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To set up for a Fourth of July party, David is making a string of red, white, and blue balloons. He places them according to the following rules:
- No red balloon is adjacent to another red balloon.
- White balloons appear in groups of exactly two, and groups of white balloons are separated by at least two non-white balloons.
- Blue balloons appear in groups of exactly three, and groups of blue balloons are separated by at least three non-blue balloons.
If David uses over 600 balloons, determine the smallest number of red balloons that he can use.
|
99 It is possible to achieve 99 red balloons with the arrangement
$$
\text { WWBBBWW } \underbrace{\text { RBBBWWRBBBWW . . RBBBWW }}_{99 \text { RBBBWW's }},
$$
which contains $99 \cdot 6+7=601$ balloons.
Now assume that one can construct a chain with 98 or fewer red balloons. Then there can be 99 blocks of non-red balloons, which in total must contain more than 502 balloons. The only valid combinations of white and blue balloons are WWBBB, BBBWW, and WWBBBWW (Any others contain the subsequence BBBWWBBB , which is invalid). The sequence ... WWR must be followed by BBBWW; otherwise two groups of white balloons would be too close. Similarly, the sequence
RWW . . . must be preceded by WWBBB. It follows that WWBBBWW can be used at most once in a valid sequence, meaning that there can be at most $98 \cdot 5+7=497$ non-red balloons. Contradiction. Therefore the minimum is 99 red balloons. (Better if the party's outdoors; then we'd have 99 red balloons floating in the summer sky. :-p)
|
99
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
To set up for a Fourth of July party, David is making a string of red, white, and blue balloons. He places them according to the following rules:
- No red balloon is adjacent to another red balloon.
- White balloons appear in groups of exactly two, and groups of white balloons are separated by at least two non-white balloons.
- Blue balloons appear in groups of exactly three, and groups of blue balloons are separated by at least three non-blue balloons.
If David uses over 600 balloons, determine the smallest number of red balloons that he can use.
|
99 It is possible to achieve 99 red balloons with the arrangement
$$
\text { WWBBBWW } \underbrace{\text { RBBBWWRBBBWW . . RBBBWW }}_{99 \text { RBBBWW's }},
$$
which contains $99 \cdot 6+7=601$ balloons.
Now assume that one can construct a chain with 98 or fewer red balloons. Then there can be 99 blocks of non-red balloons, which in total must contain more than 502 balloons. The only valid combinations of white and blue balloons are WWBBB, BBBWW, and WWBBBWW (Any others contain the subsequence BBBWWBBB , which is invalid). The sequence ... WWR must be followed by BBBWW; otherwise two groups of white balloons would be too close. Similarly, the sequence
RWW . . . must be preceded by WWBBB. It follows that WWBBBWW can be used at most once in a valid sequence, meaning that there can be at most $98 \cdot 5+7=497$ non-red balloons. Contradiction. Therefore the minimum is 99 red balloons. (Better if the party's outdoors; then we'd have 99 red balloons floating in the summer sky. :-p)
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n4. [5]",
"solution_match": "\nAnswer: "
}
|
978282dc-6912-5d96-a4ed-00fc57c27963
| 608,670
|
You have a length of string and 7 beads in the 7 colors of the rainbow. You place the beads on the string as follows - you randomly pick a bead that you haven't used yet, then randomly add it to either the left end or the right end of the string. What is the probability that, at the end, the colors of the beads are the colors of the rainbow in order? (The string cannot be flipped, so the red bead must appear on the left side and the violet bead on the right side.)
|
$\frac{1}{5040}$ The threading method does not depend on the colors of the beads, so at the end all configurations are equally likely. Since there are $7!=5040$ configurations in total, the probability of any particular configuration is $\frac{1}{5040}$.
|
\frac{1}{5040}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
You have a length of string and 7 beads in the 7 colors of the rainbow. You place the beads on the string as follows - you randomly pick a bead that you haven't used yet, then randomly add it to either the left end or the right end of the string. What is the probability that, at the end, the colors of the beads are the colors of the rainbow in order? (The string cannot be flipped, so the red bead must appear on the left side and the violet bead on the right side.)
|
$\frac{1}{5040}$ The threading method does not depend on the colors of the beads, so at the end all configurations are equally likely. Since there are $7!=5040$ configurations in total, the probability of any particular configuration is $\frac{1}{5040}$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n5. [5]",
"solution_match": "\nAnswer: "
}
|
0f0b4124-c0b1-5f56-b6fe-46ea6d29fcb5
| 608,671
|
How many different numbers are obtainable from five 5 s by first concatenating some of the 5 s, then multiplying them together? For example, we could do $5 \cdot 55 \cdot 55,555 \cdot 55$, or 55555 , but not $5 \cdot 5$ or 2525 .
|
7 If we do 55555 , then we're done.
Note that $5,55,555$, and 5555 all have completely distinguishable prime factorizations. This means that if we are given a product of them, we can obtain the individual terms. The number of 5555's is the exponent of 101 , the number of 555 's is the exponent of 37 , the number of 55 's is the exponent of 11 minus the exponent of 101 , and the number of 5 's is just whatever we need to get the proper exponent of 5 . Then the answer is the number of ways we can split the five 5 's into groups of at least one. This is the number of unordered partitions of 5 , which is 7 .
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
7
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many different numbers are obtainable from five 5 s by first concatenating some of the 5 s, then multiplying them together? For example, we could do $5 \cdot 55 \cdot 55,555 \cdot 55$, or 55555 , but not $5 \cdot 5$ or 2525 .
|
7 If we do 55555 , then we're done.
Note that $5,55,555$, and 5555 all have completely distinguishable prime factorizations. This means that if we are given a product of them, we can obtain the individual terms. The number of 5555's is the exponent of 101 , the number of 555 's is the exponent of 37 , the number of 55 's is the exponent of 11 minus the exponent of 101 , and the number of 5 's is just whatever we need to get the proper exponent of 5 . Then the answer is the number of ways we can split the five 5 's into groups of at least one. This is the number of unordered partitions of 5 , which is 7 .
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n6. [5]",
"solution_match": "\nAnswer: "
}
|
26f9406a-23bd-54ee-901f-c64db96abecf
| 608,672
|
What are the last 8 digits of
$$
11 \times 101 \times 1001 \times 10001 \times 100001 \times 1000001 \times 111 ?
$$
|
19754321 Multiply terms in a clever order.
$$
\begin{aligned}
11 \cdot 101 \cdot 10001 & =11,111,111 \\
111 \cdot 1001 \cdot 1000001 & =111,111,111,111
\end{aligned}
$$
The last eight digits of $11,111,111 \cdot 111,111,111,111$ are 87654321 . We then just need to compute the last 8 digits of $87654321 \cdot 100001=87654321+\ldots 32100000$, which are 19754321 .
|
19754321
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
What are the last 8 digits of
$$
11 \times 101 \times 1001 \times 10001 \times 100001 \times 1000001 \times 111 ?
$$
|
19754321 Multiply terms in a clever order.
$$
\begin{aligned}
11 \cdot 101 \cdot 10001 & =11,111,111 \\
111 \cdot 1001 \cdot 1000001 & =111,111,111,111
\end{aligned}
$$
The last eight digits of $11,111,111 \cdot 111,111,111,111$ are 87654321 . We then just need to compute the last 8 digits of $87654321 \cdot 100001=87654321+\ldots 32100000$, which are 19754321 .
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n7. [6]",
"solution_match": "\nAnswer: "
}
|
14a69bb0-23f0-54f9-a4ae-4c11aca4e3d4
| 608,673
|
Indecisive Andy starts out at the midpoint of the 1-unit-long segment $\overline{H T}$. He flips 2010 coins. On each flip, if the coin is heads, he moves halfway towards endpoint $H$, and if the coin is tails, he moves halfway towards endpoint $T$. After his 2010 moves, what is the expected distance between Andy and the midpoint of $\overline{H T}$ ?
|
| $\frac{1}{4}$ |
| :---: |
| Let Andy's position be $x$ units from the H end after 2009 flips. If Any moves towards | the $H$ end, he ends up at $\frac{x}{2}$, a distance of $\frac{1-x}{2}$ from the midpoint. If Andy moves towards the $T$ end, he ends up at $\frac{1+x}{2}$, a distance of $\frac{x}{2}$ from the midpoint. His expected distance from the midpoint is then
$$
\frac{\frac{1-x}{2}+\frac{x}{2}}{2}=\frac{1}{4} .
$$
Since this does not depend on $x, \frac{1}{4}$ is the answer.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND
|
\frac{1}{4}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Indecisive Andy starts out at the midpoint of the 1-unit-long segment $\overline{H T}$. He flips 2010 coins. On each flip, if the coin is heads, he moves halfway towards endpoint $H$, and if the coin is tails, he moves halfway towards endpoint $T$. After his 2010 moves, what is the expected distance between Andy and the midpoint of $\overline{H T}$ ?
|
| $\frac{1}{4}$ |
| :---: |
| Let Andy's position be $x$ units from the H end after 2009 flips. If Any moves towards | the $H$ end, he ends up at $\frac{x}{2}$, a distance of $\frac{1-x}{2}$ from the midpoint. If Andy moves towards the $T$ end, he ends up at $\frac{1+x}{2}$, a distance of $\frac{x}{2}$ from the midpoint. His expected distance from the midpoint is then
$$
\frac{\frac{1-x}{2}+\frac{x}{2}}{2}=\frac{1}{4} .
$$
Since this does not depend on $x, \frac{1}{4}$ is the answer.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n9. [6]",
"solution_match": "\nAnswer: "
}
|
8b1dad3a-c108-5e1c-bf14-c30d0ff433bd
| 608,675
|
Let $A B C$ be a triangle with $A B=8, B C=15$, and $A C=17$. Point $X$ is chosen at random on line segment $A B$. Point $Y$ is chosen at random on line segment $B C$. Point $Z$ is chosen at random on line segment $C A$. What is the expected area of triangle $X Y Z$ ?
|
15 Let $\mathbb{E}(X)$ denote the expected value of $X$, and let $[S]$ denote the area of $S$. Then
$$
\begin{aligned}
\mathbb{E}([\triangle X Y Z]) & =\mathbb{E}([\triangle A B C]-[\triangle X Y B]-[\triangle Z Y C]-[\triangle X B Z]) \\
& =[\triangle A B C]-\mathbb{E}([\triangle X Y B])-\mathbb{E}([\triangle Z Y C])-[\triangle X B Z])
\end{aligned}
$$
where the last step follows from linearity of expectation ${ }^{1}$. But $[\triangle X Y B]=\frac{1}{2} \cdot B X \cdot B Y \cdot \sin (B)$. The $\frac{1}{2} \sin (B)$ term is constant, and $B X$ and $B Y$ are both independent with expected values $\frac{A B}{2}$ and $\frac{B C}{2}$, respectively. Thus $\mathbb{E}([\triangle X Y B])=\frac{1}{8} A B \cdot B C \cdot \sin (B)=\frac{1}{4}[\triangle A B C]$. Similarly, $\mathbb{E}([\triangle Z Y C])=$ $\mathbb{E}([\triangle Z B X])=\frac{1}{4}[\triangle A B C]$.
Then we have $\mathbb{E}([\triangle X Y Z])=\left(1-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}\right)[\triangle A B C]=\frac{1}{4}[\triangle A B C]=15$.
Note: We can also solve this problem (and the more general case of polygons) by noting that the area of $X Y Z$ is linear in the coordinates of $X, Y$, and $Z$, so the expected area of $X Y Z$ is the same as the area of $X^{\prime} Y^{\prime} Z^{\prime}$, where $X^{\prime}$ is the expected location of $X, Y^{\prime}$ is the expected location of $Y$, and $Z^{\prime}$ is the expected location of $Z$. In our case, this corresponds to the midpoints of the three sides $A B$, $B C$, and $C A$.
|
15
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a triangle with $A B=8, B C=15$, and $A C=17$. Point $X$ is chosen at random on line segment $A B$. Point $Y$ is chosen at random on line segment $B C$. Point $Z$ is chosen at random on line segment $C A$. What is the expected area of triangle $X Y Z$ ?
|
15 Let $\mathbb{E}(X)$ denote the expected value of $X$, and let $[S]$ denote the area of $S$. Then
$$
\begin{aligned}
\mathbb{E}([\triangle X Y Z]) & =\mathbb{E}([\triangle A B C]-[\triangle X Y B]-[\triangle Z Y C]-[\triangle X B Z]) \\
& =[\triangle A B C]-\mathbb{E}([\triangle X Y B])-\mathbb{E}([\triangle Z Y C])-[\triangle X B Z])
\end{aligned}
$$
where the last step follows from linearity of expectation ${ }^{1}$. But $[\triangle X Y B]=\frac{1}{2} \cdot B X \cdot B Y \cdot \sin (B)$. The $\frac{1}{2} \sin (B)$ term is constant, and $B X$ and $B Y$ are both independent with expected values $\frac{A B}{2}$ and $\frac{B C}{2}$, respectively. Thus $\mathbb{E}([\triangle X Y B])=\frac{1}{8} A B \cdot B C \cdot \sin (B)=\frac{1}{4}[\triangle A B C]$. Similarly, $\mathbb{E}([\triangle Z Y C])=$ $\mathbb{E}([\triangle Z B X])=\frac{1}{4}[\triangle A B C]$.
Then we have $\mathbb{E}([\triangle X Y Z])=\left(1-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}\right)[\triangle A B C]=\frac{1}{4}[\triangle A B C]=15$.
Note: We can also solve this problem (and the more general case of polygons) by noting that the area of $X Y Z$ is linear in the coordinates of $X, Y$, and $Z$, so the expected area of $X Y Z$ is the same as the area of $X^{\prime} Y^{\prime} Z^{\prime}$, where $X^{\prime}$ is the expected location of $X, Y^{\prime}$ is the expected location of $Y$, and $Z^{\prime}$ is the expected location of $Z$. In our case, this corresponds to the midpoints of the three sides $A B$, $B C$, and $C A$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n10. [7]",
"solution_match": "\nAnswer: "
}
|
fa0747de-c257-53b6-b863-85d9138ee6c3
| 608,676
|
From the point $(x, y)$, a legal move is a move to $\left(\frac{x}{3}+u, \frac{y}{3}+v\right)$, where $u$ and $v$ are real numbers such that $u^{2}+v^{2} \leq 1$. What is the area of the set of points that can be reached from $(0,0)$ in a finite number of legal moves?
|
| $\frac{9 \pi}{4}$ |
| :---: |
| We claim that the set of points is the disc with radius $\frac{3}{2}$ centered at the origin, which | clearly has area $\frac{9 \pi}{4}$.
First, we show that the set is contained in this disc. This is because if we are currently at a distance of $r$ from the origin, then we can't end up at a distance of greater than $\frac{r}{3}+1$ from the origin after a single move. Since $\frac{r}{3}+1<\frac{3}{2}$ if $r<\frac{3}{2}$, we will always end up in the disc of radius $\frac{3}{2}$ if we start in it. Since the origin is inside this disc, any finite number of moves will leave us inside this disc.
Next, we show that all points in this disc can be reached in a finite number of moves. Indeed, after one move we can get all points within a distance of 1 . After two moves, we can get all points within a distance of $\frac{4}{3}$. After three moves, we can get all points within a distance of $\frac{13}{9}$. In general, after $n$ moves we can get all points within a distance of $\frac{3}{2}-\frac{1}{2 \cdot 3^{k-1}}$. This means that for any distance $d<\frac{3}{2}$, we will eventually get all points within a distance of $d$, so all points in the disc of radius $\frac{3}{2}$ can be reached after some number of moves.
|
\frac{9 \pi}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
From the point $(x, y)$, a legal move is a move to $\left(\frac{x}{3}+u, \frac{y}{3}+v\right)$, where $u$ and $v$ are real numbers such that $u^{2}+v^{2} \leq 1$. What is the area of the set of points that can be reached from $(0,0)$ in a finite number of legal moves?
|
| $\frac{9 \pi}{4}$ |
| :---: |
| We claim that the set of points is the disc with radius $\frac{3}{2}$ centered at the origin, which | clearly has area $\frac{9 \pi}{4}$.
First, we show that the set is contained in this disc. This is because if we are currently at a distance of $r$ from the origin, then we can't end up at a distance of greater than $\frac{r}{3}+1$ from the origin after a single move. Since $\frac{r}{3}+1<\frac{3}{2}$ if $r<\frac{3}{2}$, we will always end up in the disc of radius $\frac{3}{2}$ if we start in it. Since the origin is inside this disc, any finite number of moves will leave us inside this disc.
Next, we show that all points in this disc can be reached in a finite number of moves. Indeed, after one move we can get all points within a distance of 1 . After two moves, we can get all points within a distance of $\frac{4}{3}$. After three moves, we can get all points within a distance of $\frac{13}{9}$. In general, after $n$ moves we can get all points within a distance of $\frac{3}{2}-\frac{1}{2 \cdot 3^{k-1}}$. This means that for any distance $d<\frac{3}{2}$, we will eventually get all points within a distance of $d$, so all points in the disc of radius $\frac{3}{2}$ can be reached after some number of moves.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n11. [7]",
"solution_match": "\nAnswer: "
}
|
f06267ed-f0f3-5f30-8538-14d9eea20736
| 608,677
|
A triangle in the $x y$-plane is such that when projected onto the $x$-axis, $y$-axis, and the line $y=x$, the results are line segments whose endpoints are $(1,0)$ and $(5,0),(0,8)$ and $(0,13)$, and $(5,5)$ and $(7.5,7.5)$, respectively. What is the triangle's area?
|
$\frac{17}{2}$ Sketch the lines $x=1, x=5, y=8, y=13, y=10-x$, and $y=15-x$. The triangle has to be contained in the hexagonal region contained in all these lines. If all the projections are correct, every other vertex of the hexagon must be a vertex of the triangle, which gives us two possibilities for the triangle. One of these triangles has vertices at $(2,8),(1,13)$, and $(5,10)$, and has an area of $\frac{17}{2}$. It is easy to check that the other triangle has the same area, so the answer is unique.
|
\frac{17}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A triangle in the $x y$-plane is such that when projected onto the $x$-axis, $y$-axis, and the line $y=x$, the results are line segments whose endpoints are $(1,0)$ and $(5,0),(0,8)$ and $(0,13)$, and $(5,5)$ and $(7.5,7.5)$, respectively. What is the triangle's area?
|
$\frac{17}{2}$ Sketch the lines $x=1, x=5, y=8, y=13, y=10-x$, and $y=15-x$. The triangle has to be contained in the hexagonal region contained in all these lines. If all the projections are correct, every other vertex of the hexagon must be a vertex of the triangle, which gives us two possibilities for the triangle. One of these triangles has vertices at $(2,8),(1,13)$, and $(5,10)$, and has an area of $\frac{17}{2}$. It is easy to check that the other triangle has the same area, so the answer is unique.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n13. [8]",
"solution_match": "\nAnswer: "
}
|
1ceda535-bbe8-537c-914e-fbf1f5556a82
| 608,679
|
In how many ways can you fill a $3 \times 3$ table with the numbers 1 through 9 (each used once) such that all pairs of adjacent numbers (sharing one side) are relatively prime?
|
2016 The numbers can be separated into four sets. Numbers in the set $A=\{1,5,7\}$ can be placed next to anything. The next two sets are $B=\{2,4,8\}$ and $C=\{3,9\}$. The number 6 , which forms the final set $D$, can only be placed next to elements of $A$. The elements of each group can be interchanged without violating the condition, so without loss of generality, we can pretend we have three 1 's, three 2's, two 3 's, and one 6 , as long as we multiply our answer by $3!3!2$ ! at the end. The available arrangements are, grouped by the position of the 6 , are:
When 6 is in contact with three numbers:
| 1 | 2 | 3 |
| :--- | :--- | :--- |
| 6 | 1 | 2 |
| 1 | 2 | 3 |
When 6 is in contact with two numbers:
| 6 | 1 | 2 | | | |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 2 | 3 | 6 1 2 <br> 1 1 3 <br> 2 3 1 | 2 | 3 |
The next two can be flipped diagonally to create different arrangements:
| 6 | 1 | 2 | | | |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 2 | 3 | 6 1 | 2 | |
| 1 | 2 | 3 | | | |
| 1 | 3 | 2 | 3 | 1 | 2 |
Those seven arrangements can be rotated 90, 180, and 270 degrees about the center to generate a total of 28 arrangements. $28 \cdot 3!3!2!=2016$.
|
2016
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In how many ways can you fill a $3 \times 3$ table with the numbers 1 through 9 (each used once) such that all pairs of adjacent numbers (sharing one side) are relatively prime?
|
2016 The numbers can be separated into four sets. Numbers in the set $A=\{1,5,7\}$ can be placed next to anything. The next two sets are $B=\{2,4,8\}$ and $C=\{3,9\}$. The number 6 , which forms the final set $D$, can only be placed next to elements of $A$. The elements of each group can be interchanged without violating the condition, so without loss of generality, we can pretend we have three 1 's, three 2's, two 3 's, and one 6 , as long as we multiply our answer by $3!3!2$ ! at the end. The available arrangements are, grouped by the position of the 6 , are:
When 6 is in contact with three numbers:
| 1 | 2 | 3 |
| :--- | :--- | :--- |
| 6 | 1 | 2 |
| 1 | 2 | 3 |
When 6 is in contact with two numbers:
| 6 | 1 | 2 | | | |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 2 | 3 | 6 1 2 <br> 1 1 3 <br> 2 3 1 | 2 | 3 |
The next two can be flipped diagonally to create different arrangements:
| 6 | 1 | 2 | | | |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 2 | 3 | 6 1 | 2 | |
| 1 | 2 | 3 | | | |
| 1 | 3 | 2 | 3 | 1 | 2 |
Those seven arrangements can be rotated 90, 180, and 270 degrees about the center to generate a total of 28 arrangements. $28 \cdot 3!3!2!=2016$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n14. [8]",
"solution_match": "\nAnswer: "
}
|
a70b307a-e8e3-5cb0-85cb-644d5d84a0a0
| 608,680
|
Pick a random integer between 0 and 4095, inclusive. Write it in base 2 (without any leading zeroes). What is the expected number of consecutive digits that are not the same (that is, the expected number of occurrences of either 01 or 10 in the base 2 representation)?
|
$\frac{20481}{4096}$ Note that every number in the range can be written as a 12-digit binary string. For $i=1,2, \ldots 11$, let $R_{i}$ be a random variable which is 1 if the $i$ th and $(i+1)$ st digits differ in a randomly chosen number in the range. By linearity of expectation, $E\left(\sum_{i} R_{i}\right)=\sum E\left(R_{i}\right)$.
Since we choose every binary string of length 12 with equal probability, the sum of the expectations is $\frac{11}{2}$. However, this is not the expected number of 01 s and $10 s$ - we need to subtract the occasions where the leading digit is zero. There is a $\frac{1}{2}$ chance that the number starts with a 0 , in which case we must ignore the first digit change - unless the number was 0 , in which case there are no digit changes. Therefore, our answer is $\frac{11}{2}-\frac{1}{2}+\frac{1}{4096}=\frac{20481}{4096}$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
\frac{20481}{4096}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Pick a random integer between 0 and 4095, inclusive. Write it in base 2 (without any leading zeroes). What is the expected number of consecutive digits that are not the same (that is, the expected number of occurrences of either 01 or 10 in the base 2 representation)?
|
$\frac{20481}{4096}$ Note that every number in the range can be written as a 12-digit binary string. For $i=1,2, \ldots 11$, let $R_{i}$ be a random variable which is 1 if the $i$ th and $(i+1)$ st digits differ in a randomly chosen number in the range. By linearity of expectation, $E\left(\sum_{i} R_{i}\right)=\sum E\left(R_{i}\right)$.
Since we choose every binary string of length 12 with equal probability, the sum of the expectations is $\frac{11}{2}$. However, this is not the expected number of 01 s and $10 s$ - we need to subtract the occasions where the leading digit is zero. There is a $\frac{1}{2}$ chance that the number starts with a 0 , in which case we must ignore the first digit change - unless the number was 0 , in which case there are no digit changes. Therefore, our answer is $\frac{11}{2}-\frac{1}{2}+\frac{1}{4096}=\frac{20481}{4096}$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n15. [8]",
"solution_match": "\nAnswer: "
}
|
cb543222-8e09-5302-888b-82ee31ba4e14
| 608,681
|
Jessica has three marbles colored red, green, and blue. She randomly selects a non-empty subset of them (such that each subset is equally likely) and puts them in a bag. You then draw three marbles from the bag with replacement. The colors you see are red, blue, red. What is the probability that the only marbles in the bag are red and blue?
|
| $\frac{27}{35}$ |
| :---: |
| There are two possible sets of marbles in the bag, $\{$ red,blue\} and \{red,blue,green\}. | Initially, both these sets are equally likely to be in the bag. However, the probability of red, blue, red being drawn from a set $S$ of marbles is proportional to $|S|^{-3}$, as long as red and blue are both in $S$. By Bayes's Rule, we must weight the probability of these two sets by $|S|^{-3}$. The answer is $\frac{(1 / 2)^{3}}{(1 / 2)^{3}+(1 / 3)^{3}}$.
|
\frac{27}{35}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Jessica has three marbles colored red, green, and blue. She randomly selects a non-empty subset of them (such that each subset is equally likely) and puts them in a bag. You then draw three marbles from the bag with replacement. The colors you see are red, blue, red. What is the probability that the only marbles in the bag are red and blue?
|
| $\frac{27}{35}$ |
| :---: |
| There are two possible sets of marbles in the bag, $\{$ red,blue\} and \{red,blue,green\}. | Initially, both these sets are equally likely to be in the bag. However, the probability of red, blue, red being drawn from a set $S$ of marbles is proportional to $|S|^{-3}$, as long as red and blue are both in $S$. By Bayes's Rule, we must weight the probability of these two sets by $|S|^{-3}$. The answer is $\frac{(1 / 2)^{3}}{(1 / 2)^{3}+(1 / 3)^{3}}$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n16. [9]",
"solution_match": "\nAnswer: "
}
|
eb126528-b0af-521a-b548-fbe677bcdf1b
| 608,682
|
An ant starts at the origin, facing in the positive $x$-direction. Each second, it moves 1 unit forward, then turns counterclockwise by $\sin ^{-1}\left(\frac{3}{5}\right)$ degrees. What is the least upper bound on the distance between the ant and the origin? (The least upper bound is the smallest real number $r$ that is at least as big as every distance that the ant ever is from the origin.)
|
$\sqrt{10}$ We claim that the points the ant visits lie on a circle of radius $\frac{\sqrt{10}}{2}$. We show this by saying that the ant stays a constant distance $\frac{\sqrt{10}}{2}$ from the point $\left(\frac{1}{2}, \frac{3}{2}\right)$.
Suppose the ant moves on a plane $P$. Consider a transformation of the plane $P^{\prime}$ such that after the first move, the ant is at the origin of $P^{\prime}$ and facing in the direction of the $x^{\prime}$ axis (on $P^{\prime}$ ). The transformation to get from $P$ to $P^{\prime}$ can be gotten by rotating $P$ about the origin counterclockwise through an angle $\sin ^{-1}\left(\frac{3}{5}\right)$ and then translating it 1 unit to the right. Observe that the point $\left(\frac{1}{2}, \frac{3}{2}\right)$ is fixed under this transformation, which can be shown through the expression $\left(\frac{1}{2}+\frac{3}{2} i\right)\left(\frac{4}{5}+\frac{3}{5} i\right)+1=\frac{1}{2}+\frac{3}{2} i$. It follows that at every point the ant stops, it will always be the same distance from $\left(\frac{1}{2}, \frac{3}{2}\right)$. Since it starts at $(0,0)$, this fixed distance is $\frac{\sqrt{10}}{2}$.
Since $\sin ^{-1}\left(\frac{3}{5}\right)$ is not a rational multiple of $\pi$, the points the ant stops at form a dense subset of the circle in question. As a result, the least upper bound on the distance between the ant and the origin is the diameter of the circle, which is $\sqrt{10}$.
|
\sqrt{10}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
An ant starts at the origin, facing in the positive $x$-direction. Each second, it moves 1 unit forward, then turns counterclockwise by $\sin ^{-1}\left(\frac{3}{5}\right)$ degrees. What is the least upper bound on the distance between the ant and the origin? (The least upper bound is the smallest real number $r$ that is at least as big as every distance that the ant ever is from the origin.)
|
$\sqrt{10}$ We claim that the points the ant visits lie on a circle of radius $\frac{\sqrt{10}}{2}$. We show this by saying that the ant stays a constant distance $\frac{\sqrt{10}}{2}$ from the point $\left(\frac{1}{2}, \frac{3}{2}\right)$.
Suppose the ant moves on a plane $P$. Consider a transformation of the plane $P^{\prime}$ such that after the first move, the ant is at the origin of $P^{\prime}$ and facing in the direction of the $x^{\prime}$ axis (on $P^{\prime}$ ). The transformation to get from $P$ to $P^{\prime}$ can be gotten by rotating $P$ about the origin counterclockwise through an angle $\sin ^{-1}\left(\frac{3}{5}\right)$ and then translating it 1 unit to the right. Observe that the point $\left(\frac{1}{2}, \frac{3}{2}\right)$ is fixed under this transformation, which can be shown through the expression $\left(\frac{1}{2}+\frac{3}{2} i\right)\left(\frac{4}{5}+\frac{3}{5} i\right)+1=\frac{1}{2}+\frac{3}{2} i$. It follows that at every point the ant stops, it will always be the same distance from $\left(\frac{1}{2}, \frac{3}{2}\right)$. Since it starts at $(0,0)$, this fixed distance is $\frac{\sqrt{10}}{2}$.
Since $\sin ^{-1}\left(\frac{3}{5}\right)$ is not a rational multiple of $\pi$, the points the ant stops at form a dense subset of the circle in question. As a result, the least upper bound on the distance between the ant and the origin is the diameter of the circle, which is $\sqrt{10}$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n17. [9]",
"solution_match": "\nAnswer: "
}
|
6718f4cd-364e-51ac-91ca-7875a88e5c25
| 608,683
|
A 5-dimensional ant starts at one vertex of a 5 -dimensional hypercube of side length 1 . A move is when the ant travels from one vertex to another vertex at a distance of $\sqrt{2}$ away. How many ways can the ant make 5 moves and end up on the same vertex it started at?
|
6240 We let the cube lie in $\mathbb{R}^{5}$ with each corner with coordinates 1 or 0 . Assume the ant starts at $(0,0,0,0,0)$. Every move the ant adds or subtracts 1 to two of the places. Note that this means the ant can only land on a vertex with the sum of its coordinates an even number. Every move the ant has $\binom{5}{2}=10$ choices.
From any vertex there are 10 two-move sequences that put the ant at the same vertex it started at.
There are 6 two-move sequences to move from one vertex to a different, chosen vertex. If your chosen vertex differs from your current vertex by 2 of the 5 coordinates, your first move corrects for one of these two. There are 2 ways to choose which coordinate to correct for on the first move, and there are 3 ways to choose the second coordinate you change, yielding 6 sequences. If your chosen vertex differs from your current vertex by 4 of the 5 coordinates, each move corrects for two of these four. This yields $\binom{4}{2}=6$ sequences.
Finally, there are 60 three-move sequences that put the ant at the same vertex it started at. There are 10 ways to choose the first move, and there are 6 ways to make two moves to return to your original position.
The motion of the ant can be split into two cases.
Case 1: After the 3rd move the ant is on the vertex it started at. There are $(60)(10)=600$ different possible paths.
Case 2: After the third move the ant is on a vertex different from the one it started on. There are $\left(10^{3}-60\right)(6)=(940)(6)=5640$ different possible paths.
So there are 6240 total possible paths.
|
6240
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A 5-dimensional ant starts at one vertex of a 5 -dimensional hypercube of side length 1 . A move is when the ant travels from one vertex to another vertex at a distance of $\sqrt{2}$ away. How many ways can the ant make 5 moves and end up on the same vertex it started at?
|
6240 We let the cube lie in $\mathbb{R}^{5}$ with each corner with coordinates 1 or 0 . Assume the ant starts at $(0,0,0,0,0)$. Every move the ant adds or subtracts 1 to two of the places. Note that this means the ant can only land on a vertex with the sum of its coordinates an even number. Every move the ant has $\binom{5}{2}=10$ choices.
From any vertex there are 10 two-move sequences that put the ant at the same vertex it started at.
There are 6 two-move sequences to move from one vertex to a different, chosen vertex. If your chosen vertex differs from your current vertex by 2 of the 5 coordinates, your first move corrects for one of these two. There are 2 ways to choose which coordinate to correct for on the first move, and there are 3 ways to choose the second coordinate you change, yielding 6 sequences. If your chosen vertex differs from your current vertex by 4 of the 5 coordinates, each move corrects for two of these four. This yields $\binom{4}{2}=6$ sequences.
Finally, there are 60 three-move sequences that put the ant at the same vertex it started at. There are 10 ways to choose the first move, and there are 6 ways to make two moves to return to your original position.
The motion of the ant can be split into two cases.
Case 1: After the 3rd move the ant is on the vertex it started at. There are $(60)(10)=600$ different possible paths.
Case 2: After the third move the ant is on a vertex different from the one it started on. There are $\left(10^{3}-60\right)(6)=(940)(6)=5640$ different possible paths.
So there are 6240 total possible paths.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n19. [10]",
"solution_match": "\nAnswer: "
}
|
e46adcf6-ce2f-5a69-94f7-f94b344bbfed
| 608,685
|
Find the volume of the set of points $(x, y, z)$ satisfying
$$
\begin{aligned}
x, y, z & \geq 0 \\
x+y & \leq 1 \\
y+z & \leq 1 \\
z+x & \leq 1
\end{aligned}
$$
|
$\frac{1}{4}$ Without loss of generality, assume that $x \geq y$ - half the volume of the solid is on this side of the plane $x=y$. For each value of $c$ from 0 to $\frac{1}{2}$, the region of the intersection of this half of the solid with the plane $y=c$ is a trapezoid. The trapezoid has height $1-2 c$ and average base $\frac{1}{2}$, so it has an area of $\frac{1}{2}-c$.
The total volume of this region is $\frac{1}{2}$ times the average area of the trapezoids, which is $\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}$. Double that to get the total volume, which is $\frac{1}{4}$.
|
\frac{1}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Find the volume of the set of points $(x, y, z)$ satisfying
$$
\begin{aligned}
x, y, z & \geq 0 \\
x+y & \leq 1 \\
y+z & \leq 1 \\
z+x & \leq 1
\end{aligned}
$$
|
$\frac{1}{4}$ Without loss of generality, assume that $x \geq y$ - half the volume of the solid is on this side of the plane $x=y$. For each value of $c$ from 0 to $\frac{1}{2}$, the region of the intersection of this half of the solid with the plane $y=c$ is a trapezoid. The trapezoid has height $1-2 c$ and average base $\frac{1}{2}$, so it has an area of $\frac{1}{2}-c$.
The total volume of this region is $\frac{1}{2}$ times the average area of the trapezoids, which is $\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}$. Double that to get the total volume, which is $\frac{1}{4}$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n20. [10]",
"solution_match": "\nAnswer: "
}
|
cd99ace7-06c9-56fa-8910-aba3021b0fc5
| 608,686
|
You are the general of an army. You and the opposing general both have an equal number of troops to distribute among three battlefields. Whoever has more troops on a battlefield always wins (you win ties). An order is an ordered triple of non-negative real numbers $(x, y, z)$ such that $x+y+z=1$, and corresponds to sending a fraction $x$ of the troops to the first field, $y$ to the second, and $z$ to the third. Suppose that you give the order $\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{2}\right)$ and that the other general issues an order chosen uniformly at random from all possible orders. What is the probability that you win two out of the three battles?
|
| $\frac{5}{8}$ |
| :---: |
| Let $x$ | be the portion of soldiers the opposing general sends to the first battlefield, and $y$ the portion he sends to the second. Then $1-x-y$ is the portion he sends to the third. Then $x \geq 0$, $y \geq 0$, and $x+y \leq 1$. Furthermore, you win if one of the three conditions is satisfied: $x \leq \frac{1}{4}$ and $y \leq \frac{1}{4}, x \leq \frac{1}{4}$ and $1-x-y \leq \frac{1}{2}$, or $y \leq \frac{1}{4}$ and $1-x-y \leq \frac{1}{2}$. This is illustrated in the picture below.
Guts Round
This triangle is a linear projection of the region of feasible orders, so it preserves area and probability ratios. The probability that you win, then is given by the portion of the triangle that satisfies one of the three above constraints - in other words, the area of the shaded region divided by the area of the entire triangle. We can easily calculate this to be $\frac{\frac{5}{16}}{\frac{1}{2}}=\frac{5}{8}$.
|
\frac{5}{8}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
You are the general of an army. You and the opposing general both have an equal number of troops to distribute among three battlefields. Whoever has more troops on a battlefield always wins (you win ties). An order is an ordered triple of non-negative real numbers $(x, y, z)$ such that $x+y+z=1$, and corresponds to sending a fraction $x$ of the troops to the first field, $y$ to the second, and $z$ to the third. Suppose that you give the order $\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{2}\right)$ and that the other general issues an order chosen uniformly at random from all possible orders. What is the probability that you win two out of the three battles?
|
| $\frac{5}{8}$ |
| :---: |
| Let $x$ | be the portion of soldiers the opposing general sends to the first battlefield, and $y$ the portion he sends to the second. Then $1-x-y$ is the portion he sends to the third. Then $x \geq 0$, $y \geq 0$, and $x+y \leq 1$. Furthermore, you win if one of the three conditions is satisfied: $x \leq \frac{1}{4}$ and $y \leq \frac{1}{4}, x \leq \frac{1}{4}$ and $1-x-y \leq \frac{1}{2}$, or $y \leq \frac{1}{4}$ and $1-x-y \leq \frac{1}{2}$. This is illustrated in the picture below.
