we have exactly one $1(\bmod 3)$-divisor for every triple $(a, b, c)$ satisfying the inequality constraints. This gives a total of $(10+1)(2+1)(1+1)=66$.
|
66
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the number of positive integer divisors of 12 ! that leave a remainder of 1 when divided by 3 .
|
66 First we factor $12!=2^{10} 3^{5} 5^{2} 7^{1} 11^{1}$, and note that $2,5,11 \equiv-1(\bmod 3)$ while $7 \equiv 1$ $(\bmod 3)$. The desired divisors are precisely $2^{a} 5^{b} 7^{c} 11^{d}$ with $0 \leq a \leq 10,0 \leq b \leq 2,0 \leq c \leq 1,0 \leq d \leq 1$, and $a+b+d$ even. But then for any choice of $a, b$, exactly one $d \in\{0,1\}$ makes $a+b+d$ even, so we have exactly one $1(\bmod 3)$-divisor for every triple $(a, b, c)$ satisfying the inequality constraints. This gives a total of $(10+1)(2+1)(1+1)=66$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-gen-solutions.jsonl",
"problem_match": "\n6. [5]",
"solution_match": "\nAnswer: "
}
|
967f8e3d-a0ee-50ea-89b1-0acda6952017
| 609,152
|
Find the largest real number $\lambda$ such that $a^{2}+b^{2}+c^{2}+d^{2} \geq a b+\lambda b c+c d$ for all real numbers $a, b, c, d$.
|
$\left.\begin{array}{|c}\frac{3}{2} \\ \text { Let } \\ \\ \hline\end{array} a, b, c, d\right)=\left(a^{2}+b^{2}+c^{2}+d^{2}\right)-(a b+\lambda b c+c d)$. For fixed $(b, c, d), f$ is minimized at $a=\frac{b}{2}$, and for fixed $(a, b, c), f$ is minimized at $d=\frac{c}{2}$, so simply we want the largest $\lambda$ such that $f\left(\frac{b}{2}, b, c, \frac{c}{2}\right)=\frac{3}{4}\left(b^{2}+c^{2}\right)-\lambda b c$ is always nonnegative. By AM-GM, this holds if and only if $\lambda \leq 2 \frac{3}{4}=\frac{3}{2}$.
|
\frac{3}{2}
|
Yes
|
Yes
|
math-word-problem
|
Inequalities
|
Find the largest real number $\lambda$ such that $a^{2}+b^{2}+c^{2}+d^{2} \geq a b+\lambda b c+c d$ for all real numbers $a, b, c, d$.
|
$\left.\begin{array}{|c}\frac{3}{2} \\ \text { Let } \\ \\ \hline\end{array} a, b, c, d\right)=\left(a^{2}+b^{2}+c^{2}+d^{2}\right)-(a b+\lambda b c+c d)$. For fixed $(b, c, d), f$ is minimized at $a=\frac{b}{2}$, and for fixed $(a, b, c), f$ is minimized at $d=\frac{c}{2}$, so simply we want the largest $\lambda$ such that $f\left(\frac{b}{2}, b, c, \frac{c}{2}\right)=\frac{3}{4}\left(b^{2}+c^{2}\right)-\lambda b c$ is always nonnegative. By AM-GM, this holds if and only if $\lambda \leq 2 \frac{3}{4}=\frac{3}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-gen-solutions.jsonl",
"problem_match": "\n7. [6]",
"solution_match": "\nAnswer: "
}
|
45b25f4e-2c16-5605-b583-053b42247214
| 609,153
|
How many of the first 1000 positive integers can be written as the sum of finitely many distinct numbers from the sequence $3^{0}, 3^{1}, 3^{2}, \ldots$ ?
|
105 We want to find which integers have only 0's and 1's in their base 3 representation. Note that $1000_{10}=1101001_{3}$. We can construct a bijection from all such numbers to the binary strings, by mapping $x_{3} \leftrightarrow x_{2}$. Since $1101001_{2}=105_{10}$, we conclude that the answer is 105 .
|
105
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many of the first 1000 positive integers can be written as the sum of finitely many distinct numbers from the sequence $3^{0}, 3^{1}, 3^{2}, \ldots$ ?
|
105 We want to find which integers have only 0's and 1's in their base 3 representation. Note that $1000_{10}=1101001_{3}$. We can construct a bijection from all such numbers to the binary strings, by mapping $x_{3} \leftrightarrow x_{2}$. Since $1101001_{2}=105_{10}$, we conclude that the answer is 105 .
