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Joe B. now clears the board. How many ways can he place 3 white rooks and 3 black rooks on the board so that no two rooks of opposite color can attack each other?
|
608 Consider first placing the white rooks. They will occupy either 3 columns and 1 row, 3 columns and 2 rows, 3 columns and 3 rows, 2 rows and 2 columns, 2 columns and 3 rows, or 1 column and 3 rows. First note that placing the black rooks is impossible in the second, third, and fifth cases. The first case can happen in 4.4 ways, and each leads to a unique way to place the black rooks. In the fourth case, we can choose the row with 2 rooks in 4 ways, place the rooks in $\binom{4}{2}$ ways, choose the column of the other rook in 2 ways, and place it in 3 ways, for a total of $4 \cdot\binom{4}{2} \cdot 2 \cdot 3=144$ ways to place the white rooks in this configuration; the black rooks can then be placed in any of 4 possible locations, and there are 4 ways to do this, leading to 576 possibilities in this case. Finally, the sixth case is analogous to the first, giving 16 possibilities. Summing, we find $16+576+16=608$ total placements.
|
608
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Joe B. now clears the board. How many ways can he place 3 white rooks and 3 black rooks on the board so that no two rooks of opposite color can attack each other?
|
608 Consider first placing the white rooks. They will occupy either 3 columns and 1 row, 3 columns and 2 rows, 3 columns and 3 rows, 2 rows and 2 columns, 2 columns and 3 rows, or 1 column and 3 rows. First note that placing the black rooks is impossible in the second, third, and fifth cases. The first case can happen in 4.4 ways, and each leads to a unique way to place the black rooks. In the fourth case, we can choose the row with 2 rooks in 4 ways, place the rooks in $\binom{4}{2}$ ways, choose the column of the other rook in 2 ways, and place it in 3 ways, for a total of $4 \cdot\binom{4}{2} \cdot 2 \cdot 3=144$ ways to place the white rooks in this configuration; the black rooks can then be placed in any of 4 possible locations, and there are 4 ways to do this, leading to 576 possibilities in this case. Finally, the sixth case is analogous to the first, giving 16 possibilities. Summing, we find $16+576+16=608$ total placements.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-gen2-solutions.jsonl",
"problem_match": "\n3. [5]",
"solution_match": "\nAnswer: "
}
|
bff2f2e7-08dd-5ded-ae51-abed97a3f940
| 608,415
|
Joe B. is frustrated with chess. He breaks the board, leaving a $4 \times 4$ board, and throws 3 black knights and 3 white kings at the board. Miraculously, they all land in distinct squares! What is the expected number of checks in the resulting position? (Note that a knight can administer multiple checks and a king can be checked by multiple knights.)
|
$\quad \frac{9}{5}$ We first compute the expected number of checks between a single knight-king pair. If the king is located at any of the 4 corners, the knight has 2 possible checks. If the king is located in one of the 8 squares on the side of the board but not in the corner, the knight has 3 possible checks. If the king is located in any of the 4 central squares, the knight has 4 possible checks. Summing up, $4 \cdot 2+8 \cdot 3+4 \cdot 4=48$ of the $16 \cdot 15$ knight-king positions yield a single check, so each pair yields $\frac{48}{16 \cdot 15}=\frac{1}{5}$ expected checks. Now, note that each of the 9 knight-king pairs is in each of $16 \cdot 15$ possible positions with equal probability, so by linearity of expectation the answer is $9 \cdot \frac{1}{5}=\frac{9}{5}$.
|
\frac{9}{5}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Joe B. is frustrated with chess. He breaks the board, leaving a $4 \times 4$ board, and throws 3 black knights and 3 white kings at the board. Miraculously, they all land in distinct squares! What is the expected number of checks in the resulting position? (Note that a knight can administer multiple checks and a king can be checked by multiple knights.)
|
$\quad \frac{9}{5}$ We first compute the expected number of checks between a single knight-king pair. If the king is located at any of the 4 corners, the knight has 2 possible checks. If the king is located in one of the 8 squares on the side of the board but not in the corner, the knight has 3 possible checks. If the king is located in any of the 4 central squares, the knight has 4 possible checks. Summing up, $4 \cdot 2+8 \cdot 3+4 \cdot 4=48$ of the $16 \cdot 15$ knight-king positions yield a single check, so each pair yields $\frac{48}{16 \cdot 15}=\frac{1}{5}$ expected checks. Now, note that each of the 9 knight-king pairs is in each of $16 \cdot 15$ possible positions with equal probability, so by linearity of expectation the answer is $9 \cdot \frac{1}{5}=\frac{9}{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-gen2-solutions.jsonl",
"problem_match": "\n4. [6]",
"solution_match": "\nAnswer: "
}
|
47f62c7d-8912-5191-a08c-4a48aaab6c23
| 608,416
|
Suppose that at some point Joe B. has placed 2 black knights on the original board, but gets bored of chess. He now decides to cover the 34 remaining squares with 17 dominos so that no two overlap and the dominos cover the entire rest of the board. For how many initial arrangements of the two pieces is this possible?
|
324 Color the squares of the board red and blue in a checkerboard pattern, and observe that any domino will cover exactly one red square and one blue square. Therefore, if the two knights cover squares of the same color, this is impossible. We now claim that it is always possible if they cover squares of opposite colors, which will give an answer of $18^{2}=324$. Consider the rectangle $R$ with the knights at its corners. Because the knights cover differently colored squares, $R$ must have one side length odd and one side length even. Therefore, the 4 lines bounding $R$ cut the original board into $R$ and up to 8 other rectangles, which can be put together into rectangles with at least one side even. These rectangles can be tiled, and it is easy to see that $R$ can be tiled, proving the claim.
|
324
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Suppose that at some point Joe B. has placed 2 black knights on the original board, but gets bored of chess. He now decides to cover the 34 remaining squares with 17 dominos so that no two overlap and the dominos cover the entire rest of the board. For how many initial arrangements of the two pieces is this possible?
|
324 Color the squares of the board red and blue in a checkerboard pattern, and observe that any domino will cover exactly one red square and one blue square. Therefore, if the two knights cover squares of the same color, this is impossible. We now claim that it is always possible if they cover squares of opposite colors, which will give an answer of $18^{2}=324$. Consider the rectangle $R$ with the knights at its corners. Because the knights cover differently colored squares, $R$ must have one side length odd and one side length even. Therefore, the 4 lines bounding $R$ cut the original board into $R$ and up to 8 other rectangles, which can be put together into rectangles with at least one side even. These rectangles can be tiled, and it is easy to see that $R$ can be tiled, proving the claim.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-gen2-solutions.jsonl",
"problem_match": "\n5. [7]",
"solution_match": "\nAnswer: "
}
|
2085f2db-cdff-569c-9f57-4a2568936e7d
| 608,417
|
Find the sum of all solutions for $x$ :
$$
\begin{aligned}
x y & =1 \\
x+y & =3
\end{aligned}
$$
|
3 Substitute $3-x$ in for $y$ into the first equation:
$$
x(3-x)=1 \Leftrightarrow x^{2}-3 x+1=0
$$
This equation has two distinct roots, each of which corresponds to a possible solution $x$. The sum of the roots of the quadratic equation $a x^{2}+b x+c=0$ is $\frac{-b}{a}$, which in this case is 3 .
|
3
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find the sum of all solutions for $x$ :
$$
\begin{aligned}
x y & =1 \\
x+y & =3
\end{aligned}
$$
|
3 Substitute $3-x$ in for $y$ into the first equation:
$$
x(3-x)=1 \Leftrightarrow x^{2}-3 x+1=0
$$
This equation has two distinct roots, each of which corresponds to a possible solution $x$. The sum of the roots of the quadratic equation $a x^{2}+b x+c=0$ is $\frac{-b}{a}$, which in this case is 3 .
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n1. [5]",
"solution_match": "\nAnswer: "
}
|
54b00ac2-f426-52c7-bdaa-c8e4e819fdb2
| 608,418
|
Evaluate the sum
$$
1-2+3-4+\cdots+2007-2008
$$
|
-1004 Every odd integer term can be paired with the next even integer, and this pair sums to -1 . There are 1004 such pairs, so the total sum is -1004 .
|
-1004
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Evaluate the sum
$$
1-2+3-4+\cdots+2007-2008
$$
|
-1004 Every odd integer term can be paired with the next even integer, and this pair sums to -1 . There are 1004 such pairs, so the total sum is -1004 .
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n2. [5]",
"solution_match": "\nAnswer: "
}
|
8199f195-8863-5f34-abb3-8d3b37d72f4b
| 608,419
|
What is the largest $x$ such that $x^{2}$ divides $24 \cdot 35 \cdot 46 \cdot 57$ ?
|
12 We factor the product as $2^{4} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 19 \cdot 23$. If $x^{2}$ divides this product, $x$ can have at most 2 factors of 2,1 factor of 3 , and no factors of any other prime. So $2^{2} \cdot 3=12$ is the largest value of $x$.
## $1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
12
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
What is the largest $x$ such that $x^{2}$ divides $24 \cdot 35 \cdot 46 \cdot 57$ ?
|
12 We factor the product as $2^{4} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 19 \cdot 23$. If $x^{2}$ divides this product, $x$ can have at most 2 factors of 2,1 factor of 3 , and no factors of any other prime. So $2^{2} \cdot 3=12$ is the largest value of $x$.
## $1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n3. [5]",
"solution_match": "\nAnswer: "
}
|
e8ca1e29-d08c-5737-8ae4-cdeddab4a2b2
| 608,420
|
What is the smallest prime divisor of $5^{7^{10^{7^{10}}}}+1$ ?
|
2 Notice that 5 to any power is odd, so this number is even. Then 2 is a prime divisor. It also happens to be the smallest prime.
|
2
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
What is the smallest prime divisor of $5^{7^{10^{7^{10}}}}+1$ ?
|
2 Notice that 5 to any power is odd, so this number is even. Then 2 is a prime divisor. It also happens to be the smallest prime.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n4. [6]",
"solution_match": "\nAnswer: "
}
|
60bde09b-fe04-5e94-bbfa-2d45d54a9f2a
| 608,421
|
What is the sum of all integers $x$ such that $|x+2| \leq 10$ ?
|
$\quad-42$ The inequality $|x+2| \leq 10$ holds if and only if $x+2 \leq 10$ and $x+2 \geq-10$. So $x$ must be in the range $-12 \leq x \leq 8$. If we add up the integers in this range, each positive integer cancels with its additive inverse, so the sum is equal to $-12-11-10-9=-42$.
|
-42
|
Yes
|
Yes
|
math-word-problem
|
Inequalities
|
What is the sum of all integers $x$ such that $|x+2| \leq 10$ ?
|
$\quad-42$ The inequality $|x+2| \leq 10$ holds if and only if $x+2 \leq 10$ and $x+2 \geq-10$. So $x$ must be in the range $-12 \leq x \leq 8$. If we add up the integers in this range, each positive integer cancels with its additive inverse, so the sum is equal to $-12-11-10-9=-42$.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n5. [6]",
"solution_match": "\nAnswer: "
}
|
89ef5376-64f3-5974-a1bf-03146e90e81e
| 608,422
|
Sarah is deciding whether to visit Russia or Washington, DC for the holidays. She makes her decision by rolling a regular 6 -sided die. If she gets a 1 or 2 , she goes to DC . If she rolls a 3,4 , or 5 , she goes to Russia. If she rolls a 6 , she rolls again. What is the probability that she goes to DC?
|
$\quad \frac{2}{5}$ On each roll, the probability that Sarah decides to go to Russia is $3 / 2$ times the probability she decides to go to DC. So, the total probability that she goes to Russia is $3 / 2$ times the total probability that she goes to DC. Since these probabilities sum to 1 (they are the only two eventual outcomes) Sarah goes to DC with probability $\frac{2}{5}$ and Russia with probability $\frac{3}{5}$.
## $1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
\frac{2}{5}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Sarah is deciding whether to visit Russia or Washington, DC for the holidays. She makes her decision by rolling a regular 6 -sided die. If she gets a 1 or 2 , she goes to DC . If she rolls a 3,4 , or 5 , she goes to Russia. If she rolls a 6 , she rolls again. What is the probability that she goes to DC?
|
$\quad \frac{2}{5}$ On each roll, the probability that Sarah decides to go to Russia is $3 / 2$ times the probability she decides to go to DC. So, the total probability that she goes to Russia is $3 / 2$ times the total probability that she goes to DC. Since these probabilities sum to 1 (they are the only two eventual outcomes) Sarah goes to DC with probability $\frac{2}{5}$ and Russia with probability $\frac{3}{5}$.
## $1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n6. [6]",
"solution_match": "\nAnswer: "
}
|
efd189d4-fcce-5a0d-b55d-031f8cae0f9b
| 608,423
|
Compute $2009^{2}-2008^{2}$.
|
4017 Factoring this product with difference of squares, we find it equals:
$$
(2009+2008)(2009-2008)=(4017)(1)=4017
$$
|
4017
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Compute $2009^{2}-2008^{2}$.
|
4017 Factoring this product with difference of squares, we find it equals:
$$
(2009+2008)(2009-2008)=(4017)(1)=4017
$$
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n7. [7]",
"solution_match": "\nAnswer: "
}
|
5c5d7f9d-6462-54b2-bb02-4b099e20af5f
| 608,424
|
Alice rolls two octahedral dice with the numbers $2,3,4,5,6,7,8,9$. What's the probability the two dice sum to 11 ?
|
$\frac{1}{8}$ No matter what comes up on the first die, there is exactly one number that could appear on the second die to make the sum 11 , because 2 can be paired with 9,3 with 8 , and so on. So, there is a $\frac{1}{8}$ chance of getting the correct number on the second die.
|
\frac{1}{8}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Alice rolls two octahedral dice with the numbers $2,3,4,5,6,7,8,9$. What's the probability the two dice sum to 11 ?
|
$\frac{1}{8}$ No matter what comes up on the first die, there is exactly one number that could appear on the second die to make the sum 11 , because 2 can be paired with 9,3 with 8 , and so on. So, there is a $\frac{1}{8}$ chance of getting the correct number on the second die.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n8. [7]",
"solution_match": "\nAnswer: "
}
|
5a167cfe-9d6b-5f2d-ad84-f449ecabab62
| 608,425
|
Let $a_{0}=\frac{6}{7}$, and
$$
a_{n+1}= \begin{cases}2 a_{n} & \text { if } a_{n}<\frac{1}{2} \\ 2 a_{n}-1 & \text { if } a_{n} \geq \frac{1}{2}\end{cases}
$$
Find $a_{2008}$.
|
$\sqrt{\frac{5}{7}}$ We calculate the first few $a_{i}$ :
$$
a_{1}=\frac{5}{7}, a_{2}=\frac{3}{7}, a_{3}=\frac{6}{7}=a_{0}
$$
So this sequence repeats every three terms, so $a_{2007}=a_{0}=\frac{6}{7}$. Then $a_{2008}=\frac{5}{7}$.
```
$1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
```
|
\frac{5}{7}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a_{0}=\frac{6}{7}$, and
$$
a_{n+1}= \begin{cases}2 a_{n} & \text { if } a_{n}<\frac{1}{2} \\ 2 a_{n}-1 & \text { if } a_{n} \geq \frac{1}{2}\end{cases}
$$
Find $a_{2008}$.
|
$\sqrt{\frac{5}{7}}$ We calculate the first few $a_{i}$ :
$$
a_{1}=\frac{5}{7}, a_{2}=\frac{3}{7}, a_{3}=\frac{6}{7}=a_{0}
$$
So this sequence repeats every three terms, so $a_{2007}=a_{0}=\frac{6}{7}$. Then $a_{2008}=\frac{5}{7}$.
```
$1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
```
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n9. [7]",
"solution_match": "\nAnswer: "
}
|
a608faad-84c3-598d-90c4-44ab93c31715
| 608,426
|
Find the sum of all positive integers $n$ such that $n$ divides $n^{2}+n+2$.
|
3 Since $n$ always divides $n^{2}+n$, the only $n$ that work are divisors of 2 , because if $n$ divides $a$ and $n$ divides $b$, then $n$ divides $a+b$. So the solutions are 1 and 2 which sum to 3 .
|
3
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the sum of all positive integers $n$ such that $n$ divides $n^{2}+n+2$.
|
3 Since $n$ always divides $n^{2}+n$, the only $n$ that work are divisors of 2 , because if $n$ divides $a$ and $n$ divides $b$, then $n$ divides $a+b$. So the solutions are 1 and 2 which sum to 3 .
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n10. [8]",
"solution_match": "\nAnswer: "
}
|
91985085-b857-5d05-a0eb-6ad636ad2aa4
| 608,427
|
Al has a rectangle of integer side lengths $a$ and $b$, and area 1000 . What is the smallest perimeter it could have?
|
130 To minimize the sum of the side lengths, we need to keep the height and width as close as possible, because the square has the smallest perimeter of all rectangles with a fixed area. So, 40 and 25 multiply to 1000 and are as close as possible - the $40 \times 25$ rectangle has perimeter 130 .
|
130
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Al has a rectangle of integer side lengths $a$ and $b$, and area 1000 . What is the smallest perimeter it could have?
|
130 To minimize the sum of the side lengths, we need to keep the height and width as close as possible, because the square has the smallest perimeter of all rectangles with a fixed area. So, 40 and 25 multiply to 1000 and are as close as possible - the $40 \times 25$ rectangle has perimeter 130 .
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n11. [8]",
"solution_match": "\nAnswer: "
}
|
5f5f4954-fb67-5bc2-b7c9-eb59f01e12aa
| 608,428
|
Solve the following system of equations for $w$.
$$
\begin{aligned}
& 2 w+x+y+z=1 \\
& w+2 x+y+z=2 \\
& w+x+2 y+z=2 \\
& w+x+y+2 z=1
\end{aligned}
$$
|
$\quad \frac{-1}{5}$ Add all the equations together to find that $5 x+5 y+5 z+5 w=6$, or $x+y+z+w=\frac{6}{5}$. We can now subtract this equation from the first equation to see that $w=\frac{-1}{5}$.
|
\frac{-1}{5}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Solve the following system of equations for $w$.
$$
\begin{aligned}
& 2 w+x+y+z=1 \\
& w+2 x+y+z=2 \\
& w+x+2 y+z=2 \\
& w+x+y+2 z=1
\end{aligned}
$$
|
$\quad \frac{-1}{5}$ Add all the equations together to find that $5 x+5 y+5 z+5 w=6$, or $x+y+z+w=\frac{6}{5}$. We can now subtract this equation from the first equation to see that $w=\frac{-1}{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n12. [8]",
"solution_match": "\nAnswer: "
}
|
fc72b9d4-1b07-51ea-a537-8acac70cab14
| 608,429
|
Find the number of distinct primes dividing $1 \cdot 2 \cdot 3 \cdots 9 \cdot 10$.
|
4 A prime divides this product if and only if it divides one of the multiplicands, so prime divisors of this product must be less than 10 . There are 4 primes less than 10 , namely, $2,3,5$, and 7 .
|
4
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the number of distinct primes dividing $1 \cdot 2 \cdot 3 \cdots 9 \cdot 10$.
|
4 A prime divides this product if and only if it divides one of the multiplicands, so prime divisors of this product must be less than 10 . There are 4 primes less than 10 , namely, $2,3,5$, and 7 .
