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A right triangle has a hypotenuse of length 2 , and one of its legs has length 1 . The altitude to its hypotenuse is drawn. What is the area of the rectangle whose diagonal is this altitude?
|
Call the triangle $A B C$, with $A C=2$ and $B C=1$. By the Pythagorean theorem, $A B=\sqrt{3}$. Call the point at which the altitude intersects the hypotenuse $D$. Let
$E \neq B$ be the vertex of the rectangle on $A B$ and $F \neq B$ be the vertex of the rectangle on $B C$. Triangle $B D C$ is similar to triangle $A B C$, so $B D=\frac{\sqrt{3}}{2}$. Triangle $D B F$ is similar to triangle $A B C$, so $D F=\frac{\sqrt{3}}{4}$ and $B F=\frac{3}{4}$. The area of the rectangle is thus $\frac{\sqrt{3}}{4} \frac{3}{4}=\frac{3 \sqrt{3}}{16}$.
|
\frac{3 \sqrt{3}}{16}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A right triangle has a hypotenuse of length 2 , and one of its legs has length 1 . The altitude to its hypotenuse is drawn. What is the area of the rectangle whose diagonal is this altitude?
|
Call the triangle $A B C$, with $A C=2$ and $B C=1$. By the Pythagorean theorem, $A B=\sqrt{3}$. Call the point at which the altitude intersects the hypotenuse $D$. Let
$E \neq B$ be the vertex of the rectangle on $A B$ and $F \neq B$ be the vertex of the rectangle on $B C$. Triangle $B D C$ is similar to triangle $A B C$, so $B D=\frac{\sqrt{3}}{2}$. Triangle $D B F$ is similar to triangle $A B C$, so $D F=\frac{\sqrt{3}}{4}$ and $B F=\frac{3}{4}$. The area of the rectangle is thus $\frac{\sqrt{3}}{4} \frac{3}{4}=\frac{3 \sqrt{3}}{16}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-gen1-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nSolution: "
}
|
ebf0920f-a62e-53aa-82cf-e1ab6a35f065
| 610,870
|
How many times does 24 divide into 100 !?
|
We first determine the number of times 2 and 3 divide into $100!=1 \cdot 2 \cdot 3 \cdots 100$. Let $\langle N\rangle_{n}$ be the number of times $n$ divides into $N$ (i.e. we want to find $\langle 100!\rangle_{24}$ ). Since 2 only divides into even integers, $\langle 100!\rangle_{2}=\langle 2 \cdot 4 \cdot 6 \cdots 100\rangle$. Factoring out 2 once from each of these multiples, we get that $\langle 100!\rangle_{2}=\left\langle 2^{50} \cdot 1 \cdot 2 \cdot 3 \cdots 50\right\rangle_{2}$. Repeating this process, we find that $\langle 100!\rangle_{2}=\left\langle 20^{50+25+12+6+3+1} \cdot 1\right\rangle_{2}=97$. Similarly, $\langle 100!\rangle_{3}=\left\langle 3^{33+11+3+1}\right\rangle_{3}=48$. Now $24=2^{3} \cdot 3$, so for each factor of 24 in 100 ! there needs to be three multiples of 2 and one multiple of 3 in 100 !. Thus $\langle 100!\rangle_{24}=\left(\left[\langle 100!\rangle_{2} / 3\right]+\langle 100!\rangle_{3}\right)=32$, where $[N]$ is the greatest integer less than or equal to $N$.
|
32
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many times does 24 divide into 100 !?