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-gen-solutions.jsonl",
"problem_match": "\n8. [6]",
"solution_match": "\nAnswer: "
}
|
8a0c6cbd-069a-5c57-bb14-d3bf9c56d746
| 609,154
|
Let $A B C$ be a triangle and $D$ a point on $B C$ such that $A B=\sqrt{2}, A C=\sqrt{3}, \angle B A D=30^{\circ}$, and $\angle C A D=45^{\circ}$. Find $A D$.
|
$\quad \frac{\sqrt{6}}{2}$ OR $\frac{\sqrt{3}}{\sqrt{2}}$ Note that $[B A D]+[C A D]=[A B C]$. If $\alpha_{1}=\angle B A D, \alpha_{2}=\angle C A D$, then we deduce $\frac{\sin \left(\alpha_{1}+\alpha_{2}\right)}{A D}=\frac{\sin \alpha_{1}}{A C}+\frac{\sin \alpha_{2}}{A B}$ upon division by $A B \cdot A C \cdot A D$. Now
$$
A D=\frac{\sin \left(30^{\circ}+45^{\circ}\right)}{\frac{\sin 30^{\circ}}{\sqrt{3}}+\frac{\sin 45^{\circ}}{\sqrt{2}}}
$$
But $\sin \left(30^{\circ}+45^{\circ}\right)=\sin 30^{\circ} \cos 45^{\circ}+\sin 45^{\circ} \cos 30^{\circ}=\sin 30^{\circ} \frac{1}{\sqrt{2}}+\sin 45^{\circ} \frac{\sqrt{3}}{2}=\frac{\sqrt{6}}{2}\left(\frac{\sin 30^{\circ}}{\sqrt{3}}+\frac{\sin 45^{\circ}}{\sqrt{2}}\right)$, so our answer is $\frac{\sqrt{6}}{2}$.
|
\frac{\sqrt{6}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a triangle and $D$ a point on $B C$ such that $A B=\sqrt{2}, A C=\sqrt{3}, \angle B A D=30^{\circ}$, and $\angle C A D=45^{\circ}$. Find $A D$.
|
$\quad \frac{\sqrt{6}}{2}$ OR $\frac{\sqrt{3}}{\sqrt{2}}$ Note that $[B A D]+[C A D]=[A B C]$. If $\alpha_{1}=\angle B A D, \alpha_{2}=\angle C A D$, then we deduce $\frac{\sin \left(\alpha_{1}+\alpha_{2}\right)}{A D}=\frac{\sin \alpha_{1}}{A C}+\frac{\sin \alpha_{2}}{A B}$ upon division by $A B \cdot A C \cdot A D$. Now
$$
A D=\frac{\sin \left(30^{\circ}+45^{\circ}\right)}{\frac{\sin 30^{\circ}}{\sqrt{3}}+\frac{\sin 45^{\circ}}{\sqrt{2}}}
$$
But $\sin \left(30^{\circ}+45^{\circ}\right)=\sin 30^{\circ} \cos 45^{\circ}+\sin 45^{\circ} \cos 30^{\circ}=\sin 30^{\circ} \frac{1}{\sqrt{2}}+\sin 45^{\circ} \frac{\sqrt{3}}{2}=\frac{\sqrt{6}}{2}\left(\frac{\sin 30^{\circ}}{\sqrt{3}}+\frac{\sin 45^{\circ}}{\sqrt{2}}\right)$, so our answer is $\frac{\sqrt{6}}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-gen-solutions.jsonl",
"problem_match": "\n9. [7]",
"solution_match": "\nAnswer: "
}
|
e5162864-e4c4-53c4-a387-b213edd5aa6a
| 609,155
|
Evaluate $2+5+8+\cdots+101$.