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n13. [9]",
"solution_match": "\nAnswer: "
}
|
75744558-966a-57e0-9844-41bbd2fa7877
| 608,430
|
You have a $2 \times 3$ grid filled with integers between 1 and 9 . The numbers in each row and column are distinct, the first row sums to 23 , and the columns sum to 14,16 , and 17 respectively.
| | 14 | 16 | 17 |
| :---: | :---: | :---: | :---: |
| 23 | $a$ | $b$ | $c$ |
| | $x$ | $y$ | $z$ |
What is $x+2 y+3 z ?$
|
49 The sum of all 6 numbers is $14+16+17=47$, so $x+y+z=47-23=24$. If three distinct digits sum to 24 , they must be 7,8 , and 9 , because any other triple of digits would have a smaller sum. So, we try placing these digits in for $x, y$, and $z$, and the only arrangement that does not force equal digits in any row or column is $x=8, y=7, z=9$. In this case, $x+2 y+3 z=49$.
|
49
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
You have a $2 \times 3$ grid filled with integers between 1 and 9 . The numbers in each row and column are distinct, the first row sums to 23 , and the columns sum to 14,16 , and 17 respectively.
| | 14 | 16 | 17 |
| :---: | :---: | :---: | :---: |
| 23 | $a$ | $b$ | $c$ |
| | $x$ | $y$ | $z$ |
What is $x+2 y+3 z ?$
|
49 The sum of all 6 numbers is $14+16+17=47$, so $x+y+z=47-23=24$. If three distinct digits sum to 24 , they must be 7,8 , and 9 , because any other triple of digits would have a smaller sum. So, we try placing these digits in for $x, y$, and $z$, and the only arrangement that does not force equal digits in any row or column is $x=8, y=7, z=9$. In this case, $x+2 y+3 z=49$.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n14. [9]",
"solution_match": "\nAnswer: "
}
|
984aaf5a-de5e-57e1-bba8-69e8b8da46bc
| 608,431
|
If $p$ and $q$ are positive integers and $\frac{2008}{2009}<\frac{p}{q}<\frac{2009}{2010}$, what is the minimum value of $p$ ?
|
4017 By multiplying out the fraction inequalities, we find that $2008 q+1 \leq 2009 p$ and $2010 p+\leq 2009 q$. Adding 2009 times the first inequality to 2008 times the second, we find that $2008 \cdot 2009 q+4017 \leq 2008 \cdot 2009 q+p$, or $p \geq 4017$. This minumum is attained when $\frac{p}{q}=\frac{4017}{4019}$.
|
4017
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
If $p$ and $q$ are positive integers and $\frac{2008}{2009}<\frac{p}{q}<\frac{2009}{2010}$, what is the minimum value of $p$ ?
|
4017 By multiplying out the fraction inequalities, we find that $2008 q+1 \leq 2009 p$ and $2010 p+\leq 2009 q$. Adding 2009 times the first inequality to 2008 times the second, we find that $2008 \cdot 2009 q+4017 \leq 2008 \cdot 2009 q+p$, or $p \geq 4017$. This minumum is attained when $\frac{p}{q}=\frac{4017}{4019}$.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n16. [10]",
"solution_match": "\nAnswer: "
}
|
ba16553b-38b5-5c64-9832-3a5f919fcf6f
| 608,433
|
Determine the last two digits of $17^{17}$, written in base 10 .
|
77 We are asked to find the remainder when $17^{17}$ is divided by 100 . Write the power as $(7+10)^{17}$ and expand with the binomial theorem:
$$
(7+10)^{17}=7^{17}+17 \cdot 7^{16} \cdot 10+\ldots
$$
We can ignore terms with more than one factor of 10 because these terms are divisible by 100 , so adding them does not change the last two digits. Now, $7^{4}=2401$, which has remainder $1 \bmod 100$, so $7^{17}$ has last two digits 07 and $7^{16} \cdot 10$ has last two digits 70 . We add these together.
|
77
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Determine the last two digits of $17^{17}$, written in base 10 .
|
77 We are asked to find the remainder when $17^{17}$ is divided by 100 . Write the power as $(7+10)^{17}$ and expand with the binomial theorem:
$$
(7+10)^{17}=7^{17}+17 \cdot 7^{16} \cdot 10+\ldots
$$
We can ignore terms with more than one factor of 10 because these terms are divisible by 100 , so adding them does not change the last two digits. Now, $7^{4}=2401$, which has remainder $1 \bmod 100$, so $7^{17}$ has last two digits 07 and $7^{16} \cdot 10$ has last two digits 70 . We add these together.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n17. [10]",
"solution_match": "\nAnswer: "
}
|
127cb9ca-88fd-5d43-89a4-1f3790ed73ac
| 608,434
|
Find the coefficient of $x^{6}$ in the expansion of
$$
(x+1)^{6} \cdot \sum_{i=0}^{6} x^{i}
$$
|
64 Each term of $(x+1)^{6}$ can be multiplied by a unique power $x^{i}, 0 \leq i \leq 6$ to get a sixth degree term. So the answer is the sum of the coefficients of the terms of $(x+1)^{6}$, which is the same as substituting $x=1$ into this power to get $2^{6}=64$.
## $1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
64
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find the coefficient of $x^{6}$ in the expansion of
$$
(x+1)^{6} \cdot \sum_{i=0}^{6} x^{i}
$$
|
64 Each term of $(x+1)^{6}$ can be multiplied by a unique power $x^{i}, 0 \leq i \leq 6$ to get a sixth degree term. So the answer is the sum of the coefficients of the terms of $(x+1)^{6}$, which is the same as substituting $x=1$ into this power to get $2^{6}=64$.
## $1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n18. [10]",
"solution_match": "\nAnswer: "
}
|
af892f99-e656-50dd-bcb4-4e8e9d3723a3
| 608,435
|
Let $P$ be a polynomial with $P(1)=P(2)=\cdots=P(2007)=0$ and $P(0)=2009$ !. $P(x)$ has leading coefficient 1 and degree 2008. Find the largest root of $P(x)$.
|
$4034072 P(0)$ is the constant term of $P(x)$, which is the product of all the roots of the polynomial, because its degree is even. So the product of all 2008 roots is 2009 ! and the product of the first 2007 is 2007!, which means the last root is $\frac{2009!}{2007!}=2009 \cdot 2008=4034072$.
|
4034072
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $P$ be a polynomial with $P(1)=P(2)=\cdots=P(2007)=0$ and $P(0)=2009$ !. $P(x)$ has leading coefficient 1 and degree 2008. Find the largest root of $P(x)$.
|
$4034072 P(0)$ is the constant term of $P(x)$, which is the product of all the roots of the polynomial, because its degree is even. So the product of all 2008 roots is 2009 ! and the product of the first 2007 is 2007!, which means the last root is $\frac{2009!}{2007!}=2009 \cdot 2008=4034072$.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n19. [11]",
"solution_match": "\nAnswer: "
}
|
d6731ffd-ba81-5eb7-9463-1e36c6b6862b
| 608,436
|
You have a die with faces labelled 1 through 6 . On each face, you draw an arrow to an adjacent face, such that if you start on a face and follow the arrows, after 6 steps you will have passed through every face once and will be back on your starting face. How many ways are there to draw the arrows so that this is true?
|
32 There are 4 choices for where to go from face 1 . Consider the 4 faces adjacent to 1 . We can visit either 1,2 , or 3 of them before visiting the face opposite 1 . If we only visit one of these adjacent faces, we have 4 choices for which one, then we visit face 6 , opposite face 1 , then we visit the remaining 3 faces in one of two orders - for a total of 8 ways. If we visit 2 adjacent faces first, there is 8 choices for these two faces, then 2 choices for the path back from face 6 to face 1 . Lastly, there are 8 ways to visit three of the adjacent faces before visiting the opposite face. These choices give 32 total.
|
32
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
You have a die with faces labelled 1 through 6 . On each face, you draw an arrow to an adjacent face, such that if you start on a face and follow the arrows, after 6 steps you will have passed through every face once and will be back on your starting face. How many ways are there to draw the arrows so that this is true?
|
32 There are 4 choices for where to go from face 1 . Consider the 4 faces adjacent to 1 . We can visit either 1,2 , or 3 of them before visiting the face opposite 1 . If we only visit one of these adjacent faces, we have 4 choices for which one, then we visit face 6 , opposite face 1 , then we visit the remaining 3 faces in one of two orders - for a total of 8 ways. If we visit 2 adjacent faces first, there is 8 choices for these two faces, then 2 choices for the path back from face 6 to face 1 . Lastly, there are 8 ways to visit three of the adjacent faces before visiting the opposite face. These choices give 32 total.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n20. [11]",
"solution_match": "\nAnswer: "
}
|
4f2da00e-6a44-5eb5-b53f-7bfc9e538ef3
| 608,437
|
Call a number overweight if it has at least three positive integer divisors (including 1 and the number), and call a number obese if it has at least four positive integer divisors (including 1 and the number). How many positive integers between 1 and 200 are overweight, but not obese?
|
6 A positive integer is overweight, but not obese, if it has exactly 3 factors - this can only happen if that integer is the square of a prime. (If two primes, $p$ and $q$, divide the number, then $p, q$, $p q$, and 1 all divide it, making it at least obese). So, the integers less than 200 which are squares of a prime are the squares of $2,3,5,7,11$, and 13 .
## $1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
6
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Call a number overweight if it has at least three positive integer divisors (including 1 and the number), and call a number obese if it has at least four positive integer divisors (including 1 and the number). How many positive integers between 1 and 200 are overweight, but not obese?
|
6 A positive integer is overweight, but not obese, if it has exactly 3 factors - this can only happen if that integer is the square of a prime. (If two primes, $p$ and $q$, divide the number, then $p, q$, $p q$, and 1 all divide it, making it at least obese). So, the integers less than 200 which are squares of a prime are the squares of $2,3,5,7,11$, and 13 .
## $1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n21. [11]",
"solution_match": "\nAnswer: "
}
|
9743009f-952e-53af-9f0a-8551b65af99f
| 608,438
|
Sandra the Maverick has 5 pairs of shoes in a drawer, each pair a different color. Every day for 5 days, Sandra takes two shoes out and throws them out the window. If they are the same color, she treats herself to a practice problem from a past HMMT. What is the expected value (average number) of practice problems she gets to do?
|
$\quad \frac{5}{9}$ On any given day, there is a $\frac{1}{9}$ chance that the second shoe that Sandra chooses makes a pair with the first shoe she chose. Thus the average number of problems she does in a day is $\frac{1}{9}$, so, by the linearity of expectation, she does $\frac{5}{9}$ problems total, on average.
|
\frac{5}{9}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Sandra the Maverick has 5 pairs of shoes in a drawer, each pair a different color. Every day for 5 days, Sandra takes two shoes out and throws them out the window. If they are the same color, she treats herself to a practice problem from a past HMMT. What is the expected value (average number) of practice problems she gets to do?
|
$\quad \frac{5}{9}$ On any given day, there is a $\frac{1}{9}$ chance that the second shoe that Sandra chooses makes a pair with the first shoe she chose. Thus the average number of problems she does in a day is $\frac{1}{9}$, so, by the linearity of expectation, she does $\frac{5}{9}$ problems total, on average.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n22. [12]",
"solution_match": "\nAnswer: "
}
|
12587297-0d5b-5b7a-9a13-e5d51973cf87
| 608,439
|
If $x$ and $y$ are real numbers such that $\frac{(x-4)^{2}}{4}+\frac{y^{2}}{9}=1$, find the largest possible value of $\frac{x^{2}}{4}+\frac{y^{2}}{9}$.
|
9 The first equation is an ellipse with major axis parallel to the y-axis. If the second expression is set equal to a certain value $c$, then it is also the equation of an ellipse with major axis parallel to the y-axis; further, it is similar to the first ellipse. So the largest value of $c$ occurs when both ellipse are tangent on the x-axis, at $x=6, y=0$, which gives 9 as the largest value of $c$.
|
9
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
If $x$ and $y$ are real numbers such that $\frac{(x-4)^{2}}{4}+\frac{y^{2}}{9}=1$, find the largest possible value of $\frac{x^{2}}{4}+\frac{y^{2}}{9}$.
|
9 The first equation is an ellipse with major axis parallel to the y-axis. If the second expression is set equal to a certain value $c$, then it is also the equation of an ellipse with major axis parallel to the y-axis; further, it is similar to the first ellipse. So the largest value of $c$ occurs when both ellipse are tangent on the x-axis, at $x=6, y=0$, which gives 9 as the largest value of $c$.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n23. [12]",
"solution_match": "\nAnswer: "
}
|
f55c0357-ad47-56ad-b0e9-c5884510cb78
| 608,440
|
Let $f(x)=\frac{1}{1-x}$. Let $f^{k+1}(x)=f\left(f^{k}(x)\right)$, with $f^{1}(x)=f(x)$. What is $f^{2008}(2008)$ ?
|
$\frac{-1}{2007}$ Notice that, if $x \neq 0,1$, then $f^{2}(x)=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}$, which means that $f^{3}(x)=$ $\frac{1}{1-\frac{x-1}{x}}=x$. So $f^{n}$ is periodic with period $n=3$, which means that $f^{2007}(x)=x$ so $f^{2008}(2008)=$ $f(2008)=\frac{-1}{2007}$.
$1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
\frac{-1}{2007}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $f(x)=\frac{1}{1-x}$. Let $f^{k+1}(x)=f\left(f^{k}(x)\right)$, with $f^{1}(x)=f(x)$. What is $f^{2008}(2008)$ ?
|
$\frac{-1}{2007}$ Notice that, if $x \neq 0,1$, then $f^{2}(x)=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}$, which means that $f^{3}(x)=$ $\frac{1}{1-\frac{x-1}{x}}=x$. So $f^{n}$ is periodic with period $n=3$, which means that $f^{2007}(x)=x$ so $f^{2008}(2008)=$ $f(2008)=\frac{-1}{2007}$.
$1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n24. [12]",
"solution_match": "\nAnswer: "
}
|
9cf2b5ae-0a00-57bb-a770-d6b927b1b8d6
| 608,441
|
Evaluate the sum
$$
\cos \left(\frac{2 \pi}{18}\right)+\cos \left(\frac{4 \pi}{18}\right)+\cdots+\cos \left(\frac{34 \pi}{18}\right)
$$
|
-1 If $k<18$, then we can pair $\cos \left(\frac{k \pi}{18}\right)$ with $\cos \left(\frac{(18-k) \pi}{18}\right)$, and these two terms sum to 0 . If $k>18$, then the pair $\cos \left(\frac{k \pi}{18}\right)$ and $\cos \left(\frac{(36-k) \pi}{18}\right)$ also sums to 0 . So, the only term in this series that is left over is $\cos \left(\frac{18 \pi}{18}\right)=-1$.
|
-1
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Evaluate the sum
$$
\cos \left(\frac{2 \pi}{18}\right)+\cos \left(\frac{4 \pi}{18}\right)+\cdots+\cos \left(\frac{34 \pi}{18}\right)
$$
|
-1 If $k<18$, then we can pair $\cos \left(\frac{k \pi}{18}\right)$ with $\cos \left(\frac{(18-k) \pi}{18}\right)$, and these two terms sum to 0 . If $k>18$, then the pair $\cos \left(\frac{k \pi}{18}\right)$ and $\cos \left(\frac{(36-k) \pi}{18}\right)$ also sums to 0 . So, the only term in this series that is left over is $\cos \left(\frac{18 \pi}{18}\right)=-1$.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n25. [13]",
"solution_match": "\nAnswer: "
}
|
8764eee9-5ba5-58af-b873-db2891a8e6d5
| 608,442
|
John M. is sitting at ( 0,0 ), looking across the aisle at his friends sitting at $(i, j)$ for each $1 \leq i \leq 10$ and $0 \leq j \leq 5$. Unfortunately, John can only see a friend if the line connecting them doesn't pass through any other friend. How many friends can John see?
|
36 The simplest method is to draw a picture and count which friends he can see. John can see the friend on point $(i, j)$ if and only if $i$ and $j$ are relatively prime.
|
36
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
John M. is sitting at ( 0,0 ), looking across the aisle at his friends sitting at $(i, j)$ for each $1 \leq i \leq 10$ and $0 \leq j \leq 5$. Unfortunately, John can only see a friend if the line connecting them doesn't pass through any other friend. How many friends can John see?
|
36 The simplest method is to draw a picture and count which friends he can see. John can see the friend on point $(i, j)$ if and only if $i$ and $j$ are relatively prime.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n26. [13]",
"solution_match": "\nAnswer: "
}
|
543daa52-59b6-5307-8a93-562ecdeeae19
| 608,443
|
$A B C D E$ is a regular pentagon inscribed in a circle of radius 1 . What is the area of the set of points inside the circle that are farther from $A$ than they are from any other vertex?
|
$\quad \frac{\pi}{5}$ Draw the perpendicular bisectors of all the sides and diagonals of the pentagon with one endpoint at $A$. These lines all intersect in the center of the circle, because they are the set of points equidistant from two points on the circle. Now, a given point is farther from $A$ than from point $X$ if it is on the $X$ side of the perpendicular bisector of segment $A X$. So, we want to find the area of the set of all points which are separated from $A$ by all of these perpendicular bisectors, which turns out to be a single $72^{\circ}$ sector of the circle, which has area $\frac{\pi}{5}$.
## $1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
\frac{\pi}{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
$A B C D E$ is a regular pentagon inscribed in a circle of radius 1 . What is the area of the set of points inside the circle that are farther from $A$ than they are from any other vertex?
|
$\quad \frac{\pi}{5}$ Draw the perpendicular bisectors of all the sides and diagonals of the pentagon with one endpoint at $A$. These lines all intersect in the center of the circle, because they are the set of points equidistant from two points on the circle. Now, a given point is farther from $A$ than from point $X$ if it is on the $X$ side of the perpendicular bisector of segment $A X$. So, we want to find the area of the set of all points which are separated from $A$ by all of these perpendicular bisectors, which turns out to be a single $72^{\circ}$ sector of the circle, which has area $\frac{\pi}{5}$.
## $1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n27. [13]",
"solution_match": "\nAnswer: "
}
|
16ec2892-38a1-570f-9560-a0ef724757ee
| 608,444
|
Johnny the grad student is typing all the integers from 1 to $\infty$, in order. The 2 on his computer is broken however, so he just skips any number with a 2 . What's the 2008th number he types?
|
3781 Write 2008 in base 9 as 2671, and interpret the result as a base 10 number such that the base 9 digits $2,3, \ldots 8$ correspond to the base 10 digits $3,4, \ldots 9$. This gives an answer of 3781 .
|
3781
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Johnny the grad student is typing all the integers from 1 to $\infty$, in order. The 2 on his computer is broken however, so he just skips any number with a 2 . What's the 2008th number he types?
|
3781 Write 2008 in base 9 as 2671, and interpret the result as a base 10 number such that the base 9 digits $2,3, \ldots 8$ correspond to the base 10 digits $3,4, \ldots 9$. This gives an answer of 3781 .
