problem
stringlengths 14
7.96k
| solution
stringlengths 3
10k
| answer
stringlengths 1
91
| problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
value | question_type
stringclasses 1
value | problem_type
stringclasses 8
values | problem_raw
stringlengths 14
7.96k
| solution_raw
stringlengths 3
10k
| metadata
dict | uuid
stringlengths 36
36
| id
int64 22.6k
612k
|
|---|---|---|---|---|---|---|---|---|---|---|---|
Let $S$ be a subset of $\{1,2,3, \ldots, 12\}$ such that it is impossible to partition $S$ into $k$ disjoint subsets, each of whose elements sum to the same value, for any integer $k \geq 2$. Find the maximum possible sum of the elements of $S$.
|
77 We note that the maximum possible sum is 78 (the entire set). However, this could be partitioned into 2 subsets with sum 39: $\{1,2,3,10,11,12\}$ and $\{4,5,6,7,8,9\}$. The next largest possible sum is 77 (the entire set except 1). If $k \geq 2$ subsets each had equal sum, then they would have to be 7 subsets with sum 11 each or 11 subsets with sum 7 each. However, the subset containing 12 will have sum greater than 11 ; hence there is no way to partition the subset $\{2, \ldots, 12\}$ into equal subsets.
|
77
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $S$ be a subset of $\{1,2,3, \ldots, 12\}$ such that it is impossible to partition $S$ into $k$ disjoint subsets, each of whose elements sum to the same value, for any integer $k \geq 2$. Find the maximum possible sum of the elements of $S$.
|
77 We note that the maximum possible sum is 78 (the entire set). However, this could be partitioned into 2 subsets with sum 39: $\{1,2,3,10,11,12\}$ and $\{4,5,6,7,8,9\}$. The next largest possible sum is 77 (the entire set except 1). If $k \geq 2$ subsets each had equal sum, then they would have to be 7 subsets with sum 11 each or 11 subsets with sum 7 each. However, the subset containing 12 will have sum greater than 11 ; hence there is no way to partition the subset $\{2, \ldots, 12\}$ into equal subsets.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n22. [12]",
"solution_match": "\nAnswer: "
}
|
a4aa2ad2-5c19-5817-bff1-8b25869ea3cd
| 609,178
|
Find the number of subsets $S$ of $\{1,2, \ldots 6\}$ satisfying the following conditions:
- $S$ is non-empty.
- No subset of $S$ has the property that the sum of its elements is 10 .
|
34 We do casework based on the largest element of $S$. Call a set $n$-free if none of its subsets have elements summing to $n$.
Case 1: The largest element of $S$ is 6 . Then $4 \notin S$. If $5 \notin S$, then we wish to find all 4 -free subsets of $\{1,2,3\}$ (note that $1+2+3=6<10$ ). We just cannot include both 1,3 , so we have $2\left(2^{2}-1\right)=6$ choices here.
If $5 \in S$, then we want 4,5 -free subsets of $\{1,2,3\}$. The only 4 -but-not- 5 -free subset is $\{2,3\}$, so we have $6-1$ choices here, for a case total of $6+5=11$.
Case 2: The largest element of $S$ is 5 . We seek 5,10 -free subsets of $\{1,2,3,4\}$. We just cannot have both 1,4 or both 2,3 (note that getting 10 requires the whole set), so we have $\left(2^{2}-1\right)^{2}=9$ subsets in this case.
Case 3: The largest element of $S$ is at most 4. (So we want a 4 -free subset of $\{1,2,3,4\}$.) The only way to sum to 10 with $1,2,3,4$ is by using all the terms, so we simply discount the empty set and $\{1,2,3,4\}$, for a total of $2^{4}-2=14$ subsets.
In conclusion, the total number of subsets is $11+9+14=34$.
|
34
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Find the number of subsets $S$ of $\{1,2, \ldots 6\}$ satisfying the following conditions:
- $S$ is non-empty.
- No subset of $S$ has the property that the sum of its elements is 10 .
|
34 We do casework based on the largest element of $S$. Call a set $n$-free if none of its subsets have elements summing to $n$.
Case 1: The largest element of $S$ is 6 . Then $4 \notin S$. If $5 \notin S$, then we wish to find all 4 -free subsets of $\{1,2,3\}$ (note that $1+2+3=6<10$ ). We just cannot include both 1,3 , so we have $2\left(2^{2}-1\right)=6$ choices here.
If $5 \in S$, then we want 4,5 -free subsets of $\{1,2,3\}$. The only 4 -but-not- 5 -free subset is $\{2,3\}$, so we have $6-1$ choices here, for a case total of $6+5=11$.
Case 2: The largest element of $S$ is 5 . We seek 5,10 -free subsets of $\{1,2,3,4\}$. We just cannot have both 1,4 or both 2,3 (note that getting 10 requires the whole set), so we have $\left(2^{2}-1\right)^{2}=9$ subsets in this case.
Case 3: The largest element of $S$ is at most 4. (So we want a 4 -free subset of $\{1,2,3,4\}$.) The only way to sum to 10 with $1,2,3,4$ is by using all the terms, so we simply discount the empty set and $\{1,2,3,4\}$, for a total of $2^{4}-2=14$ subsets.
In conclusion, the total number of subsets is $11+9+14=34$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n24. [12]",
"solution_match": "\nAnswer: "
}
|
7e6a7648-f424-5060-b896-548f04fd3deb
| 609,180
|
Let $a, b$ be positive reals with $a>b>\frac{1}{2} a$. Place two squares of side lengths $a, b$ next to each other, such that the larger square has lower left corner at $(0,0)$ and the smaller square has lower left corner at $(a, 0)$. Draw the line passing through $(0, a)$ and $(a+b, 0)$. The region in the two squares lying above the line has area 2013. If $(a, b)$ is the unique pair maximizing $a+b$, compute $\frac{a}{b}$.
|
$\quad \frac{5}{3}$ Let $t=\frac{a}{b} \in(1,2)$; we will rewrite the sum $a+b$ as a function of $t$. The area condition easily translates to $\frac{a^{2}-a b+2 b^{2}}{2}=2013$, or $b^{2}\left(t^{2}-t+2\right)=4026 \Longleftrightarrow b=\sqrt{\frac{4026}{t^{2}-t+2}}$. Thus $a+b$ is a function $f(t)=(1+t) \sqrt{\frac{4026}{t^{2}-t+2}}$ of $t$, and our answer is simply the value of $t$ maximizing $f$, or equivalently $g(t)=\frac{f^{2}}{4026}=\frac{(1+t)^{2}}{t^{2}-t+2}$, over the interval $(1,2)$. (In general, such maximizers/maximums need not exist, but we shall prove there's a unique maximum here.)
We claim that $\lambda=\frac{16}{7}$ is the maximum of $\frac{(1+t)^{2}}{t^{2}-t+2}$. Indeed,
$$
\begin{aligned}
\lambda-g(t) & =\frac{(\lambda-1) t^{2}-(\lambda+2) t+(2 \lambda-1)}{t^{2}-t+2} \\
& =\frac{1}{7} \frac{9 t^{2}-30 t+25}{t^{2}-t+2}=\frac{1}{7} \frac{(3 t-5)^{2}}{\left(t-\frac{1}{2}\right)^{2}+\frac{7}{4}} \geq 0
\end{aligned}
$$
for all reals $t$ (not just $t \in(1,2)$ ), with equality at $t=\frac{5}{3} \in(1,2)$.
Comment. To motivate the choice of $\lambda$, suppose $\lambda$ were the maximum of $f$, attained at $t_{0} \in(1,2)$; then $h(t)=\lambda\left(t^{2}-t+2\right)-(t+1)^{2}$ is quadratic and nonnegative on $(1,2)$, but zero at $t=t_{0}$. If $g$ is a nontrivial quadratic (nonzero leading coefficient), then $t_{0}$ must be a double root, so $g$ has determinant 0 . Of course, $g$ could also be constant or linear over $(1,2)$, but we can easily rule out both of these possibilities.
Alternatively, we can simply take a derivative of $f$ to find critical points.
|
\frac{5}{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $a, b$ be positive reals with $a>b>\frac{1}{2} a$. Place two squares of side lengths $a, b$ next to each other, such that the larger square has lower left corner at $(0,0)$ and the smaller square has lower left corner at $(a, 0)$. Draw the line passing through $(0, a)$ and $(a+b, 0)$. The region in the two squares lying above the line has area 2013. If $(a, b)$ is the unique pair maximizing $a+b$, compute $\frac{a}{b}$.
|
$\quad \frac{5}{3}$ Let $t=\frac{a}{b} \in(1,2)$; we will rewrite the sum $a+b$ as a function of $t$. The area condition easily translates to $\frac{a^{2}-a b+2 b^{2}}{2}=2013$, or $b^{2}\left(t^{2}-t+2\right)=4026 \Longleftrightarrow b=\sqrt{\frac{4026}{t^{2}-t+2}}$. Thus $a+b$ is a function $f(t)=(1+t) \sqrt{\frac{4026}{t^{2}-t+2}}$ of $t$, and our answer is simply the value of $t$ maximizing $f$, or equivalently $g(t)=\frac{f^{2}}{4026}=\frac{(1+t)^{2}}{t^{2}-t+2}$, over the interval $(1,2)$. (In general, such maximizers/maximums need not exist, but we shall prove there's a unique maximum here.)
We claim that $\lambda=\frac{16}{7}$ is the maximum of $\frac{(1+t)^{2}}{t^{2}-t+2}$. Indeed,
$$
\begin{aligned}
\lambda-g(t) & =\frac{(\lambda-1) t^{2}-(\lambda+2) t+(2 \lambda-1)}{t^{2}-t+2} \\
& =\frac{1}{7} \frac{9 t^{2}-30 t+25}{t^{2}-t+2}=\frac{1}{7} \frac{(3 t-5)^{2}}{\left(t-\frac{1}{2}\right)^{2}+\frac{7}{4}} \geq 0
\end{aligned}
$$
for all reals $t$ (not just $t \in(1,2)$ ), with equality at $t=\frac{5}{3} \in(1,2)$.
Comment. To motivate the choice of $\lambda$, suppose $\lambda$ were the maximum of $f$, attained at $t_{0} \in(1,2)$; then $h(t)=\lambda\left(t^{2}-t+2\right)-(t+1)^{2}$ is quadratic and nonnegative on $(1,2)$, but zero at $t=t_{0}$. If $g$ is a nontrivial quadratic (nonzero leading coefficient), then $t_{0}$ must be a double root, so $g$ has determinant 0 . Of course, $g$ could also be constant or linear over $(1,2)$, but we can easily rule out both of these possibilities.
Alternatively, we can simply take a derivative of $f$ to find critical points.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n25. [13]",
"solution_match": "\nAnswer: "
}
|
bb93f658-fc06-5486-b0e9-0a617f58eec8
| 609,181
|
Triangle $A B C$ has $A B=4, B C=3$, and a right angle at $B$. Circles $\omega_{1}$ and $\omega_{2}$ of equal radii are drawn such that $\omega_{1}$ is tangent to $A B$ and $A C, \omega_{2}$ is tangent to $B C$ and $A C$, and $\omega_{1}$ is tangent to $\omega_{2}$. Find the radius of $\omega_{1}$.
 of $\omega_{1}$ and $\omega_{2}$ respectively. Suppose $\omega_{i}$ hits $A C$ at $B_{i}$ for $i=1,2$, so that $O_{1} O_{2}=B_{1} B_{2}=2 r$.
Extend angle bisector $A O_{1}$ to hit $B C$ at $P$. By the angle bisector theorem and triangle similarity $\triangle A B_{1} O_{1} \sim \triangle A B P$, we deduce $\frac{r}{A B_{1}}=\frac{B P}{A B}=\frac{3}{4+5}$. Similarly, $\frac{r}{C B_{2}}=\frac{4}{3+5}$, so
$$
5=A C=A B_{1}+B_{1} B_{2}+B_{2} C=3 r+2 r+2 r=7 r
$$
or $r=\frac{5}{7}$.
|
Use the same notation as in the previous solution, and let $\alpha=\frac{1}{2} \angle A$. By constructing right triangles with hypotenuses $A O_{1}, O_{1} O_{2}$, and $O_{2} C$ and legs parallel to $A B$ and $B C$, we obtain
$$
4=A B=r \cot \alpha+2 r \cos \angle A+r .
$$
But $\cot \alpha=\frac{1+\cos 2 \alpha}{\sin 2 \alpha}=\frac{1+\frac{4}{5}}{\frac{3}{5}}=3$ and $\cos \angle A=\frac{4}{5}$, so the above equation simplifies to
$$
4=r\left(3+\frac{8}{5}+1\right)=\frac{28 r}{5}
$$
or $r=\frac{5}{7}$.
|
\frac{5}{7}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Triangle $A B C$ has $A B=4, B C=3$, and a right angle at $B$. Circles $\omega_{1}$ and $\omega_{2}$ of equal radii are drawn such that $\omega_{1}$ is tangent to $A B$ and $A C, \omega_{2}$ is tangent to $B C$ and $A C$, and $\omega_{1}$ is tangent to $\omega_{2}$. Find the radius of $\omega_{1}$.
 of $\omega_{1}$ and $\omega_{2}$ respectively. Suppose $\omega_{i}$ hits $A C$ at $B_{i}$ for $i=1,2$, so that $O_{1} O_{2}=B_{1} B_{2}=2 r$.
Extend angle bisector $A O_{1}$ to hit $B C$ at $P$. By the angle bisector theorem and triangle similarity $\triangle A B_{1} O_{1} \sim \triangle A B P$, we deduce $\frac{r}{A B_{1}}=\frac{B P}{A B}=\frac{3}{4+5}$. Similarly, $\frac{r}{C B_{2}}=\frac{4}{3+5}$, so
$$
5=A C=A B_{1}+B_{1} B_{2}+B_{2} C=3 r+2 r+2 r=7 r
$$
or $r=\frac{5}{7}$.
|
Use the same notation as in the previous solution, and let $\alpha=\frac{1}{2} \angle A$. By constructing right triangles with hypotenuses $A O_{1}, O_{1} O_{2}$, and $O_{2} C$ and legs parallel to $A B$ and $B C$, we obtain
$$
4=A B=r \cot \alpha+2 r \cos \angle A+r .
$$
But $\cot \alpha=\frac{1+\cos 2 \alpha}{\sin 2 \alpha}=\frac{1+\frac{4}{5}}{\frac{3}{5}}=3$ and $\cos \angle A=\frac{4}{5}$, so the above equation simplifies to
$$
4=r\left(3+\frac{8}{5}+1\right)=\frac{28 r}{5}
$$
or $r=\frac{5}{7}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n28. [15]",
"solution_match": "\nSolution 2. "
}
|
9e961579-cb86-55f0-b421-1d56f3ae355f
| 609,183
|
Let $\triangle X Y Z$ be a right triangle with $\angle X Y Z=90^{\circ}$. Suppose there exists an infinite sequence of equilateral triangles $X_{0} Y_{0} T_{0}, X_{1} Y_{1} T_{1}, \ldots$ such that $X_{0}=X, Y_{0}=Y, X_{i}$ lies on the segment $X Z$ for all $i \geq 0, Y_{i}$ lies on the segment $Y Z$ for all $i \geq 0, X_{i} Y_{i}$ is perpendicular to $Y Z$ for all $i \geq 0, T_{i}$ and $Y$ are separated by line $X Z$ for all $i \geq 0$, and $X_{i}$ lies on segment $Y_{i-1} T_{i-1}$ for $i \geq 1$.
Let $\mathcal{P}$ denote the union of the equilateral triangles. If the area of $\mathcal{P}$ is equal to the area of $X Y Z$, find $\frac{X Y}{Y Z}$.
|
1 For any region $R$, let $[R]$ denote its area.
Let $a=X Y, b=Y Z, r a=X_{1} Y_{1}$. Then $[\mathcal{P}]=\left[X Y T_{0}\right]\left(1+r^{2}+r^{4}+\cdots\right),[X Y Z]=\left[X Y Y_{1} X_{1}\right](1+$ $\left.r^{2}+r^{4}+\cdots\right), Y Y_{1}=r a \sqrt{3}$, and $b=r a \sqrt{3}\left(1+r+r^{2}+\cdots\right)$ (although we can also get this by similar triangles).
Hence $\frac{a^{2} \sqrt{3}}{4}=\frac{1}{2}(r a+a)(r a \sqrt{3})$, or $2 r(r+1)=1 \Longrightarrow r=\frac{\sqrt{3}-1}{2}$. Thus $\frac{X Y}{Y Z}=\frac{a}{b}=\frac{1-r}{r \sqrt{3}}=1$.
|
1
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $\triangle X Y Z$ be a right triangle with $\angle X Y Z=90^{\circ}$. Suppose there exists an infinite sequence of equilateral triangles $X_{0} Y_{0} T_{0}, X_{1} Y_{1} T_{1}, \ldots$ such that $X_{0}=X, Y_{0}=Y, X_{i}$ lies on the segment $X Z$ for all $i \geq 0, Y_{i}$ lies on the segment $Y Z$ for all $i \geq 0, X_{i} Y_{i}$ is perpendicular to $Y Z$ for all $i \geq 0, T_{i}$ and $Y$ are separated by line $X Z$ for all $i \geq 0$, and $X_{i}$ lies on segment $Y_{i-1} T_{i-1}$ for $i \geq 1$.
Let $\mathcal{P}$ denote the union of the equilateral triangles. If the area of $\mathcal{P}$ is equal to the area of $X Y Z$, find $\frac{X Y}{Y Z}$.
|
1 For any region $R$, let $[R]$ denote its area.
Let $a=X Y, b=Y Z, r a=X_{1} Y_{1}$. Then $[\mathcal{P}]=\left[X Y T_{0}\right]\left(1+r^{2}+r^{4}+\cdots\right),[X Y Z]=\left[X Y Y_{1} X_{1}\right](1+$ $\left.r^{2}+r^{4}+\cdots\right), Y Y_{1}=r a \sqrt{3}$, and $b=r a \sqrt{3}\left(1+r+r^{2}+\cdots\right)$ (although we can also get this by similar triangles).
Hence $\frac{a^{2} \sqrt{3}}{4}=\frac{1}{2}(r a+a)(r a \sqrt{3})$, or $2 r(r+1)=1 \Longrightarrow r=\frac{\sqrt{3}-1}{2}$. Thus $\frac{X Y}{Y Z}=\frac{a}{b}=\frac{1-r}{r \sqrt{3}}=1$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n29. [15]",
"solution_match": "\nAnswer: "
}
|
e8537c93-0579-50ed-ae01-679c7214cd4b
| 609,184
|
Find the number of ordered triples of integers $(a, b, c)$ with $1 \leq a, b, c \leq 100$ and $a^{2} b+b^{2} c+c^{2} a=$ $a b^{2}+b c^{2}+c a^{2}$.
|
29800 This factors as $(a-b)(b-c)(c-a)=0$. By the inclusion-exclusion principle, we get $3 \cdot 100^{2}-3 \cdot 100+100=29800$.
|
29800
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find the number of ordered triples of integers $(a, b, c)$ with $1 \leq a, b, c \leq 100$ and $a^{2} b+b^{2} c+c^{2} a=$ $a b^{2}+b c^{2}+c a^{2}$.
|
29800 This factors as $(a-b)(b-c)(c-a)=0$. By the inclusion-exclusion principle, we get $3 \cdot 100^{2}-3 \cdot 100+100=29800$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n30. [15]",
"solution_match": "\nAnswer: "
}
|
d3b955d7-8f8d-5f96-82fd-c00486d11d7c
| 609,185
|
Chords $\overline{A B}$ and $\overline{C D}$ of circle $\omega$ intersect at $E$ such that $A E=8, B E=2, C D=10$, and $\angle A E C=90^{\circ}$. Let $R$ be a rectangle inside $\omega$ with sides parallel to $\overline{A B}$ and $\overline{C D}$, such that no point in the interior of $R$ lies on $\overline{A B}, \overline{C D}$, or the boundary of $\omega$. What is the maximum possible area of $R$ ?
|
$\quad 26+6 \sqrt{17}$ By power of a point, $(C E)(E D)=(A E)(E B)=16$, and $C E+E D=C D=$ 10. Thus $C E, E D$ are 2, 8 . Without loss of generality, assume $C E=8$ and $D E=2$.
Assume our circle is centered at the origin, with points $A=(-3,5), B=(-3,-5), C=(5,-3)$, $D=(-5,-3)$, and the equation of the circle is $x^{2}+y^{2}=34$. Clearly the largest possible rectangle must lie in the first quadrant, and if we let $(x, y)$ be the upper-right corner of the rectangle, then the area of the rectangle is $(x+3)(y+3)=9+6(x+y)+x y \leq 9+12 \sqrt{\frac{x^{2}+y^{2}}{2}}+\frac{x^{2}+y^{2}}{2}=26+6 \sqrt{17}$, where equality holds if and only if $x=y=\sqrt{17}$.
|
26+6 \sqrt{17}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Chords $\overline{A B}$ and $\overline{C D}$ of circle $\omega$ intersect at $E$ such that $A E=8, B E=2, C D=10$, and $\angle A E C=90^{\circ}$. Let $R$ be a rectangle inside $\omega$ with sides parallel to $\overline{A B}$ and $\overline{C D}$, such that no point in the interior of $R$ lies on $\overline{A B}, \overline{C D}$, or the boundary of $\omega$. What is the maximum possible area of $R$ ?
|
$\quad 26+6 \sqrt{17}$ By power of a point, $(C E)(E D)=(A E)(E B)=16$, and $C E+E D=C D=$ 10. Thus $C E, E D$ are 2, 8 . Without loss of generality, assume $C E=8$ and $D E=2$.
Assume our circle is centered at the origin, with points $A=(-3,5), B=(-3,-5), C=(5,-3)$, $D=(-5,-3)$, and the equation of the circle is $x^{2}+y^{2}=34$. Clearly the largest possible rectangle must lie in the first quadrant, and if we let $(x, y)$ be the upper-right corner of the rectangle, then the area of the rectangle is $(x+3)(y+3)=9+6(x+y)+x y \leq 9+12 \sqrt{\frac{x^{2}+y^{2}}{2}}+\frac{x^{2}+y^{2}}{2}=26+6 \sqrt{17}$, where equality holds if and only if $x=y=\sqrt{17}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n31. [17]",
"solution_match": "\nAnswer: "
}
|
4ac02246-7203-5706-afdc-b8d373bfd46c
| 609,186
|
Suppose that $x$ and $y$ are chosen randomly and uniformly from $(0,1)$. What is the probability that $\left\lfloor\sqrt{\frac{x}{y}}\right\rfloor$ is even? Hint: $\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$.
|
$1-\frac{\pi^{2}}{24}$ OR $\frac{24-\pi^{2}}{24}$ Note that for every positive integer $n$, the probability that $\left\lfloor\sqrt{\frac{x}{y}}\right\rfloor=n$ is just the area of the triangle formed between $(0,0),\left(1, \frac{1}{n^{2}}\right),\left(1, \frac{1}{(n+1)^{2}}\right)$, which is just $\frac{1}{2}\left(\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right)$.
Thus the probability that $\left\lfloor\sqrt{\frac{x}{y}}\right\rfloor$ is odd is
$$
\begin{aligned}
\sum_{k=1}^{\infty} \frac{1}{2}\left(\frac{1}{(2 k-1)^{2}}-\frac{1}{(2 k)^{2}}\right) & =\frac{1}{2} \sum_{k=1}^{\infty}\left(\frac{1}{(2 k-1)^{2}}+\frac{1}{(2 k)^{2}}\right)-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{2}} \\
& =\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2}}-\frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^{2}} \\
& =\frac{\pi^{2}}{12}-\frac{\pi^{2}}{24}=\frac{\pi^{2}}{24}
\end{aligned}
$$
Thus our answer is just $1-\frac{\pi^{2}}{24}$.
|
1-\frac{\pi^{2}}{24}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose that $x$ and $y$ are chosen randomly and uniformly from $(0,1)$. What is the probability that $\left\lfloor\sqrt{\frac{x}{y}}\right\rfloor$ is even? Hint: $\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$.
|
$1-\frac{\pi^{2}}{24}$ OR $\frac{24-\pi^{2}}{24}$ Note that for every positive integer $n$, the probability that $\left\lfloor\sqrt{\frac{x}{y}}\right\rfloor=n$ is just the area of the triangle formed between $(0,0),\left(1, \frac{1}{n^{2}}\right),\left(1, \frac{1}{(n+1)^{2}}\right)$, which is just $\frac{1}{2}\left(\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right)$.
Thus the probability that $\left\lfloor\sqrt{\frac{x}{y}}\right\rfloor$ is odd is
$$
\begin{aligned}
\sum_{k=1}^{\infty} \frac{1}{2}\left(\frac{1}{(2 k-1)^{2}}-\frac{1}{(2 k)^{2}}\right) & =\frac{1}{2} \sum_{k=1}^{\infty}\left(\frac{1}{(2 k-1)^{2}}+\frac{1}{(2 k)^{2}}\right)-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{2}} \\
& =\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2}}-\frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^{2}} \\
& =\frac{\pi^{2}}{12}-\frac{\pi^{2}}{24}=\frac{\pi^{2}}{24}
\end{aligned}
$$
Thus our answer is just $1-\frac{\pi^{2}}{24}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n32. [17]",
"solution_match": "\nAnswer: "
}
|
52bdb7fe-66ab-5c56-b9ee-2324fc094a78
| 609,187
|
On each side of a 6 by 8 rectangle, construct an equilateral triangle with that side as one edge such that the interior of the triangle intersects the interior of the rectangle. What is the total area of all regions that are contained in exactly 3 of the 4 equilateral triangles?
|
$\frac{96 \sqrt{3}-154}{\sqrt{3}}$ OR $\frac{288-154 \sqrt{3}}{3}$ OR $96-\frac{154}{\sqrt{3}}$ OR $96-\frac{154 \sqrt{3}}{3}$ Let the rectangle be $A B C D$ with $A B=8$ and $B C=6$. Let the four equilateral triangles be $A B P_{1}, B C P_{2}, C D P_{3}$, and $D A P_{4}$ (for convenience, call them the $P_{1^{-}}, P_{2^{-}}, P_{3^{-}}, P_{4^{-}}$triangles). Let $W=A P_{1} \cap D P_{3}, X=A P_{1} \cap D P_{4}$, and $Y=D P_{4} \cap C P_{2}$. Reflect $X, Y$ over the line $P_{2} P_{4}$ (the line halfway between $A B$ and $D C$ ) to points $X^{\prime}, Y^{\prime}$.
First we analyze the basic configuration of the diagram. Since $A B=8<2 \cdot 6 \frac{\sqrt{3}}{2}$, the $P_{2^{-}}, P_{4^{-}}$triangles intersect. Furthermore, $A P_{1} \perp B P_{2}$, so if $T=B P_{2} \cap A P_{1}$, then $B P_{2}=6<4 \sqrt{3}=B T$. Therefore $P_{2}$ lies inside triangle $P_{1} B A$, and by symmetry, also triangle $P_{3} D C$.
It follows that the area we wish to compute is the union of two (congruent) concave hexagons, one of which is $W X Y P_{2} Y^{\prime} X^{\prime}$. (The other is its reflection over $Y Y^{\prime}$, the mid-line of $A D$ and $B C$.) So we seek
$$
2\left[W X Y P_{2} Y^{\prime} X^{\prime}\right]=2\left(\left[W X P_{4} X^{\prime}\right]-\left[P_{2} Y P_{4} Y^{\prime}\right]\right)
$$
It's easy to see that $\left[W X P_{4} X^{\prime}\right]=\frac{1}{3}\left[A D P_{4}\right]=\frac{1}{3} \frac{6^{2} \sqrt{3}}{4}=3 \sqrt{3}$, since $W X P_{4} X^{\prime}$ and its reflections over lines $D W X^{\prime}$ and $A W X$ partition $\triangle A D P_{4}$.
It remains to consider $P_{2} Y P_{4} Y^{\prime}$, a rhombus with (perpendicular) diagonals $P_{2} P_{4}$ and $Y Y^{\prime}$. If $O$ denotes the intersection of these two diagonals (also the center of $A B C D$ ), then $O P_{2}$ is $P_{2} B \frac{\sqrt{3}}{2}-\frac{1}{2} A B=3 \sqrt{3}-4$, the difference between the lengths of the $P_{2}$-altitude in $\triangle C B P_{2}$ and the distance between the parallel lines $Y Y^{\prime}, C B$. Easy angle chasing gives $O Y=\frac{O P_{2}}{\sqrt{3}}$, so
$$
\left[P_{2} Y P_{4} Y^{\prime}\right]=4 \cdot \frac{O P_{2} \cdot O Y}{2}=\frac{2}{\sqrt{3}} O P_{2}^{2}=\frac{2}{\sqrt{3}}(3 \sqrt{3}-4)^{2}=\frac{86-48 \sqrt{3}}{\sqrt{3}}
$$
and our desired area is
$$
2\left[W X P_{4} X^{\prime}\right]-2\left[P_{2} Y P_{4} Y^{\prime}\right]=6 \sqrt{3}-\frac{172-96 \sqrt{3}}{\sqrt{3}}=\frac{96 \sqrt{3}-154}{\sqrt{3}}
$$
or $\frac{288-154 \sqrt{3}}{3}$.
|
\frac{96 \sqrt{3}-154}{\sqrt{3}}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
On each side of a 6 by 8 rectangle, construct an equilateral triangle with that side as one edge such that the interior of the triangle intersects the interior of the rectangle. What is the total area of all regions that are contained in exactly 3 of the 4 equilateral triangles?
|
$\frac{96 \sqrt{3}-154}{\sqrt{3}}$ OR $\frac{288-154 \sqrt{3}}{3}$ OR $96-\frac{154}{\sqrt{3}}$ OR $96-\frac{154 \sqrt{3}}{3}$ Let the rectangle be $A B C D$ with $A B=8$ and $B C=6$. Let the four equilateral triangles be $A B P_{1}, B C P_{2}, C D P_{3}$, and $D A P_{4}$ (for convenience, call them the $P_{1^{-}}, P_{2^{-}}, P_{3^{-}}, P_{4^{-}}$triangles). Let $W=A P_{1} \cap D P_{3}, X=A P_{1} \cap D P_{4}$, and $Y=D P_{4} \cap C P_{2}$. Reflect $X, Y$ over the line $P_{2} P_{4}$ (the line halfway between $A B$ and $D C$ ) to points $X^{\prime}, Y^{\prime}$.
First we analyze the basic configuration of the diagram. Since $A B=8<2 \cdot 6 \frac{\sqrt{3}}{2}$, the $P_{2^{-}}, P_{4^{-}}$triangles intersect. Furthermore, $A P_{1} \perp B P_{2}$, so if $T=B P_{2} \cap A P_{1}$, then $B P_{2}=6<4 \sqrt{3}=B T$. Therefore $P_{2}$ lies inside triangle $P_{1} B A$, and by symmetry, also triangle $P_{3} D C$.
It follows that the area we wish to compute is the union of two (congruent) concave hexagons, one of which is $W X Y P_{2} Y^{\prime} X^{\prime}$. (The other is its reflection over $Y Y^{\prime}$, the mid-line of $A D$ and $B C$.) So we seek
$$
2\left[W X Y P_{2} Y^{\prime} X^{\prime}\right]=2\left(\left[W X P_{4} X^{\prime}\right]-\left[P_{2} Y P_{4} Y^{\prime}\right]\right)
$$
It's easy to see that $\left[W X P_{4} X^{\prime}\right]=\frac{1}{3}\left[A D P_{4}\right]=\frac{1}{3} \frac{6^{2} \sqrt{3}}{4}=3 \sqrt{3}$, since $W X P_{4} X^{\prime}$ and its reflections over lines $D W X^{\prime}$ and $A W X$ partition $\triangle A D P_{4}$.
It remains to consider $P_{2} Y P_{4} Y^{\prime}$, a rhombus with (perpendicular) diagonals $P_{2} P_{4}$ and $Y Y^{\prime}$. If $O$ denotes the intersection of these two diagonals (also the center of $A B C D$ ), then $O P_{2}$ is $P_{2} B \frac{\sqrt{3}}{2}-\frac{1}{2} A B=3 \sqrt{3}-4$, the difference between the lengths of the $P_{2}$-altitude in $\triangle C B P_{2}$ and the distance between the parallel lines $Y Y^{\prime}, C B$. Easy angle chasing gives $O Y=\frac{O P_{2}}{\sqrt{3}}$, so
$$
\left[P_{2} Y P_{4} Y^{\prime}\right]=4 \cdot \frac{O P_{2} \cdot O Y}{2}=\frac{2}{\sqrt{3}} O P_{2}^{2}=\frac{2}{\sqrt{3}}(3 \sqrt{3}-4)^{2}=\frac{86-48 \sqrt{3}}{\sqrt{3}}
$$
and our desired area is
$$
2\left[W X P_{4} X^{\prime}\right]-2\left[P_{2} Y P_{4} Y^{\prime}\right]=6 \sqrt{3}-\frac{172-96 \sqrt{3}}{\sqrt{3}}=\frac{96 \sqrt{3}-154}{\sqrt{3}}
$$
or $\frac{288-154 \sqrt{3}}{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n33. [17]",
"solution_match": "\nAnswer: "
}
|
4cc0fd26-5383-5746-80b9-7911e99f6415
| 609,188
|
Find the number of positive integers less than 1000000 that are divisible by some perfect cube greater than 1. Your score will be $\max \left\{0,\left\lfloor 20-200\left|1-\frac{k}{S}\right|\right\rfloor\right\}$, where $k$ is your answer and $S$ is the actual answer.
|
168089 Using the following code, we get the answer (denoted by the variable ans):
ans $=0$
for $n$ in xrange $(1,1000000)$ :
```
divisible_by_cube = True
for i in xrange(2,101):
if n%(i*i*i)==0:
divisible_by_cube = False
break
if divisible_by_cube: ans = ans + 1
print ans
```
This gives the output
168089
Alternatively, let $N=1000000$ and denote by $P$ the set of primes. Then by PIE, the number of $n \in(0, N)$ divisible by a nontrivial cube, or equivalently, by $p^{3}$ for some $p \in P$, is
$$
\sum_{p \in P}\left\lfloor\frac{N-1}{p^{3}}\right\rfloor-\sum_{p<q \in P}\left\lfloor\frac{N-1}{p^{3} q^{3}}\right\rfloor \pm \cdots
$$
which deviates from
$$
\sum_{p \in P} \frac{N-1}{p^{3}}-\sum_{p<q \in P} \frac{N-1}{p^{3} q^{3}} \pm \cdots=(N-1)\left(1-\prod_{p \in P}\left(1-p^{-3}\right)\right)
$$
by at most the sum of
- $N^{1 / 3} \sup _{t \in \mathbb{R}}|t-\lfloor t\rfloor|=N^{1 / 3}$, for terms $\left\lfloor\frac{N-1}{p_{1}^{3} \cdots p_{r}^{3}}\right\rfloor$ with $p_{1} \cdots p_{r} \leq(N-1)^{1 / 3}$, and
- $(N-1) \sum_{k>(N-1)^{1 / 3}} k^{-3}<(N-1)\left[(N-1)^{-1}+\int_{(N-1)^{1 / 3}}^{\infty} x^{-3} d x\right]=1+(N-1) \frac{(N-1)^{-2 / 3}}{2}=$ $O\left(N^{1 / 3}\right)$, for the remaining terms.
So we are really interested in $10^{6}-10^{6} \prod_{p \in P}\left(1-p^{-3}\right)$ (which, for completeness, is $168092.627 \ldots$. . There are a few simple ways to approximate this:
- We can use a partial product of $\prod_{p \in P}\left(1-p^{-3}\right)$. Using just $1-2^{-3}=0.875$ gives an answer of 125000 (this is also just the number of $x \leq N$ divisible by $\left.2^{3}=8\right),\left(1-2^{-3}\right)\left(1-3^{-3}\right) \approx 0.843$ gives 157000 (around the number of $x$ divisible by $2^{3}$ or $3^{3}$ ), etc. This will give a lower bound, of course, so we can guess a bit higher. For instance, while 157000 gives a score of around 7 , rounding up to 160000 gives $\approx 10$.
- We can note that $\prod_{p \in P}\left(1-p^{-3}\right)=\zeta(3)^{-1}$ is the inverse of $1+2^{-3}+3^{-3}+\cdots$. This is a bit less efficient, but successive partial sums (starting with $1+2^{-3}$ ) give around 111000, 139000, 150000, 157000 , etc. Again, this gives a lower bound, so we can guess a little higher.
- We can optimize the previous approach with integral approximation after the $r$ th term: $\zeta(3)$ is the sum of $1+2^{-3}+\cdots+r^{-3}$ plus something between $\int_{r+1}^{\infty} x^{-3} d x=\frac{1}{2}(r+1)^{-2}$ and $\int_{r}^{\infty} x^{-3} d x=$ $\frac{1}{2} r^{-2}$. Then starting with $r=1$, we get intervals of around $(111000,334000),(152000,200000)$, $(161000,179000),(165000,173000)$, etc. Then we can take something like the average of the two endpoints as our guess; such a strategy gets a score of around 10 for $r=2$ already, and $\approx 17$ for $r=3$.
|
168089
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the number of positive integers less than 1000000 that are divisible by some perfect cube greater than 1. Your score will be $\max \left\{0,\left\lfloor 20-200\left|1-\frac{k}{S}\right|\right\rfloor\right\}$, where $k$ is your answer and $S$ is the actual answer.
|
168089 Using the following code, we get the answer (denoted by the variable ans):
ans $=0$
for $n$ in xrange $(1,1000000)$ :
```
divisible_by_cube = True
for i in xrange(2,101):
if n%(i*i*i)==0:
divisible_by_cube = False
break
if divisible_by_cube: ans = ans + 1
print ans
```
This gives the output
168089
Alternatively, let $N=1000000$ and denote by $P$ the set of primes. Then by PIE, the number of $n \in(0, N)$ divisible by a nontrivial cube, or equivalently, by $p^{3}$ for some $p \in P$, is
$$
\sum_{p \in P}\left\lfloor\frac{N-1}{p^{3}}\right\rfloor-\sum_{p<q \in P}\left\lfloor\frac{N-1}{p^{3} q^{3}}\right\rfloor \pm \cdots
$$
which deviates from
$$
\sum_{p \in P} \frac{N-1}{p^{3}}-\sum_{p<q \in P} \frac{N-1}{p^{3} q^{3}} \pm \cdots=(N-1)\left(1-\prod_{p \in P}\left(1-p^{-3}\right)\right)
$$
by at most the sum of
- $N^{1 / 3} \sup _{t \in \mathbb{R}}|t-\lfloor t\rfloor|=N^{1 / 3}$, for terms $\left\lfloor\frac{N-1}{p_{1}^{3} \cdots p_{r}^{3}}\right\rfloor$ with $p_{1} \cdots p_{r} \leq(N-1)^{1 / 3}$, and
- $(N-1) \sum_{k>(N-1)^{1 / 3}} k^{-3}<(N-1)\left[(N-1)^{-1}+\int_{(N-1)^{1 / 3}}^{\infty} x^{-3} d x\right]=1+(N-1) \frac{(N-1)^{-2 / 3}}{2}=$ $O\left(N^{1 / 3}\right)$, for the remaining terms.
So we are really interested in $10^{6}-10^{6} \prod_{p \in P}\left(1-p^{-3}\right)$ (which, for completeness, is $168092.627 \ldots$. . There are a few simple ways to approximate this:
- We can use a partial product of $\prod_{p \in P}\left(1-p^{-3}\right)$. Using just $1-2^{-3}=0.875$ gives an answer of 125000 (this is also just the number of $x \leq N$ divisible by $\left.2^{3}=8\right),\left(1-2^{-3}\right)\left(1-3^{-3}\right) \approx 0.843$ gives 157000 (around the number of $x$ divisible by $2^{3}$ or $3^{3}$ ), etc. This will give a lower bound, of course, so we can guess a bit higher. For instance, while 157000 gives a score of around 7 , rounding up to 160000 gives $\approx 10$.
- We can note that $\prod_{p \in P}\left(1-p^{-3}\right)=\zeta(3)^{-1}$ is the inverse of $1+2^{-3}+3^{-3}+\cdots$. This is a bit less efficient, but successive partial sums (starting with $1+2^{-3}$ ) give around 111000, 139000, 150000, 157000 , etc. Again, this gives a lower bound, so we can guess a little higher.