|
We first determine the number of times 2 and 3 divide into $100!=1 \cdot 2 \cdot 3 \cdots 100$. Let $\langle N\rangle_{n}$ be the number of times $n$ divides into $N$ (i.e. we want to find $\langle 100!\rangle_{24}$ ). Since 2 only divides into even integers, $\langle 100!\rangle_{2}=\langle 2 \cdot 4 \cdot 6 \cdots 100\rangle$. Factoring out 2 once from each of these multiples, we get that $\langle 100!\rangle_{2}=\left\langle 2^{50} \cdot 1 \cdot 2 \cdot 3 \cdots 50\right\rangle_{2}$. Repeating this process, we find that $\langle 100!\rangle_{2}=\left\langle 20^{50+25+12+6+3+1} \cdot 1\right\rangle_{2}=97$. Similarly, $\langle 100!\rangle_{3}=\left\langle 3^{33+11+3+1}\right\rangle_{3}=48$. Now $24=2^{3} \cdot 3$, so for each factor of 24 in 100 ! there needs to be three multiples of 2 and one multiple of 3 in 100 !. Thus $\langle 100!\rangle_{24}=\left(\left[\langle 100!\rangle_{2} / 3\right]+\langle 100!\rangle_{3}\right)=32$, where $[N]$ is the greatest integer less than or equal to $N$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-gen1-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nSolution: "
}
|
8b601ce7-6679-5373-84c2-80ca2ca37bd6
| 610,871
|
Boris was given a Connect Four game set for his birthday, but his color-blindness makes it hard to play the game. Still, he enjoys the shapes he can make by dropping checkers into the set. If the number of shapes possible modulo (horizontal) flips about the vertical axis of symmetry is expressed as $9(1+2+\cdots+n)$, find $n$. (Note: the board is a vertical grid with seven columns and eight rows. A checker is placed into the grid by dropping it from the top of a column, and it falls until it hits either the bottom of the grid or another checker already in that column. Also, $9(1+2+\cdots+n)$ is the number of shapes possible, with two shapes that are horizontal flips of each other counted as one. In other words, the shape that consists solely of 3 checkers in the rightmost row and the shape that consists solely of 3 checkers in the leftmost row are to be considered the same shape.)
|
There are $9^{7}$ total shapes possible, since each of the 7 columns can contain anywhere from 0 to 8 checkers. The number of shapes symmetric with respect to a horizontal flip is the number of shapes of the leftmost four columns, since the configuration of these four columns uniquely determines the configuration of the remaining columns if it is known the shape is symmetric: $9^{4}$. Now we know there are $9^{7}-9^{4}$ non-symmetric shapes, so there are $\frac{9^{7}-9^{4}}{2}$ non-symmetric shapes modulo flips. Thus the total number of shapes modulo flips is $9^{4}+\frac{9^{7}-9^{4}}{2}=9^{4}\left(1+\frac{9^{3}-1}{2}\right)=9^{4}\left(\frac{9^{3}+1}{2}\right)=\left(\frac{3^{8}\left(3^{6}-1\right)}{2}=9 \frac{3^{6}\left(3^{6}+1\right)}{2}=9\left(1+2+\cdots+3^{6}\right)\right.$, so $n=3^{6}=729$.
|
729
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Boris was given a Connect Four game set for his birthday, but his color-blindness makes it hard to play the game. Still, he enjoys the shapes he can make by dropping checkers into the set. If the number of shapes possible modulo (horizontal) flips about the vertical axis of symmetry is expressed as $9(1+2+\cdots+n)$, find $n$. (Note: the board is a vertical grid with seven columns and eight rows. A checker is placed into the grid by dropping it from the top of a column, and it falls until it hits either the bottom of the grid or another checker already in that column. Also, $9(1+2+\cdots+n)$ is the number of shapes possible, with two shapes that are horizontal flips of each other counted as one. In other words, the shape that consists solely of 3 checkers in the rightmost row and the shape that consists solely of 3 checkers in the leftmost row are to be considered the same shape.)
|
There are $9^{7}$ total shapes possible, since each of the 7 columns can contain anywhere from 0 to 8 checkers. The number of shapes symmetric with respect to a horizontal flip is the number of shapes of the leftmost four columns, since the configuration of these four columns uniquely determines the configuration of the remaining columns if it is known the shape is symmetric: $9^{4}$. Now we know there are $9^{7}-9^{4}$ non-symmetric shapes, so there are $\frac{9^{7}-9^{4}}{2}$ non-symmetric shapes modulo flips. Thus the total number of shapes modulo flips is $9^{4}+\frac{9^{7}-9^{4}}{2}=9^{4}\left(1+\frac{9^{3}-1}{2}\right)=9^{4}\left(\frac{9^{3}+1}{2}\right)=\left(\frac{3^{8}\left(3^{6}-1\right)}{2}=9 \frac{3^{6}\left(3^{6}+1\right)}{2}=9\left(1+2+\cdots+3^{6}\right)\right.$, so $n=3^{6}=729$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-gen1-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nSolution: "
}
|
5da2b0e3-8ee7-5160-981b-03412e242bbf
| 610,842
|
Call three sides of an opaque cube adjacent if someone can see them all at once. Draw a plane through the centers of each triple of adjacent sides of a cube with edge length 1. Find the volume of the closed figure bounded by the resulting planes.