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n28. [14]",
"solution_match": "\nAnswer: "
}
|
2fe3edb5-e17d-5b26-8aea-b919c490721a
| 608,445
|
Alice has an equilateral triangle $A B C$ of area 1. Put $D$ on $B C, E$ on $C A$, and $F$ on $A B$, with $B D=D C, C E=2 E A$, and $2 A F=F B$. Note that $A D, B E$, and $C F$ pass through a single point $M$. What is the area of triangle $E M C$ ?
|
$\frac{1}{6}$ Triangles $A C F$ and $B C F$ share a height, so the ratio of their areas is $A F / B F=1 / 2$. By the same method, the ratio of the areas of $A M F$ and $B M F$ is $1 / 2$. So, the ratio of the areas of
$A C M$ and $B C M$ is also $1 / 2$. Similarly, the ratio of the areas of $A B M$ and $B C M$ is $1 / 2$. But the sum of the areas of $A C M, B C M$, and $A B M$ is 1 , so the area of $A C M$ is $\frac{1}{4}$. Then the area of $E M C$ is $2 / 3$ the area of $A C M$, because they share heights, so their areas are in the same ratio as their bases. The area of $E M C$ is then $\frac{2 \cdot 1}{3 \cdot 4}=\frac{1}{6}$.
## $1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
\frac{1}{6}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Alice has an equilateral triangle $A B C$ of area 1. Put $D$ on $B C, E$ on $C A$, and $F$ on $A B$, with $B D=D C, C E=2 E A$, and $2 A F=F B$. Note that $A D, B E$, and $C F$ pass through a single point $M$. What is the area of triangle $E M C$ ?
|
$\frac{1}{6}$ Triangles $A C F$ and $B C F$ share a height, so the ratio of their areas is $A F / B F=1 / 2$. By the same method, the ratio of the areas of $A M F$ and $B M F$ is $1 / 2$. So, the ratio of the areas of
$A C M$ and $B C M$ is also $1 / 2$. Similarly, the ratio of the areas of $A B M$ and $B C M$ is $1 / 2$. But the sum of the areas of $A C M, B C M$, and $A B M$ is 1 , so the area of $A C M$ is $\frac{1}{4}$. Then the area of $E M C$ is $2 / 3$ the area of $A C M$, because they share heights, so their areas are in the same ratio as their bases. The area of $E M C$ is then $\frac{2 \cdot 1}{3 \cdot 4}=\frac{1}{6}$.
## $1^{\text {st }}$ HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n30. [14]",
"solution_match": "\nAnswer: "
}
|
a386d8e7-f869-5587-9b49-68357802cf7b
| 608,447
|
Find the sum of all primes $p$ for which there exists a prime $q$ such that $p^{2}+p q+q^{2}$ is a square.
|
83 and 5 both work, because $3^{2}+3 \cdot 5+5^{2}=49$. Now, say $p^{2}+p q+q^{2}=k^{2}$, for a positive integer $k$. Then $(p+q)^{2}-k^{2}=p q$, or:
$$
(p+q+k)(p+q-k)=p q
$$
Since $p+q+k$ is a divisor of $p q$, and it is greater than $p$ and $q, p+q+k=p q$. Then $p+q-k=1$. So:
$$
2 p+2 q=p q+1 \Leftrightarrow p q-2 p-2 q+4=3 \Leftrightarrow(p-2)(q-2)=3
$$
This shows that one of $p$ and $q$ is 3 and the other is 5 .
|
8
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the sum of all primes $p$ for which there exists a prime $q$ such that $p^{2}+p q+q^{2}$ is a square.
|
83 and 5 both work, because $3^{2}+3 \cdot 5+5^{2}=49$. Now, say $p^{2}+p q+q^{2}=k^{2}$, for a positive integer $k$. Then $(p+q)^{2}-k^{2}=p q$, or:
$$
(p+q+k)(p+q-k)=p q
$$
Since $p+q+k$ is a divisor of $p q$, and it is greater than $p$ and $q, p+q+k=p q$. Then $p+q-k=1$. So:
$$
2 p+2 q=p q+1 \Leftrightarrow p q-2 p-2 q+4=3 \Leftrightarrow(p-2)(q-2)=3
$$
This shows that one of $p$ and $q$ is 3 and the other is 5 .
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n31. [15]",
"solution_match": "\nAnswer: "
}
|
6e589b93-b652-5bff-936c-a035cd31a223
| 608,448
|
Pirate ships Somy and Lia are having a tough time. At the end of the year, they are both one pillage short of the minimum required for maintaining membership in the Pirate Guild, so they decide to pillage each other to bring their counts up. Somy by tradition only pillages $28 \cdot 3^{k}$ coins for integers $k$, and Lia by tradition only pillages $82 \cdot 3^{j}$ coins for integers $j$. Note that each pillage can have a different $k$ or $j$. Soma and Lia work out a system where Somy pillages Lia $n$ times, Lia pillages Somy $n$ times, and after both sets of pillages Somy and Lia are financially even.
What is the smallest $n$ can be?
|
2 Clearly, $n=1$ cannot be acheived, because $28 \cdot 3^{k}$ is never a multiple of 82 . However, two pillages is enough: Somy pillages 28 and $28 \cdot 81$ from Lia, and Lia pillages 81 and $81 \cdot 27$ from Somy. As is easily checked, both pillage $28 \cdot 82$.
|
2
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Pirate ships Somy and Lia are having a tough time. At the end of the year, they are both one pillage short of the minimum required for maintaining membership in the Pirate Guild, so they decide to pillage each other to bring their counts up. Somy by tradition only pillages $28 \cdot 3^{k}$ coins for integers $k$, and Lia by tradition only pillages $82 \cdot 3^{j}$ coins for integers $j$. Note that each pillage can have a different $k$ or $j$. Soma and Lia work out a system where Somy pillages Lia $n$ times, Lia pillages Somy $n$ times, and after both sets of pillages Somy and Lia are financially even.
What is the smallest $n$ can be?
|
2 Clearly, $n=1$ cannot be acheived, because $28 \cdot 3^{k}$ is never a multiple of 82 . However, two pillages is enough: Somy pillages 28 and $28 \cdot 81$ from Lia, and Lia pillages 81 and $81 \cdot 27$ from Somy. As is easily checked, both pillage $28 \cdot 82$.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n32. [15]",
"solution_match": "\nAnswer: "
}
|
24219ac8-201a-5ea6-9500-e23bf7d17b67
| 608,449
|
The polynomial $a x^{2}-b x+c$ has two distinct roots $p$ and $q$, with $a, b$, and $c$ positive integers and with $0<p, q<1$. Find the minimum possible value of $a$.
|
5 Let $x$ and $y$ be the roots. Then:
$$
\begin{gathered}
\frac{b}{a}=x+y<2 \Rightarrow b<2 a \\
\frac{c}{a}=x y<1 \Rightarrow c<a \Rightarrow a>1 \\
b^{2}>4 a c>4 c^{2} \Rightarrow b>2 c
\end{gathered}
$$
Evaluated at 1 , the polynomial must be greater than 0 , so $a+c>b$. Then:
$$
\begin{gathered}
2 c<b<a+c \\
2 c+1 \leq b \leq a+c-1 \\
a \geq c+2 \geq 3
\end{gathered}
$$
If $a=3$, then $c=1$ and $b=3$, by the above bounds, but this polynomial has complex roots. Similarly, if $a=4$, then $c=1$ and $b$ is forced to be either 3 or 4 , again giving either 0 or 1 distinct real roots. So $a \geq 5$. But the polynomial $5 x^{2}-5 x+1$ satisfies the condition, so 5 is the answer.
|
5
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
The polynomial $a x^{2}-b x+c$ has two distinct roots $p$ and $q$, with $a, b$, and $c$ positive integers and with $0<p, q<1$. Find the minimum possible value of $a$.
|
5 Let $x$ and $y$ be the roots. Then:
$$
\begin{gathered}
\frac{b}{a}=x+y<2 \Rightarrow b<2 a \\
\frac{c}{a}=x y<1 \Rightarrow c<a \Rightarrow a>1 \\
b^{2}>4 a c>4 c^{2} \Rightarrow b>2 c
\end{gathered}
$$
Evaluated at 1 , the polynomial must be greater than 0 , so $a+c>b$. Then:
$$
\begin{gathered}
2 c<b<a+c \\
2 c+1 \leq b \leq a+c-1 \\
a \geq c+2 \geq 3
\end{gathered}
$$
If $a=3$, then $c=1$ and $b=3$, by the above bounds, but this polynomial has complex roots. Similarly, if $a=4$, then $c=1$ and $b$ is forced to be either 3 or 4 , again giving either 0 or 1 distinct real roots. So $a \geq 5$. But the polynomial $5 x^{2}-5 x+1$ satisfies the condition, so 5 is the answer.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n33. [15]",
"solution_match": "\nAnswer: "
}
|
113d6c87-4c16-5ca1-92cb-a81bd6e7ccf9
| 608,450
|
Find $\max \{\operatorname{Perimeter}(T)\}$ for $T$ a triangle contained in a regular septagon (7-sided figure) of unit edge length. Write your answer $N$ to 2 places after the decimal. If the correct answer rounded to 2 decimal places is $A$, you will receive 0 points if $N<A$ and $\lfloor\max \{0,25-50 \cdot(N-A)\}\rfloor$ points otherwise.
|
5.85086 Let the septagon be $A_{0} A_{1} \ldots A_{6}$.
If $x$ is a point that can move along the x -axis, the distance from $x$ to a fixed point $p$ is a convex function in the x -coordinate. Therefore, the sum of the distances from $x$ to two other points is convex too, so if $x$ is constrained to lie on a closed line segment, its maximum value is attained at an endpoint. Therefore, the triangle of maximal perimeter has vertices at the vertices of the pentagon. The triangle with the largest such perimeter has almost evenly spaced vertices, so triangle $A_{0} A_{2} A_{4}$ has the maximal perimeter. It has area $5.85 \ldots$
|
5.85
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Find $\max \{\operatorname{Perimeter}(T)\}$ for $T$ a triangle contained in a regular septagon (7-sided figure) of unit edge length. Write your answer $N$ to 2 places after the decimal. If the correct answer rounded to 2 decimal places is $A$, you will receive 0 points if $N<A$ and $\lfloor\max \{0,25-50 \cdot(N-A)\}\rfloor$ points otherwise.
|
5.85086 Let the septagon be $A_{0} A_{1} \ldots A_{6}$.
If $x$ is a point that can move along the x -axis, the distance from $x$ to a fixed point $p$ is a convex function in the x -coordinate. Therefore, the sum of the distances from $x$ to two other points is convex too, so if $x$ is constrained to lie on a closed line segment, its maximum value is attained at an endpoint. Therefore, the triangle of maximal perimeter has vertices at the vertices of the pentagon. The triangle with the largest such perimeter has almost evenly spaced vertices, so triangle $A_{0} A_{2} A_{4}$ has the maximal perimeter. It has area $5.85 \ldots$
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n35. [25]",
"solution_match": "\nAnswer: "
}
|
13707b53-a24c-57c8-8b5c-501aab8d258f
| 608,451
|
How many numbers less than $1,000,000$ are the product of exactly 2 distinct primes? You will receive $\left\lfloor 25-50 \cdot\left|\frac{N}{A}-1\right|\right\rfloor$ points, if you submit $N$ and the correct answer is $A$.
|
209867 While it is difficult to compute this answer without writing a program or using a calculator, it can be approximated using the fact that the number of primes less than a positive integer $n$ is about $\frac{n}{\log n}$.
|
209867
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many numbers less than $1,000,000$ are the product of exactly 2 distinct primes? You will receive $\left\lfloor 25-50 \cdot\left|\frac{N}{A}-1\right|\right\rfloor$ points, if you submit $N$ and the correct answer is $A$.
|
209867 While it is difficult to compute this answer without writing a program or using a calculator, it can be approximated using the fact that the number of primes less than a positive integer $n$ is about $\frac{n}{\log n}$.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-guts-solutions.jsonl",
"problem_match": "\n36. [25]",
"solution_match": "\nAnswer: "
}
|
880be5cb-b987-5650-a288-48f46df76925
| 608,452
|
(a) Decompose 1 into unit fractions.
|
$\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$
(b) Decompose $\frac{1}{4}$ into unit fractions.
|
\frac{1}{2}+\frac{1}{3}+\frac{1}{6}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
(a) Decompose 1 into unit fractions.
|
$\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$
(b) Decompose $\frac{1}{4}$ into unit fractions.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nAnswer: "
}
|
380238ba-8311-5821-b6a7-1da82897f127
| 608,453
|
An aside: the sum of all the unit fractions
It is possible to show that, given any real M , there exists a positive integer $k$ large enough that:
$$
\sum_{n=1}^{k} \frac{1}{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3} \ldots>M
$$
Note that this statement means that the infinite harmonic series, $\sum_{n=1}^{\infty} \frac{1}{n}$, grows without bound, or diverges. For the specific example $\mathrm{M}=5$, find a value of $k$, not necessarily the smallest, such that the inequality holds. Justify your answer.
|
Note that $\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{1}{2 n}+\ldots+\frac{1}{2 n}=\frac{1}{2}$. Therefore, if we apply this to $n=1,2,4,8,16,32,64,128$, we get
$$
\left(\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\ldots+\left(\frac{1}{129}+\ldots+\frac{1}{256}\right)>\frac{1}{2}+\ldots+\frac{1}{2}=4
$$
so, adding in $\frac{1}{1}$, we get
$$
\sum_{n=1}^{256} \frac{1}{n}>5
$$
so $k=256$ will suffice.
|
256
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
An aside: the sum of all the unit fractions
It is possible to show that, given any real M , there exists a positive integer $k$ large enough that:
$$
\sum_{n=1}^{k} \frac{1}{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3} \ldots>M
$$
Note that this statement means that the infinite harmonic series, $\sum_{n=1}^{\infty} \frac{1}{n}$, grows without bound, or diverges. For the specific example $\mathrm{M}=5$, find a value of $k$, not necessarily the smallest, such that the inequality holds. Justify your answer.
|
Note that $\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{1}{2 n}+\ldots+\frac{1}{2 n}=\frac{1}{2}$. Therefore, if we apply this to $n=1,2,4,8,16,32,64,128$, we get
$$
\left(\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\ldots+\left(\frac{1}{129}+\ldots+\frac{1}{256}\right)>\frac{1}{2}+\ldots+\frac{1}{2}=4
$$
so, adding in $\frac{1}{1}$, we get
$$
\sum_{n=1}^{256} \frac{1}{n}>5
$$
so $k=256$ will suffice.
|
{
"resource_path": "HarvardMIT/segmented/en-121-2008-nov-team-solutions.jsonl",
"problem_match": "\n## 5. ",
"solution_match": "\nSolution: "
}
|
72de498a-6f83-5176-987e-bb92815dcf72
| 608,457
|
If $a$ and $b$ are positive integers such that $a^{2}-b^{4}=2009$, find $a+b$.
|
We can factor the equation as $\left(a-b^{2}\right)\left(a+b^{2}\right)=41 \cdot 49$, from which it is evident that $a=45$ and $b=2$ is a possible solution. By examining the factors of 2009 , one can see that there are no other solutions.
|
47
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
If $a$ and $b$ are positive integers such that $a^{2}-b^{4}=2009$, find $a+b$.
|
We can factor the equation as $\left(a-b^{2}\right)\left(a+b^{2}\right)=41 \cdot 49$, from which it is evident that $a=45$ and $b=2$ is a possible solution. By examining the factors of 2009 , one can see that there are no other solutions.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"problem_match": "\n1. [3]",
"solution_match": "\nSolution: "
}
|
c9e70559-4001-5700-9eaa-db1758d4ee9c
| 608,464
|
Let $S$ be the sum of all the real coefficients of the expansion of $(1+i x)^{2009}$. What is $\log _{2}(S)$ ?
|
The sum of all the coefficients is $(1+i)^{2009}$, and the sum of the real coefficients is the real part of this, which is $\frac{1}{2}\left((1+i)^{2009}+(1-i)^{2009}\right)=2^{1004}$. Thus $\log _{2}(S)=1004$.
|
1004
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $S$ be the sum of all the real coefficients of the expansion of $(1+i x)^{2009}$. What is $\log _{2}(S)$ ?
|
The sum of all the coefficients is $(1+i)^{2009}$, and the sum of the real coefficients is the real part of this, which is $\frac{1}{2}\left((1+i)^{2009}+(1-i)^{2009}\right)=2^{1004}$. Thus $\log _{2}(S)=1004$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"problem_match": "\n2. [3]",
"solution_match": "\nSolution: "
}
|
46629c9d-1ef8-5d70-a9e2-577b14204a83
| 608,465
|
If $\tan x+\tan y=4$ and $\cot x+\cot y=5$, compute $\tan (x+y)$.
|
We have $\cot x+\cot y=\frac{\tan x+\tan y}{\tan x \tan y}$, so $\tan x \tan y=\frac{4}{5}$. Thus, by the tan sum formula, $\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=20$.
|
20
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
If $\tan x+\tan y=4$ and $\cot x+\cot y=5$, compute $\tan (x+y)$.
|
We have $\cot x+\cot y=\frac{\tan x+\tan y}{\tan x \tan y}$, so $\tan x \tan y=\frac{4}{5}$. Thus, by the tan sum formula, $\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=20$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"problem_match": "\n3. [4]",
"solution_match": "\nSolution: "
}
|
c4537555-15e5-5219-a428-981bae856bf1
| 22,628
|
Suppose $a, b$ and $c$ are integers such that the greatest common divisor of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x+1$ (in the ring of polynomials in $x$ with integer coefficients), and the least common multiple of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x^{3}-4 x^{2}+x+6$. Find $a+b+c$.
|
Since $x+1$ divides $x^{2}+a x+b$ and the constant term is $b$, we have $x^{2}+a x+b=(x+1)(x+b)$, and similarly $x^{2}+b x+c=(x+1)(x+c)$. Therefore, $a=b+1=c+2$. Furthermore, the least common multiple of the two polynomials is $(x+1)(x+b)(x+b-1)=x^{3}-4 x^{2}+x+6$, so $b=-2$. Thus $a=-1$ and $c=-3$, and $a+b+c=-6$.
|
-6
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose $a, b$ and $c$ are integers such that the greatest common divisor of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x+1$ (in the ring of polynomials in $x$ with integer coefficients), and the least common multiple of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x^{3}-4 x^{2}+x+6$. Find $a+b+c$.
|
Since $x+1$ divides $x^{2}+a x+b$ and the constant term is $b$, we have $x^{2}+a x+b=(x+1)(x+b)$, and similarly $x^{2}+b x+c=(x+1)(x+c)$. Therefore, $a=b+1=c+2$. Furthermore, the least common multiple of the two polynomials is $(x+1)(x+b)(x+b-1)=x^{3}-4 x^{2}+x+6$, so $b=-2$. Thus $a=-1$ and $c=-3$, and $a+b+c=-6$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"problem_match": "\n4. [4]",
"solution_match": "\nSolution: "
}
|
311e8683-241a-540a-bfe8-c4226dfcbf62
| 608,466
|
Let $a, b$, and $c$ be the 3 roots of $x^{3}-x+1=0$. Find $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$.
|
We can substitute $x=y-1$ to obtain a polynomial having roots $a+1, b+1, c+1$, namely, $(y-1)^{3}-(y-1)+1=y^{3}-3 y^{2}+2 y+1$. The sum of the reciprocals of the roots of this polynomial is, by Viete's formulas, $\frac{2}{-1}=-2$.
|
-2
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a, b$, and $c$ be the 3 roots of $x^{3}-x+1=0$. Find $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$.