- We can optimize the previous approach with integral approximation after the $r$ th term: $\zeta(3)$ is the sum of $1+2^{-3}+\cdots+r^{-3}$ plus something between $\int_{r+1}^{\infty} x^{-3} d x=\frac{1}{2}(r+1)^{-2}$ and $\int_{r}^{\infty} x^{-3} d x=$ $\frac{1}{2} r^{-2}$. Then starting with $r=1$, we get intervals of around $(111000,334000),(152000,200000)$, $(161000,179000),(165000,173000)$, etc. Then we can take something like the average of the two endpoints as our guess; such a strategy gets a score of around 10 for $r=2$ already, and $\approx 17$ for $r=3$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-guts-solutions.jsonl",
"problem_match": "\n34. [20]",
"solution_match": "\nAnswer: "
}
|
946a57f2-dad6-5539-903c-b7d13f007604
| 609,189
|
Tim the Beaver can make three different types of geometrical figures: squares, regular hexagons, and regular octagons. Tim makes a random sequence $F_{0}, F_{1}, F_{2}, F_{3}, \ldots$ of figures as follows:
- $F_{0}$ is a square.
- For every positive integer $i, F_{i}$ is randomly chosen to be one of the 2 figures distinct from $F_{i-1}$ (each chosen with equal probability $\frac{1}{2}$ ).
- Tim takes 4 seconds to make squares, 6 to make hexagons, and 8 to make octagons. He makes one figure after another, with no breaks in between.
Suppose that exactly 17 seconds after he starts making $F_{0}$, Tim is making a figure with $n$ sides. What is the expected value of $n$ ?
|
7 We write $F_{i}=n$ as shorthand for "the $i$ th figure is an $n$-sided polygon."
If $F_{1}=8$, the $F_{2}=6$ or $F_{2}=4$. If $F_{2}=6$, Tim is making a 6 -gon at time 13 (probability contribution $1 / 4)$. If $F_{2}=4, F_{3}=6$ or $F_{3}=8$ will take the time 13 mark ( $1 / 8$ contribution each).
If $F_{1}=6, F_{2}=8$ or $F_{2}=4$. If $F_{2}=8$, it takes the 13 mark $\left(1 / 4\right.$ contribution). If $F_{2}=4, F_{3}=6$ or $F_{3}=8$ will take the 13 mark ( $1 / 8$ contribution each).
Thus, the expected value of the number of sides at time 13 is $0(4)+\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{8}\right)(6)+\left(\frac{1}{8}+\frac{1}{4}+\frac{1}{8}\right)(8)=7$.
|
7
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Tim the Beaver can make three different types of geometrical figures: squares, regular hexagons, and regular octagons. Tim makes a random sequence $F_{0}, F_{1}, F_{2}, F_{3}, \ldots$ of figures as follows:
- $F_{0}$ is a square.
- For every positive integer $i, F_{i}$ is randomly chosen to be one of the 2 figures distinct from $F_{i-1}$ (each chosen with equal probability $\frac{1}{2}$ ).
- Tim takes 4 seconds to make squares, 6 to make hexagons, and 8 to make octagons. He makes one figure after another, with no breaks in between.
Suppose that exactly 17 seconds after he starts making $F_{0}$, Tim is making a figure with $n$ sides. What is the expected value of $n$ ?
|
7 We write $F_{i}=n$ as shorthand for "the $i$ th figure is an $n$-sided polygon."
If $F_{1}=8$, the $F_{2}=6$ or $F_{2}=4$. If $F_{2}=6$, Tim is making a 6 -gon at time 13 (probability contribution $1 / 4)$. If $F_{2}=4, F_{3}=6$ or $F_{3}=8$ will take the time 13 mark ( $1 / 8$ contribution each).
If $F_{1}=6, F_{2}=8$ or $F_{2}=4$. If $F_{2}=8$, it takes the 13 mark $\left(1 / 4\right.$ contribution). If $F_{2}=4, F_{3}=6$ or $F_{3}=8$ will take the 13 mark ( $1 / 8$ contribution each).
Thus, the expected value of the number of sides at time 13 is $0(4)+\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{8}\right)(6)+\left(\frac{1}{8}+\frac{1}{4}+\frac{1}{8}\right)(8)=7$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-team-solutions.jsonl",
"problem_match": "\n1. [3]",
"solution_match": "\nAnswer: "
}
|
4db389c9-1097-542c-b1ad-fc71386afcad
| 609,192
|
Gary plays the following game with a fair $n$-sided die whose faces are labeled with the positive integers between 1 and $n$, inclusive: if $n=1$, he stops; otherwise he rolls the die, and starts over with a $k$-sided die, where $k$ is the number his $n$-sided die lands on. (In particular, if he gets $k=1$, he will stop rolling the die.) If he starts out with a 6 -sided die, what is the expected number of rolls he makes?
|
$\quad \frac{197}{60}$ If we let $a_{n}$ be the expected number of rolls starting with an $n$-sided die, we see immediately that $a_{1}=0$, and $a_{n}=1+\frac{1}{n} \sum_{i=1}^{n} a_{i}$ for $n>1$. Thus $a_{2}=2$, and for $n \geq 3$, $a_{n}=1+\frac{1}{n} a_{n}+\frac{n-1}{n}\left(a_{n-1}-1\right)$, or $a_{n}=a_{n-1}+\frac{1}{n-1}$. Thus $a_{n}=1+\sum_{i=1}^{n-1} \frac{1}{i}$ for $n \geq 2$, so $a_{6}=$ $1+\frac{60+30+20+15+12}{60}=\frac{197}{60}$.
|
\frac{197}{60}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Gary plays the following game with a fair $n$-sided die whose faces are labeled with the positive integers between 1 and $n$, inclusive: if $n=1$, he stops; otherwise he rolls the die, and starts over with a $k$-sided die, where $k$ is the number his $n$-sided die lands on. (In particular, if he gets $k=1$, he will stop rolling the die.) If he starts out with a 6 -sided die, what is the expected number of rolls he makes?
|
$\quad \frac{197}{60}$ If we let $a_{n}$ be the expected number of rolls starting with an $n$-sided die, we see immediately that $a_{1}=0$, and $a_{n}=1+\frac{1}{n} \sum_{i=1}^{n} a_{i}$ for $n>1$. Thus $a_{2}=2$, and for $n \geq 3$, $a_{n}=1+\frac{1}{n} a_{n}+\frac{n-1}{n}\left(a_{n-1}-1\right)$, or $a_{n}=a_{n-1}+\frac{1}{n-1}$. Thus $a_{n}=1+\sum_{i=1}^{n-1} \frac{1}{i}$ for $n \geq 2$, so $a_{6}=$ $1+\frac{60+30+20+15+12}{60}=\frac{197}{60}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-team-solutions.jsonl",
"problem_match": "\n2. [4]",
"solution_match": "\nAnswer: "
}
|
47ca4026-c7b5-5704-ad14-fec41debe2ab
| 609,193
|
The digits $1,2,3,4,5,6$ are randomly chosen (without replacement) to form the three-digit numbers $M=\overline{A B C}$ and $N=\overline{D E F}$. For example, we could have $M=413$ and $N=256$. Find the expected value of $M \cdot N$.
|
143745 By linearity of expectation and symmetry,
$$
\mathbb{E}[M N]=\mathbb{E}[(100 A+10 B+C)(100 D+10 E+F)]=111^{2} \cdot \mathbb{E}[A D]
$$
Since
$$
\mathbb{E}[A D]=\frac{(1+2+3+4+5+6)^{2}-\left(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}\right)}{6 \cdot 5}=\frac{350}{30}
$$
our answer is $111 \cdot 35 \cdot 37=111 \cdot 1295=143745$.
|
143745
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
The digits $1,2,3,4,5,6$ are randomly chosen (without replacement) to form the three-digit numbers $M=\overline{A B C}$ and $N=\overline{D E F}$. For example, we could have $M=413$ and $N=256$. Find the expected value of $M \cdot N$.
|
143745 By linearity of expectation and symmetry,
$$
\mathbb{E}[M N]=\mathbb{E}[(100 A+10 B+C)(100 D+10 E+F)]=111^{2} \cdot \mathbb{E}[A D]
$$
Since
$$
\mathbb{E}[A D]=\frac{(1+2+3+4+5+6)^{2}-\left(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}\right)}{6 \cdot 5}=\frac{350}{30}
$$
our answer is $111 \cdot 35 \cdot 37=111 \cdot 1295=143745$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-team-solutions.jsonl",
"problem_match": "\n3. [6]",
"solution_match": "\nAnswer: "
}
|
65e4f3f5-8626-5192-9537-2dfbdeefdfd5
| 609,194
|
Consider triangle $A B C$ with side lengths $A B=4, B C=7$, and $A C=8$. Let $M$ be the midpoint of segment $A B$, and let $N$ be the point on the interior of segment $A C$ that also lies on the circumcircle of triangle $M B C$. Compute $B N$.
|
$\frac{\sqrt{210}}{4}$ OR $\frac{\sqrt{105}}{2 \sqrt{2}}$ Let $\angle B A C=\theta$. Then, $\cos \theta=\frac{4^{2}+8^{2}-7^{2}}{2 \cdot 4 \cdot 8}$. Since $A M=\frac{4}{2}=2$, and power of a point gives $A M \cdot A B=A N \cdot A C$, we have $A N=\frac{2 \cdot 4}{8}=1$, so $N C=8-1=7$. Law of cosines on triangle $B A N$ gives
$$
B N^{2}=4^{2}+1^{2}-2 \cdot 4 \cdot 1 \cdot \frac{4^{2}+8^{2}-7^{2}}{2 \cdot 4 \cdot 8}=17-\frac{16+15}{8}=15-\frac{15}{8}=\frac{105}{8}
$$
so $B N=\frac{\sqrt{210}}{4}$.
|
\frac{\sqrt{210}}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Consider triangle $A B C$ with side lengths $A B=4, B C=7$, and $A C=8$. Let $M$ be the midpoint of segment $A B$, and let $N$ be the point on the interior of segment $A C$ that also lies on the circumcircle of triangle $M B C$. Compute $B N$.
|
$\frac{\sqrt{210}}{4}$ OR $\frac{\sqrt{105}}{2 \sqrt{2}}$ Let $\angle B A C=\theta$. Then, $\cos \theta=\frac{4^{2}+8^{2}-7^{2}}{2 \cdot 4 \cdot 8}$. Since $A M=\frac{4}{2}=2$, and power of a point gives $A M \cdot A B=A N \cdot A C$, we have $A N=\frac{2 \cdot 4}{8}=1$, so $N C=8-1=7$. Law of cosines on triangle $B A N$ gives
$$
B N^{2}=4^{2}+1^{2}-2 \cdot 4 \cdot 1 \cdot \frac{4^{2}+8^{2}-7^{2}}{2 \cdot 4 \cdot 8}=17-\frac{16+15}{8}=15-\frac{15}{8}=\frac{105}{8}
$$
so $B N=\frac{\sqrt{210}}{4}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-team-solutions.jsonl",
"problem_match": "\n4. [4]",
"solution_match": "\nAnswer: "
}
|
19d3ee4d-921a-555a-9c12-e62955a39268
| 609,195
|
In triangle $A B C, \angle B A C=60^{\circ}$. Let $\omega$ be a circle tangent to segment $A B$ at point $D$ and segment $A C$ at point $E$. Suppose $\omega$ intersects segment $B C$ at points $F$ and $G$ such that $F$ lies in between $B$ and $G$. Given that $A D=F G=4$ and $B F=\frac{1}{2}$, find the length of $C G$.
|
$\boxed{\frac{16}{5}}$ Let $x=CG$. First, by power of a point, $BD = \sqrt{BF(BF+FG)}=\frac{3}{2}$, and $C E=\sqrt{x(x+4)}$. By the law of cosines, we have
$$
\left(x+\frac{9}{2}\right)^{2}=\left(\frac{11}{2}\right)^{2}+(4+\sqrt{x(x+4)})^{2}-\frac{11}{2}(4+\sqrt{x(x+4)})
$$
which rearranges to $2(5 x-4)=5 \sqrt{x(x+4)}$. Squaring and noting $x>\frac{4}{5}$ gives $(5 x-16)(15 x-4)=$ $0 \Longrightarrow x=\frac{16}{5}$.
|
\frac{16}{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
In triangle $A B C, \angle B A C=60^{\circ}$. Let $\omega$ be a circle tangent to segment $A B$ at point $D$ and segment $A C$ at point $E$. Suppose $\omega$ intersects segment $B C$ at points $F$ and $G$ such that $F$ lies in between $B$ and $G$. Given that $A D=F G=4$ and $B F=\frac{1}{2}$, find the length of $C G$.
|
$\boxed{\frac{16}{5}}$ Let $x=CG$. First, by power of a point, $BD = \sqrt{BF(BF+FG)}=\frac{3}{2}$, and $C E=\sqrt{x(x+4)}$. By the law of cosines, we have
$$
\left(x+\frac{9}{2}\right)^{2}=\left(\frac{11}{2}\right)^{2}+(4+\sqrt{x(x+4)})^{2}-\frac{11}{2}(4+\sqrt{x(x+4)})
$$
which rearranges to $2(5 x-4)=5 \sqrt{x(x+4)}$. Squaring and noting $x>\frac{4}{5}$ gives $(5 x-16)(15 x-4)=$ $0 \Longrightarrow x=\frac{16}{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-team-solutions.jsonl",
"problem_match": "\n5. [4]",
"solution_match": "\nAnswer: "
}
|
d1f2b098-a863-5c5f-bd38-8a37211da957
| 609,196
|
Points $A, B, C$ lie on a circle $\omega$ such that $B C$ is a diameter. $A B$ is extended past $B$ to point $B^{\prime}$ and $A C$ is extended past $C$ to point $C^{\prime}$ such that line $B^{\prime} C^{\prime}$ is parallel to $B C$ and tangent to $\omega$ at point $D$. If $B^{\prime} D=4$ and $C^{\prime} D=6$, compute $B C$.
|
$\frac{24}{5}$ Let $x=A B$ and $y=A C$, and define $t>0$ such that $B B^{\prime}=t x$ and $C C^{\prime}=t y$. Then $10=B^{\prime} C^{\prime}=(1+t) \sqrt{x^{2}+y^{2}}, 4^{2}=t(1+t) x^{2}$, and $6^{2}=t(1+t) y^{2}$ (by power of a point), so $52=4^{2}+6^{2}=t(1+t)\left(x^{2}+y^{2}\right)$ gives $\frac{13}{25}=\frac{52}{10^{2}}=\frac{t(1+t)}{(1+t)^{2}}=\frac{t}{1+t} \Longrightarrow t=\frac{13}{12}$. Hence $B C=\sqrt{x^{2}+y^{2}}=$ $\frac{10}{1+t}=\frac{10}{25 / 12}=\frac{24}{5}$.
|
\frac{24}{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Points $A, B, C$ lie on a circle $\omega$ such that $B C$ is a diameter. $A B$ is extended past $B$ to point $B^{\prime}$ and $A C$ is extended past $C$ to point $C^{\prime}$ such that line $B^{\prime} C^{\prime}$ is parallel to $B C$ and tangent to $\omega$ at point $D$. If $B^{\prime} D=4$ and $C^{\prime} D=6$, compute $B C$.
|
$\frac{24}{5}$ Let $x=A B$ and $y=A C$, and define $t>0$ such that $B B^{\prime}=t x$ and $C C^{\prime}=t y$. Then $10=B^{\prime} C^{\prime}=(1+t) \sqrt{x^{2}+y^{2}}, 4^{2}=t(1+t) x^{2}$, and $6^{2}=t(1+t) y^{2}$ (by power of a point), so $52=4^{2}+6^{2}=t(1+t)\left(x^{2}+y^{2}\right)$ gives $\frac{13}{25}=\frac{52}{10^{2}}=\frac{t(1+t)}{(1+t)^{2}}=\frac{t}{1+t} \Longrightarrow t=\frac{13}{12}$. Hence $B C=\sqrt{x^{2}+y^{2}}=$ $\frac{10}{1+t}=\frac{10}{25 / 12}=\frac{24}{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-team-solutions.jsonl",
"problem_match": "\n6. [6]",
"solution_match": "\nAnswer: "
}
|
ef25d398-5e58-55c4-93a3-d500f8008c31
| 609,197
|
In equilateral triangle $A B C$, a circle $\omega$ is drawn such that it is tangent to all three sides of the triangle. A line is drawn from $A$ to point $D$ on segment $B C$ such that $A D$ intersects $\omega$ at points $E$ and $F$. If $E F=4$ and $A B=8$, determine $|A E-F D|$.
|
$\frac{4}{\sqrt{5}}$ OR $\frac{4 \sqrt{5}}{5}$ Without loss of generality, $A, E, F, D$ lie in that order. Let $x=A E, y=D F$.
By power of a point, $x(x+4)=4^{2} \Longrightarrow x=2 \sqrt{5}-2$, and $y(y+4)=(x+4+y)^{2}-(4 \sqrt{3})^{2} \Longrightarrow y=$ $\frac{48-(x+4)^{2}}{2(x+2)}=\frac{12-(1+\sqrt{5})^{2}}{\sqrt{5}}$. It readily follows that $x-y=\frac{4}{\sqrt{5}}=\frac{4 \sqrt{5}}{5}$.
|
\frac{4 \sqrt{5}}{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
In equilateral triangle $A B C$, a circle $\omega$ is drawn such that it is tangent to all three sides of the triangle. A line is drawn from $A$ to point $D$ on segment $B C$ such that $A D$ intersects $\omega$ at points $E$ and $F$. If $E F=4$ and $A B=8$, determine $|A E-F D|$.
|
$\frac{4}{\sqrt{5}}$ OR $\frac{4 \sqrt{5}}{5}$ Without loss of generality, $A, E, F, D$ lie in that order. Let $x=A E, y=D F$.
By power of a point, $x(x+4)=4^{2} \Longrightarrow x=2 \sqrt{5}-2$, and $y(y+4)=(x+4+y)^{2}-(4 \sqrt{3})^{2} \Longrightarrow y=$ $\frac{48-(x+4)^{2}}{2(x+2)}=\frac{12-(1+\sqrt{5})^{2}}{\sqrt{5}}$. It readily follows that $x-y=\frac{4}{\sqrt{5}}=\frac{4 \sqrt{5}}{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-team-solutions.jsonl",
"problem_match": "\n7. [7]",
"solution_match": "\nAnswer: "
}
|
e96d993f-d7c5-5f21-87f4-5880569790f9
| 609,198
|
Define the sequence $\left\{x_{i}\right\}_{i \geq 0}$ by $x_{0}=x_{1}=x_{2}=1$ and $x_{k}=\frac{x_{k-1}+x_{k-2}+1}{x_{k-3}}$ for $k>2$. Find $x_{2013}$.
|
9 We have $x_{3}=\frac{1+1+1}{1}=3, x_{4}=\frac{3+1+1}{1}=5, x_{5}=\frac{5+3+1}{1}=9, x_{6}=\frac{9+5+1}{3}=5$. By the symmetry of our recurrence (or just further computation-it doesn't matter much), $x_{7}=3$ and $x_{8}=x_{9}=x_{10}=1$, so our sequence has period 8. Thus $x_{2013}=x_{13}=x_{5}=9$.
|
9
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Define the sequence $\left\{x_{i}\right\}_{i \geq 0}$ by $x_{0}=x_{1}=x_{2}=1$ and $x_{k}=\frac{x_{k-1}+x_{k-2}+1}{x_{k-3}}$ for $k>2$. Find $x_{2013}$.
|
9 We have $x_{3}=\frac{1+1+1}{1}=3, x_{4}=\frac{3+1+1}{1}=5, x_{5}=\frac{5+3+1}{1}=9, x_{6}=\frac{9+5+1}{3}=5$. By the symmetry of our recurrence (or just further computation-it doesn't matter much), $x_{7}=3$ and $x_{8}=x_{9}=x_{10}=1$, so our sequence has period 8. Thus $x_{2013}=x_{13}=x_{5}=9$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-team-solutions.jsonl",
"problem_match": "\n8. [2]",
"solution_match": "\nAnswer: "
}
|
fb661391-36c6-5411-85f0-52922ee3560d
| 609,199
|
For an integer $n \geq 0$, let $f(n)$ be the smallest possible value of $|x+y|$, where $x$ and $y$ are integers such that $3 x-2 y=n$. Evaluate $f(0)+f(1)+f(2)+\cdots+f(2013)$.
|
2416 First, we can use $3 x-2 y=n$ to get $x=\frac{n+2 y}{3}$. Thus $|x+y|=\left|\frac{n+5 y}{3}\right|$. Given a certain $n$, the only restriction on $y$ is that $3|n+2 y \Longleftrightarrow 3| n+5 y$. Hence the set of possible $x+y$ equals the set of integers of the form $\frac{n+5 y}{3}$, which in turn equals the set of integers congruent to $3^{-1} n \equiv 2 n(\bmod 5)$. (Prove this!)
Thus $f(n)=|x+y|$ is minimized when $x+y$ equals the least absolute remainder $(2 n)_{5}$ when $2 n$ is divided by 5 , i.e. the number between -2 and 2 (inclusive) congruent to $2 n$ modulo 5 . We immediately find $f(n)=f(n+5 m)$ for all integers $m$, and the following initial values of $f: f(0)=\left|(0)_{5}\right|=0$; $f(1)=\left|(2)_{5}\right|=2 ; f(2)=\left|(4)_{5}\right|=1 ; f(3)=\left|(6)_{5}\right|=1$; and $f(4)=\left|(8)_{5}\right|=2$.
Since $2013=403 \cdot 5-2$, it follows that $f(0)+f(1)+\cdots+f(2013)=403[f(0)+f(1)+\cdots+f(4)]-f(2014)=$ $403 \cdot 6-2=2416$.
|
2416
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
For an integer $n \geq 0$, let $f(n)$ be the smallest possible value of $|x+y|$, where $x$ and $y$ are integers such that $3 x-2 y=n$. Evaluate $f(0)+f(1)+f(2)+\cdots+f(2013)$.
|
2416 First, we can use $3 x-2 y=n$ to get $x=\frac{n+2 y}{3}$. Thus $|x+y|=\left|\frac{n+5 y}{3}\right|$. Given a certain $n$, the only restriction on $y$ is that $3|n+2 y \Longleftrightarrow 3| n+5 y$. Hence the set of possible $x+y$ equals the set of integers of the form $\frac{n+5 y}{3}$, which in turn equals the set of integers congruent to $3^{-1} n \equiv 2 n(\bmod 5)$. (Prove this!)
Thus $f(n)=|x+y|$ is minimized when $x+y$ equals the least absolute remainder $(2 n)_{5}$ when $2 n$ is divided by 5 , i.e. the number between -2 and 2 (inclusive) congruent to $2 n$ modulo 5 . We immediately find $f(n)=f(n+5 m)$ for all integers $m$, and the following initial values of $f: f(0)=\left|(0)_{5}\right|=0$; $f(1)=\left|(2)_{5}\right|=2 ; f(2)=\left|(4)_{5}\right|=1 ; f(3)=\left|(6)_{5}\right|=1$; and $f(4)=\left|(8)_{5}\right|=2$.
Since $2013=403 \cdot 5-2$, it follows that $f(0)+f(1)+\cdots+f(2013)=403[f(0)+f(1)+\cdots+f(4)]-f(2014)=$ $403 \cdot 6-2=2416$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-team-solutions.jsonl",
"problem_match": "\n9. [7]",
"solution_match": "\nAnswer: "
}
|
fd5e4bf5-2e2e-5a36-81fc-153bbefbb38c
| 609,200
|
Let $\omega=\cos \frac{2 \pi}{727}+i \sin \frac{2 \pi}{727}$. The imaginary part of the complex number
$$
\prod_{k=8}^{13}\left(1+\omega^{3^{k-1}}+\omega^{2 \cdot 3^{k-1}}\right)
$$
is equal to $\sin \alpha$ for some angle $\alpha$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, inclusive. Find $\alpha$.
|
$\frac{12 \pi}{727}$ Note that $727=3^{6}-2$. Our product telescopes to $\frac{1-\omega^{3^{13}}}{1-\omega^{3^{7}}}=\frac{1-\omega^{12}}{1-\omega^{6}}=1+\omega^{6}$, which has imaginary part $\sin \frac{12 \pi}{727}$, giving $\alpha=\frac{12 \pi}{727}$.
|
\frac{12 \pi}{727}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $\omega=\cos \frac{2 \pi}{727}+i \sin \frac{2 \pi}{727}$. The imaginary part of the complex number
$$
\prod_{k=8}^{13}\left(1+\omega^{3^{k-1}}+\omega^{2 \cdot 3^{k-1}}\right)
$$
is equal to $\sin \alpha$ for some angle $\alpha$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, inclusive. Find $\alpha$.
|
$\frac{12 \pi}{727}$ Note that $727=3^{6}-2$. Our product telescopes to $\frac{1-\omega^{3^{13}}}{1-\omega^{3^{7}}}=\frac{1-\omega^{12}}{1-\omega^{6}}=1+\omega^{6}$, which has imaginary part $\sin \frac{12 \pi}{727}$, giving $\alpha=\frac{12 \pi}{727}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-team-solutions.jsonl",
"problem_match": "\n10. [7]",
"solution_match": "\nAnswer: "
}
|
936f27ce-d238-534e-a8f6-6df58d7803c2
| 609,201
|
Two cars are driving directly towards each other such that one is twice as fast as the other. The distance between their starting points is 4 miles. When the two cars meet, how many miles is the faster car from its starting point?
|
| $\frac{8}{3}$ |
| :---: |
| Note that the faster car traveled twice the distance of the slower car, and together, | the two cars traveled the total distance between the starting points, which is 4 miles. Let the distance that the faster car traveled be $x$. Then, $x+\frac{x}{2}=4 \Longrightarrow x=\frac{8}{3}$. Thus, the faster car traveled $\frac{8}{3}$ miles from the starting point.
|
\frac{8}{3}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Two cars are driving directly towards each other such that one is twice as fast as the other. The distance between their starting points is 4 miles. When the two cars meet, how many miles is the faster car from its starting point?
|
| $\frac{8}{3}$ |
| :---: |
| Note that the faster car traveled twice the distance of the slower car, and together, | the two cars traveled the total distance between the starting points, which is 4 miles. Let the distance that the faster car traveled be $x$. Then, $x+\frac{x}{2}=4 \Longrightarrow x=\frac{8}{3}$. Thus, the faster car traveled $\frac{8}{3}$ miles from the starting point.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-thm-solutions.jsonl",
"problem_match": "\n1. [2]",
"solution_match": "\nAnswer: "
}
|
26771bf6-8922-5c05-a170-d34dd4221f50
| 609,202
|
You are standing at a pole and a snail is moving directly away from the pole at $1 \mathrm{~cm} / \mathrm{s}$. When the snail is 1 meter away, you start "Round 1 ". In Round $n(n \geq 1)$, you move directly toward the snail at $n+1 \mathrm{~cm} / \mathrm{s}$. When you reach the snail, you immediately turn around and move back to the starting pole at $n+1 \mathrm{~cm} / \mathrm{s}$. When you reach the pole, you immediately turn around and Round $n+1$ begins.
At the start of Round 100, how many meters away is the snail?
|
5050 Suppose the snail is $x_{n}$ meters away at the start of round $n$, so $x_{1}=1$, and the runner takes $\frac{100 x_{n}}{(n+1)-1}=\frac{100 x_{n}}{n}$ seconds to catch up to the snail. But the runner takes the same amount of time to run back to the start, so during round $n$, the snail moves a distance of $x_{n+1}-x_{n}=\frac{200 x_{n}}{n} \frac{1}{100}=\frac{2 x_{n}}{n}$.
Finally, we have $x_{100}=\frac{101}{99} x_{99}=\frac{101}{99} \frac{100}{98} x_{98}=\cdots=\frac{101!/ 2!}{99!} x_{1}=5050$.
|
5050
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
You are standing at a pole and a snail is moving directly away from the pole at $1 \mathrm{~cm} / \mathrm{s}$. When the snail is 1 meter away, you start "Round 1 ". In Round $n(n \geq 1)$, you move directly toward the snail at $n+1 \mathrm{~cm} / \mathrm{s}$. When you reach the snail, you immediately turn around and move back to the starting pole at $n+1 \mathrm{~cm} / \mathrm{s}$. When you reach the pole, you immediately turn around and Round $n+1$ begins.
At the start of Round 100, how many meters away is the snail?
|
5050 Suppose the snail is $x_{n}$ meters away at the start of round $n$, so $x_{1}=1$, and the runner takes $\frac{100 x_{n}}{(n+1)-1}=\frac{100 x_{n}}{n}$ seconds to catch up to the snail. But the runner takes the same amount of time to run back to the start, so during round $n$, the snail moves a distance of $x_{n+1}-x_{n}=\frac{200 x_{n}}{n} \frac{1}{100}=\frac{2 x_{n}}{n}$.
Finally, we have $x_{100}=\frac{101}{99} x_{99}=\frac{101}{99} \frac{100}{98} x_{98}=\cdots=\frac{101!/ 2!}{99!} x_{1}=5050$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-thm-solutions.jsonl",
"problem_match": "\n2. [4]",
"solution_match": "\nAnswer: "
}
|
5622fdf0-3123-521f-b010-5ac1ec50c7d5
| 609,203
|
Let $A B C$ be a triangle with $A B=5, B C=4$, and $C A=3$. Initially, there is an ant at each vertex. The ants start walking at a rate of 1 unit per second, in the direction $A \rightarrow B \rightarrow C \rightarrow A$ (so the ant starting at $A$ moves along ray $\overrightarrow{A B}$, etc.). For a positive real number $t$ less than 3 , let $A(t)$ be the area of the triangle whose vertices are the positions of the ants after $t$ seconds have elapsed. For what positive real number $t$ less than 3 is $A(t)$ minimized?
|
| $\frac{47}{24}$ | We instead maximize the area of the remaining triangles. This area (using $\frac{1}{2} x y \sin \theta$ ) |
| :---: | :---: | is $\frac{1}{2}(t)(5-t) \frac{3}{5}+\frac{1}{2}(t)(3-t) \frac{4}{5}+\frac{1}{2}(t)(4-t) 1=\frac{1}{10}\left(-12 t^{2}+47 t\right)$, which has a maximum at $t=\frac{47}{24} \in(0,3)$.
|
\frac{47}{24}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a triangle with $A B=5, B C=4$, and $C A=3$. Initially, there is an ant at each vertex. The ants start walking at a rate of 1 unit per second, in the direction $A \rightarrow B \rightarrow C \rightarrow A$ (so the ant starting at $A$ moves along ray $\overrightarrow{A B}$, etc.). For a positive real number $t$ less than 3 , let $A(t)$ be the area of the triangle whose vertices are the positions of the ants after $t$ seconds have elapsed. For what positive real number $t$ less than 3 is $A(t)$ minimized?
|
| $\frac{47}{24}$ | We instead maximize the area of the remaining triangles. This area (using $\frac{1}{2} x y \sin \theta$ ) |
| :---: | :---: | is $\frac{1}{2}(t)(5-t) \frac{3}{5}+\frac{1}{2}(t)(3-t) \frac{4}{5}+\frac{1}{2}(t)(4-t) 1=\frac{1}{10}\left(-12 t^{2}+47 t\right)$, which has a maximum at $t=\frac{47}{24} \in(0,3)$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-thm-solutions.jsonl",
"problem_match": "\n3. [5]",
"solution_match": "\nAnswer: "
}
|
4fe11913-5f76-5cb3-87cb-2363f860af47
| 609,204
|
There are 2 runners on the perimeter of a regular hexagon, initially located at adjacent vertices. Every second, each of the runners independently moves either one vertex to the left, with probability $\frac{1}{2}$, or one vertex to the right, also with probability $\frac{1}{2}$. Find the probability that after a 2013 second run (in which the runners switch vertices 2013 times each), the runners end up at adjacent vertices once again.
|
$\quad \frac{2}{3}+\frac{1}{3}\left(\frac{1}{4}\right)^{2013}$ OR $\frac{2^{4027}+1}{3 \cdot 2^{4026}}$ OR $\frac{2}{3}+\frac{1}{3}\left(\frac{1}{2}\right)^{4026}$ OR $\frac{2}{3}+\frac{1}{3}\left(\frac{1}{64}\right)^{671}$ Label the runners $A$ and $B$ and arbitrarily fix an orientation of the hexagon. Let $p_{t}(i)$ be the probability that $A$ is $i(\bmod 6)$ vertices to the right of $B$ at time $t$, so without loss of generality $p_{0}(1)=1$ and $p_{0}(2)=\cdots=p_{0}(6)=0$. Then for $t>0, p_{t}(i)=\frac{1}{4} p_{t-1}(i-2)+\frac{1}{2} p_{t-1}(i)+\frac{1}{4} p_{t-1}(i+2)$.
In particular, $p_{t}(2)=p_{t}(4)=p_{t}(6)=0$ for all $t$, so we may restrict our attention to $p_{t}(1), p_{t}(3), p_{t}(5)$. Thus $p_{t}(1)+p_{t}(3)+p_{t}(5)=1$ for all $t \geq 0$, and we deduce $p_{t}(i)=\frac{1}{4}+\frac{1}{4} p_{t-1}(i)$ for $i=1,3,5$.
Finally, let $f(t)=p_{t}(1)+p_{t}(5)$ denote the probability that $A, B$ are 1 vertex apart at time $t$, so $f(t)=\frac{1}{2}+\frac{1}{4} f(t-1) \Longrightarrow f(t)-\frac{2}{3}=\frac{1}{4}\left[f(t-1)-\frac{2}{3}\right]$, and we conclude that $f(2013)=\frac{2}{3}+\frac{1}{3}\left(\frac{1}{4}\right)^{2013}$.
|
\frac{2}{3}+\frac{1}{3}\left(\frac{1}{4}\right)^{2013}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
There are 2 runners on the perimeter of a regular hexagon, initially located at adjacent vertices. Every second, each of the runners independently moves either one vertex to the left, with probability $\frac{1}{2}$, or one vertex to the right, also with probability $\frac{1}{2}$. Find the probability that after a 2013 second run (in which the runners switch vertices 2013 times each), the runners end up at adjacent vertices once again.
|
$\quad \frac{2}{3}+\frac{1}{3}\left(\frac{1}{4}\right)^{2013}$ OR $\frac{2^{4027}+1}{3 \cdot 2^{4026}}$ OR $\frac{2}{3}+\frac{1}{3}\left(\frac{1}{2}\right)^{4026}$ OR $\frac{2}{3}+\frac{1}{3}\left(\frac{1}{64}\right)^{671}$ Label the runners $A$ and $B$ and arbitrarily fix an orientation of the hexagon. Let $p_{t}(i)$ be the probability that $A$ is $i(\bmod 6)$ vertices to the right of $B$ at time $t$, so without loss of generality $p_{0}(1)=1$ and $p_{0}(2)=\cdots=p_{0}(6)=0$. Then for $t>0, p_{t}(i)=\frac{1}{4} p_{t-1}(i-2)+\frac{1}{2} p_{t-1}(i)+\frac{1}{4} p_{t-1}(i+2)$.
In particular, $p_{t}(2)=p_{t}(4)=p_{t}(6)=0$ for all $t$, so we may restrict our attention to $p_{t}(1), p_{t}(3), p_{t}(5)$. Thus $p_{t}(1)+p_{t}(3)+p_{t}(5)=1$ for all $t \geq 0$, and we deduce $p_{t}(i)=\frac{1}{4}+\frac{1}{4} p_{t-1}(i)$ for $i=1,3,5$.
Finally, let $f(t)=p_{t}(1)+p_{t}(5)$ denote the probability that $A, B$ are 1 vertex apart at time $t$, so $f(t)=\frac{1}{2}+\frac{1}{4} f(t-1) \Longrightarrow f(t)-\frac{2}{3}=\frac{1}{4}\left[f(t-1)-\frac{2}{3}\right]$, and we conclude that $f(2013)=\frac{2}{3}+\frac{1}{3}\left(\frac{1}{4}\right)^{2013}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-thm-solutions.jsonl",
"problem_match": "\n4. [7]",
"solution_match": "\nAnswer: "
}
|
7161de33-c4bc-5818-8aa6-b199cda01b7b
| 609,205
|
Let $A B C$ be a triangle with $A B=13, B C=14, C A=15$. Company XYZ wants to locate their base at the point $P$ in the plane minimizing the total distance to their workers, who are located at vertices $A, B$, and $C$. There are 1,5 , and 4 workers at $A, B$, and $C$, respectively. Find the minimum possible total distance Company XYZ's workers have to travel to get to $P$.
|
$\quad 69$ We want to minimize $1 \cdot P A+5 \cdot P B+4 \cdot P C$. By the triangle inequality, $(P A+$ $P B)+4(P B+P C) \geq A B+4 B C=13+56=69$, with equality precisely when $P=[A B] \cap[B C]=B$.
|
69
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a triangle with $A B=13, B C=14, C A=15$. Company XYZ wants to locate their base at the point $P$ in the plane minimizing the total distance to their workers, who are located at vertices $A, B$, and $C$. There are 1,5 , and 4 workers at $A, B$, and $C$, respectively. Find the minimum possible total distance Company XYZ's workers have to travel to get to $P$.
|
$\quad 69$ We want to minimize $1 \cdot P A+5 \cdot P B+4 \cdot P C$. By the triangle inequality, $(P A+$ $P B)+4(P B+P C) \geq A B+4 B C=13+56=69$, with equality precisely when $P=[A B] \cap[B C]=B$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-thm-solutions.jsonl",
"problem_match": "\n5. [7]",
"solution_match": "\nAnswer: "
}
|
045f93e6-7a5e-58a1-ae55-19003af99c41
| 609,206
|
Express -2013 in base -4 .
|
$200203-2013 \equiv 3(\bmod 4)$, so the last digit is 3 ; now $\frac{-2013-3}{-4}=504 \equiv 0$, so the next digit (to the left) is 0 ; then $\frac{504-0}{-4}=-126 \equiv 2 ; \frac{-126-2}{-4}=32 \equiv 0 ; \frac{32-0}{-4}=-8 \equiv 0 ; \frac{-8-0}{-4}=2$.
Thus $-2013_{10}=200203_{-4}$.
|
200203_{-4}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Express -2013 in base -4 .
|
$200203-2013 \equiv 3(\bmod 4)$, so the last digit is 3 ; now $\frac{-2013-3}{-4}=504 \equiv 0$, so the next digit (to the left) is 0 ; then $\frac{504-0}{-4}=-126 \equiv 2 ; \frac{-126-2}{-4}=32 \equiv 0 ; \frac{32-0}{-4}=-8 \equiv 0 ; \frac{-8-0}{-4}=2$.
Thus $-2013_{10}=200203_{-4}$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-thm-solutions.jsonl",
"problem_match": "\n7. [3]",
"solution_match": "\nAnswer: "
}
|
db248666-a2ff-5460-8d16-870f36af09d6
| 609,208
|
Let $b(n)$ be the number of digits in the base -4 representation of $n$. Evaluate $\sum_{i=1}^{2013} b(i)$.
|
12345 We have the following:
- $b(n)=1$ for $n$ between 1 and 3 .
- $b(n)=3$ for $n$ between $4^{2}-3 \cdot 4=4$ and $3 \cdot 4^{2}+3=51$. (Since $a \cdot 4^{2}-b \cdot 4+c$ takes on $3 \cdot 4 \cdot 4$ distinct values over $1 \leq a \leq 3,0 \leq b \leq 3,0 \leq c \leq 3$, with minimum 4 and maximum 51.)
- $b(n)=5$ for $n$ between $4^{4}-3 \cdot 4^{3}-3 \cdot 4=52$ and $3 \cdot 4^{4}+3 \cdot 4^{2}+3=819$.
- $b(n)=7$ for $n$ between $4^{6}-3 \cdot 4^{5}-3 \cdot 4^{3}-3 \cdot 4^{1}=820$ and $3 \cdot 4^{6}+3 \cdot 4^{4}+3 \cdot 4^{2}+3>2013$.
Thus
$$
\sum_{i=1}^{2013} b(i)=7(2013)-2(819+51+3)=14091-2(873)=14091-1746=12345 .
$$
|
12345
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $b(n)$ be the number of digits in the base -4 representation of $n$. Evaluate $\sum_{i=1}^{2013} b(i)$.
|
12345 We have the following:
- $b(n)=1$ for $n$ between 1 and 3 .
- $b(n)=3$ for $n$ between $4^{2}-3 \cdot 4=4$ and $3 \cdot 4^{2}+3=51$. (Since $a \cdot 4^{2}-b \cdot 4+c$ takes on $3 \cdot 4 \cdot 4$ distinct values over $1 \leq a \leq 3,0 \leq b \leq 3,0 \leq c \leq 3$, with minimum 4 and maximum 51.)
- $b(n)=5$ for $n$ between $4^{4}-3 \cdot 4^{3}-3 \cdot 4=52$ and $3 \cdot 4^{4}+3 \cdot 4^{2}+3=819$.