|
1751 There are $\frac{102}{3}=34$ terms with average $\frac{2+101}{2}$, so their sum is $17 \cdot 103=1751$.
|
1751
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Evaluate $2+5+8+\cdots+101$.
|
1751 There are $\frac{102}{3}=34$ terms with average $\frac{2+101}{2}$, so their sum is $17 \cdot 103=1751$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n1. [5]",
"solution_match": "\nAnswer: "
}
|
28619230-152c-54bd-95c7-4d93d573e9d0
| 609,157
|
Two fair six-sided dice are rolled. What is the probability that their sum is at least 10 ?
|
$\boxed{\frac{1}{6}}$ There are 3, 2, 1 outcomes with sum 10, 11, 12, so the probability is $\frac{3+2+1}{6^2}=\frac{1}{6}$.
|
\frac{1}{6}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Two fair six-sided dice are rolled. What is the probability that their sum is at least 10 ?
|
$\boxed{\frac{1}{6}}$ There are 3, 2, 1 outcomes with sum 10, 11, 12, so the probability is $\frac{3+2+1}{6^2}=\frac{1}{6}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n2. [5]",
"solution_match": "\nAnswer: "
}
|
e03145b3-aa1f-5150-9052-e7908bce25ab
| 609,158
|
A square is inscribed in a circle of radius 1. Find the perimeter of the square.
|
$4 \sqrt{2}$ OR $\frac{8}{\sqrt{2}}$ The square has diagonal length 2 , so side length $\sqrt{2}$ and perimeter $4 \sqrt{2}$.
|
4 \sqrt{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A square is inscribed in a circle of radius 1. Find the perimeter of the square.
|
$4 \sqrt{2}$ OR $\frac{8}{\sqrt{2}}$ The square has diagonal length 2 , so side length $\sqrt{2}$ and perimeter $4 \sqrt{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n3. [5]",
"solution_match": "\nAnswer: "
}
|
7b899ef6-88f0-5a7a-bc66-3cbfc27138a0
| 609,159
|
Find the minimum possible value of $\left(x^{2}+6 x+2\right)^{2}$ over all real numbers $x$.
|
0 This is $\left((x+3)^{2}-7\right)^{2} \geq 0$, with equality at $x+3= \pm \sqrt{7}$.
|
0
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find the minimum possible value of $\left(x^{2}+6 x+2\right)^{2}$ over all real numbers $x$.
|
0 This is $\left((x+3)^{2}-7\right)^{2} \geq 0$, with equality at $x+3= \pm \sqrt{7}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n4. [6]",
"solution_match": "\nAnswer: "
}
|
2088139e-5e11-5c76-bd89-1afb9db15c8b
| 609,160
|
A right triangle has area 5 and a hypotenuse of length 5 . Find its perimeter.
|
$\quad 5+3 \sqrt{5}$ If $x, y$ denote the legs, then $x y=10$ and $x^{2}+y^{2}=25$, so $x+y+\sqrt{x^{2}+y^{2}}=$ $\sqrt{\left(x^{2}+y^{2}\right)+2 x y}+5=\sqrt{45}+5=5+3 \sqrt{5}$.
|
5+3 \sqrt{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A right triangle has area 5 and a hypotenuse of length 5 . Find its perimeter.
|
$\quad 5+3 \sqrt{5}$ If $x, y$ denote the legs, then $x y=10$ and $x^{2}+y^{2}=25$, so $x+y+\sqrt{x^{2}+y^{2}}=$ $\sqrt{\left(x^{2}+y^{2}\right)+2 x y}+5=\sqrt{45}+5=5+3 \sqrt{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n6. [6]",
"solution_match": "\nAnswer: "
}
|
214ca2fa-b671-594c-819f-ee0346db55a6
| 609,162
|
Marty and three other people took a math test. Everyone got a non-negative integer score. The average score was 20. Marty was told the average score and concluded that everyone else scored below average. What was the minimum possible score Marty could have gotten in order to definitively reach this conclusion?
|
61 Suppose for the sake of contradiction Marty obtained a score of 60 or lower. Since the mean is 20 , the total score of the 4 test takers must be 80 . Then there exists the possibility of 2 students getting 0 , and the last student getting a score of 20 or higher. If so, Marty could not have concluded with certainty that everyone else scored below average.