|
We can substitute $x=y-1$ to obtain a polynomial having roots $a+1, b+1, c+1$, namely, $(y-1)^{3}-(y-1)+1=y^{3}-3 y^{2}+2 y+1$. The sum of the reciprocals of the roots of this polynomial is, by Viete's formulas, $\frac{2}{-1}=-2$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"problem_match": "\n5. [4]",
"solution_match": "\nSolution: "
}
|
24ac4e28-232f-5185-af08-8ec1c7ebf67a
| 608,467
|
Let $x$ and $y$ be positive real numbers and $\theta$ an angle such that $\theta \neq \frac{\pi}{2} n$ for any integer $n$. Suppose
$$
\frac{\sin \theta}{x}=\frac{\cos \theta}{y}
$$
and
$$
\frac{\cos ^{4} \theta}{x^{4}}+\frac{\sin ^{4} \theta}{y^{4}}=\frac{97 \sin 2 \theta}{x^{3} y+y^{3} x}
$$
Compute $\frac{x}{y}+\frac{y}{x}$.
|
From the first relation, there exists a real number $k$ such that $x=k \sin \theta$ and $y=k \cos \theta$. Then we have
$$
\frac{\cos ^{4} \theta}{\sin ^{4} \theta}+\frac{\sin ^{4} \theta}{\cos ^{4} \theta}=\frac{194 \sin \theta \cos \theta}{\sin \theta \cos \theta\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}=194
$$
Notice that if $t=\frac{x}{y}+\frac{y}{x}$ then $\left(t^{2}-2\right)^{2}-2=\frac{\cos ^{4} \theta}{\sin ^{4} \theta}+\frac{\sin ^{4} \theta}{\cos ^{4} \theta}=194$ and so $t=4$.
|
4
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $x$ and $y$ be positive real numbers and $\theta$ an angle such that $\theta \neq \frac{\pi}{2} n$ for any integer $n$. Suppose
$$
\frac{\sin \theta}{x}=\frac{\cos \theta}{y}
$$
and
$$
\frac{\cos ^{4} \theta}{x^{4}}+\frac{\sin ^{4} \theta}{y^{4}}=\frac{97 \sin 2 \theta}{x^{3} y+y^{3} x}
$$
Compute $\frac{x}{y}+\frac{y}{x}$.
|
From the first relation, there exists a real number $k$ such that $x=k \sin \theta$ and $y=k \cos \theta$. Then we have
$$
\frac{\cos ^{4} \theta}{\sin ^{4} \theta}+\frac{\sin ^{4} \theta}{\cos ^{4} \theta}=\frac{194 \sin \theta \cos \theta}{\sin \theta \cos \theta\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}=194
$$
Notice that if $t=\frac{x}{y}+\frac{y}{x}$ then $\left(t^{2}-2\right)^{2}-2=\frac{\cos ^{4} \theta}{\sin ^{4} \theta}+\frac{\sin ^{4} \theta}{\cos ^{4} \theta}=194$ and so $t=4$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"problem_match": "\n6. [5]",
"solution_match": "\nSolution: "
}
|
4737e054-d350-57dd-9690-6ea83af586a3
| 608,468
|
If $a, b, x$, and $y$ are real numbers such that $a x+b y=3, a x^{2}+b y^{2}=7, a x^{3}+b y^{3}=16$, and $a x^{4}+b y^{4}=42$, find $a x^{5}+b y^{5}$.
|
We have $a x^{3}+b y^{3}=16$, so $\left(a x^{3}+b y^{3}\right)(x+y)=16(x+y)$ and thus
$$
a x^{4}+b y^{4}+x y\left(a x^{2}+b y^{2}\right)=16(x+y)
$$
It follows that
$$
42+7 x y=16(x+y)
$$
From $a x^{2}+b y^{2}=7$, we have $\left(a x^{2}+b y^{2}\right)(x+y)=7(x+y)$ so $a x^{3}+b y^{3}+x y\left(a x^{2}+b y^{2}\right)=7(x+y)$. This simplifies to
$$
16+3 x y=7(x+y)
$$
We can now solve for $x+y$ and $x y$ from (1) and (2) to find $x+y=-14$ and $x y=-38$. Thus we have $\left(a x^{4}+b y^{4}\right)(x+y)=42(x+y)$, and so $a x^{5}+b y^{5}+x y\left(a x^{3}+b y^{3}\right)=42(x+y)$. Finally, it follows that $a x^{5}+b y^{5}=42(x+y)-16 x y=20$ as desired.
|
20
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
If $a, b, x$, and $y$ are real numbers such that $a x+b y=3, a x^{2}+b y^{2}=7, a x^{3}+b y^{3}=16$, and $a x^{4}+b y^{4}=42$, find $a x^{5}+b y^{5}$.
|
We have $a x^{3}+b y^{3}=16$, so $\left(a x^{3}+b y^{3}\right)(x+y)=16(x+y)$ and thus
$$
a x^{4}+b y^{4}+x y\left(a x^{2}+b y^{2}\right)=16(x+y)
$$
It follows that
$$
42+7 x y=16(x+y)
$$
From $a x^{2}+b y^{2}=7$, we have $\left(a x^{2}+b y^{2}\right)(x+y)=7(x+y)$ so $a x^{3}+b y^{3}+x y\left(a x^{2}+b y^{2}\right)=7(x+y)$. This simplifies to
$$
16+3 x y=7(x+y)
$$
We can now solve for $x+y$ and $x y$ from (1) and (2) to find $x+y=-14$ and $x y=-38$. Thus we have $\left(a x^{4}+b y^{4}\right)(x+y)=42(x+y)$, and so $a x^{5}+b y^{5}+x y\left(a x^{3}+b y^{3}\right)=42(x+y)$. Finally, it follows that $a x^{5}+b y^{5}=42(x+y)-16 x y=20$ as desired.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"problem_match": "\n8. [7]",
"solution_match": "\nSolution: "
}
|
dee3aec6-6fa7-5c7d-a199-09722a2944ce
| 608,470
|
Let $f(x)=x^{4}+14 x^{3}+52 x^{2}+56 x+16$. Let $z_{1}, z_{2}, z_{3}, z_{4}$ be the four roots of $f$. Find the smallest possible value of $\left|z_{a} z_{b}+z_{c} z_{d}\right|$ where $\{a, b, c, d\}=\{1,2,3,4\}$.
|
Note that $\frac{1}{16} f(2 x)=x^{4}+7 x^{3}+13 x^{2}+7 x+1$. Because the coefficients of this polynomial are symmetric, if $r$ is a root of $f(x)$ then $\frac{4}{r}$ is as well. Further, $f(-1)=-1$ and $f(-2)=16$ so $f(x)$ has two distinct roots on $(-2,0)$ and two more roots on $(-\infty,-2)$. Now, if $\sigma$ is a permutation of $\{1,2,3,4\}$ :
$\left|z_{\sigma(1)} z_{\sigma(2)}+z_{\sigma(3)} z_{\sigma(4)}\right| \leq \frac{1}{2}\left(z_{\sigma(1)} z_{\sigma(2)}+z_{\sigma(3)} z_{\sigma(4)}+z_{\sigma(4)} z_{\sigma(3)}+z_{\sigma(2)} z_{\sigma(1)}\right)$
Let the roots be ordered $z_{1} \leq z_{2} \leq z_{3} \leq z_{4}$, then by rearrangement the last expression is at least:
$\frac{1}{2}\left(z_{1} z_{4}+z_{2} z_{3}+z_{3} z_{2}+z_{4} z_{1}\right)$
Since the roots come in pairs $z_{1} z_{4}=z_{2} z_{3}=4$, our expression is minimized when $\sigma(1)=1, \sigma(2)=$ $4, \sigma(3)=3, \sigma(4)=2$ and its minimum value is 8 .
|
8
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $f(x)=x^{4}+14 x^{3}+52 x^{2}+56 x+16$. Let $z_{1}, z_{2}, z_{3}, z_{4}$ be the four roots of $f$. Find the smallest possible value of $\left|z_{a} z_{b}+z_{c} z_{d}\right|$ where $\{a, b, c, d\}=\{1,2,3,4\}$.
|
Note that $\frac{1}{16} f(2 x)=x^{4}+7 x^{3}+13 x^{2}+7 x+1$. Because the coefficients of this polynomial are symmetric, if $r$ is a root of $f(x)$ then $\frac{4}{r}$ is as well. Further, $f(-1)=-1$ and $f(-2)=16$ so $f(x)$ has two distinct roots on $(-2,0)$ and two more roots on $(-\infty,-2)$. Now, if $\sigma$ is a permutation of $\{1,2,3,4\}$ :
$\left|z_{\sigma(1)} z_{\sigma(2)}+z_{\sigma(3)} z_{\sigma(4)}\right| \leq \frac{1}{2}\left(z_{\sigma(1)} z_{\sigma(2)}+z_{\sigma(3)} z_{\sigma(4)}+z_{\sigma(4)} z_{\sigma(3)}+z_{\sigma(2)} z_{\sigma(1)}\right)$
Let the roots be ordered $z_{1} \leq z_{2} \leq z_{3} \leq z_{4}$, then by rearrangement the last expression is at least:
$\frac{1}{2}\left(z_{1} z_{4}+z_{2} z_{3}+z_{3} z_{2}+z_{4} z_{1}\right)$
Since the roots come in pairs $z_{1} z_{4}=z_{2} z_{3}=4$, our expression is minimized when $\sigma(1)=1, \sigma(2)=$ $4, \sigma(3)=3, \sigma(4)=2$ and its minimum value is 8 .
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-alg-solutions.jsonl",
"problem_match": "\n9. [7]",
"solution_match": "\nSolution: "
}
|
739a5b39-5cb5-58e3-90b4-59e52725b098
| 608,471
|
Let $f$ be a differentiable real-valued function defined on the positive real numbers. The tangent lines to the graph of $f$ always meet the $y$-axis 1 unit lower than where they meet the function. If $f(1)=0$, what is $f(2)$ ?
|
The tangent line to $f$ at $x$ meets the $y$-axis at $f(x)-1$ for any $x$, so the slope of the tangent line is $f^{\prime}(x)=\frac{1}{x}$, and so $f(x)=\ln (x)+C$ for some $a$. Since $f(1)=0$, we have $C=0$, and so $f(x)=\ln (x)$. Thus $f(2)=\ln (2)$.
|
\ln (2)
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Let $f$ be a differentiable real-valued function defined on the positive real numbers. The tangent lines to the graph of $f$ always meet the $y$-axis 1 unit lower than where they meet the function. If $f(1)=0$, what is $f(2)$ ?
|
The tangent line to $f$ at $x$ meets the $y$-axis at $f(x)-1$ for any $x$, so the slope of the tangent line is $f^{\prime}(x)=\frac{1}{x}$, and so $f(x)=\ln (x)+C$ for some $a$. Since $f(1)=0$, we have $C=0$, and so $f(x)=\ln (x)$. Thus $f(2)=\ln (2)$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"problem_match": "\n1. [3]",
"solution_match": "\nSolution: "
}
|
4f95e57f-3078-5933-a4c5-9beef3c882bc
| 608,473
|
The differentiable function $F: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $F(0)=-1$ and
$$
\frac{d}{d x} F(x)=\sin (\sin (\sin (\sin (x)))) \cdot \cos (\sin (\sin (x))) \cdot \cos (\sin (x)) \cdot \cos (x)
$$
Find $F(x)$ as a function of $x$.
|
$-\cos (\sin (\sin (\sin (x))))$
|
-\cos (\sin (\sin (\sin (x))))
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
The differentiable function $F: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $F(0)=-1$ and
$$
\frac{d}{d x} F(x)=\sin (\sin (\sin (\sin (x)))) \cdot \cos (\sin (\sin (x))) \cdot \cos (\sin (x)) \cdot \cos (x)
$$
Find $F(x)$ as a function of $x$.
|
$-\cos (\sin (\sin (\sin (x))))$
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"problem_match": "\n2. [3]",
"solution_match": "\nAnswer: "
}
|
9dd1469b-331b-5a88-9234-f9973b9523e2
| 608,474
|
Compute $e^{A}$ where $A$ is defined as
$$
\int_{3 / 4}^{4 / 3} \frac{2 x^{2}+x+1}{x^{3}+x^{2}+x+1} d x
$$
|
We can use partial fractions to decompose the integrand to $\frac{1}{x+1}+\frac{x}{x^{2}+1}$, and then integrate the addends separately by substituting $u=x+1$ for the former and $u=x^{2}+1$ for latter, to obtain $\ln (x+1)+\left.\frac{1}{2} \ln \left(x^{2}+1\right)\right|_{3 / 4} ^{4 / 3}=\left.\ln \left((x+1) \sqrt{x^{2}+1}\right)\right|_{3 / 4} ^{4 / 3}=\ln \frac{16}{9}$. Thus $e^{A}=16 / 9$.
Alternate solution: Substituting $u=1 / x$, we find
$$
A=\int_{4 / 3}^{3 / 4} \frac{2 u+u^{2}+u^{3}}{1+u+u^{2}+u^{3}}\left(-\frac{1}{u^{2}}\right) d u=\int_{3 / 4}^{4 / 3} \frac{2 / u+1+u}{1+u+u^{2}+u^{3}} d u
$$
Adding this to the original integral, we find
$$
2 A=\int_{3 / 4}^{4 / 3} \frac{2 / u+2+2 u+2 u^{2}}{1+u+u^{2}+u^{3}} d u=\int_{3 / 4}^{4 / 3} \frac{2}{u} d u
$$
Thus $A=\ln \frac{16}{9}$ and $e^{A}=\frac{16}{9}$.
|
\frac{16}{9}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Compute $e^{A}$ where $A$ is defined as
$$
\int_{3 / 4}^{4 / 3} \frac{2 x^{2}+x+1}{x^{3}+x^{2}+x+1} d x
$$
|
We can use partial fractions to decompose the integrand to $\frac{1}{x+1}+\frac{x}{x^{2}+1}$, and then integrate the addends separately by substituting $u=x+1$ for the former and $u=x^{2}+1$ for latter, to obtain $\ln (x+1)+\left.\frac{1}{2} \ln \left(x^{2}+1\right)\right|_{3 / 4} ^{4 / 3}=\left.\ln \left((x+1) \sqrt{x^{2}+1}\right)\right|_{3 / 4} ^{4 / 3}=\ln \frac{16}{9}$. Thus $e^{A}=16 / 9$.
Alternate solution: Substituting $u=1 / x$, we find
$$
A=\int_{4 / 3}^{3 / 4} \frac{2 u+u^{2}+u^{3}}{1+u+u^{2}+u^{3}}\left(-\frac{1}{u^{2}}\right) d u=\int_{3 / 4}^{4 / 3} \frac{2 / u+1+u}{1+u+u^{2}+u^{3}} d u
$$
Adding this to the original integral, we find
$$
2 A=\int_{3 / 4}^{4 / 3} \frac{2 / u+2+2 u+2 u^{2}}{1+u+u^{2}+u^{3}} d u=\int_{3 / 4}^{4 / 3} \frac{2}{u} d u
$$
Thus $A=\ln \frac{16}{9}$ and $e^{A}=\frac{16}{9}$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"problem_match": "\n3. [4]",
"solution_match": "\nSolution: "
}
|
26d41061-1439-5a33-927f-71cfc68510dc
| 608,475
|
Let $P$ be a fourth degree polynomial, with derivative $P^{\prime}$, such that $P(1)=P(3)=P(5)=P^{\prime}(7)=0$. Find the real number $x \neq 1,3,5$ such that $P(x)=0$.
|
Observe that 7 is not a root of $P$. If $r_{1}, r_{2}, r_{3}, r_{4}$ are the roots of $P$, then $\frac{P^{\prime}(7)}{P(7)}=$ $\sum_{i} \frac{1}{7-r_{i}}=0$. Thus $r_{4}=7-\left(\sum_{i \neq 4} \frac{1}{7-r_{i}}\right)^{-1}=7+\left(\frac{1}{6}+\frac{1}{4}+\frac{1}{2}\right)^{-1}=7+12 / 11=89 / 11$.
|
\frac{89}{11}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $P$ be a fourth degree polynomial, with derivative $P^{\prime}$, such that $P(1)=P(3)=P(5)=P^{\prime}(7)=0$. Find the real number $x \neq 1,3,5$ such that $P(x)=0$.
|
Observe that 7 is not a root of $P$. If $r_{1}, r_{2}, r_{3}, r_{4}$ are the roots of $P$, then $\frac{P^{\prime}(7)}{P(7)}=$ $\sum_{i} \frac{1}{7-r_{i}}=0$. Thus $r_{4}=7-\left(\sum_{i \neq 4} \frac{1}{7-r_{i}}\right)^{-1}=7+\left(\frac{1}{6}+\frac{1}{4}+\frac{1}{2}\right)^{-1}=7+12 / 11=89 / 11$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"problem_match": "\n4. [4]",
"solution_match": "\nSolution: "
}
|
5688be7b-fe29-5461-a9b8-ef84a4763f86
| 608,476
|
Compute
$$
\lim _{h \rightarrow 0} \frac{\sin \left(\frac{\pi}{3}+4 h\right)-4 \sin \left(\frac{\pi}{3}+3 h\right)+6 \sin \left(\frac{\pi}{3}+2 h\right)-4 \sin \left(\frac{\pi}{3}+h\right)+\sin \left(\frac{\pi}{3}\right)}{h^{4}}
$$
|
The derivative of a function is defined as $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$. Iterating this formula four times yields
$$
f^{(4)}(x)=\lim _{h \rightarrow 0} \frac{f(x+4 h)-4 f(x+3 h)+6 f(x+2 h)-4 f(x+h)+f(x)}{h^{4}} .
$$
Substituting $f=\sin$ and $x=\pi / 3$, the expression is equal to $\sin ^{(4)}(\pi / 3)=\sin (\pi / 3)=\frac{\sqrt{3}}{2}$.
|
\frac{\sqrt{3}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Compute
$$
\lim _{h \rightarrow 0} \frac{\sin \left(\frac{\pi}{3}+4 h\right)-4 \sin \left(\frac{\pi}{3}+3 h\right)+6 \sin \left(\frac{\pi}{3}+2 h\right)-4 \sin \left(\frac{\pi}{3}+h\right)+\sin \left(\frac{\pi}{3}\right)}{h^{4}}
$$
|
The derivative of a function is defined as $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$. Iterating this formula four times yields
$$
f^{(4)}(x)=\lim _{h \rightarrow 0} \frac{f(x+4 h)-4 f(x+3 h)+6 f(x+2 h)-4 f(x+h)+f(x)}{h^{4}} .
$$
Substituting $f=\sin$ and $x=\pi / 3$, the expression is equal to $\sin ^{(4)}(\pi / 3)=\sin (\pi / 3)=\frac{\sqrt{3}}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"problem_match": "\n5. [4]",
"solution_match": "\nSolution: "
}
|
a4733cd5-f057-5bba-90bd-aa369e91c8f0
| 608,477
|
Let $p_{0}(x), p_{1}(x), p_{2}(x), \ldots$ be polynomials such that $p_{0}(x)=x$ and for all positive integers $n$, $\frac{d}{d x} p_{n}(x)=p_{n-1}(x)$. Define the function $p(x):[0, \infty) \rightarrow \mathbb{R} x$ by $p(x)=p_{n}(x)$ for all $x \in[n, n+1]$. Given that $p(x)$ is continuous on $[0, \infty)$, compute
$$
\sum_{n=0}^{\infty} p_{n}(2009)
$$
|
$e^{2010}-e^{2009}-1$
|
e^{2010}-e^{2009}-1
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Let $p_{0}(x), p_{1}(x), p_{2}(x), \ldots$ be polynomials such that $p_{0}(x)=x$ and for all positive integers $n$, $\frac{d}{d x} p_{n}(x)=p_{n-1}(x)$. Define the function $p(x):[0, \infty) \rightarrow \mathbb{R} x$ by $p(x)=p_{n}(x)$ for all $x \in[n, n+1]$. Given that $p(x)$ is continuous on $[0, \infty)$, compute
$$
\sum_{n=0}^{\infty} p_{n}(2009)
$$
|
$e^{2010}-e^{2009}-1$
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"problem_match": "\n6. [5]",
"solution_match": "\nAnswer: "
}
|
684b7be4-b7aa-5782-bc78-dd1c20a7cfd5
| 608,478
|
A line in the plane is called strange if it passes through $(a, 0)$ and $(0,10-a)$ for some $a$ in the interval $[0,10]$. A point in the plane is called charming if it lies in the first quadrant and also lies below some strange line. What is the area of the set of all charming points?