- $b(n)=7$ for $n$ between $4^{6}-3 \cdot 4^{5}-3 \cdot 4^{3}-3 \cdot 4^{1}=820$ and $3 \cdot 4^{6}+3 \cdot 4^{4}+3 \cdot 4^{2}+3>2013$.
Thus
$$
\sum_{i=1}^{2013} b(i)=7(2013)-2(819+51+3)=14091-2(873)=14091-1746=12345 .
$$
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-thm-solutions.jsonl",
"problem_match": "\n8. [5]",
"solution_match": "\nAnswer: "
}
|
5bbc4168-2b41-553a-932d-5c1be300a53f
| 609,209
|
Let $N$ be the largest positive integer that can be expressed as a 2013-digit base -4 number. What is the remainder when $N$ is divided by 210 ?
|
51 The largest is $\sum_{i=0}^{1006} 3 \cdot 4^{2 i}=3 \frac{16^{1007}-1}{16-1}=\frac{16^{1007}-1}{5}$.
This is $1(\bmod 2), 0(\bmod 3), 3 \cdot 1007 \equiv 21 \equiv 1(\bmod 5)$, and $3\left(2^{1007}-1\right) \equiv 3\left(2^{8}-1\right) \equiv 3\left(2^{2}-1\right) \equiv 2$ $(\bmod 7)$, so we need $1(\bmod 10)$ and $9(\bmod 21)$, which is $9+2 \cdot 21=51(\bmod 210)$.
|
51
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $N$ be the largest positive integer that can be expressed as a 2013-digit base -4 number. What is the remainder when $N$ is divided by 210 ?
|
51 The largest is $\sum_{i=0}^{1006} 3 \cdot 4^{2 i}=3 \frac{16^{1007}-1}{16-1}=\frac{16^{1007}-1}{5}$.
This is $1(\bmod 2), 0(\bmod 3), 3 \cdot 1007 \equiv 21 \equiv 1(\bmod 5)$, and $3\left(2^{1007}-1\right) \equiv 3\left(2^{8}-1\right) \equiv 3\left(2^{2}-1\right) \equiv 2$ $(\bmod 7)$, so we need $1(\bmod 10)$ and $9(\bmod 21)$, which is $9+2 \cdot 21=51(\bmod 210)$.
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-thm-solutions.jsonl",
"problem_match": "\n9. [7]",
"solution_match": "\nAnswer: "
}
|
614083e2-77e1-5b77-b862-459aa2604ff9
| 609,210
|
Find the sum of all positive integers $n$ such that there exists an integer $b$ with $|b| \neq 4$ such that the base -4 representation of $n$ is the same as the base $b$ representation of $n$.
|
1026 All 1 digit numbers, $0,1,2,3$, are solutions when, say, $b=5$. (Of course, $d \in$ $\{0,1,2,3\}$ works for any base $b$ of absolute value greater than $d$ but not equal to 4.)
Consider now positive integers $n=\left(a_{d} \ldots a_{1} a_{0}\right)_{4}$ with more than one digit, so $d \geq 1, a_{d} \neq 0$, and $0 \leq a_{k} \leq 3$ for $k=0,1, \ldots, d$. Then $n$ has the same representation in base $b$ if and only if $|b|>\max a_{k}$ and $\sum_{k=0}^{d} a_{k}(-4)^{k}=\sum_{k=0}^{d} a_{k} b^{k}$, or equivalently, $\sum_{k=0}^{d} a_{k}\left(b^{k}-(-4)^{k}\right)=0$.
First we prove that $b \leq 3$. Indeed, if $b \geq 4$, then $b \neq 4 \Longrightarrow b \geq 5$, so $b^{k}-(-4)^{k}$ is positive for all $k \geq 1$ (and zero for $k=0$ ). But then $\sum_{k=0}^{d} a_{k}\left(b^{k}-(-4)^{k}\right) \geq a_{d}\left(b^{d}-(-4)^{d}\right)$ must be positive, and cannot vanish.
Next, we show $b \geq 2$. Assume otherwise for the sake of contradiction; $b$ cannot be $0, \pm 1$ (these bases don't make sense in general) or -4 , so we may label two distinct negative integers $-r,-s$ with $r-1 \geq s \geq 2$ such that $\{r, s\}=\{4,-b\}, s>\max a_{k}$, and $\sum_{k=0}^{d} a_{k}\left((-r)^{k}-(-s)^{k}\right)=0$, which, combined with the fact that $r^{k}-s^{k} \geq 0$ (equality only at $k=0$ ), yields
$$
\begin{aligned}
r^{d}-s^{d} \leq a_{d}\left(r^{d}-s^{d}\right) & =\sum_{k=0}^{d-1}(-1)^{d-1-k} a_{k}\left(r^{k}-s^{k}\right) \\
& \leq \sum_{k=0}^{d-1}(s-1)\left(r^{k}-s^{k}\right)=(s-1) \frac{r^{d}-1}{r-1}-\left(s^{d}-1\right)
\end{aligned}
$$
Hence $r^{d}-1 \leq(s-1) \frac{r^{d}-1}{r-1}<(r-1) \frac{r^{d}-1}{r-1}=r^{d}-1$, which is absurd.
Thus $b \geq 2$, and since $b \leq 3$ we must either have $b=2$ or $b=3$. In particular, all $a_{k}$ must be at most $b-1$. We now rewrite our condition as
$$
a_{d}\left(4^{d}-(-b)^{d}\right)=\sum_{k=0}^{d-1}(-1)^{d-1-k} a_{k}\left(4^{k}-(-b)^{k}\right)
$$
Since $4^{k}-(-b)^{k} \geq 0$ for $k \geq 0$, with equality only at $k=0$, we deduce
$$
a_{d}\left(4^{d}-(-b)^{d}\right) \leq \sum_{k \equiv d-1}(b-1)\left(4^{k}-(-b)^{k}\right)
$$
If $d-1$ is even ( $d$ is odd), this gives
$$
a_{d}\left(4^{d}+b^{d}\right) \leq(b-1) \frac{4^{d+1}-4^{0}}{4^{2}-1}-(b-1) \frac{b^{d+1}-b^{0}}{b^{2}-1}
$$
so $4^{d}<(b-1) \frac{4^{d+1}}{15} \Longrightarrow b>1+\frac{15}{4}$, which is impossible.
Thus $d-1$ is odd ( $d$ is even), and we get
$$
a_{d}\left(4^{d}-b^{d}\right) \leq(b-1) \frac{4^{d+1}-4^{1}}{4^{2}-1}+(b-1) \frac{b^{d+1}-b^{1}}{b^{2}-1} \Longleftrightarrow \frac{b^{d}-1}{4^{d}-1} \geq \frac{a_{d}-\frac{4}{15}(b-1)}{a_{d}+\frac{b}{b+1}}
$$
If $b=2$, then $a_{d}=1$, so $\frac{1}{2^{d}+1}=\frac{2^{d}-1}{4^{d}-1} \geq \frac{11}{25}$, which is clearly impossible $(d \geq 2)$.
If $b=3$ and $a_{d}=2$, then $\frac{9^{d / 2}-1}{16^{d / 2}-1} \leq \frac{8}{15}$. Since $d$ is even, it's easy to check this holds only for $d / 2=1$, with equality, so $a_{k}=b-1$ if $k \equiv d-1(\bmod 2)$. Thus $\left(a_{d}, \ldots, a_{0}\right)=\left(2,2, a_{0}\right)$, yielding solutions $(22 x)_{3}$ (which do work; note that the last digit doesn't matter).
Otherwise, if $b=3$ and $a_{d}=14$, then $\frac{9^{d / 2}-1}{16^{d / 2}-1} \leq \frac{4}{15}$. It's easy to check $d / 2 \in\{1,2\}$.
If $d / 2=1$, we're solving $16 a_{2}-4 a_{1}+a_{0}=9 a_{2}+3 a_{1}+a_{0} \Longleftrightarrow a_{2}=a_{1}$. We thus obtain the working solution $(11 x)_{3}$. (Note that $110=\frac{1}{2} 220$ in bases $-4,3$.)
If $d / 2=2$, we want $256 a_{4}-64 a_{3}+16 a_{2}-4 a_{1}+a_{0}=81 a_{4}+27 a_{3}+9 a_{2}+3 a_{1}+a_{0}$, or $175=91 a_{3}-7 a_{2}+7 a_{1}$, which simplifies to $25=13 a_{3}-a_{2}+a_{1}$. This gives the working solutions $(1210 x)_{3},(1221 x)_{3}$. (Note that $12100=110^{2}$ and $12210=110^{2}+110$ in bases $-4,3$.)
The list of all nontrivial ( $\geq 2$-digit) solutions (in base -4 and $b$ ) is then $11 x, 22 x, 1210 x, 1221 x$, where $b=3$ and $x \in\{0,1,2\}$. In base 10 , they are $12+x, 2 \cdot 12+x, 12^{2}+x, 12^{2}+12+x$, with sum $3\left(2 \cdot 12^{2}+4 \cdot 12\right)+4(0+1+2)=1020$.
Finally, we need to include the trivial solutions $n=1,2,3$, for a total sum of 1026 .
|
1026
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the sum of all positive integers $n$ such that there exists an integer $b$ with $|b| \neq 4$ such that the base -4 representation of $n$ is the same as the base $b$ representation of $n$.
|
1026 All 1 digit numbers, $0,1,2,3$, are solutions when, say, $b=5$. (Of course, $d \in$ $\{0,1,2,3\}$ works for any base $b$ of absolute value greater than $d$ but not equal to 4.)
Consider now positive integers $n=\left(a_{d} \ldots a_{1} a_{0}\right)_{4}$ with more than one digit, so $d \geq 1, a_{d} \neq 0$, and $0 \leq a_{k} \leq 3$ for $k=0,1, \ldots, d$. Then $n$ has the same representation in base $b$ if and only if $|b|>\max a_{k}$ and $\sum_{k=0}^{d} a_{k}(-4)^{k}=\sum_{k=0}^{d} a_{k} b^{k}$, or equivalently, $\sum_{k=0}^{d} a_{k}\left(b^{k}-(-4)^{k}\right)=0$.
First we prove that $b \leq 3$. Indeed, if $b \geq 4$, then $b \neq 4 \Longrightarrow b \geq 5$, so $b^{k}-(-4)^{k}$ is positive for all $k \geq 1$ (and zero for $k=0$ ). But then $\sum_{k=0}^{d} a_{k}\left(b^{k}-(-4)^{k}\right) \geq a_{d}\left(b^{d}-(-4)^{d}\right)$ must be positive, and cannot vanish.
Next, we show $b \geq 2$. Assume otherwise for the sake of contradiction; $b$ cannot be $0, \pm 1$ (these bases don't make sense in general) or -4 , so we may label two distinct negative integers $-r,-s$ with $r-1 \geq s \geq 2$ such that $\{r, s\}=\{4,-b\}, s>\max a_{k}$, and $\sum_{k=0}^{d} a_{k}\left((-r)^{k}-(-s)^{k}\right)=0$, which, combined with the fact that $r^{k}-s^{k} \geq 0$ (equality only at $k=0$ ), yields
$$
\begin{aligned}
r^{d}-s^{d} \leq a_{d}\left(r^{d}-s^{d}\right) & =\sum_{k=0}^{d-1}(-1)^{d-1-k} a_{k}\left(r^{k}-s^{k}\right) \\
& \leq \sum_{k=0}^{d-1}(s-1)\left(r^{k}-s^{k}\right)=(s-1) \frac{r^{d}-1}{r-1}-\left(s^{d}-1\right)
\end{aligned}
$$
Hence $r^{d}-1 \leq(s-1) \frac{r^{d}-1}{r-1}<(r-1) \frac{r^{d}-1}{r-1}=r^{d}-1$, which is absurd.
Thus $b \geq 2$, and since $b \leq 3$ we must either have $b=2$ or $b=3$. In particular, all $a_{k}$ must be at most $b-1$. We now rewrite our condition as
$$
a_{d}\left(4^{d}-(-b)^{d}\right)=\sum_{k=0}^{d-1}(-1)^{d-1-k} a_{k}\left(4^{k}-(-b)^{k}\right)
$$
Since $4^{k}-(-b)^{k} \geq 0$ for $k \geq 0$, with equality only at $k=0$, we deduce
$$
a_{d}\left(4^{d}-(-b)^{d}\right) \leq \sum_{k \equiv d-1}(b-1)\left(4^{k}-(-b)^{k}\right)
$$
If $d-1$ is even ( $d$ is odd), this gives
$$
a_{d}\left(4^{d}+b^{d}\right) \leq(b-1) \frac{4^{d+1}-4^{0}}{4^{2}-1}-(b-1) \frac{b^{d+1}-b^{0}}{b^{2}-1}
$$
so $4^{d}<(b-1) \frac{4^{d+1}}{15} \Longrightarrow b>1+\frac{15}{4}$, which is impossible.
Thus $d-1$ is odd ( $d$ is even), and we get
$$
a_{d}\left(4^{d}-b^{d}\right) \leq(b-1) \frac{4^{d+1}-4^{1}}{4^{2}-1}+(b-1) \frac{b^{d+1}-b^{1}}{b^{2}-1} \Longleftrightarrow \frac{b^{d}-1}{4^{d}-1} \geq \frac{a_{d}-\frac{4}{15}(b-1)}{a_{d}+\frac{b}{b+1}}
$$
If $b=2$, then $a_{d}=1$, so $\frac{1}{2^{d}+1}=\frac{2^{d}-1}{4^{d}-1} \geq \frac{11}{25}$, which is clearly impossible $(d \geq 2)$.
If $b=3$ and $a_{d}=2$, then $\frac{9^{d / 2}-1}{16^{d / 2}-1} \leq \frac{8}{15}$. Since $d$ is even, it's easy to check this holds only for $d / 2=1$, with equality, so $a_{k}=b-1$ if $k \equiv d-1(\bmod 2)$. Thus $\left(a_{d}, \ldots, a_{0}\right)=\left(2,2, a_{0}\right)$, yielding solutions $(22 x)_{3}$ (which do work; note that the last digit doesn't matter).
Otherwise, if $b=3$ and $a_{d}=14$, then $\frac{9^{d / 2}-1}{16^{d / 2}-1} \leq \frac{4}{15}$. It's easy to check $d / 2 \in\{1,2\}$.
If $d / 2=1$, we're solving $16 a_{2}-4 a_{1}+a_{0}=9 a_{2}+3 a_{1}+a_{0} \Longleftrightarrow a_{2}=a_{1}$. We thus obtain the working solution $(11 x)_{3}$. (Note that $110=\frac{1}{2} 220$ in bases $-4,3$.)
If $d / 2=2$, we want $256 a_{4}-64 a_{3}+16 a_{2}-4 a_{1}+a_{0}=81 a_{4}+27 a_{3}+9 a_{2}+3 a_{1}+a_{0}$, or $175=91 a_{3}-7 a_{2}+7 a_{1}$, which simplifies to $25=13 a_{3}-a_{2}+a_{1}$. This gives the working solutions $(1210 x)_{3},(1221 x)_{3}$. (Note that $12100=110^{2}$ and $12210=110^{2}+110$ in bases $-4,3$.)
The list of all nontrivial ( $\geq 2$-digit) solutions (in base -4 and $b$ ) is then $11 x, 22 x, 1210 x, 1221 x$, where $b=3$ and $x \in\{0,1,2\}$. In base 10 , they are $12+x, 2 \cdot 12+x, 12^{2}+x, 12^{2}+12+x$, with sum $3\left(2 \cdot 12^{2}+4 \cdot 12\right)+4(0+1+2)=1020$.
Finally, we need to include the trivial solutions $n=1,2,3$, for a total sum of 1026 .
|
{
"resource_path": "HarvardMIT/segmented/en-171-2013-nov-thm-solutions.jsonl",
"problem_match": "\n10. [8]",
"solution_match": "\nAnswer: "
}
|
2c03ef60-16c1-59f2-b035-a870f62ddfb5
| 609,211
|
Find the integer closest to
$$
\frac{1}{\sqrt[4]{5^{4}+1}-\sqrt[4]{5^{4}-1}}
$$
|
250 Let $x=\left(5^{4}+1\right)^{1 / 4}$ and $y=\left(5^{4}-1\right)^{1 / 4}$. Note that $x$ and $y$ are both approximately 5. We have
$$
\begin{aligned}
\frac{1}{x-y} & =\frac{(x+y)\left(x^{2}+y^{2}\right)}{(x-y)(x+y)\left(x^{2}+y^{2}\right)}=\frac{(x+y)\left(x^{2}+y^{2}\right)}{x^{4}-y^{4}} \\
& =\frac{(x+y)\left(x^{2}+y^{2}\right)}{2} \approx \frac{(5+5)\left(5^{2}+5^{2}\right)}{2}=250
\end{aligned}
$$
Note: To justify the $\approx$, note that $1=x^{4}-5^{4}$ implies
$$
0<x-5=\frac{1}{(x+5)\left(x^{2}+5^{2}\right)}<\frac{1}{(5+5)\left(5^{2}+5^{2}\right)}=\frac{1}{500}
$$
and similarly $1=5^{4}-y^{4}$ implies
$$
0<5-y=\frac{1}{(5+y)\left(5^{2}+y^{2}\right)}<\frac{1}{(4+4)\left(4^{2}+4^{2}\right)}=\frac{1}{256}
$$
Similarly,
$$
0<x^{2}-5^{2}=\frac{1}{x^{2}+5^{2}}<\frac{1}{2 \cdot 5^{2}}=\frac{1}{50}
$$
and
$$
0<5^{2}-y^{2}=\frac{1}{5^{2}+y^{2}}<\frac{1}{5^{2}+4.5^{2}}<\frac{1}{45}
$$
Now
$$
|x+y-10|=|(x-5)-(5-y)|<\max (|x-5|,|5-y|)<\frac{1}{256}
$$
and similarly $\left|x^{2}+y^{2}-2 \cdot 5^{2}\right|<\frac{1}{45}$. It's easy to check that $(10-1 / 256)(50-1 / 45)>499.5$ and $(10+1 / 256)(50+1 / 45)<500.5$, so we're done.
|
250
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find the integer closest to
$$
\frac{1}{\sqrt[4]{5^{4}+1}-\sqrt[4]{5^{4}-1}}
$$
|
250 Let $x=\left(5^{4}+1\right)^{1 / 4}$ and $y=\left(5^{4}-1\right)^{1 / 4}$. Note that $x$ and $y$ are both approximately 5. We have
$$
\begin{aligned}
\frac{1}{x-y} & =\frac{(x+y)\left(x^{2}+y^{2}\right)}{(x-y)(x+y)\left(x^{2}+y^{2}\right)}=\frac{(x+y)\left(x^{2}+y^{2}\right)}{x^{4}-y^{4}} \\
& =\frac{(x+y)\left(x^{2}+y^{2}\right)}{2} \approx \frac{(5+5)\left(5^{2}+5^{2}\right)}{2}=250
\end{aligned}
$$
Note: To justify the $\approx$, note that $1=x^{4}-5^{4}$ implies
$$
0<x-5=\frac{1}{(x+5)\left(x^{2}+5^{2}\right)}<\frac{1}{(5+5)\left(5^{2}+5^{2}\right)}=\frac{1}{500}
$$
and similarly $1=5^{4}-y^{4}$ implies
$$
0<5-y=\frac{1}{(5+y)\left(5^{2}+y^{2}\right)}<\frac{1}{(4+4)\left(4^{2}+4^{2}\right)}=\frac{1}{256}
$$
Similarly,
$$
0<x^{2}-5^{2}=\frac{1}{x^{2}+5^{2}}<\frac{1}{2 \cdot 5^{2}}=\frac{1}{50}
$$
and
$$
0<5^{2}-y^{2}=\frac{1}{5^{2}+y^{2}}<\frac{1}{5^{2}+4.5^{2}}<\frac{1}{45}
$$
Now
$$
|x+y-10|=|(x-5)-(5-y)|<\max (|x-5|,|5-y|)<\frac{1}{256}
$$
and similarly $\left|x^{2}+y^{2}-2 \cdot 5^{2}\right|<\frac{1}{45}$. It's easy to check that $(10-1 / 256)(50-1 / 45)>499.5$ and $(10+1 / 256)(50+1 / 45)<500.5$, so we're done.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-alg-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nAnswer: "
}
|
eaddb876-cc5d-5b3b-b02a-43adc7f5cf7c
| 609,212
|
Let
$$
A=\frac{1}{6}\left(\left(\log _{2}(3)\right)^{3}-\left(\log _{2}(6)\right)^{3}-\left(\log _{2}(12)\right)^{3}+\left(\log _{2}(24)\right)^{3}\right)
$$
Compute $2^{A}$.
|
72 Let $a=\log _{2}(3)$, so $2^{a}=3$ and $A=\frac{1}{6}\left[a^{3}-(a+1)^{3}-(a+2)^{3}+(a+3)^{3}\right]$. But $(x+1)^{3}-x^{3}=3 x^{2}+3 x+1$, so $A=\frac{1}{6}\left[3(a+2)^{2}+3(a+2)-3 a^{2}-3 a\right]=\frac{1}{2}[4 a+4+2]=2 a+3$. Thus $2^{A}=\left(2^{a}\right)^{2}\left(2^{3}\right)=9 \cdot 8=72$.
|
72
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let
$$
A=\frac{1}{6}\left(\left(\log _{2}(3)\right)^{3}-\left(\log _{2}(6)\right)^{3}-\left(\log _{2}(12)\right)^{3}+\left(\log _{2}(24)\right)^{3}\right)
$$
Compute $2^{A}$.
|
72 Let $a=\log _{2}(3)$, so $2^{a}=3$ and $A=\frac{1}{6}\left[a^{3}-(a+1)^{3}-(a+2)^{3}+(a+3)^{3}\right]$. But $(x+1)^{3}-x^{3}=3 x^{2}+3 x+1$, so $A=\frac{1}{6}\left[3(a+2)^{2}+3(a+2)-3 a^{2}-3 a\right]=\frac{1}{2}[4 a+4+2]=2 a+3$. Thus $2^{A}=\left(2^{a}\right)^{2}\left(2^{3}\right)=9 \cdot 8=72$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-alg-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nAnswer: "
}
|
cb14e64d-83f3-5153-9a01-91fb1a5d671e
| 609,213
|
Let $b$ and $c$ be real numbers, and define the polynomial $P(x)=x^{2}+b x+c$. Suppose that $P(P(1))=$ $P(P(2))=0$, and that $P(1) \neq P(2)$. Find $P(0)$.
|
$-\frac{3}{2}$ OR -1.5 OR $-1 \frac{1}{2}$ Since $P(P(1))=P(P(2))=0$, but $P(1) \neq P(2)$, it follows that $P(1)=1+b+c$ and $P(2)=4+2 b+c$ are the distinct roots of the polynomial $P(x)$. Thus, $P(x)$ factors:
$$
\begin{aligned}
P(x) & =x^{2}+b x+c \\
& =(x-(1+b+c))(x-(4+2 b+c)) \\
& =x^{2}-(5+3 b+2 c) x+(1+b+c)(4+2 b+c)
\end{aligned}
$$
It follows that $-(5+3 b+2 c)=b$, and that $c=(1+b+c)(4+2 b+c)$. From the first equation, we find $2 b+c=-5 / 2$. Plugging in $c=-5 / 2-2 b$ into the second equation yields
$$
-5 / 2-2 b=(1+(-5 / 2)-b)(4+(-5 / 2))
$$
Solving this equation yields $b=-\frac{1}{2}$, so $c=-5 / 2-2 b=-\frac{3}{2}$.
|
-\frac{3}{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $b$ and $c$ be real numbers, and define the polynomial $P(x)=x^{2}+b x+c$. Suppose that $P(P(1))=$ $P(P(2))=0$, and that $P(1) \neq P(2)$. Find $P(0)$.
|
$-\frac{3}{2}$ OR -1.5 OR $-1 \frac{1}{2}$ Since $P(P(1))=P(P(2))=0$, but $P(1) \neq P(2)$, it follows that $P(1)=1+b+c$ and $P(2)=4+2 b+c$ are the distinct roots of the polynomial $P(x)$. Thus, $P(x)$ factors:
$$
\begin{aligned}
P(x) & =x^{2}+b x+c \\
& =(x-(1+b+c))(x-(4+2 b+c)) \\
& =x^{2}-(5+3 b+2 c) x+(1+b+c)(4+2 b+c)
\end{aligned}
$$
It follows that $-(5+3 b+2 c)=b$, and that $c=(1+b+c)(4+2 b+c)$. From the first equation, we find $2 b+c=-5 / 2$. Plugging in $c=-5 / 2-2 b$ into the second equation yields
$$
-5 / 2-2 b=(1+(-5 / 2)-b)(4+(-5 / 2))
$$
Solving this equation yields $b=-\frac{1}{2}$, so $c=-5 / 2-2 b=-\frac{3}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-alg-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nAnswer: "
}
|
26976e03-0c87-54ea-8d59-53ff96b2bc0d
| 609,214
|
Find the sum of all real numbers $x$ such that $5 x^{4}-10 x^{3}+10 x^{2}-5 x-11=0$.
|
1 Rearrange the equation to $x^{5}+(1-x)^{5}-12=0$. It's easy to see this has two real roots, and that $r$ is a root if and only if $1-r$ is a root, so the answer must be 1 .
Alternate solution: Note that $5 x^{4}-10 x^{3}+10 x^{2}-5 x-11=5 x(x-1)\left(x^{2}-x+1\right)-11=5 u(u+1)-11$, where $u=x^{2}-x$. Of course, $x^{2}-x=u$ has real roots if and only if $u \geq-\frac{1}{4}$, and distinct real roots if and only if $u>-\frac{1}{4}$. But the roots of $5 u(u+1)-11$ are $\frac{-5 \pm \sqrt{5^{2}+4(5)(11)}}{2 \cdot 5}=\frac{-5 \pm 7 \sqrt{5}}{10}$, one of which is greater than $-\frac{1}{4}$ and the other less than $-\frac{1}{4}$. For the larger root, $x^{2}-x=u$ has exactly two distinct real roots, which sum up to 1 by Vieta's.
|
1
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find the sum of all real numbers $x$ such that $5 x^{4}-10 x^{3}+10 x^{2}-5 x-11=0$.
|
1 Rearrange the equation to $x^{5}+(1-x)^{5}-12=0$. It's easy to see this has two real roots, and that $r$ is a root if and only if $1-r$ is a root, so the answer must be 1 .
Alternate solution: Note that $5 x^{4}-10 x^{3}+10 x^{2}-5 x-11=5 x(x-1)\left(x^{2}-x+1\right)-11=5 u(u+1)-11$, where $u=x^{2}-x$. Of course, $x^{2}-x=u$ has real roots if and only if $u \geq-\frac{1}{4}$, and distinct real roots if and only if $u>-\frac{1}{4}$. But the roots of $5 u(u+1)-11$ are $\frac{-5 \pm \sqrt{5^{2}+4(5)(11)}}{2 \cdot 5}=\frac{-5 \pm 7 \sqrt{5}}{10}$, one of which is greater than $-\frac{1}{4}$ and the other less than $-\frac{1}{4}$. For the larger root, $x^{2}-x=u$ has exactly two distinct real roots, which sum up to 1 by Vieta's.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-alg-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nAnswer: "
}
|
5886f27c-b51c-566a-bd59-c283921c9f25
| 609,215
|
Given that $w$ and $z$ are complex numbers such that $|w+z|=1$ and $\left|w^{2}+z^{2}\right|=14$, find the smallest possible value of $\left|w^{3}+z^{3}\right|$. Here, $|\cdot|$ denotes the absolute value of a complex number, given by $|a+b i|=\sqrt{a^{2}+b^{2}}$ whenever $a$ and $b$ are real numbers.
|
$\frac{41}{2}$ OR 20.5 OR $20 \frac{1}{2}$ We can rewrite $\left|w^{3}+z^{3}\right|=|w+z|\left|w^{2}-w z+z^{2}\right|=\left|w^{2}-w z+z^{2}\right|=$ $\left|\frac{3}{2}\left(w^{2}+z^{2}\right)-\frac{1}{2}(w+z)^{2}\right|$.
By the triangle inequality, $\left|\frac{3}{2}\left(w^{2}+z^{2}\right)-\frac{1}{2}(w+z)^{2}+\frac{1}{2}(w+z)^{2}\right| \leq\left|\frac{3}{2}\left(w^{2}+z^{2}\right)-\frac{1}{2}(w+z)^{2}\right|+\left|\frac{1}{2}(w+z)^{2}\right|$. By rearranging and simplifying, we get $\left|w^{3}+z^{3}\right|=\left|\frac{3}{2}\left(w^{2}+z^{2}\right)-\frac{1}{2}(w+z)^{2}\right| \geq \frac{3}{2}\left|w^{2}+z^{2}\right|-\frac{1}{2}|w+z|^{2}=$ $\frac{3}{2}(14)-\frac{1}{2}=\frac{41}{2}$.
To achieve $41 / 2$, it suffices to take $w, z$ satisfying $w+z=1$ and $w^{2}+z^{2}=14$.
|
\frac{41}{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Given that $w$ and $z$ are complex numbers such that $|w+z|=1$ and $\left|w^{2}+z^{2}\right|=14$, find the smallest possible value of $\left|w^{3}+z^{3}\right|$. Here, $|\cdot|$ denotes the absolute value of a complex number, given by $|a+b i|=\sqrt{a^{2}+b^{2}}$ whenever $a$ and $b$ are real numbers.
|
$\frac{41}{2}$ OR 20.5 OR $20 \frac{1}{2}$ We can rewrite $\left|w^{3}+z^{3}\right|=|w+z|\left|w^{2}-w z+z^{2}\right|=\left|w^{2}-w z+z^{2}\right|=$ $\left|\frac{3}{2}\left(w^{2}+z^{2}\right)-\frac{1}{2}(w+z)^{2}\right|$.
By the triangle inequality, $\left|\frac{3}{2}\left(w^{2}+z^{2}\right)-\frac{1}{2}(w+z)^{2}+\frac{1}{2}(w+z)^{2}\right| \leq\left|\frac{3}{2}\left(w^{2}+z^{2}\right)-\frac{1}{2}(w+z)^{2}\right|+\left|\frac{1}{2}(w+z)^{2}\right|$. By rearranging and simplifying, we get $\left|w^{3}+z^{3}\right|=\left|\frac{3}{2}\left(w^{2}+z^{2}\right)-\frac{1}{2}(w+z)^{2}\right| \geq \frac{3}{2}\left|w^{2}+z^{2}\right|-\frac{1}{2}|w+z|^{2}=$ $\frac{3}{2}(14)-\frac{1}{2}=\frac{41}{2}$.
To achieve $41 / 2$, it suffices to take $w, z$ satisfying $w+z=1$ and $w^{2}+z^{2}=14$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-alg-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nAnswer: "
}
|
02ebf693-3300-56be-bb03-c2698fed047b
| 609,216
|
Find the largest real number $c$ such that
$$
\sum_{i=1}^{101} x_{i}^{2} \geq c M^{2}
$$
whenever $x_{1}, \ldots, x_{101}$ are real numbers such that $x_{1}+\cdots+x_{101}=0$ and $M$ is the median of $x_{1}, \ldots, x_{101}$.
|
$\frac{5151}{50}$ OR 103.02 OR $103 \frac{1}{50}$ Suppose without loss of generality that $x_{1} \leq \cdots \leq x_{101}$ and $M=x_{51} \geq 0$.
Note that $f(t)=t^{2}$ is a convex function over the reals, so we may "smooth" to the case $x_{1}=\cdots=$ $x_{5} 0 \leq x_{51}=\cdots=x_{101}$ (the $x_{51}=\cdots$ is why we needed to assume $x_{51} \geq 0$ ). Indeed, by Jensen's inequality, the map $x_{1}, x_{2}, \ldots, x_{50} \rightarrow \frac{x_{1}+\cdots+x_{50}}{50}, \ldots, \frac{x_{1}+\cdots+x_{50}}{50}$ will decrease or fix the LHS, while preserving the ordering condition and the zero-sum condition.
Similarly, we may without loss of generality replace $x_{51}, \ldots, x_{101}$ with their average (which will decrease or fix the LHS, but also either fix or increase the RHS). But this simplified problem has $x_{1}=\cdots=$
$x_{50}=-51 r$ and $x_{51}=\cdots=x_{101}=50 r$ for some $r \geq 0$, and by homogeneity, $C$ works if and only if $C \leq \frac{50(51)^{2}+51(50)^{2}}{50^{2}}=\frac{51(101)}{50}=\frac{5151}{50}$.
Comment: One may also use the Cauchy-Schwarz inequality or the QM-AM inequality instead of Jensen's inequality.
Comment: For this particular problem, there is another solution using the identity $101 \sum x_{i}^{2}-\left(\sum x_{i}\right)^{2}=$ $\sum\left(x_{j}-x_{i}\right)^{2}$. Indeed, we may set $u=x_{51}-\left(x_{1}+\cdots+x_{50}\right) / 50$ and $v=\left(x_{52}+\cdots+x_{101}\right) / 50-x_{51}$, and use the fact that $(u-v)^{2} \leq u^{2}+v^{2}$.
|
\frac{5151}{50}
|
Yes
|
Yes
|
math-word-problem
|
Inequalities
|
Find the largest real number $c$ such that
$$
\sum_{i=1}^{101} x_{i}^{2} \geq c M^{2}
$$
whenever $x_{1}, \ldots, x_{101}$ are real numbers such that $x_{1}+\cdots+x_{101}=0$ and $M$ is the median of $x_{1}, \ldots, x_{101}$.
|
$\frac{5151}{50}$ OR 103.02 OR $103 \frac{1}{50}$ Suppose without loss of generality that $x_{1} \leq \cdots \leq x_{101}$ and $M=x_{51} \geq 0$.
Note that $f(t)=t^{2}$ is a convex function over the reals, so we may "smooth" to the case $x_{1}=\cdots=$ $x_{5} 0 \leq x_{51}=\cdots=x_{101}$ (the $x_{51}=\cdots$ is why we needed to assume $x_{51} \geq 0$ ). Indeed, by Jensen's inequality, the map $x_{1}, x_{2}, \ldots, x_{50} \rightarrow \frac{x_{1}+\cdots+x_{50}}{50}, \ldots, \frac{x_{1}+\cdots+x_{50}}{50}$ will decrease or fix the LHS, while preserving the ordering condition and the zero-sum condition.
Similarly, we may without loss of generality replace $x_{51}, \ldots, x_{101}$ with their average (which will decrease or fix the LHS, but also either fix or increase the RHS). But this simplified problem has $x_{1}=\cdots=$
$x_{50}=-51 r$ and $x_{51}=\cdots=x_{101}=50 r$ for some $r \geq 0$, and by homogeneity, $C$ works if and only if $C \leq \frac{50(51)^{2}+51(50)^{2}}{50^{2}}=\frac{51(101)}{50}=\frac{5151}{50}$.
Comment: One may also use the Cauchy-Schwarz inequality or the QM-AM inequality instead of Jensen's inequality.
Comment: For this particular problem, there is another solution using the identity $101 \sum x_{i}^{2}-\left(\sum x_{i}\right)^{2}=$ $\sum\left(x_{j}-x_{i}\right)^{2}$. Indeed, we may set $u=x_{51}-\left(x_{1}+\cdots+x_{50}\right) / 50$ and $v=\left(x_{52}+\cdots+x_{101}\right) / 50-x_{51}$, and use the fact that $(u-v)^{2} \leq u^{2}+v^{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-alg-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nAnswer: "
}
|
a878091a-6518-52c0-909e-57f1959862db
| 609,217
|
For an integer $n$, let $f_{9}(n)$ denote the number of positive integers $d \leq 9$ dividing $n$. Suppose that $m$ is a positive integer and $b_{1}, b_{2}, \ldots, b_{m}$ are real numbers such that $f_{9}(n)=\sum_{j=1}^{m} b_{j} f_{9}(n-j)$ for all $n>m$. Find the smallest possible value of $m$.
|
28 Let $M=9$. Consider the generating function
$$
F(x)=\sum_{n \geq 1} f_{M}(n) x^{n}=\sum_{d=1}^{M} \sum_{k \geq 1} x^{d k}=\sum_{d=1}^{M} \frac{x^{d}}{1-x^{d}}
$$
Observe that $f_{M}(n)=f_{M}(n+M!)$ for all $n \geq 1$ (in fact, all $n \leq 0$ as well). Thus $f_{M}(n)$ satisfies a degree $m$ linear recurrence if and only if it eventually satisfies a degree $m$ linear recurrence. But the latter occurs if and only if $P(x) F(x)$ is a polynomial for some degree $m$ polynomial $P(x)$. (Why?)
Suppose $P(x) F(x)=Q(x)$ is a polynomial for some polynomial $P$ of degree $m$. We show that $x^{s}-1 \mid$ $P(x)$ for $s=1,2, \ldots, M$, or equivalently that $P(\omega)=0$ for all primitive $s$ th roots of unity $1 \leq s \leq M)$. Fix a primitive $s$ th root of unity $\omega$, and define a function
$$
F_{\omega}(z)=\left(1-\omega^{-1} z\right) \sum_{s \nmid d \leq M} \frac{z^{d}}{1-z^{d}}+\sum_{s \mid d \leq M} \frac{z^{d}}{1+\left(\omega^{-1} z\right)+\cdots+\left(\omega^{-1} z\right)^{d-1}}
$$
for all $z$ where all denominators are nonzero (in particular, this includes $z=\omega$ ).
Yet $F_{\omega}(z)-F(z)\left(1-\omega^{-1} z\right)=0$ for all complex $z$ such that $z^{1}, z^{2}, \ldots, z^{M} \neq 1$, so $P(z) F_{\omega}(z)-Q(z)(1-$ $\left.\omega^{-1} z\right)=0$ holds for all such $z$ as well. In particular, the rational function $P(x) F_{\omega}(x)-Q(x)\left(1-\omega^{-1} x\right)$
has infinitely many roots, so must be identically zero once we clear denominators. But no denominator vanishes at $x=\omega$, so we may plug in $x=\omega$ to the polynomial identity and then divide out by the original (nonzero) denominators to get $0=P(\omega) F_{\omega}(\omega)-Q(\omega)\left(1-\omega^{-1} \omega\right)=P(\omega) F_{\omega}(\omega)$. However,
$$
F_{\omega}(\omega)=\sum_{s \mid d \leq M} \frac{\omega^{d}}{1+\left(\omega^{-1} \omega\right)+\cdots+\left(\omega^{-1} \omega\right)^{d-1}}=\sum_{s \mid d \leq M} \frac{1}{d}
$$
is a positive integer multiple of $1 / d$, and therefore nonzero. Thus $P(\omega)=0$, as desired.
Conversely, if $x^{s}-1 \mid P(x)$ for $s=1,2, \ldots, M$, then $P(x)$ will clearly suffice. So we just want the degree of the least common multiple of the $x^{s}-1$ for $s=1,2, \ldots, M$, or just the number of roots of unity of order at most $M$, which is $\sum_{s=1}^{M} \phi(s)=1+1+2+2+4+2+6+4+6=28$.
Comment: Only at the beginning do we treat $F(x)$ strictly as a formal power series; later once we get the rational function representation $\sum_{d=1}^{6} \frac{x^{d}}{1-x^{d}}$, we can work with polynomial identities in general and don't have to worry about convergence issues for $|x| \geq 1$.
|
28
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
For an integer $n$, let $f_{9}(n)$ denote the number of positive integers $d \leq 9$ dividing $n$. Suppose that $m$ is a positive integer and $b_{1}, b_{2}, \ldots, b_{m}$ are real numbers such that $f_{9}(n)=\sum_{j=1}^{m} b_{j} f_{9}(n-j)$ for all $n>m$. Find the smallest possible value of $m$.
|
28 Let $M=9$. Consider the generating function
$$
F(x)=\sum_{n \geq 1} f_{M}(n) x^{n}=\sum_{d=1}^{M} \sum_{k \geq 1} x^{d k}=\sum_{d=1}^{M} \frac{x^{d}}{1-x^{d}}
$$
Observe that $f_{M}(n)=f_{M}(n+M!)$ for all $n \geq 1$ (in fact, all $n \leq 0$ as well). Thus $f_{M}(n)$ satisfies a degree $m$ linear recurrence if and only if it eventually satisfies a degree $m$ linear recurrence. But the latter occurs if and only if $P(x) F(x)$ is a polynomial for some degree $m$ polynomial $P(x)$. (Why?)