|
The volume of the figure is half the volume of the cube (which can be seen by cutting the cube into 8 equal cubes and realizing that the planes cut each of these cubes in half), namely $\frac{1}{2}$.
|
\frac{1}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Call three sides of an opaque cube adjacent if someone can see them all at once. Draw a plane through the centers of each triple of adjacent sides of a cube with edge length 1. Find the volume of the closed figure bounded by the resulting planes.
|
The volume of the figure is half the volume of the cube (which can be seen by cutting the cube into 8 equal cubes and realizing that the planes cut each of these cubes in half), namely $\frac{1}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-gen2-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution: "
}
|
fafe0cb4-4e76-5010-bea5-6917e986b506
| 610,873
|
Some students are taking a math contest, in which each student takes one of four tests. One third of the students take one test, one fourth take another test, one fifth take the next test, and 26 students take the last test. How many students are taking the contest in total?
|
Call the total number of students $n$. We know $n=\frac{n}{3}+\frac{n}{4}+\frac{n}{5}+26$, so $n=120$.
|
120
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Some students are taking a math contest, in which each student takes one of four tests. One third of the students take one test, one fourth take another test, one fifth take the next test, and 26 students take the last test. How many students are taking the contest in total?
|
Call the total number of students $n$. We know $n=\frac{n}{3}+\frac{n}{4}+\frac{n}{5}+26$, so $n=120$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-gen2-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution: "
}
|
26d3c88c-d5f5-58a5-b780-4fbf3d142d18
| 610,875
|
You take a wrong turn on the way to MIT and end up in Transylvania, where $99 \%$ of the inhabitants are vampires and the rest are regular humans. For obvious reasons, you want
to be able to figure out who's who. On average, nine-tenths of the vampires are correctly identified as vampires and nine-tenths of humans are correct identified as humans. What is the probability that someone identified as a human is actually a human?
|
Consider a sample of 1000 inhabitants. On average, 990 are vampires and 10 are people. 99 vampires are identified as human and 9 humans are identified as human. So out of the 108 who pass, only $\frac{1}{12}$ are human.
|
\frac{1}{12}
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
You take a wrong turn on the way to MIT and end up in Transylvania, where $99 \%$ of the inhabitants are vampires and the rest are regular humans. For obvious reasons, you want
to be able to figure out who's who. On average, nine-tenths of the vampires are correctly identified as vampires and nine-tenths of humans are correct identified as humans. What is the probability that someone identified as a human is actually a human?
|
Consider a sample of 1000 inhabitants. On average, 990 are vampires and 10 are people. 99 vampires are identified as human and 9 humans are identified as human. So out of the 108 who pass, only $\frac{1}{12}$ are human.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-gen2-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nSolution: "
}
|
39dd32fd-1aba-5d30-aa41-a49e935566b3
| 610,877
|
A real numbers $x$ is randomly chosen in the interval $\left[-15 \frac{1}{2}, 15 \frac{1}{2}\right]$. Find the probability that the closest integer to $x$ is odd.
|
By using a graphical method, we can see that, for real $x$ on $\left[-n-\frac{1}{2}, n+\frac{1}{2}\right]$, $n$ an even integer, the probability that the closest integer to $x$ is odd is $\frac{n}{2 n+1}$. The desired probability is $\left(\frac{15}{31}\right)$.
|
\frac{15}{31}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
A real numbers $x$ is randomly chosen in the interval $\left[-15 \frac{1}{2}, 15 \frac{1}{2}\right]$. Find the probability that the closest integer to $x$ is odd.
|
By using a graphical method, we can see that, for real $x$ on $\left[-n-\frac{1}{2}, n+\frac{1}{2}\right]$, $n$ an even integer, the probability that the closest integer to $x$ is odd is $\frac{n}{2 n+1}$. The desired probability is $\left(\frac{15}{31}\right)$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-gen2-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nSolution: "
}
|
c3f3ffe2-84fc-5119-a2b5-a159f2218a4b
| 610,878
|
A point on a circle inscribed in a square is 1 and 2 units from the two closest sides of the square. Find the area of the square.