Guts Round
This triangle is a linear projection of the region of feasible orders, so it preserves area and probability ratios. The probability that you win, then is given by the portion of the triangle that satisfies one of the three above constraints - in other words, the area of the shaded region divided by the area of the entire triangle. We can easily calculate this to be $\frac{\frac{5}{16}}{\frac{1}{2}}=\frac{5}{8}$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n22. [12]",
"solution_match": "\nAnswer: "
}
|
d7224d0b-3569-5a31-b052-405b8c711533
| 608,688
|
In the country of Francisca, there are 2010 cities, some of which are connected by roads. Between any two cities, there is a unique path which runs along the roads and which does not pass through any city twice. What is the maximum possible number of cities in Francisca which have at least 3 roads running out of them?
|
1004 The restrictions on how roads connect cities directly imply that the graph of the cities of Francisca with the roads as edges is a tree. Therefore the sum of the degrees of all the vertices is $2009 \cdot 2=4018$. Suppose that $b$ vertices have degree $\geq 3$. The other $2010-b$ vertices must have a degree of at least 1 , so $3 b+(2010-b) \leq 4018$ or $2 b \leq 2008$. So $b$ is at most 1004 . We can achieve $b=1004$ with the following graph:
|
1004
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In the country of Francisca, there are 2010 cities, some of which are connected by roads. Between any two cities, there is a unique path which runs along the roads and which does not pass through any city twice. What is the maximum possible number of cities in Francisca which have at least 3 roads running out of them?
|
1004 The restrictions on how roads connect cities directly imply that the graph of the cities of Francisca with the roads as edges is a tree. Therefore the sum of the degrees of all the vertices is $2009 \cdot 2=4018$. Suppose that $b$ vertices have degree $\geq 3$. The other $2010-b$ vertices must have a degree of at least 1 , so $3 b+(2010-b) \leq 4018$ or $2 b \leq 2008$. So $b$ is at most 1004 . We can achieve $b=1004$ with the following graph:
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n23. [12]",
"solution_match": "\nAnswer: "
}
|
18c7b78f-9179-5655-8028-69b7ce4a32c4
| 608,689
|
Define a sequence of polynomials as follows: let $a_{1}=3 x^{2}-x$, let $a_{2}=3 x^{2}-7 x+3$, and for $n \geq 1$, let $a_{n+2}=\frac{5}{2} a_{n+1}-a_{n}$. As $n$ tends to infinity, what is the limit of the sum of the roots of $a_{n}$ ?
|
$\frac{13}{3}$ By using standard methods for solving linear recurrences ${ }^{2}$, we see that this recurrence has a characteristic polynomial of $x^{2}-\frac{5}{2} x+1=\left(x-\frac{1}{2}\right)(x-2)$, hence $a_{n}(x)=c(x) \cdot 2^{n}+d(x) \cdot 2^{-n}$ for some polynomials $c$ and $d$. Plugging in $n=1$ and $n=2$ gives
$$
2 c(x)+\frac{1}{2} d(x)=3 x^{2}-x
$$
and
$$
4 c(x)+\frac{1}{4} d(x)=3 x^{2}-7 x+3
$$
Subtracting the first equation from two times the second equation gives $6 c(x)=3 x^{2}-13 x+6$, so $c(x)=\frac{3 x^{2}-13 x+6}{6}$. As $n$ grows large, the $c(x) 2^{n}$ term dominates compared to the $d(x) 2^{-n}$ term, so the roots of $a_{n}(x)$ converge to the roots of $c(x)$. Thus the roots of $a_{n}(x)$ converge to the roots of $3 x^{2}-13 x+6$, which by Vieta's formula ${ }^{3}$ have a sum of $\frac{13}{3}$.
|
\frac{13}{3}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Define a sequence of polynomials as follows: let $a_{1}=3 x^{2}-x$, let $a_{2}=3 x^{2}-7 x+3$, and for $n \geq 1$, let $a_{n+2}=\frac{5}{2} a_{n+1}-a_{n}$. As $n$ tends to infinity, what is the limit of the sum of the roots of $a_{n}$ ?
|
$\frac{13}{3}$ By using standard methods for solving linear recurrences ${ }^{2}$, we see that this recurrence has a characteristic polynomial of $x^{2}-\frac{5}{2} x+1=\left(x-\frac{1}{2}\right)(x-2)$, hence $a_{n}(x)=c(x) \cdot 2^{n}+d(x) \cdot 2^{-n}$ for some polynomials $c$ and $d$. Plugging in $n=1$ and $n=2$ gives
$$
2 c(x)+\frac{1}{2} d(x)=3 x^{2}-x
$$
and
$$
4 c(x)+\frac{1}{4} d(x)=3 x^{2}-7 x+3
$$
Subtracting the first equation from two times the second equation gives $6 c(x)=3 x^{2}-13 x+6$, so $c(x)=\frac{3 x^{2}-13 x+6}{6}$. As $n$ grows large, the $c(x) 2^{n}$ term dominates compared to the $d(x) 2^{-n}$ term, so the roots of $a_{n}(x)$ converge to the roots of $c(x)$. Thus the roots of $a_{n}(x)$ converge to the roots of $3 x^{2}-13 x+6$, which by Vieta's formula ${ }^{3}$ have a sum of $\frac{13}{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n24. [12]",
"solution_match": "\nAnswer: "
}
|
319889b3-c1f1-5890-95b6-7d897bcc38fa
| 608,690
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ have the property that $f(\{1,2,3\})$ and $f(f(\{1,2,3\}))$ are disjoint?
|
94 Let $f(\{1,2,3\})$ be $A$. Then $A \cap f(A)=\emptyset$, so $A$ must be a subset of $\{4,5\}$. If $B=\{4,5\}$, there are $2^{3}-2$ ways to assign each element in $\{1,2,3\}$ to a value in $\{4,5\}$, and 9 ways to assign each element of $\{4,5\}$ to a value in $\{1,2,3\}$, for a total of 54 choices of $f$. If $A=\{4\}$, there is 1 possible value for each element of $\{1,2,3\}, 4$ ways to assign $\{4\}$ with a value from $\{1,2,3,5\}$, and 5 ways to assign a value to $\{5\}$. Similarly, if $A=\{5\}$, there are $4 \cdot 5=20$ choices for $f$. In total, there are $54+20 \cdot 2=94$ possible functions.
[^1]26. [15] Express the following in closed form, as a function of $x$ :
$\sin ^{2}(x)+\sin ^{2}(2 x) \cos ^{2}(x)+\sin ^{2}(4 x) \cos ^{2}(2 x) \cos ^{2}(x)+\cdots+\sin ^{2}\left(2^{2010} x\right) \cos ^{2}\left(2^{2009} x\right) \cdots \cos ^{2}(2 x) \cos ^{2}(x)$.
|
94
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many functions $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ have the property that $f(\{1,2,3\})$ and $f(f(\{1,2,3\}))$ are disjoint?
|
94 Let $f(\{1,2,3\})$ be $A$. Then $A \cap f(A)=\emptyset$, so $A$ must be a subset of $\{4,5\}$. If $B=\{4,5\}$, there are $2^{3}-2$ ways to assign each element in $\{1,2,3\}$ to a value in $\{4,5\}$, and 9 ways to assign each element of $\{4,5\}$ to a value in $\{1,2,3\}$, for a total of 54 choices of $f$. If $A=\{4\}$, there is 1 possible value for each element of $\{1,2,3\}, 4$ ways to assign $\{4\}$ with a value from $\{1,2,3,5\}$, and 5 ways to assign a value to $\{5\}$. Similarly, if $A=\{5\}$, there are $4 \cdot 5=20$ choices for $f$. In total, there are $54+20 \cdot 2=94$ possible functions.
[^1]26. [15] Express the following in closed form, as a function of $x$ :
$\sin ^{2}(x)+\sin ^{2}(2 x) \cos ^{2}(x)+\sin ^{2}(4 x) \cos ^{2}(2 x) \cos ^{2}(x)+\cdots+\sin ^{2}\left(2^{2010} x\right) \cos ^{2}\left(2^{2009} x\right) \cdots \cos ^{2}(2 x) \cos ^{2}(x)$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n25. [15]",
"solution_match": "\nAnswer: "
}
|
9470ab36-7bfa-50ec-98bb-6fd2f04424eb
| 608,691
|
Suppose that there are real numbers $a, b, c \geq 1$ and that there are positive reals $x, y, z$ such that
$$
\begin{aligned}
a^{x}+b^{y}+c^{z} & =4 \\
x a^{x}+y b^{y}+z c^{z} & =6 \\
x^{2} a^{x}+y^{2} b^{y}+z^{2} c^{z} & =9
\end{aligned}
$$
What is the maximum possible value of $c$ ?
|
$\sqrt[3]{4}$ The Cauchy-Schwarz inequality states that given 2 sequences of $n$ real numbers $x_{1}, x_{2}, \ldots, x_{n}$ and $y_{1}, y_{2}, \ldots, y_{n}$, then $\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\right)\left(y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}\right) \geq\left(x_{1} y_{1}+x_{2} y_{2}+\ldots+x_{n} y_{n}\right)^{2}$ with equality holding if and only if $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}=\ldots=\frac{x_{n}}{y_{n}}$. Applying this to $\left\{a^{x / 2}, b^{y / 2}, c^{z / 2}\right\}$ and $\left\{x a^{x / 2}, y b^{y / 2}, z c^{z / 2}\right\}$ yields $\left(a^{x}+b^{y}+c^{z}\right)\left(x^{2} a^{x}+y^{2} b^{y}+z^{2} c^{z}\right) \geq\left(x a^{x}+y b^{y}+z b^{z}\right)^{2}$ with equality holding if and only if $x=y=z$.
However, equality does hold (both sides evaluate to 36 ), so $x=y=z$. The second equation then becomes $x\left(a^{x}+b^{x}+c^{x}\right)=6$, which implies $x=\frac{3}{2}$. Then we have $a^{3 / 2}+b^{3 / 2}+c^{3 / 2}=4$. To maximize $c$, we minimize $a$ and $b$ by setting $a=b=1$. Then $c^{3 / 2}=2$ or $c=\sqrt[3]{4}$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
\sqrt[3]{4}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose that there are real numbers $a, b, c \geq 1$ and that there are positive reals $x, y, z$ such that
$$
\begin{aligned}
a^{x}+b^{y}+c^{z} & =4 \\
x a^{x}+y b^{y}+z c^{z} & =6 \\
x^{2} a^{x}+y^{2} b^{y}+z^{2} c^{z} & =9
\end{aligned}
$$
What is the maximum possible value of $c$ ?
|
$\sqrt[3]{4}$ The Cauchy-Schwarz inequality states that given 2 sequences of $n$ real numbers $x_{1}, x_{2}, \ldots, x_{n}$ and $y_{1}, y_{2}, \ldots, y_{n}$, then $\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\right)\left(y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}\right) \geq\left(x_{1} y_{1}+x_{2} y_{2}+\ldots+x_{n} y_{n}\right)^{2}$ with equality holding if and only if $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}=\ldots=\frac{x_{n}}{y_{n}}$. Applying this to $\left\{a^{x / 2}, b^{y / 2}, c^{z / 2}\right\}$ and $\left\{x a^{x / 2}, y b^{y / 2}, z c^{z / 2}\right\}$ yields $\left(a^{x}+b^{y}+c^{z}\right)\left(x^{2} a^{x}+y^{2} b^{y}+z^{2} c^{z}\right) \geq\left(x a^{x}+y b^{y}+z b^{z}\right)^{2}$ with equality holding if and only if $x=y=z$.
However, equality does hold (both sides evaluate to 36 ), so $x=y=z$. The second equation then becomes $x\left(a^{x}+b^{x}+c^{x}\right)=6$, which implies $x=\frac{3}{2}$. Then we have $a^{3 / 2}+b^{3 / 2}+c^{3 / 2}=4$. To maximize $c$, we minimize $a$ and $b$ by setting $a=b=1$. Then $c^{3 / 2}=2$ or $c=\sqrt[3]{4}$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n27. [15]",
"solution_match": "\nAnswer: "
}
|
b39c5b70-6181-5e3e-bc2f-6ccd277dd202
| 608,692
|
Danielle Bellatrix Robinson is organizing a poker tournament with 9 people. The tournament will have 4 rounds, and in each round the 9 players are split into 3 groups of 3 . During the tournament, each player plays every other player exactly once. How many different ways can Danielle divide the 9 people into three groups in each round to satisfy these requirements?
|
20160 We first split the 9 people up arbitrarily into groups of 3 . There are $\frac{\binom{9}{3}\binom{6}{3}\binom{3}{3}}{3!}=280$ ways of doing this. Without loss of generality, label the people 1 through 9 so that the first round groups are $\{1,2,3\},\{4,5,6\}$, and $\{7,8,9\}$. We will use this numbering to count the number of ways DBR can divide the 9 players in rounds 2,3 , and 4 .
In round 2 , because players 1,2 , and 3 are together in the first round, they must be in separate groups, and likewise for $\{4,5,6\}$ and $\{7,8,9\}$. Disregarding ordering of the three groups in a single round, we will first place 1,2 , and 3 into their groups, then count the number of ways to place $\{4,5,6\}$ and $\{7,8,9\}$ in the groups with them. We do this by placing one member from each of $\{4,5,6\}$ and $\{7,8,9\}$ into each group. There are $(3!)^{2}$ ways to do this. Now, because of symmetry, we can use the round 2 grouping $\{1,4,7\},\{2,5,8\},\{3,6,9\}$ to list out the remaining possibilities for round 3 and 4 groupings.
Casework shows that there are 2 ways to group the players in the remaining two rounds. We multiply $280 \cdot(3!)^{2} \cdot 2$ to get 20160.
|
20160
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Danielle Bellatrix Robinson is organizing a poker tournament with 9 people. The tournament will have 4 rounds, and in each round the 9 players are split into 3 groups of 3 . During the tournament, each player plays every other player exactly once. How many different ways can Danielle divide the 9 people into three groups in each round to satisfy these requirements?
|
20160 We first split the 9 people up arbitrarily into groups of 3 . There are $\frac{\binom{9}{3}\binom{6}{3}\binom{3}{3}}{3!}=280$ ways of doing this. Without loss of generality, label the people 1 through 9 so that the first round groups are $\{1,2,3\},\{4,5,6\}$, and $\{7,8,9\}$. We will use this numbering to count the number of ways DBR can divide the 9 players in rounds 2,3 , and 4 .
In round 2 , because players 1,2 , and 3 are together in the first round, they must be in separate groups, and likewise for $\{4,5,6\}$ and $\{7,8,9\}$. Disregarding ordering of the three groups in a single round, we will first place 1,2 , and 3 into their groups, then count the number of ways to place $\{4,5,6\}$ and $\{7,8,9\}$ in the groups with them. We do this by placing one member from each of $\{4,5,6\}$ and $\{7,8,9\}$ into each group. There are $(3!)^{2}$ ways to do this. Now, because of symmetry, we can use the round 2 grouping $\{1,4,7\},\{2,5,8\},\{3,6,9\}$ to list out the remaining possibilities for round 3 and 4 groupings.
Casework shows that there are 2 ways to group the players in the remaining two rounds. We multiply $280 \cdot(3!)^{2} \cdot 2$ to get 20160.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n28. [18]",
"solution_match": "\nAnswer: "
}
|
3ebd378f-6426-5178-80cf-2b9acc557fae
| 608,693
|
Compute the remainder when
$$
\sum_{k=1}^{30303} k^{k}
$$
is divided by 101 .
|
29 The main idea is the following lemma:
Lemma. For any non-negative integer $n$ and prime $p, \sum_{k=n+1}^{n+p^{2}-p} k^{k} \equiv 1(\bmod p)$.
Proof. Note that $a^{b}$ depends only on the value of $a(\bmod p)$ and the value of $b(\bmod p-1)$. Since $p$ and $p-1$ are relatively prime, the Chinese Remainder Theorem ${ }^{4}$ implies that any $p^{2}-p$ consecutive integers will take on each possible pair of a residue $\bmod p$ and a residue $\bmod p-1$. In other words, if we let $(a, b)=(k(\bmod p), k(\bmod p-1))$, then as $k$ ranges through $p^{2}-p$ consecutive integers, $(a, b)$ will range through all $p^{2}-p$ ordered pairs of residues $\bmod p$ and residues mod $p-1$. This implies that
$$
\sum_{k=n+1}^{n+p^{2}-p} k^{k} \equiv \sum_{b=1}^{p-1} \sum_{a=1}^{p} a^{b} .
$$
It is well-known that $\sum_{a=1}^{p} a^{b}=\left\{\begin{array}{rl}-1 & p-1 \mid b \\ 0 & p-1 \nmid b\end{array}\right.$. We will sketch a proof here. When $p-1 \mid b$, the result follows from Fermat's Little Theorem ${ }^{5}$. When $p-1 \nmid b$, it suffices to consider the case when $b \mid p-1$, since the $b$ th powers $\bmod p$ are the same as the $\operatorname{gcd}(b, p-1)$ th powers $\bmod p$, and there are an equal number of every non-zero $b$ th power. But in this case, the $b$ th powers are just the solutions to $x^{\frac{p-1}{b}}-1$, which add up to zero by Vieta's formulas.
Now, using the formula for $\sum a^{b}$, we get that
$$
\sum_{b=1}^{p-1} \sum_{a=1}^{p} a^{b} \equiv-1 \quad(\bmod p)
$$
which completes the lemma.
We now apply the lemma with $p=101$ and $n=3,10103$, and 20103 to get that $\sum_{k=1}^{30303} k^{k} \equiv$ $\left(\sum_{k=1}^{3} k^{k}\right)-3$. But $\sum_{k=1}^{3} k^{k}=1^{1}+2^{2}+3^{3}=1+4+27=32$, so the answer is $32-3=29$.
|
29
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Compute the remainder when
$$
\sum_{k=1}^{30303} k^{k}
$$
is divided by 101 .
|
29 The main idea is the following lemma:
Lemma. For any non-negative integer $n$ and prime $p, \sum_{k=n+1}^{n+p^{2}-p} k^{k} \equiv 1(\bmod p)$.
Proof. Note that $a^{b}$ depends only on the value of $a(\bmod p)$ and the value of $b(\bmod p-1)$. Since $p$ and $p-1$ are relatively prime, the Chinese Remainder Theorem ${ }^{4}$ implies that any $p^{2}-p$ consecutive integers will take on each possible pair of a residue $\bmod p$ and a residue $\bmod p-1$. In other words, if we let $(a, b)=(k(\bmod p), k(\bmod p-1))$, then as $k$ ranges through $p^{2}-p$ consecutive integers, $(a, b)$ will range through all $p^{2}-p$ ordered pairs of residues $\bmod p$ and residues mod $p-1$. This implies that
$$
\sum_{k=n+1}^{n+p^{2}-p} k^{k} \equiv \sum_{b=1}^{p-1} \sum_{a=1}^{p} a^{b} .
$$
It is well-known that $\sum_{a=1}^{p} a^{b}=\left\{\begin{array}{rl}-1 & p-1 \mid b \\ 0 & p-1 \nmid b\end{array}\right.$. We will sketch a proof here. When $p-1 \mid b$, the result follows from Fermat's Little Theorem ${ }^{5}$. When $p-1 \nmid b$, it suffices to consider the case when $b \mid p-1$, since the $b$ th powers $\bmod p$ are the same as the $\operatorname{gcd}(b, p-1)$ th powers $\bmod p$, and there are an equal number of every non-zero $b$ th power. But in this case, the $b$ th powers are just the solutions to $x^{\frac{p-1}{b}}-1$, which add up to zero by Vieta's formulas.
Now, using the formula for $\sum a^{b}$, we get that
$$
\sum_{b=1}^{p-1} \sum_{a=1}^{p} a^{b} \equiv-1 \quad(\bmod p)
$$
which completes the lemma.
We now apply the lemma with $p=101$ and $n=3,10103$, and 20103 to get that $\sum_{k=1}^{30303} k^{k} \equiv$ $\left(\sum_{k=1}^{3} k^{k}\right)-3$. But $\sum_{k=1}^{3} k^{k}=1^{1}+2^{2}+3^{3}=1+4+27=32$, so the answer is $32-3=29$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n29. [18]",
"solution_match": "\nAnswer: "
}
|
f9e0562f-3fec-5cf1-9cab-e05ac441bdf6
| 608,694
|
A monomial term $x_{i_{1}} x_{i_{2}} \ldots x_{i_{k}}$ in the variables $x_{1}, x_{2}, \ldots x_{8}$ is square-free if $i_{1}, i_{2}, \ldots i_{k}$ are distinct. (A constant term such as 1 is considered square-free.) What is the sum of the coefficients of the squarefree terms in the following product?
$$
\prod_{1 \leq i<j \leq 8}\left(1+x_{i} x_{j}\right)
$$
|
764 Let $a_{n}$ be the sum of the coefficients of the square-terms in the product $\prod_{1 \leq i<j \leq n}(1+$ $x_{i} x_{j}$ ). Square-free terms in this product come in two types: either they include $x_{n}$, or they do not. The sum of the coefficients of the terms that include $x_{n}$ is $(n-1) a_{n-2}$, since we can choose any of the $n-1$ other variables to be paired with $x_{n}$, and then choose any square-free term from the product taken over the other $n-2$ variables. The sum of the coefficients of the terms that do not include $x_{n}$ are $a_{n-1}$, because they all come from the product over the other $n-1$ variables. Therefore, $a_{n}=a_{n-1}+(n-1) a_{n-2}$.
We use this recursion to find $a_{8}$. As base cases, $a_{0}$ and $a_{1}$ are both 1 . Then $a_{2}=2, a_{3}=4, a_{4}=10$, $a_{5}=26, a_{6}=76, a_{7}=232$, and finally, $a_{8}=764$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
764
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A monomial term $x_{i_{1}} x_{i_{2}} \ldots x_{i_{k}}$ in the variables $x_{1}, x_{2}, \ldots x_{8}$ is square-free if $i_{1}, i_{2}, \ldots i_{k}$ are distinct. (A constant term such as 1 is considered square-free.) What is the sum of the coefficients of the squarefree terms in the following product?
$$
\prod_{1 \leq i<j \leq 8}\left(1+x_{i} x_{j}\right)
$$
|
764 Let $a_{n}$ be the sum of the coefficients of the square-terms in the product $\prod_{1 \leq i<j \leq n}(1+$ $x_{i} x_{j}$ ). Square-free terms in this product come in two types: either they include $x_{n}$, or they do not. The sum of the coefficients of the terms that include $x_{n}$ is $(n-1) a_{n-2}$, since we can choose any of the $n-1$ other variables to be paired with $x_{n}$, and then choose any square-free term from the product taken over the other $n-2$ variables. The sum of the coefficients of the terms that do not include $x_{n}$ are $a_{n-1}$, because they all come from the product over the other $n-1$ variables. Therefore, $a_{n}=a_{n-1}+(n-1) a_{n-2}$.
We use this recursion to find $a_{8}$. As base cases, $a_{0}$ and $a_{1}$ are both 1 . Then $a_{2}=2, a_{3}=4, a_{4}=10$, $a_{5}=26, a_{6}=76, a_{7}=232$, and finally, $a_{8}=764$.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n30. [18]",
"solution_match": "\nAnswer: "
}
|
2669b615-13d8-5827-903f-44c166afd22f
| 608,695
|
In the Democratic Republic of Irun, 5 people are voting in an election among 5 candidates. If each person votes for a single candidate at random, what is the expected number of candidates that will be voted for?
|
| $\frac{2101}{625}$ |
| :---: |
| The probability that a chosen candidate will receive no votes at all is $\left(\frac{4}{5}\right)^{5}$, or the | probability that every person will vote for someone other than that one candidate. Then the probability that a chosen candidate will receive at least one vote is $1-\left(\frac{4}{5}\right)^{5}=\frac{2101}{3125}$. By linearity of expectations, the final answer is this probability times the number of candidates, or $5 \cdot \frac{2101}{3125}=\frac{2101}{625}$.
|
\frac{2101}{625}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In the Democratic Republic of Irun, 5 people are voting in an election among 5 candidates. If each person votes for a single candidate at random, what is the expected number of candidates that will be voted for?
|
| $\frac{2101}{625}$ |
| :---: |
| The probability that a chosen candidate will receive no votes at all is $\left(\frac{4}{5}\right)^{5}$, or the | probability that every person will vote for someone other than that one candidate. Then the probability that a chosen candidate will receive at least one vote is $1-\left(\frac{4}{5}\right)^{5}=\frac{2101}{3125}$. By linearity of expectations, the final answer is this probability times the number of candidates, or $5 \cdot \frac{2101}{3125}=\frac{2101}{625}$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n31. [21]",
"solution_match": "\nAnswer: "
}
|
2b3046e7-dec4-5e92-b61b-6a96875f75c2
| 608,696
|
There are 101 people participating in a Secret Santa gift exchange. As usual each person is randomly assigned another person for whom (s)he has to get a gift, such that each person gives and receives exactly one gift and no one gives a gift to themself. What is the probability that the first person neither gives gifts to or receives gifts from the second or third person? Express your answer as a decimal rounded to five decimal places.
|
0.96039 Let $D_{k}$ denote the number of derangements of $\{1,2, \ldots, k\}$. (A derangement is a permutation in which no element appears in its original position.)
Call the first three people $A, B$, and $C$. Let $X \rightarrow Y$ denote that $X$ gives a gift to $Y$ and let $X \nrightarrow Y$ denote that $X$ gives a gift to anyone other than $Y$. We are fine unless we have $A \rightarrow B, B \rightarrow A$, $A \rightarrow C$, or $C \rightarrow A$. We will compute the number of ways for various things to occur, then combine it into what we want.
There are $D_{n-1}$ ways to have $A \rightarrow B \nrightarrow A$. This is because if $A \rightarrow B$, we can treat $A$ and $B$ as a single vertex, and since $B \nrightarrow A$, we have a derangement. Similarly, there are $D_{n-1}$ ways to have $B \rightarrow A \nrightarrow B$. Thirdly, there are $D_{n-2}$ ways to have $A \rightarrow B \rightarrow A$, since $D_{n-2}$ says how many ways the other $n-2$ people can exchange their gifts. So there are $2 D_{n-1}+D_{n-2}$ ways to have a conflict between $A$ and $B$.
Similarly, there are $2 D_{n-1}+D_{n-2}$ ways to have a conflict between $A$ and $C$.
Using similar arguments, we can show that there are $D_{n-2}$ ways for $B \rightarrow A \rightarrow C \nrightarrow B$ to occur and $D_{n-3}$ ways for $B \rightarrow A \rightarrow C \rightarrow B$ to occur. We get the same results when $B$ and $C$ are reversed. This gives $2 D_{n-2}+2 D_{n-3}$ ways for a conflict to occur within all three people.
By the Principle of Inclusion-Exclusion, the total number of ways to have a conflict is
(\# conflicts between $A$ and $B)+(\#$ conflicts between $A$ and $C)-(\#$ conflicts between $A, B$, and $C)$,
which evaluates to $4 D_{n-1}-2 D_{n-3}$.
Approximating $D_{n}$ as $\frac{n!}{e}$ (the actual formula is this quantity rounded to the nearest integer, so this is a great approximation), we find that the probability of no conflicts is
$1-\frac{4 D_{n-1}-2 D_{n-3}}{D_{n}} \approx 1-4\left(\frac{(n-1)!/ e}{n!/ e}\right)-2\left(\frac{(n-3)!/ e}{n!/ e}\right)=\frac{n(n-1)(n-2)-4(n-1)(n-2)-2}{n(n-1)(n-2)}$.
Substitute $m=n-1$ (this makes $m=100$, so the expression is easier to evaluate) to get a probability of
$$
\begin{gathered}
\frac{m^{3}-m-4 m^{2}+4 m-2}{m^{3}-m}=\frac{m^{3}-4 m^{2}+3 m-2}{m^{3}-m}=\frac{1,000,000-40,000+300-2}{100 \cdot 9999}=\frac{960298}{100 \cdot 9999} \\
=0.960208 \cdot(1.000100010001 \ldots)=0.960208+0.0000960208+\ldots=0.9603940 \ldots
\end{gathered}
$$
To five decimal places, the desired probability is 0.96039 .
|
0.96039
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
There are 101 people participating in a Secret Santa gift exchange. As usual each person is randomly assigned another person for whom (s)he has to get a gift, such that each person gives and receives exactly one gift and no one gives a gift to themself. What is the probability that the first person neither gives gifts to or receives gifts from the second or third person? Express your answer as a decimal rounded to five decimal places.
|
0.96039 Let $D_{k}$ denote the number of derangements of $\{1,2, \ldots, k\}$. (A derangement is a permutation in which no element appears in its original position.)
Call the first three people $A, B$, and $C$. Let $X \rightarrow Y$ denote that $X$ gives a gift to $Y$ and let $X \nrightarrow Y$ denote that $X$ gives a gift to anyone other than $Y$. We are fine unless we have $A \rightarrow B, B \rightarrow A$, $A \rightarrow C$, or $C \rightarrow A$. We will compute the number of ways for various things to occur, then combine it into what we want.
There are $D_{n-1}$ ways to have $A \rightarrow B \nrightarrow A$. This is because if $A \rightarrow B$, we can treat $A$ and $B$ as a single vertex, and since $B \nrightarrow A$, we have a derangement. Similarly, there are $D_{n-1}$ ways to have $B \rightarrow A \nrightarrow B$. Thirdly, there are $D_{n-2}$ ways to have $A \rightarrow B \rightarrow A$, since $D_{n-2}$ says how many ways the other $n-2$ people can exchange their gifts. So there are $2 D_{n-1}+D_{n-2}$ ways to have a conflict between $A$ and $B$.
Similarly, there are $2 D_{n-1}+D_{n-2}$ ways to have a conflict between $A$ and $C$.
Using similar arguments, we can show that there are $D_{n-2}$ ways for $B \rightarrow A \rightarrow C \nrightarrow B$ to occur and $D_{n-3}$ ways for $B \rightarrow A \rightarrow C \rightarrow B$ to occur. We get the same results when $B$ and $C$ are reversed. This gives $2 D_{n-2}+2 D_{n-3}$ ways for a conflict to occur within all three people.
By the Principle of Inclusion-Exclusion, the total number of ways to have a conflict is
(\# conflicts between $A$ and $B)+(\#$ conflicts between $A$ and $C)-(\#$ conflicts between $A, B$, and $C)$,
which evaluates to $4 D_{n-1}-2 D_{n-3}$.
Approximating $D_{n}$ as $\frac{n!}{e}$ (the actual formula is this quantity rounded to the nearest integer, so this is a great approximation), we find that the probability of no conflicts is
$1-\frac{4 D_{n-1}-2 D_{n-3}}{D_{n}} \approx 1-4\left(\frac{(n-1)!/ e}{n!/ e}\right)-2\left(\frac{(n-3)!/ e}{n!/ e}\right)=\frac{n(n-1)(n-2)-4(n-1)(n-2)-2}{n(n-1)(n-2)}$.
Substitute $m=n-1$ (this makes $m=100$, so the expression is easier to evaluate) to get a probability of
$$
\begin{gathered}
\frac{m^{3}-m-4 m^{2}+4 m-2}{m^{3}-m}=\frac{m^{3}-4 m^{2}+3 m-2}{m^{3}-m}=\frac{1,000,000-40,000+300-2}{100 \cdot 9999}=\frac{960298}{100 \cdot 9999} \\
=0.960208 \cdot(1.000100010001 \ldots)=0.960208+0.0000960208+\ldots=0.9603940 \ldots
\end{gathered}
$$
To five decimal places, the desired probability is 0.96039 .
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n32. [21]",
"solution_match": "\nAnswer: "
}
|
8f0404eb-42a4-5c0d-bfbd-220dd093da66
| 608,697
|
Let $a_{1}=3$, and for $n>1$, let $a_{n}$ be the largest real number such that
$$
4\left(a_{n-1}^{2}+a_{n}^{2}\right)=10 a_{n-1} a_{n}-9
$$
What is the largest positive integer less than $a_{8}$ ?
|
335 Let $t_{n}$ be the larger real such that $a_{n}=t_{n}+\frac{1}{t_{n}}$. Then $t_{1}=\frac{3+\sqrt{5}}{2}$. We claim that $t_{n}=2 t_{n-1}$. Writing the recurrence as a quadratic polynomial in $a_{n}$, we have:
$$
4 a_{n}^{2}-10 a_{n-1} a_{n}+4 a_{n-1}^{2}+9=0
$$
Using the quadratic formula, we see that $a_{n}=\frac{5}{4} a_{n-1}+\frac{3}{4} \sqrt{a_{n-1}^{2}-4}$. (We ignore the negative square root, since $a_{n}$ is the largest real number satisfying the polynomial.) Substituting $t_{n-1}+\frac{1}{t_{n-1}}$ for $a_{n-1}$, we see that $\sqrt{a_{n-1}^{2}-4}=\sqrt{t_{n-1}^{2}-2+\frac{1}{t_{n-1}^{2}}}$, so we have:
$$
a_{n}=\frac{5}{4}\left(t_{n-1}+\frac{1}{t_{n-1}}\right)+\frac{3}{4} \sqrt{\left(t_{n-1}-\frac{1}{t_{n-1}}\right)^{2}}=2 t_{n-1}+\frac{1}{2 t_{n-1}}
$$
so $t_{n}=2 t_{n-1}$, as claimed. Then $a_{8}=\frac{128(3+\sqrt{5})}{2}+\frac{2}{128(3+\sqrt{5})}$. The second term is vanishingly small, so $\left\lfloor a_{8}\right\rfloor=\lfloor 64(3+\sqrt{5})\rfloor$. We approximate $\sqrt{5}$ to two decimal places as 2.24 , making this expression $\lfloor 335.36\rfloor=335$. Since our value of $\sqrt{5}$ is correct to within 0.005 , the decimal is correct to within 0.32 , which means the final answer is exact.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
335
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a_{1}=3$, and for $n>1$, let $a_{n}$ be the largest real number such that
$$
4\left(a_{n-1}^{2}+a_{n}^{2}\right)=10 a_{n-1} a_{n}-9
$$
What is the largest positive integer less than $a_{8}$ ?
|
335 Let $t_{n}$ be the larger real such that $a_{n}=t_{n}+\frac{1}{t_{n}}$. Then $t_{1}=\frac{3+\sqrt{5}}{2}$. We claim that $t_{n}=2 t_{n-1}$. Writing the recurrence as a quadratic polynomial in $a_{n}$, we have:
$$
4 a_{n}^{2}-10 a_{n-1} a_{n}+4 a_{n-1}^{2}+9=0
$$
Using the quadratic formula, we see that $a_{n}=\frac{5}{4} a_{n-1}+\frac{3}{4} \sqrt{a_{n-1}^{2}-4}$. (We ignore the negative square root, since $a_{n}$ is the largest real number satisfying the polynomial.) Substituting $t_{n-1}+\frac{1}{t_{n-1}}$ for $a_{n-1}$, we see that $\sqrt{a_{n-1}^{2}-4}=\sqrt{t_{n-1}^{2}-2+\frac{1}{t_{n-1}^{2}}}$, so we have:
$$
a_{n}=\frac{5}{4}\left(t_{n-1}+\frac{1}{t_{n-1}}\right)+\frac{3}{4} \sqrt{\left(t_{n-1}-\frac{1}{t_{n-1}}\right)^{2}}=2 t_{n-1}+\frac{1}{2 t_{n-1}}
$$
so $t_{n}=2 t_{n-1}$, as claimed. Then $a_{8}=\frac{128(3+\sqrt{5})}{2}+\frac{2}{128(3+\sqrt{5})}$. The second term is vanishingly small, so $\left\lfloor a_{8}\right\rfloor=\lfloor 64(3+\sqrt{5})\rfloor$. We approximate $\sqrt{5}$ to two decimal places as 2.24 , making this expression $\lfloor 335.36\rfloor=335$. Since our value of $\sqrt{5}$ is correct to within 0.005 , the decimal is correct to within 0.32 , which means the final answer is exact.
$13^{\text {th }}$ ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 - GUTS ROUND
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n33. [21]",
"solution_match": "\nAnswer: "
}
|
a9d74bd9-08c9-5ed9-b956-21ee46062ec9
| 608,698
|
3000 people each go into one of three rooms randomly. What is the most likely value for the maximum number of people in any of the rooms? Your score for this problem will be 0 if you write down a number less than or equal to 1000 . Otherwise, it will be $25-27 \frac{|A-C|}{\min (A, C)-1000}$.
|
1019 To get a rough approximation, we can use the fact that a sum of identical random variables converges to a Gaussian distribution ${ }^{6}$ in this case with a mean of 1000 and a variance of $3000 \cdot \frac{2}{9}=667$. Since $\sqrt{667} \approx 26,1026$ is a good guess, as Gaussians tend to differ from their mean by approximately their variance.
The actual answer was computed with the following python program:
```
facts = [0]*3001
facts[0]=1
for a in range(1,3001):
facts[a]=a*facts[a-1]
def binom(n,k):
return facts[n]/(facts[k]*facts[n-k])
```
[^3]```
maxes = [0]*3001
M = 1075
for a in range(0,3001):
for b in range(0,3001-a):
c = 3000-a-b
m = max (a,max (b,c))
if m < M:
maxes[m] += facts[3000]/(facts[a]*facts[b]*facts[c])
print [a,b]
best = 1000
for a in range(1000,1050):
print maxes[a],a
if maxes[best] <= maxes[a]:
best = a
print maxes[best]
print best
```
We can use arguments involving the Chernoff bound ${ }^{7}$ to show that the answer is necessarily less than 1075. Alternately, if we wanted to be really careful, we could just set $M=3001$, but then we'd have to wait a while for the script to finish.
|
1019
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
3000 people each go into one of three rooms randomly. What is the most likely value for the maximum number of people in any of the rooms? Your score for this problem will be 0 if you write down a number less than or equal to 1000 . Otherwise, it will be $25-27 \frac{|A-C|}{\min (A, C)-1000}$.
|
1019 To get a rough approximation, we can use the fact that a sum of identical random variables converges to a Gaussian distribution ${ }^{6}$ in this case with a mean of 1000 and a variance of $3000 \cdot \frac{2}{9}=667$. Since $\sqrt{667} \approx 26,1026$ is a good guess, as Gaussians tend to differ from their mean by approximately their variance.
The actual answer was computed with the following python program:
```
facts = [0]*3001
facts[0]=1
for a in range(1,3001):
facts[a]=a*facts[a-1]
def binom(n,k):
return facts[n]/(facts[k]*facts[n-k])
```
[^3]```
maxes = [0]*3001
M = 1075
for a in range(0,3001):
for b in range(0,3001-a):
c = 3000-a-b
m = max (a,max (b,c))
if m < M:
maxes[m] += facts[3000]/(facts[a]*facts[b]*facts[c])
print [a,b]
best = 1000
for a in range(1000,1050):
print maxes[a],a
if maxes[best] <= maxes[a]:
best = a
print maxes[best]
print best
```
We can use arguments involving the Chernoff bound ${ }^{7}$ to show that the answer is necessarily less than 1075. Alternately, if we wanted to be really careful, we could just set $M=3001$, but then we'd have to wait a while for the script to finish.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n34. [25]",
"solution_match": "\nAnswer: "
}
|
483a6ab2-38bc-57ea-8a7b-0662b7735ce7
| 608,699
|
Call an positive integer almost-square if it can be written as $a \cdot b$, where $a$ and $b$ are integers and $a \leq b \leq \frac{4}{3} a$. How many almost-square positive integers are less than or equal to 1000000 ? Your score will be equal to $25-65 \frac{|A-C|}{\min (A, C)}$.
|
130348 To get a good estimate for the number of almost-square integers, note that any number of the form $a \cdot b$, with $b \leq \frac{4}{3} a$, will be by definition almost-square. Let's assume that it's relatively unlikely that a number is almost-square in more than one way. Then the number of almostsquare numbers less than $n$ will be approximately
$$
\sum_{a=1}^{\sqrt{n}} \sum_{b=a}^{\frac{4}{3} a} 1=\frac{1}{3} \sum_{a=1}^{\sqrt{n}} a=\frac{1}{6} \sqrt{n}(\sqrt{n}+1)
$$
which is about $\frac{n}{6}$. So, $\frac{n}{6}$ will be a fairly good estimate for the number of almost-square numbers less than $n$, making 160000 a reasonable guess.