Suppose $P(x) F(x)=Q(x)$ is a polynomial for some polynomial $P$ of degree $m$. We show that $x^{s}-1 \mid$ $P(x)$ for $s=1,2, \ldots, M$, or equivalently that $P(\omega)=0$ for all primitive $s$ th roots of unity $1 \leq s \leq M)$. Fix a primitive $s$ th root of unity $\omega$, and define a function
$$
F_{\omega}(z)=\left(1-\omega^{-1} z\right) \sum_{s \nmid d \leq M} \frac{z^{d}}{1-z^{d}}+\sum_{s \mid d \leq M} \frac{z^{d}}{1+\left(\omega^{-1} z\right)+\cdots+\left(\omega^{-1} z\right)^{d-1}}
$$
for all $z$ where all denominators are nonzero (in particular, this includes $z=\omega$ ).
Yet $F_{\omega}(z)-F(z)\left(1-\omega^{-1} z\right)=0$ for all complex $z$ such that $z^{1}, z^{2}, \ldots, z^{M} \neq 1$, so $P(z) F_{\omega}(z)-Q(z)(1-$ $\left.\omega^{-1} z\right)=0$ holds for all such $z$ as well. In particular, the rational function $P(x) F_{\omega}(x)-Q(x)\left(1-\omega^{-1} x\right)$
has infinitely many roots, so must be identically zero once we clear denominators. But no denominator vanishes at $x=\omega$, so we may plug in $x=\omega$ to the polynomial identity and then divide out by the original (nonzero) denominators to get $0=P(\omega) F_{\omega}(\omega)-Q(\omega)\left(1-\omega^{-1} \omega\right)=P(\omega) F_{\omega}(\omega)$. However,
$$
F_{\omega}(\omega)=\sum_{s \mid d \leq M} \frac{\omega^{d}}{1+\left(\omega^{-1} \omega\right)+\cdots+\left(\omega^{-1} \omega\right)^{d-1}}=\sum_{s \mid d \leq M} \frac{1}{d}
$$
is a positive integer multiple of $1 / d$, and therefore nonzero. Thus $P(\omega)=0$, as desired.
Conversely, if $x^{s}-1 \mid P(x)$ for $s=1,2, \ldots, M$, then $P(x)$ will clearly suffice. So we just want the degree of the least common multiple of the $x^{s}-1$ for $s=1,2, \ldots, M$, or just the number of roots of unity of order at most $M$, which is $\sum_{s=1}^{M} \phi(s)=1+1+2+2+4+2+6+4+6=28$.
Comment: Only at the beginning do we treat $F(x)$ strictly as a formal power series; later once we get the rational function representation $\sum_{d=1}^{6} \frac{x^{d}}{1-x^{d}}$, we can work with polynomial identities in general and don't have to worry about convergence issues for $|x| \geq 1$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-alg-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nAnswer: "
}
|
8bfe39fc-9c21-5efd-b43b-3a5923282517
| 609,220
|
There are 100 students who want to sign up for the class Introduction to Acting. There are three class sections for Introduction to Acting, each of which will fit exactly 20 students. The 100 students, including Alex and Zhu, are put in a lottery, and 60 of them are randomly selected to fill up the classes. What is the probability that Alex and Zhu end up getting into the same section for the class?
|
$19 / 165$ There is a $\frac{60}{100}=\frac{3}{5}$ chance that Alex is in the class. If Alex is in the class, the probability that Zhu is in his section is $\frac{19}{99}$. So the answer is $\frac{3}{5} \cdot \frac{19}{99}=\frac{19}{165}$.
|
\frac{19}{165}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
There are 100 students who want to sign up for the class Introduction to Acting. There are three class sections for Introduction to Acting, each of which will fit exactly 20 students. The 100 students, including Alex and Zhu, are put in a lottery, and 60 of them are randomly selected to fill up the classes. What is the probability that Alex and Zhu end up getting into the same section for the class?
|
$19 / 165$ There is a $\frac{60}{100}=\frac{3}{5}$ chance that Alex is in the class. If Alex is in the class, the probability that Zhu is in his section is $\frac{19}{99}$. So the answer is $\frac{3}{5} \cdot \frac{19}{99}=\frac{19}{165}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nAnswer: "
}
|
24f3dbe7-a7b3-57b1-8803-62b1089fbf40
| 609,221
|
There are 10 people who want to choose a committee of 5 people among them. They do this by first electing a set of $1,2,3$, or 4 committee leaders, who then choose among the remaining people to complete the 5 -person committee. In how many ways can the committee be formed, assuming that people are distinguishable? (Two committees that have the same members but different sets of leaders are considered to be distinct.)
|
7560 There are $\binom{10}{5}$ ways to choose the 5 -person committee. After choosing the committee, there are $2^{5}-2=30$ ways to choose the leaders. So the answer is $30 \cdot\binom{10}{5}=7560$.
|
7560
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
There are 10 people who want to choose a committee of 5 people among them. They do this by first electing a set of $1,2,3$, or 4 committee leaders, who then choose among the remaining people to complete the 5 -person committee. In how many ways can the committee be formed, assuming that people are distinguishable? (Two committees that have the same members but different sets of leaders are considered to be distinct.)
|
7560 There are $\binom{10}{5}$ ways to choose the 5 -person committee. After choosing the committee, there are $2^{5}-2=30$ ways to choose the leaders. So the answer is $30 \cdot\binom{10}{5}=7560$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nAnswer: "
}
|
b1c46a2b-7634-5ec1-b9af-9ee9a8c5a14c
| 609,222
|
Bob writes a random string of 5 letters, where each letter is either $A, B, C$, or $D$. The letter in each position is independently chosen, and each of the letters $A, B, C, D$ is chosen with equal probability. Given that there are at least two $A$ 's in the string, find the probability that there are at least three $A$ 's in the string.
|
$\quad \frac{53}{188}$ There are $\binom{5}{2} 3^{3}=270$ strings with $2 A$ 's. There are $\binom{5}{3} 3^{2}=90$ strings with 3 A's.
There are $\binom{5}{4} 3^{1}=15$ strings with $4 A$ 's. There is $\binom{5}{5} 3^{0}=1$ string with $5 A$ 's.
The desired probability is $\frac{90+15+1}{270+90+15+1}=\frac{53}{188}$.
|
\frac{53}{188}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Bob writes a random string of 5 letters, where each letter is either $A, B, C$, or $D$. The letter in each position is independently chosen, and each of the letters $A, B, C, D$ is chosen with equal probability. Given that there are at least two $A$ 's in the string, find the probability that there are at least three $A$ 's in the string.
|
$\quad \frac{53}{188}$ There are $\binom{5}{2} 3^{3}=270$ strings with $2 A$ 's. There are $\binom{5}{3} 3^{2}=90$ strings with 3 A's.
There are $\binom{5}{4} 3^{1}=15$ strings with $4 A$ 's. There is $\binom{5}{5} 3^{0}=1$ string with $5 A$ 's.
The desired probability is $\frac{90+15+1}{270+90+15+1}=\frac{53}{188}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nAnswer: "
}
|
2116d55a-e26e-5c3f-837b-887309c2ae5f
| 609,223
|
Find the number of triples of sets $(A, B, C)$ such that:
(a) $A, B, C \subseteq\{1,2,3, \ldots, 8\}$.
(b) $|A \cap B|=|B \cap C|=|C \cap A|=2$.
(c) $|A|=|B|=|C|=4$.
Here, $|S|$ denotes the number of elements in the set $S$.
|
45360 We consider the sets drawn in a Venn diagram.
Note that each element that is in at least one of the subsets lies in these seven possible spaces. We split by casework, with the cases based on $N=\left|R_{7}\right|=|A \cap B \cap C|$.
Case 1: $N=2$
Because we are given that $\left|R_{4}\right|+N=\left|R_{5}\right|+N=\left|R_{6}\right|+N=2$, we must have $\left|R_{4}\right|=\left|R_{5}\right|=\left|R_{6}\right|=0$. But we also know that $\left|R_{1}\right|+\left|R_{5}\right|+\left|R_{6}\right|+N=4$, so $\left|R_{1}\right|=2$. Similarly, $\left|R_{2}\right|=\left|R_{3}\right|=2$. Since these regions are distinguishable, we multiply through and obtain $\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}=2520$ ways.
Case 2: $N=1$
In this case, we can immediately deduce $\left|R_{4}\right|=\left|R_{5}\right|=\left|R_{6}\right|=1$. From this, it follows that $\left|R_{1}\right|=$ $4-1-1-1=1$, and similarly, $\left|R_{2}\right|=\left|R_{3}\right|=1$. All seven regions each contain one integer, so there are a total of $(8)(7) \ldots(2)=40320$ ways.
Case 3: $N=0$
Because $\left|R_{4}\right|+N=\left|R_{5}\right|+N=\left|R_{6}\right|+N=2$, we must have $\left|R_{4}\right|=\left|R_{5}\right|=\left|R_{6}\right|=2$. Since $\left|R_{1}\right|+\left|R_{5}\right|+\left|R_{6}\right|+N=4$, we immediately see that $\left|R_{1}\right|=0$. Similarly, $\left|R_{2}\right|=\left|R_{3}\right|=0$. The number of ways to fill $R_{4}, R_{5}, R_{6}$ is $\binom{8}{2}\binom{6}{2}\binom{4}{2}=2520$.
This clearly exhausts all the possibilities, so adding gives us $40320+2520+2520=45360$ ways.
|
45360
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Find the number of triples of sets $(A, B, C)$ such that:
(a) $A, B, C \subseteq\{1,2,3, \ldots, 8\}$.
(b) $|A \cap B|=|B \cap C|=|C \cap A|=2$.
(c) $|A|=|B|=|C|=4$.
Here, $|S|$ denotes the number of elements in the set $S$.
|
45360 We consider the sets drawn in a Venn diagram.
Note that each element that is in at least one of the subsets lies in these seven possible spaces. We split by casework, with the cases based on $N=\left|R_{7}\right|=|A \cap B \cap C|$.
Case 1: $N=2$
Because we are given that $\left|R_{4}\right|+N=\left|R_{5}\right|+N=\left|R_{6}\right|+N=2$, we must have $\left|R_{4}\right|=\left|R_{5}\right|=\left|R_{6}\right|=0$. But we also know that $\left|R_{1}\right|+\left|R_{5}\right|+\left|R_{6}\right|+N=4$, so $\left|R_{1}\right|=2$. Similarly, $\left|R_{2}\right|=\left|R_{3}\right|=2$. Since these regions are distinguishable, we multiply through and obtain $\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}=2520$ ways.
Case 2: $N=1$
In this case, we can immediately deduce $\left|R_{4}\right|=\left|R_{5}\right|=\left|R_{6}\right|=1$. From this, it follows that $\left|R_{1}\right|=$ $4-1-1-1=1$, and similarly, $\left|R_{2}\right|=\left|R_{3}\right|=1$. All seven regions each contain one integer, so there are a total of $(8)(7) \ldots(2)=40320$ ways.
Case 3: $N=0$
Because $\left|R_{4}\right|+N=\left|R_{5}\right|+N=\left|R_{6}\right|+N=2$, we must have $\left|R_{4}\right|=\left|R_{5}\right|=\left|R_{6}\right|=2$. Since $\left|R_{1}\right|+\left|R_{5}\right|+\left|R_{6}\right|+N=4$, we immediately see that $\left|R_{1}\right|=0$. Similarly, $\left|R_{2}\right|=\left|R_{3}\right|=0$. The number of ways to fill $R_{4}, R_{5}, R_{6}$ is $\binom{8}{2}\binom{6}{2}\binom{4}{2}=2520$.
This clearly exhausts all the possibilities, so adding gives us $40320+2520+2520=45360$ ways.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nAnswer: "
}
|
8272e690-49c4-588a-bf04-ca5e6cdb86b7
| 609,224
|
Eli, Joy, Paul, and Sam want to form a company; the company will have 16 shares to split among the 4 people. The following constraints are imposed:
- Every person must get a positive integer number of shares, and all 16 shares must be given out.
- No one person can have more shares than the other three people combined.
Assuming that shares are indistinguishable, but people are distinguishable, in how many ways can the shares be given out?
|
315 We are finding the number of integer solutions to $a+b+c+d=16$ with $1 \leq$ $a, b, c, d \leq 8$. We count the number of solutions to $a+b+c+d=16$ over positive integers, and subtract the number of solutions in which at least one variable is larger than 8 . If at least one variable is larger than 8 , exactly one of the variables is larger than 8 . We have 4 choices for this variable. The number of solutions to $a+b+c+d=16$ over positive integers, where $a>8$, is just the number of solutions to $a^{\prime}+b+c+d=8$ over positive integers, since we can substitute $a^{\prime}=a-8$. Thus, by the stars and bars formula (the number of positive integer solutions to $x_{1}+\cdots+x_{m}=n$ is $\binom{n-1}{m-1}$ ), the answer is $\binom{16-1}{4-1}-\binom{4}{1}\binom{(16-8)-1}{4-1}=35 \cdot 13-4 \cdot 35=315$.
|
315
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Eli, Joy, Paul, and Sam want to form a company; the company will have 16 shares to split among the 4 people. The following constraints are imposed:
- Every person must get a positive integer number of shares, and all 16 shares must be given out.
- No one person can have more shares than the other three people combined.
Assuming that shares are indistinguishable, but people are distinguishable, in how many ways can the shares be given out?
|
315 We are finding the number of integer solutions to $a+b+c+d=16$ with $1 \leq$ $a, b, c, d \leq 8$. We count the number of solutions to $a+b+c+d=16$ over positive integers, and subtract the number of solutions in which at least one variable is larger than 8 . If at least one variable is larger than 8 , exactly one of the variables is larger than 8 . We have 4 choices for this variable. The number of solutions to $a+b+c+d=16$ over positive integers, where $a>8$, is just the number of solutions to $a^{\prime}+b+c+d=8$ over positive integers, since we can substitute $a^{\prime}=a-8$. Thus, by the stars and bars formula (the number of positive integer solutions to $x_{1}+\cdots+x_{m}=n$ is $\binom{n-1}{m-1}$ ), the answer is $\binom{16-1}{4-1}-\binom{4}{1}\binom{(16-8)-1}{4-1}=35 \cdot 13-4 \cdot 35=315$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nAnswer: "
}
|
c582ffce-dd92-5c1d-a34f-522fc9bc5c32
| 609,225
|
We have a calculator with two buttons that displays an integer $x$. Pressing the first button replaces $x$ by $\left\lfloor\frac{x}{2}\right\rfloor$, and pressing the second button replaces $x$ by $4 x+1$. Initially, the calculator displays 0 . How many integers less than or equal to 2014 can be achieved through a sequence of arbitrary button presses? (It is permitted for the number displayed to exceed 2014 during the sequence. Here, $\lfloor y\rfloor$ denotes the greatest integer less than or equal to the real number $y$.)
|
233 We consider the integers from this process written in binary. The first operation truncates the rightmost digit, while the second operation appends 01 to the right.
We cannot have a number with a substring 11. For simplicity, call a string valid if it has no consecutive $1^{\prime} s$. Note that any number generated by this process is valid, as truncating the rightmost digit and appending 01 to the right of the digits clearly preserve validity.
Since we can effectively append a zero by applying the second operation and then the first operation, we see that we can achieve all valid strings.
Note that 2014 has eleven digits when written in binary, and any valid binary string with eleven digits is at most $10111111111=1535$. Therefore, our problem reduces to finding the number of eleven-digit valid strings. Let $F_{n}$ denote the number of valid strings of length $n$. For any valid string of length $n$, we can create a valid string of length $n+1$ by appending a 0 , or we can create a valid string of length $n+2$ by appending 01 . This process is clearly reversible, so our recursion is given by $F_{n}=F_{n-1}+F_{n-2}$, with $F_{1}=2, F_{2}=3$. This yields a sequence of Fibonacci numbers starting from 2, and some computation shows that our answer is $F_{11}=233$.
|
233
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
We have a calculator with two buttons that displays an integer $x$. Pressing the first button replaces $x$ by $\left\lfloor\frac{x}{2}\right\rfloor$, and pressing the second button replaces $x$ by $4 x+1$. Initially, the calculator displays 0 . How many integers less than or equal to 2014 can be achieved through a sequence of arbitrary button presses? (It is permitted for the number displayed to exceed 2014 during the sequence. Here, $\lfloor y\rfloor$ denotes the greatest integer less than or equal to the real number $y$.)
|
233 We consider the integers from this process written in binary. The first operation truncates the rightmost digit, while the second operation appends 01 to the right.
We cannot have a number with a substring 11. For simplicity, call a string valid if it has no consecutive $1^{\prime} s$. Note that any number generated by this process is valid, as truncating the rightmost digit and appending 01 to the right of the digits clearly preserve validity.
Since we can effectively append a zero by applying the second operation and then the first operation, we see that we can achieve all valid strings.
Note that 2014 has eleven digits when written in binary, and any valid binary string with eleven digits is at most $10111111111=1535$. Therefore, our problem reduces to finding the number of eleven-digit valid strings. Let $F_{n}$ denote the number of valid strings of length $n$. For any valid string of length $n$, we can create a valid string of length $n+1$ by appending a 0 , or we can create a valid string of length $n+2$ by appending 01 . This process is clearly reversible, so our recursion is given by $F_{n}=F_{n-1}+F_{n-2}$, with $F_{1}=2, F_{2}=3$. This yields a sequence of Fibonacci numbers starting from 2, and some computation shows that our answer is $F_{11}=233$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nAnswer: "
}
|
e2ea0bd5-f358-57a2-90d0-10689f90fce3
| 609,226
|
Six distinguishable players are participating in a tennis tournament. Each player plays one match of tennis against every other player. There are no ties in this tournament- each tennis match results in a win for one player and a loss for the other. Suppose that whenever $A$ and $B$ are players in the tournament such that $A$ wins strictly more matches than $B$ over the course of the tournament, it is also true that $A$ wins the match against $B$ in the tournament. In how many ways could the tournament have gone?
|
2048 We first group the players by wins, so let $G_{1}$ be the set of all players with the most wins, $G_{2}$ be the set of all players with the second most wins, $\ldots, G_{n}$ be the set of all players with the least wins. By the condition in the problem, everyone in group $G_{i}$ must beat everyone in group $G_{j}$ for all $i<j$. Now, consider the mini-tournament consisting of the matches among players inside a single group $G_{i}$. Each must have the same number of wins, say $x_{i}$. But the total number of games is $\binom{\left|G_{i}\right|}{2}$ and each game corresponds to exactly one win, so we must have $\binom{\left|G_{i}\right|}{2}=\left|G_{i}\right| x_{i} \Longrightarrow\left|G_{i}\right|=2 x_{i}+1$. Therefore, the number of players in each $G_{i}$ is odd.
We now have $\sum\left|G_{i}\right|=6$ and all $\left|G_{i}\right|$ are odd, so we can now do casework on the possibilities.
Case 1: $G_{i}$ 's have sizes 5 and 1. In this case, there are 2 ways to permute the groups (i.e. either $\left|G_{1}\right|=5,\left|G_{2}\right|=1$ or $\left|G_{1}\right|=1,\left|G_{2}\right|=5$ ). There are 6 ways to distribute the players into the two groups. There are 24 possible mini-tournaments in the group of size 5 ; to prove this, we label the players $p_{1}, \ldots, p_{5}$ and note that each player has 2 wins. Without loss of generality, let $p_{1}$ beat $p_{2}$ and $p_{3}$, and also without loss of generality let $p_{2}$ beat $p_{3}$. It's easy to verify that there are 2 possible mini-tournaments, depending on whether $p_{4}$ beats $p_{5}$ or $p_{5}$ beats $p_{4}$. Since there are $\binom{4}{2} \cdot 2=12$ ways to pick the two players $p_{1}$ defeats and choose which one beats the other, there are indeed $12 \cdot 2=24$ tournaments. Then the total number of possible tournaments in this case is $2 \cdot 6 \cdot 24=288$.
Case 2: The sizes are 3,3 . In this case, there are $\binom{6}{3}=20$ ways to distribute the players into the groups, and 2 possible mini-tournaments in either group, so the total here is $20 \cdot 2 \cdot 2=80$.
Case 3: The sizes are $3,1,1,1$. In this case, there are 4 ways to permute the groups, $\binom{6}{3} \cdot 6=120$ ways to distribute the players into groups, and 2 possible mini-tournaments in the group of size 3 , for a total of $4 \cdot 120 \cdot 2=960$.
Case 4: The sizes are $1,1,1,1,1,1$. There are 720 ways to distribute the players into groups.
The final answer is $288+80+960+720=2048$.
|
2048
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Six distinguishable players are participating in a tennis tournament. Each player plays one match of tennis against every other player. There are no ties in this tournament- each tennis match results in a win for one player and a loss for the other. Suppose that whenever $A$ and $B$ are players in the tournament such that $A$ wins strictly more matches than $B$ over the course of the tournament, it is also true that $A$ wins the match against $B$ in the tournament. In how many ways could the tournament have gone?
|
2048 We first group the players by wins, so let $G_{1}$ be the set of all players with the most wins, $G_{2}$ be the set of all players with the second most wins, $\ldots, G_{n}$ be the set of all players with the least wins. By the condition in the problem, everyone in group $G_{i}$ must beat everyone in group $G_{j}$ for all $i<j$. Now, consider the mini-tournament consisting of the matches among players inside a single group $G_{i}$. Each must have the same number of wins, say $x_{i}$. But the total number of games is $\binom{\left|G_{i}\right|}{2}$ and each game corresponds to exactly one win, so we must have $\binom{\left|G_{i}\right|}{2}=\left|G_{i}\right| x_{i} \Longrightarrow\left|G_{i}\right|=2 x_{i}+1$. Therefore, the number of players in each $G_{i}$ is odd.
We now have $\sum\left|G_{i}\right|=6$ and all $\left|G_{i}\right|$ are odd, so we can now do casework on the possibilities.
Case 1: $G_{i}$ 's have sizes 5 and 1. In this case, there are 2 ways to permute the groups (i.e. either $\left|G_{1}\right|=5,\left|G_{2}\right|=1$ or $\left|G_{1}\right|=1,\left|G_{2}\right|=5$ ). There are 6 ways to distribute the players into the two groups. There are 24 possible mini-tournaments in the group of size 5 ; to prove this, we label the players $p_{1}, \ldots, p_{5}$ and note that each player has 2 wins. Without loss of generality, let $p_{1}$ beat $p_{2}$ and $p_{3}$, and also without loss of generality let $p_{2}$ beat $p_{3}$. It's easy to verify that there are 2 possible mini-tournaments, depending on whether $p_{4}$ beats $p_{5}$ or $p_{5}$ beats $p_{4}$. Since there are $\binom{4}{2} \cdot 2=12$ ways to pick the two players $p_{1}$ defeats and choose which one beats the other, there are indeed $12 \cdot 2=24$ tournaments. Then the total number of possible tournaments in this case is $2 \cdot 6 \cdot 24=288$.
Case 2: The sizes are 3,3 . In this case, there are $\binom{6}{3}=20$ ways to distribute the players into the groups, and 2 possible mini-tournaments in either group, so the total here is $20 \cdot 2 \cdot 2=80$.
Case 3: The sizes are $3,1,1,1$. In this case, there are 4 ways to permute the groups, $\binom{6}{3} \cdot 6=120$ ways to distribute the players into groups, and 2 possible mini-tournaments in the group of size 3 , for a total of $4 \cdot 120 \cdot 2=960$.
Case 4: The sizes are $1,1,1,1,1,1$. There are 720 ways to distribute the players into groups.
The final answer is $288+80+960+720=2048$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nAnswer: "
}
|
b4952ad0-4e34-5522-9f71-89bcf669579b
| 609,227
|
The integers $1,2, \ldots, 64$ are written in the squares of a $8 \times 8$ chess board, such that for each $1 \leq i<64$, the numbers $i$ and $i+1$ are in squares that share an edge. What is the largest possible sum that can appear along one of the diagonals?
|
432 Our answer is $26+52+54+56+58+60+62+64$.
One possible configuration:
WLOG, we seek to maximize the sum of the numbers on the main diagonal (top left to bottom right). If we color the squares in a checker-board pattern and use the fact that $a$ and $a+1$ lie on different colored squares, we notice that all numbers appearing on the main diagonal must be of the same parity.
Consider the smallest value $m$ on the main diagonal. All numbers from 1 to $m-1$ must lie on one side of the diagonal since the main diagonal disconnects the board into two regions, and by assumption, all
| 26 | 25 | 24 | 23 | 18 | 17 | 8 | 7 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 27 | 52 | 53 | 22 | 19 | 16 | 9 | 6 |
| 28 | 51 | 54 | 21 | 20 | 15 | 10 | 5 |
| 29 | 50 | 55 | 56 | 57 | 14 | 11 | 4 |
| 30 | 49 | 44 | 43 | 58 | 13 | 12 | 3 |
| 31 | 48 | 45 | 42 | 59 | 60 | 61 | 2 |
| 32 | 47 | 46 | 41 | 40 | 39 | 62 | 1 |
| 33 | 34 | 35 | 36 | 37 | 38 | 63 | 64 |
numbers less than $m$ cannot lie on the main diagonal. Therefore, $m \leq 29$ (one more than the seventh triangular number) But if $m=29$, then the sum of the numbers on the main diagonal is at most $29+51+53+55+57+59+61+63=428$, as these numbers must be odd. Similarly, $m=27$ is also not optimal.
This leaves $m=28$ as a possibility. But if this were the case, the only way it beats our answer is if we have $28+52+54+\ldots+64$, which would require $52,54, \ldots, 64$ to appear sequentially along the diagonal, forcing 28 to be in one of the corners.
Now label the squares (row, column) with $(1,1)$ being the top left and $(8,8)$ being the bottom right. Assume WLOG that 28 occupies $(1,1)$. Since 62 and 64 are in $(7,7)$ and $(8,8)$, respectively, we must have 63 in $(7,8)$ or $(8,7)$, and WLOG, assume it's in $(8,7)$. Since 61 is next to 60 , it is not difficult to see that $(7,8)$ must be occupied by 1 (all numbers $a$ between 2 and 60 must have $a-1$ and $a+1$ as neighbors). Since 1 is above the main diagonal, all numbers from 1 to 27 must also be above the main diagonal. Since there are 28 squares above the main diagonal, there is exactly one number above the main diagonal greater than 28 .
Notice that 61 must occupy $(7,6)$ or $(6,7)$. If it occupies $(7,6)$, then we are stuck at $(8,6)$, since it must contain a number between 2 and 59 , which is impossible. Therefore, 61 must occupy $(6,7)$, and no more numbers greater than 28 can be above the main diagonal. This forces $59,57,55$, and 53 to occupy $(6,5),(5,4),(4,3),(3,2)$, respectively. But we see that 27 occupies $(1,2)$ and 29 occupies $(2,1)$, leaving nowhere for 51 .
This is a contradiction, so our answer is therefore optimal.
Alternate solution: Another method of proving that $m \leq 26$ is to note that each side of the diagonal has 28 squares, 16 of which are one color and 12 of which are the other color. As the path has to alternate colors, one can make at most $13+12=25$ steps before moving on the diagonal.
|
432
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
The integers $1,2, \ldots, 64$ are written in the squares of a $8 \times 8$ chess board, such that for each $1 \leq i<64$, the numbers $i$ and $i+1$ are in squares that share an edge. What is the largest possible sum that can appear along one of the diagonals?
|
432 Our answer is $26+52+54+56+58+60+62+64$.
One possible configuration:
WLOG, we seek to maximize the sum of the numbers on the main diagonal (top left to bottom right). If we color the squares in a checker-board pattern and use the fact that $a$ and $a+1$ lie on different colored squares, we notice that all numbers appearing on the main diagonal must be of the same parity.
Consider the smallest value $m$ on the main diagonal. All numbers from 1 to $m-1$ must lie on one side of the diagonal since the main diagonal disconnects the board into two regions, and by assumption, all
| 26 | 25 | 24 | 23 | 18 | 17 | 8 | 7 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 27 | 52 | 53 | 22 | 19 | 16 | 9 | 6 |
| 28 | 51 | 54 | 21 | 20 | 15 | 10 | 5 |
| 29 | 50 | 55 | 56 | 57 | 14 | 11 | 4 |
| 30 | 49 | 44 | 43 | 58 | 13 | 12 | 3 |
| 31 | 48 | 45 | 42 | 59 | 60 | 61 | 2 |
| 32 | 47 | 46 | 41 | 40 | 39 | 62 | 1 |
| 33 | 34 | 35 | 36 | 37 | 38 | 63 | 64 |
numbers less than $m$ cannot lie on the main diagonal. Therefore, $m \leq 29$ (one more than the seventh triangular number) But if $m=29$, then the sum of the numbers on the main diagonal is at most $29+51+53+55+57+59+61+63=428$, as these numbers must be odd. Similarly, $m=27$ is also not optimal.
This leaves $m=28$ as a possibility. But if this were the case, the only way it beats our answer is if we have $28+52+54+\ldots+64$, which would require $52,54, \ldots, 64$ to appear sequentially along the diagonal, forcing 28 to be in one of the corners.
Now label the squares (row, column) with $(1,1)$ being the top left and $(8,8)$ being the bottom right. Assume WLOG that 28 occupies $(1,1)$. Since 62 and 64 are in $(7,7)$ and $(8,8)$, respectively, we must have 63 in $(7,8)$ or $(8,7)$, and WLOG, assume it's in $(8,7)$. Since 61 is next to 60 , it is not difficult to see that $(7,8)$ must be occupied by 1 (all numbers $a$ between 2 and 60 must have $a-1$ and $a+1$ as neighbors). Since 1 is above the main diagonal, all numbers from 1 to 27 must also be above the main diagonal. Since there are 28 squares above the main diagonal, there is exactly one number above the main diagonal greater than 28 .
Notice that 61 must occupy $(7,6)$ or $(6,7)$. If it occupies $(7,6)$, then we are stuck at $(8,6)$, since it must contain a number between 2 and 59 , which is impossible. Therefore, 61 must occupy $(6,7)$, and no more numbers greater than 28 can be above the main diagonal. This forces $59,57,55$, and 53 to occupy $(6,5),(5,4),(4,3),(3,2)$, respectively. But we see that 27 occupies $(1,2)$ and 29 occupies $(2,1)$, leaving nowhere for 51 .
This is a contradiction, so our answer is therefore optimal.
Alternate solution: Another method of proving that $m \leq 26$ is to note that each side of the diagonal has 28 squares, 16 of which are one color and 12 of which are the other color. As the path has to alternate colors, one can make at most $13+12=25$ steps before moving on the diagonal.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nAnswer: "
}
|
ca3d5115-5cc5-56d7-9a36-753d1af22454
| 609,228
|
There is a heads up coin on every integer of the number line. Lucky is initially standing on the zero point of the number line facing in the positive direction. Lucky performs the following procedure: he looks at the coin (or lack thereof) underneath him, and then,
- If the coin is heads up, Lucky flips it to tails up, turns around, and steps forward a distance of one unit.
- If the coin is tails up, Lucky picks up the coin and steps forward a distance of one unit facing the same direction.
- If there is no coin, Lucky places a coin heads up underneath him and steps forward a distance of one unit facing the same direction.
He repeats this procedure until there are 20 coins anywhere that are tails up. How many times has Lucky performed the procedure when the process stops?
|
6098 We keep track of the following quantities: Let $N$ be the sum of $2^{k}$, where $k$ ranges over all nonnegative integers such that position $-1-k$ on the number line contains a tails-up coin. Let $M$ be the sum of $2^{k}$, where $k$ ranges over all nonnegative integers such that position $k$ contains a tails-up coin.
We also make the following definitions: A "right event" is the event that Lucky crosses from the negative integers on the number line to the non-negative integers. A "left event" is the event that Lucky crosses from the non-negative integers on the number line to the negative integers.
We now make the following claims:
(a) Every time a right event or left event occurs, every point on the number line contains a coin.
(b) Suppose that $n$ is a positive integer. When the $n$th left event occurs, the value of $M$ is equal to $n$. When the $n$th right event occurs, the value of $N$ is equal to $n$.
(c) For a nonzero integer $n$, denote by $\nu_{2}(n)$ the largest integer $k$ such that $2^{k}$ divides $n$. The number of steps that elapse between the $(n-1)$ st right event and the $n$th left event is equal to $2 \nu_{2}(n)+1$. The number of steps that elapse between the $n$th left event and the $n$th right event is also equal to $2 \nu_{2}(n)+1$. (If $n-1=0$, then the " $(n-1)$ st right event" refers to the beginning of the simulation.)
(d) The man stops as soon as the 1023 rd right event occurs. (Note that $1023=2^{10}-1$.)
In other words, Lucky is keeping track of two numbers $M$ and $N$, which are obtained by interpreting the coins on the number line as binary strings, and alternately incrementing each of them by one. We will prove claim 2 ; the other claims follow from very similar reasoning and their proofs will be omitted.
Clearly, left and right events alternate. That is, a left event occurs, then a right event, then a left event, and so on. So it's enough to prove that, between each right event and the following left event, the value of $M$ is incremented by 1 , and that between each left event and the following right event, the value of $N$ is incremented by 1 . We will show the first statement; the second follows from symmetry.
Suppose that a right event has just occurred. Then, by claim 1, every space on the number line cotnains a coin. So, there is some nonnegative integer $\ell$ for which positions $0, \ldots, \ell-1$ on the number line contain a tails up coin, and position $\ell$ contains a heads up coin. Since Lucky is standing at position 0 facing rightward, the following sequence of steps will occur:
(a) Lucky will take $\ell$ steps to the right, eventually reaching position $\ell$. During this process, he will pick up the coins at positions $0, \ldots, \ell-1$.
(b) Then, Lucky turn the coin at position $\ell$ to a tails up coin and turn around.
(c) Finally, Lucky will take $\ell+1$ steps to the left, eventually reaching position -1 (at which point a left event occurs). During this process, he will place a heads up coin at positions $0, \ldots, \ell-1$.
During this sequence, the tails up coins at positions $0, \ldots, \ell-1$ have been changed to heads up coins, and the heads up coin at position $\ell$ has been changed to a tails up coin. So the value of $M$ has been incremented by
$$
2^{\ell}-\sum_{i=0}^{\ell-1} 2^{i}=1
$$
as desired.
Now, it remains to compute the answer to the question. By claims 3 and 4 , the total number of steps taken by the simulation is
$$
2 \sum_{n=1}^{1023}\left(2 \nu_{2}(n)+1\right)
$$
This can be rewritten as
$$
4 \sum_{n=1}^{1023} \nu_{2}(n)+2 \cdot 1023=4 \nu_{2}(1023!)+2046
$$
We can compute $\nu_{2}(1023!)=1013$ using Legendre's formula for the highest power of 2 dividing a factorial. This results in the final answer 6098.
|
6098
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
There is a heads up coin on every integer of the number line. Lucky is initially standing on the zero point of the number line facing in the positive direction. Lucky performs the following procedure: he looks at the coin (or lack thereof) underneath him, and then,
- If the coin is heads up, Lucky flips it to tails up, turns around, and steps forward a distance of one unit.
- If the coin is tails up, Lucky picks up the coin and steps forward a distance of one unit facing the same direction.
- If there is no coin, Lucky places a coin heads up underneath him and steps forward a distance of one unit facing the same direction.
He repeats this procedure until there are 20 coins anywhere that are tails up. How many times has Lucky performed the procedure when the process stops?
|
6098 We keep track of the following quantities: Let $N$ be the sum of $2^{k}$, where $k$ ranges over all nonnegative integers such that position $-1-k$ on the number line contains a tails-up coin. Let $M$ be the sum of $2^{k}$, where $k$ ranges over all nonnegative integers such that position $k$ contains a tails-up coin.
We also make the following definitions: A "right event" is the event that Lucky crosses from the negative integers on the number line to the non-negative integers. A "left event" is the event that Lucky crosses from the non-negative integers on the number line to the negative integers.
We now make the following claims:
(a) Every time a right event or left event occurs, every point on the number line contains a coin.
(b) Suppose that $n$ is a positive integer. When the $n$th left event occurs, the value of $M$ is equal to $n$. When the $n$th right event occurs, the value of $N$ is equal to $n$.
(c) For a nonzero integer $n$, denote by $\nu_{2}(n)$ the largest integer $k$ such that $2^{k}$ divides $n$. The number of steps that elapse between the $(n-1)$ st right event and the $n$th left event is equal to $2 \nu_{2}(n)+1$. The number of steps that elapse between the $n$th left event and the $n$th right event is also equal to $2 \nu_{2}(n)+1$. (If $n-1=0$, then the " $(n-1)$ st right event" refers to the beginning of the simulation.)
(d) The man stops as soon as the 1023 rd right event occurs. (Note that $1023=2^{10}-1$.)
In other words, Lucky is keeping track of two numbers $M$ and $N$, which are obtained by interpreting the coins on the number line as binary strings, and alternately incrementing each of them by one. We will prove claim 2 ; the other claims follow from very similar reasoning and their proofs will be omitted.
Clearly, left and right events alternate. That is, a left event occurs, then a right event, then a left event, and so on. So it's enough to prove that, between each right event and the following left event, the value of $M$ is incremented by 1 , and that between each left event and the following right event, the value of $N$ is incremented by 1 . We will show the first statement; the second follows from symmetry.
Suppose that a right event has just occurred. Then, by claim 1, every space on the number line cotnains a coin. So, there is some nonnegative integer $\ell$ for which positions $0, \ldots, \ell-1$ on the number line contain a tails up coin, and position $\ell$ contains a heads up coin. Since Lucky is standing at position 0 facing rightward, the following sequence of steps will occur:
(a) Lucky will take $\ell$ steps to the right, eventually reaching position $\ell$. During this process, he will pick up the coins at positions $0, \ldots, \ell-1$.
(b) Then, Lucky turn the coin at position $\ell$ to a tails up coin and turn around.
(c) Finally, Lucky will take $\ell+1$ steps to the left, eventually reaching position -1 (at which point a left event occurs). During this process, he will place a heads up coin at positions $0, \ldots, \ell-1$.
During this sequence, the tails up coins at positions $0, \ldots, \ell-1$ have been changed to heads up coins, and the heads up coin at position $\ell$ has been changed to a tails up coin. So the value of $M$ has been incremented by
$$
2^{\ell}-\sum_{i=0}^{\ell-1} 2^{i}=1
$$
as desired.
Now, it remains to compute the answer to the question. By claims 3 and 4 , the total number of steps taken by the simulation is
$$
2 \sum_{n=1}^{1023}\left(2 \nu_{2}(n)+1\right)
$$
This can be rewritten as
$$
4 \sum_{n=1}^{1023} \nu_{2}(n)+2 \cdot 1023=4 \nu_{2}(1023!)+2046
$$
We can compute $\nu_{2}(1023!)=1013$ using Legendre's formula for the highest power of 2 dividing a factorial. This results in the final answer 6098.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nAnswer: "
}
|
ca7623b0-f251-5ae0-b828-f35520e8730a
| 609,229
|
An up-right path from $(a, b) \in \mathbb{R}^{2}$ to $(c, d) \in \mathbb{R}^{2}$ is a finite sequence $\left(x_{1}, y_{1}\right), \ldots,\left(x_{k}, y_{k}\right)$ of points in $\mathbb{R}^{2}$ such that $(a, b)=\left(x_{1}, y_{1}\right),(c, d)=\left(x_{k}, y_{k}\right)$, and for each $1 \leq i<k$ we have that either $\left(x_{i+1}, y_{i+1}\right)=\left(x_{i}+1, y_{i}\right)$ or $\left(x_{i+1}, y_{i+1}\right)=\left(x_{i}, y_{i}+1\right)$. Two up-right paths are said to intersect if they share any point.
Find the number of pairs $(A, B)$ where $A$ is an up-right path from $(0,0)$ to $(4,4), B$ is an up-right path from $(2,0)$ to $(6,4)$, and $A$ and $B$ do not intersect.
|
1750 The number of up-right paths from $(0,0)$ to $(4,4)$ is $\binom{8}{4}$ because any such upright path is identical to a sequence of 4 U's and 4 R's, where $U$ corresponds to a step upwards and R corresponds to a step rightwards. Therefore, the total number of pairs of (possibly intersecting) up-right paths from $(0,0)$ to $(4,4)$ and $(2,0)$ to $(6,4)$ is $\binom{8}{4}^{2}$.
We will now count the number of intersecting pairs of up-right paths and subtract it to get the answer. Consider an up-right path $A$ from $(0,0)$ to $(4,4)$ and an up-right path $B$ from $(2,0)$ to $(6,4)$. If they intersect, take the point $(x, y)$ where they first meet each other, and switch the parts of the paths after $(x, y)$ to make an up-right path $A^{\prime}$ from $(0,0)$ to $(6,4)$ and an up-right path $B^{\prime}$ from $(2,0)$ to $(4,4)$.
Conversely, given an up-right path $A^{\prime}$ from $(0,0)$ to $(6,4)$ and an up-right path $B^{\prime}$ from $(2,0)$ to $(4,4)$, they must intersect somewhere, so we can again take their first intersection point and switch the ends to get the original up-right path $A$ from $(0,0)$ to $(4,4)$ and up-right path $B$ from $(2,0)$ to $(6,4)$, where $A$ and $B$ intersect.