|
Call the point in question $A$, the center of the circle $O$, and its radius $r$. Consider a right triangle $B O A$ with hypotenuse $O A$ : $O A$ has length $r$, and $B O$ and $B A$ have lengths $r-1$ and $r-2$. By the Pythagorean theorem, $(r-1)^{2}+(r-2)^{2}=r^{2} \Rightarrow$ $r^{2}-6 r+5=0 \Rightarrow r=5$ since $r>4$. The area of the square is $(2 r)^{2}=100$.
|
100
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A point on a circle inscribed in a square is 1 and 2 units from the two closest sides of the square. Find the area of the square.
|
Call the point in question $A$, the center of the circle $O$, and its radius $r$. Consider a right triangle $B O A$ with hypotenuse $O A$ : $O A$ has length $r$, and $B O$ and $B A$ have lengths $r-1$ and $r-2$. By the Pythagorean theorem, $(r-1)^{2}+(r-2)^{2}=r^{2} \Rightarrow$ $r^{2}-6 r+5=0 \Rightarrow r=5$ since $r>4$. The area of the square is $(2 r)^{2}=100$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-gen2-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nSolution: "
}
|
7a8b04a4-68de-5685-8577-f83a5333ba05
| 610,879
|
Two circles are concentric. The area of the ring between them is $A$. In terms of $A$, find the length of the longest chord contained entirely within the ring.
|
Let the radii of the circles be $r$ and $R>r$, so $A=\pi\left(R^{2}-r^{2}\right)$. By the Pythagorean theorem, the length of the chord is $2 \sqrt{R^{2}-r^{2}}=2 \sqrt{\frac{A}{\pi}}$.
|
2 \sqrt{\frac{A}{\pi}}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Two circles are concentric. The area of the ring between them is $A$. In terms of $A$, find the length of the longest chord contained entirely within the ring.
|
Let the radii of the circles be $r$ and $R>r$, so $A=\pi\left(R^{2}-r^{2}\right)$. By the Pythagorean theorem, the length of the chord is $2 \sqrt{R^{2}-r^{2}}=2 \sqrt{\frac{A}{\pi}}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-gen2-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nSolution: "
}
|
9b3f126e-ad6a-51c1-9361-47bcfed94f18
| 610,880
|
Find the volume of the tetrahedron with vertices $(5,8,10),(10,10,17),(4,45,46),(2,5,4)$.
|
Each vertex $(x, y, z)$ obeys $x+y=z+3$, so all the vertices are coplanar and the volume of the tetrahedron is 0 .
|
0
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Find the volume of the tetrahedron with vertices $(5,8,10),(10,10,17),(4,45,46),(2,5,4)$.
|
Each vertex $(x, y, z)$ obeys $x+y=z+3$, so all the vertices are coplanar and the volume of the tetrahedron is 0 .
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-gen2-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nSolution: "
}
|
fb87f771-76e8-5b70-b648-03dab7458516
| 610,881
|
Call three sides of an opaque cube adjacent if someone can see them all at once. Draw a plane through the centers of each triple of adjacent sides of a cube with edge length 1 . Find the volume of the closed figure bounded by the resulting planes.
|
The volume of the figure is half the volume of the cube (which can be seen by cutting the cube into 8 equal cubes and realizing that the planes cut each of these cubes in half), namely $\frac{1}{2}$.
|
\frac{1}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Call three sides of an opaque cube adjacent if someone can see them all at once. Draw a plane through the centers of each triple of adjacent sides of a cube with edge length 1 . Find the volume of the closed figure bounded by the resulting planes.
|
The volume of the figure is half the volume of the cube (which can be seen by cutting the cube into 8 equal cubes and realizing that the planes cut each of these cubes in half), namely $\frac{1}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-geo-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution: "
}
|
57f40acf-2dab-5b32-9499-e54602c0d58f
| 610,883
|
Square $A B C D$ is drawn. Isosceles Triangle $C D E$ is drawn with $E$ a right angle. Square $D E F G$ is drawn. Isosceles triangle $F G H$ is drawn with $H$ a right angle. This process is repeated infinitely so that no two figures overlap each other. If square $A B C D$ has area 1 , compute the area of the entire figure.