We can do better, though. For example, we summed $\frac{a}{3}$ all the way up to $\sqrt{n}$, but we are really overcounting here because when $a$ is close to $\sqrt{n}, a \cdot b$ will be less than $n$ only when $b \leq \frac{n}{a}$, as opposed to $b \leq \frac{4 a}{3}$. So we should really be taking the sum
[^4]\[
$$
\begin{aligned}
& \sum_{a=1}^{\sqrt{\frac{3 n}{4}}} \sum_{b=a}^{\frac{4 a}{3}} 1+\sum_{a=\sqrt{\frac{3 n}{4}}}^{\sqrt{n}} \sum_{b=a}^{\frac{n}{a}} 1 \\
& \quad=\sum_{a=1}^{\sqrt{\frac{3 n}{4}}} \frac{a}{3}+\sum_{a=\sqrt{\frac{3 n}{4}}}^{\sqrt{n}}\left(\frac{n}{a}-a\right) \\
& \quad \approx \frac{1}{6} \frac{3 n}{4}+n\left(\log (\sqrt{n})-\log \left(\sqrt{\frac{3 n}{4}}\right)\right)-\left(\frac{n}{2}-\frac{3 n}{8}\right) \\
& \quad=\frac{n}{8}+n \frac{\log (4)-\log (3)}{2}-\frac{n}{8} \\
& \quad=n \frac{\log (4)-\log (3)}{2}
\end{aligned}
$$
\]
In the process of taking the sum, we saw that we had something between $\frac{n}{8}$ and $\frac{n}{6}$, so we could also guess something between 166000 and 125000 , which would give us about 145000 , an even better answer. If we actually calculate $\frac{\log (4)-\log (3)}{2}$, we see that it's about 0.14384 , so 143840 would be the best guess if we were to use this strategy. In reality, we would want to round down a bit in both cases, since we are overcounting (because numbers could be square-free in multiple ways), so we should probably answer something like 140000 .
A final refinement to our calculation (and perhaps easier than the previous one), is to assume that the products $a \cdot b$ that we consider are randomly distributed between 1 and $n$, and to compute the expected number of distinct numbers we end up with. This is the same type of problem as number 31 on this contest, and we compute that if we randomly distribute $k$ numbers between 1 and $n$ then we expect to end up with $n\left(1-\left(1-\frac{1}{n}\right)^{k}\right)$ distinct numbers. When $k=n \frac{\log (4)-\log (3)}{2}$, we get that this equals
$$
\begin{aligned}
n\left(1-\left(\left(1-\frac{1}{n}\right)^{n}\right)^{\frac{\log (4)-\log (3)}{2}}\right) & =n\left(1-\sqrt{e^{\log (3)-\log (4)}}\right) \\
& =n\left(1-\sqrt{\frac{3}{4}}\right) \\
& =n\left(1-\frac{\sqrt{3}}{2}\right) \\
& \approx 0.134 n
\end{aligned}
$$
Giving us an answer of 134000 , which is very close to the correct answer.
The actual answer was found by computer, using the following $\mathrm{C}++$ program:
```
#include <stdio.h>
using namespace std;
bool isAlmostSquare(int n){
for(int k=1;k*k<=n;k++)
if(n%k==0 && 3*(n/k) <= 4*k) return true;
return false;
}
```
```
int main(){
int c = 0;
for(int n=1;n<=1000000;n++)
if(isAlmostSquare(n)) c++;
printf("%d\n",c);
return 0;
}
```
|
130348
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Call an positive integer almost-square if it can be written as $a \cdot b$, where $a$ and $b$ are integers and $a \leq b \leq \frac{4}{3} a$. How many almost-square positive integers are less than or equal to 1000000 ? Your score will be equal to $25-65 \frac{|A-C|}{\min (A, C)}$.
|
130348 To get a good estimate for the number of almost-square integers, note that any number of the form $a \cdot b$, with $b \leq \frac{4}{3} a$, will be by definition almost-square. Let's assume that it's relatively unlikely that a number is almost-square in more than one way. Then the number of almostsquare numbers less than $n$ will be approximately
$$
\sum_{a=1}^{\sqrt{n}} \sum_{b=a}^{\frac{4}{3} a} 1=\frac{1}{3} \sum_{a=1}^{\sqrt{n}} a=\frac{1}{6} \sqrt{n}(\sqrt{n}+1)
$$
which is about $\frac{n}{6}$. So, $\frac{n}{6}$ will be a fairly good estimate for the number of almost-square numbers less than $n$, making 160000 a reasonable guess.
We can do better, though. For example, we summed $\frac{a}{3}$ all the way up to $\sqrt{n}$, but we are really overcounting here because when $a$ is close to $\sqrt{n}, a \cdot b$ will be less than $n$ only when $b \leq \frac{n}{a}$, as opposed to $b \leq \frac{4 a}{3}$. So we should really be taking the sum
[^4]\[
$$
\begin{aligned}
& \sum_{a=1}^{\sqrt{\frac{3 n}{4}}} \sum_{b=a}^{\frac{4 a}{3}} 1+\sum_{a=\sqrt{\frac{3 n}{4}}}^{\sqrt{n}} \sum_{b=a}^{\frac{n}{a}} 1 \\
& \quad=\sum_{a=1}^{\sqrt{\frac{3 n}{4}}} \frac{a}{3}+\sum_{a=\sqrt{\frac{3 n}{4}}}^{\sqrt{n}}\left(\frac{n}{a}-a\right) \\
& \quad \approx \frac{1}{6} \frac{3 n}{4}+n\left(\log (\sqrt{n})-\log \left(\sqrt{\frac{3 n}{4}}\right)\right)-\left(\frac{n}{2}-\frac{3 n}{8}\right) \\
& \quad=\frac{n}{8}+n \frac{\log (4)-\log (3)}{2}-\frac{n}{8} \\
& \quad=n \frac{\log (4)-\log (3)}{2}
\end{aligned}
$$
\]
In the process of taking the sum, we saw that we had something between $\frac{n}{8}$ and $\frac{n}{6}$, so we could also guess something between 166000 and 125000 , which would give us about 145000 , an even better answer. If we actually calculate $\frac{\log (4)-\log (3)}{2}$, we see that it's about 0.14384 , so 143840 would be the best guess if we were to use this strategy. In reality, we would want to round down a bit in both cases, since we are overcounting (because numbers could be square-free in multiple ways), so we should probably answer something like 140000 .
A final refinement to our calculation (and perhaps easier than the previous one), is to assume that the products $a \cdot b$ that we consider are randomly distributed between 1 and $n$, and to compute the expected number of distinct numbers we end up with. This is the same type of problem as number 31 on this contest, and we compute that if we randomly distribute $k$ numbers between 1 and $n$ then we expect to end up with $n\left(1-\left(1-\frac{1}{n}\right)^{k}\right)$ distinct numbers. When $k=n \frac{\log (4)-\log (3)}{2}$, we get that this equals
$$
\begin{aligned}
n\left(1-\left(\left(1-\frac{1}{n}\right)^{n}\right)^{\frac{\log (4)-\log (3)}{2}}\right) & =n\left(1-\sqrt{e^{\log (3)-\log (4)}}\right) \\
& =n\left(1-\sqrt{\frac{3}{4}}\right) \\
& =n\left(1-\frac{\sqrt{3}}{2}\right) \\
& \approx 0.134 n
\end{aligned}
$$
Giving us an answer of 134000 , which is very close to the correct answer.
The actual answer was found by computer, using the following $\mathrm{C}++$ program:
```
#include <stdio.h>
using namespace std;
bool isAlmostSquare(int n){
for(int k=1;k*k<=n;k++)
if(n%k==0 && 3*(n/k) <= 4*k) return true;
return false;
}
```
```
int main(){
int c = 0;
for(int n=1;n<=1000000;n++)
if(isAlmostSquare(n)) c++;
printf("%d\n",c);
return 0;
}
```
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-guts-solutions.jsonl",
"problem_match": "\n35. [25]",
"solution_match": "\nAnswer: "
}
|
e508ba5d-e331-5350-97d5-ff8505eff44f
| 608,700
|
How many ways are there to place pawns on an $8 \times 8$ chessboard, so that there is at most 1 pawn in each horizontal row? Express your answer in the form $p_{1}^{e_{1}} \cdot p_{2}^{e_{2}} \cdots$, where the $p_{i}$ are distinct primes and the $e_{i}$ are positive integers.
|
$3^{16}$ If there is at most 1 pawn in each row, then each row of the chessboard may have either 0 or 1 pawn somewhere in the row. There is 1 case if there are no pawns in the row. There are 8 possible cases if there is 1 pawn in the row, one case for each square in the row. Hence for each row, there are 9 possible pawn arrangements. There are 8 rows, thus we have $9^{8}=3^{16}$.
|
3^{16}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many ways are there to place pawns on an $8 \times 8$ chessboard, so that there is at most 1 pawn in each horizontal row? Express your answer in the form $p_{1}^{e_{1}} \cdot p_{2}^{e_{2}} \cdots$, where the $p_{i}$ are distinct primes and the $e_{i}$ are positive integers.
|
$3^{16}$ If there is at most 1 pawn in each row, then each row of the chessboard may have either 0 or 1 pawn somewhere in the row. There is 1 case if there are no pawns in the row. There are 8 possible cases if there is 1 pawn in the row, one case for each square in the row. Hence for each row, there are 9 possible pawn arrangements. There are 8 rows, thus we have $9^{8}=3^{16}$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"problem_match": "\n1. [10]",
"solution_match": "\nAnswer: "
}
|
38102933-3d22-5b18-aeb2-c95e004691b6
| 608,712
|
In the following figure, a regular hexagon of side length 1 is attached to a semicircle of diameter 1. What is the longest distance between any two points in the figure?
|
$\frac{1+\sqrt{13}}{2}$ Inspection shows that one point must be on the semicircle and the other must be on the side of the hexagon directly opposite the edge with the semicircle, the bottom edge of the hexagon in the above diagram. Let $O$ be the center of the semicircle and let $M$ be the midpoint of the bottom edge.
We will determine the longest distance between points in the figure by comparing the lengths of all the segments with one endpoint on the bottom edge and the other endpoint on the semicircle. Fix a point $A$ on the bottom edge of the hexagon. Suppose that $B$ is chosen on the semicircle such that $A B$ is as long as possible. Let $C$ be the circle centered at $A$ with radius $A B$. If $C$ is not tangent to the semicircle, then part of the semicircle is outside $C$, so we could pick a $B^{\prime}$ on the semicircle such that $A B^{\prime}$ is longer than $A B$. So $C$ must be tangent to the semicircle, and $A B$ must pass through $O$.
Then $O B$ is always $\frac{1}{2}$, no matter which $A$ we choose on the bottom edge. All that remains is maximizing $A O$. This length is the hypotenuse of a right triangle with the fixed height $M O$, so it is maximized when $A M$ is as large as possible - when $A$ is an endpoint of the bottom edge. Note that $M O=2 \cdot \frac{\sqrt{3}}{2}$, and that $A M$ can be at most $\frac{1}{2}$, so $A O$ can be at most $\frac{\sqrt{13}}{2}$. So the maximum distance between two points in the diagram is $A O+O B=\frac{1+\sqrt{13}}{2}$.
|
\frac{1+\sqrt{13}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
In the following figure, a regular hexagon of side length 1 is attached to a semicircle of diameter 1. What is the longest distance between any two points in the figure?
|
$\frac{1+\sqrt{13}}{2}$ Inspection shows that one point must be on the semicircle and the other must be on the side of the hexagon directly opposite the edge with the semicircle, the bottom edge of the hexagon in the above diagram. Let $O$ be the center of the semicircle and let $M$ be the midpoint of the bottom edge.
We will determine the longest distance between points in the figure by comparing the lengths of all the segments with one endpoint on the bottom edge and the other endpoint on the semicircle. Fix a point $A$ on the bottom edge of the hexagon. Suppose that $B$ is chosen on the semicircle such that $A B$ is as long as possible. Let $C$ be the circle centered at $A$ with radius $A B$. If $C$ is not tangent to the semicircle, then part of the semicircle is outside $C$, so we could pick a $B^{\prime}$ on the semicircle such that $A B^{\prime}$ is longer than $A B$. So $C$ must be tangent to the semicircle, and $A B$ must pass through $O$.
Then $O B$ is always $\frac{1}{2}$, no matter which $A$ we choose on the bottom edge. All that remains is maximizing $A O$. This length is the hypotenuse of a right triangle with the fixed height $M O$, so it is maximized when $A M$ is as large as possible - when $A$ is an endpoint of the bottom edge. Note that $M O=2 \cdot \frac{\sqrt{3}}{2}$, and that $A M$ can be at most $\frac{1}{2}$, so $A O$ can be at most $\frac{\sqrt{13}}{2}$. So the maximum distance between two points in the diagram is $A O+O B=\frac{1+\sqrt{13}}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"problem_match": "\n2. [10]",
"solution_match": "\nAnswer: "
}
|
8794ba5e-a0a2-560a-b5c8-8a9063347ed4
| 608,713
|
Let $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}$, where each $a_{i}$ is either 1 or -1 . Let $r$ be a root of $p$. If $|r|>\frac{15}{8}$, what is the minimum possible value of $n$ ?
|
4 We claim that $n=4$ is the answer. First, we show that $n>3$. Suppose that $n \leq 3$. Let $r$ be the root of the polynomial with $|r| \geq \frac{15}{8}$. Then, by the Triangle Inequality, we have:
$$
\begin{gathered}
\left|a_{n} r^{n}\right|=\left|a_{n-1} r^{n-1}+a_{n-2} r^{n-2}+\ldots+a_{0}\right| \leq\left|a_{n-1} r^{n-1}\right|+\left|a_{n-2} r^{n-2}\right|+\ldots+\left|a_{0}\right| \\
|r|^{n} \leq\left|r^{n-1}\right|+\left|r^{n-2}\right|+\ldots+|1|=\frac{|r|^{n}-1}{|r|-1} \\
|r|^{n+1}-2|r|^{n}+1 \leq 0 \Rightarrow 1 \leq|r|^{n}(2-|r|)
\end{gathered}
$$
The right-hand side is increasing in $n$, for $|r|>1$, so it is bounded by $|r|^{3}(2-|r|)$. This expression is decreasing in $r$ for $r \geq \frac{3}{2}$. When $|r|=\frac{15}{8}$, then the right-hand side is less than 1 , which violates the inequalities. Therefore $n>3$. Now, we claim that there is a 4 th degree polynomial with a root $r$ with $|r| \geq \frac{15}{8}$. Let $p(x)=x^{4}-x^{3}-x^{2}-x-1$. Then $p\left(\frac{15}{8}\right)<0$ and $p(2)>2$. By the Intermediate Value Theorem, $p(x)$ has such a root $r$.
|
4
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}$, where each $a_{i}$ is either 1 or -1 . Let $r$ be a root of $p$. If $|r|>\frac{15}{8}$, what is the minimum possible value of $n$ ?
|
4 We claim that $n=4$ is the answer. First, we show that $n>3$. Suppose that $n \leq 3$. Let $r$ be the root of the polynomial with $|r| \geq \frac{15}{8}$. Then, by the Triangle Inequality, we have:
$$
\begin{gathered}
\left|a_{n} r^{n}\right|=\left|a_{n-1} r^{n-1}+a_{n-2} r^{n-2}+\ldots+a_{0}\right| \leq\left|a_{n-1} r^{n-1}\right|+\left|a_{n-2} r^{n-2}\right|+\ldots+\left|a_{0}\right| \\
|r|^{n} \leq\left|r^{n-1}\right|+\left|r^{n-2}\right|+\ldots+|1|=\frac{|r|^{n}-1}{|r|-1} \\
|r|^{n+1}-2|r|^{n}+1 \leq 0 \Rightarrow 1 \leq|r|^{n}(2-|r|)
\end{gathered}
$$
The right-hand side is increasing in $n$, for $|r|>1$, so it is bounded by $|r|^{3}(2-|r|)$. This expression is decreasing in $r$ for $r \geq \frac{3}{2}$. When $|r|=\frac{15}{8}$, then the right-hand side is less than 1 , which violates the inequalities. Therefore $n>3$. Now, we claim that there is a 4 th degree polynomial with a root $r$ with $|r| \geq \frac{15}{8}$. Let $p(x)=x^{4}-x^{3}-x^{2}-x-1$. Then $p\left(\frac{15}{8}\right)<0$ and $p(2)>2$. By the Intermediate Value Theorem, $p(x)$ has such a root $r$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"problem_match": "\n3. [15]",
"solution_match": "\nAnswer: "
}
|
058ffa1f-c37f-5dd6-aa31-43a6de2fef23
| 608,714
|
Find all 4-digit integers of the form $a a b b$ (when written in base 10) that are perfect squares.
|
7744 Let $x$ be an integer such that $x^{2}$ is of the desired form. Then $1100 a+11 b=x^{2}$. Then $x^{2}$ is divisible by 11 , which means $x$ is divisible by 11 . Then for some integer, $y, x=11 y$. Then $1100 a+11 b=11^{2} y^{2} \Rightarrow 100 a+b=11 y^{2}$. This means that $100 a+b \equiv 0(\bmod 11) \Rightarrow a+b \equiv 0(\bmod 11)$. Because $a$ and $b$ must be nonzero digits, we have $2 \leq a, b \leq 9$, so we can write $b=11-a$.
Replacing $b$ in the equation derived above, we obtain $99 a+11=11 y^{2} \Rightarrow 9 a+1=y^{2}$. We check the possible values of $a$ from 2 to 9 , and only $a=7$ yields a perfect square. When $a=7, b=4$, so the only perfect square of for $a a b b$ is 7744 .
|
7744
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find all 4-digit integers of the form $a a b b$ (when written in base 10) that are perfect squares.
|
7744 Let $x$ be an integer such that $x^{2}$ is of the desired form. Then $1100 a+11 b=x^{2}$. Then $x^{2}$ is divisible by 11 , which means $x$ is divisible by 11 . Then for some integer, $y, x=11 y$. Then $1100 a+11 b=11^{2} y^{2} \Rightarrow 100 a+b=11 y^{2}$. This means that $100 a+b \equiv 0(\bmod 11) \Rightarrow a+b \equiv 0(\bmod 11)$. Because $a$ and $b$ must be nonzero digits, we have $2 \leq a, b \leq 9$, so we can write $b=11-a$.
Replacing $b$ in the equation derived above, we obtain $99 a+11=11 y^{2} \Rightarrow 9 a+1=y^{2}$. We check the possible values of $a$ from 2 to 9 , and only $a=7$ yields a perfect square. When $a=7, b=4$, so the only perfect square of for $a a b b$ is 7744 .
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"problem_match": "\n4. [20]",
"solution_match": "\nAnswer: "
}
|
39c39f5e-e463-5260-a2ab-b141566ab97d
| 608,715
|
Compute
$$
\sum_{n=1}^{98} \frac{2}{\sqrt{n}+\sqrt{n+2}}+\frac{1}{\sqrt{n+1}+\sqrt{n+2}}
$$
|
$\quad 3 \sqrt{11}-2 \sqrt{2}+19$ Rationalizing the denominator of both terms in the summation yields $\sqrt{n+2}-\sqrt{n}+\sqrt{n+2}-\sqrt{n+1}=2 \sqrt{n+2}-(\sqrt{n}+\sqrt{n+1})$. Then the sum $\sum_{n=1}^{98} 2 \sqrt{n+2}-(\sqrt{n}+\sqrt{n+1})$ telescopes. All terms cancel except for $-(\sqrt{1}+\sqrt{2})-\sqrt{2}+2 \sqrt{99}+2 \sqrt{100}-\sqrt{99}=3 \sqrt{11}-2 \sqrt{2}+19$.
|
3 \sqrt{11}-2 \sqrt{2}+19
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Compute
$$
\sum_{n=1}^{98} \frac{2}{\sqrt{n}+\sqrt{n+2}}+\frac{1}{\sqrt{n+1}+\sqrt{n+2}}
$$
|
$\quad 3 \sqrt{11}-2 \sqrt{2}+19$ Rationalizing the denominator of both terms in the summation yields $\sqrt{n+2}-\sqrt{n}+\sqrt{n+2}-\sqrt{n+1}=2 \sqrt{n+2}-(\sqrt{n}+\sqrt{n+1})$. Then the sum $\sum_{n=1}^{98} 2 \sqrt{n+2}-(\sqrt{n}+\sqrt{n+1})$ telescopes. All terms cancel except for $-(\sqrt{1}+\sqrt{2})-\sqrt{2}+2 \sqrt{99}+2 \sqrt{100}-\sqrt{99}=3 \sqrt{11}-2 \sqrt{2}+19$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"problem_match": "\n5. [25]",
"solution_match": "\nAnswer: "
}
|
94d17d17-f7b8-5370-bafa-d09aa694c92b
| 608,716
|
Into how many regions can a circle be cut by 10 parabolas?
|
201 We will consider the general case of $n$ parabolas, for which the answer is $2 n^{2}+1$.
We will start with some rough intuition, then fill in the details afterwards. The intuition is that, if we make the parabolas steep enough, we can basically treat them as two parallel lines. Furthermore, the number of regions is given in terms of the number of intersections of the parabolas that occur within the circle, since every time two parabolas cross a new region is created. Since two pairs of parallel lines intersect in 4 points, and pairs of parabolas also intersect in 4 points, as long as we can always make all 4 points of intersection lie inside the circle, the parallel lines case is the best we can do.
In other words, the answer is the same as the answer if we were trying to add ten pairs of parallel lines. We can compute the answer for pairs of parallel lines as follows - when we add the $k$ th set of parallel lines, there are already $2 k-2$ lines that the two new lines can intersect, meaning that each of the lines adds $2 k-1$ new regions ${ }^{1}$. This means that we add $4 k-2$ regions when adding the $k$ th set of lines, making the answer $1+2+6+10+14+\cdots+(4 n-2)=1+2(1+3+5+7+\cdots+(2 n-1))=1+2 \cdot n^{2}=2 n^{2}+1$.
Now that we have sketched out the solution, we will fill in the details more rigorously. First, if there are $n$ parabolas inside the circle, and the they intersect in $K$ points total, then we claim that the number of regions the circle is divided into will be at most $K+n+r+1$, where $r$ is the number of parabolas that intersect the circle itself in exactly four points.
We will prove this by induction. In the base case of $n=0$, we are just saying that the circle itself consists of exactly one region.
To prove the inductive step, suppose that we have $n$ parabolas with $K$ points of intersection. We want to show that if we add an additional parabola, and this parabola intersects the other parabolas in $p$ points, then this new parabola adds either $p+1$ or $p+2$ regions to the circle, and that we get $p+2$ regions if and only if the parabola intersects the circle in exactly four points.
We will do this by considering how many regions the parabola cuts through, following its path from when it initially enters the circle to when it exits the circle for the last time. When it initially enters the circle, it cuts through one region, thereby increasing the number of regions by on ${ }^{2}$. Then, for each other parabola that this parabola crosses, we cut through one additional region. It is also possible for the parabola to leave and then re-enter the circle, which happens if and only if the parabola intersects
[^0]the circle in four points, and also adds one additional region. Therefore, the number of regions is either $p+1$ or $p+2$, and it is $p+2$ if and only if the parabola intersects the circle in four points. This completes the induction and proves the claim.
So, we are left with trying to maximize $K+n+r+1$. Since a pair of parabolas intersects in at most 4 points, and there are $\binom{n}{2}$ pairs of parabolas, we have $K \leq 4\binom{n}{2}=2 n^{2}-2 n$. Also, $r \leq n$, so $K+n+r+1 \leq 2 n^{2}+1$. On the other hand, as explained in the paragraphs giving the intuition, we can attain $2 n^{2}+1$ by making the parabolas sufficiently steep that they act like pairs of parallel lines. Therefore, the answer is $2 n^{2}+1$, as claimed.
|
201
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Into how many regions can a circle be cut by 10 parabolas?
|
201 We will consider the general case of $n$ parabolas, for which the answer is $2 n^{2}+1$.
We will start with some rough intuition, then fill in the details afterwards. The intuition is that, if we make the parabolas steep enough, we can basically treat them as two parallel lines. Furthermore, the number of regions is given in terms of the number of intersections of the parabolas that occur within the circle, since every time two parabolas cross a new region is created. Since two pairs of parallel lines intersect in 4 points, and pairs of parabolas also intersect in 4 points, as long as we can always make all 4 points of intersection lie inside the circle, the parallel lines case is the best we can do.
In other words, the answer is the same as the answer if we were trying to add ten pairs of parallel lines. We can compute the answer for pairs of parallel lines as follows - when we add the $k$ th set of parallel lines, there are already $2 k-2$ lines that the two new lines can intersect, meaning that each of the lines adds $2 k-1$ new regions ${ }^{1}$. This means that we add $4 k-2$ regions when adding the $k$ th set of lines, making the answer $1+2+6+10+14+\cdots+(4 n-2)=1+2(1+3+5+7+\cdots+(2 n-1))=1+2 \cdot n^{2}=2 n^{2}+1$.
Now that we have sketched out the solution, we will fill in the details more rigorously. First, if there are $n$ parabolas inside the circle, and the they intersect in $K$ points total, then we claim that the number of regions the circle is divided into will be at most $K+n+r+1$, where $r$ is the number of parabolas that intersect the circle itself in exactly four points.
We will prove this by induction. In the base case of $n=0$, we are just saying that the circle itself consists of exactly one region.
To prove the inductive step, suppose that we have $n$ parabolas with $K$ points of intersection. We want to show that if we add an additional parabola, and this parabola intersects the other parabolas in $p$ points, then this new parabola adds either $p+1$ or $p+2$ regions to the circle, and that we get $p+2$ regions if and only if the parabola intersects the circle in exactly four points.
We will do this by considering how many regions the parabola cuts through, following its path from when it initially enters the circle to when it exits the circle for the last time. When it initially enters the circle, it cuts through one region, thereby increasing the number of regions by on ${ }^{2}$. Then, for each other parabola that this parabola crosses, we cut through one additional region. It is also possible for the parabola to leave and then re-enter the circle, which happens if and only if the parabola intersects
[^0]the circle in four points, and also adds one additional region. Therefore, the number of regions is either $p+1$ or $p+2$, and it is $p+2$ if and only if the parabola intersects the circle in four points. This completes the induction and proves the claim.
So, we are left with trying to maximize $K+n+r+1$. Since a pair of parabolas intersects in at most 4 points, and there are $\binom{n}{2}$ pairs of parabolas, we have $K \leq 4\binom{n}{2}=2 n^{2}-2 n$. Also, $r \leq n$, so $K+n+r+1 \leq 2 n^{2}+1$. On the other hand, as explained in the paragraphs giving the intuition, we can attain $2 n^{2}+1$ by making the parabolas sufficiently steep that they act like pairs of parallel lines. Therefore, the answer is $2 n^{2}+1$, as claimed.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"problem_match": "\n6. [25]",
"solution_match": "\nAnswer: "
}
|
8f5c8dda-d1e6-5c28-a0ea-6b9198b63abc
| 608,717
|
Evaluate
$$
\sum_{k=1}^{2010} \cos ^{2}(k)
$$
|
$1005+\frac{\sin (4021)-\sin (1)}{4 \sin (1)}$ We use the identity $\cos ^{2}(k)=\frac{1+\cos (2 x)}{2}$. Then our expression evalutes to $1005+\frac{(\cos (2)+\ldots+\cos (4020))}{2}$.
To evaluate $\cos (2)+\cdots+\cos (4020)$, let $y=\cos (2)+\cdots+\cos (4020) \Rightarrow y(\sin (1))=\cos (2) \sin (1)+$ $\ldots+\cos (4020) \sin (1)$. Observe that for any $x, \cos (x) \sin (1)=\frac{\sin (x+1)-\sin (x-1)}{2}$. Then $y(\sin (1))=$ $\frac{\sin (3)-\sin (1)}{2}+\frac{\sin (5)-\sin (3)}{2}+\ldots+\frac{\sin 4021-\sin (4019)}{2}$. This is clearly a telescoping sum; we get $y(\sin (1))=$ $\frac{\sin (4021)-\sin (1)}{2}$. Then we have the desired $y=\frac{\sin (4021)-\sin (1)}{2 \sin (1)}$. Then our original expression evaluates to $1005+\frac{\sin (4021)-\sin (1)}{4 \sin (1)}$.
|
1005+\frac{\sin (4021)-\sin (1)}{4 \sin (1)}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Evaluate
$$
\sum_{k=1}^{2010} \cos ^{2}(k)
$$
|
$1005+\frac{\sin (4021)-\sin (1)}{4 \sin (1)}$ We use the identity $\cos ^{2}(k)=\frac{1+\cos (2 x)}{2}$. Then our expression evalutes to $1005+\frac{(\cos (2)+\ldots+\cos (4020))}{2}$.
To evaluate $\cos (2)+\cdots+\cos (4020)$, let $y=\cos (2)+\cdots+\cos (4020) \Rightarrow y(\sin (1))=\cos (2) \sin (1)+$ $\ldots+\cos (4020) \sin (1)$. Observe that for any $x, \cos (x) \sin (1)=\frac{\sin (x+1)-\sin (x-1)}{2}$. Then $y(\sin (1))=$ $\frac{\sin (3)-\sin (1)}{2}+\frac{\sin (5)-\sin (3)}{2}+\ldots+\frac{\sin 4021-\sin (4019)}{2}$. This is clearly a telescoping sum; we get $y(\sin (1))=$ $\frac{\sin (4021)-\sin (1)}{2}$. Then we have the desired $y=\frac{\sin (4021)-\sin (1)}{2 \sin (1)}$. Then our original expression evaluates to $1005+\frac{\sin (4021)-\sin (1)}{4 \sin (1)}$.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"problem_match": "\n7. [30]",
"solution_match": "\nAnswer: "
}
|
40b6850c-84df-5142-ac0b-7d54a7cdd799
| 608,718
|
Consider the following two-player game. Player 1 starts with a number, $N$. He then subtracts a proper divisor of $N$ from $N$ and gives the result to player 2 (a proper divisor of $N$ is a positive divisor of $N$ that is not equal to 1 or $N)$. Player 2 does the same thing with the number she gets from player 1 , and gives the result back to player 1. The two players continue until a player is given a prime number, at which point that player loses. For how many values of $N$ between 2 and 100 inclusive does player 1 have a winning strategy?
|
47 We claim that player 1 has a winning strategy if and only if $N$ is even and not an odd power of 2 .
First we show that if you are stuck with an odd number, then you are guaranteed to lose. Suppose you have an odd number $a b$, where $a$ and $b$ are odd numbers, and you choose to subtract $a$. You pass your opponent the number $a(b-1)$. This cannot be a power of 2 (otherwise $a$ is a power of 2 and hence $a=1$, which is not allowed), so your opponent can find an odd proper divisor of $a(b-1)$ (such as $a$ ), and you will have a smaller odd number. Eventually you will get to an odd prime and lose.
Now consider even numbers that aren't powers of 2. As with before, you can find an odd proper divisor of $N$ and pass your opponent an odd number, so you are guaranteed to win.
Finally consider powers of 2 . If you have the number $N=2^{k}$, it would be unwise to choose a proper divisor other than $2^{k-1}$; otherwise you would give your opponent an even number that isn't a power of 2 . Therefore if $k$ is odd, you will end up with 2 and lose. If $k$ is even, though, your opponent will end up with 2 and you will win.
Therefore player 1 has a winning strategy for all even numbers except for odd powers of 2 .
|
47
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Consider the following two-player game. Player 1 starts with a number, $N$. He then subtracts a proper divisor of $N$ from $N$ and gives the result to player 2 (a proper divisor of $N$ is a positive divisor of $N$ that is not equal to 1 or $N)$. Player 2 does the same thing with the number she gets from player 1 , and gives the result back to player 1. The two players continue until a player is given a prime number, at which point that player loses. For how many values of $N$ between 2 and 100 inclusive does player 1 have a winning strategy?
|
47 We claim that player 1 has a winning strategy if and only if $N$ is even and not an odd power of 2 .
First we show that if you are stuck with an odd number, then you are guaranteed to lose. Suppose you have an odd number $a b$, where $a$ and $b$ are odd numbers, and you choose to subtract $a$. You pass your opponent the number $a(b-1)$. This cannot be a power of 2 (otherwise $a$ is a power of 2 and hence $a=1$, which is not allowed), so your opponent can find an odd proper divisor of $a(b-1)$ (such as $a$ ), and you will have a smaller odd number. Eventually you will get to an odd prime and lose.
Now consider even numbers that aren't powers of 2. As with before, you can find an odd proper divisor of $N$ and pass your opponent an odd number, so you are guaranteed to win.
Finally consider powers of 2 . If you have the number $N=2^{k}$, it would be unwise to choose a proper divisor other than $2^{k-1}$; otherwise you would give your opponent an even number that isn't a power of 2 . Therefore if $k$ is odd, you will end up with 2 and lose. If $k$ is even, though, your opponent will end up with 2 and you will win.
Therefore player 1 has a winning strategy for all even numbers except for odd powers of 2 .
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"problem_match": "\n8. [30]",
"solution_match": "\nAnswer: "
}
|
b10da2c7-420d-5ceb-8bca-0984a9fd10cb
| 608,719
|
Let $S$ be the set of ordered pairs of integers $(x, y)$ with $1 \leq x \leq 5$ and $1 \leq y \leq 3$. How many subsets $R$ of $S$ have the property that all the points of $R$ lie on the graph of a single cubic? A cubic is a polynomial of the form $y=a x^{3}+b x^{2}+c x+d$, where $a, b, c$, and $d$ are real numbers (meaning that $a$ is allowed to be 0 ).
|
796 We observe that $R$ must contain at most 1 point from each column of $S$, because no function can contain more than 1 point with the same $x$-coordinate. Therefore, $|R| \leq 5(|R|$ denotes the number of elements of $R$ ). Note that 4 points determine a cubic, so if $R$ is any subset of points in distinct columns and $|R| \leq 4$, then $R$ has the desired property. There are $4^{5}$ ways to choose at most 1
point from each column and $3^{5}$ ways to choose exactly 1 point from each column. There are therefore $4^{5}-3^{5}=781$ subsets $R$ of $S$ such that $|R| \leq 4$ and all points of $R$ lie in distinct columns. As noted, these sets all automatically have the desired property.
Now we consider all sets $R$ of size 5 . As before, each point in $R$ must come from a different column. Let us shift our origin to $(3,2)$, and let $p$ be the polynomial containing all 5 points of $R$. Then $R=\{(-2, p(-2)),(-1, p(-1)),(0, p(0)),(1, p(1)),(2, p(2))\}$.
By the method of finite differences $3^{3}$ or alternately by Lagrange Interpolation there is a unique polynomial $p$ of degree less than 5 going through 5 specified points, and this polynomial is of degree less than 4 if and only if $p(-2)-4 p(-1)+6 p(0)-4 p(1)+p(2)=0$.
Then $p(-2)+p(2)+6 p(0)=4(p(-1)+p(1))$, where $p(-2)+p(2) \in\{-2-1,0,1,2\}, p(-1)+p(1) \in$ $\{-2,-1,0,1,2\}$, and $p(0) \in\{-1,0,1\}$. We know that $6 p(0)$ and $4(p(-1)+p(1))$ are necessarily even, thus we must have $p(-2)+p(2) \in\{-2,0,2\}$ in order for the equation to be satisfied.
Let $(a, b, c)=(p(-2)+p(2), 6 p(0), 4(p(-1)+p(1)))$. The possible values of $(a, b, c)$ that are solutions to $a+b=c$ are then $\{(-2,-6,-8),(-2,6,4),(0,0,0),(2,-6,-4),(2,6,8)\}$.
If $(a, b, c)=(-2,-6,-8)$, then we need $p(-2)+p(2)=-2, p(0)=-1, p(-1)+p(1)=-2$. There is only 1 possible solution to each of these equations: $(p(-2), p(2))=(-1,-1)$ for the first one, $p(0)=-1$ for the second, and $(p(1))=(-1,-1)$ for the third. Hence there is 1 possible subset $R$ for the case $(a, b, c)=(-2,-6,-8)$.
If $(a, b, c)=(-2,6,4)$, then there is again 1 possible solution to $p(-2)+p(2)=1$. There are two solutions to $p(-1)+p(1)=1:(p(-1), p(1))=(0,1),(1,0)$. Also, $p(0)$ can only be 1 , so there are 2 possible subsets for this case.
If $(a, b, c)=(0,0,0)$, then there there are 3 possible solutions to $p(-2)+p(2)=0:(p(-2), p(2))=$ $(-1,1),(0,0),(1,-1)$. Similarly, there are 3 possible solutions to $p(-1)+p(1)=0$. Also, $p(0)$ can only be 0 , so there are 9 possible subsets for this case.
If $(a, b, c)=(2,-6,-4)$, then there is 1 possible solution to $p(-2)+p(2)=2:(p(-2), p(2))=(1,1)$. There are 2 possible solutions to $p(-1)+p(1)=-1:(p(-1), p(1))=(0,-1),(-1,0)$. Also, $p(0)$ can only be -1 , so there are 2 possible subsets for this case.
If $(a, b, c)=(2,6,8)$, then there is 1 possible solution to $p(-2)+p(2)=2$, as shown above. There is 1 solution to $p(-1)+p(1)=2:(p(-1), p(1))=(1,1)$. Also, $p(0)$ can only be 1 , so there is 1 possible subset for this case.
Then there are $1+2+9+2+1=15$ total possible subsets of size 5 that can be fit to a polynomial of degree less than 4 . Hence there are $781+15=796$ possible subsets total.
|
796
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $S$ be the set of ordered pairs of integers $(x, y)$ with $1 \leq x \leq 5$ and $1 \leq y \leq 3$. How many subsets $R$ of $S$ have the property that all the points of $R$ lie on the graph of a single cubic? A cubic is a polynomial of the form $y=a x^{3}+b x^{2}+c x+d$, where $a, b, c$, and $d$ are real numbers (meaning that $a$ is allowed to be 0 ).
|
796 We observe that $R$ must contain at most 1 point from each column of $S$, because no function can contain more than 1 point with the same $x$-coordinate. Therefore, $|R| \leq 5(|R|$ denotes the number of elements of $R$ ). Note that 4 points determine a cubic, so if $R$ is any subset of points in distinct columns and $|R| \leq 4$, then $R$ has the desired property. There are $4^{5}$ ways to choose at most 1
point from each column and $3^{5}$ ways to choose exactly 1 point from each column. There are therefore $4^{5}-3^{5}=781$ subsets $R$ of $S$ such that $|R| \leq 4$ and all points of $R$ lie in distinct columns. As noted, these sets all automatically have the desired property.
Now we consider all sets $R$ of size 5 . As before, each point in $R$ must come from a different column. Let us shift our origin to $(3,2)$, and let $p$ be the polynomial containing all 5 points of $R$. Then $R=\{(-2, p(-2)),(-1, p(-1)),(0, p(0)),(1, p(1)),(2, p(2))\}$.