Consequently, the number of intersecting pairs of up-right paths is exactly equal to the number of pairs of up-right paths from $(0,0)$ to $(6,4)$ and $(2,0)$ to $(4,4)$, which is $\binom{10}{4}\binom{6}{4}$. The number of pairs that do not intersect is therefore $\binom{8}{4}^{2}-\binom{10}{4}\binom{6}{4}=4900-3150=1750$.
|
1750
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
An up-right path from $(a, b) \in \mathbb{R}^{2}$ to $(c, d) \in \mathbb{R}^{2}$ is a finite sequence $\left(x_{1}, y_{1}\right), \ldots,\left(x_{k}, y_{k}\right)$ of points in $\mathbb{R}^{2}$ such that $(a, b)=\left(x_{1}, y_{1}\right),(c, d)=\left(x_{k}, y_{k}\right)$, and for each $1 \leq i<k$ we have that either $\left(x_{i+1}, y_{i+1}\right)=\left(x_{i}+1, y_{i}\right)$ or $\left(x_{i+1}, y_{i+1}\right)=\left(x_{i}, y_{i}+1\right)$. Two up-right paths are said to intersect if they share any point.
Find the number of pairs $(A, B)$ where $A$ is an up-right path from $(0,0)$ to $(4,4), B$ is an up-right path from $(2,0)$ to $(6,4)$, and $A$ and $B$ do not intersect.
|
1750 The number of up-right paths from $(0,0)$ to $(4,4)$ is $\binom{8}{4}$ because any such upright path is identical to a sequence of 4 U's and 4 R's, where $U$ corresponds to a step upwards and R corresponds to a step rightwards. Therefore, the total number of pairs of (possibly intersecting) up-right paths from $(0,0)$ to $(4,4)$ and $(2,0)$ to $(6,4)$ is $\binom{8}{4}^{2}$.
We will now count the number of intersecting pairs of up-right paths and subtract it to get the answer. Consider an up-right path $A$ from $(0,0)$ to $(4,4)$ and an up-right path $B$ from $(2,0)$ to $(6,4)$. If they intersect, take the point $(x, y)$ where they first meet each other, and switch the parts of the paths after $(x, y)$ to make an up-right path $A^{\prime}$ from $(0,0)$ to $(6,4)$ and an up-right path $B^{\prime}$ from $(2,0)$ to $(4,4)$.
Conversely, given an up-right path $A^{\prime}$ from $(0,0)$ to $(6,4)$ and an up-right path $B^{\prime}$ from $(2,0)$ to $(4,4)$, they must intersect somewhere, so we can again take their first intersection point and switch the ends to get the original up-right path $A$ from $(0,0)$ to $(4,4)$ and up-right path $B$ from $(2,0)$ to $(6,4)$, where $A$ and $B$ intersect.
Consequently, the number of intersecting pairs of up-right paths is exactly equal to the number of pairs of up-right paths from $(0,0)$ to $(6,4)$ and $(2,0)$ to $(4,4)$, which is $\binom{10}{4}\binom{6}{4}$. The number of pairs that do not intersect is therefore $\binom{8}{4}^{2}-\binom{10}{4}\binom{6}{4}=4900-3150=1750$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-comb-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nAnswer: "
}
|
03468f54-2097-5331-9a60-9dbe8857bb29
| 609,230
|
Point $P$ and line $\ell$ are such that the distance from $P$ to $\ell$ is 12 . Given that $T$ is a point on $\ell$ such that $P T=13$, find the radius of the circle passing through $P$ and tangent to $\ell$ at $T$.
|
$169 / 24$ Let $O$ be the center of the given circle, $Q$ be the foot of the altitude from $P$ to $\ell$, and $M$ be the midpoint of $P T$. Then since $O M \perp P T$ and $\angle O T P=\angle T P Q, \triangle O M P \sim \triangle T Q P$. Thus the $O P=T P \cdot \frac{P M}{P Q}=13 \cdot \frac{13 / 2}{12}=\frac{169}{24}$
|
\frac{169}{24}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Point $P$ and line $\ell$ are such that the distance from $P$ to $\ell$ is 12 . Given that $T$ is a point on $\ell$ such that $P T=13$, find the radius of the circle passing through $P$ and tangent to $\ell$ at $T$.
|
$169 / 24$ Let $O$ be the center of the given circle, $Q$ be the foot of the altitude from $P$ to $\ell$, and $M$ be the midpoint of $P T$. Then since $O M \perp P T$ and $\angle O T P=\angle T P Q, \triangle O M P \sim \triangle T Q P$. Thus the $O P=T P \cdot \frac{P M}{P Q}=13 \cdot \frac{13 / 2}{12}=\frac{169}{24}$
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-geo-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nAnswer: "
}
|
9e97967f-5e56-51c8-9aa4-55fadb0c0eb3
| 609,232
|
$A B C$ is a triangle such that $B C=10, C A=12$. Let $M$ be the midpoint of side $A C$. Given that $B M$ is parallel to the external bisector of $\angle A$, find area of triangle $A B C$. (Lines $A B$ and $A C$ form two angles, one of which is $\angle B A C$. The external bisector of $\angle A$ is the line that bisects the other angle.)
|
$8 \sqrt{14}$ Since $B M$ is parallel to the external bisector of $\angle A=\angle B A M$, it is perpendicular to the angle bisector of $\angle B A M$. Thus $B A=B M=\frac{1}{2} B C=6$. By Heron's formula, the area of $\triangle A B C$ is therefore $\sqrt{(14)(8)(4)(2)}=8 \sqrt{14}$.
|
8 \sqrt{14}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
$A B C$ is a triangle such that $B C=10, C A=12$. Let $M$ be the midpoint of side $A C$. Given that $B M$ is parallel to the external bisector of $\angle A$, find area of triangle $A B C$. (Lines $A B$ and $A C$ form two angles, one of which is $\angle B A C$. The external bisector of $\angle A$ is the line that bisects the other angle.)
|
$8 \sqrt{14}$ Since $B M$ is parallel to the external bisector of $\angle A=\angle B A M$, it is perpendicular to the angle bisector of $\angle B A M$. Thus $B A=B M=\frac{1}{2} B C=6$. By Heron's formula, the area of $\triangle A B C$ is therefore $\sqrt{(14)(8)(4)(2)}=8 \sqrt{14}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-geo-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nAnswer: "
}
|
47793833-2f1b-5197-93d1-4e14ab07c0c9
| 609,233
|
In quadrilateral $A B C D, \angle D A C=98^{\circ}, \angle D B C=82^{\circ}, \angle B C D=70^{\circ}$, and $B C=A D$. Find $\angle A C D$.
|
28 Let $B^{\prime}$ be the reflection of $B$ across $C D$. Note that $A D=B C$, and $\angle D A C+\angle C B^{\prime} D=$ $180^{\circ}$, so $A C B^{\prime} D$ is a cyclic trapezoid. Thus, $A C B^{\prime} D$ is an isosceles trapezoid, so $\angle A C B^{\prime}=98^{\circ}$. Note that $\angle D C B^{\prime}=\angle B C D=70^{\circ}$, so $\angle A C D=\angle A C B^{\prime}-\angle D C B^{\prime}=98^{\circ}-70^{\circ}=28^{\circ}$.
|
28
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
In quadrilateral $A B C D, \angle D A C=98^{\circ}, \angle D B C=82^{\circ}, \angle B C D=70^{\circ}$, and $B C=A D$. Find $\angle A C D$.
|
28 Let $B^{\prime}$ be the reflection of $B$ across $C D$. Note that $A D=B C$, and $\angle D A C+\angle C B^{\prime} D=$ $180^{\circ}$, so $A C B^{\prime} D$ is a cyclic trapezoid. Thus, $A C B^{\prime} D$ is an isosceles trapezoid, so $\angle A C B^{\prime}=98^{\circ}$. Note that $\angle D C B^{\prime}=\angle B C D=70^{\circ}$, so $\angle A C D=\angle A C B^{\prime}-\angle D C B^{\prime}=98^{\circ}-70^{\circ}=28^{\circ}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-geo-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nAnswer: "
}
|
d8668fed-027e-508f-b9e6-677c462d81eb
| 609,234
|
Let $\mathcal{C}$ be a circle in the $x y$ plane with radius 1 and center $(0,0,0)$, and let $P$ be a point in space with coordinates $(3,4,8)$. Find the largest possible radius of a sphere that is contained entirely in the slanted cone with base $\mathcal{C}$ and vertex $P$.
|
$\quad 3-\sqrt{5}$ Consider the plane passing through $P$ that is perpendicular to the plane of the circle. The intersection of the plane with the cone and sphere is a cross section consisting of a circle inscribed in a triangle with a vertex $P$. By symmetry, this circle is a great circle of the sphere, and hence has the same radius. The other two vertices of the triangle are the points of intersection between the plane and the unit circle, so the other two vertices are $\left(\frac{3}{5}, \frac{4}{5}, 0\right),\left(-\frac{3}{5},-\frac{4}{5}, 0\right)$.
Using the formula $A=r s$ and using the distance formula to find the side lengths, we find that $r=\frac{2 A}{2 s}=\frac{2 * 8}{2+10+4 \sqrt{5}}=3-\sqrt{5}$.
|
3-\sqrt{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $\mathcal{C}$ be a circle in the $x y$ plane with radius 1 and center $(0,0,0)$, and let $P$ be a point in space with coordinates $(3,4,8)$. Find the largest possible radius of a sphere that is contained entirely in the slanted cone with base $\mathcal{C}$ and vertex $P$.
|
$\quad 3-\sqrt{5}$ Consider the plane passing through $P$ that is perpendicular to the plane of the circle. The intersection of the plane with the cone and sphere is a cross section consisting of a circle inscribed in a triangle with a vertex $P$. By symmetry, this circle is a great circle of the sphere, and hence has the same radius. The other two vertices of the triangle are the points of intersection between the plane and the unit circle, so the other two vertices are $\left(\frac{3}{5}, \frac{4}{5}, 0\right),\left(-\frac{3}{5},-\frac{4}{5}, 0\right)$.
Using the formula $A=r s$ and using the distance formula to find the side lengths, we find that $r=\frac{2 A}{2 s}=\frac{2 * 8}{2+10+4 \sqrt{5}}=3-\sqrt{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-geo-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nAnswer: "
}
|
99bcd91e-2666-58e4-8096-c1757e35f62f
| 609,235
|
In quadrilateral $A B C D$, we have $A B=5, B C=6, C D=5, D A=4$, and $\angle A B C=90^{\circ}$. Let $A C$ and $B D$ meet at $E$. Compute $\frac{B E}{E D}$.
|
$\sqrt{3}$ We find that $A C=\sqrt{61}$, and applying the law of cosines to triangle $A C D$ tells us that $\angle A D C=120$. Then $\frac{B E}{E D}$ is the ratio of the areas of triangles $A B C$ and $A D C$, which is $\frac{(5)(6)}{(4)(5) \frac{\sqrt{3}}{2}}=\sqrt{3}$.
|
\sqrt{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
In quadrilateral $A B C D$, we have $A B=5, B C=6, C D=5, D A=4$, and $\angle A B C=90^{\circ}$. Let $A C$ and $B D$ meet at $E$. Compute $\frac{B E}{E D}$.
|
$\sqrt{3}$ We find that $A C=\sqrt{61}$, and applying the law of cosines to triangle $A C D$ tells us that $\angle A D C=120$. Then $\frac{B E}{E D}$ is the ratio of the areas of triangles $A B C$ and $A D C$, which is $\frac{(5)(6)}{(4)(5) \frac{\sqrt{3}}{2}}=\sqrt{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-geo-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nAnswer: "
}
|
5a41b5d5-2fe1-5795-b435-eabf301109a7
| 609,236
|
Triangle $A B C$ has sides $A B=14, B C=13$, and $C A=15$. It is inscribed in circle $\Gamma$, which has center $O$. Let $M$ be the midpoint of $A B$, let $B^{\prime}$ be the point on $\Gamma$ diametrically opposite $B$, and let $X$ be the intersection of $A O$ and $M B^{\prime}$. Find the length of $A X$.
|
$65 / 12$ Since $B^{\prime} B$ is a diameter, $\angle B^{\prime} A B=90^{\circ}$, so $B^{\prime} A \| O M$, so $\frac{O M}{B^{\prime} A}=\frac{B M}{B A}=\frac{1}{2}$. Thus $\frac{A X}{X O}=\frac{B^{\prime} A}{O M}=2$, so $A X=\frac{2}{3} R$, where $R=\frac{a b c}{4 A}=\frac{(13)(14)(15)}{4(84)}=\frac{65}{8}$ is the circumradius of $A B C$. Putting it all together gives $A X=\frac{65}{12}$.
|
\frac{65}{12}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Triangle $A B C$ has sides $A B=14, B C=13$, and $C A=15$. It is inscribed in circle $\Gamma$, which has center $O$. Let $M$ be the midpoint of $A B$, let $B^{\prime}$ be the point on $\Gamma$ diametrically opposite $B$, and let $X$ be the intersection of $A O$ and $M B^{\prime}$. Find the length of $A X$.
|
$65 / 12$ Since $B^{\prime} B$ is a diameter, $\angle B^{\prime} A B=90^{\circ}$, so $B^{\prime} A \| O M$, so $\frac{O M}{B^{\prime} A}=\frac{B M}{B A}=\frac{1}{2}$. Thus $\frac{A X}{X O}=\frac{B^{\prime} A}{O M}=2$, so $A X=\frac{2}{3} R$, where $R=\frac{a b c}{4 A}=\frac{(13)(14)(15)}{4(84)}=\frac{65}{8}$ is the circumradius of $A B C$. Putting it all together gives $A X=\frac{65}{12}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-geo-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nAnswer: "
}
|
0f6f2667-2bbe-51f8-8987-9b298b57d2fc
| 609,237
|
Let $A B C$ be a triangle with sides $A B=6, B C=10$, and $C A=8$. Let $M$ and $N$ be the midpoints of $B A$ and $B C$, respectively. Choose the point $Y$ on ray $C M$ so that the circumcircle of triangle $A M Y$ is tangent to $A N$. Find the area of triangle $N A Y$.
|
$600 / 73$ Let $G=A N \cap C M$ be the centroid of $A B C$. Then $G A=\frac{2}{3} G N=\frac{10}{3}$ and $G M=$ $\frac{1}{3} C M=\frac{1}{3} \sqrt{8^{2}+3^{2}}=\frac{\sqrt{73}}{3}$. By power of a point, $(G M)(G Y)=G A^{2}$, so $G Y=\frac{G A^{2}}{G Y}=\frac{(10 / 3)^{2}}{\frac{\sqrt{73}}{3}}=\frac{100}{3 \sqrt{73}}$. Thus
$$
\begin{aligned}
{[N A Y] } & =[G A M] \cdot \frac{[G A Y]}{[G A M]} \cdot \frac{[N A Y]}{[G A Y]} \\
& =\frac{1}{6}[A B C] \cdot \frac{G Y}{G M} \cdot \frac{N A}{G A} \\
& =4 \cdot \frac{100}{73} \cdot \frac{3}{2}=\frac{600}{73}
\end{aligned}
$$
|
\frac{600}{73}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a triangle with sides $A B=6, B C=10$, and $C A=8$. Let $M$ and $N$ be the midpoints of $B A$ and $B C$, respectively. Choose the point $Y$ on ray $C M$ so that the circumcircle of triangle $A M Y$ is tangent to $A N$. Find the area of triangle $N A Y$.
|
$600 / 73$ Let $G=A N \cap C M$ be the centroid of $A B C$. Then $G A=\frac{2}{3} G N=\frac{10}{3}$ and $G M=$ $\frac{1}{3} C M=\frac{1}{3} \sqrt{8^{2}+3^{2}}=\frac{\sqrt{73}}{3}$. By power of a point, $(G M)(G Y)=G A^{2}$, so $G Y=\frac{G A^{2}}{G Y}=\frac{(10 / 3)^{2}}{\frac{\sqrt{73}}{3}}=\frac{100}{3 \sqrt{73}}$. Thus
$$
\begin{aligned}
{[N A Y] } & =[G A M] \cdot \frac{[G A Y]}{[G A M]} \cdot \frac{[N A Y]}{[G A Y]} \\
& =\frac{1}{6}[A B C] \cdot \frac{G Y}{G M} \cdot \frac{N A}{G A} \\
& =4 \cdot \frac{100}{73} \cdot \frac{3}{2}=\frac{600}{73}
\end{aligned}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-geo-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nAnswer: "
}
|
f93275be-232d-55d6-98ce-82fc7b87faa1
| 609,238
|
Two circles are said to be orthogonal if they intersect in two points, and their tangents at either point of intersection are perpendicular. Two circles $\omega_{1}$ and $\omega_{2}$ with radii 10 and 13 , respectively, are externally tangent at point $P$. Another circle $\omega_{3}$ with radius $2 \sqrt{2}$ passes through $P$ and is orthogonal to both $\omega_{1}$ and $\omega_{2}$. A fourth circle $\omega_{4}$, orthogonal to $\omega_{3}$, is externally tangent to $\omega_{1}$ and $\omega_{2}$. Compute the radius of $\omega_{4}$.
|
$\frac{92}{61}$ Let $\omega_{i}$ have center $O_{i}$ and radius $r_{i}$. Since $\omega_{3}$ is orthogonal to $\omega_{1}, \omega_{2}, \omega_{4}$, it has equal power $r_{3}^{2}$ to each of them. Thus $O_{3}$ is the radical center of $\omega_{1}, \omega_{2}, \omega_{4}$, which is equidistant to the three sides of $\triangle O_{1} O_{2} O_{4}$ and therefore its incenter.
For distinct $i, j \in\{1,2,4\}, \omega_{i} \cap \omega_{j}$ lies on the circles with diameters $O_{3} O_{i}$ and $O_{3} O_{j}$, and hence $\omega_{3}$ itself. It follows that $\omega_{3}$ is the incircle of $\triangle O_{1} O_{2} O_{4}$, so $8=r_{3}^{2}=\frac{r_{1} r_{2} r_{4}}{r_{1}+r_{2}+r_{4}}=\frac{130 r_{4}}{23+r_{4}} \Longrightarrow r_{4}=\frac{92}{61}$.
Comment: The condition $P \in \omega_{3}$ is unnecessary.
|
\frac{92}{61}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Two circles are said to be orthogonal if they intersect in two points, and their tangents at either point of intersection are perpendicular. Two circles $\omega_{1}$ and $\omega_{2}$ with radii 10 and 13 , respectively, are externally tangent at point $P$. Another circle $\omega_{3}$ with radius $2 \sqrt{2}$ passes through $P$ and is orthogonal to both $\omega_{1}$ and $\omega_{2}$. A fourth circle $\omega_{4}$, orthogonal to $\omega_{3}$, is externally tangent to $\omega_{1}$ and $\omega_{2}$. Compute the radius of $\omega_{4}$.
|
$\frac{92}{61}$ Let $\omega_{i}$ have center $O_{i}$ and radius $r_{i}$. Since $\omega_{3}$ is orthogonal to $\omega_{1}, \omega_{2}, \omega_{4}$, it has equal power $r_{3}^{2}$ to each of them. Thus $O_{3}$ is the radical center of $\omega_{1}, \omega_{2}, \omega_{4}$, which is equidistant to the three sides of $\triangle O_{1} O_{2} O_{4}$ and therefore its incenter.
For distinct $i, j \in\{1,2,4\}, \omega_{i} \cap \omega_{j}$ lies on the circles with diameters $O_{3} O_{i}$ and $O_{3} O_{j}$, and hence $\omega_{3}$ itself. It follows that $\omega_{3}$ is the incircle of $\triangle O_{1} O_{2} O_{4}$, so $8=r_{3}^{2}=\frac{r_{1} r_{2} r_{4}}{r_{1}+r_{2}+r_{4}}=\frac{130 r_{4}}{23+r_{4}} \Longrightarrow r_{4}=\frac{92}{61}$.
Comment: The condition $P \in \omega_{3}$ is unnecessary.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-geo-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nAnswer: "
}
|
c964eef9-6c4a-55b4-98fa-b1b9f7e8d220
| 609,239
|
Let $A B C$ be a triangle with $A B=13, B C=14$, and $C A=15$. Let $\Gamma$ be the circumcircle of $A B C$, let $O$ be its circumcenter, and let $M$ be the midpoint of minor arc $\widehat{B C}$. Circle $\omega_{1}$ is internally tangent to $\Gamma$ at $A$, and circle $\omega_{2}$, centered at $M$, is externally tangent to $\omega_{1}$ at a point $T$. Ray $A T$ meets segment $B C$ at point $S$, such that $B S-C S=4 / 15$. Find the radius of $\omega_{2}$.
|
$\quad 1235 / 108$ Let $N$ be the midpoint of $B C$. Notice that $B S-C S=\frac{4}{15}$ means that $N S=\frac{2}{15}$. let lines $M N$ and $A S$ meet at $P$, and let $D$ be the foot of the altitude from $A$ to $B C$. Then $B D=5$ and $A D=12$, so $D N=2$ and $D S=\frac{32}{15}$. Thus $N P=A D \frac{S N}{S D}=12 \frac{2 / 15}{32 / 15}=\frac{3}{4}$. Now $O B=R=\frac{a b c}{4 A}=\frac{(13)(14)(15)}{4(84)}=\frac{65}{8}$, so $O N=\sqrt{O B^{2}-B N^{2}}=\sqrt{\left(\frac{65}{8}\right)^{2}-7^{2}}=\frac{33}{8}$. Thus $O P=\frac{27}{8}$ and $P M=O M-O P=\frac{19}{4}$. By Monge's theorem, the exsimilicenter of $\omega_{1}$ and $\Gamma$ (which is $A$ ), the insimilicenter of $\omega_{1}$ and $\omega_{2}$ (which is $T$ ), and the insimilicenter of $\omega_{2}$ and $\Gamma$ (call this $P^{\prime}$ ) are collinear. But notice that this means $P^{\prime}=O M \cap A T=P$. From this we get
$$
\frac{\text { radius of } \omega_{2}}{R}=\frac{M P}{O P}=\frac{38}{27}
$$
Thus the radius of $\omega_{2}$ is $\frac{65}{8} \cdot \frac{38}{27}=\frac{1235}{108}$.
|
\frac{1235}{108}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a triangle with $A B=13, B C=14$, and $C A=15$. Let $\Gamma$ be the circumcircle of $A B C$, let $O$ be its circumcenter, and let $M$ be the midpoint of minor arc $\widehat{B C}$. Circle $\omega_{1}$ is internally tangent to $\Gamma$ at $A$, and circle $\omega_{2}$, centered at $M$, is externally tangent to $\omega_{1}$ at a point $T$. Ray $A T$ meets segment $B C$ at point $S$, such that $B S-C S=4 / 15$. Find the radius of $\omega_{2}$.
|
$\quad 1235 / 108$ Let $N$ be the midpoint of $B C$. Notice that $B S-C S=\frac{4}{15}$ means that $N S=\frac{2}{15}$. let lines $M N$ and $A S$ meet at $P$, and let $D$ be the foot of the altitude from $A$ to $B C$. Then $B D=5$ and $A D=12$, so $D N=2$ and $D S=\frac{32}{15}$. Thus $N P=A D \frac{S N}{S D}=12 \frac{2 / 15}{32 / 15}=\frac{3}{4}$. Now $O B=R=\frac{a b c}{4 A}=\frac{(13)(14)(15)}{4(84)}=\frac{65}{8}$, so $O N=\sqrt{O B^{2}-B N^{2}}=\sqrt{\left(\frac{65}{8}\right)^{2}-7^{2}}=\frac{33}{8}$. Thus $O P=\frac{27}{8}$ and $P M=O M-O P=\frac{19}{4}$. By Monge's theorem, the exsimilicenter of $\omega_{1}$ and $\Gamma$ (which is $A$ ), the insimilicenter of $\omega_{1}$ and $\omega_{2}$ (which is $T$ ), and the insimilicenter of $\omega_{2}$ and $\Gamma$ (call this $P^{\prime}$ ) are collinear. But notice that this means $P^{\prime}=O M \cap A T=P$. From this we get
$$
\frac{\text { radius of } \omega_{2}}{R}=\frac{M P}{O P}=\frac{38}{27}
$$
Thus the radius of $\omega_{2}$ is $\frac{65}{8} \cdot \frac{38}{27}=\frac{1235}{108}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-geo-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nAnswer: "
}
|
97ed65a6-3ea0-56ba-9e9f-a9594f22536a
| 609,240
|
Compute the prime factorization of 159999.
|
$3 \cdot 7 \cdot 19 \cdot 401$ We have $159999=160000-1=20^{4}-1=(20-1)(20+1)\left(20^{2}+1\right)=$ 3•7•19•401.
|
3 \cdot 7 \cdot 19 \cdot 401
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Compute the prime factorization of 159999.
|
$3 \cdot 7 \cdot 19 \cdot 401$ We have $159999=160000-1=20^{4}-1=(20-1)(20+1)\left(20^{2}+1\right)=$ 3•7•19•401.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n1. [4]",
"solution_match": "\nAnswer: "
}
|
9a7f2edc-9d5b-5fec-b490-75c344e8a0ae
| 609,241
|
Let $x_{1}, \ldots, x_{100}$ be defined so that for each $i, x_{i}$ is a (uniformly) random integer between 1 and 6 inclusive. Find the expected number of integers in the set $\left\{x_{1}, x_{1}+x_{2}, \ldots, x_{1}+x_{2}+\ldots+x_{100}\right\}$ that are multiples of 6 .
|
$\boxed{\frac{50}{3}}$ Note that for any $i$, the probability that $x_1 + x_2 + ... + x_i$ is a multiple of 6 is $\frac{1}{6}$ because exactly 1 value out of 6 possible values of $x_{i}$ works. Because these 100 events are independent, the expected value is $100 \cdot \frac{1}{6}=\frac{50}{3}$.
|
\frac{50}{3}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $x_{1}, \ldots, x_{100}$ be defined so that for each $i, x_{i}$ is a (uniformly) random integer between 1 and 6 inclusive. Find the expected number of integers in the set $\left\{x_{1}, x_{1}+x_{2}, \ldots, x_{1}+x_{2}+\ldots+x_{100}\right\}$ that are multiples of 6 .
|
$\boxed{\frac{50}{3}}$ Note that for any $i$, the probability that $x_1 + x_2 + ... + x_i$ is a multiple of 6 is $\frac{1}{6}$ because exactly 1 value out of 6 possible values of $x_{i}$ works. Because these 100 events are independent, the expected value is $100 \cdot \frac{1}{6}=\frac{50}{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n2. [4]",
"solution_match": "\nAnswer: "
}
|
c15104e4-1e22-5505-952d-72bd28c60ded
| 609,242
|
Let $A B C D E F$ be a regular hexagon. Let $P$ be the circle inscribed in $\triangle B D F$. Find the ratio of the area of circle $P$ to the area of rectangle $A B D E$.
|
$\quad \frac{\pi \sqrt{3}}{12}$ Let the side length of the hexagon be $s$. The length of $B D$ is $s \sqrt{3}$, so the area of rectangle $A B D E$ is $s^{2} \sqrt{3}$. Equilateral triangle $B D F$ has side length $s \sqrt{3}$. The inradius of an equilateral triangle is $\sqrt{3} / 6$ times the length of its side, and so has length $\frac{s}{2}$. Thus, the area of circle $P$ is $\frac{\pi s^{2}}{4}$, so the ratio is $\frac{\pi s^{2} / 4}{s^{2} \sqrt{3}}=\frac{\pi \sqrt{3}}{12}$.
|
\frac{\pi \sqrt{3}}{12}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C D E F$ be a regular hexagon. Let $P$ be the circle inscribed in $\triangle B D F$. Find the ratio of the area of circle $P$ to the area of rectangle $A B D E$.
|
$\quad \frac{\pi \sqrt{3}}{12}$ Let the side length of the hexagon be $s$. The length of $B D$ is $s \sqrt{3}$, so the area of rectangle $A B D E$ is $s^{2} \sqrt{3}$. Equilateral triangle $B D F$ has side length $s \sqrt{3}$. The inradius of an equilateral triangle is $\sqrt{3} / 6$ times the length of its side, and so has length $\frac{s}{2}$. Thus, the area of circle $P$ is $\frac{\pi s^{2}}{4}$, so the ratio is $\frac{\pi s^{2} / 4}{s^{2} \sqrt{3}}=\frac{\pi \sqrt{3}}{12}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n3. [4]",
"solution_match": "\nAnswer: "
}
|
ff682ecc-2285-5a7b-9422-c2bbac3fb399
| 609,243
|
Let $D$ be the set of divisors of 100 . Let $Z$ be the set of integers between 1 and 100, inclusive. Mark chooses an element $d$ of $D$ and an element $z$ of $Z$ uniformly at random. What is the probability that $d$ divides $z$ ?
|
$\frac{217}{900}$ As $100=2^{2} \cdot 5^{2}$, there are $3 \cdot 3=9$ divisors of 100 , so there are 900 possible pairs of $d$ and $z$ that can be chosen.
If $d$ is chosen, then there are $\frac{100}{d}$ possible values of $z$ such that $d$ divides $z$, so the total number of valid pairs of $d$ and $z$ is $\sum_{d \mid 100} \frac{100}{d}=\sum_{d \mid 100} d=\left(1+2+2^{2}\right)\left(1+5+5^{2}\right)=7 \cdot 31=217$. The answer is therefore $\frac{217}{900}$.
|
\frac{217}{900}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $D$ be the set of divisors of 100 . Let $Z$ be the set of integers between 1 and 100, inclusive. Mark chooses an element $d$ of $D$ and an element $z$ of $Z$ uniformly at random. What is the probability that $d$ divides $z$ ?
|
$\frac{217}{900}$ As $100=2^{2} \cdot 5^{2}$, there are $3 \cdot 3=9$ divisors of 100 , so there are 900 possible pairs of $d$ and $z$ that can be chosen.
If $d$ is chosen, then there are $\frac{100}{d}$ possible values of $z$ such that $d$ divides $z$, so the total number of valid pairs of $d$ and $z$ is $\sum_{d \mid 100} \frac{100}{d}=\sum_{d \mid 100} d=\left(1+2+2^{2}\right)\left(1+5+5^{2}\right)=7 \cdot 31=217$. The answer is therefore $\frac{217}{900}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n4. [4]",
"solution_match": "\nAnswer: "
}
|
b65d7ee7-b007-5baa-8008-b1524c8eb154
| 609,244
|
If four fair six-sided dice are rolled, what is the probability that the lowest number appearing on any die is exactly 3 ?
|
$175 / 1296$ The probability that all the die rolls are at least 3 is $\frac{4^{4}}{6}$. The probability they are all at least 4 is $\frac{3^{4}}{6}$. The probability of being in the former category but not the latter is thus $\frac{4}{6}^{4}-\frac{3}{6}^{4}=\frac{256-81}{1296}=\frac{175}{1296}$.
|
\frac{175}{1296}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
If four fair six-sided dice are rolled, what is the probability that the lowest number appearing on any die is exactly 3 ?
|
$175 / 1296$ The probability that all the die rolls are at least 3 is $\frac{4^{4}}{6}$. The probability they are all at least 4 is $\frac{3^{4}}{6}$. The probability of being in the former category but not the latter is thus $\frac{4}{6}^{4}-\frac{3}{6}^{4}=\frac{256-81}{1296}=\frac{175}{1296}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n5. [5]",
"solution_match": "\nAnswer: "
}
|
bd94bff9-34ea-5fbe-801a-15f47a0ec8c0
| 609,245
|
The Evil League of Evil is plotting to poison the city's water supply. They plan to set out from their headquarters at $(5,1)$ and put poison in two pipes, one along the line $y=x$ and one along the line $x=7$. However, they need to get the job done quickly before Captain Hammer catches them. What's the shortest distance they can travel to visit both pipes and then return to their headquarters?
|
$4 \sqrt{5}$ After they go to $y=x$, we reflect the remainder of their path in $y=x$, along with the second pipe and their headquarters. Now, they must go from $(5,1)$ to $y=7$ crossing $y=x$,
and then go to $(1,5)$. When they reach $y=7$, we reflect the remainder of their path again, so now their reflected headquarters is at $(1,9)$. Thus, they just go from $(5,1)$ to $(1,9)$ in some path that inevitably crosses $y=x$ and $y=7$. The shortest path they can take is a straight line with length $\sqrt{4^{2}+8^{2}}=4 \sqrt{5}$.
Comment. These ideas can be used to prove that the orthic triangle of an acute triangle has the smallest possible perimeter of all inscribed triangles.
Also, see if you can find an alternative solution using Minkowski's inequality!
|
4 \sqrt{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
The Evil League of Evil is plotting to poison the city's water supply. They plan to set out from their headquarters at $(5,1)$ and put poison in two pipes, one along the line $y=x$ and one along the line $x=7$. However, they need to get the job done quickly before Captain Hammer catches them. What's the shortest distance they can travel to visit both pipes and then return to their headquarters?
|
$4 \sqrt{5}$ After they go to $y=x$, we reflect the remainder of their path in $y=x$, along with the second pipe and their headquarters. Now, they must go from $(5,1)$ to $y=7$ crossing $y=x$,
and then go to $(1,5)$. When they reach $y=7$, we reflect the remainder of their path again, so now their reflected headquarters is at $(1,9)$. Thus, they just go from $(5,1)$ to $(1,9)$ in some path that inevitably crosses $y=x$ and $y=7$. The shortest path they can take is a straight line with length $\sqrt{4^{2}+8^{2}}=4 \sqrt{5}$.
Comment. These ideas can be used to prove that the orthic triangle of an acute triangle has the smallest possible perimeter of all inscribed triangles.
Also, see if you can find an alternative solution using Minkowski's inequality!
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n7. [5]",
"solution_match": "\nAnswer: "
}
|
57f5c52f-8d5f-510a-af94-1eb8054c1c63
| 609,247
|
The numbers $2^{0}, 2^{1}, \cdots, 2^{15}, 2^{16}=65536$ are written on a blackboard. You repeatedly take two numbers on the blackboard, subtract one from the other, erase them both, and write the result of the subtraction on the blackboard. What is the largest possible number that can remain on the blackboard when there is only one number left?
|
131069 If we reverse the order of the numbers in the final subtraction we perform, then the final number will be negated. Thus, the possible final numbers come in pairs with opposite signs. Therefore, the largest possible number is the negative of the smallest possible number. To get the smallest possible number, clearly we can take the smallest number originally on the board and subtract all of the other numbers from it (you can make this rigorous pretty easily if needed), so the smallest possible number is $1-\sum_{k=1}^{16} 2^{k}=1-131070=-131069$, and thus the largest possible number is 131069 .
|
131069
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
The numbers $2^{0}, 2^{1}, \cdots, 2^{15}, 2^{16}=65536$ are written on a blackboard. You repeatedly take two numbers on the blackboard, subtract one from the other, erase them both, and write the result of the subtraction on the blackboard. What is the largest possible number that can remain on the blackboard when there is only one number left?
|
131069 If we reverse the order of the numbers in the final subtraction we perform, then the final number will be negated. Thus, the possible final numbers come in pairs with opposite signs. Therefore, the largest possible number is the negative of the smallest possible number. To get the smallest possible number, clearly we can take the smallest number originally on the board and subtract all of the other numbers from it (you can make this rigorous pretty easily if needed), so the smallest possible number is $1-\sum_{k=1}^{16} 2^{k}=1-131070=-131069$, and thus the largest possible number is 131069 .
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n8. [5]",
"solution_match": "\nAnswer: "
}
|
dab14527-8f72-5613-8382-2e652cd19c89
| 609,248
|
Compute the side length of the largest cube contained in the region
$$
\left\{(x, y, z): x^{2}+y^{2}+z^{2} \leq 25 \text { and } x \geq 0\right\}
$$
of three-dimensional space.
|
$\sqrt{\frac{5 \sqrt{6}}{3}}$ The given region is a hemisphere, so the largest cube that can fit inside it has one face centered at the origin and the four vertices of the opposite face on the spherical surface. Let the side length of this cube be $s$. Then, the radius of the circle is the hypotenuse of a triangle with side lengths $s$ and $\frac{\sqrt{2}}{2} s$. So, by the Pythagorean Theorem, the radius equals $\frac{\sqrt{6}}{2} s$. Since the radius of the hemisphere is 5 , the side length of the cube is $\frac{5 \sqrt{6}}{3}$.
|
\frac{5 \sqrt{6}}{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Compute the side length of the largest cube contained in the region
$$
\left\{(x, y, z): x^{2}+y^{2}+z^{2} \leq 25 \text { and } x \geq 0\right\}
$$
of three-dimensional space.
|
$\sqrt{\frac{5 \sqrt{6}}{3}}$ The given region is a hemisphere, so the largest cube that can fit inside it has one face centered at the origin and the four vertices of the opposite face on the spherical surface. Let the side length of this cube be $s$. Then, the radius of the circle is the hypotenuse of a triangle with side lengths $s$ and $\frac{\sqrt{2}}{2} s$. So, by the Pythagorean Theorem, the radius equals $\frac{\sqrt{6}}{2} s$. Since the radius of the hemisphere is 5 , the side length of the cube is $\frac{5 \sqrt{6}}{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n9. [6]",
"solution_match": "\nAnswer: "
}
|
ad3387bf-3eaa-5834-885f-d4e89199ac75
| 609,249
|
Find the number of nonempty sets $\mathcal{F}$ of subsets of the set $\{1, \ldots, 2014\}$ such that:
(a) For any subsets $S_{1}, S_{2} \in \mathcal{F}, S_{1} \cap S_{2} \in \mathcal{F}$.
(b) If $S \in \mathcal{F}, T \subseteq\{1, \ldots, 2014\}$, and $S \subseteq T$, then $T \in \mathcal{F}$.
|
$2^{2014}$ For a subset $S$ of $\{1, \ldots, 2014\}$, let $\mathcal{F}_{S}$ be the set of all sets $T$ such that $S \subseteq T \subseteq$ $\{1, \ldots, 2014\}$. It can be checked that the sets $\mathcal{F}_{S}$ satisfy the conditions 1 and 2 . We claim that the $\mathcal{F}_{S}$ are the only sets of subsets of $\{1, \ldots, 2014\}$ satisfying the conditions 1 and 2 . (Thus, the answer is the number of subsets $S$ of $\{1, \ldots, 2014\}$, which is $2^{2014}$.)
Suppose that $\mathcal{F}$ satisfies the conditions 1 and 2 , and let $S$ be the intersection of all the sets of $\mathcal{F}$. We claim that $\mathcal{F}=\mathcal{F}_{S}$. First, by definition of $S$, all elements $T \in \mathcal{F}$ are supersets of $S$, so $\mathcal{F} \subseteq \mathcal{F}_{S}$. On the other hand, by iterating condition 1 , it follows that $S$ is an element of $\mathcal{F}$, so by condition 2 any set $T$ with $S \subseteq T \subseteq\{1, \ldots, 2014\}$ is an element of $\mathcal{F}$. So $\mathcal{F} \supseteq \mathcal{F}_{S}$. Thus $\mathcal{F}=\mathcal{F}_{S}$.
|
2^{2014}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Find the number of nonempty sets $\mathcal{F}$ of subsets of the set $\{1, \ldots, 2014\}$ such that:
(a) For any subsets $S_{1}, S_{2} \in \mathcal{F}, S_{1} \cap S_{2} \in \mathcal{F}$.
(b) If $S \in \mathcal{F}, T \subseteq\{1, \ldots, 2014\}$, and $S \subseteq T$, then $T \in \mathcal{F}$.
|
$2^{2014}$ For a subset $S$ of $\{1, \ldots, 2014\}$, let $\mathcal{F}_{S}$ be the set of all sets $T$ such that $S \subseteq T \subseteq$ $\{1, \ldots, 2014\}$. It can be checked that the sets $\mathcal{F}_{S}$ satisfy the conditions 1 and 2 . We claim that the $\mathcal{F}_{S}$ are the only sets of subsets of $\{1, \ldots, 2014\}$ satisfying the conditions 1 and 2 . (Thus, the answer is the number of subsets $S$ of $\{1, \ldots, 2014\}$, which is $2^{2014}$.)