With a score of 61 , any of the other three students must have scored points lower or equal to 19 points. Thus Marty is able to conclude that everyone else scored below average.
|
61
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
Marty and three other people took a math test. Everyone got a non-negative integer score. The average score was 20. Marty was told the average score and concluded that everyone else scored below average. What was the minimum possible score Marty could have gotten in order to definitively reach this conclusion?
|
61 Suppose for the sake of contradiction Marty obtained a score of 60 or lower. Since the mean is 20 , the total score of the 4 test takers must be 80 . Then there exists the possibility of 2 students getting 0 , and the last student getting a score of 20 or higher. If so, Marty could not have concluded with certainty that everyone else scored below average.
With a score of 61 , any of the other three students must have scored points lower or equal to 19 points. Thus Marty is able to conclude that everyone else scored below average.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n7. [7]",
"solution_match": "\nAnswer: "
}
|
55f1c3c2-a78a-58c7-b449-4dcd477fa589
| 609,163
|
Find the remainder when $1^{2}+3^{2}+5^{2}+\cdots+99^{2}$ is divided by 1000 .
|
650 We have $S=\sum_{i=0}^{49}(2 i+1)^{2}=\sum_{i=0}^{49} 4 i^{2}+4 i+1=4 \cdot \frac{49 \cdot 50 \cdot 99}{6}+4 \cdot \frac{49 \cdot 50}{2}+50 \equiv$ $700+900+50(\bmod 1000) \equiv 650(\bmod 1000)$.
|
650
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the remainder when $1^{2}+3^{2}+5^{2}+\cdots+99^{2}$ is divided by 1000 .
|
650 We have $S=\sum_{i=0}^{49}(2 i+1)^{2}=\sum_{i=0}^{49} 4 i^{2}+4 i+1=4 \cdot \frac{49 \cdot 50 \cdot 99}{6}+4 \cdot \frac{49 \cdot 50}{2}+50 \equiv$ $700+900+50(\bmod 1000) \equiv 650(\bmod 1000)$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n9. [7]",
"solution_match": "\nAnswer: "
}
|
51dd08d5-ac62-5633-bf7c-edff6135311b
| 609,165
|
How many pairs of real numbers $(x, y)$ satisfy the equation
$$
y^{4}-y^{2}=x y^{3}-x y=x^{3} y-x y=x^{4}-x^{2}=0 ?
$$
|
9 We can see that if they solve the first and fourth equations, they are automatically solutions to the second and third equations. Hence, the solutions are just the $3^{2}=9$ points where $x, y$ can be any of $-1,0,1$.
|
9
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
How many pairs of real numbers $(x, y)$ satisfy the equation
$$
y^{4}-y^{2}=x y^{3}-x y=x^{3} y-x y=x^{4}-x^{2}=0 ?
$$
|
9 We can see that if they solve the first and fourth equations, they are automatically solutions to the second and third equations. Hence, the solutions are just the $3^{2}=9$ points where $x, y$ can be any of $-1,0,1$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n10. [8]",
"solution_match": "\nAnswer: "
}
|
f1b261e5-ffa1-5148-9321-787d44986e4c
| 609,166
|
David has a unit triangular array of 10 points, 4 on each side. A looping path is a sequence $A_{1}, A_{2}, \ldots, A_{10}$ containing each of the 10 points exactly once, such that $A_{i}$ and $A_{i+1}$ are adjacent (exactly 1 unit apart) for $i=1,2, \ldots, 10$. (Here $A_{11}=A_{1}$.) Find the number of looping paths in this array.
|
60 There are $10 \cdot 2$ times as many loop sequences as loops. To count the number of loops, first focus on the three corners of the array: their edges are uniquely determined. It's now easy to see there are 3 loops (they form " $V$-shapes"), so the answer is $10 \cdot 2 \cdot 3=60$.