By the method of finite differences $3^{3}$ or alternately by Lagrange Interpolation there is a unique polynomial $p$ of degree less than 5 going through 5 specified points, and this polynomial is of degree less than 4 if and only if $p(-2)-4 p(-1)+6 p(0)-4 p(1)+p(2)=0$.
Then $p(-2)+p(2)+6 p(0)=4(p(-1)+p(1))$, where $p(-2)+p(2) \in\{-2-1,0,1,2\}, p(-1)+p(1) \in$ $\{-2,-1,0,1,2\}$, and $p(0) \in\{-1,0,1\}$. We know that $6 p(0)$ and $4(p(-1)+p(1))$ are necessarily even, thus we must have $p(-2)+p(2) \in\{-2,0,2\}$ in order for the equation to be satisfied.
Let $(a, b, c)=(p(-2)+p(2), 6 p(0), 4(p(-1)+p(1)))$. The possible values of $(a, b, c)$ that are solutions to $a+b=c$ are then $\{(-2,-6,-8),(-2,6,4),(0,0,0),(2,-6,-4),(2,6,8)\}$.
If $(a, b, c)=(-2,-6,-8)$, then we need $p(-2)+p(2)=-2, p(0)=-1, p(-1)+p(1)=-2$. There is only 1 possible solution to each of these equations: $(p(-2), p(2))=(-1,-1)$ for the first one, $p(0)=-1$ for the second, and $(p(1))=(-1,-1)$ for the third. Hence there is 1 possible subset $R$ for the case $(a, b, c)=(-2,-6,-8)$.
If $(a, b, c)=(-2,6,4)$, then there is again 1 possible solution to $p(-2)+p(2)=1$. There are two solutions to $p(-1)+p(1)=1:(p(-1), p(1))=(0,1),(1,0)$. Also, $p(0)$ can only be 1 , so there are 2 possible subsets for this case.
If $(a, b, c)=(0,0,0)$, then there there are 3 possible solutions to $p(-2)+p(2)=0:(p(-2), p(2))=$ $(-1,1),(0,0),(1,-1)$. Similarly, there are 3 possible solutions to $p(-1)+p(1)=0$. Also, $p(0)$ can only be 0 , so there are 9 possible subsets for this case.
If $(a, b, c)=(2,-6,-4)$, then there is 1 possible solution to $p(-2)+p(2)=2:(p(-2), p(2))=(1,1)$. There are 2 possible solutions to $p(-1)+p(1)=-1:(p(-1), p(1))=(0,-1),(-1,0)$. Also, $p(0)$ can only be -1 , so there are 2 possible subsets for this case.
If $(a, b, c)=(2,6,8)$, then there is 1 possible solution to $p(-2)+p(2)=2$, as shown above. There is 1 solution to $p(-1)+p(1)=2:(p(-1), p(1))=(1,1)$. Also, $p(0)$ can only be 1 , so there is 1 possible subset for this case.
Then there are $1+2+9+2+1=15$ total possible subsets of size 5 that can be fit to a polynomial of degree less than 4 . Hence there are $781+15=796$ possible subsets total.
|
{
"resource_path": "HarvardMIT/segmented/en-132-2010-feb-team2-solutions.jsonl",
"problem_match": "\n9. [35]",
"solution_match": "\nAnswer: "
}
|
40492d81-50bf-5d46-95f4-c5d970e61c31
| 608,720
|
Jacob flips five coins, exactly three of which land heads. What is the probability that the first two are both heads?
|
$\frac{3}{10}$ We can associate with each sequence of coin flips a unique word where $H$ represents heads, and T represents tails. For example, the word HHTTH would correspond to the coin flip sequence where the first two flips were heads, the next two were tails, and the last was heads. We are given that exactly three of the five coin flips came up heads, so our word must be some rearrangement of HHHTT. To calculate the total number of possibilities, any rearrangement corresponds to a choice of three spots to place the H flips, so there are $\binom{5}{3}=10$ possibilities. If the first two flips are both heads, then we can only rearrange the last three HTT flips, which corresponds to choosing one spot for the remaining $H$. This can be done in $\binom{3}{1}=3$ ways. Finally, the probability is the quotient of these two, so we get the answer of $\frac{3}{10}$. Alternatively, since the total number of possiblities is small, we can write out all rearrangements: HHHTT, HHTHT, HHTTH, HTHHT, HTHTH, HTTHH, THHHT, THHTH, THTHH, TTHHH. Of these ten, only in the first three do we flip heads the first two times, so we get the same answer of $\frac{3}{10}$.
|
\frac{3}{10}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Jacob flips five coins, exactly three of which land heads. What is the probability that the first two are both heads?
|
$\frac{3}{10}$ We can associate with each sequence of coin flips a unique word where $H$ represents heads, and T represents tails. For example, the word HHTTH would correspond to the coin flip sequence where the first two flips were heads, the next two were tails, and the last was heads. We are given that exactly three of the five coin flips came up heads, so our word must be some rearrangement of HHHTT. To calculate the total number of possibilities, any rearrangement corresponds to a choice of three spots to place the H flips, so there are $\binom{5}{3}=10$ possibilities. If the first two flips are both heads, then we can only rearrange the last three HTT flips, which corresponds to choosing one spot for the remaining $H$. This can be done in $\binom{3}{1}=3$ ways. Finally, the probability is the quotient of these two, so we get the answer of $\frac{3}{10}$. Alternatively, since the total number of possiblities is small, we can write out all rearrangements: HHHTT, HHTHT, HHTTH, HTHHT, HTHTH, HTTHH, THHHT, THHTH, THTHH, TTHHH. Of these ten, only in the first three do we flip heads the first two times, so we get the same answer of $\frac{3}{10}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen1-solutions.jsonl",
"problem_match": "\n1. [2]",
"solution_match": "\nAnswer: "
}
|
d3fca1d6-a3cc-5b46-9c8e-45e43c234e61
| 608,722
|
How many sequences $a_{1}, a_{2}, \ldots, a_{8}$ of zeroes and ones have $a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{7} a_{8}=5$ ?
|
9 First, note that we have seven terms in the left hand side, and each term can be either 0 or 1 , so we must have five terms equal to 1 and two terms equal to 0 . Thus, for $n \in\{1,2, \ldots, 8\}$, at least one of the $a_{n}$ must be equal to 0 . If we can find $i, j \in\{2,3, \ldots, 7\}$ such that $a_{i}=a_{j}=0$ and $i<j$, then the terms $a_{i-1} a_{i}, a_{i} a_{i+1}$, and $a_{j} a_{j+1}$ will all be equal to 0 . We did not count any term twice because $i-1<i<j$, so we would have three terms equal to 0 , which cannot happen because we can have only two. Thus, we can find at most one $n \in\{2,3, \ldots, 7\}$ such that $a_{n}=0$. We will do casework on which $n$ in this range have $a_{n}=0$.
If $n \in\{3,4,5,6\}$, then we know that the terms $a_{n-1} a_{n}=a_{n} a_{n+1}=0$, so all other terms must be 1 , so $a_{1} a_{2}=a_{2} a_{3}=\ldots=a_{n-2} a_{n-1}=1$ and $a_{n+1} a_{n+2}=\ldots=a_{7} a_{8}=1$. Because every $a_{i}$ appears in one of these equations for $i \neq n$, then we must have $a_{i}=1$ for all $i \neq n$, so we have 1 possibility for each choice of $n$ and thus 4 possibilities total for this case.
If $n=2$, then again we have $a_{1} a_{2}=a_{2} a_{3}=0$, so we must have $a_{3} a_{4}=a_{4} a_{5}=\ldots=a_{7} a_{8}=1$, so $a_{3}=a_{4}=\ldots=a_{8}=1$. However, this time $a_{1}$ is not fixed, and we see that regardless of our choice of $a_{1}$ the sum will still be equal to 5 . Thus, since there are 2 choices for $a_{1}$, then there are 2 possibilities total for this case. The case where $n=7$ is analogous, with $a_{8}$ having 2 possibilities, so we have another 2 possibilities.
Finally, if $a_{n}=1$ for $n \in\{2,3, \ldots, 7\}$, then we will have $a_{2} a_{3}=a_{3} a_{4}=\ldots=a_{6} a_{7}=1$. We already have five terms equal to 1 , so the remaining two terms $a_{1} a_{2}$ and $a_{7} a_{8}$ must be 0 . Since $a_{2}=1$, then we must have $a_{1}=0$, and since $a_{7}=1$ then $a_{8}=0$. Thus, there is only 1 possibility for this case.
Summing, we have $4+2+2+1=9$ total sequences.
|
9
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many sequences $a_{1}, a_{2}, \ldots, a_{8}$ of zeroes and ones have $a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{7} a_{8}=5$ ?
|
9 First, note that we have seven terms in the left hand side, and each term can be either 0 or 1 , so we must have five terms equal to 1 and two terms equal to 0 . Thus, for $n \in\{1,2, \ldots, 8\}$, at least one of the $a_{n}$ must be equal to 0 . If we can find $i, j \in\{2,3, \ldots, 7\}$ such that $a_{i}=a_{j}=0$ and $i<j$, then the terms $a_{i-1} a_{i}, a_{i} a_{i+1}$, and $a_{j} a_{j+1}$ will all be equal to 0 . We did not count any term twice because $i-1<i<j$, so we would have three terms equal to 0 , which cannot happen because we can have only two. Thus, we can find at most one $n \in\{2,3, \ldots, 7\}$ such that $a_{n}=0$. We will do casework on which $n$ in this range have $a_{n}=0$.
If $n \in\{3,4,5,6\}$, then we know that the terms $a_{n-1} a_{n}=a_{n} a_{n+1}=0$, so all other terms must be 1 , so $a_{1} a_{2}=a_{2} a_{3}=\ldots=a_{n-2} a_{n-1}=1$ and $a_{n+1} a_{n+2}=\ldots=a_{7} a_{8}=1$. Because every $a_{i}$ appears in one of these equations for $i \neq n$, then we must have $a_{i}=1$ for all $i \neq n$, so we have 1 possibility for each choice of $n$ and thus 4 possibilities total for this case.
If $n=2$, then again we have $a_{1} a_{2}=a_{2} a_{3}=0$, so we must have $a_{3} a_{4}=a_{4} a_{5}=\ldots=a_{7} a_{8}=1$, so $a_{3}=a_{4}=\ldots=a_{8}=1$. However, this time $a_{1}$ is not fixed, and we see that regardless of our choice of $a_{1}$ the sum will still be equal to 5 . Thus, since there are 2 choices for $a_{1}$, then there are 2 possibilities total for this case. The case where $n=7$ is analogous, with $a_{8}$ having 2 possibilities, so we have another 2 possibilities.
Finally, if $a_{n}=1$ for $n \in\{2,3, \ldots, 7\}$, then we will have $a_{2} a_{3}=a_{3} a_{4}=\ldots=a_{6} a_{7}=1$. We already have five terms equal to 1 , so the remaining two terms $a_{1} a_{2}$ and $a_{7} a_{8}$ must be 0 . Since $a_{2}=1$, then we must have $a_{1}=0$, and since $a_{7}=1$ then $a_{8}=0$. Thus, there is only 1 possibility for this case.
Summing, we have $4+2+2+1=9$ total sequences.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen1-solutions.jsonl",
"problem_match": "\n2. [3]",
"solution_match": "\nAnswer: "
}
|
2029eec9-9571-59db-b620-d5b45541236f
| 608,723
|
Triangle $A B C$ has $A B=5, B C=7$, and $C A=8$. New lines not containing but parallel to $A B$, $B C$, and $C A$ are drawn tangent to the incircle of $A B C$. What is the area of the hexagon formed by the sides of the original triangle and the newly drawn lines?
|
$\frac{31}{5} \sqrt{3}$
From the law of cosines we compute $\measuredangle A=\cos ^{-1}\left(\frac{5^{2}+8^{2}-7^{2}}{2(5)(8)}\right)=60^{\circ}$. Using brackets to denote the area of a region, we find that
$$
[A B C]=\frac{1}{2} A B \cdot A C \cdot \sin 60^{\circ}=10 \sqrt{3}
$$
The radius of the incircle can be computed by the formula
$$
r=\frac{2[A B C]}{A B+B C+C A}=\frac{20 \sqrt{3}}{20}=\sqrt{3}
$$
Now the height from $A$ to $B C$ is $\frac{2[A B C]}{B C}=\frac{20 \sqrt{3}}{7}$. Then the height from $A$ to $D E$ is $\frac{20 \sqrt{3}}{7}-2 r=\frac{6 \sqrt{3}}{7}$. Then $[A D E]=\left(\frac{6 \sqrt{3} / 7}{20 \sqrt{3} / 7}\right)^{2}[A B C]=\frac{9}{100}[A B C]$. Here, we use the fact that $\triangle A B C$ and $\triangle A D E$ are similar.
Similarly, we compute that the height from $B$ to $C A$ is $\frac{2[A B C]}{C A}=\frac{20 \sqrt{3}}{8}=\frac{5 \sqrt{3}}{2}$. Then the height from $B$ to $H J$ is $\frac{5 \sqrt{3}}{2}-2 r=\frac{\sqrt{3}}{2}$. Then $[B H J]=\left(\frac{\sqrt{3} / 2}{5 \sqrt{3} / 2}\right)^{2}[A B C]=\frac{1}{25}[A B C]$.
Finally, we compute that the height from $C$ to $A B$ is $\frac{2[A B C]}{5}=\frac{20 \sqrt{3}}{5}=4 \sqrt{3}$. Then the height from $C$ to $F G$ is $4 \sqrt{3}-2 r=2 \sqrt{3}$. Then $[C F G]=\left(\frac{2 \sqrt{3}}{4 \sqrt{3}}\right)^{2}[A B C]=\frac{1}{4}[A B C]$.
Finally we can compute the area of hexagon $D E F G H J$. We have
$$
\begin{gathered}
{[D E F G H J]=[A B C]-[A D E]-[B H J]-[C F G]=[A B C]\left(1-\frac{9}{100}-\frac{1}{25}-\frac{1}{4}\right)=[A B C]\left(\frac{31}{50}\right)=} \\
10 \sqrt{3}\left(\frac{31}{50}\right)=\frac{31}{5} \sqrt{3} .
\end{gathered}
$$
|
\frac{31}{5} \sqrt{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Triangle $A B C$ has $A B=5, B C=7$, and $C A=8$. New lines not containing but parallel to $A B$, $B C$, and $C A$ are drawn tangent to the incircle of $A B C$. What is the area of the hexagon formed by the sides of the original triangle and the newly drawn lines?
|
$\frac{31}{5} \sqrt{3}$
From the law of cosines we compute $\measuredangle A=\cos ^{-1}\left(\frac{5^{2}+8^{2}-7^{2}}{2(5)(8)}\right)=60^{\circ}$. Using brackets to denote the area of a region, we find that
$$
[A B C]=\frac{1}{2} A B \cdot A C \cdot \sin 60^{\circ}=10 \sqrt{3}
$$
The radius of the incircle can be computed by the formula
$$
r=\frac{2[A B C]}{A B+B C+C A}=\frac{20 \sqrt{3}}{20}=\sqrt{3}
$$
Now the height from $A$ to $B C$ is $\frac{2[A B C]}{B C}=\frac{20 \sqrt{3}}{7}$. Then the height from $A$ to $D E$ is $\frac{20 \sqrt{3}}{7}-2 r=\frac{6 \sqrt{3}}{7}$. Then $[A D E]=\left(\frac{6 \sqrt{3} / 7}{20 \sqrt{3} / 7}\right)^{2}[A B C]=\frac{9}{100}[A B C]$. Here, we use the fact that $\triangle A B C$ and $\triangle A D E$ are similar.
Similarly, we compute that the height from $B$ to $C A$ is $\frac{2[A B C]}{C A}=\frac{20 \sqrt{3}}{8}=\frac{5 \sqrt{3}}{2}$. Then the height from $B$ to $H J$ is $\frac{5 \sqrt{3}}{2}-2 r=\frac{\sqrt{3}}{2}$. Then $[B H J]=\left(\frac{\sqrt{3} / 2}{5 \sqrt{3} / 2}\right)^{2}[A B C]=\frac{1}{25}[A B C]$.
Finally, we compute that the height from $C$ to $A B$ is $\frac{2[A B C]}{5}=\frac{20 \sqrt{3}}{5}=4 \sqrt{3}$. Then the height from $C$ to $F G$ is $4 \sqrt{3}-2 r=2 \sqrt{3}$. Then $[C F G]=\left(\frac{2 \sqrt{3}}{4 \sqrt{3}}\right)^{2}[A B C]=\frac{1}{4}[A B C]$.
Finally we can compute the area of hexagon $D E F G H J$. We have
$$
\begin{gathered}
{[D E F G H J]=[A B C]-[A D E]-[B H J]-[C F G]=[A B C]\left(1-\frac{9}{100}-\frac{1}{25}-\frac{1}{4}\right)=[A B C]\left(\frac{31}{50}\right)=} \\
10 \sqrt{3}\left(\frac{31}{50}\right)=\frac{31}{5} \sqrt{3} .
\end{gathered}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen1-solutions.jsonl",
"problem_match": "\n3. [3]",
"solution_match": "\nAnswer: "
}
|
e153fd02-15d4-5517-ae2b-a3c28973626c
| 608,724
|
An ant starts at the point $(1,0)$. Each minute, it walks from its current position to one of the four adjacent lattice points until it reaches a point $(x, y)$ with $|x|+|y| \geq 2$. What is the probability that the ant ends at the point $(1,1)$ ?
|
$\frac{7}{24}$ From the starting point of $(1,0)$, there is a $\frac{1}{4}$ chance we will go directly to $(1,1)$, a $\frac{1}{2}$ chance we will end at $(2,0)$ or $(1,-1)$, and a $\frac{1}{4}$ chance we will go to $(0,0)$. Thus, if $p$ is the probability that we will reach $(1,1)$ from $(0,0)$, then the desired probability is equal to $\frac{1}{4}+\frac{1}{4} p$, so we need only calculate $p$. Note that we can replace the condition $|x|+|y| \geq 2$ by $|x|+|y|=2$, since in each iteration the quantity $|x|+|y|$ can increase by at most 1 . Thus, we only have to consider the eight points $(2,0),(1,1),(0,2),(-1,1),(-2,0),(-1,-1),(0,-2),(1,-1)$. Let $p_{1}, p_{2}, \ldots, p_{8}$ be the probability of
reaching each of these points from $(0,0)$, respectively. By symmetry, we see that $p_{1}=p_{3}=p_{5}=p_{7}$ and $p_{2}=p_{4}=p_{6}=p_{8}$. We also know that there are two paths from $(0,0)$ to $(1,1)$ and one path from $(0,0)$ to $(2,0)$, thus $p_{2}=2 p_{1}$. Because the sum of all probabilities is 1 , we have $p_{1}+p_{2}+\ldots+p_{8}=1$. Combining these equations, we see that $4 p_{1}+4 p_{2}=12 p_{1}=1$, so $p_{1}=\frac{1}{12}$ and $p_{2}=\frac{1}{6}$. Since $p=p_{2}=\frac{1}{6}$, then the final answer is $\frac{1}{4}+\frac{1}{4} \cdot \frac{1}{6}=\frac{7}{24}$
|
\frac{7}{24}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
An ant starts at the point $(1,0)$. Each minute, it walks from its current position to one of the four adjacent lattice points until it reaches a point $(x, y)$ with $|x|+|y| \geq 2$. What is the probability that the ant ends at the point $(1,1)$ ?
|
$\frac{7}{24}$ From the starting point of $(1,0)$, there is a $\frac{1}{4}$ chance we will go directly to $(1,1)$, a $\frac{1}{2}$ chance we will end at $(2,0)$ or $(1,-1)$, and a $\frac{1}{4}$ chance we will go to $(0,0)$. Thus, if $p$ is the probability that we will reach $(1,1)$ from $(0,0)$, then the desired probability is equal to $\frac{1}{4}+\frac{1}{4} p$, so we need only calculate $p$. Note that we can replace the condition $|x|+|y| \geq 2$ by $|x|+|y|=2$, since in each iteration the quantity $|x|+|y|$ can increase by at most 1 . Thus, we only have to consider the eight points $(2,0),(1,1),(0,2),(-1,1),(-2,0),(-1,-1),(0,-2),(1,-1)$. Let $p_{1}, p_{2}, \ldots, p_{8}$ be the probability of
reaching each of these points from $(0,0)$, respectively. By symmetry, we see that $p_{1}=p_{3}=p_{5}=p_{7}$ and $p_{2}=p_{4}=p_{6}=p_{8}$. We also know that there are two paths from $(0,0)$ to $(1,1)$ and one path from $(0,0)$ to $(2,0)$, thus $p_{2}=2 p_{1}$. Because the sum of all probabilities is 1 , we have $p_{1}+p_{2}+\ldots+p_{8}=1$. Combining these equations, we see that $4 p_{1}+4 p_{2}=12 p_{1}=1$, so $p_{1}=\frac{1}{12}$ and $p_{2}=\frac{1}{6}$. Since $p=p_{2}=\frac{1}{6}$, then the final answer is $\frac{1}{4}+\frac{1}{4} \cdot \frac{1}{6}=\frac{7}{24}$
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen1-solutions.jsonl",
"problem_match": "\n4. [4]",
"solution_match": "\nAnswer: "
}
|
53d69feb-6077-5181-866b-c78215a7a196
| 608,725
|
A polynomial $P$ is of the form $\pm x^{6} \pm x^{5} \pm x^{4} \pm x^{3} \pm x^{2} \pm x \pm 1$. Given that $P(2)=27$, what is $P(3)$ ?
|
439 We use the following lemma:
Lemma. The sign of $\pm 2^{n} \pm 2^{n-1} \pm \cdots \pm 2 \pm 1$ is the same as the sign of the $2^{n}$ term.
Proof. Without loss of generality, let $2^{n}$ be positive. (We can flip all signs.) Notice that $2^{n} \pm 2^{n-1} \pm$ $2^{n-2} \pm \cdots 2 \pm 1 \geq 2^{n}-2^{n-1}-2^{n-2}-\cdots-2-1=1$, which is positive.
We can use this lemma to uniquely determine the signs of $P$. Since our desired sum, 27, is positive, the coefficient of $x^{6}$ must be positive. Subtracting 64 , we now have that $\pm 2^{5} \pm 2^{4} \pm \ldots \pm 2 \pm 1=-37$, so the sign of $2^{5}$ must be negative. Continuing in this manner, we find that $P(x)=x^{6}-x^{5}-x^{4}+x^{3}+x^{2}-x+1$, so $P(3)=3^{6}-3^{5}-3^{4}+3^{3}+3^{2}-3+1=439$.
|
439
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
A polynomial $P$ is of the form $\pm x^{6} \pm x^{5} \pm x^{4} \pm x^{3} \pm x^{2} \pm x \pm 1$. Given that $P(2)=27$, what is $P(3)$ ?
|
439 We use the following lemma:
Lemma. The sign of $\pm 2^{n} \pm 2^{n-1} \pm \cdots \pm 2 \pm 1$ is the same as the sign of the $2^{n}$ term.
Proof. Without loss of generality, let $2^{n}$ be positive. (We can flip all signs.) Notice that $2^{n} \pm 2^{n-1} \pm$ $2^{n-2} \pm \cdots 2 \pm 1 \geq 2^{n}-2^{n-1}-2^{n-2}-\cdots-2-1=1$, which is positive.
We can use this lemma to uniquely determine the signs of $P$. Since our desired sum, 27, is positive, the coefficient of $x^{6}$ must be positive. Subtracting 64 , we now have that $\pm 2^{5} \pm 2^{4} \pm \ldots \pm 2 \pm 1=-37$, so the sign of $2^{5}$ must be negative. Continuing in this manner, we find that $P(x)=x^{6}-x^{5}-x^{4}+x^{3}+x^{2}-x+1$, so $P(3)=3^{6}-3^{5}-3^{4}+3^{3}+3^{2}-3+1=439$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen1-solutions.jsonl",
"problem_match": "\n5. [5]",
"solution_match": "\nAnswer: "
}
|
e46b046c-1db1-594a-ae1d-f2e470e92c97
| 608,726
|
What is the sum of the positive solutions to $2 x^{2}-x\lfloor x\rfloor=5$, where $\lfloor x\rfloor$ is the largest integer less than or equal to $x$ ?
|
$\frac{3+\sqrt{41}+2 \sqrt{11}}{4}$ We first note that $\lfloor x\rfloor \leq x$, so $2 x^{2}-x\lfloor x\rfloor \geq 2 x^{2}-x^{2}=x^{2}$. Since this function is increasing on the positive reals, all solutions must be at most $\sqrt{5}$. This gives us 3 possible values of $\lfloor x\rfloor: 0,1$, and 2 .
If $\lfloor x\rfloor=0$, then our equation becomes $2 x^{2}=5$, which has positive solution $x=\sqrt{\frac{5}{2}}$. This number is greater than 1 , so its floor is not 0 ; thus, there are no solutions in this case.
If $\lfloor x\rfloor=1$, then our equation becomes $2 x^{2}-x=5$. Using the quadratic formula, we find the positive solution $x=\frac{1+\sqrt{41}}{4}$. Since $3<\sqrt{41}<7$, this number is between 1 and 2 , so it satisfies the equation.
If $\lfloor x\rfloor=2$, then our equation becomes $2 x^{2}-2 x=5$. We find the positive solution $x=\frac{1+\sqrt{11}}{2}$. Since $3<\sqrt{11}<5$, this number is between 2 and 3 , so it satisfies the equation.
We then find that the sum of positive solutions is $\frac{1+\sqrt{41}}{4}+\frac{1+\sqrt{11}}{2}=\frac{3+\sqrt{41}+2 \sqrt{11}}{4}$.
|
\frac{3+\sqrt{41}+2 \sqrt{11}}{4}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
What is the sum of the positive solutions to $2 x^{2}-x\lfloor x\rfloor=5$, where $\lfloor x\rfloor$ is the largest integer less than or equal to $x$ ?
|
$\frac{3+\sqrt{41}+2 \sqrt{11}}{4}$ We first note that $\lfloor x\rfloor \leq x$, so $2 x^{2}-x\lfloor x\rfloor \geq 2 x^{2}-x^{2}=x^{2}$. Since this function is increasing on the positive reals, all solutions must be at most $\sqrt{5}$. This gives us 3 possible values of $\lfloor x\rfloor: 0,1$, and 2 .
If $\lfloor x\rfloor=0$, then our equation becomes $2 x^{2}=5$, which has positive solution $x=\sqrt{\frac{5}{2}}$. This number is greater than 1 , so its floor is not 0 ; thus, there are no solutions in this case.
If $\lfloor x\rfloor=1$, then our equation becomes $2 x^{2}-x=5$. Using the quadratic formula, we find the positive solution $x=\frac{1+\sqrt{41}}{4}$. Since $3<\sqrt{41}<7$, this number is between 1 and 2 , so it satisfies the equation.
If $\lfloor x\rfloor=2$, then our equation becomes $2 x^{2}-2 x=5$. We find the positive solution $x=\frac{1+\sqrt{11}}{2}$. Since $3<\sqrt{11}<5$, this number is between 2 and 3 , so it satisfies the equation.
We then find that the sum of positive solutions is $\frac{1+\sqrt{41}}{4}+\frac{1+\sqrt{11}}{2}=\frac{3+\sqrt{41}+2 \sqrt{11}}{4}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen1-solutions.jsonl",
"problem_match": "\n6. [5]",
"solution_match": "\nAnswer: "
}
|
63a3f3f5-12cc-520c-9f94-005e053459b3
| 608,727
|
What is the remainder when $(1+x)^{2010}$ is divided by $1+x+x^{2}$ ?
|
1 We use polynomial congruence $\bmod 1+x+x^{2}$ to find the desired remainder. Since $x^{2}+x+1 \mid x^{3}-1$, we have that $x^{3} \equiv 1\left(\bmod 1+x+x^{2}\right)$. Now:
$$
\begin{aligned}
(1+x)^{2010} & \equiv\left(-x^{2}\right)^{2010} \quad\left(\bmod 1+x+x^{2}\right) \\
& \equiv x^{4020} \quad\left(\bmod 1+x+x^{2}\right) \\
& \equiv\left(x^{3}\right)^{1340} \quad\left(\bmod 1+x+x^{2}\right) \\
& \equiv 1^{1340} \quad\left(\bmod 1+x+x^{2}\right) \\
& \equiv 1 \quad\left(\bmod 1+x+x^{2}\right)
\end{aligned}
$$
Thus, the answer is 1 .
|
1
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
What is the remainder when $(1+x)^{2010}$ is divided by $1+x+x^{2}$ ?
|
1 We use polynomial congruence $\bmod 1+x+x^{2}$ to find the desired remainder. Since $x^{2}+x+1 \mid x^{3}-1$, we have that $x^{3} \equiv 1\left(\bmod 1+x+x^{2}\right)$. Now:
$$
\begin{aligned}
(1+x)^{2010} & \equiv\left(-x^{2}\right)^{2010} \quad\left(\bmod 1+x+x^{2}\right) \\
& \equiv x^{4020} \quad\left(\bmod 1+x+x^{2}\right) \\
& \equiv\left(x^{3}\right)^{1340} \quad\left(\bmod 1+x+x^{2}\right) \\
& \equiv 1^{1340} \quad\left(\bmod 1+x+x^{2}\right) \\
& \equiv 1 \quad\left(\bmod 1+x+x^{2}\right)
\end{aligned}
$$
Thus, the answer is 1 .
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen1-solutions.jsonl",
"problem_match": "\n7. [6]",
"solution_match": "\nAnswer: "
}
|
204ad520-ba1a-5dd5-87f6-d8ff5522ec1b
| 608,728
|
Two circles with radius one are drawn in the coordinate plane, one with center $(0,1)$ and the other with center $(2, y)$, for some real number $y$ between 0 and 1 . A third circle is drawn so as to be tangent to both of the other two circles as well as the $x$ axis. What is the smallest possible radius for this third circle?
|
$3-2 \sqrt{2}$ Suppose that the smaller circle has radius $r$. Call the three circles (in order from left to right) $O_{1}, O_{2}$, and $O_{3}$. The distance between the centers of $O_{1}$ and $O_{2}$ is $1+r$, and the distance in their $y$-coordinates is $1-r$. Therefore, by the Pythagorean theorem, the difference in $x$-coordinates
is $\sqrt{(1+r)^{2}-(1-r)^{2}}=2 \sqrt{r}$, which means that $O_{2}$ has a center at $(2 \sqrt{r}, r)$. But $O_{2}$ is also tangent to $O_{3}$, which means that the difference in $x$-coordinate from the right-most point of $O_{2}$ to the center of $O_{3}$ is at most 1. Therefore, the center of $O_{3}$ has an $x$-coordinate of at most $2 \sqrt{r}+r+1$, meaning that $2 \sqrt{r}+r+1 \leq 2$. We can use the quadratic formula to see that this implies that $\sqrt{r} \leq \sqrt{2}-1$, so $r \leq 3-2 \sqrt{2}$. We can achieve equality by placing the center of $O_{3}$ at $(2, r)$ (which in this case is $(2,3-2 \sqrt{2}))$.
|
3-2 \sqrt{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Two circles with radius one are drawn in the coordinate plane, one with center $(0,1)$ and the other with center $(2, y)$, for some real number $y$ between 0 and 1 . A third circle is drawn so as to be tangent to both of the other two circles as well as the $x$ axis. What is the smallest possible radius for this third circle?
|
$3-2 \sqrt{2}$ Suppose that the smaller circle has radius $r$. Call the three circles (in order from left to right) $O_{1}, O_{2}$, and $O_{3}$. The distance between the centers of $O_{1}$ and $O_{2}$ is $1+r$, and the distance in their $y$-coordinates is $1-r$. Therefore, by the Pythagorean theorem, the difference in $x$-coordinates
is $\sqrt{(1+r)^{2}-(1-r)^{2}}=2 \sqrt{r}$, which means that $O_{2}$ has a center at $(2 \sqrt{r}, r)$. But $O_{2}$ is also tangent to $O_{3}$, which means that the difference in $x$-coordinate from the right-most point of $O_{2}$ to the center of $O_{3}$ is at most 1. Therefore, the center of $O_{3}$ has an $x$-coordinate of at most $2 \sqrt{r}+r+1$, meaning that $2 \sqrt{r}+r+1 \leq 2$. We can use the quadratic formula to see that this implies that $\sqrt{r} \leq \sqrt{2}-1$, so $r \leq 3-2 \sqrt{2}$. We can achieve equality by placing the center of $O_{3}$ at $(2, r)$ (which in this case is $(2,3-2 \sqrt{2}))$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen1-solutions.jsonl",
"problem_match": "\n8. [7]",
"solution_match": "\nAnswer: "
}
|
ba86a616-9374-50ea-81e4-ae4b5458bfed
| 608,729
|
What is the sum of all numbers between 0 and 511 inclusive that have an even number of 1 s when written in binary?
|
65408 Call a digit in the binary representation of a number a bit. We claim that for any given $i$ between 0 and 8 , there are 128 numbers with an even number of 1 s that have a 1 in the bit representing $2^{i}$. To prove this, we simply make that bit a 1 , then consider all possible configurations of the other bits, excluding the last bit (or the second-last bit if our given bit is already the last bit). The last bit will then be restricted to satisfy the parity condition on the number of 1 s . As there are 128 possible configurations of all the bits but two, we find 128 possible numbers, proving our claim.
Therefore, each bit is present as a 1 in 128 numbers in the sum, so the bit representing $2^{i}$ contributes $128 \cdot 2^{i}$ to our sum. Summing over all $0 \leq i \leq 8$, we find the answer to be $128(1+2+\ldots+128)=$ $128 \cdot 511=65408$.
|
65408
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
What is the sum of all numbers between 0 and 511 inclusive that have an even number of 1 s when written in binary?
|
65408 Call a digit in the binary representation of a number a bit. We claim that for any given $i$ between 0 and 8 , there are 128 numbers with an even number of 1 s that have a 1 in the bit representing $2^{i}$. To prove this, we simply make that bit a 1 , then consider all possible configurations of the other bits, excluding the last bit (or the second-last bit if our given bit is already the last bit). The last bit will then be restricted to satisfy the parity condition on the number of 1 s . As there are 128 possible configurations of all the bits but two, we find 128 possible numbers, proving our claim.
Therefore, each bit is present as a 1 in 128 numbers in the sum, so the bit representing $2^{i}$ contributes $128 \cdot 2^{i}$ to our sum. Summing over all $0 \leq i \leq 8$, we find the answer to be $128(1+2+\ldots+128)=$ $128 \cdot 511=65408$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen1-solutions.jsonl",
"problem_match": "\n9. [7]",
"solution_match": "\nAnswer: "
}
|
d77228a4-1d67-58a5-9f96-a778eeda57e7
| 608,730
|
You are given two diameters $A B$ and $C D$ of circle $\Omega$ with radius 1. A circle is drawn in one of the smaller sectors formed such that it is tangent to $A B$ at $E$, tangent to $C D$ at $F$, and tangent to $\Omega$ at $P$. Lines $P E$ and $P F$ intersect $\Omega$ again at $X$ and $Y$. What is the length of $X Y$, given that $A C=\frac{2}{3}$ ?
|
\( \frac{4\sqrt{2}}{3} \). Let \( O \) denote the center of circle \( \Omega \). We first prove that \( OX \perp AB \) and \( OY \perp CD \). Consider the homothety about $P$ which maps the smaller circle to $\Omega$. This homothety takes $E$ to $X$ and also takes $A B$ to the line tangent to circle $\Omega$ parallel to $A B$. Therefore, $X$ is the midpoint of the arc $A B$, and so $O X \perp A B$. Similarly, $O Y \perp C D$.
Let $\theta=\angle A O C$. By the Law of Sines, we have $A C=2 \sin \frac{\theta}{2}$. Thus, $\sin \frac{\theta}{2}=\frac{1}{3}$, and $\cos \frac{\theta}{2}=\sqrt{1-\left(\frac{1}{3}\right)^{2}}=$ $\frac{2 \sqrt{2}}{3}$. Therefore,
$$
\begin{aligned}
X Y & =2 \sin \frac{\angle X O Y}{2} \\
& =2 \sin \left(90^{\circ}-\frac{\theta}{2}\right) \\
& =2 \cos \frac{\theta}{2} \\
& =\frac{4 \sqrt{2}}{3}
\end{aligned}
$$
|
\frac{4\sqrt{2}}{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
You are given two diameters $A B$ and $C D$ of circle $\Omega$ with radius 1. A circle is drawn in one of the smaller sectors formed such that it is tangent to $A B$ at $E$, tangent to $C D$ at $F$, and tangent to $\Omega$ at $P$. Lines $P E$ and $P F$ intersect $\Omega$ again at $X$ and $Y$. What is the length of $X Y$, given that $A C=\frac{2}{3}$ ?
|
\( \frac{4\sqrt{2}}{3} \). Let \( O \) denote the center of circle \( \Omega \). We first prove that \( OX \perp AB \) and \( OY \perp CD \). Consider the homothety about $P$ which maps the smaller circle to $\Omega$. This homothety takes $E$ to $X$ and also takes $A B$ to the line tangent to circle $\Omega$ parallel to $A B$. Therefore, $X$ is the midpoint of the arc $A B$, and so $O X \perp A B$. Similarly, $O Y \perp C D$.