|
Let the area of the $n$th square drawn be $S_{n}$ and the area of the $n$th triangle be $T_{n}$. Since the hypotenuse of the $n$th triangle is of length $\sqrt{S_{n}}$, its legs are of length $l=\sqrt{\frac{S_{n}}{2}}$, so $S_{n+1}=l^{2}=\frac{S_{n}}{2}$ and $T_{n}=\frac{l^{2}}{2}=\frac{S_{n}}{4}$. Using the recursion relations, $S_{n}=\frac{1}{2^{n-1}}$ and $T_{n}=\frac{1}{2^{n+1}}$, so $S_{n}+T_{n}=\frac{1}{2^{n-1}}+\frac{1}{2^{n+1}}=\left(\frac{1}{2}+2\right) \frac{1}{2^{n}}=\frac{5}{2} \frac{1}{2^{n}}$. Thus the total area of the figure is $\sum_{n=1}^{\infty} S_{n}+T_{n}=\frac{5}{2} \sum_{n=1}^{\infty} \frac{1}{2^{n}}=\frac{5}{2}$.
|
\frac{5}{2}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Square $A B C D$ is drawn. Isosceles Triangle $C D E$ is drawn with $E$ a right angle. Square $D E F G$ is drawn. Isosceles triangle $F G H$ is drawn with $H$ a right angle. This process is repeated infinitely so that no two figures overlap each other. If square $A B C D$ has area 1 , compute the area of the entire figure.
|
Let the area of the $n$th square drawn be $S_{n}$ and the area of the $n$th triangle be $T_{n}$. Since the hypotenuse of the $n$th triangle is of length $\sqrt{S_{n}}$, its legs are of length $l=\sqrt{\frac{S_{n}}{2}}$, so $S_{n+1}=l^{2}=\frac{S_{n}}{2}$ and $T_{n}=\frac{l^{2}}{2}=\frac{S_{n}}{4}$. Using the recursion relations, $S_{n}=\frac{1}{2^{n-1}}$ and $T_{n}=\frac{1}{2^{n+1}}$, so $S_{n}+T_{n}=\frac{1}{2^{n-1}}+\frac{1}{2^{n+1}}=\left(\frac{1}{2}+2\right) \frac{1}{2^{n}}=\frac{5}{2} \frac{1}{2^{n}}$. Thus the total area of the figure is $\sum_{n=1}^{\infty} S_{n}+T_{n}=\frac{5}{2} \sum_{n=1}^{\infty} \frac{1}{2^{n}}=\frac{5}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-geo-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution: "
}
|
581852df-2660-51cd-8d94-9a8dd2b899a6
| 610,884
|
A circle has two parallel chords of length $x$ that are $x$ units apart. If the part of the circle included between the chords has area $2+\pi$, find $x$.
|
Let $C$ be the area of the circle, $S$ be the area of the square two of whose edges are the chords, and $A$ be the area of the part of the circle included between the chords. The radius of the circle is $\frac{\sqrt{2}}{2} x$, so $C=\frac{\pi}{2} x^{2}$, and $S=x^{2}$. Then the area $A$ is the area of the square plus one half of the difference between the areas of the circle and square: $A=\frac{C-S}{2}+S=\frac{C+S}{2}=\frac{1+\frac{\pi}{2}}{2} x^{2}$, so $x=\sqrt{\frac{2 A}{1+\frac{\pi}{2}}}=2$.
|
2
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A circle has two parallel chords of length $x$ that are $x$ units apart. If the part of the circle included between the chords has area $2+\pi$, find $x$.
|
Let $C$ be the area of the circle, $S$ be the area of the square two of whose edges are the chords, and $A$ be the area of the part of the circle included between the chords. The radius of the circle is $\frac{\sqrt{2}}{2} x$, so $C=\frac{\pi}{2} x^{2}$, and $S=x^{2}$. Then the area $A$ is the area of the square plus one half of the difference between the areas of the circle and square: $A=\frac{C-S}{2}+S=\frac{C+S}{2}=\frac{1+\frac{\pi}{2}}{2} x^{2}$, so $x=\sqrt{\frac{2 A}{1+\frac{\pi}{2}}}=2$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-geo-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution: "
}
|
1c9de47f-87ae-5f5f-a8da-30691fcbdd13
| 610,885
|
Equilateral triangle $A B C$ with side length 1 is drawn. A square is drawn such that its vertex at $A$ is opposite to its vertex at the midpoint of $B C$. Find the area enclosed within the intersection of the insides of the triangle and square. Hint: $\sin 75=\frac{\sqrt{2}(\sqrt{3}+1)}{4}$.