|
$50 / 3$
|
\frac{50}{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A line in the plane is called strange if it passes through $(a, 0)$ and $(0,10-a)$ for some $a$ in the interval $[0,10]$. A point in the plane is called charming if it lies in the first quadrant and also lies below some strange line. What is the area of the set of all charming points?
|
$50 / 3$
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"problem_match": "\n7. [5]",
"solution_match": "\nAnswer: "
}
|
182dd21d-c342-5b48-86ed-40d60b8fc242
| 608,479
|
Compute
$$
\int_{1}^{\sqrt{3}} x^{2 x^{2}+1}+\ln \left(x^{2 x^{2 x^{2}+1}}\right) d x
$$
|
Using the fact that $x=e^{\ln (x)}$, we evaluate the integral as follows:
$$
\begin{aligned}
\int x^{2 x^{2}+1}+\ln \left(x^{2 x^{2 x^{2}+1}}\right) d x & =\int x^{2 x^{2}+1}+x^{2 x^{2}+1} \ln \left(x^{2}\right) d x \\
& =\int e^{\ln (x)\left(2 x^{2}+1\right)}\left(1+\ln \left(x^{2}\right)\right) d x \\
& =\int x e^{x^{2} \ln \left(x^{2}\right)}\left(1+\ln \left(x^{2}\right)\right) d x
\end{aligned}
$$
Noticing that the derivative of $x^{2} \ln \left(x^{2}\right)$ is $2 x\left(1+\ln \left(x^{2}\right)\right)$, it follows that the integral evaluates to
$$
\frac{1}{2} e^{x^{2} \ln \left(x^{2}\right)}=\frac{1}{2} x^{2 x^{2}}
$$
Evaluating this from 1 to $\sqrt{3}$ we obtain the answer.
|
\frac{1}{2} \left( (\sqrt{3})^{2 (\sqrt{3})^{2}} - 1^{2 \cdot 1^{2}} \right)
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Compute
$$
\int_{1}^{\sqrt{3}} x^{2 x^{2}+1}+\ln \left(x^{2 x^{2 x^{2}+1}}\right) d x
$$
|
Using the fact that $x=e^{\ln (x)}$, we evaluate the integral as follows:
$$
\begin{aligned}
\int x^{2 x^{2}+1}+\ln \left(x^{2 x^{2 x^{2}+1}}\right) d x & =\int x^{2 x^{2}+1}+x^{2 x^{2}+1} \ln \left(x^{2}\right) d x \\
& =\int e^{\ln (x)\left(2 x^{2}+1\right)}\left(1+\ln \left(x^{2}\right)\right) d x \\
& =\int x e^{x^{2} \ln \left(x^{2}\right)}\left(1+\ln \left(x^{2}\right)\right) d x
\end{aligned}
$$
Noticing that the derivative of $x^{2} \ln \left(x^{2}\right)$ is $2 x\left(1+\ln \left(x^{2}\right)\right)$, it follows that the integral evaluates to
$$
\frac{1}{2} e^{x^{2} \ln \left(x^{2}\right)}=\frac{1}{2} x^{2 x^{2}}
$$
Evaluating this from 1 to $\sqrt{3}$ we obtain the answer.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"problem_match": "\n8. [7]",
"solution_match": "\nSolution: "
}
|
c90637ab-b8f5-5080-a430-42b7cb1ad838
| 608,480
|
let $\mathcal{R}$ be the region in the plane bounded by the graphs of $y=x$ and $y=x^{2}$. Compute the volume of the region formed by revolving $\mathcal{R}$ around the line $y=x$.
|
We integrate from 0 to 1 using the method of washers. Fix $d$ between 0 and 1. Let the line $x=d$ intersect the graph of $y=x^{2}$ at $Q$, and let the line $x=d$ intersect the graph of $y=x$ at $P$. Then $P=(d, d)$, and $Q=\left(d, d^{2}\right)$. Now drop a perpendicular from $Q$ to the line $y=x$, and let $R$ be the foot of this perpendicular. Because $P Q R$ is a $45-45-90$ triangle, $Q R=\left(d-d^{2}\right) / \sqrt{2}$. So the differential washer has a radius of $\left(d-d^{2}\right) / \sqrt{2}$ and a height of $\sqrt{2} d x$. So we integrate (from 0 to 1 ) the expression $\left[\left(x-x^{2}\right) / \sqrt{2}\right]^{2} \sqrt{2} d x$, and the answer follows.
|
\frac{\pi}{6}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
let $\mathcal{R}$ be the region in the plane bounded by the graphs of $y=x$ and $y=x^{2}$. Compute the volume of the region formed by revolving $\mathcal{R}$ around the line $y=x$.
|
We integrate from 0 to 1 using the method of washers. Fix $d$ between 0 and 1. Let the line $x=d$ intersect the graph of $y=x^{2}$ at $Q$, and let the line $x=d$ intersect the graph of $y=x$ at $P$. Then $P=(d, d)$, and $Q=\left(d, d^{2}\right)$. Now drop a perpendicular from $Q$ to the line $y=x$, and let $R$ be the foot of this perpendicular. Because $P Q R$ is a $45-45-90$ triangle, $Q R=\left(d-d^{2}\right) / \sqrt{2}$. So the differential washer has a radius of $\left(d-d^{2}\right) / \sqrt{2}$ and a height of $\sqrt{2} d x$. So we integrate (from 0 to 1 ) the expression $\left[\left(x-x^{2}\right) / \sqrt{2}\right]^{2} \sqrt{2} d x$, and the answer follows.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"problem_match": "\n9. [7]",
"solution_match": "\nSolution: "
}
|
b8a41aeb-49e2-593d-8c18-988ef93d29c6
| 608,481
|
Let $a$ and $b$ be real numbers satisfying $a>b>0$. Evaluate
$$
\int_{0}^{2 \pi} \frac{1}{a+b \cos (\theta)} d \theta
$$
Express your answer in terms of $a$ and $b$.
|
Using the geometric series formula, we can expand the integral as follows:
$$
\begin{aligned}
\int_{0}^{2 \pi} \frac{1}{a+b \cos (\theta)} d \theta & =\frac{1}{a} \int_{0}^{2 \pi} 1+\frac{b}{a} \cos (\theta)+\left(\frac{b}{a}\right)^{2} \cos ^{2}(\theta) d \theta \\
& =\frac{1}{a} \sum_{n=0}^{\infty} \int_{0}^{2 \pi}\left(\frac{b}{a}\right)^{n}\left(\frac{e^{i \theta}+e^{-i \theta}}{2}\right)^{n} d \theta \\
& =\frac{2 \pi}{a} \sum_{n=0}^{\infty}\left(\frac{b^{2}}{a^{2}}\right)^{n} \frac{\binom{2 n}{n}}{2^{2 n}} d \theta
\end{aligned}
$$
To evaluate this sum, recall that $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$ is the $n$th Catalan number. The generating function for the Catalan numbers is
$$
\sum_{n=0}^{\infty} C_{n} x^{n}=\frac{1-\sqrt{1-4 x}}{2 x}
$$
and taking the derivative of $x$ times this generating function yields $\sum\binom{2 n}{n} x^{n}=\frac{1}{\sqrt{1-4 x}}$. Thus the integral evaluates to $\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}$, as desired.
|
\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Let $a$ and $b$ be real numbers satisfying $a>b>0$. Evaluate
$$
\int_{0}^{2 \pi} \frac{1}{a+b \cos (\theta)} d \theta
$$
Express your answer in terms of $a$ and $b$.
|
Using the geometric series formula, we can expand the integral as follows:
$$
\begin{aligned}
\int_{0}^{2 \pi} \frac{1}{a+b \cos (\theta)} d \theta & =\frac{1}{a} \int_{0}^{2 \pi} 1+\frac{b}{a} \cos (\theta)+\left(\frac{b}{a}\right)^{2} \cos ^{2}(\theta) d \theta \\
& =\frac{1}{a} \sum_{n=0}^{\infty} \int_{0}^{2 \pi}\left(\frac{b}{a}\right)^{n}\left(\frac{e^{i \theta}+e^{-i \theta}}{2}\right)^{n} d \theta \\
& =\frac{2 \pi}{a} \sum_{n=0}^{\infty}\left(\frac{b^{2}}{a^{2}}\right)^{n} \frac{\binom{2 n}{n}}{2^{2 n}} d \theta
\end{aligned}
$$
To evaluate this sum, recall that $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$ is the $n$th Catalan number. The generating function for the Catalan numbers is
$$
\sum_{n=0}^{\infty} C_{n} x^{n}=\frac{1-\sqrt{1-4 x}}{2 x}
$$
and taking the derivative of $x$ times this generating function yields $\sum\binom{2 n}{n} x^{n}=\frac{1}{\sqrt{1-4 x}}$. Thus the integral evaluates to $\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}$, as desired.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-calc-solutions.jsonl",
"problem_match": "\n10. [8]",
"solution_match": "\nSolution: "
}
|
18d53dbb-23b7-5fa8-84a3-0ee17d6b0d18
| 608,482
|
How many ways can the integers from -7 to 7 be arranged in a sequence such that the absolute value of the numbers in the sequence is nondecreasing?
|
Each of the pairs $a,-a$ must occur in increasing order of $a$ for $a=1, \ldots, 7$, but $a$ can either occur before or after $-a$, for a total of $2^{7}=128$ possible sequences.
|
128
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many ways can the integers from -7 to 7 be arranged in a sequence such that the absolute value of the numbers in the sequence is nondecreasing?
|
Each of the pairs $a,-a$ must occur in increasing order of $a$ for $a=1, \ldots, 7$, but $a$ can either occur before or after $-a$, for a total of $2^{7}=128$ possible sequences.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"problem_match": "\n1. [3]",
"solution_match": "\nSolution: "
}
|
c4ea3674-8db5-5ab1-9d3f-1f2348062cc9
| 608,483
|
In how many ways can you rearrange the letters of "HMMTHMMT" such that the consecutive substring "HMMT" does not appear?
|
There are $8!/(4!2!2!)=420$ ways to order the letters. If the permuted letters contain "HMMT", there are $5 \cdot 4!/ 2!=60$ ways to order the other letters, so we subtract these. However, we have subtracted "HMMTHMMT" twice, so we add it back once to obtain 361 possibilities.
|
361
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In how many ways can you rearrange the letters of "HMMTHMMT" such that the consecutive substring "HMMT" does not appear?
|
There are $8!/(4!2!2!)=420$ ways to order the letters. If the permuted letters contain "HMMT", there are $5 \cdot 4!/ 2!=60$ ways to order the other letters, so we subtract these. However, we have subtracted "HMMTHMMT" twice, so we add it back once to obtain 361 possibilities.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"problem_match": "\n3. [4]",
"solution_match": "\nSolution: "
}
|
2b5e23c7-7971-54fa-bd19-27bcc8aa1654
| 608,485
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ satisfy $f(f(x))=f(x)$ for all $x \in\{1,2,3,4,5\}$ ?
|
A fixed point of a function $f$ is an element $a$ such that $f(a)=a$. The condition is equivalent to the property that $f$ maps every number to a fixed point. Counting by the number of fixed points of $f$, the total number of such functions is
$$
\begin{aligned}
\sum_{k=1}^{5}\binom{5}{k} k^{5-k} & =1 \cdot\left(5^{0}\right)+5 \cdot\left(1^{4}+4^{1}\right)+10 \cdot\left(2^{3}+3^{2}\right) \\
& =1+25+10 \cdot 17 \\
& =196
\end{aligned}
$$
|
196
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ satisfy $f(f(x))=f(x)$ for all $x \in\{1,2,3,4,5\}$ ?
|
A fixed point of a function $f$ is an element $a$ such that $f(a)=a$. The condition is equivalent to the property that $f$ maps every number to a fixed point. Counting by the number of fixed points of $f$, the total number of such functions is
$$
\begin{aligned}
\sum_{k=1}^{5}\binom{5}{k} k^{5-k} & =1 \cdot\left(5^{0}\right)+5 \cdot\left(1^{4}+4^{1}\right)+10 \cdot\left(2^{3}+3^{2}\right) \\
& =1+25+10 \cdot 17 \\
& =196
\end{aligned}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"problem_match": "\n4. [4]",
"solution_match": "\nSolution: "
}
|
913202d5-ccfa-5850-b77f-99b11ecbf01e
| 608,486
|
Let $s(n)$ denote the number of 1's in the binary representation of $n$. Compute
$$
\frac{1}{255} \sum_{0 \leq n<16} 2^{n}(-1)^{s(n)}
$$
|
Notice that if $n<8,(-1)^{s(n)}=(-1) \cdot(-1)^{s(n+8)}$ so the sum becomes $\frac{1}{255}\left(1-2^{8}\right) \sum_{0 \leq n<8} 2^{n}(-1)^{s(n)}=$ 45.
|
45
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $s(n)$ denote the number of 1's in the binary representation of $n$. Compute
$$
\frac{1}{255} \sum_{0 \leq n<16} 2^{n}(-1)^{s(n)}
$$
|
Notice that if $n<8,(-1)^{s(n)}=(-1) \cdot(-1)^{s(n+8)}$ so the sum becomes $\frac{1}{255}\left(1-2^{8}\right) \sum_{0 \leq n<8} 2^{n}(-1)^{s(n)}=$ 45.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"problem_match": "\n5. [4]",
"solution_match": "\nSolution: "
}
|
1887a55a-6086-5043-ab7c-9a7c6cd0a285
| 608,487
|
How many sequences of 5 positive integers $(a, b, c, d, e)$ satisfy $a b c d e \leq a+b+c+d+e \leq 10$ ?
|
We count based on how many 1's the sequence contains. If $a=b=c=d=e=1$ then this gives us 1 possibility. If $a=b=c=d=1$ and $e \neq 1$, $e$ can be $2,3,4,5,6$. Each such sequence $(1,1,1,1, e)$ can be arranged in 5 different ways, for a total of $5 \cdot 5=25$ ways in this case.
If three of the numbers are 1 , the last two can be $(2,2),(3,3),(2,3),(2,4)$, or $(2,5)$. Counting ordering, this gives a total of $2 \cdot 10+3 \cdot 20=80$ possibilities.
If two of the numbers are 1 , the other three must be equal to 2 for the product to be under 10 , and this yields 10 more possibilities.
Thus there are $1+25+80+10=116$ such sequences.
|
116
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many sequences of 5 positive integers $(a, b, c, d, e)$ satisfy $a b c d e \leq a+b+c+d+e \leq 10$ ?
|
We count based on how many 1's the sequence contains. If $a=b=c=d=e=1$ then this gives us 1 possibility. If $a=b=c=d=1$ and $e \neq 1$, $e$ can be $2,3,4,5,6$. Each such sequence $(1,1,1,1, e)$ can be arranged in 5 different ways, for a total of $5 \cdot 5=25$ ways in this case.
If three of the numbers are 1 , the last two can be $(2,2),(3,3),(2,3),(2,4)$, or $(2,5)$. Counting ordering, this gives a total of $2 \cdot 10+3 \cdot 20=80$ possibilities.
If two of the numbers are 1 , the other three must be equal to 2 for the product to be under 10 , and this yields 10 more possibilities.
Thus there are $1+25+80+10=116$ such sequences.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"problem_match": "\n6. [5]",
"solution_match": "\nSolution: "
}
|
773863aa-46db-59c4-8c11-0609fde83041
| 608,488
|
Paul fills in a $7 \times 7$ grid with the numbers 1 through 49 in a random arrangement. He then erases his work and does the same thing again (to obtain two different random arrangements of the numbers in the grid). What is the expected number of pairs of numbers that occur in either the same row as each other or the same column as each other in both of the two arrangements?
|
$147 / 2$
|
\frac{147}{2}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Paul fills in a $7 \times 7$ grid with the numbers 1 through 49 in a random arrangement. He then erases his work and does the same thing again (to obtain two different random arrangements of the numbers in the grid). What is the expected number of pairs of numbers that occur in either the same row as each other or the same column as each other in both of the two arrangements?
|
$147 / 2$
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"problem_match": "\n7. [7]",
"solution_match": "\nAnswer: "
}
|
26c91933-0d09-51bd-b2b9-4382b23780fe
| 608,489
|
There are 5 students on a team for a math competition. The math competition has 5 subject tests. Each student on the team must choose 2 distinct tests, and each test must be taken by exactly two people. In how many ways can this be done?
|
We can model the situation as a bipartite graph on 10 vertices, with 5 nodes representing the students and the other 5 representing the tests. We now simply want to count the number of bipartite graphs on these two sets such that there are two edges incident on each vertex.
Notice that in such a graph, we can start at any vertex and follow one of the edges eminating from it, then follow the other edge eminating from the second vertex, etc, and in this manner we must eventually end up back at the starting vertex, so the graph is partitioned into even cycles. Since each vertex has degree two, we cannot have a 2 -cycle, so we must have either a 10 -cycle or a 4 -cycle and a 6 -cycle.
In the former case, starting with Person $A$, there are 5 ways to choose one of his tests. This test can be taken by one of 4 other people, who can take one of 4 other tests, which can be taken by one of 3 other people, etc, so the number of 10 -cycles we obtain in this way is $5!\cdot 4!$. However, it does not matter which of the first person's tests we choose first in a given 10 -cycle, so we overcounted by a factor of 2 . Thus there are $5!\cdot 4!/ 2=1440$ possibilities in this case.
In the latter case, there are $\binom{5}{3}^{2}=100$ ways to choose which three people and which three tests are in the 6 -cycle. After choosing this, a similar argument to that above shows there are $2!\cdot 1!/ 2$ possible 4 -cycles and $3!\cdot 2!/ 2$ possible 6 -cycles, for a total of $100 \cdot 1 \cdot 6=600$ possibilities in this case.
Thus there are a total of 2040 ways they can take the tests.
|
2040
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
There are 5 students on a team for a math competition. The math competition has 5 subject tests. Each student on the team must choose 2 distinct tests, and each test must be taken by exactly two people. In how many ways can this be done?
|
We can model the situation as a bipartite graph on 10 vertices, with 5 nodes representing the students and the other 5 representing the tests. We now simply want to count the number of bipartite graphs on these two sets such that there are two edges incident on each vertex.
Notice that in such a graph, we can start at any vertex and follow one of the edges eminating from it, then follow the other edge eminating from the second vertex, etc, and in this manner we must eventually end up back at the starting vertex, so the graph is partitioned into even cycles. Since each vertex has degree two, we cannot have a 2 -cycle, so we must have either a 10 -cycle or a 4 -cycle and a 6 -cycle.