Let $\theta=\angle A O C$. By the Law of Sines, we have $A C=2 \sin \frac{\theta}{2}$. Thus, $\sin \frac{\theta}{2}=\frac{1}{3}$, and $\cos \frac{\theta}{2}=\sqrt{1-\left(\frac{1}{3}\right)^{2}}=$ $\frac{2 \sqrt{2}}{3}$. Therefore,
$$
\begin{aligned}
X Y & =2 \sin \frac{\angle X O Y}{2} \\
& =2 \sin \left(90^{\circ}-\frac{\theta}{2}\right) \\
& =2 \cos \frac{\theta}{2} \\
& =\frac{4 \sqrt{2}}{3}
\end{aligned}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen1-solutions.jsonl",
"problem_match": "\n10. [8]",
"solution_match": "\nAnswer: "
}
|
035c25cd-e077-5406-b3d4-08949de2dfb8
| 608,731
|
16 progamers are playing in a single elimination tournament. Each player has a different skill level and when two play against each other the one with the higher skill level will always win. Each round, each progamer plays a match against another and the loser is eliminated. This continues until only one remains. How many different progamers can reach the round that has 2 players remaining?
|
9 Each finalist must be better than the person he beat in the semifinals, both of the people they beat in the second round, and all 4 of the people any of those people beat in the first round. So, none of the 7 worst players can possibly make it to the finals. Any of the 9 best players can make it to the finals if the other 8 of the best 9 play each other in all rounds before the finals. So, exactly 9 people are capable of making it to the finals.
|
9
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
16 progamers are playing in a single elimination tournament. Each player has a different skill level and when two play against each other the one with the higher skill level will always win. Each round, each progamer plays a match against another and the loser is eliminated. This continues until only one remains. How many different progamers can reach the round that has 2 players remaining?
|
9 Each finalist must be better than the person he beat in the semifinals, both of the people they beat in the second round, and all 4 of the people any of those people beat in the first round. So, none of the 7 worst players can possibly make it to the finals. Any of the 9 best players can make it to the finals if the other 8 of the best 9 play each other in all rounds before the finals. So, exactly 9 people are capable of making it to the finals.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen2-solutions.jsonl",
"problem_match": "\n1. [3]",
"solution_match": "\nAnswer: "
}
|
2a80ec22-2cf7-571f-b97b-b1a5e30cdf19
| 608,732
|
16 progamers are playing in another single elimination tournament. Each round, each of the remaining progamers plays against another and the loser is eliminated. Additionally, each time a progamer wins, he will have a ceremony to celebrate. A player's first ceremony is ten seconds long, and afterward each ceremony is ten seconds longer than the last. What is the total length in seconds of all the ceremonies over the entire tournament?
|
260 At the end of the first round, each of the 8 winners has a 10 second ceremony. After the second round, the 4 winners have a 20 second ceremony. The two remaining players have 30 second ceremonies after the third round, and the winner has a 40 second ceremony after the finals. So, all of the ceremonies combined take $8 \cdot 10+4 \cdot 20+2 \cdot 30+40=260$ seconds.
|
260
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
16 progamers are playing in another single elimination tournament. Each round, each of the remaining progamers plays against another and the loser is eliminated. Additionally, each time a progamer wins, he will have a ceremony to celebrate. A player's first ceremony is ten seconds long, and afterward each ceremony is ten seconds longer than the last. What is the total length in seconds of all the ceremonies over the entire tournament?
|
260 At the end of the first round, each of the 8 winners has a 10 second ceremony. After the second round, the 4 winners have a 20 second ceremony. The two remaining players have 30 second ceremonies after the third round, and the winner has a 40 second ceremony after the finals. So, all of the ceremonies combined take $8 \cdot 10+4 \cdot 20+2 \cdot 30+40=260$ seconds.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen2-solutions.jsonl",
"problem_match": "\n2. [4]",
"solution_match": "\nAnswer: "
}
|
bc788a94-5b03-504f-bf7c-eb39ae1e5e0b
| 608,733
|
Dragoons take up $1 \times 1$ squares in the plane with sides parallel to the coordinate axes such that the interiors of the squares do not intersect. A dragoon can fire at another dragoon if the difference in the $x$-coordinates of their centers and the difference in the $y$-coordinates of their centers are both at most 6 , regardless of any dragoons in between. For example, a dragoon centered at $(4,5)$ can fire at a dragoon centered at the origin, but a dragoon centered at $(7,0)$ can not. A dragoon cannot fire at itself. What is the maximum number of dragoons that can fire at a single dragoon simultaneously?
|
168 Assign coordinates in such a way that the dragoon being fired on is centered at $(0,0)$. Any dragoon firing at it must have a center with $x$-coordinates and $y$-coordinates that are no smaller than -6 and no greater than 6 . That means that every dragoon firing at it must lie entirely in the region bounded by the lines $x=-6.5, x=6.5, y=-6.5$, and $y=6.5$. This is a square with sides of length 13 , so there is room for exactly 169 dragoons in it. One of them is the dragoon being fired on, so there are at most 168 dragoons firing at it.
|
168
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Dragoons take up $1 \times 1$ squares in the plane with sides parallel to the coordinate axes such that the interiors of the squares do not intersect. A dragoon can fire at another dragoon if the difference in the $x$-coordinates of their centers and the difference in the $y$-coordinates of their centers are both at most 6 , regardless of any dragoons in between. For example, a dragoon centered at $(4,5)$ can fire at a dragoon centered at the origin, but a dragoon centered at $(7,0)$ can not. A dragoon cannot fire at itself. What is the maximum number of dragoons that can fire at a single dragoon simultaneously?
|
168 Assign coordinates in such a way that the dragoon being fired on is centered at $(0,0)$. Any dragoon firing at it must have a center with $x$-coordinates and $y$-coordinates that are no smaller than -6 and no greater than 6 . That means that every dragoon firing at it must lie entirely in the region bounded by the lines $x=-6.5, x=6.5, y=-6.5$, and $y=6.5$. This is a square with sides of length 13 , so there is room for exactly 169 dragoons in it. One of them is the dragoon being fired on, so there are at most 168 dragoons firing at it.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen2-solutions.jsonl",
"problem_match": "\n3. [5]",
"solution_match": "\nAnswer: "
}
|
45c68351-c951-5ea9-9079-0d0e80ba30f0
| 608,734
|
A zerg player can produce one zergling every minute and a protoss player can produce one zealot every 2.1 minutes. Both players begin building their respective units immediately from the beginning of the game. In a fight, a zergling army overpowers a zealot army if the ratio of zerglings to zealots is more than 3 . What is the total amount of time (in minutes) during the game such that at that time the zergling army would overpower the zealot army?
|
1.3 At the end of the first minute, the zerg player produces a zergling and has a superior army for the 1.1 minutes before the protoss player produces the first zealot. At this point, the zealot is at least a match for the zerglings until the fourth is produced 4 minutes into the game. Then, the zerg army has the advantage for the .2 minutes before a second zealot is produced. A third zealot will be produced 6.3 minutes into the game, which will be before the zerg player accumulates the 7 zerglings needed to overwhelm the first 2 zealots. After this, the zerglings will never regain the advantage because the zerg player can never produce 3 more zerglings to counter the last zealot before another one is produced. So, the zerg player will have the military advantage for $1.1+.2=1.3$ minutes.
|
1.3
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
A zerg player can produce one zergling every minute and a protoss player can produce one zealot every 2.1 minutes. Both players begin building their respective units immediately from the beginning of the game. In a fight, a zergling army overpowers a zealot army if the ratio of zerglings to zealots is more than 3 . What is the total amount of time (in minutes) during the game such that at that time the zergling army would overpower the zealot army?
|
1.3 At the end of the first minute, the zerg player produces a zergling and has a superior army for the 1.1 minutes before the protoss player produces the first zealot. At this point, the zealot is at least a match for the zerglings until the fourth is produced 4 minutes into the game. Then, the zerg army has the advantage for the .2 minutes before a second zealot is produced. A third zealot will be produced 6.3 minutes into the game, which will be before the zerg player accumulates the 7 zerglings needed to overwhelm the first 2 zealots. After this, the zerglings will never regain the advantage because the zerg player can never produce 3 more zerglings to counter the last zealot before another one is produced. So, the zerg player will have the military advantage for $1.1+.2=1.3$ minutes.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen2-solutions.jsonl",
"problem_match": "\n4. [5]",
"solution_match": "\nAnswer: "
}
|
e51f60a1-5df1-5aec-9164-b5408ed13a44
| 608,735
|
There are 111 StarCraft progamers. The StarCraft team SKT starts with a given set of eleven progamers on it, and at the end of each season, it drops a progamer and adds a progamer (possibly the same one). At the start of the second season, SKT has to field a team of five progamers to play the opening match. How many different lineups of five players could be fielded if the order of players on the lineup matters?
|
4015440 We disregard the order of the players, multiplying our answer by $5!=120$ at the end to account for it. Clearly, SKT will be able to field at most 1 player not in the original set of eleven players. If it does not field a new player, then it has $\binom{11}{5}=462$ choices. If it does field a new player, then it has 100 choices for the new player, and $\binom{11}{4}=330$ choices for the 4 other players, giving 33000 possibilities. Thus, SKT can field at most $33000+462=33462$ unordered lineups, and multiplying this by 120 , we find the answer to be 4015440 .
## Unfair Coins
|
4015440
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
There are 111 StarCraft progamers. The StarCraft team SKT starts with a given set of eleven progamers on it, and at the end of each season, it drops a progamer and adds a progamer (possibly the same one). At the start of the second season, SKT has to field a team of five progamers to play the opening match. How many different lineups of five players could be fielded if the order of players on the lineup matters?
|
4015440 We disregard the order of the players, multiplying our answer by $5!=120$ at the end to account for it. Clearly, SKT will be able to field at most 1 player not in the original set of eleven players. If it does not field a new player, then it has $\binom{11}{5}=462$ choices. If it does field a new player, then it has 100 choices for the new player, and $\binom{11}{4}=330$ choices for the 4 other players, giving 33000 possibilities. Thus, SKT can field at most $33000+462=33462$ unordered lineups, and multiplying this by 120 , we find the answer to be 4015440 .
## Unfair Coins
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen2-solutions.jsonl",
"problem_match": "\n5. [7]",
"solution_match": "\nAnswer: "
}
|
855bfb82-438f-5491-bed0-c9cefc0265e5
| 608,736
|
When flipped, a coin has a probability $p$ of landing heads. When flipped twice, it is twice as likely to land on the same side both times as it is to land on each side once. What is the larger possible value of $p$ ?
|
$\frac{3+\sqrt{3}}{6}$ The probability that the coin will land on the same side twice is $p^{2}+(1-p)^{2}=$ $2 p^{2}-2 p+1$. The probability that the coin will land on each side once is $p(p-1)+(p-1) p=$ $2 p(1-p)=2 p-2 p^{2}$. We are told that it is twice as likely to land on the same side both times, so $2 p^{2}-2 p+1=2\left(2 p-2 p^{2}\right)=4 p-4 p^{2}$. Solving, we get $6 p^{2}-6 p+1=0$, so $p=\frac{3 \pm \sqrt{3}}{6}$. The larger solution is $p=\frac{3+\sqrt{3}}{6}$.
|
\frac{3+\sqrt{3}}{6}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
When flipped, a coin has a probability $p$ of landing heads. When flipped twice, it is twice as likely to land on the same side both times as it is to land on each side once. What is the larger possible value of $p$ ?
|
$\frac{3+\sqrt{3}}{6}$ The probability that the coin will land on the same side twice is $p^{2}+(1-p)^{2}=$ $2 p^{2}-2 p+1$. The probability that the coin will land on each side once is $p(p-1)+(p-1) p=$ $2 p(1-p)=2 p-2 p^{2}$. We are told that it is twice as likely to land on the same side both times, so $2 p^{2}-2 p+1=2\left(2 p-2 p^{2}\right)=4 p-4 p^{2}$. Solving, we get $6 p^{2}-6 p+1=0$, so $p=\frac{3 \pm \sqrt{3}}{6}$. The larger solution is $p=\frac{3+\sqrt{3}}{6}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen2-solutions.jsonl",
"problem_match": "\n6. [4]",
"solution_match": "\nAnswer: "
}
|
6fd1c393-4334-5a1e-8338-3d1c4537bda7
| 608,737
|
George has two coins, one of which is fair and the other of which always comes up heads. Jacob takes one of them at random and flips it twice. Given that it came up heads both times, what is the probability that it is the coin that always comes up heads?
|
\( \frac{4}{5} \). In general, \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( P(A|B) \) is the probability of \( A \) given \( B \) and $P(A \cap B)$ is the probability of $A$ and $B$ (See http://en.wikipedia.org/wiki/Conditional_probability for more information). If $A$ is the event of selecting the "double-headed" coin and $B$ is the event of Jacob flipping two heads, then $P(A \cap B)=\left(\frac{1}{2}\right)(1)$, since there is a $\frac{1}{2}$ chance of picking the doubleheaded coin and Jacob will always flip two heads when he has it. By conditional probability, $P(B)=$ $\left(\frac{1}{2}\right)(1)+\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)=\frac{5}{8}$, so $P(A \mid B)=\frac{1 / 2}{5 / 8}=\frac{4}{5}$.
|
\frac{4}{5}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
George has two coins, one of which is fair and the other of which always comes up heads. Jacob takes one of them at random and flips it twice. Given that it came up heads both times, what is the probability that it is the coin that always comes up heads?
|
\( \frac{4}{5} \). In general, \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( P(A|B) \) is the probability of \( A \) given \( B \) and $P(A \cap B)$ is the probability of $A$ and $B$ (See http://en.wikipedia.org/wiki/Conditional_probability for more information). If $A$ is the event of selecting the "double-headed" coin and $B$ is the event of Jacob flipping two heads, then $P(A \cap B)=\left(\frac{1}{2}\right)(1)$, since there is a $\frac{1}{2}$ chance of picking the doubleheaded coin and Jacob will always flip two heads when he has it. By conditional probability, $P(B)=$ $\left(\frac{1}{2}\right)(1)+\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)=\frac{5}{8}$, so $P(A \mid B)=\frac{1 / 2}{5 / 8}=\frac{4}{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen2-solutions.jsonl",
"problem_match": "\n7. [4]",
"solution_match": "\nAnswer: "
}
|
93ad6466-15d8-57e2-be88-b50d63f29d7d
| 74,267
|
Newton and Leibniz are playing a game with a coin that comes up heads with probability $p$. They take turns flipping the coin until one of them wins with Newton going first. Newton wins if he flips a heads and Leibniz wins if he flips a tails. Given that Newton and Leibniz each win the game half of the time, what is the probability $p$ ?
|
$\frac{\frac{3-\sqrt{5}}{2}}{}$ The probability that Newton will win on the first flip is $p$. The probability that Newton will win on the third flip is $(1-p) p^{2}$, since the first flip must be tails, the second must be heads, and the third flip must be heads. By the same logic, the probability Newton will win on the $(2 n+1)^{\text {st }}$ flip is $(1-p)^{n}(p)^{n+1}$. Thus, we have an infinite geometric sequence $p+(1-p) p^{2}+(1-p)^{2} p^{3}+\ldots$ which equals $\frac{p}{1-p(1-p)}$. We are given that this sum must equal $\frac{1}{2}$, so $1-p+p^{2}=2 p$, so $p=\frac{3-\sqrt{5}}{2}$ (the other solution is greater than 1).
|
\frac{3-\sqrt{5}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Newton and Leibniz are playing a game with a coin that comes up heads with probability $p$. They take turns flipping the coin until one of them wins with Newton going first. Newton wins if he flips a heads and Leibniz wins if he flips a tails. Given that Newton and Leibniz each win the game half of the time, what is the probability $p$ ?
|
$\frac{\frac{3-\sqrt{5}}{2}}{}$ The probability that Newton will win on the first flip is $p$. The probability that Newton will win on the third flip is $(1-p) p^{2}$, since the first flip must be tails, the second must be heads, and the third flip must be heads. By the same logic, the probability Newton will win on the $(2 n+1)^{\text {st }}$ flip is $(1-p)^{n}(p)^{n+1}$. Thus, we have an infinite geometric sequence $p+(1-p) p^{2}+(1-p)^{2} p^{3}+\ldots$ which equals $\frac{p}{1-p(1-p)}$. We are given that this sum must equal $\frac{1}{2}$, so $1-p+p^{2}=2 p$, so $p=\frac{3-\sqrt{5}}{2}$ (the other solution is greater than 1).
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen2-solutions.jsonl",
"problem_match": "\n9. [6]",
"solution_match": "\nAnswer: "
}
|
6a43fae1-a630-5670-b24e-95ee05172a6a
| 608,739
|
Justine has a coin which will come up the same as the last flip $\frac{2}{3}$ of the time and the other side $\frac{1}{3}$ of the time. She flips it and it comes up heads. She then flips it 2010 more times. What is the probability that the last flip is heads?
|
| $\frac{3^{2010}+1}{2 \cdot 3^{2010}}$ | Let the "value" of a flip be 1 if the flip is different from the previous flip and let |
| :---: | :---: | it be 0 if the flip is the same as the previous flip. The last flip will be heads if the sum of the values of all 2010 flips is even. The probability that this will happen is $\sum_{i=0}^{1005}\binom{2010}{2 i}\left(\frac{1}{3}\right)^{2 i}\left(\frac{2}{3}\right)^{2010-2 i}$.
We know that
$$
\begin{gathered}
\sum_{i=0}^{1005}\binom{2010}{2 i}\left(\frac{1}{3}\right)^{2 i}\left(\frac{2}{3}\right)^{2010-2 i}+\binom{2010}{2 i+1}\left(\frac{1}{3}\right)^{2 i+1}\left(\frac{2}{3}\right)^{2010-(2 i+1)}= \\
\sum_{k=0}^{2010}\binom{2010}{k}\left(\frac{1}{3}\right)^{k}\left(\frac{2}{3}\right)^{2010-k}=\left(\frac{1}{3}+\frac{2}{3}\right)^{2010}=1
\end{gathered}
$$
and
$$
\begin{gathered}
\sum_{i=0}^{1005}\binom{2010}{2 i}\left(\frac{1}{3}\right)^{2 i}\left(\frac{2}{3}\right)^{2010-2 i}-\binom{2010}{2 i+1}\left(\frac{1}{3}\right)^{2 i+1}\left(\frac{2}{3}\right)^{2010-(2 i+1)}= \\
\sum_{k=0}^{2010}\binom{2010}{k}\left(\frac{-1}{3}\right)^{k}\left(\frac{2}{3}\right)^{2010-k}=\left(\frac{-1}{3}+\frac{2}{3}\right)^{2010}=\left(\frac{1}{3}\right)^{2010}
\end{gathered}
$$
Summing these two values and dividing by 2 gives the answer $\frac{1+\left(\frac{1}{3}\right)^{2010}}{2}=\frac{3^{2010}+1}{2 \cdot 3^{2010}}$.
|
\frac{3^{2010}+1}{2 \cdot 3^{2010}}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Justine has a coin which will come up the same as the last flip $\frac{2}{3}$ of the time and the other side $\frac{1}{3}$ of the time. She flips it and it comes up heads. She then flips it 2010 more times. What is the probability that the last flip is heads?
|
| $\frac{3^{2010}+1}{2 \cdot 3^{2010}}$ | Let the "value" of a flip be 1 if the flip is different from the previous flip and let |
| :---: | :---: | it be 0 if the flip is the same as the previous flip. The last flip will be heads if the sum of the values of all 2010 flips is even. The probability that this will happen is $\sum_{i=0}^{1005}\binom{2010}{2 i}\left(\frac{1}{3}\right)^{2 i}\left(\frac{2}{3}\right)^{2010-2 i}$.
We know that
$$
\begin{gathered}
\sum_{i=0}^{1005}\binom{2010}{2 i}\left(\frac{1}{3}\right)^{2 i}\left(\frac{2}{3}\right)^{2010-2 i}+\binom{2010}{2 i+1}\left(\frac{1}{3}\right)^{2 i+1}\left(\frac{2}{3}\right)^{2010-(2 i+1)}= \\
\sum_{k=0}^{2010}\binom{2010}{k}\left(\frac{1}{3}\right)^{k}\left(\frac{2}{3}\right)^{2010-k}=\left(\frac{1}{3}+\frac{2}{3}\right)^{2010}=1
\end{gathered}
$$
and
$$
\begin{gathered}
\sum_{i=0}^{1005}\binom{2010}{2 i}\left(\frac{1}{3}\right)^{2 i}\left(\frac{2}{3}\right)^{2010-2 i}-\binom{2010}{2 i+1}\left(\frac{1}{3}\right)^{2 i+1}\left(\frac{2}{3}\right)^{2010-(2 i+1)}= \\
\sum_{k=0}^{2010}\binom{2010}{k}\left(\frac{-1}{3}\right)^{k}\left(\frac{2}{3}\right)^{2010-k}=\left(\frac{-1}{3}+\frac{2}{3}\right)^{2010}=\left(\frac{1}{3}\right)^{2010}
\end{gathered}
$$
Summing these two values and dividing by 2 gives the answer $\frac{1+\left(\frac{1}{3}\right)^{2010}}{2}=\frac{3^{2010}+1}{2 \cdot 3^{2010}}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-gen2-solutions.jsonl",
"problem_match": "\n10. [7]",
"solution_match": "\nAnswer: "
}
|
88d54f36-6289-5289-84ef-b7924275eed7
| 608,740
|
David, Delong, and Justin each showed up to a problem writing session at a random time during the session. If David arrived before Delong, what is the probability that he also arrived before Justin?
|
$\sqrt[2]{3}$ Let $t_{1}$ be the time that David arrives, let $t_{2}$ be the time that Delong arrives, and let $t_{3}$ be the time that Justin arrives. We can assume that all times are pairwise distinct because the probability of any two being equal is zero. Because the times were originally random and independent before we were given any information, then all orders $t_{1}<t_{2}<t_{3}, t_{1}<t_{3}<t_{2}, t_{3}<t_{1}<t_{2}, t_{2}<t_{1}<t_{3}$, $t_{2}<t_{3}<t_{1}, t_{3}<t_{2}<t_{1}$ must all be equally likely. Since we are given that $t_{1}<t_{2}$, then we only have the first three cases to consider, and $t_{1}<t_{3}$ in two cases of these three. Thus, the desired probability is $\frac{2}{3}$.
|
\frac{2}{3}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
David, Delong, and Justin each showed up to a problem writing session at a random time during the session. If David arrived before Delong, what is the probability that he also arrived before Justin?
|
$\sqrt[2]{3}$ Let $t_{1}$ be the time that David arrives, let $t_{2}$ be the time that Delong arrives, and let $t_{3}$ be the time that Justin arrives. We can assume that all times are pairwise distinct because the probability of any two being equal is zero. Because the times were originally random and independent before we were given any information, then all orders $t_{1}<t_{2}<t_{3}, t_{1}<t_{3}<t_{2}, t_{3}<t_{1}<t_{2}, t_{2}<t_{1}<t_{3}$, $t_{2}<t_{3}<t_{1}, t_{3}<t_{2}<t_{1}$ must all be equally likely. Since we are given that $t_{1}<t_{2}$, then we only have the first three cases to consider, and $t_{1}<t_{3}$ in two cases of these three. Thus, the desired probability is $\frac{2}{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n1. [5]",
"solution_match": "\nAnswer: "
}
|
5e60561f-9ab3-5ed6-b9d5-ee26ebfb066a
| 608,741
|
A circle of radius 6 is drawn centered at the origin. How many squares of side length 1 and integer coordinate vertices intersect the interior of this circle?
|
132 By symmetry, the answer is four times the number of squares in the first quadrant. Let's identify each square by its coordinates at the bottom-left corner, $(x, y)$. When $x=0$, we can have $y=0 \ldots 5$, so there are 6 squares. (Letting $y=6$ is not allowed because that square intersects only the boundary of the circle.) When $x=1$, how many squares are there? The equation of the circle is $y=\sqrt{36-x^{2}}=\sqrt{36-1^{2}}$ is between 5 and 6 , so we can again have $y=0 \ldots 5$. Likewise for $x=2$ and $x=3$. When $x=4$ we have $y=\sqrt{20}$ which is between 4 and 5 , so there are 5 squares, and when $x=5$ we have $y=\sqrt{11}$ which is between 3 and 4 , so there are 4 squares. Finally, when $x=6$, we have $y=0$, and no squares intersect the interior of the circle. This gives $6+6+6+6+5+4=33$. Since this is the number in the first quadrant, we multiply by four to get 132 .
|
132
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A circle of radius 6 is drawn centered at the origin. How many squares of side length 1 and integer coordinate vertices intersect the interior of this circle?
|
132 By symmetry, the answer is four times the number of squares in the first quadrant. Let's identify each square by its coordinates at the bottom-left corner, $(x, y)$. When $x=0$, we can have $y=0 \ldots 5$, so there are 6 squares. (Letting $y=6$ is not allowed because that square intersects only the boundary of the circle.) When $x=1$, how many squares are there? The equation of the circle is $y=\sqrt{36-x^{2}}=\sqrt{36-1^{2}}$ is between 5 and 6 , so we can again have $y=0 \ldots 5$. Likewise for $x=2$ and $x=3$. When $x=4$ we have $y=\sqrt{20}$ which is between 4 and 5 , so there are 5 squares, and when $x=5$ we have $y=\sqrt{11}$ which is between 3 and 4 , so there are 4 squares. Finally, when $x=6$, we have $y=0$, and no squares intersect the interior of the circle. This gives $6+6+6+6+5+4=33$. Since this is the number in the first quadrant, we multiply by four to get 132 .
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n2. [5]",
"solution_match": "\nAnswer: "
}
|
f3dc9647-7ea3-5fe3-97eb-f03e4278f28b
| 608,742
|
Jacob flipped a fair coin five times. In the first three flips, the coin came up heads exactly twice. In the last three flips, the coin also came up heads exactly twice. What is the probability that the third flip was heads?
|
| $\frac{4}{5}$ |
| :---: |
| How many sequences of five flips satisfy the conditions, and have the third flip be heads? | We have __H_- , so exactly one of the first two flips is heads, and exactly one of the last two flips is heads. This gives $2 \times 2=4$ possibilities. How many sequences of five flips satisfy the conditions, and have the third flip be tails? Now we have __T__, so the first two and the last two flips must all be heads. This gives only 1 possibility. So the probability that the third flip was heads is $\frac{4}{(4+1)}=\frac{4}{5}$
|
\frac{4}{5}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Jacob flipped a fair coin five times. In the first three flips, the coin came up heads exactly twice. In the last three flips, the coin also came up heads exactly twice. What is the probability that the third flip was heads?
|
| $\frac{4}{5}$ |
| :---: |
| How many sequences of five flips satisfy the conditions, and have the third flip be heads? | We have __H_- , so exactly one of the first two flips is heads, and exactly one of the last two flips is heads. This gives $2 \times 2=4$ possibilities. How many sequences of five flips satisfy the conditions, and have the third flip be tails? Now we have __T__, so the first two and the last two flips must all be heads. This gives only 1 possibility. So the probability that the third flip was heads is $\frac{4}{(4+1)}=\frac{4}{5}$
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n3. [5]",
"solution_match": "\nAnswer: "
}
|
09adf546-789a-552c-babd-9dedef7c4414
| 608,743
|
Let $x$ be a real number. Find the maximum value of $2^{x(1-x)}$.
|
$\sqrt[4]{2}$ Consider the function $2^{y}$. This is monotonically increasing, so to maximize $2^{y}$, you simply want to maximize $y$. Here, $y=x(1-x)=-x^{2}+x$ is a parabola opening downwards. The vertex of the parabola occurs at $x=(-1) /(-2)=1 / 2$, so the maximum value of the function is $2^{(1 / 2)(1 / 2)}=\sqrt[4]{2}$.
|
\sqrt[4]{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $x$ be a real number. Find the maximum value of $2^{x(1-x)}$.
|
$\sqrt[4]{2}$ Consider the function $2^{y}$. This is monotonically increasing, so to maximize $2^{y}$, you simply want to maximize $y$. Here, $y=x(1-x)=-x^{2}+x$ is a parabola opening downwards. The vertex of the parabola occurs at $x=(-1) /(-2)=1 / 2$, so the maximum value of the function is $2^{(1 / 2)(1 / 2)}=\sqrt[4]{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n4. [6]",
"solution_match": "\nAnswer: "
}
|
24a6f600-86a3-5c31-9bb3-ee803cfe01e6
| 608,744
|
How many ordered pairs $(S, T)$ of subsets of $\{1,2,3,4,5,6,7,8,9,10\}$ are there whose union contains exactly three elements?
|
3240 Let the three elements in the union be $a, b$, and $c$. We know that $a$ can be only in $S$, only in $T$, or both, so there are 3 possibilities for placing it. (Recall that $S=\{a\}, T=\{b, c\}$ is different from $S=\{b, c\}, T=\{a\}$ because $S$ and $T$ are an ordered pair.) Likewise for $b$ and $c$. The other 7 elements are in neither $S$ nor $T$, so there is only 1 possibility for placing them. This gives $3^{3}=27$ ways to pick $S$ and $T$ once you've picked the union. There are $\binom{10}{3}=120$ ways to pick the elements in the union, so we have $120 \times 27=3240$ ways total.
|
3240
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many ordered pairs $(S, T)$ of subsets of $\{1,2,3,4,5,6,7,8,9,10\}$ are there whose union contains exactly three elements?
|
3240 Let the three elements in the union be $a, b$, and $c$. We know that $a$ can be only in $S$, only in $T$, or both, so there are 3 possibilities for placing it. (Recall that $S=\{a\}, T=\{b, c\}$ is different from $S=\{b, c\}, T=\{a\}$ because $S$ and $T$ are an ordered pair.) Likewise for $b$ and $c$. The other 7 elements are in neither $S$ nor $T$, so there is only 1 possibility for placing them. This gives $3^{3}=27$ ways to pick $S$ and $T$ once you've picked the union. There are $\binom{10}{3}=120$ ways to pick the elements in the union, so we have $120 \times 27=3240$ ways total.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n6. [6]",
"solution_match": "\nAnswer: "
}
|
b4c215d3-b827-547f-9104-b0686084ab93
| 608,746
|
Let $f(x, y)=x^{2}+2 x+y^{2}+4 y$. Let $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$, and $\left(x_{4}, y_{4}\right)$ be the vertices of a square with side length one and sides parallel to the coordinate axes. What is the minimum value of $f\left(x_{1}, y_{1}\right)+f\left(x_{2}, y_{2}\right)+f\left(x_{3}, y_{3}\right)+f\left(x_{4}, y_{4}\right) ?$
|
-18 The square's corners must be at $(x, y),(x+1, y),(x+1, y+1)$, and $(x, y+1)$ for some $x$ and $y$. So,
$$
\begin{aligned}
f\left(x_{1}, y_{1}\right) & +f\left(x_{2}, y_{2}\right)+f\left(x_{3}, y_{3}\right)+f\left(x_{4}, y_{4}\right) \\
& =2\left(x^{2}+2 x\right)+2\left((x+1)^{2}+2(x+1)\right)+2\left(y^{2}+4 y\right)+2\left((y+1)^{2}+4(y+1)\right) \\
& =4 x^{2}+12 x+6+4 y^{2}+20 y+10 \\
& =(2 x+3)^{2}-3+(2 y+5)^{2}-15 \\
& \geq-18
\end{aligned}
$$
This attains its minimum value of -18 when $x=-\frac{3}{2}$ and $y=-\frac{5}{2}$.
|
-18
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $f(x, y)=x^{2}+2 x+y^{2}+4 y$. Let $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$, and $\left(x_{4}, y_{4}\right)$ be the vertices of a square with side length one and sides parallel to the coordinate axes. What is the minimum value of $f\left(x_{1}, y_{1}\right)+f\left(x_{2}, y_{2}\right)+f\left(x_{3}, y_{3}\right)+f\left(x_{4}, y_{4}\right) ?$
|
-18 The square's corners must be at $(x, y),(x+1, y),(x+1, y+1)$, and $(x, y+1)$ for some $x$ and $y$. So,
$$
\begin{aligned}
f\left(x_{1}, y_{1}\right) & +f\left(x_{2}, y_{2}\right)+f\left(x_{3}, y_{3}\right)+f\left(x_{4}, y_{4}\right) \\
& =2\left(x^{2}+2 x\right)+2\left((x+1)^{2}+2(x+1)\right)+2\left(y^{2}+4 y\right)+2\left((y+1)^{2}+4(y+1)\right) \\
& =4 x^{2}+12 x+6+4 y^{2}+20 y+10 \\
& =(2 x+3)^{2}-3+(2 y+5)^{2}-15 \\
& \geq-18
\end{aligned}
$$
This attains its minimum value of -18 when $x=-\frac{3}{2}$ and $y=-\frac{5}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n7. [7]",
"solution_match": "\nAnswer: "
}
|
d5d88d6a-53cf-5fd2-9fba-41a4f7161554
| 608,747
|
What is the sum of all four-digit numbers that are equal to the cube of the sum of their digits (leading zeros are not allowed)?
|
10745 We want to find all integers $x$ between 1000 and 9999 that are the cube of the sum of their digits. Of course, our search is only restricted to perfect cubes. The smallest such cube is $10^{3}=1000$ and the largest such cube is $21^{3}=9261$. This means we only have to check 12 different cubes, which is quite doable, but we can reduce the search even further with a little number theory.
Suppose we write our number as $x=1000 a+100 b+10 c+d$, where $a, b, c$, and $d$ are the decimal digits of $x$. Then we have
$$
(a+b+c+d)^{3} \equiv 1000 a+100 b+10 c+d \equiv a+b+c+d(\bmod 9)
$$
If we let $k=a+b+c+d$, then $k$ must be a solution to the modular equation $k^{3} \equiv k(\bmod 9)$. A quick check of the values 0 through 8 shows that the only solutions are 0,1 , and 8 .
Now, in our search, we only have to check values that are the cube of a number which is either 0 , 1 , or $8 \bmod 9$.
$$
\begin{aligned}
& 10^{3}=1000, \text { but } 1+0+0+0 \neq 10 \\
& 17^{3}=4913, \text { and } 4+9+1+3=17 \\
& 18^{3}=5832, \text { and } 5+8+3+2=18 \\
& 19^{3}=6859, \text { but } 6+8+5+9 \neq 19
\end{aligned}
$$
So the only solutions are 4913 and 5832, which sum to 10745 .
|
10745
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
What is the sum of all four-digit numbers that are equal to the cube of the sum of their digits (leading zeros are not allowed)?
|
10745 We want to find all integers $x$ between 1000 and 9999 that are the cube of the sum of their digits. Of course, our search is only restricted to perfect cubes. The smallest such cube is $10^{3}=1000$ and the largest such cube is $21^{3}=9261$. This means we only have to check 12 different cubes, which is quite doable, but we can reduce the search even further with a little number theory.
Suppose we write our number as $x=1000 a+100 b+10 c+d$, where $a, b, c$, and $d$ are the decimal digits of $x$. Then we have
$$
(a+b+c+d)^{3} \equiv 1000 a+100 b+10 c+d \equiv a+b+c+d(\bmod 9)
$$
If we let $k=a+b+c+d$, then $k$ must be a solution to the modular equation $k^{3} \equiv k(\bmod 9)$. A quick check of the values 0 through 8 shows that the only solutions are 0,1 , and 8 .
Now, in our search, we only have to check values that are the cube of a number which is either 0 , 1 , or $8 \bmod 9$.
$$
\begin{aligned}
& 10^{3}=1000, \text { but } 1+0+0+0 \neq 10 \\
& 17^{3}=4913, \text { and } 4+9+1+3=17 \\
& 18^{3}=5832, \text { and } 5+8+3+2=18 \\
& 19^{3}=6859, \text { but } 6+8+5+9 \neq 19
\end{aligned}
$$
So the only solutions are 4913 and 5832, which sum to 10745 .
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n8. [7]",
"solution_match": "\nAnswer: "
}
|
d08b51d3-5a1a-50a4-bfc8-d4ab6594b64f
| 608,748
|
How many functions $f:\{1,2, \ldots, 10\} \rightarrow\{1,2, \ldots, 10\}$ satisfy the property that $f(i)+f(j)=11$ for all values of $i$ and $j$ such that $i+j=11$.
|
100000 To construct such a function $f$, we just need to choose a value for $f(x)$ from $\{1,2, \ldots, 10\}$ for each $x \in\{1,2, \ldots, 10\}$. But the condition that $f(i)+f(j)=11$ whenever $i+j=11$ means that
$$
\begin{aligned}
f(10) & =11-f(1) \\
f(9) & =11-f(2) \\
\vdots & \\
f(6) & =11-f(5)
\end{aligned}
$$
This means that once we have chosen $f(1), f(2), f(3), f(4)$, and $f(5)$, the five remaining values of $f(6), f(7), f(8), f(9)$, and $f(10)$ are already determined. The answer is therefore just the number of ways to choose these first five values. Since there are 10 possibilities for each one, we get that the answer is $10^{5}=100000$.
|
100000
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many functions $f:\{1,2, \ldots, 10\} \rightarrow\{1,2, \ldots, 10\}$ satisfy the property that $f(i)+f(j)=11$ for all values of $i$ and $j$ such that $i+j=11$.
|
100000 To construct such a function $f$, we just need to choose a value for $f(x)$ from $\{1,2, \ldots, 10\}$ for each $x \in\{1,2, \ldots, 10\}$. But the condition that $f(i)+f(j)=11$ whenever $i+j=11$ means that
$$
\begin{aligned}
f(10) & =11-f(1) \\
f(9) & =11-f(2) \\
\vdots & \\
f(6) & =11-f(5)
\end{aligned}
$$
This means that once we have chosen $f(1), f(2), f(3), f(4)$, and $f(5)$, the five remaining values of $f(6), f(7), f(8), f(9)$, and $f(10)$ are already determined. The answer is therefore just the number of ways to choose these first five values. Since there are 10 possibilities for each one, we get that the answer is $10^{5}=100000$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n9. [7]",
"solution_match": "\nAnswer: "
}
|
e1f04314-e851-5c04-a08a-91797ed840cc
| 608,749
|
What is the smallest integer greater than 10 such that the sum of the digits in its base 17 representation is equal to the sum of the digits in its base 10 representation?
|
153 We assume that the answer is at most three digits (in base 10). Then our desired number can be expressed in the form $\overline{a b c}_{10}=\overline{d e f}_{17}$, where $a, b, c$ are digits in base 10 , and $d, e, f$ are digits in base 17 . These variables then satisfy the equations
$$
\begin{aligned}
100 a+10 b+c & =289 d+17 e+f \\
a+b+c & =d+e+f
\end{aligned}
$$
Subtracting the second equation from the first, we obtain $99 a+9 b=288 d+16 e$, or $9(11 a+b)=$ $16(18 d+e)$. From this equation, we find that $11 a+b$ must be divisible by 16 , and $18 d+e$ must be divisible by 9 . To minimize $\overline{a b c}$, we find the minimal possible value of $a$ : If $a=0$, then the only way for $11 a+b=b$ to be divisible by 16 is to set $b=0$; however, this is disallowed by the problem condition, which stipulates that the number must be greater than 10 . If we try $a=1$, then we find that the only possible value of $b$ which lets $11 a+b=b+11$ be divisible by 16 is $b=5$. Plugging these in and simplifying, we find that we must have $18 d+e=9$. The only possible solution to this is $d=0, e=9$. Now to satisfy $a+b+c=d+e+f$, we must have $1+5+c=0+9+f$, or $c=f+3$. The minimal possible solution to this is $c=3, f=0$. So our answer is $\overline{a b c}=153$, which is also equal to $090_{17}$.
|
153
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
What is the smallest integer greater than 10 such that the sum of the digits in its base 17 representation is equal to the sum of the digits in its base 10 representation?
|
153 We assume that the answer is at most three digits (in base 10). Then our desired number can be expressed in the form $\overline{a b c}_{10}=\overline{d e f}_{17}$, where $a, b, c$ are digits in base 10 , and $d, e, f$ are digits in base 17 . These variables then satisfy the equations
$$
\begin{aligned}
100 a+10 b+c & =289 d+17 e+f \\
a+b+c & =d+e+f
\end{aligned}
$$
Subtracting the second equation from the first, we obtain $99 a+9 b=288 d+16 e$, or $9(11 a+b)=$ $16(18 d+e)$. From this equation, we find that $11 a+b$ must be divisible by 16 , and $18 d+e$ must be divisible by 9 . To minimize $\overline{a b c}$, we find the minimal possible value of $a$ : If $a=0$, then the only way for $11 a+b=b$ to be divisible by 16 is to set $b=0$; however, this is disallowed by the problem condition, which stipulates that the number must be greater than 10 . If we try $a=1$, then we find that the only possible value of $b$ which lets $11 a+b=b+11$ be divisible by 16 is $b=5$. Plugging these in and simplifying, we find that we must have $18 d+e=9$. The only possible solution to this is $d=0, e=9$. Now to satisfy $a+b+c=d+e+f$, we must have $1+5+c=0+9+f$, or $c=f+3$. The minimal possible solution to this is $c=3, f=0$. So our answer is $\overline{a b c}=153$, which is also equal to $090_{17}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n10. [8]",
"solution_match": "\nAnswer: "
}
|
68ecc999-5c39-52a1-886c-c3a5980005b3
| 608,750
|
How many nondecreasing sequences $a_{1}, a_{2}, \ldots, a_{10}$ are composed entirely of at most three distinct numbers from the set $\{1,2, \ldots, 9\}$ (so $1,1,1,2,2,2,3,3,3,3$ and $2,2,2,2,5,5,5,5,5,5$ are both allowed)?
|
3357 From any sequence $a_{1}, a_{2}, \ldots, a_{10}$, construct a sequence $b_{1}, b_{2}, \ldots, b_{9}$, where $b_{i}$ counts the number of times $i$ occurs in the sequence. There is a correspondence from all possible sequences $b_{1}, b_{2}, \ldots, b_{9}$ with at most 3 nonzero terms which add to 10 , since any sequence of $a_{1}, a_{2}, \ldots, a_{10}$ will be converted to this form, and from any sequence $b_{1}, b_{2}, \ldots, b_{9}$, we can construct a unique sequence of $a$-s by listing $i b_{i}$ times (for $1 \leq i \leq 9$ ) in nondecreasing order.