|
Let $D$ be the midpoint of $B C, F \neq A$ be the point of intersection of the square and triangle lying on $A C, b$ be the length of $F C, x$ be the side length of the triangle, and $y$ be the length of $A D$. By the law of sines on triangle $C D F$, we have $\frac{2 \sin 75}{x}=\frac{\sin 45}{b}$, so $b=\frac{x \sin 45}{2 \sin 75}=\frac{\sqrt{2} x}{4 \sin 75}$. The area of the desired figure can easily be seen to be $\frac{1}{2}(x-b) y$ since it can be seen as two triangles of width $y$ and height $\frac{x-b}{2}$. This reduces to $\frac{1}{2}\left(1-\frac{\sqrt{2}}{4 \sin 75}\right) x y$. Then by the Pythagorean theorem on triangle $A B D, x^{2}=\left(\frac{x}{2}\right)^{2}+y^{2}$, so $y=\frac{\sqrt{3}}{2} x$, and the area becomes $\frac{\sqrt{3}}{4}\left(1-\frac{\sqrt{2}}{4 \sin 75}\right) x^{2}=\frac{\sqrt{3}}{4}\left(1-\frac{\sqrt{2}}{4 \sin 75}\right)=\frac{3}{4(\sqrt{3}+1)}$.
|
\frac{3}{4(\sqrt{3}+1)}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Equilateral triangle $A B C$ with side length 1 is drawn. A square is drawn such that its vertex at $A$ is opposite to its vertex at the midpoint of $B C$. Find the area enclosed within the intersection of the insides of the triangle and square. Hint: $\sin 75=\frac{\sqrt{2}(\sqrt{3}+1)}{4}$.
|
Let $D$ be the midpoint of $B C, F \neq A$ be the point of intersection of the square and triangle lying on $A C, b$ be the length of $F C, x$ be the side length of the triangle, and $y$ be the length of $A D$. By the law of sines on triangle $C D F$, we have $\frac{2 \sin 75}{x}=\frac{\sin 45}{b}$, so $b=\frac{x \sin 45}{2 \sin 75}=\frac{\sqrt{2} x}{4 \sin 75}$. The area of the desired figure can easily be seen to be $\frac{1}{2}(x-b) y$ since it can be seen as two triangles of width $y$ and height $\frac{x-b}{2}$. This reduces to $\frac{1}{2}\left(1-\frac{\sqrt{2}}{4 \sin 75}\right) x y$. Then by the Pythagorean theorem on triangle $A B D, x^{2}=\left(\frac{x}{2}\right)^{2}+y^{2}$, so $y=\frac{\sqrt{3}}{2} x$, and the area becomes $\frac{\sqrt{3}}{4}\left(1-\frac{\sqrt{2}}{4 \sin 75}\right) x^{2}=\frac{\sqrt{3}}{4}\left(1-\frac{\sqrt{2}}{4 \sin 75}\right)=\frac{3}{4(\sqrt{3}+1)}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-geo-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nSolution: "
}
|
4df1d632-1923-51c6-aeec-a82308e5788e
| 610,886
|
Point D is drawn on side $B C$ of equilateral triangle $A B C$, and $A D$ is extended past $D$ to $E$ such that angles $E A C$ and $E B C$ are equal. If $B E=5$ and $C E=12$, determine the length of $A E$.
|
By construction, $A B E C$ is a cyclic quadrilateral. Ptolemy's theorem says that for cyclic quadrilaterals, the sum of the products of the lengths of the opposite sides equals the product of the lengths of the diagonals. This yields $(B C)(A E)=(B A)(C E)+(B E)(A C)$. Since $A B C$ is equilateral, $B C=A C=A B$, so dividing out by this common value we get $A E=C E+B E=17$.
|
17
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Point D is drawn on side $B C$ of equilateral triangle $A B C$, and $A D$ is extended past $D$ to $E$ such that angles $E A C$ and $E B C$ are equal. If $B E=5$ and $C E=12$, determine the length of $A E$.
|
By construction, $A B E C$ is a cyclic quadrilateral. Ptolemy's theorem says that for cyclic quadrilaterals, the sum of the products of the lengths of the opposite sides equals the product of the lengths of the diagonals. This yields $(B C)(A E)=(B A)(C E)+(B E)(A C)$. Since $A B C$ is equilateral, $B C=A C=A B$, so dividing out by this common value we get $A E=C E+B E=17$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-geo-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nSolution:"
}
|
27ae9bd0-4f77-5dba-b8b7-0d4c5f5a502c
| 610,887
|
Parallelogram $A E C F$ is inscribed in square $A B C D$. It is reflected across diagonal $A C$ to form another parallelogram $A E^{\prime} C F^{\prime}$. The region common to both parallelograms has area $m$ and perimeter $n$. Compute the value of $\frac{m}{n^{2}}$ if $A F: A D=1: 4$.