|
60
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
David has a unit triangular array of 10 points, 4 on each side. A looping path is a sequence $A_{1}, A_{2}, \ldots, A_{10}$ containing each of the 10 points exactly once, such that $A_{i}$ and $A_{i+1}$ are adjacent (exactly 1 unit apart) for $i=1,2, \ldots, 10$. (Here $A_{11}=A_{1}$.) Find the number of looping paths in this array.
|
60 There are $10 \cdot 2$ times as many loop sequences as loops. To count the number of loops, first focus on the three corners of the array: their edges are uniquely determined. It's now easy to see there are 3 loops (they form " $V$-shapes"), so the answer is $10 \cdot 2 \cdot 3=60$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n11. [8]",
"solution_match": "\nAnswer: "
}
|
1d6ae806-ab08-572e-8fef-5acf2c5676e2
| 609,167
|
Let $S=\{1,2, \ldots, 2013\}$. Find the number of ordered triples $(A, B, C)$ of subsets of $S$ such that $A \subseteq B$ and $A \cup B \cup C=S$.
|
$5^{2013}$ OR $125^{671}$ Let $n=2013$. Each of the $n$ elements can be independently placed in 5 spots: there are $2^{3}-1$ choices with element $x$ in at least one set, and we subtract the $2^{1}$ choices with element $x$ in set $A$ but not $B$. Specifying where the elements go uniquely determines $A, B, C$, so there are $5^{n}=5^{2013}$ ordered triples.
|
5^{2013}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $S=\{1,2, \ldots, 2013\}$. Find the number of ordered triples $(A, B, C)$ of subsets of $S$ such that $A \subseteq B$ and $A \cup B \cup C=S$.
|
$5^{2013}$ OR $125^{671}$ Let $n=2013$. Each of the $n$ elements can be independently placed in 5 spots: there are $2^{3}-1$ choices with element $x$ in at least one set, and we subtract the $2^{1}$ choices with element $x$ in set $A$ but not $B$. Specifying where the elements go uniquely determines $A, B, C$, so there are $5^{n}=5^{2013}$ ordered triples.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n13. [9]",
"solution_match": "\nAnswer: "
}
|
122ac494-bb9a-55bf-a175-733722efa08f
| 609,169
|
A bug is on one exterior vertex of solid $S$, a $3 \times 3 \times 3$ cube that has its center $1 \times 1 \times 1$ cube removed, and wishes to travel to the opposite exterior vertex. Let $O$ denote the outer surface of $S$ (formed by the surface of the $3 \times 3 \times 3$ cube). Let $L(S)$ denote the length of the shortest path through $S$. (Note that such a path cannot pass through the missing center cube, which is empty space.) Let $L(O)$ denote the length of the shortest path through $O$. What is the ratio $\frac{L(S)}{L(O)}$ ?
|
$\frac{\sqrt{29}}{3 \sqrt{5}}$ OR $\frac{\sqrt{145}}{15}$ By $\left(^{*}\right)$, the shortest route in $O$ has length $2 \sqrt{1.5^{2}+3^{2}}=3 \sqrt{5}$. By $\left({ }^{* *}\right)$,
the shortest route overall (in $S$ ) has length $2 \sqrt{1.5^{2}+1^{2}+2^{2}}=\sqrt{3^{2}+2^{2}+4^{2}}=\sqrt{29}$. Therefore the desired ratio is $\frac{\sqrt{29}}{3 \sqrt{5}}=\frac{\sqrt{145}}{15}$.$\left(^{*}\right)$ Suppose we're trying to get from $(0,0,0)$ to $(3,3,3)$ through $O$. Then one minimal-length path through $O$ is $(0,0,0) \rightarrow(1.5,0,3) \rightarrow(3,3,3)$.
$\left(^{* *}\right)$ Suppose we're trying to get from $(0,0,0)$ to $(3,3,3)$ through $S$. Then the inner hole is $[1,2] \times$ $[1,2] \times[1,2]$, and one minimal-length path is $(0,0,0) \rightarrow(1.5,1,2) \rightarrow(3,3,3)$.