Suppose that $\mathcal{F}$ satisfies the conditions 1 and 2 , and let $S$ be the intersection of all the sets of $\mathcal{F}$. We claim that $\mathcal{F}=\mathcal{F}_{S}$. First, by definition of $S$, all elements $T \in \mathcal{F}$ are supersets of $S$, so $\mathcal{F} \subseteq \mathcal{F}_{S}$. On the other hand, by iterating condition 1 , it follows that $S$ is an element of $\mathcal{F}$, so by condition 2 any set $T$ with $S \subseteq T \subseteq\{1, \ldots, 2014\}$ is an element of $\mathcal{F}$. So $\mathcal{F} \supseteq \mathcal{F}_{S}$. Thus $\mathcal{F}=\mathcal{F}_{S}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n10. [6]",
"solution_match": "\nAnswer: "
}
|
3056c047-44ae-5769-947b-9242dd135abf
| 609,250
|
Two fair octahedral dice, each with the numbers 1 through 8 on their faces, are rolled. Let $N$ be the remainder when the product of the numbers showing on the two dice is divided by 8 . Find the expected value of $N$.
|
$\frac{11}{4}$ If the first die is odd, which has $\frac{1}{2}$ probability, then $N$ can be any of $0,1,2,3$, $4,5,6,7$ with equal probability, because multiplying each element of $\{0, \ldots, 7\}$ with an odd number and taking modulo 8 results in the same numbers, as all odd numbers are relatively prime to 8 . The expected value in this case is 3.5 .
If the first die is even but not a multiple of 4 , which has $\frac{1}{4}$ probability, then using similar reasoning, $N$ can be any of $0,2,4,6$ with equal probability, so the expected value is 3 .
If the first die is 4 , which has $\frac{1}{8}$ probability, then $N$ can by any of 0,4 with equal probability, so the expected value is 2 .
Finally, if the first die is 8 , which has $\frac{1}{8}$ probability, then $N=0$. The total expected value is $\frac{1}{2}(3.5)+$ $\frac{1}{4}(3)+\frac{1}{8}(2)+\frac{1}{8}(0)=\frac{11}{4}$.
|
\frac{11}{4}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Two fair octahedral dice, each with the numbers 1 through 8 on their faces, are rolled. Let $N$ be the remainder when the product of the numbers showing on the two dice is divided by 8 . Find the expected value of $N$.
|
$\frac{11}{4}$ If the first die is odd, which has $\frac{1}{2}$ probability, then $N$ can be any of $0,1,2,3$, $4,5,6,7$ with equal probability, because multiplying each element of $\{0, \ldots, 7\}$ with an odd number and taking modulo 8 results in the same numbers, as all odd numbers are relatively prime to 8 . The expected value in this case is 3.5 .
If the first die is even but not a multiple of 4 , which has $\frac{1}{4}$ probability, then using similar reasoning, $N$ can be any of $0,2,4,6$ with equal probability, so the expected value is 3 .
If the first die is 4 , which has $\frac{1}{8}$ probability, then $N$ can by any of 0,4 with equal probability, so the expected value is 2 .
Finally, if the first die is 8 , which has $\frac{1}{8}$ probability, then $N=0$. The total expected value is $\frac{1}{2}(3.5)+$ $\frac{1}{4}(3)+\frac{1}{8}(2)+\frac{1}{8}(0)=\frac{11}{4}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n11. [6]",
"solution_match": "\nAnswer: "
}
|
9c9103b2-c63d-5459-b5e5-8f477943d7e2
| 609,251
|
Find a nonzero monic polynomial $P(x)$ with integer coefficients and minimal degree such that $P(1-\sqrt[3]{2}+\sqrt[3]{4})=0$. (A polynomial is called monic if its leading coefficient is 1.)
|
$x^{3}-3 x^{2}+9 x-9$ Note that $(1-\sqrt[3]{2}+\sqrt[3]{4})(1+\sqrt[3]{2})=3$, so $1-\sqrt[3]{2}+\sqrt[3]{4}=\frac{3}{1+\sqrt[3]{2}}$.
Now, if $f(x)=x^{3}-2$, we have $f(\sqrt[3]{2})=0$, so if we let $g(x)=f(x-1)=(x-1)^{3}-2=x^{3}-3 x^{2}+3 x-3$, then $g(1+\sqrt[3]{2})=f(\sqrt[3]{2})=0$. Finally, we let $h(x)=g\left(\frac{3}{x}\right)=\frac{27}{x^{3}}-\frac{27}{x^{2}}+\frac{9}{x}-3$ so $h\left(\frac{3}{1+\sqrt[3]{2}}\right)=g(1+\sqrt[3]{2})=0$. To make this a monic polynomial, we multiply $h(x)$ by $-\frac{x^{3}}{3}$ to get $x^{3}-3 x^{2}+9 x-9$.
|
x^{3}-3 x^{2}+9 x-9
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find a nonzero monic polynomial $P(x)$ with integer coefficients and minimal degree such that $P(1-\sqrt[3]{2}+\sqrt[3]{4})=0$. (A polynomial is called monic if its leading coefficient is 1.)
|
$x^{3}-3 x^{2}+9 x-9$ Note that $(1-\sqrt[3]{2}+\sqrt[3]{4})(1+\sqrt[3]{2})=3$, so $1-\sqrt[3]{2}+\sqrt[3]{4}=\frac{3}{1+\sqrt[3]{2}}$.
Now, if $f(x)=x^{3}-2$, we have $f(\sqrt[3]{2})=0$, so if we let $g(x)=f(x-1)=(x-1)^{3}-2=x^{3}-3 x^{2}+3 x-3$, then $g(1+\sqrt[3]{2})=f(\sqrt[3]{2})=0$. Finally, we let $h(x)=g\left(\frac{3}{x}\right)=\frac{27}{x^{3}}-\frac{27}{x^{2}}+\frac{9}{x}-3$ so $h\left(\frac{3}{1+\sqrt[3]{2}}\right)=g(1+\sqrt[3]{2})=0$. To make this a monic polynomial, we multiply $h(x)$ by $-\frac{x^{3}}{3}$ to get $x^{3}-3 x^{2}+9 x-9$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n12. [6]",
"solution_match": "\nAnswer: "
}
|
991e5429-ad25-5df7-9ac9-563bcabcc7db
| 609,252
|
Let $A B C D$ be a trapezoid with $A B \| C D$ and $\angle D=90^{\circ}$. Suppose that there is a point $E$ on $C D$ such that $A E=B E$ and that triangles $A E D$ and $C E B$ are similar, but not congruent. Given that $\frac{C D}{A B}=2014$, find $\frac{B C}{A D}$.
|
$\sqrt{4027}$ Let $M$ be the midpoint of $A B$. Let $A M=M B=E D=a, M E=A D=b$, and $A E=B E=c$. Since $\triangle B E C \sim \triangle D E A$, but $\triangle B E C$ is not congruent to $\triangle D A E$, we must have $\triangle B E C \sim \triangle D E A$. Thus, $B C / B E=A D / D E=b / a$, so $B C=b c / a$, and $C E / E B=A E / E D=c / a$, so $E C=c^{2} / a$. We are given that $C D / A B=\frac{\frac{c^{2}}{a}+a}{2 a}=\frac{c^{2}}{2 a^{2}}+\frac{1}{2}=2014 \Rightarrow \frac{c^{2}}{a^{2}}=4027$. Thus, $B C / A D=\frac{b c / a}{b}=c / a=\sqrt{4027}$.
|
\sqrt{4027}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C D$ be a trapezoid with $A B \| C D$ and $\angle D=90^{\circ}$. Suppose that there is a point $E$ on $C D$ such that $A E=B E$ and that triangles $A E D$ and $C E B$ are similar, but not congruent. Given that $\frac{C D}{A B}=2014$, find $\frac{B C}{A D}$.
|
$\sqrt{4027}$ Let $M$ be the midpoint of $A B$. Let $A M=M B=E D=a, M E=A D=b$, and $A E=B E=c$. Since $\triangle B E C \sim \triangle D E A$, but $\triangle B E C$ is not congruent to $\triangle D A E$, we must have $\triangle B E C \sim \triangle D E A$. Thus, $B C / B E=A D / D E=b / a$, so $B C=b c / a$, and $C E / E B=A E / E D=c / a$, so $E C=c^{2} / a$. We are given that $C D / A B=\frac{\frac{c^{2}}{a}+a}{2 a}=\frac{c^{2}}{2 a^{2}}+\frac{1}{2}=2014 \Rightarrow \frac{c^{2}}{a^{2}}=4027$. Thus, $B C / A D=\frac{b c / a}{b}=c / a=\sqrt{4027}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n14. [8]",
"solution_match": "\nAnswer: "
}
|
155312a7-5e50-5a91-a66b-a7b4e9689705
| 609,254
|
Given a regular pentagon of area 1, a pivot line is a line not passing through any of the pentagon's vertices such that there are 3 vertices of the pentagon on one side of the line and 2 on the other. A pivot point is a point inside the pentagon with only finitely many non-pivot lines passing through it. Find the area of the region of pivot points.
|
$\frac{1}{2}(7-3 \sqrt{5})$ Let the pentagon be labeled $A B C D E$. First, no pivot point can be on the same side of $A C$ as vertex $B$. Any such point $P$ has the infinite set of non-pivot lines within the hourglass shape formed by the acute angles between lines $P A$ and $P C$. Similar logic can be applied to points on the same side of $B D$ as $C$, and so on. The set of pivot points is thus a small pentagon with sides on $A C, B D, C E, D A, E B$. The side ratio of this small pentagon to the large pentagon is
$$
\left(2 \cos \left(72^{\circ}\right)\right)^{2}=\frac{3-\sqrt{5}}{2}
$$
so the area of the small pentagon is
$$
\left(\frac{3-\sqrt{5}}{2}\right)^{2}=\frac{1}{2}(7-3 \sqrt{5})
$$
|
\frac{1}{2}(7-3 \sqrt{5})
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Given a regular pentagon of area 1, a pivot line is a line not passing through any of the pentagon's vertices such that there are 3 vertices of the pentagon on one side of the line and 2 on the other. A pivot point is a point inside the pentagon with only finitely many non-pivot lines passing through it. Find the area of the region of pivot points.
|
$\frac{1}{2}(7-3 \sqrt{5})$ Let the pentagon be labeled $A B C D E$. First, no pivot point can be on the same side of $A C$ as vertex $B$. Any such point $P$ has the infinite set of non-pivot lines within the hourglass shape formed by the acute angles between lines $P A$ and $P C$. Similar logic can be applied to points on the same side of $B D$ as $C$, and so on. The set of pivot points is thus a small pentagon with sides on $A C, B D, C E, D A, E B$. The side ratio of this small pentagon to the large pentagon is
$$
\left(2 \cos \left(72^{\circ}\right)\right)^{2}=\frac{3-\sqrt{5}}{2}
$$
so the area of the small pentagon is
$$
\left(\frac{3-\sqrt{5}}{2}\right)^{2}=\frac{1}{2}(7-3 \sqrt{5})
$$
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n15. [8]",
"solution_match": "\nAnswer: "
}
|
6cc5bd66-bf4d-5599-ac5c-b9e2663c344a
| 609,255
|
Suppose that $x$ and $y$ are positive real numbers such that $x^{2}-x y+2 y^{2}=8$. Find the maximum possible value of $x^{2}+x y+2 y^{2}$.
|
$\frac{72+32 \sqrt{2}}{7}$ Let $u=x^{2}+2 y^{2}$. By AM-GM, $u \geq \sqrt{8} x y$, so $x y \leq \frac{u}{\sqrt{8}}$. If we let $x y=k u$ where $k \leq \frac{1}{\sqrt{8}}$, then we have
$$
\begin{gathered}
u(1-k)=8 \\
u(1+k)=x^{2}+x y+2 y^{2}
\end{gathered}
$$
that is, $u(1+k)=8 \cdot \frac{1+k}{1-k}$. It is not hard to see that the maximum value of this expression occurs at $k=\frac{1}{\sqrt{8}}$, so the maximum value is $8 \cdot \frac{1+\frac{1}{\sqrt{8}}}{1-\frac{1}{\sqrt{8}}}=\frac{72+32 \sqrt{2}}{7}$.
|
\frac{72+32 \sqrt{2}}{7}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose that $x$ and $y$ are positive real numbers such that $x^{2}-x y+2 y^{2}=8$. Find the maximum possible value of $x^{2}+x y+2 y^{2}$.
|
$\frac{72+32 \sqrt{2}}{7}$ Let $u=x^{2}+2 y^{2}$. By AM-GM, $u \geq \sqrt{8} x y$, so $x y \leq \frac{u}{\sqrt{8}}$. If we let $x y=k u$ where $k \leq \frac{1}{\sqrt{8}}$, then we have
$$
\begin{gathered}
u(1-k)=8 \\
u(1+k)=x^{2}+x y+2 y^{2}
\end{gathered}
$$
that is, $u(1+k)=8 \cdot \frac{1+k}{1-k}$. It is not hard to see that the maximum value of this expression occurs at $k=\frac{1}{\sqrt{8}}$, so the maximum value is $8 \cdot \frac{1+\frac{1}{\sqrt{8}}}{1-\frac{1}{\sqrt{8}}}=\frac{72+32 \sqrt{2}}{7}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n16. [8]",
"solution_match": "\nAnswer: "
}
|
f9d605e7-fdec-509f-a494-0cebff1d0c8d
| 609,256
|
Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a function satisfying the following conditions:
(a) $f(1)=1$.
(b) $f(a) \leq f(b)$ whenever $a$ and $b$ are positive integers with $a \leq b$.
(c) $f(2 a)=f(a)+1$ for all positive integers $a$.
How many possible values can the 2014-tuple $(f(1), f(2), \ldots, f(2014))$ take?
|
1007 Note that $f(2014)=f(1007)+1$, so there must be exactly one index $1008 \leq i \leq 2014$ such that $f(i)=f(i-1)+1$, and for all $1008 \leq j \leq 2014, j \neq i$ we must have $f(j)=f(j-1)$. We first claim that each value of $i$ corresponds to exactly one 2014-tuple $(f(1), \ldots, f(2014))$. To prove this, note that $f(1024)=11$, so each $i$ uniquely determines the values of $f(1007), \ldots, f(2014)$. Then all of $f(1), \ldots, f(1006)$ can be uniquely determined from these values because for any $1 \leq k \leq 1006$, there exists a unique $n$ such that $1007 \leq k \cdot 2^{n} \leq 2014$. It's also clear that these values satisfy the condition that $f$ is nondecreasing, so we have a correspondence from each $1008 \leq i \leq 2014$ to a unique 2014-tuple.
Also, given any valid 2014-tuple $(f(1), \ldots, f(2014)$ ), we know that $f(1), \ldots, f(1006)$ can be uniquely determined by $f(1007), \ldots, f(2014)$, which yields some $1008 \leq i \leq 2014$ where $f(i)=f(i-1)+1$, so we actually have a bijection between possible values of $i$ and 2014-tuples. Therefore, the total number of possible 2014-tuples is 1007 .
|
1007
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a function satisfying the following conditions:
(a) $f(1)=1$.
(b) $f(a) \leq f(b)$ whenever $a$ and $b$ are positive integers with $a \leq b$.
(c) $f(2 a)=f(a)+1$ for all positive integers $a$.
How many possible values can the 2014-tuple $(f(1), f(2), \ldots, f(2014))$ take?
|
1007 Note that $f(2014)=f(1007)+1$, so there must be exactly one index $1008 \leq i \leq 2014$ such that $f(i)=f(i-1)+1$, and for all $1008 \leq j \leq 2014, j \neq i$ we must have $f(j)=f(j-1)$. We first claim that each value of $i$ corresponds to exactly one 2014-tuple $(f(1), \ldots, f(2014))$. To prove this, note that $f(1024)=11$, so each $i$ uniquely determines the values of $f(1007), \ldots, f(2014)$. Then all of $f(1), \ldots, f(1006)$ can be uniquely determined from these values because for any $1 \leq k \leq 1006$, there exists a unique $n$ such that $1007 \leq k \cdot 2^{n} \leq 2014$. It's also clear that these values satisfy the condition that $f$ is nondecreasing, so we have a correspondence from each $1008 \leq i \leq 2014$ to a unique 2014-tuple.
Also, given any valid 2014-tuple $(f(1), \ldots, f(2014)$ ), we know that $f(1), \ldots, f(1006)$ can be uniquely determined by $f(1007), \ldots, f(2014)$, which yields some $1008 \leq i \leq 2014$ where $f(i)=f(i-1)+1$, so we actually have a bijection between possible values of $i$ and 2014-tuples. Therefore, the total number of possible 2014-tuples is 1007 .
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n17. [11]",
"solution_match": "\nAnswer: "
}
|
544ce527-8140-5111-9caa-1c02c628d9c9
| 609,257
|
Find the number of ordered quadruples of positive integers $(a, b, c, d)$ such that $a, b, c$, and $d$ are all (not necessarily distinct) factors of 30 and $a b c d>900$.
|
1940 Since $a b c d>900 \Longleftrightarrow \frac{30}{a} \frac{30}{b} \frac{30}{c} \frac{30}{d}<900$, and there are $\binom{4}{2}^{3}$ solutions to $a b c d=2^{2} 3^{2} 5^{2}$, the answer is $\frac{1}{2}\left(8^{4}-\binom{4}{2}^{3}\right)=1940$ by symmetry.
|
1940
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the number of ordered quadruples of positive integers $(a, b, c, d)$ such that $a, b, c$, and $d$ are all (not necessarily distinct) factors of 30 and $a b c d>900$.
|
1940 Since $a b c d>900 \Longleftrightarrow \frac{30}{a} \frac{30}{b} \frac{30}{c} \frac{30}{d}<900$, and there are $\binom{4}{2}^{3}$ solutions to $a b c d=2^{2} 3^{2} 5^{2}$, the answer is $\frac{1}{2}\left(8^{4}-\binom{4}{2}^{3}\right)=1940$ by symmetry.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n18. [11]",
"solution_match": "\nAnswer: "
}
|
d3a90b05-8812-5518-8f45-065a17674f04
| 609,258
|
Let $A B C D$ be a trapezoid with $A B \| C D$. The bisectors of $\angle C D A$ and $\angle D A B$ meet at $E$, the bisectors of $\angle A B C$ and $\angle B C D$ meet at $F$, the bisectors of $\angle B C D$ and $\angle C D A$ meet at $G$, and the bisectors of $\angle D A B$ and $\angle A B C$ meet at $H$. Quadrilaterals $E A B F$ and $E D C F$ have areas 24 and 36, respectively, and triangle $A B H$ has area 25 . Find the area of triangle $C D G$.
|
$\boxed{\frac{256}{7}}$ Let $M, N$ be the midpoints of $AD, BC$ respectively. Since $AE$ and $DE$ are bisectors of supplementary angles, the triangle $A E D$ is right with right angle $E$. Then $E M$ is the median of a right triangle from the right angle, so triangles $E M A$ and $E M D$ are isosceles with vertex $M$. But then $\angle M E A=\angle E A M=\angle E A B$, so $E M \| A B$. Similarly, $F N \| B A$. Thus, both $E$ and $F$ are on the midline of this trapezoid. Let the length of $E F$ be $x$. Triangle $E F H$ has area 1 and is similar to triangle $A B H$, which has area 25 , so $A B=5 x$. Then, letting the heights of trapezoids $E A B F$ and $E D C F$ be $h$ (they are equal since $E F$ is on the midline), the area of trapezoid $E A B F$ is $\frac{6 x h}{2}=24$. So the area of trapezoid $E D C F$ is $36=\frac{9 x h}{2}$. Thus $D C=8 x$. Then, triangle $G E F$ is similar to and has $\frac{1}{64}$ times the area of triangle $C D G$. So the area of triangle $C D G$ is $\frac{64}{63}$ times the area of quadrilateral $E D C F$, or $\frac{256}{7}$.
|
\frac{256}{7}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C D$ be a trapezoid with $A B \| C D$. The bisectors of $\angle C D A$ and $\angle D A B$ meet at $E$, the bisectors of $\angle A B C$ and $\angle B C D$ meet at $F$, the bisectors of $\angle B C D$ and $\angle C D A$ meet at $G$, and the bisectors of $\angle D A B$ and $\angle A B C$ meet at $H$. Quadrilaterals $E A B F$ and $E D C F$ have areas 24 and 36, respectively, and triangle $A B H$ has area 25 . Find the area of triangle $C D G$.
|
$\boxed{\frac{256}{7}}$ Let $M, N$ be the midpoints of $AD, BC$ respectively. Since $AE$ and $DE$ are bisectors of supplementary angles, the triangle $A E D$ is right with right angle $E$. Then $E M$ is the median of a right triangle from the right angle, so triangles $E M A$ and $E M D$ are isosceles with vertex $M$. But then $\angle M E A=\angle E A M=\angle E A B$, so $E M \| A B$. Similarly, $F N \| B A$. Thus, both $E$ and $F$ are on the midline of this trapezoid. Let the length of $E F$ be $x$. Triangle $E F H$ has area 1 and is similar to triangle $A B H$, which has area 25 , so $A B=5 x$. Then, letting the heights of trapezoids $E A B F$ and $E D C F$ be $h$ (they are equal since $E F$ is on the midline), the area of trapezoid $E A B F$ is $\frac{6 x h}{2}=24$. So the area of trapezoid $E D C F$ is $36=\frac{9 x h}{2}$. Thus $D C=8 x$. Then, triangle $G E F$ is similar to and has $\frac{1}{64}$ times the area of triangle $C D G$. So the area of triangle $C D G$ is $\frac{64}{63}$ times the area of quadrilateral $E D C F$, or $\frac{256}{7}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n19. [11]",
"solution_match": "\nAnswer: "
}
|
287f7f4a-849c-56ec-8480-a15b173d95bf
| 609,259
|
A deck of 8056 cards has 2014 ranks numbered 1-2014. Each rank has four suits-hearts, diamonds, clubs, and spades. Each card has a rank and a suit, and no two cards have the same rank and the same suit. How many subsets of the set of cards in this deck have cards from an odd number of distinct ranks?
|
$\frac{1}{2}\left(16^{2014}-14^{2014}\right)$ There are $\binom{2014}{k}$ ways to pick $k$ ranks, and 15 ways to pick the suits in each rank (because there are 16 subsets of suits, and we must exclude the empty one). We therefore want to evaluate the sum $\binom{2014}{1} 15^{1}+\binom{2014}{3} 15^{3}+\cdots+\binom{2014}{2013} 15^{2013}$.
Note that $(1+15)^{2014}=1+\binom{2014}{1} 15^{1}+\binom{2014}{2} 15^{2}+\ldots+\binom{2014}{2013} 15^{2013}+15^{2014}$ and $(1-15)^{2014}=$ $1-\binom{2014}{1} 15^{1}+\binom{2014}{2} 15^{2}-\ldots-\binom{2014}{2013} 15^{2013}+15^{2014}$, so our sum is simply $\frac{(1+15)^{2014}-(1-15)^{2014}}{2}=$ $\frac{1}{2}\left(16^{2014}-14^{2014}\right)$.
|
\frac{1}{2}\left(16^{2014}-14^{2014}\right)
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A deck of 8056 cards has 2014 ranks numbered 1-2014. Each rank has four suits-hearts, diamonds, clubs, and spades. Each card has a rank and a suit, and no two cards have the same rank and the same suit. How many subsets of the set of cards in this deck have cards from an odd number of distinct ranks?
|
$\frac{1}{2}\left(16^{2014}-14^{2014}\right)$ There are $\binom{2014}{k}$ ways to pick $k$ ranks, and 15 ways to pick the suits in each rank (because there are 16 subsets of suits, and we must exclude the empty one). We therefore want to evaluate the sum $\binom{2014}{1} 15^{1}+\binom{2014}{3} 15^{3}+\cdots+\binom{2014}{2013} 15^{2013}$.
Note that $(1+15)^{2014}=1+\binom{2014}{1} 15^{1}+\binom{2014}{2} 15^{2}+\ldots+\binom{2014}{2013} 15^{2013}+15^{2014}$ and $(1-15)^{2014}=$ $1-\binom{2014}{1} 15^{1}+\binom{2014}{2} 15^{2}-\ldots-\binom{2014}{2013} 15^{2013}+15^{2014}$, so our sum is simply $\frac{(1+15)^{2014}-(1-15)^{2014}}{2}=$ $\frac{1}{2}\left(16^{2014}-14^{2014}\right)$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n20. [11]",
"solution_match": "\nAnswer: "
}
|
448bb66e-b6ed-58f0-b28c-066e341f5828
| 609,260
|
Compute the number of ordered quintuples of nonnegative integers ( $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ ) such that $0 \leq a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \leq 7$ and 5 divides $2^{a_{1}}+2^{a_{2}}+2^{a_{3}}+2^{a_{4}}+2^{a_{5}}$.
|
6528 Let $f(n)$ denote the number of $n$-tuples $\left(a_{1}, \ldots, a_{n}\right)$ such that $0 \leq a_{1}, \ldots, a_{n} \leq 7$ and $5 \mid 2^{a_{1}}+\ldots+2^{a_{n}}$. To compute $f(n+1)$ from $f(n)$, we note that given any $n$-tuple $\left(a_{1}, \ldots, a_{n}\right)$ such that $0 \leq a_{1}, \ldots, a_{n} \leq 7$ and $5 \nmid 2^{a_{1}}+\ldots+2^{a_{n}}$, there are exactly two possible values for $a_{n+1}$ such that $0 \leq a_{n+1} \leq 7$ and $5 \mid 2^{a_{1}}+\ldots+2^{a_{n+1}}$, because $2^{n} \equiv 1,2,4,3,1,2,4,3(\bmod 5)$ for $n=0,1,2,3,4,5,6,7$ respectively.
Also, given any valid $(n+1)$-tuple $\left(a_{1}, \ldots, a_{n+1}\right)$, we can remove $a_{n+1}$ to get an $n$-tuple $\left(a_{1}, \ldots, a_{n}\right)$ such that $0 \leq a_{1}, \ldots, a_{n} \leq 7$ and $5 \nmid 2^{a_{1}}+\ldots+2^{a_{n}}$, so these are in bijection. There are a total of $8^{n}$ $n$-tuples, $f(n)$ of which satisfy $5 \mid 2^{a_{1}}+\ldots+2^{a_{n}}$, so there are $8^{n}-f(n)$ for which $5 \nmid 2^{a_{1}}+\ldots+2^{a_{n}}$. Therefore, $f(n+1)=2\left(8^{n}-f(n)\right)$.
We now have $f(1)=0, f(2)=2(8-0)=16, f(3)=2(64-16)=96, f(4)=2(512-96)=832$, $f(5)=2(4096-832)=6528$.
|
6528
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Compute the number of ordered quintuples of nonnegative integers ( $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ ) such that $0 \leq a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \leq 7$ and 5 divides $2^{a_{1}}+2^{a_{2}}+2^{a_{3}}+2^{a_{4}}+2^{a_{5}}$.
|
6528 Let $f(n)$ denote the number of $n$-tuples $\left(a_{1}, \ldots, a_{n}\right)$ such that $0 \leq a_{1}, \ldots, a_{n} \leq 7$ and $5 \mid 2^{a_{1}}+\ldots+2^{a_{n}}$. To compute $f(n+1)$ from $f(n)$, we note that given any $n$-tuple $\left(a_{1}, \ldots, a_{n}\right)$ such that $0 \leq a_{1}, \ldots, a_{n} \leq 7$ and $5 \nmid 2^{a_{1}}+\ldots+2^{a_{n}}$, there are exactly two possible values for $a_{n+1}$ such that $0 \leq a_{n+1} \leq 7$ and $5 \mid 2^{a_{1}}+\ldots+2^{a_{n+1}}$, because $2^{n} \equiv 1,2,4,3,1,2,4,3(\bmod 5)$ for $n=0,1,2,3,4,5,6,7$ respectively.
Also, given any valid $(n+1)$-tuple $\left(a_{1}, \ldots, a_{n+1}\right)$, we can remove $a_{n+1}$ to get an $n$-tuple $\left(a_{1}, \ldots, a_{n}\right)$ such that $0 \leq a_{1}, \ldots, a_{n} \leq 7$ and $5 \nmid 2^{a_{1}}+\ldots+2^{a_{n}}$, so these are in bijection. There are a total of $8^{n}$ $n$-tuples, $f(n)$ of which satisfy $5 \mid 2^{a_{1}}+\ldots+2^{a_{n}}$, so there are $8^{n}-f(n)$ for which $5 \nmid 2^{a_{1}}+\ldots+2^{a_{n}}$. Therefore, $f(n+1)=2\left(8^{n}-f(n)\right)$.
We now have $f(1)=0, f(2)=2(8-0)=16, f(3)=2(64-16)=96, f(4)=2(512-96)=832$, $f(5)=2(4096-832)=6528$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n21. [14]",
"solution_match": "\nAnswer: "
}
|
4050e216-0619-56f6-80ba-254340240fe9
| 609,261
|
Let $\omega$ be a circle, and let $A B C D$ be a quadrilateral inscribed in $\omega$. Suppose that $B D$ and $A C$ intersect at a point $E$. The tangent to $\omega$ at $B$ meets line $A C$ at a point $F$, so that $C$ lies between $E$ and $F$. Given that $A E=6, E C=4, B E=2$, and $B F=12$, find $D A$.
|
$2 \sqrt{42}$ By power of a point, we have $E D \cdot E B=E A \cdot E C$, whence $E D=12$. Additionally, by power of a point, we have $144=F B^{2}=F C \cdot F A=F C(F C+10)$, so $F C=8$. Note that $\angle F B C=$ $\angle F A B$ and $\angle C F B=\angle A F B$, so $\triangle F B C \sim \triangle F A B$. Thus, $A B / B C=F A / F B=18 / 12=3 / 2$, so $A B=3 k$ and $B C=2 k$ for some $k$. Since $\triangle B E C \sim \triangle A E D$, we have $A D / B C=A E / B E=3$, so $A D=3 B C=6 k$. By Stewart's theorem on $\triangle E B F$, we have
$$
(4)(8)(12)+(2 k)^{2}(12)=(2)^{2}(8)+(12)^{2}(4) \Longrightarrow 8+k^{2}=8 / 12+12
$$
whence $k^{2}=14 / 3$. Thus,
$$
D A=6 k=6 \sqrt{14 / 3}=6 \frac{\sqrt{42}}{3}=2 \sqrt{42} .
$$
|
2 \sqrt{42}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $\omega$ be a circle, and let $A B C D$ be a quadrilateral inscribed in $\omega$. Suppose that $B D$ and $A C$ intersect at a point $E$. The tangent to $\omega$ at $B$ meets line $A C$ at a point $F$, so that $C$ lies between $E$ and $F$. Given that $A E=6, E C=4, B E=2$, and $B F=12$, find $D A$.
|
$2 \sqrt{42}$ By power of a point, we have $E D \cdot E B=E A \cdot E C$, whence $E D=12$. Additionally, by power of a point, we have $144=F B^{2}=F C \cdot F A=F C(F C+10)$, so $F C=8$. Note that $\angle F B C=$ $\angle F A B$ and $\angle C F B=\angle A F B$, so $\triangle F B C \sim \triangle F A B$. Thus, $A B / B C=F A / F B=18 / 12=3 / 2$, so $A B=3 k$ and $B C=2 k$ for some $k$. Since $\triangle B E C \sim \triangle A E D$, we have $A D / B C=A E / B E=3$, so $A D=3 B C=6 k$. By Stewart's theorem on $\triangle E B F$, we have
$$
(4)(8)(12)+(2 k)^{2}(12)=(2)^{2}(8)+(12)^{2}(4) \Longrightarrow 8+k^{2}=8 / 12+12
$$
whence $k^{2}=14 / 3$. Thus,
$$
D A=6 k=6 \sqrt{14 / 3}=6 \frac{\sqrt{42}}{3}=2 \sqrt{42} .
$$
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n22. [14]",
"solution_match": "\nAnswer: "
}
|
db0d4586-1c8e-5c87-8dc2-58859ad3eebf
| 609,262
|
Let $S=\{-100,-99,-98, \ldots, 99,100\}$. Choose a 50 -element subset $T$ of $S$ at random. Find the expected number of elements of the set $\{|x|: x \in T\}$.
|
$\square$ Let us solve a more generalized version of the problem: Let $S$ be a set with $2 n+1$ elements, and partition $S$ into sets $A_{0}, A_{1}, \ldots, A_{n}$ such that $\left|A_{0}\right|=1$ and $\left|A_{1}\right|=\left|A_{2}\right|=\cdots=\left|A_{n}\right|=2$. (In this problem, we have $A_{0}=\{0\}$ and $A_{k}=\{k,-k\}$ for $k=1,2, \ldots, 100$.) Let $T$ be a randomly chosen $m$-element subset of $S$. What is the expected number of $A_{k}$ 's that have a representative in $T$ ?
For $k=0,1, \ldots, n$, let $w_{k}=1$ if $T \cap A_{k} \neq \emptyset$ and 0 otherwise, so that the number of $A_{k}$ 's that have a representative in $T$ is equal to $\sum_{k=0}^{n} w_{k}$. It follows that the expected number of $A_{k}$ 's that have a representative in $T$ is equal to
$$
\mathrm{E}\left[w_{0}+w_{1}+\cdots+w_{n}\right]=\mathrm{E}\left[w_{0}\right]+\mathrm{E}\left[w_{1}\right]+\cdots+\mathrm{E}\left[w_{n}\right]=\mathrm{E}\left[w_{0}\right]+n \mathrm{E}\left[w_{1}\right]
$$
since $\mathrm{E}\left[w_{1}\right]=\mathrm{E}\left[w_{2}\right]=\cdots=\mathrm{E}\left[w_{n}\right]$ by symmetry.
Now $\mathrm{E}\left[w_{0}\right]$ is equal to the probability that $T \cap A_{0} \neq \emptyset$, that is, the probability that the single element of $A_{0}$ is in $T$, which is $T\left|/|S|=m /(2 n+1)\right.$. Similarly, $\mathrm{E}\left[w_{1}\right]$ is the probability that $T \cap A_{1} \neq \emptyset$, that is, the probability that at least one of the two elements of $A_{1}$ is in $T$. Since there are $\binom{2 n-1}{m} m$-element subsets of $S$ that exclude both elements of $A_{1}$, and there are $\binom{2 n+1}{m} m$-element subsets of $S$ in total, we have that
$$
\mathrm{E}\left[w_{1}\right]=1-\frac{\binom{2 n-1}{m}}{\binom{2 n+1}{m}}=1-\frac{(2 n-m)(2 n-m+1)}{2 n(2 n+1)}
$$
Putting this together, we find that the expected number of $A_{k}$ 's that have a representative in $T$ is
$$
\frac{m}{2 n+1}+n-\frac{(2 n-m+1)(2 n-m)}{2(2 n+1)}
$$
In this particular problem, we have $n=100$ and $m=50$, so substituting these values gives our answer of $\frac{8825}{201}$.
|
\frac{8825}{201}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $S=\{-100,-99,-98, \ldots, 99,100\}$. Choose a 50 -element subset $T$ of $S$ at random. Find the expected number of elements of the set $\{|x|: x \in T\}$.
|
$\square$ Let us solve a more generalized version of the problem: Let $S$ be a set with $2 n+1$ elements, and partition $S$ into sets $A_{0}, A_{1}, \ldots, A_{n}$ such that $\left|A_{0}\right|=1$ and $\left|A_{1}\right|=\left|A_{2}\right|=\cdots=\left|A_{n}\right|=2$. (In this problem, we have $A_{0}=\{0\}$ and $A_{k}=\{k,-k\}$ for $k=1,2, \ldots, 100$.) Let $T$ be a randomly chosen $m$-element subset of $S$. What is the expected number of $A_{k}$ 's that have a representative in $T$ ?
For $k=0,1, \ldots, n$, let $w_{k}=1$ if $T \cap A_{k} \neq \emptyset$ and 0 otherwise, so that the number of $A_{k}$ 's that have a representative in $T$ is equal to $\sum_{k=0}^{n} w_{k}$. It follows that the expected number of $A_{k}$ 's that have a representative in $T$ is equal to
$$
\mathrm{E}\left[w_{0}+w_{1}+\cdots+w_{n}\right]=\mathrm{E}\left[w_{0}\right]+\mathrm{E}\left[w_{1}\right]+\cdots+\mathrm{E}\left[w_{n}\right]=\mathrm{E}\left[w_{0}\right]+n \mathrm{E}\left[w_{1}\right]
$$
since $\mathrm{E}\left[w_{1}\right]=\mathrm{E}\left[w_{2}\right]=\cdots=\mathrm{E}\left[w_{n}\right]$ by symmetry.
Now $\mathrm{E}\left[w_{0}\right]$ is equal to the probability that $T \cap A_{0} \neq \emptyset$, that is, the probability that the single element of $A_{0}$ is in $T$, which is $T\left|/|S|=m /(2 n+1)\right.$. Similarly, $\mathrm{E}\left[w_{1}\right]$ is the probability that $T \cap A_{1} \neq \emptyset$, that is, the probability that at least one of the two elements of $A_{1}$ is in $T$. Since there are $\binom{2 n-1}{m} m$-element subsets of $S$ that exclude both elements of $A_{1}$, and there are $\binom{2 n+1}{m} m$-element subsets of $S$ in total, we have that
$$
\mathrm{E}\left[w_{1}\right]=1-\frac{\binom{2 n-1}{m}}{\binom{2 n+1}{m}}=1-\frac{(2 n-m)(2 n-m+1)}{2 n(2 n+1)}
$$
Putting this together, we find that the expected number of $A_{k}$ 's that have a representative in $T$ is
$$
\frac{m}{2 n+1}+n-\frac{(2 n-m+1)(2 n-m)}{2(2 n+1)}
$$
In this particular problem, we have $n=100$ and $m=50$, so substituting these values gives our answer of $\frac{8825}{201}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n23. [14]",
"solution_match": "\nAnswer: "
}
|
b196c2b8-ced2-51de-b668-69018d0c6c83
| 609,263
|
Let $A=\left\{a_{1}, a_{2}, \ldots, a_{7}\right\}$ be a set of distinct positive integers such that the mean of the elements of any nonempty subset of $A$ is an integer. Find the smallest possible value of the sum of the elements in $A$.
|
1267 For $2 \leq i \leq 6$, we claim that $a_{1} \equiv \ldots \equiv a_{7}(\bmod i)$. This is because if we consider any $i-1$ of the 7 numbers, the other $7-(i-1)=8-i$ of them must all be equal modulo $i$, because we want the sum of all subsets of size $i$ to be a multiple of $i$. However, $8-i \geq 2$, and this argument applies to any $8-i$ of the 7 integers, so in fact all of them must be equal modulo $i$.
We now have that all of the integers are equivalent modulo all of $2, \ldots, 6$, so they are equivalent modulo 60 , their least common multiple. Therefore, if the smallest integer is $k$, then the other 6 integers must be at least $k+60, k+60 \cdot 2, \ldots, k+60 \cdot 6$. This means the sum is $7 k+60 \cdot 21 \geq 7+60 \cdot 21=1267$. 1267 is achievable with $\{1,1+60, \ldots, 1+60 \cdot 6\}$, so it is the answer.
|
1267
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $A=\left\{a_{1}, a_{2}, \ldots, a_{7}\right\}$ be a set of distinct positive integers such that the mean of the elements of any nonempty subset of $A$ is an integer. Find the smallest possible value of the sum of the elements in $A$.
|
1267 For $2 \leq i \leq 6$, we claim that $a_{1} \equiv \ldots \equiv a_{7}(\bmod i)$. This is because if we consider any $i-1$ of the 7 numbers, the other $7-(i-1)=8-i$ of them must all be equal modulo $i$, because we want the sum of all subsets of size $i$ to be a multiple of $i$. However, $8-i \geq 2$, and this argument applies to any $8-i$ of the 7 integers, so in fact all of them must be equal modulo $i$.
We now have that all of the integers are equivalent modulo all of $2, \ldots, 6$, so they are equivalent modulo 60 , their least common multiple. Therefore, if the smallest integer is $k$, then the other 6 integers must be at least $k+60, k+60 \cdot 2, \ldots, k+60 \cdot 6$. This means the sum is $7 k+60 \cdot 21 \geq 7+60 \cdot 21=1267$. 1267 is achievable with $\{1,1+60, \ldots, 1+60 \cdot 6\}$, so it is the answer.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n24. [14]",
"solution_match": "\nAnswer: "
}
|
09eb2d86-340c-55ce-bf57-91b7752f9e3e
| 609,264
|
Let $A B C$ be an equilateral triangle of side length 6 inscribed in a circle $\omega$. Let $A_{1}, A_{2}$ be the points (distinct from $A$ ) where the lines through $A$ passing through the two trisection points of $B C$ meet $\omega$. Define $B_{1}, B_{2}, C_{1}, C_{2}$ similarly. Given that $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ appear on $\omega$ in that order, find the area of hexagon $A_{1} A_{2} B_{1} B_{2} C_{1} C_{2}$.
|
$\frac{846 \sqrt{3}}{49}$ Let $A^{\prime}$ be the point on $B C$ such that $2 B A^{\prime}=A^{\prime} C$. By law of cosines on triangle $A A^{\prime} B$, we find that $A A^{\prime}=2 \sqrt{7}$. By power of a point, $A^{\prime} A_{1}=\frac{2 * 4}{2 \sqrt{7}}=\frac{4}{\sqrt{7}}$. Using side length ratios, $A_{1} A_{2}=2 \frac{A A_{1}}{A A^{\prime}}=2 \frac{2 \sqrt{7}+\frac{4}{\sqrt{7}}}{2 \sqrt{7}}=\frac{18}{7}$.