Our goal now is to count the number of possible sequences $b_{1}, b_{2}, \ldots, b_{9}$ meeting our conditions. We casework on the number of nonzero terms in the sequence:
Case 1: The sequence has exactly one nonzero term.
Then exactly one of $b_{1}, b_{2}, \ldots, b_{9}$ is equal to 10 , and all the rest are equal to 0 . This gives us 9 possible sequences in this case.
Case 2: The sequence has exactly two nonzero terms.
There are $\binom{9}{2}=36$ ways to choose the two terms $b_{i}, b_{j}(i<j)$ which are nonzero. From here, we have 9 choices for the value of $b_{i}$, namely 1 through 9 (since both $b_{i}$ and $b_{j}$ must be nonzero), and $b_{j}$ will be fixed, so this case gives us $36 \cdot 9=324$ possible sequences.
Case 3: The sequence has exactly three nonzero terms.
There are $\binom{9}{3}=84$ ways to choose the three terms $b_{i}, b_{j}, b_{k}(i<j<k)$ which are nonzero. Letting
$c_{i}=b_{i}-1, c_{j}=b_{j}-1, c_{k}=b_{k}-1$, we have that $c_{i}, c_{j}, c_{k}$ are nonnegative integers which sum to 7 . There are $\binom{9}{2}=36$ solutions to this equation (consider placing two dividers in the nine spaces between the ten elements), giving $84 \cdot 36=3024$ possbilities in this case.
We then have $9+324+3024=3357$ possible sequences.
|
3357
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many nondecreasing sequences $a_{1}, a_{2}, \ldots, a_{10}$ are composed entirely of at most three distinct numbers from the set $\{1,2, \ldots, 9\}$ (so $1,1,1,2,2,2,3,3,3,3$ and $2,2,2,2,5,5,5,5,5,5$ are both allowed)?
|
3357 From any sequence $a_{1}, a_{2}, \ldots, a_{10}$, construct a sequence $b_{1}, b_{2}, \ldots, b_{9}$, where $b_{i}$ counts the number of times $i$ occurs in the sequence. There is a correspondence from all possible sequences $b_{1}, b_{2}, \ldots, b_{9}$ with at most 3 nonzero terms which add to 10 , since any sequence of $a_{1}, a_{2}, \ldots, a_{10}$ will be converted to this form, and from any sequence $b_{1}, b_{2}, \ldots, b_{9}$, we can construct a unique sequence of $a$-s by listing $i b_{i}$ times (for $1 \leq i \leq 9$ ) in nondecreasing order.
Our goal now is to count the number of possible sequences $b_{1}, b_{2}, \ldots, b_{9}$ meeting our conditions. We casework on the number of nonzero terms in the sequence:
Case 1: The sequence has exactly one nonzero term.
Then exactly one of $b_{1}, b_{2}, \ldots, b_{9}$ is equal to 10 , and all the rest are equal to 0 . This gives us 9 possible sequences in this case.
Case 2: The sequence has exactly two nonzero terms.
There are $\binom{9}{2}=36$ ways to choose the two terms $b_{i}, b_{j}(i<j)$ which are nonzero. From here, we have 9 choices for the value of $b_{i}$, namely 1 through 9 (since both $b_{i}$ and $b_{j}$ must be nonzero), and $b_{j}$ will be fixed, so this case gives us $36 \cdot 9=324$ possible sequences.
Case 3: The sequence has exactly three nonzero terms.
There are $\binom{9}{3}=84$ ways to choose the three terms $b_{i}, b_{j}, b_{k}(i<j<k)$ which are nonzero. Letting
$c_{i}=b_{i}-1, c_{j}=b_{j}-1, c_{k}=b_{k}-1$, we have that $c_{i}, c_{j}, c_{k}$ are nonnegative integers which sum to 7 . There are $\binom{9}{2}=36$ solutions to this equation (consider placing two dividers in the nine spaces between the ten elements), giving $84 \cdot 36=3024$ possbilities in this case.
We then have $9+324+3024=3357$ possible sequences.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n11. [8]",
"solution_match": "\nAnswer: "
}
|
70f382ad-48cf-5c1f-8e7b-2f76599d4bf3
| 608,751
|
An ant starts at the origin of a coordinate plane. Each minute, it either walks one unit to the right or one unit up, but it will never move in the same direction more than twice in the row. In how many different ways can it get to the point $(5,5)$ ?
|
84 We can change the ant's sequence of moves to a sequence $a_{1}, a_{2}, \ldots, a_{10}$, with $a_{i}=0$ if the $i$-th step is up, and $a_{i}=1$ if the $i$-th step is right. We define a subsequence of moves $a_{i}, a_{i+1}, \ldots, a_{j}$, ( $i \leq j$ ) as an up run if all terms of the subsequence are equal to 0 , and $a_{i-1}$ and $a_{j+1}$ either do not exist or are not equal to 0 , and define a right run similarly. In a sequence of moves, up runs and right runs alternate, so the number of up rights can differ from the number of right runs by at most one.
Now let $f(n)$ denote the number of sequences $a_{1}, a_{2}, \ldots, a_{n}$ where $a_{i} \in\{1,2\}$ for $1 \leq i \leq n$, and $a_{1}+a_{2}+\cdots+a_{n}=5$. (In essence, we are splitting the possible 5 up moves into up runs, and we are doing the same with the right moves). We can easily compute that $f(3)=3, f(4)=4, f(5)=1$, and $f(n)=0$ otherwise.
For each possible pair of numbers of up runs and right runs, we have two choices of which type of run is first. Our answer is then $2\left(f(3)^{2}+f(3) f(4)+f(4)^{2}+f(4) f(5)+f(5)^{2}\right)=2(9+12+16+4+1)=84$.
|
84
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
An ant starts at the origin of a coordinate plane. Each minute, it either walks one unit to the right or one unit up, but it will never move in the same direction more than twice in the row. In how many different ways can it get to the point $(5,5)$ ?
|
84 We can change the ant's sequence of moves to a sequence $a_{1}, a_{2}, \ldots, a_{10}$, with $a_{i}=0$ if the $i$-th step is up, and $a_{i}=1$ if the $i$-th step is right. We define a subsequence of moves $a_{i}, a_{i+1}, \ldots, a_{j}$, ( $i \leq j$ ) as an up run if all terms of the subsequence are equal to 0 , and $a_{i-1}$ and $a_{j+1}$ either do not exist or are not equal to 0 , and define a right run similarly. In a sequence of moves, up runs and right runs alternate, so the number of up rights can differ from the number of right runs by at most one.
Now let $f(n)$ denote the number of sequences $a_{1}, a_{2}, \ldots, a_{n}$ where $a_{i} \in\{1,2\}$ for $1 \leq i \leq n$, and $a_{1}+a_{2}+\cdots+a_{n}=5$. (In essence, we are splitting the possible 5 up moves into up runs, and we are doing the same with the right moves). We can easily compute that $f(3)=3, f(4)=4, f(5)=1$, and $f(n)=0$ otherwise.
For each possible pair of numbers of up runs and right runs, we have two choices of which type of run is first. Our answer is then $2\left(f(3)^{2}+f(3) f(4)+f(4)^{2}+f(4) f(5)+f(5)^{2}\right)=2(9+12+16+4+1)=84$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n12. $[\\mathbf{8}]$",
"solution_match": "\nAnswer: "
}
|
193fe2d6-83c8-53e2-aff5-fba1ee54e625
| 608,752
|
How many sequences of ten binary digits are there in which neither two zeroes nor three ones ever appear in a row?
|
28 Let $a_{n}$ be the number of binary sequences of length $n$ satisfying the conditions and ending in 0 , let $b_{n}$ be the number ending in 01, and let $c_{n}$ be the number ending in 11. From the legal sequences of length $201,11,10$, we find that $a_{2}=b_{2}=c_{2}=1$. We now establish a recursion by building sequences of length $n+1$ from sequences of length $n$. We can add a 0 to a sequence of length $n$ if and only if it ended with a 1 , so $a_{n+1}=b_{n}+c_{n}$. We can have a sequence of length $n+1$ ending with 01 only by adding a 1 to a sequence of length $n$ ending in 0 , so $b_{n+1}=a_{n}$. We can have a sequence of length $n+1$ ending with 11 only by adding a 1 to a sequence of length $n$ ending in 01 , so $c_{n+1}=b_{n}$. We can now run the recursion:
| $n$ | $a_{n}$ | $b_{n}$ | $c_{n}$ |
| :---: | :---: | :---: | :---: |
| 2 | 1 | 1 | 1 |
| 3 | 2 | 1 | 1 |
| 4 | 2 | 2 | 1 |
| 5 | 3 | 2 | 2 |
| 6 | 4 | 3 | 2 |
| 7 | 5 | 4 | 3 |
| 8 | 7 | 5 | 4 |
| 9 | 9 | 7 | 5 |
| 10 | 12 | 9 | 7 |
Our answer is then $12+9+7=28$.
|
28
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many sequences of ten binary digits are there in which neither two zeroes nor three ones ever appear in a row?
|
28 Let $a_{n}$ be the number of binary sequences of length $n$ satisfying the conditions and ending in 0 , let $b_{n}$ be the number ending in 01, and let $c_{n}$ be the number ending in 11. From the legal sequences of length $201,11,10$, we find that $a_{2}=b_{2}=c_{2}=1$. We now establish a recursion by building sequences of length $n+1$ from sequences of length $n$. We can add a 0 to a sequence of length $n$ if and only if it ended with a 1 , so $a_{n+1}=b_{n}+c_{n}$. We can have a sequence of length $n+1$ ending with 01 only by adding a 1 to a sequence of length $n$ ending in 0 , so $b_{n+1}=a_{n}$. We can have a sequence of length $n+1$ ending with 11 only by adding a 1 to a sequence of length $n$ ending in 01 , so $c_{n+1}=b_{n}$. We can now run the recursion:
| $n$ | $a_{n}$ | $b_{n}$ | $c_{n}$ |
| :---: | :---: | :---: | :---: |
| 2 | 1 | 1 | 1 |
| 3 | 2 | 1 | 1 |
| 4 | 2 | 2 | 1 |
| 5 | 3 | 2 | 2 |
| 6 | 4 | 3 | 2 |
| 7 | 5 | 4 | 3 |
| 8 | 7 | 5 | 4 |
| 9 | 9 | 7 | 5 |
| 10 | 12 | 9 | 7 |
Our answer is then $12+9+7=28$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n13. [8]",
"solution_match": "\nAnswer: "
}
|
4f814a99-b20f-53c2-8bb4-3650bc57ce4e
| 608,753
|
The positive integer $i$ is chosen at random such that the probability of a positive integer $k$ being chosen is $\frac{3}{2}$ times the probability of $k+1$ being chosen. What is the probability that the $i^{\text {th }}$ digit after the decimal point of the decimal expansion of $\frac{1}{7}$ is a 2 ?
|
$\frac{108}{665}$ First we note that the probability that $n$ is picked is $\frac{1}{2} \times\left(\frac{2}{3}\right)^{n}$, because this is the sequence whose terms decrease by a factor of $\frac{2}{3}$ each time and whose sum is 1 (recall that probabilities must sum to 1).
Now note that $\frac{1}{7}=.142857142857 \ldots$, meaning that 2 occurs at digits $3,9,15,21$, etc. We can then calculate the probability that we ever pick 2 as
$$
\begin{aligned}
\sum_{k=0}^{\infty} \frac{1}{2} \cdot\left(\frac{2}{3}\right)^{6 k+3} & =\frac{4}{27} \sum_{k=0}^{\infty}\left(\frac{2}{3}\right)^{6 k} \\
& =\frac{4}{27} \cdot \frac{1}{1-\left(\frac{2}{3}\right)^{6}} \\
& =\frac{4}{27} \cdot \frac{729}{729-64} \\
& =\frac{4}{27} \cdot \frac{729}{665} \\
& =\frac{108}{665}
\end{aligned}
$$
|
\frac{108}{665}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
The positive integer $i$ is chosen at random such that the probability of a positive integer $k$ being chosen is $\frac{3}{2}$ times the probability of $k+1$ being chosen. What is the probability that the $i^{\text {th }}$ digit after the decimal point of the decimal expansion of $\frac{1}{7}$ is a 2 ?
|
$\frac{108}{665}$ First we note that the probability that $n$ is picked is $\frac{1}{2} \times\left(\frac{2}{3}\right)^{n}$, because this is the sequence whose terms decrease by a factor of $\frac{2}{3}$ each time and whose sum is 1 (recall that probabilities must sum to 1).
Now note that $\frac{1}{7}=.142857142857 \ldots$, meaning that 2 occurs at digits $3,9,15,21$, etc. We can then calculate the probability that we ever pick 2 as
$$
\begin{aligned}
\sum_{k=0}^{\infty} \frac{1}{2} \cdot\left(\frac{2}{3}\right)^{6 k+3} & =\frac{4}{27} \sum_{k=0}^{\infty}\left(\frac{2}{3}\right)^{6 k} \\
& =\frac{4}{27} \cdot \frac{1}{1-\left(\frac{2}{3}\right)^{6}} \\
& =\frac{4}{27} \cdot \frac{729}{729-64} \\
& =\frac{4}{27} \cdot \frac{729}{665} \\
& =\frac{108}{665}
\end{aligned}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n14. [8]",
"solution_match": "\nAnswer: "
}
|
11ce4360-35f5-56ca-9c87-5cddbcfa7d5f
| 608,754
|
Distinct points $A, B, C, D$ are given such that triangles $A B C$ and $A B D$ are equilateral and both are of side length 10 . Point $E$ lies inside triangle $A B C$ such that $E A=8$ and $E B=3$, and point $F$ lies inside triangle $A B D$ such that $F D=8$ and $F B=3$. What is the area of quadrilateral $A E F D$ ?
|
$\frac{\frac{91 \sqrt{3}}{4}}{}$ Since $A E B \cong D F B$, we have $\angle E B A=\angle F B D$. Thus, $\angle E B F=\angle E B A+\angle A B F=$ $\angle F B D+\angle A B F=\angle A B D=60^{\circ}$. Since $E B=B F=3$, this means that $E B F$ is an equilateral triangle of side length 3. Now we have $[A E F D]=[A E B D]-[E B F]-[F B D]=[A E B]+[A B D]-[E B F]-$ $[F B D]=[A B D]-[E B F]=\frac{\sqrt{3}}{4}\left(10^{2}-3^{2}\right)=\frac{91 \sqrt{3}}{4}$.
|
\frac{91 \sqrt{3}}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Distinct points $A, B, C, D$ are given such that triangles $A B C$ and $A B D$ are equilateral and both are of side length 10 . Point $E$ lies inside triangle $A B C$ such that $E A=8$ and $E B=3$, and point $F$ lies inside triangle $A B D$ such that $F D=8$ and $F B=3$. What is the area of quadrilateral $A E F D$ ?
|
$\frac{\frac{91 \sqrt{3}}{4}}{}$ Since $A E B \cong D F B$, we have $\angle E B A=\angle F B D$. Thus, $\angle E B F=\angle E B A+\angle A B F=$ $\angle F B D+\angle A B F=\angle A B D=60^{\circ}$. Since $E B=B F=3$, this means that $E B F$ is an equilateral triangle of side length 3. Now we have $[A E F D]=[A E B D]-[E B F]-[F B D]=[A E B]+[A B D]-[E B F]-$ $[F B D]=[A B D]-[E B F]=\frac{\sqrt{3}}{4}\left(10^{2}-3^{2}\right)=\frac{91 \sqrt{3}}{4}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n15. [8]",
"solution_match": "\nAnswer: "
}
|
b151839e-0177-5406-ab3f-7a6e55108cc5
| 608,755
|
Triangle $A B C$ is given in the plane. Let $A D$ be the angle bisector of $\angle B A C$; let $B E$ be the altitude from $B$ to $A D$, and let $F$ be the midpoint of $A B$. Given that $A B=28, B C=33, C A=37$, what is the length of $E F$ ?
|
$14 \triangle A B E$ is a right triangle, and $F$ is the midpoint of the hypotenuse (and therefore the circumcenter), so $E F=B F=A F=14$.
|
14
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Triangle $A B C$ is given in the plane. Let $A D$ be the angle bisector of $\angle B A C$; let $B E$ be the altitude from $B$ to $A D$, and let $F$ be the midpoint of $A B$. Given that $A B=28, B C=33, C A=37$, what is the length of $E F$ ?
|
$14 \triangle A B E$ is a right triangle, and $F$ is the midpoint of the hypotenuse (and therefore the circumcenter), so $E F=B F=A F=14$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n16. [9]",
"solution_match": "\nAnswer: "
}
|
2444cd11-8625-5c58-8526-b13a7c44a24e
| 608,756
|
A triangle with side lengths $5,7,8$ is inscribed in a circle $C$. The diameters of $C$ parallel to the sides of lengths 5 and 8 divide $C$ into four sectors. What is the area of either of the two smaller ones?
|
$\sqrt{\frac{49}{18} \pi}$ Let $\triangle P Q R$ have sides $p=7, q=5, r=8$. Of the four sectors determined by the diameters of $C$ that are parallel to $P Q$ and $P R$, two have angles equal to $P$ and the other two have angles equal to $\pi-P$. We first find $P$ using the law of cosines: $49=25+64-2(5)(8) \cos P$ implies $\cos P=\frac{1}{2}$ implies $P=\frac{\pi}{3}$. Thus the two smaller sectors will have angle $\frac{\pi}{3}$. Next we find the circumradius of $\triangle P Q R$ using the formula $R=\frac{p q r}{4[P Q R]}$, where $[P Q R]$ is the area of $\triangle P Q R$. By Heron's Formula we have $[P Q R]=\sqrt{10(5)(3)(2)}=10 \sqrt{3}$; thus $R=\frac{5 \cdot 7 \cdot 8}{4(10 \sqrt{3})}=\frac{7}{\sqrt{3}}$. The area of a smaller sector is thus $\frac{\pi / 3}{2 \pi}\left(\pi R^{2}\right)=\frac{\pi}{6}\left(\frac{7}{\sqrt{3}}\right)^{2}=\frac{49}{18} \pi$
|
\frac{49}{18} \pi
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A triangle with side lengths $5,7,8$ is inscribed in a circle $C$. The diameters of $C$ parallel to the sides of lengths 5 and 8 divide $C$ into four sectors. What is the area of either of the two smaller ones?
|
$\sqrt{\frac{49}{18} \pi}$ Let $\triangle P Q R$ have sides $p=7, q=5, r=8$. Of the four sectors determined by the diameters of $C$ that are parallel to $P Q$ and $P R$, two have angles equal to $P$ and the other two have angles equal to $\pi-P$. We first find $P$ using the law of cosines: $49=25+64-2(5)(8) \cos P$ implies $\cos P=\frac{1}{2}$ implies $P=\frac{\pi}{3}$. Thus the two smaller sectors will have angle $\frac{\pi}{3}$. Next we find the circumradius of $\triangle P Q R$ using the formula $R=\frac{p q r}{4[P Q R]}$, where $[P Q R]$ is the area of $\triangle P Q R$. By Heron's Formula we have $[P Q R]=\sqrt{10(5)(3)(2)}=10 \sqrt{3}$; thus $R=\frac{5 \cdot 7 \cdot 8}{4(10 \sqrt{3})}=\frac{7}{\sqrt{3}}$. The area of a smaller sector is thus $\frac{\pi / 3}{2 \pi}\left(\pi R^{2}\right)=\frac{\pi}{6}\left(\frac{7}{\sqrt{3}}\right)^{2}=\frac{49}{18} \pi$
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n17. [9]",
"solution_match": "\nAnswer: "
}
|
b1ddf8c1-48a7-5c20-9262-8f927b99438b
| 608,757
|
Jeff has a 50 point quiz at 11 am . He wakes up at a random time between 10 am and noon, then arrives at class 15 minutes later. If he arrives on time, he will get a perfect score, but if he arrives more than 30 minutes after the quiz starts, he will get a 0 , but otherwise, he loses a point for each minute he's late (he can lose parts of one point if he arrives a nonintegral number of minutes late). What is Jeff's expected score on the quiz?
|
$\sqrt{\frac{55}{2}}$ If Jeff wakes up between 10:00 and 10:45, he gets 50. If he wakes up between 10:45 and 11:15, and he wakes up $k$ minutes after 10:45, then he gets $50-k$ points. Finally, if he wakes up between 11:15 and 12:00 he gets 0 points. So he has a $\frac{3}{8}$ probability of 50 , a $\frac{3}{8}$ probability of 0 , and a $\frac{1}{4}$ probability of a number chosen uniformly between 20 and 50 (for an average of 35). Thus his expected score is $\frac{3}{8} \times 50+\frac{1}{4} \times 35=\frac{75+35}{4}=\frac{110}{4}=\frac{55}{2}$.
|
\frac{55}{2}
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
Jeff has a 50 point quiz at 11 am . He wakes up at a random time between 10 am and noon, then arrives at class 15 minutes later. If he arrives on time, he will get a perfect score, but if he arrives more than 30 minutes after the quiz starts, he will get a 0 , but otherwise, he loses a point for each minute he's late (he can lose parts of one point if he arrives a nonintegral number of minutes late). What is Jeff's expected score on the quiz?
|
$\sqrt{\frac{55}{2}}$ If Jeff wakes up between 10:00 and 10:45, he gets 50. If he wakes up between 10:45 and 11:15, and he wakes up $k$ minutes after 10:45, then he gets $50-k$ points. Finally, if he wakes up between 11:15 and 12:00 he gets 0 points. So he has a $\frac{3}{8}$ probability of 50 , a $\frac{3}{8}$ probability of 0 , and a $\frac{1}{4}$ probability of a number chosen uniformly between 20 and 50 (for an average of 35). Thus his expected score is $\frac{3}{8} \times 50+\frac{1}{4} \times 35=\frac{75+35}{4}=\frac{110}{4}=\frac{55}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n18. [9]",
"solution_match": "\nAnswer: "
}
|
7442f7d5-b879-522f-99a8-ac675385f758
| 608,758
|
How many 8 -digit numbers begin with 1 , end with 3 , and have the property that each successive digit is either one more or two more than the previous digit, considering 0 to be one more than 9 ?
|
21 Given an 8-digit number $a$ that satifies the conditions in the problem, let $a_{i}$ denote the difference between its $(i+1)$ th and $i$ th digit. Since $i \in\{1,2\}$ for all $1 \leq i \leq 7$, we have $7 \leq$ $a_{1}+a_{2}+\cdots+a_{7} \leq 14$. The difference between the last digit and the first digit of $m$ is $3-1 \equiv 2$ (mod 10), which means $a_{1}+\cdots+a_{7}=12$. Thus, exactly five of the $a_{i}$ s equal to 2 and the remaining two equal to 1 . The number of permutations of five 2 s and two 1 s is $\binom{7}{2}=21$.
|
21
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many 8 -digit numbers begin with 1 , end with 3 , and have the property that each successive digit is either one more or two more than the previous digit, considering 0 to be one more than 9 ?
|
21 Given an 8-digit number $a$ that satifies the conditions in the problem, let $a_{i}$ denote the difference between its $(i+1)$ th and $i$ th digit. Since $i \in\{1,2\}$ for all $1 \leq i \leq 7$, we have $7 \leq$ $a_{1}+a_{2}+\cdots+a_{7} \leq 14$. The difference between the last digit and the first digit of $m$ is $3-1 \equiv 2$ (mod 10), which means $a_{1}+\cdots+a_{7}=12$. Thus, exactly five of the $a_{i}$ s equal to 2 and the remaining two equal to 1 . The number of permutations of five 2 s and two 1 s is $\binom{7}{2}=21$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n19. [11]",
"solution_match": "\nAnswer: "
}
|
25178fca-af5c-5fbf-a4cb-0426dd5d3873
| 608,759
|
Given a permutation $\pi$ of the set $\{1,2, \ldots, 10\}$, define a rotated cycle as a set of three integers $i, j, k$ such that $i<j<k$ and $\pi(j)<\pi(k)<\pi(i)$. What is the total number of rotated cycles over all permutations $\pi$ of the set $\{1,2, \ldots, 10\}$ ?
|
72576000 Let us consider a triple $(i, j, k)$ with $i<j<k$ and determine how many permutations rotate it. There are $\binom{10}{3}$ choices for the values of $\pi(i), \pi(j), \pi(k)$ and the choice of this set of three determines the values of $\pi(i), \pi(j), \pi(k)$. The other 7 values then have 7 ! ways to be arranged (any permutation of them will work), so exactly $\binom{10}{3} 7$ ! permutations rotate $(i, j, k)$. Therefore, as there are $\binom{10}{3}$ such triples, the total number of rotated triples is $\binom{10}{3}^{2} \cdot 7!=72576000$.
|
72576000
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Given a permutation $\pi$ of the set $\{1,2, \ldots, 10\}$, define a rotated cycle as a set of three integers $i, j, k$ such that $i<j<k$ and $\pi(j)<\pi(k)<\pi(i)$. What is the total number of rotated cycles over all permutations $\pi$ of the set $\{1,2, \ldots, 10\}$ ?
|
72576000 Let us consider a triple $(i, j, k)$ with $i<j<k$ and determine how many permutations rotate it. There are $\binom{10}{3}$ choices for the values of $\pi(i), \pi(j), \pi(k)$ and the choice of this set of three determines the values of $\pi(i), \pi(j), \pi(k)$. The other 7 values then have 7 ! ways to be arranged (any permutation of them will work), so exactly $\binom{10}{3} 7$ ! permutations rotate $(i, j, k)$. Therefore, as there are $\binom{10}{3}$ such triples, the total number of rotated triples is $\binom{10}{3}^{2} \cdot 7!=72576000$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n20. [11]",
"solution_match": "\nAnswer: "
}
|
c3f0f22f-139f-5ff7-9eda-90e22ec7cd2a
| 608,760
|
George, Jeff, Brian, and Travis decide to play a game of hot potato. They begin by arranging themselves clockwise in a circle in that order. George and Jeff both start with a hot potato. On his turn, a player gives a hot potato (if he has one) to a randomly chosen player among the other three (if a player has two hot potatoes on his turn, he only passes one). If George goes first, and play proceedes clockwise, what is the probability that Travis has a hot potato after each player takes one turn?
|
| $\frac{5}{27}$ |
| :---: |
| Notice that Travis can only have the hot potato at the end if he has two potatoes | before his turn. A little bit of casework shows that this can only happen when
Case 1: George gives Travis his potato, while Jeff gives Brian his potato, which in then goes to Travis. The probability of this occuring is $\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$
Case 2: George gives Travis his potato, while Jeff gives Travis his potato. The probability of this occuring is $\left(\frac{1}{3}\right)^{2}=\frac{1}{9}$
Case 3: George gives Brian his potato, Jeff gives Travis his potato, and then Brian gives Travis his potato. The probability of this occuring is $\frac{1}{27}$
Because these events are all disjoint, the probability that Travis ends up with the hot potato is $\frac{5}{27}$
|
\frac{5}{27}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
George, Jeff, Brian, and Travis decide to play a game of hot potato. They begin by arranging themselves clockwise in a circle in that order. George and Jeff both start with a hot potato. On his turn, a player gives a hot potato (if he has one) to a randomly chosen player among the other three (if a player has two hot potatoes on his turn, he only passes one). If George goes first, and play proceedes clockwise, what is the probability that Travis has a hot potato after each player takes one turn?
|
| $\frac{5}{27}$ |
| :---: |
| Notice that Travis can only have the hot potato at the end if he has two potatoes | before his turn. A little bit of casework shows that this can only happen when
Case 1: George gives Travis his potato, while Jeff gives Brian his potato, which in then goes to Travis. The probability of this occuring is $\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$
Case 2: George gives Travis his potato, while Jeff gives Travis his potato. The probability of this occuring is $\left(\frac{1}{3}\right)^{2}=\frac{1}{9}$
Case 3: George gives Brian his potato, Jeff gives Travis his potato, and then Brian gives Travis his potato. The probability of this occuring is $\frac{1}{27}$
Because these events are all disjoint, the probability that Travis ends up with the hot potato is $\frac{5}{27}$
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n21. [11]",
"solution_match": "\nAnswer: "
}
|
c76328d5-e3ad-50b2-8ae5-89963f49600b
| 608,761
|
Let $g_{1}(x)=\frac{1}{3}\left(1+x+x^{2}+\cdots\right)$ for all values of $x$ for which the right hand side converges. Let $g_{n}(x)=g_{1}\left(g_{n-1}(x)\right)$ for all integers $n \geq 2$. What is the largest integer $r$ such that $g_{r}(x)$ is defined for some real number $x$ ?
|
5 Notice that the series is geometric with ratio $x$, so it converges if $-1<x<1$. Also notice that where $g_{1}(x)$ is defined, it is equal to $\frac{1}{3(1-x)}$. The image of $g_{1}(x)$ is then the interval $\left(\frac{1}{6}, \infty\right)$. The image of $g_{2}(x)$ is simply the values of $g_{1}(x)$ for $x$ in $\left(\frac{1}{6}, 1\right)$, which is the interval $\left(\frac{2}{5}, \infty\right)$. Similarly, the image of $g_{3}(x)$ is $\left(\frac{5}{9}, \infty\right)$, the image of $g_{4}(x)$ is $\left(\frac{3}{4}, \infty\right)$, and the image of $g_{5}(x)$ is $\left(\frac{4}{3}, \infty\right)$. As this does not intersect the interval $(-1,1), g_{6}(x)$ is not defined for any $x$, so the answer is 5 .
|
5
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $g_{1}(x)=\frac{1}{3}\left(1+x+x^{2}+\cdots\right)$ for all values of $x$ for which the right hand side converges. Let $g_{n}(x)=g_{1}\left(g_{n-1}(x)\right)$ for all integers $n \geq 2$. What is the largest integer $r$ such that $g_{r}(x)$ is defined for some real number $x$ ?
|
5 Notice that the series is geometric with ratio $x$, so it converges if $-1<x<1$. Also notice that where $g_{1}(x)$ is defined, it is equal to $\frac{1}{3(1-x)}$. The image of $g_{1}(x)$ is then the interval $\left(\frac{1}{6}, \infty\right)$. The image of $g_{2}(x)$ is simply the values of $g_{1}(x)$ for $x$ in $\left(\frac{1}{6}, 1\right)$, which is the interval $\left(\frac{2}{5}, \infty\right)$. Similarly, the image of $g_{3}(x)$ is $\left(\frac{5}{9}, \infty\right)$, the image of $g_{4}(x)$ is $\left(\frac{3}{4}, \infty\right)$, and the image of $g_{5}(x)$ is $\left(\frac{4}{3}, \infty\right)$. As this does not intersect the interval $(-1,1), g_{6}(x)$ is not defined for any $x$, so the answer is 5 .
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n22. [12]",
"solution_match": "\nAnswer: "
}
|
3db9e146-e691-5da4-9394-04c13c7f3cfc
| 608,762
|
Let $a_{1}, a_{2}, \ldots$ be an infinite sequence of positive integers such that for integers $n>2, a_{n}=$ $3 a_{n-1}-2 a_{n-2}$. How many such sequences $\left\{a_{n}\right\}$ are there such that $a_{2010} \leq 2^{2012}$ ?
|
$36 \cdot 2^{2009}+36$ Consider the characteristic polynomial for the recurrence $a_{n+2}-3 a_{n+1}+$ $2 a_{n}=0$, which is $x^{2}-3 x+2$. The roots are at 2 and 1 , so we know that numbers $a_{i}$ must be of the form $a_{i}=a 2^{i-1}+b$ for integers $a$ and $b$. Therefore $a_{2010}$ must equal to $a 2^{2009}+b$, where $a$ and $b$ are both integers. If the expression is always positive, it is sufficient to say $a_{1}$ is positive and $a$ is nonnegative, or $a+b>0$, and $a \geq 0$.
For a given value of $a, 1-a \leq b \leq 2^{2012}-a 2^{2009}$, so there are $2^{2012}-a 2^{2009}+a$ possible values of $b$ for each $a$ (where the quantity is positive). $a$ can take any value between 0 and $2^{3}$, we sum over all such a in this range, to attain $9 \cdot 2^{2012}-(1+2+3+4+5+6+7+8) 2^{2009}+(1+2+3+4+5+6+7+8)$, or $36\left(2^{2009}\right)+36$, which is our answer.
|
36 \cdot 2^{2009} + 36
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a_{1}, a_{2}, \ldots$ be an infinite sequence of positive integers such that for integers $n>2, a_{n}=$ $3 a_{n-1}-2 a_{n-2}$. How many such sequences $\left\{a_{n}\right\}$ are there such that $a_{2010} \leq 2^{2012}$ ?
|
$36 \cdot 2^{2009}+36$ Consider the characteristic polynomial for the recurrence $a_{n+2}-3 a_{n+1}+$ $2 a_{n}=0$, which is $x^{2}-3 x+2$. The roots are at 2 and 1 , so we know that numbers $a_{i}$ must be of the form $a_{i}=a 2^{i-1}+b$ for integers $a$ and $b$. Therefore $a_{2010}$ must equal to $a 2^{2009}+b$, where $a$ and $b$ are both integers. If the expression is always positive, it is sufficient to say $a_{1}$ is positive and $a$ is nonnegative, or $a+b>0$, and $a \geq 0$.