In the former case, starting with Person $A$, there are 5 ways to choose one of his tests. This test can be taken by one of 4 other people, who can take one of 4 other tests, which can be taken by one of 3 other people, etc, so the number of 10 -cycles we obtain in this way is $5!\cdot 4!$. However, it does not matter which of the first person's tests we choose first in a given 10 -cycle, so we overcounted by a factor of 2 . Thus there are $5!\cdot 4!/ 2=1440$ possibilities in this case.
In the latter case, there are $\binom{5}{3}^{2}=100$ ways to choose which three people and which three tests are in the 6 -cycle. After choosing this, a similar argument to that above shows there are $2!\cdot 1!/ 2$ possible 4 -cycles and $3!\cdot 2!/ 2$ possible 6 -cycles, for a total of $100 \cdot 1 \cdot 6=600$ possibilities in this case.
Thus there are a total of 2040 ways they can take the tests.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"problem_match": "\n8. [7]",
"solution_match": "\nSolution: "
}
|
e7a518fd-4576-5eba-bbc1-fc139f68180d
| 608,490
|
The squares of a $3 \times 3$ grid are filled with positive integers such that 1 is the label of the upperleftmost square, 2009 is the label of the lower-rightmost square, and the label of each square divides the one directly to the right of it and the one directly below it. How many such labelings are possible?
|
We factor 2009 as $7^{2} \cdot 41$ and place the 41 's and the 7 's in the squares separately. The number of ways to fill the grid with 1 's and 41 's so that the divisibility property is satisfied is equal to the number of nondecreasing sequences $a_{1}, a_{2}, a_{3}$ where each $a_{i} \in\{0,1,2,3\}$ and the sequence is not $0,0,0$ and not $1,1,1$ (here $a_{i}$ corresponds to the number of 41 's in the $i$ th column.) Thus there are $\binom{3+4-1}{3}-2=18$ ways to choose which squares are divisible by 41 .
To count the arrangements of divisibility by 7 and 49 , we consider three cases.
If 49 divides the middle square, then each of the squares to the right and below it are divisible 49. The two squares in the top row (besides the upper left) can be $(1,1),(1,7),(1,49),(7,7),(7,49)$, or $(49,49)$ (in terms of the highest power of 7 dividing the square). The same is true, independently, for the two blank squares on the left column, for a total of $6^{2}=36$ possibilities in this case.
If 1 is the highest power of 7 dividing the middle square, there are also 36 possibilities by a similar argument.
If 7 is the highest power of 7 dividing the middle square, there are 8 possibilities for the upper right three squares. Thus there are 64 possibilities in this case.
Thus there are a total of 136 options for the divisibility of each number by 7 and $7^{2}$, and 18 options for the divisibility of the numbers by 41 . Since each number divides 2009 , this uniquely determines the numbers, and so there are a total of $18 \cdot 136=2448$ possibilities.
|
2448
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
The squares of a $3 \times 3$ grid are filled with positive integers such that 1 is the label of the upperleftmost square, 2009 is the label of the lower-rightmost square, and the label of each square divides the one directly to the right of it and the one directly below it. How many such labelings are possible?
|
We factor 2009 as $7^{2} \cdot 41$ and place the 41 's and the 7 's in the squares separately. The number of ways to fill the grid with 1 's and 41 's so that the divisibility property is satisfied is equal to the number of nondecreasing sequences $a_{1}, a_{2}, a_{3}$ where each $a_{i} \in\{0,1,2,3\}$ and the sequence is not $0,0,0$ and not $1,1,1$ (here $a_{i}$ corresponds to the number of 41 's in the $i$ th column.) Thus there are $\binom{3+4-1}{3}-2=18$ ways to choose which squares are divisible by 41 .
To count the arrangements of divisibility by 7 and 49 , we consider three cases.
If 49 divides the middle square, then each of the squares to the right and below it are divisible 49. The two squares in the top row (besides the upper left) can be $(1,1),(1,7),(1,49),(7,7),(7,49)$, or $(49,49)$ (in terms of the highest power of 7 dividing the square). The same is true, independently, for the two blank squares on the left column, for a total of $6^{2}=36$ possibilities in this case.
If 1 is the highest power of 7 dividing the middle square, there are also 36 possibilities by a similar argument.
If 7 is the highest power of 7 dividing the middle square, there are 8 possibilities for the upper right three squares. Thus there are 64 possibilities in this case.
Thus there are a total of 136 options for the divisibility of each number by 7 and $7^{2}$, and 18 options for the divisibility of the numbers by 41 . Since each number divides 2009 , this uniquely determines the numbers, and so there are a total of $18 \cdot 136=2448$ possibilities.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"problem_match": "\n9. [5]",
"solution_match": "\nSolution: "
}
|
a5286597-e393-5193-9d0f-7463d4067fbb
| 608,491
|
Given a rearrangement of the numbers from 1 to $n$, each pair of consecutive elements $a$ and $b$ of the sequence can be either increasing (if $a<b$ ) or decreasing (if $b<a$ ). How many rearrangements of the numbers from 1 to $n$ have exactly two increasing pairs of consecutive elements?
|
$3^{n}-(n+1) \cdot 2^{n}+n(n+1) / 2$ or equivalent
|
3^{n}-(n+1) \cdot 2^{n}+n(n+1) / 2
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Given a rearrangement of the numbers from 1 to $n$, each pair of consecutive elements $a$ and $b$ of the sequence can be either increasing (if $a<b$ ) or decreasing (if $b<a$ ). How many rearrangements of the numbers from 1 to $n$ have exactly two increasing pairs of consecutive elements?
|
$3^{n}-(n+1) \cdot 2^{n}+n(n+1) / 2$ or equivalent
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-comb-solutions.jsonl",
"problem_match": "\n10. [8]",
"solution_match": "\nAnswer: "
}
|
e9cd37d3-4ab5-56ec-b148-497025134ed2
| 608,492
|
Suppose $N$ is a 6-digit number having base-10 representation $\underline{a} \underline{b} \underline{c} \underline{d} \underline{e} \underline{f}$. If $N$ is $6 / 7$ of the number having base-10 representation $\underline{d} \underline{e} \underline{f} \underline{a} \underline{b} \underline{c}$, find $N$.
|
We have $7(a b c d e f)_{10}=6(\text { defabc })_{10}$, so $699400 a+69940 b+6994 c=599300 d+59930 e+$ $5993 f$. We can factor this equation as $6994(100 a+10 b+c)=5993(100 d+10 e+f)$, which yields $538(a b c)_{10}=461(d e f)_{10}$. Since $\operatorname{gcd}(538,461)=1$, we must have $(a b c)_{10}=461$ and $(d e f)_{10}=538$.
|
461538
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Suppose $N$ is a 6-digit number having base-10 representation $\underline{a} \underline{b} \underline{c} \underline{d} \underline{e} \underline{f}$. If $N$ is $6 / 7$ of the number having base-10 representation $\underline{d} \underline{e} \underline{f} \underline{a} \underline{b} \underline{c}$, find $N$.
|
We have $7(a b c d e f)_{10}=6(\text { defabc })_{10}$, so $699400 a+69940 b+6994 c=599300 d+59930 e+$ $5993 f$. We can factor this equation as $6994(100 a+10 b+c)=5993(100 d+10 e+f)$, which yields $538(a b c)_{10}=461(d e f)_{10}$. Since $\operatorname{gcd}(538,461)=1$, we must have $(a b c)_{10}=461$ and $(d e f)_{10}=538$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"problem_match": "\n2. [2]",
"solution_match": "\nSolution: "
}
|
1a60c1a0-5511-5ff8-a754-03ed5fe71864
| 608,493
|
A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below, so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of the resulting trapezoid.
|
$55 / 2$
|
\frac{55}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below, so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of the resulting trapezoid.
|
$55 / 2$
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"problem_match": "\n3. [3]",
"solution_match": "\nAnswer: "
}
|
3e93655e-6e6f-5858-b64e-50dc500d3e49
| 608,494
|
Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. What is the expected number of cards that will be strictly between the two jokers?
|
$52 / 3$
|
\frac{52}{3}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. What is the expected number of cards that will be strictly between the two jokers?
|
$52 / 3$
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"problem_match": "\n5. [4]",
"solution_match": "\nAnswer: "
}
|
15c4dfb2-9ad2-5032-900c-f31860f6a125
| 608,495
|
A kite is a quadrilateral whose diagonals are perpendicular. Let kite $A B C D$ be such that $\angle B=$ $\angle D=90^{\circ}$. Let $M$ and $N$ be the points of tangency of the incircle of $A B C D$ to $A B$ and $B C$ respectively. Let $\omega$ be the circle centered at $C$ and tangent to $A B$ and $A D$. Construct another kite $A B^{\prime} C^{\prime} D^{\prime}$ that is similar to $A B C D$ and whose incircle is $\omega$. Let $N^{\prime}$ be the point of tangency of $B^{\prime} C^{\prime}$ to $\omega$. If $M N^{\prime} \| A C$, then what is the ratio of $A B: B C$ ?
|
$\frac{1+\sqrt{5}}{2}$
|
\frac{1+\sqrt{5}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A kite is a quadrilateral whose diagonals are perpendicular. Let kite $A B C D$ be such that $\angle B=$ $\angle D=90^{\circ}$. Let $M$ and $N$ be the points of tangency of the incircle of $A B C D$ to $A B$ and $B C$ respectively. Let $\omega$ be the circle centered at $C$ and tangent to $A B$ and $A D$. Construct another kite $A B^{\prime} C^{\prime} D^{\prime}$ that is similar to $A B C D$ and whose incircle is $\omega$. Let $N^{\prime}$ be the point of tangency of $B^{\prime} C^{\prime}$ to $\omega$. If $M N^{\prime} \| A C$, then what is the ratio of $A B: B C$ ?
|
$\frac{1+\sqrt{5}}{2}$
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen1-solutions.jsonl",
"problem_match": "\n10. [6]",
"solution_match": "\nAnswer: "
}
|
fca939b9-6b29-52d5-b4d7-08fff4f35d77
| 608,496
|
How many ways can the integers from -7 to 7 be arranged in a sequence such that the absolute values of the numbers in the sequence are nonincreasing?
|
Each of the pairs $a,-a$ must occur in increasing order of $a$ for $a=1, \ldots, 7$, but $a$ can either occur before or after $-a$, for a total of $2^{7}=128$ possible sequences.
|
128
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many ways can the integers from -7 to 7 be arranged in a sequence such that the absolute values of the numbers in the sequence are nonincreasing?
|
Each of the pairs $a,-a$ must occur in increasing order of $a$ for $a=1, \ldots, 7$, but $a$ can either occur before or after $-a$, for a total of $2^{7}=128$ possible sequences.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen2-solutions.jsonl",
"problem_match": "\n1. [2]",
"solution_match": "\nSolution: "
}
|
408b43c9-cbe3-558b-aeee-d5d205b120b3
| 608,497
|
A torus (donut) having inner radius 2 and outer radius 4 sits on a flat table. What is the radius of the largest spherical ball that can be placed on top of the center torus so that the ball still touches the horizontal plane? (If the $x-y$ plane is the table, the torus is formed by revolving the circle in the $x-z$ plane centered at $(3,0,1)$ with radius 1 about the $z$ axis. The spherical ball has its center on the $z$-axis and rests on either the table or the donut.)
|
Let $r$ be the radius of the sphere. One can see that it satisfies $(r+1)^{2}=(r-1)^{2}+3^{2}$ by the Pythagorean Theorem, so $r=9 / 4$.
|
\frac{9}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A torus (donut) having inner radius 2 and outer radius 4 sits on a flat table. What is the radius of the largest spherical ball that can be placed on top of the center torus so that the ball still touches the horizontal plane? (If the $x-y$ plane is the table, the torus is formed by revolving the circle in the $x-z$ plane centered at $(3,0,1)$ with radius 1 about the $z$ axis. The spherical ball has its center on the $z$-axis and rests on either the table or the donut.)
|
Let $r$ be the radius of the sphere. One can see that it satisfies $(r+1)^{2}=(r-1)^{2}+3^{2}$ by the Pythagorean Theorem, so $r=9 / 4$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen2-solutions.jsonl",
"problem_match": "\n4. [3]",
"solution_match": "\nSolution: "
}
|
cae66c6e-8cae-5034-99f1-149c88a04651
| 608,499
|
Suppose $a, b$ and $c$ are integers such that the greatest common divisor of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x+1$ (in the set of polynomials in $x$ with integer coefficients), and the least common multiple of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x^{3}-4 x^{2}+x+6$. Find $a+b+c$.
|
Since $x+1$ divides $x^{2}+a x+b$ and the constant term is $b$, we have $x^{2}+a x+b=(x+1)(x+b)$, and similarly $x^{2}+b x+c=(x+1)(x+c)$. Therefore, $a=b+1=c+2$. Furthermore, the least common
multiple of the two polynomials is $(x+1)(x+b)(x+b-1)=x^{3}-4 x^{2}+x+6$, so $b=-2$. Thus $a=-1$ and $c=-3$, and $a+b+c=-6$.
|
-6
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose $a, b$ and $c$ are integers such that the greatest common divisor of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x+1$ (in the set of polynomials in $x$ with integer coefficients), and the least common multiple of $x^{2}+a x+b$ and $x^{2}+b x+c$ is $x^{3}-4 x^{2}+x+6$. Find $a+b+c$.
|
Since $x+1$ divides $x^{2}+a x+b$ and the constant term is $b$, we have $x^{2}+a x+b=(x+1)(x+b)$, and similarly $x^{2}+b x+c=(x+1)(x+c)$. Therefore, $a=b+1=c+2$. Furthermore, the least common
multiple of the two polynomials is $(x+1)(x+b)(x+b-1)=x^{3}-4 x^{2}+x+6$, so $b=-2$. Thus $a=-1$ and $c=-3$, and $a+b+c=-6$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen2-solutions.jsonl",
"problem_match": "\n5. [4]",
"solution_match": "\nSolution: "
}
|
b4284453-9a78-5982-8f4e-84d1f57ecbb6
| 608,500
|
Let $F_{n}$ be the Fibonacci sequence, that is, $F_{0}=0, F_{1}=1$, and $F_{n+2}=F_{n+1}+F_{n}$. Compute $\sum_{n=0}^{\infty} F_{n} / 10^{n}$.
|
Write $F(x)=\sum_{n=0}^{\infty} F_{n} x^{n}$. Then the Fibonacci recursion tells us that $F(x)-x F(x)-$ $x^{2} F(x)=x$, so $F(x)=x /\left(1-x-x^{2}\right)$. Plugging in $x=1 / 10$ gives the answer.
|
\frac{10}{89}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $F_{n}$ be the Fibonacci sequence, that is, $F_{0}=0, F_{1}=1$, and $F_{n+2}=F_{n+1}+F_{n}$. Compute $\sum_{n=0}^{\infty} F_{n} / 10^{n}$.
|
Write $F(x)=\sum_{n=0}^{\infty} F_{n} x^{n}$. Then the Fibonacci recursion tells us that $F(x)-x F(x)-$ $x^{2} F(x)=x$, so $F(x)=x /\left(1-x-x^{2}\right)$. Plugging in $x=1 / 10$ gives the answer.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen2-solutions.jsonl",
"problem_match": "\n7. [5]",
"solution_match": "\nSolution: "
}
|
e0ee947d-8d99-53a5-8239-2243402b9118
| 608,501
|
Let $T$ be a right triangle with sides having lengths 3,4 , and 5 . A point $P$ is called awesome if $P$ is the center of a parallelogram whose vertices all lie on the boundary of $T$. What is the area of the set of awesome points?
|
The set of awesome points is the medial triangle, which has area $6 / 4=3 / 2$.
|
\frac{3}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $T$ be a right triangle with sides having lengths 3,4 , and 5 . A point $P$ is called awesome if $P$ is the center of a parallelogram whose vertices all lie on the boundary of $T$. What is the area of the set of awesome points?
|
The set of awesome points is the medial triangle, which has area $6 / 4=3 / 2$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-gen2-solutions.jsonl",
"problem_match": "\n10. [6]",
"solution_match": "\nSolution: "
}
|
dfd55940-70c7-5cd9-8970-a224af623044
| 608,503
|
Circle $B$ has radius $6 \sqrt{7}$. Circle $A$, centered at point $C$, has radius $\sqrt{7}$ and is contained in $B$. Let $L$ be the locus of centers $C$ such that there exists a point $D$ on the boundary of $B$ with the following property: if the tangents from $D$ to circle $A$ intersect circle $B$ again at $X$ and $Y$, then $X Y$ is also tangent to $A$. Find the area contained by the boundary of $L$.
|
The conditions imply that there exists a triangle such that $B$ is the circumcircle and $A$ is the incircle for the position of $A$. The distance between the circumcenter and incenter is given by $\sqrt{(R-2 r) R}$, where $R, r$ are the circumradius and inradius, respectively. Thus the locus of $C$ is a circle concentric to $B$ with radius $2 \sqrt{42}$. The conclusion follows.
|
2 \sqrt{42}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Circle $B$ has radius $6 \sqrt{7}$. Circle $A$, centered at point $C$, has radius $\sqrt{7}$ and is contained in $B$. Let $L$ be the locus of centers $C$ such that there exists a point $D$ on the boundary of $B$ with the following property: if the tangents from $D$ to circle $A$ intersect circle $B$ again at $X$ and $Y$, then $X Y$ is also tangent to $A$. Find the area contained by the boundary of $L$.
|
The conditions imply that there exists a triangle such that $B$ is the circumcircle and $A$ is the incircle for the position of $A$. The distance between the circumcenter and incenter is given by $\sqrt{(R-2 r) R}$, where $R, r$ are the circumradius and inradius, respectively. Thus the locus of $C$ is a circle concentric to $B$ with radius $2 \sqrt{42}$. The conclusion follows.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-geo-solutions.jsonl",
"problem_match": "\n5. [4]",
"solution_match": "\nSolution: "
}
|
1915fce9-bbb1-5b14-8749-a2905f2ca142
| 608,505
|
In $\triangle A B C, D$ is the midpoint of $B C, E$ is the foot of the perpendicular from $A$ to $B C$, and $F$ is the foot of the perpendicular from $D$ to $A C$. Given that $B E=5, E C=9$, and the area of triangle $A B C$ is 84 , compute $|E F|$.
|
There are two possibilities for the triangle $A B C$ based on whether $E$ is between $B$ and $C$ or not. We first consider the former case.
We find from the area and the Pythagorean theorem that $A E=12, A B=13$, and $A C=15$. We can then use Stewart's theorem to obtain $A D=2 \sqrt{37}$.
Since the area of $\triangle A D C$ is half that of $A B C$, we have $\frac{1}{2} A C \cdot D F=42$, so $D F=14 / 5$. Also, $D C=14 / 2=7$ so $E D=9-7=2$.
Notice that $A E D F$ is a cyclic quadrilateral. By Ptolemy's theorem, we have $E F \cdot 2 \sqrt{37}=(28 / 5) \cdot 12+$ $2 \cdot(54 / 5)$. Thus $E F=\frac{6 \sqrt{37}}{5}$ as desired.
The latter case is similar.
|
\frac{6 \sqrt{37}}{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
In $\triangle A B C, D$ is the midpoint of $B C, E$ is the foot of the perpendicular from $A$ to $B C$, and $F$ is the foot of the perpendicular from $D$ to $A C$. Given that $B E=5, E C=9$, and the area of triangle $A B C$ is 84 , compute $|E F|$.
|
There are two possibilities for the triangle $A B C$ based on whether $E$ is between $B$ and $C$ or not. We first consider the former case.