Now our hexagon can be broken down into equilateral triangle $A_{1} B_{1} C_{1}$ and three copies of triangle $A_{1} C_{1} C_{2}$. Since our hexagon has rotational symmetry, $\angle C_{2}=120$, and using law of cosines on this triangle with side lengths $\frac{18}{7}$ and 6 , a little algebra yields $A_{1} C_{2}=\frac{30}{7}$ (this is a 3-5-7 triangle with an angle 120).
The area of the hexagon is therefore $\frac{6^{2} \sqrt{3}}{4}+3 * \frac{1}{2} \frac{18}{7} \frac{30}{7} \frac{\sqrt{3}}{2}=\frac{846 \sqrt{3}}{49}$
|
\frac{846 \sqrt{3}}{49}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be an equilateral triangle of side length 6 inscribed in a circle $\omega$. Let $A_{1}, A_{2}$ be the points (distinct from $A$ ) where the lines through $A$ passing through the two trisection points of $B C$ meet $\omega$. Define $B_{1}, B_{2}, C_{1}, C_{2}$ similarly. Given that $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ appear on $\omega$ in that order, find the area of hexagon $A_{1} A_{2} B_{1} B_{2} C_{1} C_{2}$.
|
$\frac{846 \sqrt{3}}{49}$ Let $A^{\prime}$ be the point on $B C$ such that $2 B A^{\prime}=A^{\prime} C$. By law of cosines on triangle $A A^{\prime} B$, we find that $A A^{\prime}=2 \sqrt{7}$. By power of a point, $A^{\prime} A_{1}=\frac{2 * 4}{2 \sqrt{7}}=\frac{4}{\sqrt{7}}$. Using side length ratios, $A_{1} A_{2}=2 \frac{A A_{1}}{A A^{\prime}}=2 \frac{2 \sqrt{7}+\frac{4}{\sqrt{7}}}{2 \sqrt{7}}=\frac{18}{7}$.
Now our hexagon can be broken down into equilateral triangle $A_{1} B_{1} C_{1}$ and three copies of triangle $A_{1} C_{1} C_{2}$. Since our hexagon has rotational symmetry, $\angle C_{2}=120$, and using law of cosines on this triangle with side lengths $\frac{18}{7}$ and 6 , a little algebra yields $A_{1} C_{2}=\frac{30}{7}$ (this is a 3-5-7 triangle with an angle 120).
The area of the hexagon is therefore $\frac{6^{2} \sqrt{3}}{4}+3 * \frac{1}{2} \frac{18}{7} \frac{30}{7} \frac{\sqrt{3}}{2}=\frac{846 \sqrt{3}}{49}$
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n25. [17]",
"solution_match": "\nAnswer: "
}
|
57e71edd-5220-5a1f-ae16-c20e3787bc2a
| 609,265
|
For $1 \leq j \leq 2014$, define
$$
b_{j}=j^{2014} \prod_{i=1, i \neq j}^{2014}\left(i^{2014}-j^{2014}\right)
$$
where the product is over all $i \in\{1, \ldots, 2014\}$ except $i=j$. Evaluate
$$
\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{2014}}
$$
|
$\frac{1}{2014!^{2014}}$ We perform Lagrange interpolation on the polynomial $P(x)=1$ through the points $1^{2014}, 2^{2014}, \ldots, 2014^{2014}$. We have
$$
1=P(x)=\sum_{j=1}^{2014} \frac{\prod_{i=1, i \neq j}^{2014}\left(x-i^{2014}\right)}{\prod_{i=1, i \neq j}^{2014}\left(j^{2014}-i^{2014}\right)} .
$$
Thus,
$$
1=P(0)=\sum_{j=1}^{2014} \frac{\left((-1)^{2013}\right) \frac{2014!^{2014}}{j^{014}}}{(-1)^{2013} \prod_{i=1, i \neq j}^{2014}\left(i^{2014}-j^{2014}\right)}
$$
which equals
$$
2014!^{2014} \sum_{j=1}^{2014} \frac{1}{j^{2014} \prod_{i=1, i \neq j}^{2014}\left(i^{2014}-j^{2014}\right)}=2014!^{2014}\left(\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{2014}}\right)
$$
so the desired sum is $\frac{1}{2014!^{2014}}$.
|
\frac{1}{2014!^{2014}}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
For $1 \leq j \leq 2014$, define
$$
b_{j}=j^{2014} \prod_{i=1, i \neq j}^{2014}\left(i^{2014}-j^{2014}\right)
$$
where the product is over all $i \in\{1, \ldots, 2014\}$ except $i=j$. Evaluate
$$
\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{2014}}
$$
|
$\frac{1}{2014!^{2014}}$ We perform Lagrange interpolation on the polynomial $P(x)=1$ through the points $1^{2014}, 2^{2014}, \ldots, 2014^{2014}$. We have
$$
1=P(x)=\sum_{j=1}^{2014} \frac{\prod_{i=1, i \neq j}^{2014}\left(x-i^{2014}\right)}{\prod_{i=1, i \neq j}^{2014}\left(j^{2014}-i^{2014}\right)} .
$$
Thus,
$$
1=P(0)=\sum_{j=1}^{2014} \frac{\left((-1)^{2013}\right) \frac{2014!^{2014}}{j^{014}}}{(-1)^{2013} \prod_{i=1, i \neq j}^{2014}\left(i^{2014}-j^{2014}\right)}
$$
which equals
$$
2014!^{2014} \sum_{j=1}^{2014} \frac{1}{j^{2014} \prod_{i=1, i \neq j}^{2014}\left(i^{2014}-j^{2014}\right)}=2014!^{2014}\left(\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{2014}}\right)
$$
so the desired sum is $\frac{1}{2014!^{2014}}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n26. [17]",
"solution_match": "\nAnswer: "
}
|
12c743e4-bf60-5dfd-aa53-4dc9e1e9984e
| 609,266
|
Suppose that $\left(a_{1}, \ldots, a_{20}\right)$ and $\left(b_{1}, \ldots, b_{20}\right)$ are two sequences of integers such that the sequence $\left(a_{1}, \ldots, a_{20}, b_{1}, \ldots, b_{20}\right)$ contains each of the numbers $1, \ldots, 40$ exactly once. What is the maximum possible value of the sum
$$
\sum_{i=1}^{20} \sum_{j=1}^{20} \min \left(a_{i}, b_{j}\right) ?
$$
|
5530 Let $x_{k}$, for $1 \leq k \leq 40$, be the number of integers $i$ with $1 \leq i \leq 20$ such that $a_{i} \geq k$. Let $y_{k}$, for $1 \leq k \leq 40$, be the number of integers $j$ with $1 \leq j \leq 20$ such that $b_{j} \geq k$. It follows from the problem statement that $x_{k}+y_{k}$ is the number of elements of the set $\{1, \ldots, 40\}$ which are greater than or equal to 40 , which is just $41-k$.
Note that if $1 \leq i, j \leq 20$, and $1 \leq k \leq 40$, then $\min \left(a_{i}, b_{j}\right) \geq k$ if and only if $a_{i} \geq k$ and $b_{j} \geq k$. So for a fixed $k$ with $1 \leq k \leq 40$, the number of pairs $(i, j)$ with $1 \leq i, j \leq 20 \operatorname{such}$ that $\min \left(a_{i}, b_{j}\right) \geq k$ is equal to $x_{k} y_{k}$. So we can rewrite
$$
\sum_{i=1}^{20} \sum_{j=1}^{20} \min \left(a_{i}, b_{j}\right)=\sum_{k=1}^{40} x_{k} y_{k}
$$
Since $x_{k}+y_{k}=41-k$ for $1 \leq k \leq 40$, we have
$$
x_{k} y_{k} \leq\left\lfloor\frac{41-k}{2}\right\rfloor\left\lceil\frac{41-k}{2}\right\rceil
$$
by a convexity argument. So
$$
\sum_{i=1}^{20} \sum_{j=1}^{20} \min \left(a_{i}, b_{j}\right) \leq \sum_{k=1}^{40}\left\lfloor\frac{41-k}{2}\right\rfloor\left\lceil\frac{41-k}{2}\right\rceil=5530 .
$$
Equality holds when $\left(a_{1}, \ldots, a_{20}\right)=(2,4, \ldots, 38,40)$ and $\left(b_{1}, \ldots, b_{20}\right)=(1,3, \ldots, 37,39)$.
|
5530
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Suppose that $\left(a_{1}, \ldots, a_{20}\right)$ and $\left(b_{1}, \ldots, b_{20}\right)$ are two sequences of integers such that the sequence $\left(a_{1}, \ldots, a_{20}, b_{1}, \ldots, b_{20}\right)$ contains each of the numbers $1, \ldots, 40$ exactly once. What is the maximum possible value of the sum
$$
\sum_{i=1}^{20} \sum_{j=1}^{20} \min \left(a_{i}, b_{j}\right) ?
$$
|
5530 Let $x_{k}$, for $1 \leq k \leq 40$, be the number of integers $i$ with $1 \leq i \leq 20$ such that $a_{i} \geq k$. Let $y_{k}$, for $1 \leq k \leq 40$, be the number of integers $j$ with $1 \leq j \leq 20$ such that $b_{j} \geq k$. It follows from the problem statement that $x_{k}+y_{k}$ is the number of elements of the set $\{1, \ldots, 40\}$ which are greater than or equal to 40 , which is just $41-k$.
Note that if $1 \leq i, j \leq 20$, and $1 \leq k \leq 40$, then $\min \left(a_{i}, b_{j}\right) \geq k$ if and only if $a_{i} \geq k$ and $b_{j} \geq k$. So for a fixed $k$ with $1 \leq k \leq 40$, the number of pairs $(i, j)$ with $1 \leq i, j \leq 20 \operatorname{such}$ that $\min \left(a_{i}, b_{j}\right) \geq k$ is equal to $x_{k} y_{k}$. So we can rewrite
$$
\sum_{i=1}^{20} \sum_{j=1}^{20} \min \left(a_{i}, b_{j}\right)=\sum_{k=1}^{40} x_{k} y_{k}
$$
Since $x_{k}+y_{k}=41-k$ for $1 \leq k \leq 40$, we have
$$
x_{k} y_{k} \leq\left\lfloor\frac{41-k}{2}\right\rfloor\left\lceil\frac{41-k}{2}\right\rceil
$$
by a convexity argument. So
$$
\sum_{i=1}^{20} \sum_{j=1}^{20} \min \left(a_{i}, b_{j}\right) \leq \sum_{k=1}^{40}\left\lfloor\frac{41-k}{2}\right\rfloor\left\lceil\frac{41-k}{2}\right\rceil=5530 .
$$
Equality holds when $\left(a_{1}, \ldots, a_{20}\right)=(2,4, \ldots, 38,40)$ and $\left(b_{1}, \ldots, b_{20}\right)=(1,3, \ldots, 37,39)$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n27. [17]",
"solution_match": "\nAnswer: "
}
|
d9c60a67-9837-5094-95a9-a99eb9ebe513
| 609,267
|
Let $f(n)$ and $g(n)$ be polynomials of degree 2014 such that $f(n)+(-1)^{n} g(n)=2^{n}$ for $n=$ $1,2, \ldots, 4030$. Find the coefficient of $x^{2014}$ in $g(x)$.
|
$\frac{3^{2014}}{2^{2014} \cdot 2014!}$ Define the polynomial functions $h_{1}$ and $h_{2}$ by $h_{1}(x)=f(2 x)+g(2 x)$ and $h_{2}(x)=f(2 x-1)-g(2 x-1)$. Then, the problem conditions tell us that $h_{1}(x)=2^{2 x}$ and $h_{2}(x)=2^{2 x-1}$ for $x=1,2, \ldots, 2015$.
By the Lagrange interpolation formula, the polynomial $h_{1}$ is given by
$$
h_{1}(x)=\sum_{i=1}^{2015} 2^{2 i} \prod_{\substack{j=1 \\ i \neq j}}^{2015} \frac{x-j}{i-j}
$$
So the coefficient of $x^{2014}$ in $h_{1}(x)$ is
$$
\sum_{i=1}^{2015} 2^{2 i} \prod_{\substack{j=1 \\ i \neq j}}^{2015} \frac{1}{i-j}=\frac{1}{2014!} \sum_{i=1}^{2015} 2^{2 i}(-1)^{2015-i}\binom{2014}{i-1}=\frac{4 \cdot 3^{2014}}{2014!}
$$
where the last equality follows from the binomial theorem. By a similar argument, the coefficient of $x^{2014}$ in $h_{2}(x)$ is $\frac{2 \cdot 3^{2014}}{2014!}$.
We can write $g(x)=\frac{1}{2}\left(h_{1}(x / 2)-h_{2}((x+1) / 2)\right)$. So, the coefficient of $x^{2014}$ in $g(x)$ is
$$
\frac{1}{2}\left(\frac{4 \cdot 3^{2014}}{2^{2014} \cdot 2014!}-\frac{2 \cdot 3^{2014}}{2^{2014} \cdot 2014!}\right)=\frac{3^{2014}}{2^{2014} \cdot 2014!}
$$
|
\frac{3^{2014}}{2^{2014} \cdot 2014!}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $f(n)$ and $g(n)$ be polynomials of degree 2014 such that $f(n)+(-1)^{n} g(n)=2^{n}$ for $n=$ $1,2, \ldots, 4030$. Find the coefficient of $x^{2014}$ in $g(x)$.
|
$\frac{3^{2014}}{2^{2014} \cdot 2014!}$ Define the polynomial functions $h_{1}$ and $h_{2}$ by $h_{1}(x)=f(2 x)+g(2 x)$ and $h_{2}(x)=f(2 x-1)-g(2 x-1)$. Then, the problem conditions tell us that $h_{1}(x)=2^{2 x}$ and $h_{2}(x)=2^{2 x-1}$ for $x=1,2, \ldots, 2015$.
By the Lagrange interpolation formula, the polynomial $h_{1}$ is given by
$$
h_{1}(x)=\sum_{i=1}^{2015} 2^{2 i} \prod_{\substack{j=1 \\ i \neq j}}^{2015} \frac{x-j}{i-j}
$$
So the coefficient of $x^{2014}$ in $h_{1}(x)$ is
$$
\sum_{i=1}^{2015} 2^{2 i} \prod_{\substack{j=1 \\ i \neq j}}^{2015} \frac{1}{i-j}=\frac{1}{2014!} \sum_{i=1}^{2015} 2^{2 i}(-1)^{2015-i}\binom{2014}{i-1}=\frac{4 \cdot 3^{2014}}{2014!}
$$
where the last equality follows from the binomial theorem. By a similar argument, the coefficient of $x^{2014}$ in $h_{2}(x)$ is $\frac{2 \cdot 3^{2014}}{2014!}$.
We can write $g(x)=\frac{1}{2}\left(h_{1}(x / 2)-h_{2}((x+1) / 2)\right)$. So, the coefficient of $x^{2014}$ in $g(x)$ is
$$
\frac{1}{2}\left(\frac{4 \cdot 3^{2014}}{2^{2014} \cdot 2014!}-\frac{2 \cdot 3^{2014}}{2^{2014} \cdot 2014!}\right)=\frac{3^{2014}}{2^{2014} \cdot 2014!}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n28. [17]",
"solution_match": "\nAnswer: "
}
|
15b4dfe7-8b09-52af-861a-45cb9f921a84
| 609,268
|
Natalie has a copy of the unit interval $[0,1]$ that is colored white. She also has a black marker, and she colors the interval in the following manner: at each step, she selects a value $x \in[0,1]$ uniformly at random, and
(a) If $x \leq \frac{1}{2}$ she colors the interval $\left[x, x+\frac{1}{2}\right]$ with her marker.
(b) If $x>\frac{1}{2}$ she colors the intervals $[x, 1]$ and $\left[0, x-\frac{1}{2}\right]$ with her marker.
What is the expected value of the number of steps Natalie will need to color the entire interval black?
|
5 The first choice always wipes out half the interval. So we calculate the expected value of the amount of time needed to wipe out the other half.
Solution 1 (non-calculus):
We assume the interval has $2 n$ points and we start with the last $n$ colored black. We let $f(k)$ be the expected value of the number of turns we need if there are $k$ white points left. So we must calculate $f(n)$.
We observe that
$$
\begin{gathered}
f(k)=1+\frac{(n-k+1) \cdot 0+(n-k+1) \cdot f(k)+2 \sum_{i=1}^{k-1} f(i)}{2 n} \\
f(k) \frac{n+k-1}{2 n}=1+\frac{\sum_{i=1}^{k-1} f(i)}{n} \\
f(k+1) \frac{n+k}{2 n}=1+\frac{\sum_{i=1}^{k} f(i)}{n} \\
f(k+1)=f(k) \frac{n+k+1}{n+k} \\
f(k)=f(1) \frac{n+k}{n+1}
\end{gathered}
$$
And note that $f(1)=2$ so $f(n)=\frac{4 n}{n+1}$ and $\lim _{n \rightarrow \infty} f(n)=4$.
Therefore adding the first turn, the expected value is 5 .
Solution 2 (calculus):
We let $f(x)$ be the expected value with length $x$ uncolored. Like above, $\lim _{x \rightarrow 0} f(x)=2$.
Similarly we have the recursion
$$
\begin{gathered}
f(x)=1+\left(\frac{1}{2}-x\right) f(x)+2 \int_{0}^{x} f(y) d y \\
f^{\prime}(x)=0+\frac{1}{2} f^{\prime}(x)-f(x)-x f^{\prime}(x)+2 f(x) \\
\frac{f^{\prime}(x)}{f(x)}=\frac{1}{x+\frac{1}{2}}
\end{gathered}
$$
And solving yields $f(x)=c\left(\frac{1}{2}+x\right)$ and since $\lim _{x \rightarrow 0} f(x)=2, c=4$. So $f(x)=2+4 x$ and $f\left(\frac{1}{2}\right)=4$. Therefore adding the first turn, our expected value is 5 .
|
5
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Natalie has a copy of the unit interval $[0,1]$ that is colored white. She also has a black marker, and she colors the interval in the following manner: at each step, she selects a value $x \in[0,1]$ uniformly at random, and
(a) If $x \leq \frac{1}{2}$ she colors the interval $\left[x, x+\frac{1}{2}\right]$ with her marker.
(b) If $x>\frac{1}{2}$ she colors the intervals $[x, 1]$ and $\left[0, x-\frac{1}{2}\right]$ with her marker.
What is the expected value of the number of steps Natalie will need to color the entire interval black?
|
5 The first choice always wipes out half the interval. So we calculate the expected value of the amount of time needed to wipe out the other half.
Solution 1 (non-calculus):
We assume the interval has $2 n$ points and we start with the last $n$ colored black. We let $f(k)$ be the expected value of the number of turns we need if there are $k$ white points left. So we must calculate $f(n)$.
We observe that
$$
\begin{gathered}
f(k)=1+\frac{(n-k+1) \cdot 0+(n-k+1) \cdot f(k)+2 \sum_{i=1}^{k-1} f(i)}{2 n} \\
f(k) \frac{n+k-1}{2 n}=1+\frac{\sum_{i=1}^{k-1} f(i)}{n} \\
f(k+1) \frac{n+k}{2 n}=1+\frac{\sum_{i=1}^{k} f(i)}{n} \\
f(k+1)=f(k) \frac{n+k+1}{n+k} \\
f(k)=f(1) \frac{n+k}{n+1}
\end{gathered}
$$
And note that $f(1)=2$ so $f(n)=\frac{4 n}{n+1}$ and $\lim _{n \rightarrow \infty} f(n)=4$.
Therefore adding the first turn, the expected value is 5 .
Solution 2 (calculus):
We let $f(x)$ be the expected value with length $x$ uncolored. Like above, $\lim _{x \rightarrow 0} f(x)=2$.
Similarly we have the recursion
$$
\begin{gathered}
f(x)=1+\left(\frac{1}{2}-x\right) f(x)+2 \int_{0}^{x} f(y) d y \\
f^{\prime}(x)=0+\frac{1}{2} f^{\prime}(x)-f(x)-x f^{\prime}(x)+2 f(x) \\
\frac{f^{\prime}(x)}{f(x)}=\frac{1}{x+\frac{1}{2}}
\end{gathered}
$$
And solving yields $f(x)=c\left(\frac{1}{2}+x\right)$ and since $\lim _{x \rightarrow 0} f(x)=2, c=4$. So $f(x)=2+4 x$ and $f\left(\frac{1}{2}\right)=4$. Therefore adding the first turn, our expected value is 5 .
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n29. [20]",
"solution_match": "\nAnswer: "
}
|
632aa6e3-dd54-5a02-9115-88e65feace2b
| 609,269
|
Let $A B C$ be a triangle with circumcenter $O$, incenter $I, \angle B=45^{\circ}$, and $O I \| B C$. Find $\cos \angle C$.
|
$1-\frac{\sqrt{2}}{2}$ Let $M$ be the midpoint of $B C$, and $D$ the foot of the perpendicular of $I$ with $B C$. Because $O I \| B C$, we have $O M=I D$. Since $\angle B O C=2 \angle A$, the length of $O M$ is $O A \cos \angle B O M=$ $O A \cos A=R \cos A$, and the length of $I D$ is $r$, where $R$ and $r$ are the circumradius and inradius of $\triangle A B C$, respectively.
Thus, $r=R \cos A$, so $1+\cos A=(R+r) / R$. By Carnot's theorem, $(R+r) / R=\cos A+\cos B+\cos C$, so we have $\cos B+\cos C=1$. Since $\cos B=\frac{\sqrt{2}}{2}$, we have $\cos C=1-\frac{\sqrt{2}}{2}$.
|
1-\frac{\sqrt{2}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a triangle with circumcenter $O$, incenter $I, \angle B=45^{\circ}$, and $O I \| B C$. Find $\cos \angle C$.
|
$1-\frac{\sqrt{2}}{2}$ Let $M$ be the midpoint of $B C$, and $D$ the foot of the perpendicular of $I$ with $B C$. Because $O I \| B C$, we have $O M=I D$. Since $\angle B O C=2 \angle A$, the length of $O M$ is $O A \cos \angle B O M=$ $O A \cos A=R \cos A$, and the length of $I D$ is $r$, where $R$ and $r$ are the circumradius and inradius of $\triangle A B C$, respectively.
Thus, $r=R \cos A$, so $1+\cos A=(R+r) / R$. By Carnot's theorem, $(R+r) / R=\cos A+\cos B+\cos C$, so we have $\cos B+\cos C=1$. Since $\cos B=\frac{\sqrt{2}}{2}$, we have $\cos C=1-\frac{\sqrt{2}}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n30. [20]",
"solution_match": "\nAnswer: "
}
|
11befaf3-36d5-5345-8c53-3183d7e20f78
| 609,270
|
An up-right path from $(a, b) \in \mathbb{R}^{2}$ to $(c, d) \in \mathbb{R}^{2}$ is a finite sequence $\left(x_{1}, y_{1}\right), \ldots,\left(x_{k}, y_{k}\right)$ of points in $\mathbb{R}^{2}$ such that $(a, b)=\left(x_{1}, y_{1}\right),(c, d)=\left(x_{k}, y_{k}\right)$, and for each $1 \leq i<k$ we have that either $\left(x_{i+1}, y_{i+1}\right)=\left(x_{i}+1, y_{i}\right)$ or $\left(x_{i+1}, y_{i+1}\right)=\left(x_{i}, y_{i}+1\right)$.
Let $S$ be the set of all up-right paths from $(-400,-400)$ to $(400,400)$. What fraction of the paths in $S$ do not contain any point $(x, y)$ such that $|x|,|y| \leq 10$ ? Express your answer as a decimal number between 0 and 1.
If $C$ is the actual answer to this question and $A$ is your answer, then your score on this problem is $\lceil\max \{25(1-10|C-A|), 0\}\rceil$.
|
$0.2937156494680644 \ldots$ Note that any up-right path must pass through exactly one point of the form $(n,-n)$ (i.e. a point on the upper-left to lower-right diagonal), and the number of such paths is $\binom{800}{400-n}^{2}$ because there are $\binom{800}{400-n}$ up-right paths from $(-400,-400)$ to $(n,-n)$ and another $\binom{800}{400-n}$ from $(n,-n)$ to $(400,400)$. An up-right path contains a point $(x, y)$ with $|x|,|y| \leq 10$ if and only if $-10 \leq n \leq 10$, so the probability that this happens is
$$
\frac{\sum_{n=-10}^{10}\binom{800}{400-n}^{2}}{\sum_{n=-400}^{400}\binom{800}{400-n}^{2}}=\frac{\sum_{n=-10}^{10}\binom{800}{400-n}^{2}}{\binom{1600}{800}}
$$
To estimate this, recall that if we normalize $\binom{800}{n}$ to be a probability density function, then it will be approximately normal with mean 400 and variance $800 \cdot \frac{1}{4}=200$. If this is squared, then it is proportional to a normal distribution with half the variance and the same mean, because the probability density function of a normal distribution is proportional to $e^{-\frac{(x-\mu)^{2}}{2 \sigma^{2}}}$, where $\mu$ is the mean and $\sigma^{2}$ is the variance.
Therefore, the $\binom{800}{n}^{2}$ probability density function is roughly proportional to a normal distribution with mean 400 and variance 100 , or standard deviation 10 . So $\sum_{n=-10}^{10}\binom{800}{400-n}^{2}$ represents roughly one standard deviation. Recall that approximately 68 percent of a normal distribution lies within one standard deviation of the mean (look up the 68-95-99.7 rule to read more), so a good guess would be around . 32 . This guess can be improved by noting that we're actually summing 21 values instead of 20 , so you'd have approximately $.68 \cdot \frac{21}{20} \approx .71$ of the normal distribution, giving an answer of .29 .
|
0.2937156494680644
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
An up-right path from $(a, b) \in \mathbb{R}^{2}$ to $(c, d) \in \mathbb{R}^{2}$ is a finite sequence $\left(x_{1}, y_{1}\right), \ldots,\left(x_{k}, y_{k}\right)$ of points in $\mathbb{R}^{2}$ such that $(a, b)=\left(x_{1}, y_{1}\right),(c, d)=\left(x_{k}, y_{k}\right)$, and for each $1 \leq i<k$ we have that either $\left(x_{i+1}, y_{i+1}\right)=\left(x_{i}+1, y_{i}\right)$ or $\left(x_{i+1}, y_{i+1}\right)=\left(x_{i}, y_{i}+1\right)$.
Let $S$ be the set of all up-right paths from $(-400,-400)$ to $(400,400)$. What fraction of the paths in $S$ do not contain any point $(x, y)$ such that $|x|,|y| \leq 10$ ? Express your answer as a decimal number between 0 and 1.
If $C$ is the actual answer to this question and $A$ is your answer, then your score on this problem is $\lceil\max \{25(1-10|C-A|), 0\}\rceil$.
|
$0.2937156494680644 \ldots$ Note that any up-right path must pass through exactly one point of the form $(n,-n)$ (i.e. a point on the upper-left to lower-right diagonal), and the number of such paths is $\binom{800}{400-n}^{2}$ because there are $\binom{800}{400-n}$ up-right paths from $(-400,-400)$ to $(n,-n)$ and another $\binom{800}{400-n}$ from $(n,-n)$ to $(400,400)$. An up-right path contains a point $(x, y)$ with $|x|,|y| \leq 10$ if and only if $-10 \leq n \leq 10$, so the probability that this happens is
$$
\frac{\sum_{n=-10}^{10}\binom{800}{400-n}^{2}}{\sum_{n=-400}^{400}\binom{800}{400-n}^{2}}=\frac{\sum_{n=-10}^{10}\binom{800}{400-n}^{2}}{\binom{1600}{800}}
$$
To estimate this, recall that if we normalize $\binom{800}{n}$ to be a probability density function, then it will be approximately normal with mean 400 and variance $800 \cdot \frac{1}{4}=200$. If this is squared, then it is proportional to a normal distribution with half the variance and the same mean, because the probability density function of a normal distribution is proportional to $e^{-\frac{(x-\mu)^{2}}{2 \sigma^{2}}}$, where $\mu$ is the mean and $\sigma^{2}$ is the variance.
Therefore, the $\binom{800}{n}^{2}$ probability density function is roughly proportional to a normal distribution with mean 400 and variance 100 , or standard deviation 10 . So $\sum_{n=-10}^{10}\binom{800}{400-n}^{2}$ represents roughly one standard deviation. Recall that approximately 68 percent of a normal distribution lies within one standard deviation of the mean (look up the 68-95-99.7 rule to read more), so a good guess would be around . 32 . This guess can be improved by noting that we're actually summing 21 values instead of 20 , so you'd have approximately $.68 \cdot \frac{21}{20} \approx .71$ of the normal distribution, giving an answer of .29 .
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n33. [25]",
"solution_match": "\nAnswer: "
}
|
28622cef-46fa-5416-bd73-79c9bf1ec8f3
| 609,272
|
Consider a number line, with a lily pad placed at each integer point. A frog is standing at the lily pad at the point 0 on the number line, and wants to reach the lily pad at the point 2014 on the number line. If the frog stands at the point $n$ on the number line, it can jump directly to either point
$n+2$ or point $n+3$ on the number line. Each of the lily pads at the points $1, \cdots, 2013$ on the number line has, independently and with probability $1 / 2$, a snake. Let $p$ be the probability that the frog can make some sequence of jumps to reach the lily pad at the point 2014 on the number line, without ever landing on a lily pad containing a snake. What is $p^{1 / 2014}$ ? Express your answer as a decimal number. If $C$ is the actual answer to this question and $A$ is your answer, then your score on this problem is $\lceil\max \{25(1-20|C-A|), 0\}\rceil$.
|
0.9102805441016536
First, we establish a rough upper bound for the probability $p$. Let $q$ be the probability that the frog can reach the lily pad at the point 2014 on the number line if it is allowed to jump from a point $n$ on the number line to the point $n+1$, in addition to the points $n+2$ and $n+3$. Clearly, $p \leq q$. Furthermore, $p$ is approximated by $q$; it should be easy to convince one's self that jumps from a point $n$ to the point $n+1$ are only useful for reaching the lily pad at point 2014 in very few situations.
Now we compute $q$. We note that, if the frog can jump from points $n$ to points $n+1, n+2$, and $n+3$, then it can reach the lily pad at the point 2014 on the number line if and only if each snake-free lily pad is at most 3 units away from the closest snake-free lily pad on the left.
Define the sequence $\left\{a_{m}\right\}_{m=1}^{\infty}$ by $a_{0}=1, a_{1}=1, a_{2}=2$, and $a_{m+3}=a_{m+2}+a_{m+1}+a_{m}$ for $m \geq 0$. Then, it can be shown by induction that $a_{m}$ is the number of possible arrangements of snakes on lily pads at points $1, \cdots, m-1$ so that the frog can make some sequence of jumps (of size 1,2 , or 3 ) from the lily pad at point 0 to the lily pad at point $m$ without landing on a lily pad containing a snake. It follows that $q=a_{2014} / 2^{2013}$. So
$$
p^{1 / 2014} \approx q^{1 / 2014}=\left(a_{2014}\right)^{1 / 2014} / 2^{2013 / 2014} \approx\left(a_{2014}\right)^{1 / 2014} / 2
$$
Analyzing the recurrence relation $a_{m+3}=a_{m+2}+a_{m+1}+a_{m}$ yields that $\left(a_{2014}\right)^{1 / 2014}$ is approximately equal to the largest real root $r$ of the characteristic polynomial equation $r^{3}-r^{2}-r-1=0$. So to roughly approximate $p$, it suffices to find the largest real root of this equation.
For this, we apply Newton's method, or one of many other methods for computing the roots of a polynomial. With an initial guess of 2 , one iteration of Newton's method yields $r \approx 13 / 7$, so $p \approx r / 2 \approx$ $13 / 14 \approx 0.928571$. A second iteration yields $r \approx 1777 / 966$, so $p \approx r / 2 \approx 1777 / 1932 \approx 0.919772$. (It turns out that the value of $r$ is $1.839286 \ldots$, yielding $p \approx r / 2=0.919643 \ldots$..)
Using tools from probability theory, we can get an even better estimate for $p$. We model the problem using a discrete-time Markov chain. The state of the Markov chain at time $n$, for $n=0,1, \ldots, 2013$, indicates which of the lily pads at positions $n-2, n-1, n$ are reachable by the frog. It is clear that the state of the Markov chain at time $n$ only depends (randomly) on its state at time $n-1$. There are $2^{3}=8$ possible states for this Markov chain, because each of the lily pads at positions $n-2, n-1, n$ can be either reachable or unreachable by the frog. Number each state using the number $1+d_{2}+2 d_{1}+4 d_{0}$, where $d_{i}$ is 1 if the lily pad at point $n-i$ is reachable, and 0 otherwise. So, for example, at time $n=0$, the lily pad at point $n$ is reachable $\left(d_{0}=1\right)$ whereas the lily pads at points $n-1$ and $n-2$ are unreachable $\left(d_{1}=d_{2}=0\right)$, so the Markov chain is in state number $1+d_{2}+2 d_{1}+4 d_{0}=5$.
The transition matrix $M$ for the Markov chain can now be computed directly from the conditions of the problem. It is equal to
$$
M:=\left[\begin{array}{cccccccc}
1 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 \\
0 & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 \\
0 & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 \\
0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} \\
0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2}
\end{array}\right] .
$$
(The verification of this transition matrix is left as an exercise for the reader.) So the state vector $v$ for the Markov chain at time 2013 is
$$
v:=M^{2014}[0,1,0,0,0,0,0,0]^{t}
$$
Now, the lily pad at point 2014 is reachable by the frog if and only if the Markov chain is in state $3,4,5,6,7$, or 8 at time 2013. This happens with probability
$$
p=[0,0,1,1,1,1,1,1] v
$$
By expanding $[0,1,0,0,0,0,0,0]^{t}$ in an eigenbasis for $M$, we find that $p^{1 / 2014}$ is approximately equal to the second-largest real eigenvalue of the matrix $M$. The characteristic polynomial of $M$ is
$$
\operatorname{det}(\lambda I-M)=-\frac{\lambda^{3}}{8}+\frac{3 \lambda^{4}}{8}+\frac{\lambda^{6}}{4}-\frac{3 \lambda^{7}}{2}+\lambda^{8}
$$
so its eigenvalues are the roots of this polynomial. The largest real root of this characteristic polynomial is $\lambda=1$, and the second-largest real root is $0.9105247383471604 \ldots$ (which can be found, again, using Newton's method, after factoring out $(\lambda-1) \lambda^{3}$ from the polynomial), which is a good approximation for $p$.
|
0.9102805441016536
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Consider a number line, with a lily pad placed at each integer point. A frog is standing at the lily pad at the point 0 on the number line, and wants to reach the lily pad at the point 2014 on the number line. If the frog stands at the point $n$ on the number line, it can jump directly to either point
$n+2$ or point $n+3$ on the number line. Each of the lily pads at the points $1, \cdots, 2013$ on the number line has, independently and with probability $1 / 2$, a snake. Let $p$ be the probability that the frog can make some sequence of jumps to reach the lily pad at the point 2014 on the number line, without ever landing on a lily pad containing a snake. What is $p^{1 / 2014}$ ? Express your answer as a decimal number. If $C$ is the actual answer to this question and $A$ is your answer, then your score on this problem is $\lceil\max \{25(1-20|C-A|), 0\}\rceil$.
|
0.9102805441016536
First, we establish a rough upper bound for the probability $p$. Let $q$ be the probability that the frog can reach the lily pad at the point 2014 on the number line if it is allowed to jump from a point $n$ on the number line to the point $n+1$, in addition to the points $n+2$ and $n+3$. Clearly, $p \leq q$. Furthermore, $p$ is approximated by $q$; it should be easy to convince one's self that jumps from a point $n$ to the point $n+1$ are only useful for reaching the lily pad at point 2014 in very few situations.
Now we compute $q$. We note that, if the frog can jump from points $n$ to points $n+1, n+2$, and $n+3$, then it can reach the lily pad at the point 2014 on the number line if and only if each snake-free lily pad is at most 3 units away from the closest snake-free lily pad on the left.
Define the sequence $\left\{a_{m}\right\}_{m=1}^{\infty}$ by $a_{0}=1, a_{1}=1, a_{2}=2$, and $a_{m+3}=a_{m+2}+a_{m+1}+a_{m}$ for $m \geq 0$. Then, it can be shown by induction that $a_{m}$ is the number of possible arrangements of snakes on lily pads at points $1, \cdots, m-1$ so that the frog can make some sequence of jumps (of size 1,2 , or 3 ) from the lily pad at point 0 to the lily pad at point $m$ without landing on a lily pad containing a snake. It follows that $q=a_{2014} / 2^{2013}$. So
$$
p^{1 / 2014} \approx q^{1 / 2014}=\left(a_{2014}\right)^{1 / 2014} / 2^{2013 / 2014} \approx\left(a_{2014}\right)^{1 / 2014} / 2
$$
Analyzing the recurrence relation $a_{m+3}=a_{m+2}+a_{m+1}+a_{m}$ yields that $\left(a_{2014}\right)^{1 / 2014}$ is approximately equal to the largest real root $r$ of the characteristic polynomial equation $r^{3}-r^{2}-r-1=0$. So to roughly approximate $p$, it suffices to find the largest real root of this equation.
For this, we apply Newton's method, or one of many other methods for computing the roots of a polynomial. With an initial guess of 2 , one iteration of Newton's method yields $r \approx 13 / 7$, so $p \approx r / 2 \approx$ $13 / 14 \approx 0.928571$. A second iteration yields $r \approx 1777 / 966$, so $p \approx r / 2 \approx 1777 / 1932 \approx 0.919772$. (It turns out that the value of $r$ is $1.839286 \ldots$, yielding $p \approx r / 2=0.919643 \ldots$..)
Using tools from probability theory, we can get an even better estimate for $p$. We model the problem using a discrete-time Markov chain. The state of the Markov chain at time $n$, for $n=0,1, \ldots, 2013$, indicates which of the lily pads at positions $n-2, n-1, n$ are reachable by the frog. It is clear that the state of the Markov chain at time $n$ only depends (randomly) on its state at time $n-1$. There are $2^{3}=8$ possible states for this Markov chain, because each of the lily pads at positions $n-2, n-1, n$ can be either reachable or unreachable by the frog. Number each state using the number $1+d_{2}+2 d_{1}+4 d_{0}$, where $d_{i}$ is 1 if the lily pad at point $n-i$ is reachable, and 0 otherwise. So, for example, at time $n=0$, the lily pad at point $n$ is reachable $\left(d_{0}=1\right)$ whereas the lily pads at points $n-1$ and $n-2$ are unreachable $\left(d_{1}=d_{2}=0\right)$, so the Markov chain is in state number $1+d_{2}+2 d_{1}+4 d_{0}=5$.
The transition matrix $M$ for the Markov chain can now be computed directly from the conditions of the problem. It is equal to
$$
M:=\left[\begin{array}{cccccccc}
1 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 \\
0 & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 \\
0 & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 \\
0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} \\
0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2}
\end{array}\right] .
$$
(The verification of this transition matrix is left as an exercise for the reader.) So the state vector $v$ for the Markov chain at time 2013 is
$$
v:=M^{2014}[0,1,0,0,0,0,0,0]^{t}
$$
Now, the lily pad at point 2014 is reachable by the frog if and only if the Markov chain is in state $3,4,5,6,7$, or 8 at time 2013. This happens with probability
$$
p=[0,0,1,1,1,1,1,1] v
$$
By expanding $[0,1,0,0,0,0,0,0]^{t}$ in an eigenbasis for $M$, we find that $p^{1 / 2014}$ is approximately equal to the second-largest real eigenvalue of the matrix $M$. The characteristic polynomial of $M$ is
$$
\operatorname{det}(\lambda I-M)=-\frac{\lambda^{3}}{8}+\frac{3 \lambda^{4}}{8}+\frac{\lambda^{6}}{4}-\frac{3 \lambda^{7}}{2}+\lambda^{8}
$$
so its eigenvalues are the roots of this polynomial. The largest real root of this characteristic polynomial is $\lambda=1$, and the second-largest real root is $0.9105247383471604 \ldots$ (which can be found, again, using Newton's method, after factoring out $(\lambda-1) \lambda^{3}$ from the polynomial), which is a good approximation for $p$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n34. [25]",
"solution_match": "\nAnswer: "
}
|
64366be1-d05d-58d8-a449-079ae78c2d27
| 609,273
|
How many times does the letter "e" occur in all problem statements in this year's HMMT February competition?