For a given value of $a, 1-a \leq b \leq 2^{2012}-a 2^{2009}$, so there are $2^{2012}-a 2^{2009}+a$ possible values of $b$ for each $a$ (where the quantity is positive). $a$ can take any value between 0 and $2^{3}$, we sum over all such a in this range, to attain $9 \cdot 2^{2012}-(1+2+3+4+5+6+7+8) 2^{2009}+(1+2+3+4+5+6+7+8)$, or $36\left(2^{2009}\right)+36$, which is our answer.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n23. [12]",
"solution_match": "\nAnswer: "
}
|
97ed73a5-4b7e-5b83-9bd9-346da57633e9
| 608,763
|
Let $P(x)$ be a polynomial of degree at most 3 such that $P(x)=\frac{1}{1+x+x^{2}}$ for $x=1,2,3,4$. What is $P(5)$ ?
|
$\frac{-3}{91}$ The forward difference of a polynomial $P$ is $\Delta P(x)=P(x+1)-P(x)$, which is a new polynomial with degree reduced by one. Therefore, if we apply this operation three times we'll get a constant function, and we can work back up to get a value of $P(5)$. Practically, we create the following table of differences:
$$
\begin{array}{ccccccc}
\frac{1}{3} & & \frac{1}{7} & & \frac{1}{13} & & \frac{1}{21} \\
& \frac{-4}{21} & & \frac{-6}{91} & & \frac{-8}{273} & \\
& \frac{34}{273} & & \frac{10}{273} & & \\
& & \frac{-24}{273} & & &
\end{array}
$$
Then extend it to be the following table:
So our answer is $\frac{-9}{273}=\frac{-3}{91}$
|
\frac{-3}{91}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $P(x)$ be a polynomial of degree at most 3 such that $P(x)=\frac{1}{1+x+x^{2}}$ for $x=1,2,3,4$. What is $P(5)$ ?
|
$\frac{-3}{91}$ The forward difference of a polynomial $P$ is $\Delta P(x)=P(x+1)-P(x)$, which is a new polynomial with degree reduced by one. Therefore, if we apply this operation three times we'll get a constant function, and we can work back up to get a value of $P(5)$. Practically, we create the following table of differences:
$$
\begin{array}{ccccccc}
\frac{1}{3} & & \frac{1}{7} & & \frac{1}{13} & & \frac{1}{21} \\
& \frac{-4}{21} & & \frac{-6}{91} & & \frac{-8}{273} & \\
& \frac{34}{273} & & \frac{10}{273} & & \\
& & \frac{-24}{273} & & &
\end{array}
$$
Then extend it to be the following table:
So our answer is $\frac{-9}{273}=\frac{-3}{91}$
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n24. [12]",
"solution_match": "\nAnswer: "
}
|
e3b294bc-8f32-5e97-97f4-5a43b91426e4
| 608,764
|
Triangle $A B C$ is given with $A B=13, B C=14, C A=15$. Let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively. Let $G$ be the foot of the altitude from $A$ in triangle $A F E$. Find $A G$.
|
| $\frac{396}{65}$ |
| :---: | By Heron's formula we have $[A B C]=\sqrt{21(8)(7)(6)}=84$. Let $D$ be the foot of the altitude from $A$ to $B C$; then $A D=2 \cdot \frac{84}{14}=12$. Notice that because $\angle B F C=\angle B E C, B F E C$ is cyclic, so $\angle A F E=90-\angle E F C=90-\angle E B C=\angle C$. Therefore, we have $\triangle A E F \sim \triangle A B C$, so $\frac{A G}{A D}=\frac{A E}{A B} ; \frac{1}{2}(B E)(A C)=84 \Longrightarrow B E=\frac{56}{5} \Longrightarrow A E=\sqrt{13^{2}-\left(\frac{56}{5}\right)^{2}}=\sqrt{\frac{65^{2}-56^{2}}{5^{2}}}=\frac{33}{5}$. Then $A G=A D \cdot \frac{A E}{A B}=12 \cdot \frac{33 / 5}{13}=\frac{396}{65}$.
|
\frac{396}{65}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Triangle $A B C$ is given with $A B=13, B C=14, C A=15$. Let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively. Let $G$ be the foot of the altitude from $A$ in triangle $A F E$. Find $A G$.
|
| $\frac{396}{65}$ |
| :---: | By Heron's formula we have $[A B C]=\sqrt{21(8)(7)(6)}=84$. Let $D$ be the foot of the altitude from $A$ to $B C$; then $A D=2 \cdot \frac{84}{14}=12$. Notice that because $\angle B F C=\angle B E C, B F E C$ is cyclic, so $\angle A F E=90-\angle E F C=90-\angle E B C=\angle C$. Therefore, we have $\triangle A E F \sim \triangle A B C$, so $\frac{A G}{A D}=\frac{A E}{A B} ; \frac{1}{2}(B E)(A C)=84 \Longrightarrow B E=\frac{56}{5} \Longrightarrow A E=\sqrt{13^{2}-\left(\frac{56}{5}\right)^{2}}=\sqrt{\frac{65^{2}-56^{2}}{5^{2}}}=\frac{33}{5}$. Then $A G=A D \cdot \frac{A E}{A B}=12 \cdot \frac{33 / 5}{13}=\frac{396}{65}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n25. [14]",
"solution_match": "\nAnswer: "
}
|
9c43d0ca-5b45-5e90-bc43-adb7fc9b4403
| 608,765
|
$w, x, y, z$ are real numbers such that
$$
\begin{aligned}
w+x+y+z & =5 \\
2 w+4 x+8 y+16 z & =7 \\
3 w+9 x+27 y+81 z & =11 \\
4 w+16 x+64 y+256 z & =1
\end{aligned}
$$
What is the value of $5 w+25 x+125 y+625 z ?$
|
-60 We note this system of equations is equivalent to evaluating the polynomial (in $a$ ) $P(a)=w a+x a^{2}+y a^{3}+z a^{4}$ at $1,2,3$, and 4 . We know that $P(0)=0, P(1)=5, P(2)=7, P(3)=11$, $P(4)=1$. The finite difference of a polynomial $f$ is $f(n+1)-f(n)$, which is a polynomial with degree one less than the degree of $f$. The second, third, etc finite differences come from applying this operation repeatedly. The fourth finite difference of this polynomial is constant because this is a fourth degree polynomial. Repeatedly applying finite differences, we get
and we see that the fourth finite difference is -21 . We can extend this table, knowing that the fourth finite difference is always -21 , and we find that that $P(5)=-60$. The complete table is
| 0 | | 5 | | 7 | | 11 | | 1 | | -60 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | 5 | 2 | | 4 | | -10 | | -61 | | |
| | | -3 | | 2 | | -14 | | -51 | | |
| | | 5 | | -16 | | -37 | | | | |
| | | | | -21 | | -21 | | | | |
|
-60
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
$w, x, y, z$ are real numbers such that
$$
\begin{aligned}
w+x+y+z & =5 \\
2 w+4 x+8 y+16 z & =7 \\
3 w+9 x+27 y+81 z & =11 \\
4 w+16 x+64 y+256 z & =1
\end{aligned}
$$
What is the value of $5 w+25 x+125 y+625 z ?$
|
-60 We note this system of equations is equivalent to evaluating the polynomial (in $a$ ) $P(a)=w a+x a^{2}+y a^{3}+z a^{4}$ at $1,2,3$, and 4 . We know that $P(0)=0, P(1)=5, P(2)=7, P(3)=11$, $P(4)=1$. The finite difference of a polynomial $f$ is $f(n+1)-f(n)$, which is a polynomial with degree one less than the degree of $f$. The second, third, etc finite differences come from applying this operation repeatedly. The fourth finite difference of this polynomial is constant because this is a fourth degree polynomial. Repeatedly applying finite differences, we get
and we see that the fourth finite difference is -21 . We can extend this table, knowing that the fourth finite difference is always -21 , and we find that that $P(5)=-60$. The complete table is
| 0 | | 5 | | 7 | | 11 | | 1 | | -60 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | 5 | 2 | | 4 | | -10 | | -61 | | |
| | | -3 | | 2 | | -14 | | -51 | | |
| | | 5 | | -16 | | -37 | | | | |
| | | | | -21 | | -21 | | | | |
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n26. [14]",
"solution_match": "\nAnswer: "
}
|
56910d98-308b-5b52-bc69-7bcf4afd49bf
| 608,766
|
Let $f(x)=-x^{2}+10 x-20$. Find the sum of all $2^{2010}$ solutions to $\underbrace{f(f(\ldots(x) \ldots))}_{2010 f \mathrm{~s}}=2$.
|
$5 \cdot 2^{2010}$ Define $g(x)=f(f(\ldots(x) \ldots))$. We calculate:
$f(10-x)=-(10-x)^{2}+10(10-x)-20=-100+20 x-x^{2}+100-10 x-20=-x^{2}+10 x-20=f(x)$.
This implies that $g(10-x)=g(x)$. So if $g(x)=2$, then $g(10-x)=2$. Moreover, we can calculate $f(5)=-25+50-20=5$, so $g(5)=5 \neq 2$. Thus the possible solutions to $g(x)=2$ can be grouped into pairs, $\left(x_{1}, 10-x_{1}\right),\left(x_{2}, 10-x_{2}\right), \ldots$ The sum of the members of each pair is 10 , and there are $2^{2009}$ pairs, so the sum is
$$
10 \cdot 2^{2009}=5 \cdot 2^{2010}
$$
|
5 \cdot 2^{2010}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $f(x)=-x^{2}+10 x-20$. Find the sum of all $2^{2010}$ solutions to $\underbrace{f(f(\ldots(x) \ldots))}_{2010 f \mathrm{~s}}=2$.
|
$5 \cdot 2^{2010}$ Define $g(x)=f(f(\ldots(x) \ldots))$. We calculate:
$f(10-x)=-(10-x)^{2}+10(10-x)-20=-100+20 x-x^{2}+100-10 x-20=-x^{2}+10 x-20=f(x)$.
This implies that $g(10-x)=g(x)$. So if $g(x)=2$, then $g(10-x)=2$. Moreover, we can calculate $f(5)=-25+50-20=5$, so $g(5)=5 \neq 2$. Thus the possible solutions to $g(x)=2$ can be grouped into pairs, $\left(x_{1}, 10-x_{1}\right),\left(x_{2}, 10-x_{2}\right), \ldots$ The sum of the members of each pair is 10 , and there are $2^{2009}$ pairs, so the sum is
$$
10 \cdot 2^{2009}=5 \cdot 2^{2010}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n27. [14]",
"solution_match": "\nAnswer: "
}
|
64d1aa0b-bcee-5eb9-b7be-f7121d057ede
| 608,767
|
In the game of set, each card has four attributes, each of which takes on one of three values. A set deck consists of one card for each of the 81 possible four-tuples of attributes. Given a collection of 3 cards, call an attribute good for that collection if the three cards either all take on the same value of that attribute or take on all three different values of that attribute. Call a collection of 3 cards two-good if exactly two attributes are good for that collection. How many two-good collections of 3 cards are there? The order in which the cards appear does not matter.
|
25272 In counting the number of sets of 3 cards, we first want to choose which of our two attributes will be good and which of our two attributes will not be good. There are $\binom{4}{2}=6$ such choices.
Now consider the two attributes which are not good, attribute X and attribute Y. Since these are not good, some value should appear exactly twice. Suppose the value $a$ appears twice and $b$ appears once for attribute $X$ and that the value $c$ appears twice and $d$ appears once for attribute $Y$. There are three choices for $a$ and then two choices for $b$; similarly, there are three choices for $c$ and then two choices for $d$. This gives $3 \cdot 2 \cdot 3 \cdot 2=36$ choices of $a, b, c$, and $d$.
There are two cases to consider. The first is that there are two cards which both have $a$ and $c$, while the other card has both $b$ and $d$. The second case is that only one card has both $a$ and $c$, while one card has $a$ and $d$ and the other has $b$ and $c$.
Case 1:
| Card $1 \quad$ Card 2 | Card 3 | |
| :---: | :---: | :---: |
| - Good attribute 1 - | | |
| - Good attribute $2-$ | | |
| a | a | b |
| c | c | d |
The three cards need to be distinct. Card 3 is necessarily distinct from Card 1 and Card 2, but we need to ensure that Card 1 and Card 2 are distinct from each other. There are 9 choices for the two good attributes of Card 1, and then 8 choices for the two good attributes of Card 2. But we also want to divide by 2 since we do not care about the order of Card 1 and Card 2. So there are $\frac{9 \cdot 8}{2}=36$ choices for the good attributes on Card 1 and Card 2. Then, the values of the good attributes of Card 1 and Card 2 uniquely determine the values of the good attributes of Card 3.
Case 2:
Card 1 Card 2 Card 3
- Good attribute 1 -
- Good attribute 2 -
| $a$ | $a$ | $b$ |
| :--- | :--- | :--- |
| $c$ | $d$ | $c$ |
Card 1, Card 2, and Card 3 will all be distinct no matter what the values of the good attributes are, because the values of attributes $X$ and $Y$ are unique to each card. So there are 9 possibilities for the the values of the good attributes on card 1, and then there are 9 more possibilities for the values of the good attribute on Card 2. We do not have to divide by 2 this time, since Card 1 and Card 2 have distinct values in $X$ and $Y$. So there are $9^{2}=81$ possibilities here.
So our final answer is $6 \cdot 6^{2} \cdot(36+81)=25272$.
|
25272
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In the game of set, each card has four attributes, each of which takes on one of three values. A set deck consists of one card for each of the 81 possible four-tuples of attributes. Given a collection of 3 cards, call an attribute good for that collection if the three cards either all take on the same value of that attribute or take on all three different values of that attribute. Call a collection of 3 cards two-good if exactly two attributes are good for that collection. How many two-good collections of 3 cards are there? The order in which the cards appear does not matter.
|
25272 In counting the number of sets of 3 cards, we first want to choose which of our two attributes will be good and which of our two attributes will not be good. There are $\binom{4}{2}=6$ such choices.
Now consider the two attributes which are not good, attribute X and attribute Y. Since these are not good, some value should appear exactly twice. Suppose the value $a$ appears twice and $b$ appears once for attribute $X$ and that the value $c$ appears twice and $d$ appears once for attribute $Y$. There are three choices for $a$ and then two choices for $b$; similarly, there are three choices for $c$ and then two choices for $d$. This gives $3 \cdot 2 \cdot 3 \cdot 2=36$ choices of $a, b, c$, and $d$.
There are two cases to consider. The first is that there are two cards which both have $a$ and $c$, while the other card has both $b$ and $d$. The second case is that only one card has both $a$ and $c$, while one card has $a$ and $d$ and the other has $b$ and $c$.
Case 1:
| Card $1 \quad$ Card 2 | Card 3 | |
| :---: | :---: | :---: |
| - Good attribute 1 - | | |
| - Good attribute $2-$ | | |
| a | a | b |
| c | c | d |
The three cards need to be distinct. Card 3 is necessarily distinct from Card 1 and Card 2, but we need to ensure that Card 1 and Card 2 are distinct from each other. There are 9 choices for the two good attributes of Card 1, and then 8 choices for the two good attributes of Card 2. But we also want to divide by 2 since we do not care about the order of Card 1 and Card 2. So there are $\frac{9 \cdot 8}{2}=36$ choices for the good attributes on Card 1 and Card 2. Then, the values of the good attributes of Card 1 and Card 2 uniquely determine the values of the good attributes of Card 3.
Case 2:
Card 1 Card 2 Card 3
- Good attribute 1 -
- Good attribute 2 -
| $a$ | $a$ | $b$ |
| :--- | :--- | :--- |
| $c$ | $d$ | $c$ |
Card 1, Card 2, and Card 3 will all be distinct no matter what the values of the good attributes are, because the values of attributes $X$ and $Y$ are unique to each card. So there are 9 possibilities for the the values of the good attributes on card 1, and then there are 9 more possibilities for the values of the good attribute on Card 2. We do not have to divide by 2 this time, since Card 1 and Card 2 have distinct values in $X$ and $Y$. So there are $9^{2}=81$ possibilities here.
So our final answer is $6 \cdot 6^{2} \cdot(36+81)=25272$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n28. [17]",
"solution_match": "\nAnswer: "
}
|
d9f3ac7e-9583-5036-8bb4-68ee33eb5bd5
| 608,768
|
In the game of Galactic Dominion, players compete to amass cards, each of which is worth a certain number of points. Say you are playing a version of this game with only two kinds of cards, planet cards and hegemon cards. Each planet card is worth 2010 points, and each hegemon card is worth four points per planet card held. You start with no planet cards and no hegemon cards, and, on each turn, starting at turn one, you take either a planet card or a hegemon card, whichever is worth more points given the hand you currently hold. Define a sequence $\left\{a_{n}\right\}$ for all positive integers $n$ by setting $a_{n}$ to be 0 if on turn $n$ you take a planet card and 1 if you take a hegemon card. What is the smallest value of $N$ such that the sequence $a_{N}, a_{N+1}, \ldots$ is necessarily periodic (meaning that there is a positive integer $k$ such that $a_{n+k}=a_{n}$ for all $\left.n \geq N\right)$ ?
|
503 If you have $P$ planets and $H$ hegemons, buying a planet gives you $2010+4 H$ points while buying a hegemon gives you $4 P$ points. Thus you buy a hegemon whenever $P-H \geq 502.5$, and you buy a planet whenever $P-H \leq 502.5$. Therefore $a_{i}=1$ for $1 \leq i \leq 503$. Starting at $i=504$ (at which point you have bought 503 planets) you must alternate buying planets and hegemons. The sequence $\left\{a_{i}\right\}_{i \geq 503}$ is periodic with period 2 .
|
503
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In the game of Galactic Dominion, players compete to amass cards, each of which is worth a certain number of points. Say you are playing a version of this game with only two kinds of cards, planet cards and hegemon cards. Each planet card is worth 2010 points, and each hegemon card is worth four points per planet card held. You start with no planet cards and no hegemon cards, and, on each turn, starting at turn one, you take either a planet card or a hegemon card, whichever is worth more points given the hand you currently hold. Define a sequence $\left\{a_{n}\right\}$ for all positive integers $n$ by setting $a_{n}$ to be 0 if on turn $n$ you take a planet card and 1 if you take a hegemon card. What is the smallest value of $N$ such that the sequence $a_{N}, a_{N+1}, \ldots$ is necessarily periodic (meaning that there is a positive integer $k$ such that $a_{n+k}=a_{n}$ for all $\left.n \geq N\right)$ ?
|
503 If you have $P$ planets and $H$ hegemons, buying a planet gives you $2010+4 H$ points while buying a hegemon gives you $4 P$ points. Thus you buy a hegemon whenever $P-H \geq 502.5$, and you buy a planet whenever $P-H \leq 502.5$. Therefore $a_{i}=1$ for $1 \leq i \leq 503$. Starting at $i=504$ (at which point you have bought 503 planets) you must alternate buying planets and hegemons. The sequence $\left\{a_{i}\right\}_{i \geq 503}$ is periodic with period 2 .
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n29. [17]",
"solution_match": "\nAnswer: "
}
|
f99406ea-824a-580a-aa29-2e6da2166f97
| 608,769
|
In the game of projective set, each card contains some nonempty subset of six distinguishable dots. A projective set deck consists of one card for each of the 63 possible nonempty subsets of dots. How many collections of five cards have an even number of each dot? The order in which the cards appear does not matter.
|
109368 We'll first count sets of cards where the order does matter. Suppose we choose the first four cards. Then there is exactly one card that can make each dot appear twice. However, this card could be empty or it could be one of the cards we've already chosen, so we have to subtract for these two cases. First, there are $63 \cdot 62 \cdot 61 \cdot 60$ ways to choose the first four cards. Let's now count how many ways there are that the fifth card could be empty.
The fifth card is empty if and only if the first four cards already have an even number of each dot. Suppose we choose the first two cards. There is a possible fourth card if the third card is not either of the first two or the card that completes a set. If that is the case, then the fourth card is unique. This comes to $63 \cdot 62 \cdot 60$ cases.
Now consider how many ways there are for the fifth card to be a duplicate. This is just the number of ways for three cards to have an even number of each dot, then have two copies of the same card in the other two slots, one of which needs to be the fifth slot. The number of ways for three cards to have an even number of each dot is just the number of ways to choose two cards. Therefore, we'll choose two cards ( $63 \cdot 62$ ways), choose the slot in the first four positions for the duplicate card ( 4 ways), and the duplicate card, which can't be any of the nonduplicated cards, so there are 60 choices. Therefore, there are $63 \cdot 62 \cdot 4 \cdot 60$ ways for the fifth card to be the same as one of the first four.
This means that the number of five card sets where the order does matter is $63 \cdot 62 \cdot 61 \cdot 60-63 \cdot 62 \cdot 60-$ $63 \cdot 62 \cdot 4 \cdot 60$, so our final answer is $\frac{63 \cdot 62 \cdot 61 \cdot 60-63 \cdot 62 \cdot 60-63 \cdot 62 \cdot 4 \cdot 60}{120}=\frac{63 \cdot 62 \cdot(61-1-4)}{2}=63 \cdot 31 \cdot 56=109368$.
|
109368
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In the game of projective set, each card contains some nonempty subset of six distinguishable dots. A projective set deck consists of one card for each of the 63 possible nonempty subsets of dots. How many collections of five cards have an even number of each dot? The order in which the cards appear does not matter.
|
109368 We'll first count sets of cards where the order does matter. Suppose we choose the first four cards. Then there is exactly one card that can make each dot appear twice. However, this card could be empty or it could be one of the cards we've already chosen, so we have to subtract for these two cases. First, there are $63 \cdot 62 \cdot 61 \cdot 60$ ways to choose the first four cards. Let's now count how many ways there are that the fifth card could be empty.
The fifth card is empty if and only if the first four cards already have an even number of each dot. Suppose we choose the first two cards. There is a possible fourth card if the third card is not either of the first two or the card that completes a set. If that is the case, then the fourth card is unique. This comes to $63 \cdot 62 \cdot 60$ cases.
Now consider how many ways there are for the fifth card to be a duplicate. This is just the number of ways for three cards to have an even number of each dot, then have two copies of the same card in the other two slots, one of which needs to be the fifth slot. The number of ways for three cards to have an even number of each dot is just the number of ways to choose two cards. Therefore, we'll choose two cards ( $63 \cdot 62$ ways), choose the slot in the first four positions for the duplicate card ( 4 ways), and the duplicate card, which can't be any of the nonduplicated cards, so there are 60 choices. Therefore, there are $63 \cdot 62 \cdot 4 \cdot 60$ ways for the fifth card to be the same as one of the first four.
This means that the number of five card sets where the order does matter is $63 \cdot 62 \cdot 61 \cdot 60-63 \cdot 62 \cdot 60-$ $63 \cdot 62 \cdot 4 \cdot 60$, so our final answer is $\frac{63 \cdot 62 \cdot 61 \cdot 60-63 \cdot 62 \cdot 60-63 \cdot 62 \cdot 4 \cdot 60}{120}=\frac{63 \cdot 62 \cdot(61-1-4)}{2}=63 \cdot 31 \cdot 56=109368$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n30. [17]",
"solution_match": "\nAnswer: "
}
|
fb5f2dad-1500-5989-8e97-47be89e6884f
| 608,770
|
What is the perimeter of the triangle formed by the points of tangency of the incircle of a 5-7-8 triangle with its sides?
|
$\frac{9 \sqrt{21}}{7}+3$ Let $\triangle A B C$ be a triangle with sides $a=7, b=5$, and $c=8$. Let the incircle of $\triangle A B C$ be tangent to sides $B C, C A$, and $A B$ at points $D, E$, and $F$. By the law of cosines (using the form $\left.\cos (A)=\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)$, we have
$$
\begin{aligned}
& \cos (A)=\frac{8^{2}+5^{2}-7^{2}}{2(5)(8)}=\frac{1}{2} \\
& \cos (B)=\frac{8^{2}+7^{2}-5^{2}}{2(7)(8)}=\frac{11}{14} \\
& \cos (C)=\frac{5^{2}+7^{2}-8^{2}}{2(7)(5)}=\frac{1}{7}
\end{aligned}
$$
Now we observe that $A E F, B D F$, and $C D E$ are all isosceles. Let us call the lengths of the legs of these triangles $s, t$, and $u$, respectively. Then we know that $s+t=8, t+u=7$, and $u+s=5$, so $s=3, t=5$, and $u=2$.
Our final observation is that an isosceles angle with legs of length $l$ and whose non-equal angle is $\theta$ has a base of length $l \sqrt{2(1-\cos (\theta))}$. This can be proven using the law of cosines or the Pythagorean theorem.
Using this, we can calculate that
$$
\begin{aligned}
D E & =2 \sqrt{2(1-\cos (C))} \\
& =2 \sqrt{\frac{12}{7}} \\
E F & =3 \sqrt{2(1-\cos (A))} \\
& =3 \\
F D & =5 \sqrt{2(1-\cos (B))} \\
& =5 \sqrt{\frac{3}{7}}
\end{aligned}
$$
and then
$$
\begin{aligned}
D E+E F+F D & =2 \sqrt{\frac{12}{7}}+3+5 \sqrt{\frac{3}{7}} \\
& =3+9 \sqrt{\frac{3}{7}} \\
& =3+9 \frac{\sqrt{21}}{7}
\end{aligned}
$$
|
3+\frac{9\sqrt{21}}{7}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
What is the perimeter of the triangle formed by the points of tangency of the incircle of a 5-7-8 triangle with its sides?
|
$\frac{9 \sqrt{21}}{7}+3$ Let $\triangle A B C$ be a triangle with sides $a=7, b=5$, and $c=8$. Let the incircle of $\triangle A B C$ be tangent to sides $B C, C A$, and $A B$ at points $D, E$, and $F$. By the law of cosines (using the form $\left.\cos (A)=\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)$, we have
$$
\begin{aligned}
& \cos (A)=\frac{8^{2}+5^{2}-7^{2}}{2(5)(8)}=\frac{1}{2} \\
& \cos (B)=\frac{8^{2}+7^{2}-5^{2}}{2(7)(8)}=\frac{11}{14} \\
& \cos (C)=\frac{5^{2}+7^{2}-8^{2}}{2(7)(5)}=\frac{1}{7}
\end{aligned}
$$
Now we observe that $A E F, B D F$, and $C D E$ are all isosceles. Let us call the lengths of the legs of these triangles $s, t$, and $u$, respectively. Then we know that $s+t=8, t+u=7$, and $u+s=5$, so $s=3, t=5$, and $u=2$.
Our final observation is that an isosceles angle with legs of length $l$ and whose non-equal angle is $\theta$ has a base of length $l \sqrt{2(1-\cos (\theta))}$. This can be proven using the law of cosines or the Pythagorean theorem.
Using this, we can calculate that
$$
\begin{aligned}
D E & =2 \sqrt{2(1-\cos (C))} \\
& =2 \sqrt{\frac{12}{7}} \\
E F & =3 \sqrt{2(1-\cos (A))} \\
& =3 \\
F D & =5 \sqrt{2(1-\cos (B))} \\
& =5 \sqrt{\frac{3}{7}}
\end{aligned}
$$
and then
$$
\begin{aligned}
D E+E F+F D & =2 \sqrt{\frac{12}{7}}+3+5 \sqrt{\frac{3}{7}} \\
& =3+9 \sqrt{\frac{3}{7}} \\
& =3+9 \frac{\sqrt{21}}{7}
\end{aligned}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n31. [20]",
"solution_match": "\nAnswer: "
}
|
cb22d921-b984-5c46-b2df-a785a28fa690
| 608,771
|
Let $T$ be the set of numbers of the form $2^{a} 3^{b}$ where $a$ and $b$ are integers satisfying $0 \leq a, b \leq 5$. How many subsets $S$ of $T$ have the property that if $n$ is in $S$ then all positive integer divisors of $n$ are in $S$ ?
|
924 Consider the correspondence $(a, b) \leftrightarrow 2^{a} 3^{b}$ for non-negative integers $a$ and $b$. So we can view $T$ as the square of lattice points $(a, b)$ where $0 \leq a, b \leq 5$, and subsets of $T$ as subsets of this square.
Notice then that the integer corresponding to $\left(a_{1}, b_{1}\right)$ is a divisor of the integer corresponding to $\left(a_{2}, b_{2}\right)$ if and only if $0 \leq a_{1} \leq a_{1}$ and $0 \leq b_{1} \leq b_{2}$. This means that subsets $S \subset T$ with the desired property,
correspond to subsets of the square where if a point is in the set, then so are all points to the left and south of it.
Consider any such subset $S$. For each $0 \leq x \leq 5$, let $S_{x}$ be the maximum $y$ value of any point $(x, y) \in S$, or -1 if there is no such point. We claim the values $S_{x}$ uniquely characterize $S$. This is because each $S_{x}$ characterizes the points of the form $(x, y)$ in $S$. In particular, $(x, z)$ will be in $S$ if and only if $z \leq S_{x}$. If $(x, z) \in S$ with $z>S_{x}$, then $S_{x}$ is not the maximum value, and if $(x, z) \notin S$ with $z \leq S_{x}$, then $S$ fails to satisfy the desired property.
We now claim that $S_{x} \geq S_{y}$ for $x<y$, so the sequence $S_{0}, \ldots, S_{1}$ is decreasing. This is because if $\left(y, S_{y}\right)$ is in the set $S$, then so must be $\left(x, S_{y}\right)$. Conversely, it is easy to see that if $S_{0}, \ldots, S_{1}$ is decreasing, then $S$ is a set satisfying the desired property.
We now claim that decreasing sequences $S_{0}, \ldots, S_{5}$ are in bijective correspondence with walks going only right and down from $(-1,5)$ to $(5,-1)$. The sequence $S_{0}, \ldots, S_{5}$ simply corresponds to the walk $(-1,5) \rightarrow\left(-1, S_{0}\right) \rightarrow\left(0, S_{0}\right) \rightarrow\left(0, S_{1}\right) \rightarrow\left(1, S_{1}\right) \rightarrow \cdots \rightarrow\left(4, S_{5}\right) \rightarrow\left(5, S_{5}\right) \rightarrow(5,-1)$. Geometrically, we are tracing out the outline of the set $S$.
The number of such walks is simply $\binom{12}{6}$, since we can view it as choosing the 6 of 12 steps at which to move right. Thus the number of subsets $S$ of $T$ with the desired property is $\binom{12}{6}=924$.
|
924
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $T$ be the set of numbers of the form $2^{a} 3^{b}$ where $a$ and $b$ are integers satisfying $0 \leq a, b \leq 5$. How many subsets $S$ of $T$ have the property that if $n$ is in $S$ then all positive integer divisors of $n$ are in $S$ ?
|
924 Consider the correspondence $(a, b) \leftrightarrow 2^{a} 3^{b}$ for non-negative integers $a$ and $b$. So we can view $T$ as the square of lattice points $(a, b)$ where $0 \leq a, b \leq 5$, and subsets of $T$ as subsets of this square.
Notice then that the integer corresponding to $\left(a_{1}, b_{1}\right)$ is a divisor of the integer corresponding to $\left(a_{2}, b_{2}\right)$ if and only if $0 \leq a_{1} \leq a_{1}$ and $0 \leq b_{1} \leq b_{2}$. This means that subsets $S \subset T$ with the desired property,
correspond to subsets of the square where if a point is in the set, then so are all points to the left and south of it.
Consider any such subset $S$. For each $0 \leq x \leq 5$, let $S_{x}$ be the maximum $y$ value of any point $(x, y) \in S$, or -1 if there is no such point. We claim the values $S_{x}$ uniquely characterize $S$. This is because each $S_{x}$ characterizes the points of the form $(x, y)$ in $S$. In particular, $(x, z)$ will be in $S$ if and only if $z \leq S_{x}$. If $(x, z) \in S$ with $z>S_{x}$, then $S_{x}$ is not the maximum value, and if $(x, z) \notin S$ with $z \leq S_{x}$, then $S$ fails to satisfy the desired property.
We now claim that $S_{x} \geq S_{y}$ for $x<y$, so the sequence $S_{0}, \ldots, S_{1}$ is decreasing. This is because if $\left(y, S_{y}\right)$ is in the set $S$, then so must be $\left(x, S_{y}\right)$. Conversely, it is easy to see that if $S_{0}, \ldots, S_{1}$ is decreasing, then $S$ is a set satisfying the desired property.
We now claim that decreasing sequences $S_{0}, \ldots, S_{5}$ are in bijective correspondence with walks going only right and down from $(-1,5)$ to $(5,-1)$. The sequence $S_{0}, \ldots, S_{5}$ simply corresponds to the walk $(-1,5) \rightarrow\left(-1, S_{0}\right) \rightarrow\left(0, S_{0}\right) \rightarrow\left(0, S_{1}\right) \rightarrow\left(1, S_{1}\right) \rightarrow \cdots \rightarrow\left(4, S_{5}\right) \rightarrow\left(5, S_{5}\right) \rightarrow(5,-1)$. Geometrically, we are tracing out the outline of the set $S$.
The number of such walks is simply $\binom{12}{6}$, since we can view it as choosing the 6 of 12 steps at which to move right. Thus the number of subsets $S$ of $T$ with the desired property is $\binom{12}{6}=924$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n32. [20]",
"solution_match": "\nAnswer: "
}
|
fcfcac7f-9f19-5ead-9c45-57c12738c203
| 608,772
|
Convex quadrilateral $B C D E$ lies in the plane. Lines $E B$ and $D C$ intersect at $A$, with $A B=2$, $A C=5, A D=200, A E=500$, and $\cos \angle B A C=\frac{7}{9}$. What is the largest number of nonoverlapping circles that can lie in quadrilateral $B C D E$ such that all of them are tangent to both lines $B E$ and $C D$ ?
|
5 Let $\theta=\angle B A C$, and $\cos \theta=\frac{7}{9}$ implies $\cos \frac{\theta}{2}=\sqrt{\frac{1+\frac{7}{9}}{2}}=\frac{2 \sqrt{2}}{3} ; \sin \frac{\theta}{2}=\frac{1}{3} ; B C=$ $\sqrt{4+25-2(2)(5) \frac{7}{9}}=\frac{11}{3}$. Let $O_{1}$ be the excircle of $\triangle A B C$ tangent to lines $A B$ and $A C$, and let $r_{1}$ be its radius; let $O_{1}$ be tangent to line $A B$ at point $P_{1}$. Then $A P_{1}=\frac{A B+B C+C A}{2}$ and $\frac{r_{1}}{A P_{1}}=\tan \frac{\theta}{2}=$ $\frac{1}{2 \sqrt{2}} \Longrightarrow r_{1}=\frac{16}{3 \cdot 2 \sqrt{2}}$. Let $O_{n}$ be a circle tangent to $O_{n-1}$ and the lines $A B$ and $A C$, and let $r_{n}$ be its radius; let $O_{n}$ be tangent to line $A B$ at point $P_{n}$. Then $\frac{O_{n} P_{n}}{A O_{n}}=\sin \frac{\theta}{2}=\frac{1}{3}$; since $\triangle A P_{n} O_{n} \sim$ $\triangle A P_{n-1} O_{n-1}$ and $O_{n} O_{n-1}=r_{n}+r_{n-1}$, we have $\frac{1}{3}=\frac{O_{n} P_{n}}{A O_{n}}=\frac{r_{n}}{A O_{n-1}+O_{n-1} O_{n}}=\frac{r_{n}}{3 r_{n-1}+r_{n}+r_{n-1}} \Longrightarrow$ $r_{n}=2 r_{n-1}=2^{n-1} \frac{16}{3 \cdot 2 \sqrt{2}}$. We want the highest $n$ such that $O_{n}$ is contained inside $\triangle A D E$. Let the incircle of $\triangle A D E$ be tangent to $A D$ at $X$; then the inradius of $\triangle A D E$ is $\frac{A X}{\tan \frac{\theta}{2}}=\frac{\frac{500+200-\frac{1100}{2}}{2 \sqrt{2}}}{2}=\frac{500}{3 \cdot 2 \sqrt{2}}$. We want the highest $n$ such that $r_{n} \leq \frac{500}{3 \cdot 2 \sqrt{2}}$; thus $2^{n-1} \cdot 16 \leq 500 \Longrightarrow n=5$.
|
5
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Convex quadrilateral $B C D E$ lies in the plane. Lines $E B$ and $D C$ intersect at $A$, with $A B=2$, $A C=5, A D=200, A E=500$, and $\cos \angle B A C=\frac{7}{9}$. What is the largest number of nonoverlapping circles that can lie in quadrilateral $B C D E$ such that all of them are tangent to both lines $B E$ and $C D$ ?
|
5 Let $\theta=\angle B A C$, and $\cos \theta=\frac{7}{9}$ implies $\cos \frac{\theta}{2}=\sqrt{\frac{1+\frac{7}{9}}{2}}=\frac{2 \sqrt{2}}{3} ; \sin \frac{\theta}{2}=\frac{1}{3} ; B C=$ $\sqrt{4+25-2(2)(5) \frac{7}{9}}=\frac{11}{3}$. Let $O_{1}$ be the excircle of $\triangle A B C$ tangent to lines $A B$ and $A C$, and let $r_{1}$ be its radius; let $O_{1}$ be tangent to line $A B$ at point $P_{1}$. Then $A P_{1}=\frac{A B+B C+C A}{2}$ and $\frac{r_{1}}{A P_{1}}=\tan \frac{\theta}{2}=$ $\frac{1}{2 \sqrt{2}} \Longrightarrow r_{1}=\frac{16}{3 \cdot 2 \sqrt{2}}$. Let $O_{n}$ be a circle tangent to $O_{n-1}$ and the lines $A B$ and $A C$, and let $r_{n}$ be its radius; let $O_{n}$ be tangent to line $A B$ at point $P_{n}$. Then $\frac{O_{n} P_{n}}{A O_{n}}=\sin \frac{\theta}{2}=\frac{1}{3}$; since $\triangle A P_{n} O_{n} \sim$ $\triangle A P_{n-1} O_{n-1}$ and $O_{n} O_{n-1}=r_{n}+r_{n-1}$, we have $\frac{1}{3}=\frac{O_{n} P_{n}}{A O_{n}}=\frac{r_{n}}{A O_{n-1}+O_{n-1} O_{n}}=\frac{r_{n}}{3 r_{n-1}+r_{n}+r_{n-1}} \Longrightarrow$ $r_{n}=2 r_{n-1}=2^{n-1} \frac{16}{3 \cdot 2 \sqrt{2}}$. We want the highest $n$ such that $O_{n}$ is contained inside $\triangle A D E$. Let the incircle of $\triangle A D E$ be tangent to $A D$ at $X$; then the inradius of $\triangle A D E$ is $\frac{A X}{\tan \frac{\theta}{2}}=\frac{\frac{500+200-\frac{1100}{2}}{2 \sqrt{2}}}{2}=\frac{500}{3 \cdot 2 \sqrt{2}}$. We want the highest $n$ such that $r_{n} \leq \frac{500}{3 \cdot 2 \sqrt{2}}$; thus $2^{n-1} \cdot 16 \leq 500 \Longrightarrow n=5$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n33. [20]",
"solution_match": "\nAnswer: "
}
|
2b03e710-57b2-5d67-8fca-3150b013c262
| 608,773
|
Estimate the sum of all the prime numbers less than $1,000,000$. If the correct answer is $X$ and you write down $A$, your team will receive $\min \left(\left\lfloor\frac{25 X}{A}\right\rfloor,\left\lfloor\frac{25 A}{X}\right\rfloor\right)$ points, where $\lfloor x\rfloor$ is the largest integer less than or equal to $x$.
|
37550402023 A decent approximation to the sum of all the primes can be obtained with the following two facts. First, there are approximately $\frac{n}{\ln n}$ primes less than $n$ and second, the $n^{\text {th }}$ prime is approximately $n \ln n$. We'll approximate $\ln 1000000$ as 15 (the actual number is 13.8), so there are approximately $\frac{10^{6}}{15}$ primes. Then we want $\sum_{n=1}^{\frac{106}{15}} n \ln n$. If you know calculus, this can be approximated by the integral $\int_{1}^{\frac{10^{6}}{15}} x \ln x d x$, which has the antiderivative $\frac{1}{2} x^{2} \ln x-\frac{1}{4} x^{2}$, giving an answer of around $\frac{1}{2} \cdot \frac{10^{12}}{225} \cdot(15-\ln 15)-\frac{1}{4} \cdot \frac{10^{12}}{225}$. Estimating $\ln 15$ as about 3 , this is approximately $\frac{23 \cdot 10^{12}}{900}=2.5 \cdot 10^{10}=25,000,000,000$. The actual answer is $37,550,402,023$, so an approximation of this accuracy would get 16 points. We can arrive at the same answer without calculus by approximating $\ln n$ as $\sum_{k=1}^{n} \frac{1}{k}$, so that our sum becomes $\sum_{n=1}^{\frac{106}{15}} \sum_{k=1}^{n} \frac{n}{k}=\sum_{k=1}^{\frac{106}{15}} \sum_{n=k}^{\frac{10^{6}}{15}} \frac{n}{k} \approx \sum_{k=1}^{\frac{10^{6}}{15}} \frac{1}{2} \cdot \frac{\frac{10^{12}}{225}-k^{2}}{k}=$ $\frac{1}{2} \cdot \frac{10^{12}}{225} \cdot \ln \left(\frac{10^{6}}{15}\right)-\frac{1}{2} \sum_{k=1}^{\frac{10^{6}}{15}} k \approx \frac{1}{2} \cdot \frac{10^{12}}{225} \cdot \ln \left(\frac{10^{6}}{15}\right)-\frac{1}{4} \frac{10^{12}}{225} \approx 2.5 \cdot 10^{10}$.
|
37550402023
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Estimate the sum of all the prime numbers less than $1,000,000$. If the correct answer is $X$ and you write down $A$, your team will receive $\min \left(\left\lfloor\frac{25 X}{A}\right\rfloor,\left\lfloor\frac{25 A}{X}\right\rfloor\right)$ points, where $\lfloor x\rfloor$ is the largest integer less than or equal to $x$.
|
37550402023 A decent approximation to the sum of all the primes can be obtained with the following two facts. First, there are approximately $\frac{n}{\ln n}$ primes less than $n$ and second, the $n^{\text {th }}$ prime is approximately $n \ln n$. We'll approximate $\ln 1000000$ as 15 (the actual number is 13.8), so there are approximately $\frac{10^{6}}{15}$ primes. Then we want $\sum_{n=1}^{\frac{106}{15}} n \ln n$. If you know calculus, this can be approximated by the integral $\int_{1}^{\frac{10^{6}}{15}} x \ln x d x$, which has the antiderivative $\frac{1}{2} x^{2} \ln x-\frac{1}{4} x^{2}$, giving an answer of around $\frac{1}{2} \cdot \frac{10^{12}}{225} \cdot(15-\ln 15)-\frac{1}{4} \cdot \frac{10^{12}}{225}$. Estimating $\ln 15$ as about 3 , this is approximately $\frac{23 \cdot 10^{12}}{900}=2.5 \cdot 10^{10}=25,000,000,000$. The actual answer is $37,550,402,023$, so an approximation of this accuracy would get 16 points. We can arrive at the same answer without calculus by approximating $\ln n$ as $\sum_{k=1}^{n} \frac{1}{k}$, so that our sum becomes $\sum_{n=1}^{\frac{106}{15}} \sum_{k=1}^{n} \frac{n}{k}=\sum_{k=1}^{\frac{106}{15}} \sum_{n=k}^{\frac{10^{6}}{15}} \frac{n}{k} \approx \sum_{k=1}^{\frac{10^{6}}{15}} \frac{1}{2} \cdot \frac{\frac{10^{12}}{225}-k^{2}}{k}=$ $\frac{1}{2} \cdot \frac{10^{12}}{225} \cdot \ln \left(\frac{10^{6}}{15}\right)-\frac{1}{2} \sum_{k=1}^{\frac{10^{6}}{15}} k \approx \frac{1}{2} \cdot \frac{10^{12}}{225} \cdot \ln \left(\frac{10^{6}}{15}\right)-\frac{1}{4} \frac{10^{12}}{225} \approx 2.5 \cdot 10^{10}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n34. [25]",
"solution_match": "\nAnswer: "
}
|
27bab699-8956-5879-b0d3-07beab3a505e
| 608,774
|
A mathematician $M^{\prime}$ is called a descendent of mathematician $M$ if there is a sequence of mathematicians $M=M_{1}, M_{2}, \ldots, M_{k}=M^{\prime}$ such that $M_{i}$ was $M_{i+1}$ 's doctoral advisor for all $i$. Estimate
the number of descendents that the mathematician who has had the largest number of descendents has had, according to the Mathematical Genealogy Project. Note that the Mathematical Genealogy Project has records dating back to the 1300s. If the correct answer is $X$ and you write down $A$, your team will receive max $\left(25-\left\lfloor\frac{|X-A|}{100}\right\rfloor, 0\right)$ points, where $\lfloor x\rfloor$ is the largest integer less than or equal to $x$.
|
82310 First let's estimate how many "generations" of mathematicians there have been since 1300. If we suppose that a mathematician gets his PhD around age 30 and becomes a PhD advisor around age 60 , then we'll get a generation length of approximately 30 years. However, not all mathematicians will train more than one PhD . Let's say that only $40 \%$ of mathematicians train at least 2 PhDs. Then effectively we have only $40 \%$ of the generations, or in other words each effective generation takes 75 years. Then we have $\frac{22}{3}$ branching generations. If we assume that all of these only train 2 PhDs , then we get an answer of $2^{\frac{22}{3}} \approx 1625$. But we can ensure that our chain has at least a single person who trained 100 PhDs (this is approximately the largest number of advisees for a single mathematician), allowing us to change one factor of 2 into a factor of 100 . That gives us an answer of $1625 \cdot 50=81250$, which is very close to the actual value of 82310 .
|
82310
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A mathematician $M^{\prime}$ is called a descendent of mathematician $M$ if there is a sequence of mathematicians $M=M_{1}, M_{2}, \ldots, M_{k}=M^{\prime}$ such that $M_{i}$ was $M_{i+1}$ 's doctoral advisor for all $i$. Estimate
the number of descendents that the mathematician who has had the largest number of descendents has had, according to the Mathematical Genealogy Project. Note that the Mathematical Genealogy Project has records dating back to the 1300s. If the correct answer is $X$ and you write down $A$, your team will receive max $\left(25-\left\lfloor\frac{|X-A|}{100}\right\rfloor, 0\right)$ points, where $\lfloor x\rfloor$ is the largest integer less than or equal to $x$.
|
82310 First let's estimate how many "generations" of mathematicians there have been since 1300. If we suppose that a mathematician gets his PhD around age 30 and becomes a PhD advisor around age 60 , then we'll get a generation length of approximately 30 years. However, not all mathematicians will train more than one PhD . Let's say that only $40 \%$ of mathematicians train at least 2 PhDs. Then effectively we have only $40 \%$ of the generations, or in other words each effective generation takes 75 years. Then we have $\frac{22}{3}$ branching generations. If we assume that all of these only train 2 PhDs , then we get an answer of $2^{\frac{22}{3}} \approx 1625$. But we can ensure that our chain has at least a single person who trained 100 PhDs (this is approximately the largest number of advisees for a single mathematician), allowing us to change one factor of 2 into a factor of 100 . That gives us an answer of $1625 \cdot 50=81250$, which is very close to the actual value of 82310 .