To justify $\left(^{*}\right)$, note that we must at some point hit one of the three faces incident to $(3,3,3)$, and therefore one of the edges of those faces. Without loss of generality, the first of these edges (which must lie on a face incident to $(0,0,0))$ is $\{(t, 0,3): 0 \leq t \leq 3\}$. Then the shortest path goes directly from the origin to the edge, and then directly to $(3,3,3) ; t=1.5$ minimizes the resulting distance. (One may either appeal to the classic geometric "unfolding" argument, or just direct algebraic minimization.)
To justify $\left({ }^{* *}\right)$, consider the portion of $S$ "visible" from $(0,0,0)$. It sees 3 mutually adjacent faces of the center cube "hole" (when looking inside the solid) and its 6 edges. ( $3,3,3$ ) can also see these 6 edges. The shortest path through $S$ must be a straight line from the start vertex to some point $P$ on the surface of the center cube $(+)$, and it's easy to see, using the triangle inequality, that this point $P$ must be on one of the 6 edges. Without loss of generality, it's on the edge $\{(t, 1,2): 1 \leq t \leq 2\}$. The remaining path is a straight line to $(3,3,3)$. Then once again, $t=1.5$ minimizes the distance. (And once again, one may either appeal to the classic geometric "unfolding" argument-though this time it's a little trickier - or just direct algebraic minimization.)
Comment. ( + ) can be proven in many ways. The rough physical intuition is that "a fully stretched rubber band" with ends at $(0,0,0)$ and $(3,3,3)$ must not have any "wiggle room" (and so touches the inner cube). Perhaps more rigorously, we can try shortening a path that does not hit the center cube, for instance by "projecting the path down" towards the closest edge of the inner cube.
|
\frac{\sqrt{145}}{15}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A bug is on one exterior vertex of solid $S$, a $3 \times 3 \times 3$ cube that has its center $1 \times 1 \times 1$ cube removed, and wishes to travel to the opposite exterior vertex. Let $O$ denote the outer surface of $S$ (formed by the surface of the $3 \times 3 \times 3$ cube). Let $L(S)$ denote the length of the shortest path through $S$. (Note that such a path cannot pass through the missing center cube, which is empty space.) Let $L(O)$ denote the length of the shortest path through $O$. What is the ratio $\frac{L(S)}{L(O)}$ ?
|
$\frac{\sqrt{29}}{3 \sqrt{5}}$ OR $\frac{\sqrt{145}}{15}$ By $\left(^{*}\right)$, the shortest route in $O$ has length $2 \sqrt{1.5^{2}+3^{2}}=3 \sqrt{5}$. By $\left({ }^{* *}\right)$,
the shortest route overall (in $S$ ) has length $2 \sqrt{1.5^{2}+1^{2}+2^{2}}=\sqrt{3^{2}+2^{2}+4^{2}}=\sqrt{29}$. Therefore the desired ratio is $\frac{\sqrt{29}}{3 \sqrt{5}}=\frac{\sqrt{145}}{15}$.$\left(^{*}\right)$ Suppose we're trying to get from $(0,0,0)$ to $(3,3,3)$ through $O$. Then one minimal-length path through $O$ is $(0,0,0) \rightarrow(1.5,0,3) \rightarrow(3,3,3)$.
$\left(^{* *}\right)$ Suppose we're trying to get from $(0,0,0)$ to $(3,3,3)$ through $S$. Then the inner hole is $[1,2] \times$ $[1,2] \times[1,2]$, and one minimal-length path is $(0,0,0) \rightarrow(1.5,1,2) \rightarrow(3,3,3)$.
To justify $\left(^{*}\right)$, note that we must at some point hit one of the three faces incident to $(3,3,3)$, and therefore one of the edges of those faces. Without loss of generality, the first of these edges (which must lie on a face incident to $(0,0,0))$ is $\{(t, 0,3): 0 \leq t \leq 3\}$. Then the shortest path goes directly from the origin to the edge, and then directly to $(3,3,3) ; t=1.5$ minimizes the resulting distance. (One may either appeal to the classic geometric "unfolding" argument, or just direct algebraic minimization.)