We find from the area and the Pythagorean theorem that $A E=12, A B=13$, and $A C=15$. We can then use Stewart's theorem to obtain $A D=2 \sqrt{37}$.
Since the area of $\triangle A D C$ is half that of $A B C$, we have $\frac{1}{2} A C \cdot D F=42$, so $D F=14 / 5$. Also, $D C=14 / 2=7$ so $E D=9-7=2$.
Notice that $A E D F$ is a cyclic quadrilateral. By Ptolemy's theorem, we have $E F \cdot 2 \sqrt{37}=(28 / 5) \cdot 12+$ $2 \cdot(54 / 5)$. Thus $E F=\frac{6 \sqrt{37}}{5}$ as desired.
The latter case is similar.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-geo-solutions.jsonl",
"problem_match": "\n7. [5]",
"solution_match": "\nSolution: "
}
|
d0e5a838-1609-51aa-868f-900ca9415691
| 608,507
|
Triangle $A B C$ has side lengths $A B=231, B C=160$, and $A C=281$. Point $D$ is constructed on the opposite side of line $A C$ as point $B$ such that $A D=178$ and $C D=153$. Compute the distance from $B$ to the midpoint of segment $A D$.
|
Note that $\angle A B C$ is right since
$$
B C^{2}=160^{2}=50 \cdot 512=(A C-A B) \cdot(A C+A B)=A C^{2}-A B^{2}
$$
Construct point $B^{\prime}$ such that $A B C B^{\prime}$ is a rectangle, and construct $D^{\prime}$ on segment $B^{\prime} C$ such that $A D=A D^{\prime}$. Then
$$
B^{\prime} D^{\prime 2}=A D^{\prime 2}-A B^{\prime 2}=A D^{2}-B C^{2}=(A D-B C)(A D+B C)=18 \cdot 338=78^{2}
$$
It follows that $C D^{\prime}=B^{\prime} C-B^{\prime} D^{\prime}=153=C D$; thus, points $D$ and $D^{\prime}$ coincide, and $A B \| C D$. Let $M$ denote the midpoint of segment $A D$, and denote the orthogonal projections $M$ to lines $A B$ and $B C$ by $P$ and $Q$ respectively. Then $Q$ is the midpoint of $B C$ and $A P=39$, so that $P B=A B-A P=192$ and
$$
B M=P Q=\sqrt{80^{2}+192^{2}}=16 \sqrt{5^{2}+12^{2}}=208
$$
|
208
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Triangle $A B C$ has side lengths $A B=231, B C=160$, and $A C=281$. Point $D$ is constructed on the opposite side of line $A C$ as point $B$ such that $A D=178$ and $C D=153$. Compute the distance from $B$ to the midpoint of segment $A D$.
|
Note that $\angle A B C$ is right since
$$
B C^{2}=160^{2}=50 \cdot 512=(A C-A B) \cdot(A C+A B)=A C^{2}-A B^{2}
$$
Construct point $B^{\prime}$ such that $A B C B^{\prime}$ is a rectangle, and construct $D^{\prime}$ on segment $B^{\prime} C$ such that $A D=A D^{\prime}$. Then
$$
B^{\prime} D^{\prime 2}=A D^{\prime 2}-A B^{\prime 2}=A D^{2}-B C^{2}=(A D-B C)(A D+B C)=18 \cdot 338=78^{2}
$$
It follows that $C D^{\prime}=B^{\prime} C-B^{\prime} D^{\prime}=153=C D$; thus, points $D$ and $D^{\prime}$ coincide, and $A B \| C D$. Let $M$ denote the midpoint of segment $A D$, and denote the orthogonal projections $M$ to lines $A B$ and $B C$ by $P$ and $Q$ respectively. Then $Q$ is the midpoint of $B C$ and $A P=39$, so that $P B=A B-A P=192$ and
$$
B M=P Q=\sqrt{80^{2}+192^{2}}=16 \sqrt{5^{2}+12^{2}}=208
$$
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-geo-solutions.jsonl",
"problem_match": "\n8. [7]",
"solution_match": "\nSolution: "
}
|
5a6d7221-4f74-5c5b-ac98-0ea14597a9e6
| 608,508
|
Let $A B C$ be a triangle with $A B=16$ and $A C=5$. Suppose the bisectors of angles $\angle A B C$ and $\angle B C A$ meet at point $P$ in the triangle's interior. Given that $A P=4$, compute $B C$.
|
As the incenter of triangle $A B C$, point $P$ has many properties. Extend $A P$ past $P$ to its intersection with the circumcircle of triangle $A B C$, and call this intersection $M$. Now observe that
$$
\angle P B M=\angle P B C+\angle C B M=\angle P B C+\angle C A M=\beta+\alpha=90-\gamma
$$
where $\alpha, \beta$, and $\gamma$ are the half-angles of triangle $A B C$. Since
$$
\angle B M P=\angle B M A=\angle B C A=2 \gamma
$$
it follows that $B M=M P=C M$. Let $Q$ denote the intersection of $A M$ and $B C$, and observe that $\triangle A Q B \sim \triangle C Q M$ and $\triangle A Q C \sim \triangle B Q M$; some easy algebra gives
$$
A M / B C=(A B \cdot A C+B M \cdot C M) /(A C \cdot C M+A B \cdot B M)
$$
Writing $(a, b, c, d, x)=(B C, A C, A B, M P, A P)$, this is $(x+d) / a=\left(b c+d^{2}\right) /((b+c) d)$. Ptolemy's theorem applied to $A B C D$ gives $a(d+x)=d(b+c)$. Multiplying the two gives $(d+x)^{2}=b c+d^{2}$. We easily solve for $d=\left(b c-x^{2}\right) /(2 x)=8$ and $a=d(b+c) /(d+x)=14$.
|
14
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a triangle with $A B=16$ and $A C=5$. Suppose the bisectors of angles $\angle A B C$ and $\angle B C A$ meet at point $P$ in the triangle's interior. Given that $A P=4$, compute $B C$.
|
As the incenter of triangle $A B C$, point $P$ has many properties. Extend $A P$ past $P$ to its intersection with the circumcircle of triangle $A B C$, and call this intersection $M$. Now observe that
$$
\angle P B M=\angle P B C+\angle C B M=\angle P B C+\angle C A M=\beta+\alpha=90-\gamma
$$
where $\alpha, \beta$, and $\gamma$ are the half-angles of triangle $A B C$. Since
$$
\angle B M P=\angle B M A=\angle B C A=2 \gamma
$$
it follows that $B M=M P=C M$. Let $Q$ denote the intersection of $A M$ and $B C$, and observe that $\triangle A Q B \sim \triangle C Q M$ and $\triangle A Q C \sim \triangle B Q M$; some easy algebra gives
$$
A M / B C=(A B \cdot A C+B M \cdot C M) /(A C \cdot C M+A B \cdot B M)
$$
Writing $(a, b, c, d, x)=(B C, A C, A B, M P, A P)$, this is $(x+d) / a=\left(b c+d^{2}\right) /((b+c) d)$. Ptolemy's theorem applied to $A B C D$ gives $a(d+x)=d(b+c)$. Multiplying the two gives $(d+x)^{2}=b c+d^{2}$. We easily solve for $d=\left(b c-x^{2}\right) /(2 x)=8$ and $a=d(b+c) /(d+x)=14$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-geo-solutions.jsonl",
"problem_match": "\n9. [7]",
"solution_match": "\nSolution: "
}
|
34acb1f6-b445-576d-92a8-ab2993cb5e52
| 608,509
|
Points $A$ and $B$ lie on circle $\omega$. Point $P$ lies on the extension of segment $A B$ past $B$. Line $\ell$ passes through $P$ and is tangent to $\omega$. The tangents to $\omega$ at points $A$ and $B$ intersect $\ell$ at points $D$ and $C$ respectively. Given that $A B=7, B C=2$, and $A D=3$, compute $B P$.
|
Say that $\ell$ be tangent to $\omega$ at point $T$. Observing equal tangents, write
$$
C D=C T+D T=B C+A D=5 .
$$
Let the tangents to $\omega$ at $A$ and $B$ intersect each other at $Q$. Working from Menelaus applied to triangle $C D Q$ and line $A B$ gives
$$
\begin{aligned}
-1 & =\frac{D A}{A Q} \cdot \frac{Q B}{B C} \cdot \frac{C P}{P D} \\
& =\frac{D A}{B C} \cdot \frac{C P}{P C+C D} \\
& =\frac{3}{2} \cdot \frac{C P}{P C+5},
\end{aligned}
$$
from which $P C=10$. By power of a point, $P T^{2}=A P \cdot B P$, or $12^{2}=B P \cdot(B P+7)$, from which $B P=9$.
|
9
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Points $A$ and $B$ lie on circle $\omega$. Point $P$ lies on the extension of segment $A B$ past $B$. Line $\ell$ passes through $P$ and is tangent to $\omega$. The tangents to $\omega$ at points $A$ and $B$ intersect $\ell$ at points $D$ and $C$ respectively. Given that $A B=7, B C=2$, and $A D=3$, compute $B P$.
|
Say that $\ell$ be tangent to $\omega$ at point $T$. Observing equal tangents, write
$$
C D=C T+D T=B C+A D=5 .
$$
Let the tangents to $\omega$ at $A$ and $B$ intersect each other at $Q$. Working from Menelaus applied to triangle $C D Q$ and line $A B$ gives
$$
\begin{aligned}
-1 & =\frac{D A}{A Q} \cdot \frac{Q B}{B C} \cdot \frac{C P}{P D} \\
& =\frac{D A}{B C} \cdot \frac{C P}{P C+C D} \\
& =\frac{3}{2} \cdot \frac{C P}{P C+5},
\end{aligned}
$$
from which $P C=10$. By power of a point, $P T^{2}=A P \cdot B P$, or $12^{2}=B P \cdot(B P+7)$, from which $B P=9$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-geo-solutions.jsonl",
"problem_match": "\n10. [8]",
"solution_match": "\nSolution: "
}
|
fa04261a-1bc7-5c76-bbec-fcad5c70640b
| 608,510
|
Compute
$$
1 \cdot 2^{2}+2 \cdot 3^{2}+3 \cdot 4^{2}+\cdots+19 \cdot 20^{2}
$$
|
41230 y Solution: We can write this as $\left(1^{3}+2^{3}+\cdots+20^{3}\right)-\left(1^{2}+2^{2}+\cdots+20^{2}\right)$, which is equal to $44100-2870=41230$.
|
41230
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Compute
$$
1 \cdot 2^{2}+2 \cdot 3^{2}+3 \cdot 4^{2}+\cdots+19 \cdot 20^{2}
$$
|
41230 y Solution: We can write this as $\left(1^{3}+2^{3}+\cdots+20^{3}\right)-\left(1^{2}+2^{2}+\cdots+20^{2}\right)$, which is equal to $44100-2870=41230$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n1. [5]",
"solution_match": "\nAnswer: "
}
|
b5ae8b31-bc9b-5ee8-802a-1b80744495c5
| 608,511
|
Given that $\sin A+\sin B=1$ and $\cos A+\cos B=3 / 2$, what is the value of $\cos (A-B)$ ?
|
Squaring both equations and add them together, one obtains $1+9 / 4=2+2(\cos (A) \cos (B)+$ $\sin (A) \sin (B))=2+2 \cos (A-B)$. Thus $\cos A-B=5 / 8$.
|
\frac{5}{8}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Given that $\sin A+\sin B=1$ and $\cos A+\cos B=3 / 2$, what is the value of $\cos (A-B)$ ?
|
Squaring both equations and add them together, one obtains $1+9 / 4=2+2(\cos (A) \cos (B)+$ $\sin (A) \sin (B))=2+2 \cos (A-B)$. Thus $\cos A-B=5 / 8$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n2. [5]",
"solution_match": "\nSolution: "
}
|
d216d06e-fa09-5caa-8b05-782589d6cf32
| 608,512
|
Simplify: $i^{0}+i^{1}+\cdots+i^{2009}$.
|
By the geometric series formula, the sum is equal to $\frac{i^{2010}-1}{i-1}=\frac{-2}{i-1}=1+i$.
|
1+i
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Simplify: $i^{0}+i^{1}+\cdots+i^{2009}$.
|
By the geometric series formula, the sum is equal to $\frac{i^{2010}-1}{i-1}=\frac{-2}{i-1}=1+i$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n4. [6]",
"solution_match": "\nSolution: "
}
|
f6b894f9-28f0-594b-81de-2c7fbf315b87
| 608,514
|
In how many distinct ways can you color each of the vertices of a tetrahedron either red, blue, or green such that no face has all three vertices the same color? (Two colorings are considered the same if one coloring can be rotated in three dimensions to obtain the other.)
|
If only two colors are used, there is only one possible arrangement up to rotation, so this gives 3 possibilities. If all three colors are used, then one is used twice. There are 3 ways to choose the color that is used twice. Say this color is red. Then the red vertices are on a common edge, and the green and blue vertices are on another edge. We see that either choice of arrangement of the green and blue vertices is the same up to rotation. Thus there are 6 possibilities total.
|
6
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In how many distinct ways can you color each of the vertices of a tetrahedron either red, blue, or green such that no face has all three vertices the same color? (Two colorings are considered the same if one coloring can be rotated in three dimensions to obtain the other.)
|
If only two colors are used, there is only one possible arrangement up to rotation, so this gives 3 possibilities. If all three colors are used, then one is used twice. There are 3 ways to choose the color that is used twice. Say this color is red. Then the red vertices are on a common edge, and the green and blue vertices are on another edge. We see that either choice of arrangement of the green and blue vertices is the same up to rotation. Thus there are 6 possibilities total.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n5. [6]",
"solution_match": "\nSolution: "
}
|
09f79fab-5464-53b1-9713-24315a44f3cb
| 608,515
|
Let $A B C$ be a right triangle with hypotenuse $A C$. Let $B^{\prime}$ be the reflection of point $B$ across $A C$, and let $C^{\prime}$ be the reflection of $C$ across $A B^{\prime}$. Find the ratio of $\left[B C B^{\prime}\right]$ to $\left[B C^{\prime} B^{\prime}\right]$.
|
1 Solution: Since $C, B^{\prime}$, and $C^{\prime}$ are collinear, it is evident that $\left[B C B^{\prime}\right]=\frac{1}{2}\left[B C C^{\prime}\right]$. It immediately follows that $\left[B C B^{\prime}\right]=\left[B C^{\prime} B^{\prime}\right]$. Thus, the ratio is 1 .
$12^{\text {th }}$ HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 - GUTS ROUND
|
1
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a right triangle with hypotenuse $A C$. Let $B^{\prime}$ be the reflection of point $B$ across $A C$, and let $C^{\prime}$ be the reflection of $C$ across $A B^{\prime}$. Find the ratio of $\left[B C B^{\prime}\right]$ to $\left[B C^{\prime} B^{\prime}\right]$.
|
1 Solution: Since $C, B^{\prime}$, and $C^{\prime}$ are collinear, it is evident that $\left[B C B^{\prime}\right]=\frac{1}{2}\left[B C C^{\prime}\right]$. It immediately follows that $\left[B C B^{\prime}\right]=\left[B C^{\prime} B^{\prime}\right]$. Thus, the ratio is 1 .
$12^{\text {th }}$ HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n6. [6]",
"solution_match": "\nAnswer: "
}
|
2b94a5f7-cd9a-5129-b83e-215d07ebeea4
| 608,516
|
How many perfect squares divide $2^{3} \cdot 3^{5} \cdot 5^{7} \cdot 7^{9}$ ?
|
The number of such perfect squares is $2 \cdot 3 \cdot 4 \cdot 5$, since the exponent of each prime can be any nonnegative even number less than the given exponent.
|
2 \cdot 3 \cdot 4 \cdot 5
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many perfect squares divide $2^{3} \cdot 3^{5} \cdot 5^{7} \cdot 7^{9}$ ?
|
The number of such perfect squares is $2 \cdot 3 \cdot 4 \cdot 5$, since the exponent of each prime can be any nonnegative even number less than the given exponent.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n7. [6]",
"solution_match": "\nSolution: "
}
|
ae1fb6d6-977f-522c-b5f4-dc9758d59937
| 608,517
|
An icosidodecahedron is a convex polyhedron with 20 triangular faces and 12 pentagonal faces. How many vertices does it have?
|
Since every edge is shared by exactly two faces, there are $(20 \cdot 3+12 \cdot 5) / 2=60$ edges. Using Euler's formula $v-e+f=2$, we see that there are 30 vertices.
$12^{\text {th }}$ HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 - GUTS ROUND
|
30
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
An icosidodecahedron is a convex polyhedron with 20 triangular faces and 12 pentagonal faces. How many vertices does it have?
|
Since every edge is shared by exactly two faces, there are $(20 \cdot 3+12 \cdot 5) / 2=60$ edges. Using Euler's formula $v-e+f=2$, we see that there are 30 vertices.
$12^{\text {th }}$ HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n9. [6]",
"solution_match": "\nSolution: "
}
|
4b67d981-b00e-5a62-be13-e83c2cc1966a
| 608,519
|
There are 2008 distinct points on a circle. If you connect two of these points to form a line and then connect another two points (distinct from the first two) to form another line, what is the probability that the two lines intersect inside the circle?
|
Given four of these points, there are 3 ways in which to connect two of them and then connect the other two, and of these possibilities exactly one will intersect inside the circle. Thus $1 / 3$ of all the ways to connect two lines and then connect two others have an intersection point inside the circle.
|
\frac{1}{3}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
There are 2008 distinct points on a circle. If you connect two of these points to form a line and then connect another two points (distinct from the first two) to form another line, what is the probability that the two lines intersect inside the circle?
|
Given four of these points, there are 3 ways in which to connect two of them and then connect the other two, and of these possibilities exactly one will intersect inside the circle. Thus $1 / 3$ of all the ways to connect two lines and then connect two others have an intersection point inside the circle.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n11. [7]",
"solution_match": "\nSolution: "
}
|
a8eadfa4-37fc-5557-8145-f6cfbd91f511
| 608,521
|
Bob is writing a sequence of letters of the alphabet, each of which can be either uppercase or lowercase, according to the following two rules:
- If he had just written an uppercase letter, he can either write the same letter in lowercase after it, or the next letter of the alphabet in uppercase.
- If he had just written a lowercase letter, he can either write the same letter in uppercase after it, or the preceding letter of the alphabet in lowercase.
For instance, one such sequence is $a A a A B C D d c b B C$. How many sequences of 32 letters can he write that start at (lowercase) $a$ and end at (lowercase) $z$ ? (The alphabet contains 26 letters from $a$ to $z$.)
|
The smallest possible sequence from $a$ to $z$ is $a A B C D \ldots Z z$, which has 28 letters. To insert 4 more letters, we can either switch two (not necessarily distinct) letters to lowercase and back again (as in $a A B C c C D E F f F G H \ldots Z z$ ), or we can insert a lowercase letter after its corresponding uppercase letter, insert the previous letter of the alphabet, switch back to uppercase, and continue the sequence (as in $a A B C c b B C D E \ldots Z z$ ). There are $\binom{27}{2}=13 \cdot 27$ sequences of the former type and 25 of the latter, for a total of 376 such sequences.
## $12^{\text {th }}$ HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 - GUTS ROUND
|
376
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Bob is writing a sequence of letters of the alphabet, each of which can be either uppercase or lowercase, according to the following two rules:
- If he had just written an uppercase letter, he can either write the same letter in lowercase after it, or the next letter of the alphabet in uppercase.