If $C$ is the actual answer to this question and $A$ is your answer, then your score on this problem is $\left\lceil\max \left\{25\left(1-\left|\log _{2}(C / A)\right|\right), 0\right\}\right\rceil$.
|
1661 It is possible to arrive at a good estimate using Fermi estimation. See http: //en.wikipedia.org/wiki/Fermi_problem for more details.
For example, there are 76 problems on the HMMT this year. You might guess that the average number of words in a problem is approximately 40 , and the average number of of letters in a word is about 5 . The frequency of the letter "e" in the English language is about $10 \%$, resulting in an estimate of
$$
76 \cdot 40 \cdot 5 \cdot 0.1=1520
$$
This is remarkably close to the actual answer.
|
1661
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many times does the letter "e" occur in all problem statements in this year's HMMT February competition?
If $C$ is the actual answer to this question and $A$ is your answer, then your score on this problem is $\left\lceil\max \left\{25\left(1-\left|\log _{2}(C / A)\right|\right), 0\right\}\right\rceil$.
|
1661 It is possible to arrive at a good estimate using Fermi estimation. See http: //en.wikipedia.org/wiki/Fermi_problem for more details.
For example, there are 76 problems on the HMMT this year. You might guess that the average number of words in a problem is approximately 40 , and the average number of of letters in a word is about 5 . The frequency of the letter "e" in the English language is about $10 \%$, resulting in an estimate of
$$
76 \cdot 40 \cdot 5 \cdot 0.1=1520
$$
This is remarkably close to the actual answer.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-guts-solutions.jsonl",
"problem_match": "\n35. [25]",
"solution_match": "\nAnswer: "
}
|
552a385f-42ab-50c4-b7c3-199defed906d
| 609,274
|
Let $a_{1}, a_{2}, \ldots$ be an infinite sequence of integers such that $a_{i}$ divides $a_{i+1}$ for all $i \geq 1$, and let $b_{i}$ be the remainder when $a_{i}$ is divided by 210 . What is the maximal number of distinct terms in the sequence $b_{1}, b_{2}, \ldots$ ?
|
$\quad 127$ It is clear that the sequence $\left\{a_{i}\right\}$ will be a concatenation of sequences of the form $\left\{v_{i}\right\}_{i=1}^{N_{0}},\left\{w_{i} \cdot p_{1}\right\}_{i=1}^{N_{1}},\left\{x_{i} \cdot p_{1} p_{2}\right\}_{i=1}^{N_{2}},\left\{y_{i} \cdot p_{1} p_{2} p_{3}\right\}_{i=1}^{N_{3}}$, and $\left\{z_{i} \cdot p_{1} p_{2} p_{3} p_{4}\right\}_{i=1}^{N_{4}}$, for some permutation $\left(p_{1}, p_{2}, p_{3}, p_{4}\right)$ of $(2,3,5,7)$ and some sequences of integers $\left\{v_{i}\right\} \cdot\left\{w_{i}\right\} \cdot\left\{x_{i}\right\} \cdot\left\{y_{i}\right\} \cdot\left\{z_{i}\right\}$, each coprime with 210.
In $\left\{v_{i}\right\}_{i=1}^{N_{0}}$, there are a maximum of $\phi(210)$ distinct terms $\bmod 210$. In $\left\{w_{i} \cdot p_{1}\right\}_{i=1}^{N_{1}}$, there are a maximum of $\phi\left(\frac{210}{p_{1}}\right)$ distinct terms mod 210. In $\left\{x_{i} \cdot p_{1} p_{2}\right\}_{i=1}^{N_{2}}$, there are a maximum of $\phi\left(\frac{210}{p_{1} p_{2}}\right)$ distinct terms mod 210. In $\left\{y_{i} \cdot p_{1} p_{2} p_{3}\right\}_{i=1}^{N_{3}}$, there are a maximum of $\phi\left(\frac{210}{p_{1} p_{2} p_{3}}\right)$ distinct terms mod 210. In $\left\{z_{i} \cdot p_{1} p_{2} p_{3} p_{4}\right\}_{i=1}^{N_{4}}$, there can only be one distinct term $\bmod 210$.
Therefore we wish to maximize $\phi(210)+\phi\left(\frac{210}{p_{1}}\right)+\phi\left(\frac{210}{p_{1} p_{2}}\right)+\phi\left(\frac{210}{p_{1} p_{2} p_{3}}\right)+1$ over all permutations $\left(p_{1}, p_{2}, p_{3}, p_{4}\right)$ of $(2,3,5,7)$. It's easy to see that the maximum occurs when we take $p_{1}=2, p_{2}=$ $3, p_{3}=5, p_{4}=7$ for an answer of $\phi(210)+\phi(105)+\phi(35)+\phi(7)+1=127$. This upper bound is clearly attainable by having the $v_{i}$ 's cycle through the $\phi(210)$ integers less than 210 coprime to 210 , the $w_{i}$ 's cycle through the $\phi\left(\frac{210}{p_{1}}\right)$ integers less than $\frac{210}{p_{1}}$ coprime to $\frac{210}{p_{1}}$, etc.
|
127
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $a_{1}, a_{2}, \ldots$ be an infinite sequence of integers such that $a_{i}$ divides $a_{i+1}$ for all $i \geq 1$, and let $b_{i}$ be the remainder when $a_{i}$ is divided by 210 . What is the maximal number of distinct terms in the sequence $b_{1}, b_{2}, \ldots$ ?
|
$\quad 127$ It is clear that the sequence $\left\{a_{i}\right\}$ will be a concatenation of sequences of the form $\left\{v_{i}\right\}_{i=1}^{N_{0}},\left\{w_{i} \cdot p_{1}\right\}_{i=1}^{N_{1}},\left\{x_{i} \cdot p_{1} p_{2}\right\}_{i=1}^{N_{2}},\left\{y_{i} \cdot p_{1} p_{2} p_{3}\right\}_{i=1}^{N_{3}}$, and $\left\{z_{i} \cdot p_{1} p_{2} p_{3} p_{4}\right\}_{i=1}^{N_{4}}$, for some permutation $\left(p_{1}, p_{2}, p_{3}, p_{4}\right)$ of $(2,3,5,7)$ and some sequences of integers $\left\{v_{i}\right\} \cdot\left\{w_{i}\right\} \cdot\left\{x_{i}\right\} \cdot\left\{y_{i}\right\} \cdot\left\{z_{i}\right\}$, each coprime with 210.
In $\left\{v_{i}\right\}_{i=1}^{N_{0}}$, there are a maximum of $\phi(210)$ distinct terms $\bmod 210$. In $\left\{w_{i} \cdot p_{1}\right\}_{i=1}^{N_{1}}$, there are a maximum of $\phi\left(\frac{210}{p_{1}}\right)$ distinct terms mod 210. In $\left\{x_{i} \cdot p_{1} p_{2}\right\}_{i=1}^{N_{2}}$, there are a maximum of $\phi\left(\frac{210}{p_{1} p_{2}}\right)$ distinct terms mod 210. In $\left\{y_{i} \cdot p_{1} p_{2} p_{3}\right\}_{i=1}^{N_{3}}$, there are a maximum of $\phi\left(\frac{210}{p_{1} p_{2} p_{3}}\right)$ distinct terms mod 210. In $\left\{z_{i} \cdot p_{1} p_{2} p_{3} p_{4}\right\}_{i=1}^{N_{4}}$, there can only be one distinct term $\bmod 210$.
Therefore we wish to maximize $\phi(210)+\phi\left(\frac{210}{p_{1}}\right)+\phi\left(\frac{210}{p_{1} p_{2}}\right)+\phi\left(\frac{210}{p_{1} p_{2} p_{3}}\right)+1$ over all permutations $\left(p_{1}, p_{2}, p_{3}, p_{4}\right)$ of $(2,3,5,7)$. It's easy to see that the maximum occurs when we take $p_{1}=2, p_{2}=$ $3, p_{3}=5, p_{4}=7$ for an answer of $\phi(210)+\phi(105)+\phi(35)+\phi(7)+1=127$. This upper bound is clearly attainable by having the $v_{i}$ 's cycle through the $\phi(210)$ integers less than 210 coprime to 210 , the $w_{i}$ 's cycle through the $\phi\left(\frac{210}{p_{1}}\right)$ integers less than $\frac{210}{p_{1}}$ coprime to $\frac{210}{p_{1}}$, etc.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-team-solutions.jsonl",
"problem_match": "\n2. [15]",
"solution_match": "\nAnswer: "
}
|
747e5217-16d9-5ea9-ab3e-fe168b1e04b1
| 609,277
|
There are $n$ girls $G_{1}, \ldots, G_{n}$ and $n$ boys $B_{1}, \ldots, B_{n}$. A pair $\left(G_{i}, B_{j}\right)$ is called suitable if and only if girl $G_{i}$ is willing to marry boy $B_{j}$. Given that there is exactly one way to pair each girl with a distinct boy that she is willing to marry, what is the maximal possible number of suitable pairs?
|
$\frac{n(n+1)}{2}$ We represent the problem as a graph with vertices $G_{1}, \ldots, G_{n}, B_{1}, \ldots, B_{n}$ such that there is an edge between vertices $G_{i}$ and $B_{j}$ if and only if $\left(G_{i}, B_{j}\right)$ is suitable, so we want to maximize the number of edges while having a unique matching.
We claim the answer is $\frac{n(n+1)}{2}$. First, note that this can be achieved by having an edge between $G_{i}$ and $B_{j}$ for all pairs $j \leq i$, because the only possible matching in this case is pairing $G_{i}$ with $B_{i}$ for all $i$. To prove that this is maximal, we first assume without loss of generality that our unique matching consists of pairing $G_{i}$ with $B_{i}$ for all $i$, which takes $n$ edges. Now, note that for any $i, j$, at most one of the two edges $G_{i} B_{j}$ and $G_{j} B_{i}$ can be added, because if both were added, we could pair $G_{i}$ with $B_{j}$
and $G_{j}$ with $B_{i}$ instead to get another valid matching. Therefore, we may add at most $\binom{n}{2} \cdot 1=\frac{n(n-1)}{2}$ edges, so the maximal number of edges is $n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}$ as desired.
|
\frac{n(n+1)}{2}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
There are $n$ girls $G_{1}, \ldots, G_{n}$ and $n$ boys $B_{1}, \ldots, B_{n}$. A pair $\left(G_{i}, B_{j}\right)$ is called suitable if and only if girl $G_{i}$ is willing to marry boy $B_{j}$. Given that there is exactly one way to pair each girl with a distinct boy that she is willing to marry, what is the maximal possible number of suitable pairs?
|
$\frac{n(n+1)}{2}$ We represent the problem as a graph with vertices $G_{1}, \ldots, G_{n}, B_{1}, \ldots, B_{n}$ such that there is an edge between vertices $G_{i}$ and $B_{j}$ if and only if $\left(G_{i}, B_{j}\right)$ is suitable, so we want to maximize the number of edges while having a unique matching.
We claim the answer is $\frac{n(n+1)}{2}$. First, note that this can be achieved by having an edge between $G_{i}$ and $B_{j}$ for all pairs $j \leq i$, because the only possible matching in this case is pairing $G_{i}$ with $B_{i}$ for all $i$. To prove that this is maximal, we first assume without loss of generality that our unique matching consists of pairing $G_{i}$ with $B_{i}$ for all $i$, which takes $n$ edges. Now, note that for any $i, j$, at most one of the two edges $G_{i} B_{j}$ and $G_{j} B_{i}$ can be added, because if both were added, we could pair $G_{i}$ with $B_{j}$
and $G_{j}$ with $B_{i}$ instead to get another valid matching. Therefore, we may add at most $\binom{n}{2} \cdot 1=\frac{n(n-1)}{2}$ edges, so the maximal number of edges is $n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}$ as desired.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-team-solutions.jsonl",
"problem_match": "\n3. [15]",
"solution_match": "\nAnswer: "
}
|
3f09cc0e-670c-5ca2-a940-3671838426ed
| 609,278
|
Compute
$$
\sum_{k=0}^{100}\left\lfloor\frac{2^{100}}{2^{50}+2^{k}}\right\rfloor
$$
(Here, if $x$ is a real number, then $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$.)
|
$101 \cdot 2^{49}-50$ Let $a_{k}=\frac{2^{100}}{2^{50}+2^{k}}$. Notice that, for $k=0,1, \ldots, 49$,
$$
a_{k}+a_{100-k}=\frac{2^{100}}{2^{50}+2^{k}}+\frac{2^{100}}{2^{50}+2^{100-k}}=\frac{2^{100}}{2^{50}+2^{k}}+\frac{2^{50+k}}{2^{k}+2^{50}}=2^{50}
$$
It is clear that for $k=0,1, \ldots, 49, a_{k}, a_{100-k} \notin \mathbb{Z}$, so $\left\lfloor a_{k}\right\rfloor+\left\lfloor a_{100-k}\right\rfloor=2^{50}-1$ (since the sum of floors is an integer less than $a_{k}+a_{100-k}$ but greater than $\left.a_{k}-1+a_{100-k}-1\right)$. Thus,
$$
\sum_{k=0}^{100}\left\lfloor a_{k}\right\rfloor=50 \cdot\left(2^{50}-1\right)+2^{49}=101 \cdot 2^{49}-50
$$
|
101 \cdot 2^{49}-50
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Compute
$$
\sum_{k=0}^{100}\left\lfloor\frac{2^{100}}{2^{50}+2^{k}}\right\rfloor
$$
(Here, if $x$ is a real number, then $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$.)
|
$101 \cdot 2^{49}-50$ Let $a_{k}=\frac{2^{100}}{2^{50}+2^{k}}$. Notice that, for $k=0,1, \ldots, 49$,
$$
a_{k}+a_{100-k}=\frac{2^{100}}{2^{50}+2^{k}}+\frac{2^{100}}{2^{50}+2^{100-k}}=\frac{2^{100}}{2^{50}+2^{k}}+\frac{2^{50+k}}{2^{k}+2^{50}}=2^{50}
$$
It is clear that for $k=0,1, \ldots, 49, a_{k}, a_{100-k} \notin \mathbb{Z}$, so $\left\lfloor a_{k}\right\rfloor+\left\lfloor a_{100-k}\right\rfloor=2^{50}-1$ (since the sum of floors is an integer less than $a_{k}+a_{100-k}$ but greater than $\left.a_{k}-1+a_{100-k}-1\right)$. Thus,
$$
\sum_{k=0}^{100}\left\lfloor a_{k}\right\rfloor=50 \cdot\left(2^{50}-1\right)+2^{49}=101 \cdot 2^{49}-50
$$
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-team-solutions.jsonl",
"problem_match": "\n4. [20]",
"solution_match": "\nAnswer: "
}
|
43942d06-aed9-506f-b1a4-74dd3d78fa6b
| 609,279
|
Find the maximum possible number of diagonals of equal length in a convex hexagon.
|
7 First, we will prove that 7 is possible. Consider the following hexagon $A B C D E F$ whose vertices are located at $A(0,0), B\left(\frac{1}{2}, 1-\frac{\sqrt{3}}{2}\right), C\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), D(0,1), E\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), F\left(-\frac{1}{2}, 1-\frac{\sqrt{3}}{2}\right)$. One can easily verify that all diagonals but $B E$ and $C F$ have length 1.
Now suppose that there are at least 8 diagonals in a certain convex hexagon $A B C D E F$ whose lengths are equal. There must be a diagonal such that, with this diagonal taken out, the other 8 have equal length. There are two cases.
Case I. The diagonal is one of $A C, B D, C E, D F, E A, F B$. WLOG, assume it is $A C$. We have $E C=$ $E B=F B=F C$. Thus, $B$ and $C$ are both on the perpendicular bisector of $E F$. Since $A B C D E F$ is convex, both $B$ and $C$ must be on the same side of line $E F$, but this is impossible as one of $B$ or $C$, must be contained in triangle $C E F$. Contradiction.
Case II: The diagonal is one of $A D, B E, C F$. WLOG, assume it is $A D$. Again, we have $E C=E B=$ $F B=F C$. By the above reasoning, this is a contradiction.
Thus, 7 is the maximum number of possible diagonals.
|
7
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Find the maximum possible number of diagonals of equal length in a convex hexagon.
|
7 First, we will prove that 7 is possible. Consider the following hexagon $A B C D E F$ whose vertices are located at $A(0,0), B\left(\frac{1}{2}, 1-\frac{\sqrt{3}}{2}\right), C\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), D(0,1), E\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), F\left(-\frac{1}{2}, 1-\frac{\sqrt{3}}{2}\right)$. One can easily verify that all diagonals but $B E$ and $C F$ have length 1.
Now suppose that there are at least 8 diagonals in a certain convex hexagon $A B C D E F$ whose lengths are equal. There must be a diagonal such that, with this diagonal taken out, the other 8 have equal length. There are two cases.
Case I. The diagonal is one of $A C, B D, C E, D F, E A, F B$. WLOG, assume it is $A C$. We have $E C=$ $E B=F B=F C$. Thus, $B$ and $C$ are both on the perpendicular bisector of $E F$. Since $A B C D E F$ is convex, both $B$ and $C$ must be on the same side of line $E F$, but this is impossible as one of $B$ or $C$, must be contained in triangle $C E F$. Contradiction.
Case II: The diagonal is one of $A D, B E, C F$. WLOG, assume it is $A D$. Again, we have $E C=E B=$ $F B=F C$. By the above reasoning, this is a contradiction.
Thus, 7 is the maximum number of possible diagonals.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-team-solutions.jsonl",
"problem_match": "\n7. [30]",
"solution_match": "\nAnswer: "
}
|
870eae23-2c6d-5ea9-8d78-afc7b3c558ee
| 73,455
|
Let $A B C$ be an acute triangle with circumcenter $O$ such that $A B=4, A C=5$, and $B C=6$. Let $D$ be the foot of the altitude from $A$ to $B C$, and $E$ be the intersection of $A O$ with $B C$. Suppose that $X$ is on $B C$ between $D$ and $E$ such that there is a point $Y$ on $A D$ satisfying $X Y \| A O$ and $Y O \perp A X$. Determine the length of $B X$.
|
$96 / 41$ Let $A X$ intersect the circumcircle of $\triangle A B C$ again at $K$. Let $O Y$ intersect $A K$ and $B C$ at $T$ and $L$, respectively. We have $\angle L O A=\angle O Y X=\angle T D X=\angle L A K$, so $A L$ is tangent to the circumcircle. Furthermore, $O L \perp A K$, so $\triangle A L K$ is isosceles with $A L=A K$, so $A K$ is also tangent to the circumcircle. Since $B C$ and the tangents to the circumcircle at $A$ and $K$ all intersect at the same point $L, C L$ is a symmedian of $\triangle A C K$. Then $A K$ is a symmedian of $\triangle A B C$. Then we can use $\frac{B X}{X C}=\frac{(A B)^{2}}{(A C)^{2}}$ to compute $B X=\frac{96}{41}$.
|
\frac{96}{41}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be an acute triangle with circumcenter $O$ such that $A B=4, A C=5$, and $B C=6$. Let $D$ be the foot of the altitude from $A$ to $B C$, and $E$ be the intersection of $A O$ with $B C$. Suppose that $X$ is on $B C$ between $D$ and $E$ such that there is a point $Y$ on $A D$ satisfying $X Y \| A O$ and $Y O \perp A X$. Determine the length of $B X$.
|
$96 / 41$ Let $A X$ intersect the circumcircle of $\triangle A B C$ again at $K$. Let $O Y$ intersect $A K$ and $B C$ at $T$ and $L$, respectively. We have $\angle L O A=\angle O Y X=\angle T D X=\angle L A K$, so $A L$ is tangent to the circumcircle. Furthermore, $O L \perp A K$, so $\triangle A L K$ is isosceles with $A L=A K$, so $A K$ is also tangent to the circumcircle. Since $B C$ and the tangents to the circumcircle at $A$ and $K$ all intersect at the same point $L, C L$ is a symmedian of $\triangle A C K$. Then $A K$ is a symmedian of $\triangle A B C$. Then we can use $\frac{B X}{X C}=\frac{(A B)^{2}}{(A C)^{2}}$ to compute $B X=\frac{96}{41}$.
|
{
"resource_path": "HarvardMIT/segmented/en-172-2014-feb-team-solutions.jsonl",
"problem_match": "\n8. [35]",
"solution_match": "\nAnswer: "
}
|
2946e0b8-6359-5872-bc76-e51cd9d58a56
| 609,282
|
Two circles $\omega$ and $\gamma$ have radii 3 and 4 respectively, and their centers are 10 units apart. Let $x$ be the shortest possible distance between a point on $\omega$ and a point on $\gamma$, and let $y$ be the longest possible distance between a point on $\omega$ and a point on $\gamma$. Find the product $x y$.
|
51 Let $\ell$ be the line connecting the centers of $\omega$ and $\gamma$. Let $A$ and $B$ be the intersections of $\ell$ with $\omega$, and let $C$ and $D$ be the intersections of $\ell$ with $\gamma$, so that $A, B, C$, and $D$ are collinear, in that order. The shortest distance between a point on $\omega$ and a point on $\gamma$ is $B C=3$. The longest distance is $A D=3+10+4=17$. The product is 51 .
|
51
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Two circles $\omega$ and $\gamma$ have radii 3 and 4 respectively, and their centers are 10 units apart. Let $x$ be the shortest possible distance between a point on $\omega$ and a point on $\gamma$, and let $y$ be the longest possible distance between a point on $\omega$ and a point on $\gamma$. Find the product $x y$.
|
51 Let $\ell$ be the line connecting the centers of $\omega$ and $\gamma$. Let $A$ and $B$ be the intersections of $\ell$ with $\omega$, and let $C$ and $D$ be the intersections of $\ell$ with $\gamma$, so that $A, B, C$, and $D$ are collinear, in that order. The shortest distance between a point on $\omega$ and a point on $\gamma$ is $B C=3$. The longest distance is $A D=3+10+4=17$. The product is 51 .
|
{
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-gen-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nAnswer: "
}
|
939ff6cc-143f-5137-99a2-aeac7169cad2
| 609,289
|
Let $A B C$ be a triangle with $\angle B=90^{\circ}$. Given that there exists a point $D$ on $A C$ such that $A D=D C$ and $B D=B C$, compute the value of the ratio $\frac{A B}{B C}$.
|
$\sqrt{3} \quad D$ is the circumcenter of $A B C$ because it is the midpoint of the hypotenuse. Therefore, $D B=D A=D C$ because they are all radii of the circumcircle, so $D B C$ is an equilateral triangle, and $\angle C=60^{\circ}$. This means that $A B C$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, with $\frac{A B}{B C}=\sqrt{3}$.
|
\sqrt{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a triangle with $\angle B=90^{\circ}$. Given that there exists a point $D$ on $A C$ such that $A D=D C$ and $B D=B C$, compute the value of the ratio $\frac{A B}{B C}$.
|
$\sqrt{3} \quad D$ is the circumcenter of $A B C$ because it is the midpoint of the hypotenuse. Therefore, $D B=D A=D C$ because they are all radii of the circumcircle, so $D B C$ is an equilateral triangle, and $\angle C=60^{\circ}$. This means that $A B C$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, with $\frac{A B}{B C}=\sqrt{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-gen-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nAnswer: "
}
|
a0aa9c7c-b530-54ad-a6f5-1d21fd70345b
| 609,290
|
Compute the greatest common divisor of $4^{8}-1$ and $8^{12}-1$.
|
15 Let $d=\operatorname{gcd}(a, b)$ for some $a, b \in \mathbb{Z}^{+}$.
Then, we can write $d=a x-b y$, where $x, y \in \mathbb{Z}^{+}$, and
$$
\begin{aligned}
& 2^{a}-1 \mid 2^{a x}-1 \\
& 2^{b}-1 \mid 2^{b y}-1
\end{aligned}
$$
Multiplying the right-hand side of (2) by $2^{d}$, we get,
$$
2^{b}-1 \mid 2^{a x}-2^{d}
$$
Thus, $\operatorname{gcd}\left(2^{a}-1,2^{b}-1\right)=2^{d}-1=2^{\operatorname{gcd}(a, b)}-1$.
Using $a=16$ and $b=36$, we get
$$
\operatorname{gcd}\left(2^{16}-1,2^{36}-1\right)=2^{\operatorname{gcd}(16,36)}-1=2^{4}-1=15
$$
|
15
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Compute the greatest common divisor of $4^{8}-1$ and $8^{12}-1$.
|
15 Let $d=\operatorname{gcd}(a, b)$ for some $a, b \in \mathbb{Z}^{+}$.
Then, we can write $d=a x-b y$, where $x, y \in \mathbb{Z}^{+}$, and
$$
\begin{aligned}
& 2^{a}-1 \mid 2^{a x}-1 \\
& 2^{b}-1 \mid 2^{b y}-1
\end{aligned}
$$
Multiplying the right-hand side of (2) by $2^{d}$, we get,
$$
2^{b}-1 \mid 2^{a x}-2^{d}
$$
Thus, $\operatorname{gcd}\left(2^{a}-1,2^{b}-1\right)=2^{d}-1=2^{\operatorname{gcd}(a, b)}-1$.
Using $a=16$ and $b=36$, we get
$$
\operatorname{gcd}\left(2^{16}-1,2^{36}-1\right)=2^{\operatorname{gcd}(16,36)}-1=2^{4}-1=15
$$
|
{
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-gen-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nAnswer: "
}
|
e6f1697b-1c54-520e-b065-ea63807116db
| 609,291
|
In rectangle $A B C D$ with area 1 , point $M$ is selected on $\overline{A B}$ and points $X, Y$ are selected on $\overline{C D}$ such that $A X<A Y$. Suppose that $A M=B M$. Given that the area of triangle $M X Y$ is $\frac{1}{2014}$, compute the area of trapezoid $A X Y B$.
|
$\frac{1}{2}+\frac{1}{2014}$ OR $\frac{504}{1007}$ Notice that $[A M X]+[B Y M]=\frac{1}{2}[A B C D]=\frac{1}{2}$. Thus,
$$
[A X Y B]=[A M X]+[B Y M]+[M X Y]=\frac{1}{2}+\frac{1}{2014}=\frac{504}{1007}
$$
|
\frac{504}{1007}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
In rectangle $A B C D$ with area 1 , point $M$ is selected on $\overline{A B}$ and points $X, Y$ are selected on $\overline{C D}$ such that $A X<A Y$. Suppose that $A M=B M$. Given that the area of triangle $M X Y$ is $\frac{1}{2014}$, compute the area of trapezoid $A X Y B$.
|
$\frac{1}{2}+\frac{1}{2014}$ OR $\frac{504}{1007}$ Notice that $[A M X]+[B Y M]=\frac{1}{2}[A B C D]=\frac{1}{2}$. Thus,
$$
[A X Y B]=[A M X]+[B Y M]+[M X Y]=\frac{1}{2}+\frac{1}{2014}=\frac{504}{1007}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-gen-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nAnswer: "
}
|
bdc5cad5-bab0-52f7-be6c-e5c4d1ceb77c
| 609,292
|
Mark and William are playing a game with a stored value. On his turn, a player may either multiply the stored value by 2 and add 1 or he may multiply the stored value by 4 and add 3 . The first player to make the stored value exceed $2^{100}$ wins. The stored value starts at 1 and Mark goes first. Assuming both players play optimally, what is the maximum number of times that William can make a move?
(By optimal play, we mean that on any turn the player selects the move which leads to the best possible outcome given that the opponent is also playing optimally. If both moves lead to the same outcome, the player selects one of them arbitrarily.)
|
33 We will work in the binary system in this solution.
Let multiplying the stored value by 2 and adding 1 be Move $A$ and multiplying the stored value by 4 and adding 3 be Move $B$. Let the stored value be $S$. Then, Move $A$ affixes one 1 to $S$, while Move $B$ affixes two 1 s . The goal is to have greater than or equal to 1011 s . If any player makes the number of 1 s in $S$ congruent to $2 \bmod 3$, then no matter what the other player does, he will lose, since the number of 1 s in $S$ reaches 101 or 102 only from $99 \equiv 0(\bmod 3)$ or $100 \equiv 1(\bmod 3)$.
Mark's winning strategy: Do Move $A$. In the succeeding moves, if William does Move $B$, then Mark does Move $A$, and vice versa, which in total, affixes three 1s to $S$. This ensures that William always takes his turn while the number of 1 s in $S$ is congruent to $2 \bmod 3$. Note that Mark has to follow this strategy because once he does not, then William can follow the same strategy and make Mark lose, a contradiction to the required optimal play. Since $S$ starts out with one 1, this process gives William a maximum of 33 moves.
|
33
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Mark and William are playing a game with a stored value. On his turn, a player may either multiply the stored value by 2 and add 1 or he may multiply the stored value by 4 and add 3 . The first player to make the stored value exceed $2^{100}$ wins. The stored value starts at 1 and Mark goes first. Assuming both players play optimally, what is the maximum number of times that William can make a move?
(By optimal play, we mean that on any turn the player selects the move which leads to the best possible outcome given that the opponent is also playing optimally. If both moves lead to the same outcome, the player selects one of them arbitrarily.)
|
33 We will work in the binary system in this solution.
Let multiplying the stored value by 2 and adding 1 be Move $A$ and multiplying the stored value by 4 and adding 3 be Move $B$. Let the stored value be $S$. Then, Move $A$ affixes one 1 to $S$, while Move $B$ affixes two 1 s . The goal is to have greater than or equal to 1011 s . If any player makes the number of 1 s in $S$ congruent to $2 \bmod 3$, then no matter what the other player does, he will lose, since the number of 1 s in $S$ reaches 101 or 102 only from $99 \equiv 0(\bmod 3)$ or $100 \equiv 1(\bmod 3)$.
Mark's winning strategy: Do Move $A$. In the succeeding moves, if William does Move $B$, then Mark does Move $A$, and vice versa, which in total, affixes three 1s to $S$. This ensures that William always takes his turn while the number of 1 s in $S$ is congruent to $2 \bmod 3$. Note that Mark has to follow this strategy because once he does not, then William can follow the same strategy and make Mark lose, a contradiction to the required optimal play. Since $S$ starts out with one 1, this process gives William a maximum of 33 moves.
|
{
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-gen-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nAnswer: "
}
|
9597a45d-86df-5cb2-84c8-9f0ecf9e73c9
| 609,293
|
Let $A B C$ be a triangle with $A B=5, A C=4, B C=6$. The angle bisector of $C$ intersects side $A B$ at $X$. Points $M$ and $N$ are drawn on sides $B C$ and $A C$, respectively, such that $\overline{X M} \| \overline{A C}$ and $\overline{X N} \| \overline{B C}$. Compute the length $M N$.
|
$\frac{3 \sqrt{14}}{5}$ By Stewart's Theorem on the angle bisector,
$$
C X^{2}=A C \cdot B C\left(1-\frac{A B}{A C+B C}^{2}\right)
$$
Thus,
$$
C X^{2}=4 \cdot 6\left(1-\frac{5}{10}^{2}\right)=18
$$
Since $\overline{X M} \| \overline{A C}$ and $\overline{X N} \| \overline{B C}$, we produce equal angles. So, by similar triangles, $X M=X N=$ $\frac{4 \cdot 6}{10}=\frac{12}{5}$.
Moreover, triangles $M C X$ and $N C X$ are congruent isosceles triangles with vertices $M$ and $N$, respectively. Since $C X$ is an angle bisector, then $C X$ and $M N$ are perpendicular bisectors of each other. Therefore,
$$
M N^{2}=4\left(X N^{2}-(C X / 2)^{2}\right)=4 \cdot\left(\frac{12}{5}\right)^{2}-18=\frac{126}{25}
$$
and
$$
M N=\frac{3 \sqrt{14}}{5}
$$
|
\frac{3 \sqrt{14}}{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a triangle with $A B=5, A C=4, B C=6$. The angle bisector of $C$ intersects side $A B$ at $X$. Points $M$ and $N$ are drawn on sides $B C$ and $A C$, respectively, such that $\overline{X M} \| \overline{A C}$ and $\overline{X N} \| \overline{B C}$. Compute the length $M N$.
|
$\frac{3 \sqrt{14}}{5}$ By Stewart's Theorem on the angle bisector,
$$
C X^{2}=A C \cdot B C\left(1-\frac{A B}{A C+B C}^{2}\right)
$$
Thus,
$$
C X^{2}=4 \cdot 6\left(1-\frac{5}{10}^{2}\right)=18
$$
Since $\overline{X M} \| \overline{A C}$ and $\overline{X N} \| \overline{B C}$, we produce equal angles. So, by similar triangles, $X M=X N=$ $\frac{4 \cdot 6}{10}=\frac{12}{5}$.
Moreover, triangles $M C X$ and $N C X$ are congruent isosceles triangles with vertices $M$ and $N$, respectively. Since $C X$ is an angle bisector, then $C X$ and $M N$ are perpendicular bisectors of each other. Therefore,
$$
M N^{2}=4\left(X N^{2}-(C X / 2)^{2}\right)=4 \cdot\left(\frac{12}{5}\right)^{2}-18=\frac{126}{25}
$$
and
$$
M N=\frac{3 \sqrt{14}}{5}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-gen-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nAnswer: "
}
|
2472b4d5-207d-5cac-bb9b-3e945aff3a5e
| 609,294
|
Consider the set of 5 -tuples of positive integers at most 5 . We say the tuple ( $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ ) is perfect if for any distinct indices $i, j, k$, the three numbers $a_{i}, a_{j}, a_{k}$ do not form an arithmetic progression (in any order). Find the number of perfect 5 -tuples.
|
780 There are two situations.
- 1. The multiset is $a a b b c$; the only condition here is $c \neq \frac{1}{2}(a+b)$, for $\left(\binom{5}{3}-|S|\right) \cdot\binom{3}{1}=18$ such triples, where $S$ is the set of unordered triples $(a, b, c)$ which do not satisfy the condition, and $S=$ $\{(1,2,3),(2,3,4),(3,4,5),(1,3,5)\}$. Each one gives $\frac{5!}{2!2!}=30$ orderings, so $18 \cdot 30=540$ in this case.
- 2. There are four distinct elements in the tuple. Then, the elements must be $\{1,2,4,5\}$. All of them work, for an additional $4 \cdot 60=240$.
Therefore, there are $540+240=780$ such tuples.
|
780
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Consider the set of 5 -tuples of positive integers at most 5 . We say the tuple ( $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ ) is perfect if for any distinct indices $i, j, k$, the three numbers $a_{i}, a_{j}, a_{k}$ do not form an arithmetic progression (in any order). Find the number of perfect 5 -tuples.
|
780 There are two situations.
- 1. The multiset is $a a b b c$; the only condition here is $c \neq \frac{1}{2}(a+b)$, for $\left(\binom{5}{3}-|S|\right) \cdot\binom{3}{1}=18$ such triples, where $S$ is the set of unordered triples $(a, b, c)$ which do not satisfy the condition, and $S=$ $\{(1,2,3),(2,3,4),(3,4,5),(1,3,5)\}$. Each one gives $\frac{5!}{2!2!}=30$ orderings, so $18 \cdot 30=540$ in this case.
- 2. There are four distinct elements in the tuple. Then, the elements must be $\{1,2,4,5\}$. All of them work, for an additional $4 \cdot 60=240$.
Therefore, there are $540+240=780$ such tuples.
|
{
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-gen-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nAnswer: "
}
|
bc406492-153a-537a-88cf-b7cc02dc905f
| 609,295
|
Let $a, b, c, x$ be reals with $(a+b)(b+c)(c+a) \neq 0$ that satisfy
$$
\frac{a^{2}}{a+b}=\frac{a^{2}}{a+c}+20, \quad \frac{b^{2}}{b+c}=\frac{b^{2}}{b+a}+14, \quad \text { and } \quad \frac{c^{2}}{c+a}=\frac{c^{2}}{c+b}+x
$$
Compute $x$.
|
-34 Note that
$$
\begin{aligned}
\frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}-\frac{a^{2}}{c+a}-\frac{b^{2}}{a+b}-\frac{c^{2}}{b+c} & =\frac{a^{2}-b^{2}}{a+b}+\frac{b^{2}-c^{2}}{b+c}+\frac{c^{2}-a^{2}}{c+a} \\
& =(a-b)+(b-c)+(c-a) \\
& =0
\end{aligned}
$$
Thus, when we sum up all the given equations, we get that $20+14+x=0$. Therefore, $x=-34$.
|
-34
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a, b, c, x$ be reals with $(a+b)(b+c)(c+a) \neq 0$ that satisfy
$$
\frac{a^{2}}{a+b}=\frac{a^{2}}{a+c}+20, \quad \frac{b^{2}}{b+c}=\frac{b^{2}}{b+a}+14, \quad \text { and } \quad \frac{c^{2}}{c+a}=\frac{c^{2}}{c+b}+x
$$
Compute $x$.
|
-34 Note that
$$
\begin{aligned}
\frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}-\frac{a^{2}}{c+a}-\frac{b^{2}}{a+b}-\frac{c^{2}}{b+c} & =\frac{a^{2}-b^{2}}{a+b}+\frac{b^{2}-c^{2}}{b+c}+\frac{c^{2}-a^{2}}{c+a} \\
& =(a-b)+(b-c)+(c-a) \\
& =0
\end{aligned}
$$
Thus, when we sum up all the given equations, we get that $20+14+x=0$. Therefore, $x=-34$.
|
{
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-gen-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nAnswer: "
}
|
938ce06b-bb5a-5f69-8cc0-fc57c9b9f1db
| 609,296
|
For any positive integers $a$ and $b$, define $a \oplus b$ to be the result when adding $a$ to $b$ in binary (base 2), neglecting any carry-overs. For example, $20 \oplus 14=10100_{2} \oplus 1110_{2}=11010_{2}=26$. (The operation $\oplus$ is called the exclusive or.) Compute the sum
$$
\sum_{k=0}^{2^{2014}-1}\left(k \oplus\left\lfloor\frac{k}{2}\right\rfloor\right)
$$
Here $\lfloor x\rfloor$ is the greatest integer not exceeding $x$.
|
$\quad 2^{2013}\left(2^{2014}-1\right)$ OR $2^{4027}-2^{2013}$ Let $k=a_{2013} a_{2012} \ldots a_{0}$ in base 2 . Then $\left\lfloor\frac{k}{2}\right\rfloor=$ $\overline{0 a_{2013} \ldots a_{1}}$ in base 2. So the leftmost digit of $k \oplus\left\lfloor\frac{k}{2}\right\rfloor$ is 1 if and only if $a_{2013}=1$, and the $n$th digit from the right is 1 if and only if $a_{n} \neq a_{n-1}(1 \leq n \leq 2013)$.
In either case, the probability of each digit being 1 is $\frac{1}{2}$. Therefore, the sum of all such numbers is
$$
\frac{1}{2} \cdot 2^{2014} \cdot \underbrace{11 \ldots 11_{2}}_{2014 \text { digits }}=2^{2013}\left(2^{2014}-1\right)
$$
|
2^{2013}\left(2^{2014}-1\right)
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
For any positive integers $a$ and $b$, define $a \oplus b$ to be the result when adding $a$ to $b$ in binary (base 2), neglecting any carry-overs. For example, $20 \oplus 14=10100_{2} \oplus 1110_{2}=11010_{2}=26$. (The operation $\oplus$ is called the exclusive or.) Compute the sum
$$
\sum_{k=0}^{2^{2014}-1}\left(k \oplus\left\lfloor\frac{k}{2}\right\rfloor\right)
$$
Here $\lfloor x\rfloor$ is the greatest integer not exceeding $x$.
|
$\quad 2^{2013}\left(2^{2014}-1\right)$ OR $2^{4027}-2^{2013}$ Let $k=a_{2013} a_{2012} \ldots a_{0}$ in base 2 . Then $\left\lfloor\frac{k}{2}\right\rfloor=$ $\overline{0 a_{2013} \ldots a_{1}}$ in base 2. So the leftmost digit of $k \oplus\left\lfloor\frac{k}{2}\right\rfloor$ is 1 if and only if $a_{2013}=1$, and the $n$th digit from the right is 1 if and only if $a_{n} \neq a_{n-1}(1 \leq n \leq 2013)$.
In either case, the probability of each digit being 1 is $\frac{1}{2}$. Therefore, the sum of all such numbers is
$$
\frac{1}{2} \cdot 2^{2014} \cdot \underbrace{11 \ldots 11_{2}}_{2014 \text { digits }}=2^{2013}\left(2^{2014}-1\right)
$$
|
{
"resource_path": "HarvardMIT/segmented/en-181-2014-nov-gen-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nAnswer: "
}
|
4f0d20d3-d4c5-5ad7-983c-030d0d3af057
| 609,297
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.