To justify $\left({ }^{* *}\right)$, consider the portion of $S$ "visible" from $(0,0,0)$. It sees 3 mutually adjacent faces of the center cube "hole" (when looking inside the solid) and its 6 edges. ( $3,3,3$ ) can also see these 6 edges. The shortest path through $S$ must be a straight line from the start vertex to some point $P$ on the surface of the center cube $(+)$, and it's easy to see, using the triangle inequality, that this point $P$ must be on one of the 6 edges. Without loss of generality, it's on the edge $\{(t, 1,2): 1 \leq t \leq 2\}$. The remaining path is a straight line to $(3,3,3)$. Then once again, $t=1.5$ minimizes the distance. (And once again, one may either appeal to the classic geometric "unfolding" argument-though this time it's a little trickier - or just direct algebraic minimization.)
Comment. ( + ) can be proven in many ways. The rough physical intuition is that "a fully stretched rubber band" with ends at $(0,0,0)$ and $(3,3,3)$ must not have any "wiggle room" (and so touches the inner cube). Perhaps more rigorously, we can try shortening a path that does not hit the center cube, for instance by "projecting the path down" towards the closest edge of the inner cube.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n16. [10]",
"solution_match": "\n## Answer: "
}
|
e9b8d3bb-87bc-5859-9448-1b02a67261e7
| 609,172
|
Find the sum of $\frac{1}{n}$ over all positive integers $n$ with the property that the decimal representation of $\frac{1}{n}$ terminates.
|
$\sqrt{\frac{5}{2}}$ The decimal representation of $\frac{1}{n}$ terminates if and only if $n=2^{i} 5^{j}$ for some nonnegative integers $i, j$, so our desired sum is
$$
\sum_{i \geq 0} \sum_{j \geq 0} 2^{-i} 5^{-j}=\sum_{i \geq 0} 2^{-i} \sum_{j \geq 0} 5^{-j}=\left(1-2^{-1}\right)^{-1}\left(1-5^{-1}\right)^{-1}=\frac{2}{1} \frac{5}{4}=\frac{5}{2}
$$
|
\frac{5}{2}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the sum of $\frac{1}{n}$ over all positive integers $n$ with the property that the decimal representation of $\frac{1}{n}$ terminates.
|
$\sqrt{\frac{5}{2}}$ The decimal representation of $\frac{1}{n}$ terminates if and only if $n=2^{i} 5^{j}$ for some nonnegative integers $i, j$, so our desired sum is
$$
\sum_{i \geq 0} \sum_{j \geq 0} 2^{-i} 5^{-j}=\sum_{i \geq 0} 2^{-i} \sum_{j \geq 0} 5^{-j}=\left(1-2^{-1}\right)^{-1}\left(1-5^{-1}\right)^{-1}=\frac{2}{1} \frac{5}{4}=\frac{5}{2}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n17. [10]",
"solution_match": "\nAnswer: "
}
|
3ec560e2-8b76-59b6-8e85-ff673f294b17
| 609,173
|
The rightmost nonzero digit in the decimal expansion of 101 ! is the same as the rightmost nonzero digit of $n!$, where $n$ is an integer greater than 101. Find the smallest possible value of $n$.
|
103 101! has more factors of 2 than 5 , so its rightmost nonzero digit is one of $2,4,6,8$. Notice that if the rightmost nonzero digit of 101 ! is $2 k(1 \leq k \leq 4)$, then 102 ! has rightmost nonzero digit $102(2 k) \equiv 4 k(\bmod 10)$, and $103!$ has rightmost nonzero digit $103(4 k) \equiv 2 k(\bmod 10)$. Hence $n=103$.
|
103
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
The rightmost nonzero digit in the decimal expansion of 101 ! is the same as the rightmost nonzero digit of $n!$, where $n$ is an integer greater than 101. Find the smallest possible value of $n$.