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n35. [25]",
"solution_match": "\nAnswer: "
}
|
df0efba3-321e-5b3f-8b96-7d46e92d5658
| 608,775
|
Paul Erdős was one of the most prolific mathematicians of all time and was renowned for his many collaborations. The Erdős number of a mathematician is defined as follows. Erdős has an Erdős number of 0, a mathematician who has coauthored a paper with Erdős has an Erdős number of 1, a mathematician who has not coauthored a paper with Erdős, but has coauthored a paper with a mathematician with Erdős number 1 has an Erdős number of 2, etc. If no such chain exists between Erdős and another mathematician, that mathematician has an Erdős number of infinity. Of the mathematicians with a finite Erdős number (including those who are no longer alive), what is their average Erdős number according to the Erdős Number Project? If the correct answer is $X$ and you write down $A$, your team will receive $\max (25-\lfloor 100|X-A|\rfloor, 0)$ points where $\lfloor x\rfloor$ is the largest integer less than or equal to $x$.
|
4.65 We'll suppose that each mathematician collaborates with approximately 20 people (except for Erdős himself, of course). Furthermore, if a mathematician has Erdős number $k$, then we'd expect him to be the cause of approximately $\frac{1}{2^{k}}$ of his collaborators' Erdős numbers. This is because as we get to higher Erdős numbers, it is more likely that a collaborator has a lower Erdős number already. Therefore, we'd expect about 10 times as many people to have an Erdős number of 2 than with an Erdős number of 1 , then a ratio of $5,2.5,1.25$, and so on. This tells us that more mathematicians have an Erdős number of 5 than any other number, then 4 , then 6 , and so on. If we use this approximation, we have a ratio of mathematicians with Erdős number 1, 2, and so on of about $1: 10: 50: 125: 156: 97: 30: 4: 0.3$, which gives an average Erdős number of 4.8. This is close to the actual value of 4.65 .
|
4.65
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Paul Erdős was one of the most prolific mathematicians of all time and was renowned for his many collaborations. The Erdős number of a mathematician is defined as follows. Erdős has an Erdős number of 0, a mathematician who has coauthored a paper with Erdős has an Erdős number of 1, a mathematician who has not coauthored a paper with Erdős, but has coauthored a paper with a mathematician with Erdős number 1 has an Erdős number of 2, etc. If no such chain exists between Erdős and another mathematician, that mathematician has an Erdős number of infinity. Of the mathematicians with a finite Erdős number (including those who are no longer alive), what is their average Erdős number according to the Erdős Number Project? If the correct answer is $X$ and you write down $A$, your team will receive $\max (25-\lfloor 100|X-A|\rfloor, 0)$ points where $\lfloor x\rfloor$ is the largest integer less than or equal to $x$.
|
4.65 We'll suppose that each mathematician collaborates with approximately 20 people (except for Erdős himself, of course). Furthermore, if a mathematician has Erdős number $k$, then we'd expect him to be the cause of approximately $\frac{1}{2^{k}}$ of his collaborators' Erdős numbers. This is because as we get to higher Erdős numbers, it is more likely that a collaborator has a lower Erdős number already. Therefore, we'd expect about 10 times as many people to have an Erdős number of 2 than with an Erdős number of 1 , then a ratio of $5,2.5,1.25$, and so on. This tells us that more mathematicians have an Erdős number of 5 than any other number, then 4 , then 6 , and so on. If we use this approximation, we have a ratio of mathematicians with Erdős number 1, 2, and so on of about $1: 10: 50: 125: 156: 97: 30: 4: 0.3$, which gives an average Erdős number of 4.8. This is close to the actual value of 4.65 .
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-guts-solutions.jsonl",
"problem_match": "\n36. [25]",
"solution_match": "\nAnswer: "
}
|
736cf1dd-18ac-5cc4-8ece-c6b53fa82362
| 608,776
|
Travis is hopping around on the vertices of a cube. Each minute he hops from the vertex he's currently on to the other vertex of an edge that he is next to. After four minutes, what is the probability that he is back where he started?
|
$\boxed{\frac{7}{27}}$ Let the cube have vertices all 0 or 1 in the $x, y, z,$ coordinate system. Travis starts at $(0,0,0)$. If after 3 moves he is at $(1,1,1)$ he cannot get back to $(0,0,0)$. From any other vertex he has a $\frac{1}{3}$ chance of getting back on the final move. There is a $\frac{2}{9}$ chance he ends up at $(1,1,1)$, and thus a $\frac{7}{9}$ chance he does not end up there, and thus a $\frac{7}{27}$ chance he ends up at $(0,0,0)$.
|
\frac{7}{27}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Travis is hopping around on the vertices of a cube. Each minute he hops from the vertex he's currently on to the other vertex of an edge that he is next to. After four minutes, what is the probability that he is back where he started?
|
$\boxed{\frac{7}{27}}$ Let the cube have vertices all 0 or 1 in the $x, y, z,$ coordinate system. Travis starts at $(0,0,0)$. If after 3 moves he is at $(1,1,1)$ he cannot get back to $(0,0,0)$. From any other vertex he has a $\frac{1}{3}$ chance of getting back on the final move. There is a $\frac{2}{9}$ chance he ends up at $(1,1,1)$, and thus a $\frac{7}{9}$ chance he does not end up there, and thus a $\frac{7}{27}$ chance he ends up at $(0,0,0)$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-team-solutions.jsonl",
"problem_match": "\n1. [3]",
"solution_match": "\nAnswer: "
}
|
3f4a257d-ddbe-52ff-bd6f-2791b528d0d5
| 608,777
|
In terms of $k$, for $k>0$ how likely is he to be back where he started after $2 k$ minutes?
|
$\frac{1}{4}+\frac{3}{4}\left(\frac{1}{9}\right)^{k}$ Again, Travis starts at $(0,0,0)$. At each step, exactly one of the three coordinates will change. The parity of the sum of the three coordinates will change at each step, so after $2 k$ steps, the sum of the coordinates must be even. There are only four possibilites for Travis's position: $(0,0,0),(1,1,0),(1,0,1)$, and $(0,1,1)$. Let $p_{k}$ be the probability that Travis is at $(0,0,0)$ after $2 k$ steps. Then $1-p_{k}$ is the probability that he is on $(1,1,0),(1,0,1)$, or $(0,1,1)$. Suppose we want to compute $p_{k+1}$. There are two possibilities: we were either at $(0,0,0)$ after $2 k$ steps or not. If we were, then there is a $\frac{1}{3}$ probability that we will return (since our $(2 k+1)^{\text {th }}$ step can be arbitary, but there is a $\frac{1}{3}$ chance that we will reverse that as our $(2 k+2)^{\text {th }}$ step). If we were not at $(0,0,0)$ after our $2 k^{\text {th }}$ steps, then two of our coordinates must have been ones. There is a $\frac{2}{3}$ probability that the $(2 k+1)^{\text {th }}$ step will change one of those to a zero, and there is a $\frac{1}{3}$ step that that the $(2 k+2)^{\text {th }}$ step will change the remaining one. Hence, in this case, there is a $\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)=\frac{2}{9}$ probability that Travis ends up $(0,0,0)$ in this case. So we have:
$$
\begin{aligned}
p_{k+1} & =p_{k}\left(\frac{1}{3}\right)+\left(1-p_{k}\right)\left(\frac{2}{9}\right) \\
p_{k+1} & =\frac{1}{9} p_{k}+\frac{2}{9} \\
\left(p_{k+1}-\frac{1}{4}\right) & =\frac{1}{9}\left(p_{k}-\frac{1}{4}\right)
\end{aligned}
$$
(We get the value $\frac{1}{4}$ either by guessing that the sequence $p_{0}, p_{1}, p_{2}, \ldots$ should converge to $\frac{1}{4}$ or simply by solving the equation $-\frac{1}{9} x+x=\frac{2}{9}$.) This shows that $p_{0}-\frac{1}{4}, p_{1}-\frac{1}{4}, \ldots$ is a geometric series with ratio $\frac{1}{9}$. Since $p_{0}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}$, we get that $p_{k}-\frac{1}{4}=\frac{3}{4}\left(\frac{1}{9}\right)^{k}$, or that $p_{k}=\frac{1}{4}+\frac{3}{4}\left(\frac{1}{9}\right)^{k}$.
|
\frac{1}{4}+\frac{3}{4}\left(\frac{1}{9}\right)^{k}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In terms of $k$, for $k>0$ how likely is he to be back where he started after $2 k$ minutes?
|
$\frac{1}{4}+\frac{3}{4}\left(\frac{1}{9}\right)^{k}$ Again, Travis starts at $(0,0,0)$. At each step, exactly one of the three coordinates will change. The parity of the sum of the three coordinates will change at each step, so after $2 k$ steps, the sum of the coordinates must be even. There are only four possibilites for Travis's position: $(0,0,0),(1,1,0),(1,0,1)$, and $(0,1,1)$. Let $p_{k}$ be the probability that Travis is at $(0,0,0)$ after $2 k$ steps. Then $1-p_{k}$ is the probability that he is on $(1,1,0),(1,0,1)$, or $(0,1,1)$. Suppose we want to compute $p_{k+1}$. There are two possibilities: we were either at $(0,0,0)$ after $2 k$ steps or not. If we were, then there is a $\frac{1}{3}$ probability that we will return (since our $(2 k+1)^{\text {th }}$ step can be arbitary, but there is a $\frac{1}{3}$ chance that we will reverse that as our $(2 k+2)^{\text {th }}$ step). If we were not at $(0,0,0)$ after our $2 k^{\text {th }}$ steps, then two of our coordinates must have been ones. There is a $\frac{2}{3}$ probability that the $(2 k+1)^{\text {th }}$ step will change one of those to a zero, and there is a $\frac{1}{3}$ step that that the $(2 k+2)^{\text {th }}$ step will change the remaining one. Hence, in this case, there is a $\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)=\frac{2}{9}$ probability that Travis ends up $(0,0,0)$ in this case. So we have:
$$
\begin{aligned}
p_{k+1} & =p_{k}\left(\frac{1}{3}\right)+\left(1-p_{k}\right)\left(\frac{2}{9}\right) \\
p_{k+1} & =\frac{1}{9} p_{k}+\frac{2}{9} \\
\left(p_{k+1}-\frac{1}{4}\right) & =\frac{1}{9}\left(p_{k}-\frac{1}{4}\right)
\end{aligned}
$$
(We get the value $\frac{1}{4}$ either by guessing that the sequence $p_{0}, p_{1}, p_{2}, \ldots$ should converge to $\frac{1}{4}$ or simply by solving the equation $-\frac{1}{9} x+x=\frac{2}{9}$.) This shows that $p_{0}-\frac{1}{4}, p_{1}-\frac{1}{4}, \ldots$ is a geometric series with ratio $\frac{1}{9}$. Since $p_{0}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}$, we get that $p_{k}-\frac{1}{4}=\frac{3}{4}\left(\frac{1}{9}\right)^{k}$, or that $p_{k}=\frac{1}{4}+\frac{3}{4}\left(\frac{1}{9}\right)^{k}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-team-solutions.jsonl",
"problem_match": "\n2. [6]",
"solution_match": "\nAnswer: "
}
|
6067fd56-a48e-55ce-bbcc-810d7f5bfc3e
| 608,778
|
While Travis is having fun on cubes, Sherry is hopping in the same manner on an octahedron. An octahedron has six vertices and eight regular triangular faces. After five minutes, how likely is Sherry to be one edge away from where she started?
|
$\frac{11}{16}$ Let the starting vertex be the 'bottom' one. Then there is a 'top' vertex, and 4 'middle' ones. If $p(n)$ is the probability that Sherry is on a middle vertex after $n$ minutes, $p(0)=0$,
$p(n+1)=(1-p(n))+p(n) \cdot \frac{1}{2}$. This recurrence gives us the following equations.
$$
\begin{aligned}
p(n+1) & =1-\frac{p(n)}{2} \\
p(0) & =0 \\
p(1) & =1 \\
p(2) & =\frac{1}{2} \\
p(3) & =\frac{3}{4} \\
p(4) & =\frac{5}{8} \\
p(5) & =\frac{11}{16}
\end{aligned}
$$
|
\frac{11}{16}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
While Travis is having fun on cubes, Sherry is hopping in the same manner on an octahedron. An octahedron has six vertices and eight regular triangular faces. After five minutes, how likely is Sherry to be one edge away from where she started?
|
$\frac{11}{16}$ Let the starting vertex be the 'bottom' one. Then there is a 'top' vertex, and 4 'middle' ones. If $p(n)$ is the probability that Sherry is on a middle vertex after $n$ minutes, $p(0)=0$,
$p(n+1)=(1-p(n))+p(n) \cdot \frac{1}{2}$. This recurrence gives us the following equations.
$$
\begin{aligned}
p(n+1) & =1-\frac{p(n)}{2} \\
p(0) & =0 \\
p(1) & =1 \\
p(2) & =\frac{1}{2} \\
p(3) & =\frac{3}{4} \\
p(4) & =\frac{5}{8} \\
p(5) & =\frac{11}{16}
\end{aligned}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-team-solutions.jsonl",
"problem_match": "\n3. [3]",
"solution_match": "\nAnswer: "
}
|
55640d4f-dfdf-5dda-a1eb-47eac9c36868
| 608,779
|
In terms of $k$, for $k>0$, how likely is it that after $k$ minutes Sherry is at the vertex opposite the vertex where she started?
|
| $\frac{1}{6}+\frac{1}{3(-2)^{k}}$ |
| :---: | Take $p(n)$ from the last problem. By examining the last move of Sherry, the probability that she ends up on the original vertex is equal to the probability that she ends up on the top vertex, and both are equal to $\frac{1-p(n)}{2}$ for $n \geq 1$.
From the last problem,
$$
\begin{aligned}
p(n+1) & =1-\frac{p(n)}{2} \\
p(n+1)-\frac{2}{3} & =-\frac{1}{2}\left(p(n)-\frac{2}{3}\right)
\end{aligned}
$$
and so $p(n)-\frac{2}{3}$ is a geometric series with ratio $-\frac{1}{2}$. Since $p(0)=0$, we get $p(n)-\frac{2}{3}=-\frac{2}{3}\left(-\frac{1}{2}\right)^{n}$, or that $p(n)=\frac{2}{3}-\frac{2}{3}\left(-\frac{1}{2}\right)^{n}$.
Now, for $k \geq 1$, we have that the probability of ending up on the vertex opposite Sherry's initial vertex after $k$ minutes is $\frac{1-p(k)}{2}=\frac{1}{2}-\frac{1}{2}\left(\frac{2}{3}-\frac{2}{3}\left(-\frac{1}{2}\right)^{k}\right)=\frac{1}{6}+\frac{1}{3}\left(-\frac{1}{2}\right)^{k}=\frac{1}{6}+\frac{1}{3(-2)^{k}}$.
## Circles in Circles
|
\frac{1}{6}+\frac{1}{3(-2)^{k}}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In terms of $k$, for $k>0$, how likely is it that after $k$ minutes Sherry is at the vertex opposite the vertex where she started?
|
| $\frac{1}{6}+\frac{1}{3(-2)^{k}}$ |
| :---: | Take $p(n)$ from the last problem. By examining the last move of Sherry, the probability that she ends up on the original vertex is equal to the probability that she ends up on the top vertex, and both are equal to $\frac{1-p(n)}{2}$ for $n \geq 1$.
From the last problem,
$$
\begin{aligned}
p(n+1) & =1-\frac{p(n)}{2} \\
p(n+1)-\frac{2}{3} & =-\frac{1}{2}\left(p(n)-\frac{2}{3}\right)
\end{aligned}
$$
and so $p(n)-\frac{2}{3}$ is a geometric series with ratio $-\frac{1}{2}$. Since $p(0)=0$, we get $p(n)-\frac{2}{3}=-\frac{2}{3}\left(-\frac{1}{2}\right)^{n}$, or that $p(n)=\frac{2}{3}-\frac{2}{3}\left(-\frac{1}{2}\right)^{n}$.
Now, for $k \geq 1$, we have that the probability of ending up on the vertex opposite Sherry's initial vertex after $k$ minutes is $\frac{1-p(k)}{2}=\frac{1}{2}-\frac{1}{2}\left(\frac{2}{3}-\frac{2}{3}\left(-\frac{1}{2}\right)^{k}\right)=\frac{1}{6}+\frac{1}{3}\left(-\frac{1}{2}\right)^{k}=\frac{1}{6}+\frac{1}{3(-2)^{k}}$.
## Circles in Circles
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-team-solutions.jsonl",
"problem_match": "\n4. [6]",
"solution_match": "\nAnswer: "
}
|
9c8da399-888e-5f24-951b-0235c4ac6772
| 608,780
|
Circle $O$ has chord $A B$. A circle is tangent to $O$ at $T$ and tangent to $A B$ at $X$ such that $A X=2 X B$. What is $\frac{A T}{B T}$ ?
|
2 Let $T X$ meet circle $O$ again at $Y$. Since the homethety centered at $T$ takes $X$ to $Y$ also takes $A B$ to the tangent line of circle $O$ passing through $Y$, we have $Y$ is the midpoint of arc $A B$. This means that $\angle A T Y=\angle Y T B$. By the Angle Bisector Theorem, $\frac{A T}{B T}=\frac{A X}{B X}=2$.
|
2
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Circle $O$ has chord $A B$. A circle is tangent to $O$ at $T$ and tangent to $A B$ at $X$ such that $A X=2 X B$. What is $\frac{A T}{B T}$ ?
|
2 Let $T X$ meet circle $O$ again at $Y$. Since the homethety centered at $T$ takes $X$ to $Y$ also takes $A B$ to the tangent line of circle $O$ passing through $Y$, we have $Y$ is the midpoint of arc $A B$. This means that $\angle A T Y=\angle Y T B$. By the Angle Bisector Theorem, $\frac{A T}{B T}=\frac{A X}{B X}=2$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-team-solutions.jsonl",
"problem_match": "\n5. [4]",
"solution_match": "\nAnswer: "
}
|
e7581b75-0cc4-57e7-bfa0-cca983156716
| 608,781
|
$A B$ is a diameter of circle $O . X$ is a point on $A B$ such that $A X=3 B X$. Distinct circles $\omega_{1}$ and $\omega_{2}$ are tangent to $O$ at $T_{1}$ and $T_{2}$ and to $A B$ at $X$. The lines $T_{1} X$ and $T_{2} X$ intersect $O$ again at $S_{1}$ and $S_{2}$. What is the ratio $\frac{T_{1} T_{2}}{S_{1} S_{2}}$ ?
|
| $\frac{3}{5}$ | Since the problem only deals with ratios, we can assume that the radius of $O$ is 1 . As |
| :---: | :---: | we have proven in Problem 5, points $S_{1}$ and $S_{2}$ are midpoints of arc $A B$. Since $A B$ is a diameter, $S_{1} S_{2}$ is also a diameter, and thus $S_{1} S_{2}=2$.
Let $O_{1}, O_{2}$, and $P$ denote the center of circles $\omega_{1}, \omega_{2}$, and $O$. Since $\omega_{1}$ is tangent to $O$, we have $P O_{1}+O_{1} X=1$. But $O_{1} X \perp A B$. So $\triangle P O_{1} X$ is a right triangle, and $O_{1} X^{2}+X P^{2}=O_{1} P^{2}$. Thus, $O_{1} X^{2}+1 / 4=\left(1-O_{1} X\right)^{2}$, which means $O_{1} X=\frac{3}{8}$ and $O_{1} P=\frac{5}{8}$.
Since $T_{1} T_{2} \| O_{1} O_{2}$, we have $T_{1} T_{2}=O_{1} O_{2} \cdot \frac{P T_{1}}{P O_{1}}=2 O_{1} X \cdot \frac{P T_{1}}{P O_{1}}=2\left(\frac{3}{8}\right) \frac{1}{5 / 8}=\frac{6}{5}$. Thus $\frac{T_{1} T_{2}}{S_{1} S_{2}}=\frac{6 / 5}{2}=\frac{3}{5}$.
|
\frac{3}{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
$A B$ is a diameter of circle $O . X$ is a point on $A B$ such that $A X=3 B X$. Distinct circles $\omega_{1}$ and $\omega_{2}$ are tangent to $O$ at $T_{1}$ and $T_{2}$ and to $A B$ at $X$. The lines $T_{1} X$ and $T_{2} X$ intersect $O$ again at $S_{1}$ and $S_{2}$. What is the ratio $\frac{T_{1} T_{2}}{S_{1} S_{2}}$ ?
|
| $\frac{3}{5}$ | Since the problem only deals with ratios, we can assume that the radius of $O$ is 1 . As |
| :---: | :---: | we have proven in Problem 5, points $S_{1}$ and $S_{2}$ are midpoints of arc $A B$. Since $A B$ is a diameter, $S_{1} S_{2}$ is also a diameter, and thus $S_{1} S_{2}=2$.
Let $O_{1}, O_{2}$, and $P$ denote the center of circles $\omega_{1}, \omega_{2}$, and $O$. Since $\omega_{1}$ is tangent to $O$, we have $P O_{1}+O_{1} X=1$. But $O_{1} X \perp A B$. So $\triangle P O_{1} X$ is a right triangle, and $O_{1} X^{2}+X P^{2}=O_{1} P^{2}$. Thus, $O_{1} X^{2}+1 / 4=\left(1-O_{1} X\right)^{2}$, which means $O_{1} X=\frac{3}{8}$ and $O_{1} P=\frac{5}{8}$.
Since $T_{1} T_{2} \| O_{1} O_{2}$, we have $T_{1} T_{2}=O_{1} O_{2} \cdot \frac{P T_{1}}{P O_{1}}=2 O_{1} X \cdot \frac{P T_{1}}{P O_{1}}=2\left(\frac{3}{8}\right) \frac{1}{5 / 8}=\frac{6}{5}$. Thus $\frac{T_{1} T_{2}}{S_{1} S_{2}}=\frac{6 / 5}{2}=\frac{3}{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-team-solutions.jsonl",
"problem_match": "\n6. [6]",
"solution_match": "\nAnswer: "
}
|
3a847aec-fcef-527c-81e6-8837e226e199
| 608,782
|
$A B C$ is a right triangle with $\angle A=30^{\circ}$ and circumcircle $O$. Circles $\omega_{1}, \omega_{2}$, and $\omega_{3}$ lie outside $A B C$ and are tangent to $O$ at $T_{1}, T_{2}$, and $T_{3}$ respectively and to $A B, B C$, and $C A$ at $S_{1}, S_{2}$, and $S_{3}$, respectively. Lines $T_{1} S_{1}, T_{2} S_{2}$, and $T_{3} S_{3}$ intersect $O$ again at $A^{\prime}, B^{\prime}$, and $C^{\prime}$, respectively. What is the ratio of the area of $A^{\prime} B^{\prime} C^{\prime}$ to the area of $A B C$ ?
|
$\frac{\sqrt{3}+1}{2}$ Let $[P Q R]$ denote the area of $\triangle P Q R$. The key to this problem is following fact: $[P Q R]=\frac{1}{2} P Q \cdot P R \sin \angle Q P R$.
Assume that the radius of $O$ is 1 . Since $\angle A=30^{\circ}$, we have $B C=1$ and $A B=\sqrt{3}$. So $[A B C]=\frac{\sqrt{3}}{2}$. Let $K$ denote the center of $O$. Notice that $\angle B^{\prime} K A^{\prime}=90^{\circ}, \angle A K C^{\prime}=90^{\circ}$, and $\angle B^{\prime} K A=\angle K A B=$ $30^{\circ}$. Thus, $\angle B^{\prime} K C^{\prime}=\angle B^{\prime} K A+\angle A K C^{\prime}=120^{\circ}$ and consequently $\angle C^{\prime} K A^{\prime}=150^{\circ}$.
Therefore, $\left[A^{\prime} B^{\prime} C^{\prime}\right]=\left[A^{\prime} K B^{\prime}\right]+\left[B^{\prime} K C^{\prime}\right]+\left[C^{\prime} K A^{\prime}\right]=\frac{1}{2}+\frac{1}{2} \sin 120^{\circ}+\frac{1}{2} \sin 150^{\circ}=\frac{3}{4}+\frac{\sqrt{3}}{4}$. This gives the desired result that $\left[A^{\prime} B^{\prime} C^{\prime}\right]=\frac{\sqrt{3}+1}{2}[A B C]$.
## Linear? What's The Problem?
A function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ is said to be linear in each of its variables if it is a polynomial such that no variable appears with power higher than one in any term. For example, $1+x+x y$ is linear in $x$ and $y$, but $1+x^{2}$ is not. Similarly, $2 x+3 y z$ is linear in $x, y$, and $z$, but $x y z^{2}$ is not.
|
\frac{\sqrt{3}+1}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
$A B C$ is a right triangle with $\angle A=30^{\circ}$ and circumcircle $O$. Circles $\omega_{1}, \omega_{2}$, and $\omega_{3}$ lie outside $A B C$ and are tangent to $O$ at $T_{1}, T_{2}$, and $T_{3}$ respectively and to $A B, B C$, and $C A$ at $S_{1}, S_{2}$, and $S_{3}$, respectively. Lines $T_{1} S_{1}, T_{2} S_{2}$, and $T_{3} S_{3}$ intersect $O$ again at $A^{\prime}, B^{\prime}$, and $C^{\prime}$, respectively. What is the ratio of the area of $A^{\prime} B^{\prime} C^{\prime}$ to the area of $A B C$ ?
|
$\frac{\sqrt{3}+1}{2}$ Let $[P Q R]$ denote the area of $\triangle P Q R$. The key to this problem is following fact: $[P Q R]=\frac{1}{2} P Q \cdot P R \sin \angle Q P R$.
Assume that the radius of $O$ is 1 . Since $\angle A=30^{\circ}$, we have $B C=1$ and $A B=\sqrt{3}$. So $[A B C]=\frac{\sqrt{3}}{2}$. Let $K$ denote the center of $O$. Notice that $\angle B^{\prime} K A^{\prime}=90^{\circ}, \angle A K C^{\prime}=90^{\circ}$, and $\angle B^{\prime} K A=\angle K A B=$ $30^{\circ}$. Thus, $\angle B^{\prime} K C^{\prime}=\angle B^{\prime} K A+\angle A K C^{\prime}=120^{\circ}$ and consequently $\angle C^{\prime} K A^{\prime}=150^{\circ}$.
Therefore, $\left[A^{\prime} B^{\prime} C^{\prime}\right]=\left[A^{\prime} K B^{\prime}\right]+\left[B^{\prime} K C^{\prime}\right]+\left[C^{\prime} K A^{\prime}\right]=\frac{1}{2}+\frac{1}{2} \sin 120^{\circ}+\frac{1}{2} \sin 150^{\circ}=\frac{3}{4}+\frac{\sqrt{3}}{4}$. This gives the desired result that $\left[A^{\prime} B^{\prime} C^{\prime}\right]=\frac{\sqrt{3}+1}{2}[A B C]$.
## Linear? What's The Problem?
A function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ is said to be linear in each of its variables if it is a polynomial such that no variable appears with power higher than one in any term. For example, $1+x+x y$ is linear in $x$ and $y$, but $1+x^{2}$ is not. Similarly, $2 x+3 y z$ is linear in $x, y$, and $z$, but $x y z^{2}$ is not.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-team-solutions.jsonl",
"problem_match": "\n7. [7]",
"solution_match": "\nAnswer: "
}
|
08bb4a6c-cc6e-5cb9-b7df-298434504760
| 608,783
|
A function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ is linear in each of the $x_{i}$ and $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\frac{1}{x_{1} x_{2} \cdots x_{n}}$ when $x_{i} \in\{3,4\}$ for all $i$. In terms of $n$, what is $f(5,5, \ldots, 5) ?$
|
$\frac{1}{6^{n}}$ Let $f_{n}\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ denote the $n$-variable version of the function. We will prove that $f_{n}(5, \ldots, 5)=\frac{1}{6^{n}}$ by induction. The base case was done in the two previous problems. Suppose we know that $f_{n-1}(5,5, \ldots, 5)=\frac{1}{.6^{n-1}}$. Let $g\left(x_{1}, \ldots, x_{n-1}\right)=3 f_{n}\left(x_{1}, \ldots, x_{n-1}, 3\right)$. We have that $g$ is linear in $x_{1}, \ldots, x_{n-1}$ and $g\left(x_{1}, \ldots, x_{n-1}\right)=\frac{1}{x_{1} \cdots x_{n-1}}$ for all $x_{1}, \ldots, x_{n-1} \in\{3,4\}$. By the inductive hypothesis, we have $g(5, \ldots, 5)=\frac{1}{6^{n-1}}=f_{n-1}(5, \ldots, 5)$. Therefore, $f_{n}(5, \ldots, 5,3)=\frac{f_{n-1}(5, \ldots, 5)}{3}$. Similarly, $f_{n}(5, \ldots, 5,4)=\frac{f_{n-1}(5, \ldots, 5)}{4}$.
$$
\begin{aligned}
f_{n}(5,5, \ldots, 5,5) & =2 f_{n}(5,5, \ldots, 5,4)-f_{n}(5,5, \ldots, 5,3) \\
& =\frac{1}{2} f_{n-1}(5,5, \ldots, 5)-\frac{1}{3} f_{n-1} \\
& =\frac{1}{6} f_{n-1}(5,5, \ldots, 5) \\
& =\frac{1}{6 \cdot 6^{n-1}} \\
& =\frac{1}{6^{n}}
\end{aligned}
$$
and this proves our conjecture by induction.
|
\frac{1}{6^{n}}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
A function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ is linear in each of the $x_{i}$ and $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\frac{1}{x_{1} x_{2} \cdots x_{n}}$ when $x_{i} \in\{3,4\}$ for all $i$. In terms of $n$, what is $f(5,5, \ldots, 5) ?$
|
$\frac{1}{6^{n}}$ Let $f_{n}\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ denote the $n$-variable version of the function. We will prove that $f_{n}(5, \ldots, 5)=\frac{1}{6^{n}}$ by induction. The base case was done in the two previous problems. Suppose we know that $f_{n-1}(5,5, \ldots, 5)=\frac{1}{.6^{n-1}}$. Let $g\left(x_{1}, \ldots, x_{n-1}\right)=3 f_{n}\left(x_{1}, \ldots, x_{n-1}, 3\right)$. We have that $g$ is linear in $x_{1}, \ldots, x_{n-1}$ and $g\left(x_{1}, \ldots, x_{n-1}\right)=\frac{1}{x_{1} \cdots x_{n-1}}$ for all $x_{1}, \ldots, x_{n-1} \in\{3,4\}$. By the inductive hypothesis, we have $g(5, \ldots, 5)=\frac{1}{6^{n-1}}=f_{n-1}(5, \ldots, 5)$. Therefore, $f_{n}(5, \ldots, 5,3)=\frac{f_{n-1}(5, \ldots, 5)}{3}$. Similarly, $f_{n}(5, \ldots, 5,4)=\frac{f_{n-1}(5, \ldots, 5)}{4}$.
$$
\begin{aligned}
f_{n}(5,5, \ldots, 5,5) & =2 f_{n}(5,5, \ldots, 5,4)-f_{n}(5,5, \ldots, 5,3) \\
& =\frac{1}{2} f_{n-1}(5,5, \ldots, 5)-\frac{1}{3} f_{n-1} \\
& =\frac{1}{6} f_{n-1}(5,5, \ldots, 5) \\
& =\frac{1}{6 \cdot 6^{n-1}} \\
& =\frac{1}{6^{n}}
\end{aligned}
$$
and this proves our conjecture by induction.
|
{
"resource_path": "HarvardMIT/segmented/en-141-2010-nov-team-solutions.jsonl",
"problem_match": "\n10. [6]",
"solution_match": "\nAnswer: "
}
|
730369b7-ade8-5202-9fb4-7254e480a50d
| 608,786
|
Let $a, b$, and $c$ be positive real numbers. Determine the largest total number of real roots that the following three polynomials may have among them: $a x^{2}+b x+c, b x^{2}+c x+a$, and $c x^{2}+a x+b$.
|
4 If all the polynomials had real roots, their discriminants would all be nonnegative: $a^{2} \geq$ $4 b c, b^{2} \geq 4 c a$, and $c^{2} \geq 4 a b$. Multiplying these inequalities gives $(a b c)^{2} \geq 64(a b c)^{2}$, a contradiction. Hence one of the quadratics has no real roots. The maximum of 4 real roots is attainable: for example, the values $(a, b, c)=(1,5,6)$ give $-2,-3$ as roots to $x^{2}+5 x+6$ and $-1,-\frac{1}{5}$ as roots to $5 x^{2}+6 x+1$.
|
4
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a, b$, and $c$ be positive real numbers. Determine the largest total number of real roots that the following three polynomials may have among them: $a x^{2}+b x+c, b x^{2}+c x+a$, and $c x^{2}+a x+b$.
|
4 If all the polynomials had real roots, their discriminants would all be nonnegative: $a^{2} \geq$ $4 b c, b^{2} \geq 4 c a$, and $c^{2} \geq 4 a b$. Multiplying these inequalities gives $(a b c)^{2} \geq 64(a b c)^{2}$, a contradiction. Hence one of the quadratics has no real roots. The maximum of 4 real roots is attainable: for example, the values $(a, b, c)=(1,5,6)$ give $-2,-3$ as roots to $x^{2}+5 x+6$ and $-1,-\frac{1}{5}$ as roots to $5 x^{2}+6 x+1$.
|
{
"resource_path": "HarvardMIT/segmented/en-142-2011-feb-algcalc-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nAnswer: "
}
|
0ef53064-87e8-5497-bc43-2b17a4ecaf97
| 608,787
|
Josh takes a walk on a rectangular grid of $n$ rows and 3 columns, starting from the bottom left corner. At each step, he can either move one square to the right or simultaneously move one square to the left and one square up. In how many ways can he reach the center square of the topmost row?
|
$2^{n-1}$ Note that Josh must pass through the center square of each row. There are 2 ways to get from the center square of row $k$ to the center square of row $k+1$. So there are $2^{n-1}$ ways to get to the center square of row $n$.
|
2^{n-1}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Josh takes a walk on a rectangular grid of $n$ rows and 3 columns, starting from the bottom left corner. At each step, he can either move one square to the right or simultaneously move one square to the left and one square up. In how many ways can he reach the center square of the topmost row?
|
$2^{n-1}$ Note that Josh must pass through the center square of each row. There are 2 ways to get from the center square of row $k$ to the center square of row $k+1$. So there are $2^{n-1}$ ways to get to the center square of row $n$.
|
{
"resource_path": "HarvardMIT/segmented/en-142-2011-feb-algcalc-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nAnswer: "
}
|
bc25ceca-c89c-538f-9f5b-7d2e547e3bec
| 608,788
|
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