- If he had just written a lowercase letter, he can either write the same letter in uppercase after it, or the preceding letter of the alphabet in lowercase.
For instance, one such sequence is $a A a A B C D d c b B C$. How many sequences of 32 letters can he write that start at (lowercase) $a$ and end at (lowercase) $z$ ? (The alphabet contains 26 letters from $a$ to $z$.)
|
The smallest possible sequence from $a$ to $z$ is $a A B C D \ldots Z z$, which has 28 letters. To insert 4 more letters, we can either switch two (not necessarily distinct) letters to lowercase and back again (as in $a A B C c C D E F f F G H \ldots Z z$ ), or we can insert a lowercase letter after its corresponding uppercase letter, insert the previous letter of the alphabet, switch back to uppercase, and continue the sequence (as in $a A B C c b B C D E \ldots Z z$ ). There are $\binom{27}{2}=13 \cdot 27$ sequences of the former type and 25 of the latter, for a total of 376 such sequences.
## $12^{\text {th }}$ HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n12. [7]",
"solution_match": "\nSolution: "
}
|
c07306a3-2f21-5c88-a153-bb500f09a0c6
| 608,522
|
How many ordered quadruples $(a, b, c, d)$ of four distinct numbers chosen from the set $\{1,2,3, \ldots, 9\}$ satisfy $b<a, b<c$, and $d<c$ ?
|
Given any 4 elements $p<q<r<s$ of $\{1,2, \ldots, 9\}$, there are 5 ways of rearranging them to satisfy the inequality: prqs, psqr, qspr, qrps, and rspq. This gives a total of $\binom{9}{4} \cdot 5=630$ quadruples.
|
630
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many ordered quadruples $(a, b, c, d)$ of four distinct numbers chosen from the set $\{1,2,3, \ldots, 9\}$ satisfy $b<a, b<c$, and $d<c$ ?
|
Given any 4 elements $p<q<r<s$ of $\{1,2, \ldots, 9\}$, there are 5 ways of rearranging them to satisfy the inequality: prqs, psqr, qspr, qrps, and rspq. This gives a total of $\binom{9}{4} \cdot 5=630$ quadruples.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n13. [8]",
"solution_match": "\nSolution: "
}
|
93609cd1-66fd-5c8f-8592-d9a88bdaa79f
| 608,523
|
Compute
$$
\sum_{k=1}^{2009} k\left(\left\lfloor\frac{2009}{k}\right\rfloor-\left\lfloor\frac{2008}{k}\right\rfloor\right)
$$
|
The summand is equal to $k$ if $k$ divides 2009 and 0 otherwise. Thus the sum is equal to the sum of the divisors of 2009 , or 2394.
|
2394
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Compute
$$
\sum_{k=1}^{2009} k\left(\left\lfloor\frac{2009}{k}\right\rfloor-\left\lfloor\frac{2008}{k}\right\rfloor\right)
$$
|
The summand is equal to $k$ if $k$ divides 2009 and 0 otherwise. Thus the sum is equal to the sum of the divisors of 2009 , or 2394.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n14. [8]",
"solution_match": "\nSolution: "
}
|
5f70b34a-b725-5604-8b74-7fe0bf51b7ec
| 608,524
|
Stan has a stack of 100 blocks and starts with a score of 0 , and plays a game in which he iterates the following two-step procedure:
(a) Stan picks a stack of blocks and splits it into 2 smaller stacks each with a positive number of blocks, say $a$ and $b$. (The order in which the new piles are placed does not matter.)
(b) Stan adds the product of the two piles' sizes, $a b$, to his score.
The game ends when there are only 1-block stacks left. What is the expected value of Stan's score at the end of the game?
|
Let $E(n)$ be the expected value of the score for an $n$-block game. It suffices to show that the score is invariant regardless of how the game is played. We proceed by induction. We have $E(1)=0$ and $E(2)=1$. We require that $E(n)=E(n-k)+E(k)+(n-k) k$ for all $k$. Setting $k=1$, we hypothesize that $E(n)=n(n-1) / 2$. This satisfies the recursion and base cases so $E(100)=100 \cdot 99 / 2=4950$.
|
4950
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Stan has a stack of 100 blocks and starts with a score of 0 , and plays a game in which he iterates the following two-step procedure:
(a) Stan picks a stack of blocks and splits it into 2 smaller stacks each with a positive number of blocks, say $a$ and $b$. (The order in which the new piles are placed does not matter.)
(b) Stan adds the product of the two piles' sizes, $a b$, to his score.
The game ends when there are only 1-block stacks left. What is the expected value of Stan's score at the end of the game?
|
Let $E(n)$ be the expected value of the score for an $n$-block game. It suffices to show that the score is invariant regardless of how the game is played. We proceed by induction. We have $E(1)=0$ and $E(2)=1$. We require that $E(n)=E(n-k)+E(k)+(n-k) k$ for all $k$. Setting $k=1$, we hypothesize that $E(n)=n(n-1) / 2$. This satisfies the recursion and base cases so $E(100)=100 \cdot 99 / 2=4950$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n15. [8]",
"solution_match": "\nSolution: "
}
|
decddd25-8005-516f-b510-37183afa5187
| 608,525
|
A spider is making a web between $n>1$ distinct leaves which are equally spaced around a circle. He chooses a leaf to start at, and to make the base layer he travels to each leaf one at a time, making a straight line of silk between each consecutive pair of leaves, such that no two of the lines of silk cross each other and he visits every leaf exactly once. In how many ways can the spider make the base layer of the web? Express your answer in terms of $n$.
|
There are $n$ ways to choose a starting vertex, and at each vertex he has only two choices for where to go next: the nearest untouched leaf in the clockwise direction, and the nearest untouched leaf in the counterclockwise direction. For, if the spider visited a leaf which is not nearest in some direction, there are two untouched leaves which are separated by this line of silk, and so the silk would eventually cross itself. Thus, for the first $n-2$ choices there are 2 possibilities, and the $(n-1)$ st choice is then determined.
Note: This formula can also be derived recursively.
|
n \cdot 2^{n-2}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A spider is making a web between $n>1$ distinct leaves which are equally spaced around a circle. He chooses a leaf to start at, and to make the base layer he travels to each leaf one at a time, making a straight line of silk between each consecutive pair of leaves, such that no two of the lines of silk cross each other and he visits every leaf exactly once. In how many ways can the spider make the base layer of the web? Express your answer in terms of $n$.
|
There are $n$ ways to choose a starting vertex, and at each vertex he has only two choices for where to go next: the nearest untouched leaf in the clockwise direction, and the nearest untouched leaf in the counterclockwise direction. For, if the spider visited a leaf which is not nearest in some direction, there are two untouched leaves which are separated by this line of silk, and so the silk would eventually cross itself. Thus, for the first $n-2$ choices there are 2 possibilities, and the $(n-1)$ st choice is then determined.
Note: This formula can also be derived recursively.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n16. [9]",
"solution_match": "\nSolution: "
}
|
52d006ab-e26f-5664-aebf-27c5609d233d
| 608,526
|
How many positive integers $n \leq 2009$ have the property that $\left\lfloor\log _{2}(n)\right\rfloor$ is odd?
|
682
|
682
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many positive integers $n \leq 2009$ have the property that $\left\lfloor\log _{2}(n)\right\rfloor$ is odd?
|
682
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n17. [9]",
"solution_match": "\nAnswer: "
}
|
eec11259-1ba3-57c2-8fbd-7838dcc76ecd
| 608,527
|
If $n$ is a positive integer such that $n^{3}+2 n^{2}+9 n+8$ is the cube of an integer, find $n$.
|
Since $n^{3}<n^{3}+2 n^{2}+9 n+8<(n+2)^{3}$, we must have $n^{3}+2 n^{2}+9 n+8=(n+1)^{3}$. Thus $n^{2}=6 n+7$, so $n=7$.
$\qquad$
$12^{\text {th }}$ HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 - GUTS ROUND
|
7
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
If $n$ is a positive integer such that $n^{3}+2 n^{2}+9 n+8$ is the cube of an integer, find $n$.
|
Since $n^{3}<n^{3}+2 n^{2}+9 n+8<(n+2)^{3}$, we must have $n^{3}+2 n^{2}+9 n+8=(n+1)^{3}$. Thus $n^{2}=6 n+7$, so $n=7$.
$\qquad$
$12^{\text {th }}$ HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n18. [9]",
"solution_match": "\nSolution: "
}
|
f606fb95-b4df-5e8b-9292-02e2ddec8b3a
| 608,528
|
Shelly writes down a vector $v=(a, b, c, d)$, where $0<a<b<c<d$ are integers. Let $\sigma(v)$ denote the set of 24 vectors whose coordinates are $a, b, c$, and $d$ in some order. For instance, $\sigma(v)$ contains $(b, c, d, a)$. Shelly notes that there are 3 vectors in $\sigma(v)$ whose sum is of the form $(s, s, s, s)$ for some $s$. What is the smallest possible value of $d$ ?
|
If $k=a+b+c+d$, first you notice $4 \mid 3 k$, and $k \geq 10$. So we try $k=12$, which works with $a, b, c, d=1,2,3,6$ and not $1,2,4,5$.
|
6
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Shelly writes down a vector $v=(a, b, c, d)$, where $0<a<b<c<d$ are integers. Let $\sigma(v)$ denote the set of 24 vectors whose coordinates are $a, b, c$, and $d$ in some order. For instance, $\sigma(v)$ contains $(b, c, d, a)$. Shelly notes that there are 3 vectors in $\sigma(v)$ whose sum is of the form $(s, s, s, s)$ for some $s$. What is the smallest possible value of $d$ ?
|
If $k=a+b+c+d$, first you notice $4 \mid 3 k$, and $k \geq 10$. So we try $k=12$, which works with $a, b, c, d=1,2,3,6$ and not $1,2,4,5$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n19. [10]",
"solution_match": "\nSolution: "
}
|
1159070b-e9a0-5e25-b85a-d05b436f60c1
| 608,529
|
A positive integer is called jubilant if the number of 1's in its binary representation is even. For example, $6=110_{2}$ is a jubilant number. What is the 2009th smallest jubilant number?
|
Notice that for each pair of consecutive positive integers $2 k$ and $2 k+1$, their binary representation differs by exactly one 1 (in the units digit), so exactly one of 2 and 3 is jubilant, exactly one of 4 and 5 is jubilant, etc. It follows that there are exactly 2009 jubilant numbers less than or equal to 4019 . We now simply need to check whether 4018 or 4019 is jubilant. Since the binary representation of 4018 is 111110110010,4018 is the 2009 th jubilant number.
|
4018
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
A positive integer is called jubilant if the number of 1's in its binary representation is even. For example, $6=110_{2}$ is a jubilant number. What is the 2009th smallest jubilant number?
|
Notice that for each pair of consecutive positive integers $2 k$ and $2 k+1$, their binary representation differs by exactly one 1 (in the units digit), so exactly one of 2 and 3 is jubilant, exactly one of 4 and 5 is jubilant, etc. It follows that there are exactly 2009 jubilant numbers less than or equal to 4019 . We now simply need to check whether 4018 or 4019 is jubilant. Since the binary representation of 4018 is 111110110010,4018 is the 2009 th jubilant number.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n20. [10]",
"solution_match": "\nSolution: "
}
|
b1a656da-985c-596a-9a89-9a1f4a550818
| 608,530
|
A circle having radius $r_{1}$ centered at point $N$ is tangent to a circle of radius $r_{2}$ centered at $M$. Let $l$ and $j$ be the two common external tangent lines to the two circles. A circle centered at $P$ with radius $r_{2}$ is externally tangent to circle $N$ at the point at which $l$ coincides with circle $N$, and line $k$ is externally tangent to $P$ and $N$ such that points $M, N$, and $P$ all lie on the same side of $k$. For what ratio $r_{1} / r_{2}$ are $j$ and $k$ parallel?
|
Suppose the lines are parallel. Draw the other tangent line to $N$ and $P$ - since $M$ and $P$ have the same radius, it is tangent to all three circles. Let $j$ and $k$ meet circle $N$ at $A$ and $B$, respectively. Then by symmetry we see that $\angle A N M=\angle M N P=\angle P N B=60^{\circ}$ since $A, N$, and $B$ are collinear (perpendicular to $j$ and $k$ ). Let $D$ be the foot of the perpendicular from $M$ to $A N$. In $\triangle M D N$, we have $M N=2 D N$, so $r_{1}+r_{2}=2\left(r_{1}-r_{2}\right)$, and so $r_{1} / r_{2}=3$.
|
3
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A circle having radius $r_{1}$ centered at point $N$ is tangent to a circle of radius $r_{2}$ centered at $M$. Let $l$ and $j$ be the two common external tangent lines to the two circles. A circle centered at $P$ with radius $r_{2}$ is externally tangent to circle $N$ at the point at which $l$ coincides with circle $N$, and line $k$ is externally tangent to $P$ and $N$ such that points $M, N$, and $P$ all lie on the same side of $k$. For what ratio $r_{1} / r_{2}$ are $j$ and $k$ parallel?
|
Suppose the lines are parallel. Draw the other tangent line to $N$ and $P$ - since $M$ and $P$ have the same radius, it is tangent to all three circles. Let $j$ and $k$ meet circle $N$ at $A$ and $B$, respectively. Then by symmetry we see that $\angle A N M=\angle M N P=\angle P N B=60^{\circ}$ since $A, N$, and $B$ are collinear (perpendicular to $j$ and $k$ ). Let $D$ be the foot of the perpendicular from $M$ to $A N$. In $\triangle M D N$, we have $M N=2 D N$, so $r_{1}+r_{2}=2\left(r_{1}-r_{2}\right)$, and so $r_{1} / r_{2}=3$.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n22. [10]",
"solution_match": "\nSolution: "
}
|
6df4b15e-2d01-5fdc-9858-bfe6026b7f0e
| 608,532
|
Four points, $A, B, C$, and $D$, are chosen randomly on the circumference of a circle with independent uniform probability. What is the expected number of sides of triangle $A B C$ for which the projection of $D$ onto the line containing the side lies between the two vertices?
|
$3 / 2$
|
\frac{3}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Four points, $A, B, C$, and $D$, are chosen randomly on the circumference of a circle with independent uniform probability. What is the expected number of sides of triangle $A B C$ for which the projection of $D$ onto the line containing the side lies between the two vertices?
|
$3 / 2$
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n25. [12]",
"solution_match": "\nAnswer: "
}
|
6938ab92-2f0f-565f-8870-e95edf43a8d5
| 608,535
|
Define the sequence $\left\{x_{i}\right\}_{i \geq 0}$ by $x_{0}=2009$ and $x_{n}=-\frac{2009}{n} \sum_{k=0}^{n-1} x_{k}$ for all $n \geq 1$. Compute the value of $\sum_{n=0}^{2009} 2^{n} x_{n}$.
|
We have
$$
-\frac{n x_{n}}{2009}=x_{n-1}+x_{n-2}+\ldots+x_{0}=x_{n-1}+\frac{(n-1) x_{n-1}}{2009}
$$
, which yields the recursion $x_{n}=\frac{n-2010}{n} x_{n-1}$. Unwinding this recursion, we find $x_{n}=(-1)^{n} \cdot 2009$. $\binom{2008}{n}$. Thus
$$
\begin{aligned}
\sum_{k=0}^{2009} 2^{n} x_{n} & =\sum_{k=0}^{2009}(-2)^{n} \cdot 2009 \cdot\binom{2008}{n} \\
& =2009 \sum_{k=0}^{2008}(-2)^{n}\binom{2008}{n} \\
& =2009(-2+1)^{2008}
\end{aligned}
$$
as desired.
|
2009
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Define the sequence $\left\{x_{i}\right\}_{i \geq 0}$ by $x_{0}=2009$ and $x_{n}=-\frac{2009}{n} \sum_{k=0}^{n-1} x_{k}$ for all $n \geq 1$. Compute the value of $\sum_{n=0}^{2009} 2^{n} x_{n}$.
|
We have
$$
-\frac{n x_{n}}{2009}=x_{n-1}+x_{n-2}+\ldots+x_{0}=x_{n-1}+\frac{(n-1) x_{n-1}}{2009}
$$
, which yields the recursion $x_{n}=\frac{n-2010}{n} x_{n-1}$. Unwinding this recursion, we find $x_{n}=(-1)^{n} \cdot 2009$. $\binom{2008}{n}$. Thus
$$
\begin{aligned}
\sum_{k=0}^{2009} 2^{n} x_{n} & =\sum_{k=0}^{2009}(-2)^{n} \cdot 2009 \cdot\binom{2008}{n} \\
& =2009 \sum_{k=0}^{2008}(-2)^{n}\binom{2008}{n} \\
& =2009(-2+1)^{2008}
\end{aligned}
$$
as desired.
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n26. [12]",
"solution_match": "\nSolution: "
}
|
58491cdb-ec17-5162-a6e1-7a31adb88775
| 608,536
|
Circle $\Omega$ has radius 5. Points $A$ and $B$ lie on $\Omega$ such that chord $A B$ has length 6 . A unit circle $\omega$ is tangent to chord $A B$ at point $T$. Given that $\omega$ is also internally tangent to $\Omega$, find $A T \cdot B T$.
|
Let $M$ be the midpoint of chord $A B$ and let $O$ be the center of $\Omega$. Since $A M=B M=3$, Pythagoras on triangle $A M O$ gives $O M=4$. Now let $\omega$ be centered at $P$ and say that $\omega$ and $\Omega$ are tangent at $Q$. Because the diameter of $\omega$ exceeds 1 , points $P$ and $Q$ lie on the same side of $A B$. By tangency, $O, P$, and $Q$ are collinear, so that $O P=O Q-P Q=4$. Let $H$ be the orthogonal projection of $P$ onto $O M$; then $O H=O M-H M=O M-P T=3$. Pythagoras on $O H P$ gives $H P^{2}=7$. Finally,
$$
A T \cdot B T=A M^{2}-M T^{2}=A M^{2}-H P^{2}=9-7=2
$$
$12^{\text {th }}$ HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 - GUTS ROUND
|
2
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Circle $\Omega$ has radius 5. Points $A$ and $B$ lie on $\Omega$ such that chord $A B$ has length 6 . A unit circle $\omega$ is tangent to chord $A B$ at point $T$. Given that $\omega$ is also internally tangent to $\Omega$, find $A T \cdot B T$.
|
Let $M$ be the midpoint of chord $A B$ and let $O$ be the center of $\Omega$. Since $A M=B M=3$, Pythagoras on triangle $A M O$ gives $O M=4$. Now let $\omega$ be centered at $P$ and say that $\omega$ and $\Omega$ are tangent at $Q$. Because the diameter of $\omega$ exceeds 1 , points $P$ and $Q$ lie on the same side of $A B$. By tangency, $O, P$, and $Q$ are collinear, so that $O P=O Q-P Q=4$. Let $H$ be the orthogonal projection of $P$ onto $O M$; then $O H=O M-H M=O M-P T=3$. Pythagoras on $O H P$ gives $H P^{2}=7$. Finally,
$$
A T \cdot B T=A M^{2}-M T^{2}=A M^{2}-H P^{2}=9-7=2
$$
$12^{\text {th }}$ HARVARD-MIT MATHEMATICS TOURNAMENT, 21 FEBRUARY 2009 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-122-2009-feb-guts-solutions.jsonl",
"problem_match": "\n27. [12]",
"solution_match": "\nSolution: "
}
|
78d9f7e5-ab34-58c9-817c-369b695cb263
| 608,537
|
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