|
103 101! has more factors of 2 than 5 , so its rightmost nonzero digit is one of $2,4,6,8$. Notice that if the rightmost nonzero digit of 101 ! is $2 k(1 \leq k \leq 4)$, then 102 ! has rightmost nonzero digit $102(2 k) \equiv 4 k(\bmod 10)$, and $103!$ has rightmost nonzero digit $103(4 k) \equiv 2 k(\bmod 10)$. Hence $n=103$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n18. [10]",
"solution_match": "\nAnswer: "
}
|
24334d52-be25-54e5-bdb2-53b4245fb5a3
| 609,174
|
Let $p, q, r, s$ be distinct primes such that $p q-r s$ is divisible by 30 . Find the minimum possible value of $p+q+r+s$.
|
54 The key is to realize none of the primes can be 2,3 , or 5 , or else we would have to use one of them twice. Hence $p, q, r, s$ must lie among $7,11,13,17,19,23,29, \ldots$. These options give remainders of $1(\bmod 2)$ (obviously), $1,-1,1,-1,1,-1,-1, \ldots$ modulo 3 , and $2,1,3,2,4,3,4, \ldots$ modulo 5 . We automatically have $2 \mid p q-r s$, and we have $3 \mid p q-r s$ if and only if $p q r s \equiv(p q)^{2} \equiv 1$ $(\bmod 3)$, i.e. there are an even number of $-1(\bmod 3)$ 's among $p, q, r, s$.
If $\{p, q, r, s\}=\{7,11,13,17\}$, then we cannot have $5 \mid p q-r s$, or else $12 \equiv p q r s \equiv(p q)^{2}(\bmod 5)$ is a quadratic residue. Our next smallest choice (in terms of $p+q+r+s$ ) is $\{7,11,17,19\}$, which works: $7 \cdot 17-11 \cdot 19 \equiv 2^{2}-4 \equiv 0(\bmod 5)$. This gives an answer of $7+17+11+19=54$.
|
54
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $p, q, r, s$ be distinct primes such that $p q-r s$ is divisible by 30 . Find the minimum possible value of $p+q+r+s$.
|
54 The key is to realize none of the primes can be 2,3 , or 5 , or else we would have to use one of them twice. Hence $p, q, r, s$ must lie among $7,11,13,17,19,23,29, \ldots$. These options give remainders of $1(\bmod 2)$ (obviously), $1,-1,1,-1,1,-1,-1, \ldots$ modulo 3 , and $2,1,3,2,4,3,4, \ldots$ modulo 5 . We automatically have $2 \mid p q-r s$, and we have $3 \mid p q-r s$ if and only if $p q r s \equiv(p q)^{2} \equiv 1$ $(\bmod 3)$, i.e. there are an even number of $-1(\bmod 3)$ 's among $p, q, r, s$.
If $\{p, q, r, s\}=\{7,11,13,17\}$, then we cannot have $5 \mid p q-r s$, or else $12 \equiv p q r s \equiv(p q)^{2}(\bmod 5)$ is a quadratic residue. Our next smallest choice (in terms of $p+q+r+s$ ) is $\{7,11,17,19\}$, which works: $7 \cdot 17-11 \cdot 19 \equiv 2^{2}-4 \equiv 0(\bmod 5)$. This gives an answer of $7+17+11+19=54$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n19. [11]",
"solution_match": "\nAnswer: "
}
|
d581ef1d-1256-5422-ac52-9be16b059897
| 609,175
|
There exist unique nonnegative integers $A, B$ between 0 and 9 , inclusive, such that
$$
(1001 \cdot A+110 \cdot B)^{2}=57,108,249
$$
Find $10 \cdot A+B$.
|
75 We only need to bound for $A B 00$; in other words, $A B^{2} \leq 5710$ but $(A B+1)^{2} \geq 5710$. A quick check gives $A B=75$. (Lots of ways to get this...)
|
75
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
There exist unique nonnegative integers $A, B$ between 0 and 9 , inclusive, such that
$$
(1001 \cdot A+110 \cdot B)^{2}=57,108,249
$$
Find $10 \cdot A+B$.
|
75 We only need to bound for $A B 00$; in other words, $A B^{2} \leq 5710$ but $(A B+1)^{2} \geq 5710$. A quick check gives $A B=75$. (Lots of ways to get this...)
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n20. [11]",
"solution_match": "\nAnswer: "
}
|
d6a8893d-8104-5252-9eac-7cd80ecdae8a
| 609,176
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.