|
By symmetry, the region is a rhombus, AXCY, centered at the center of the square, $O$. Consider isoceles right triangle $A C D$. By the technique of mass points, we find that $D O: Y O=7: 1$. Therefore, the rhombus is composed of four triangles, whose sides are in the ratio $1: 7: 5 \sqrt{2}$. The perimeter of the rhombus is $20 \sqrt{2} N$, and the area is $14 N^{2}$. The required ratio is thus $\frac{7}{400}$.
|
\frac{7}{400}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Parallelogram $A E C F$ is inscribed in square $A B C D$. It is reflected across diagonal $A C$ to form another parallelogram $A E^{\prime} C F^{\prime}$. The region common to both parallelograms has area $m$ and perimeter $n$. Compute the value of $\frac{m}{n^{2}}$ if $A F: A D=1: 4$.
|
By symmetry, the region is a rhombus, AXCY, centered at the center of the square, $O$. Consider isoceles right triangle $A C D$. By the technique of mass points, we find that $D O: Y O=7: 1$. Therefore, the rhombus is composed of four triangles, whose sides are in the ratio $1: 7: 5 \sqrt{2}$. The perimeter of the rhombus is $20 \sqrt{2} N$, and the area is $14 N^{2}$. The required ratio is thus $\frac{7}{400}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-geo-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nSolution: "
}
|
2b4899d7-37c5-502e-b5f0-a1591798f644
| 610,888
|
$A$ is the center of a semicircle, with radius $A D$ lying on the base. $B$ lies on the base between $A$ and $D$, and $E$ is on the circular portion of the semicircle such that $E B A$ is a right angle. Extend $E A$ through $A$ to $C$, and put $F$ on line $C D$ such that $E B F$ is a line. Now $E A=1, A C=\sqrt{2}, B F=\frac{2-\sqrt{2}}{4}, C F=\frac{2 \sqrt{5}+\sqrt{10}}{4}$, and $D F=\frac{2 \sqrt{5}-\sqrt{10}}{4}$. Find $D E$.
|
Let $\theta=\angle A E D$ and $x=D E$. By the law of cosines on triangle $A D E$, we have $1=1+x^{2}-2 x \cos \theta \Rightarrow 2 x \cos \theta=x^{2}$. Then by the law of cosines on triangle $C D E$ (note that $C D=\sqrt{5})$, we have $5=(1+\sqrt{2})^{2}+x^{2}-2(1+\sqrt{2}) x \cos \theta=(1+\sqrt{2})^{2}+x^{2}-(1+\sqrt{2}) x^{2}$. Solving the quadratic equation gives $x=\sqrt{2-\sqrt{2}}$.
|
\sqrt{2-\sqrt{2}}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
$A$ is the center of a semicircle, with radius $A D$ lying on the base. $B$ lies on the base between $A$ and $D$, and $E$ is on the circular portion of the semicircle such that $E B A$ is a right angle. Extend $E A$ through $A$ to $C$, and put $F$ on line $C D$ such that $E B F$ is a line. Now $E A=1, A C=\sqrt{2}, B F=\frac{2-\sqrt{2}}{4}, C F=\frac{2 \sqrt{5}+\sqrt{10}}{4}$, and $D F=\frac{2 \sqrt{5}-\sqrt{10}}{4}$. Find $D E$.
|
Let $\theta=\angle A E D$ and $x=D E$. By the law of cosines on triangle $A D E$, we have $1=1+x^{2}-2 x \cos \theta \Rightarrow 2 x \cos \theta=x^{2}$. Then by the law of cosines on triangle $C D E$ (note that $C D=\sqrt{5})$, we have $5=(1+\sqrt{2})^{2}+x^{2}-2(1+\sqrt{2}) x \cos \theta=(1+\sqrt{2})^{2}+x^{2}-(1+\sqrt{2}) x^{2}$. Solving the quadratic equation gives $x=\sqrt{2-\sqrt{2}}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-geo-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nSolution: "
}
|
50c5f75a-3315-51a1-8604-e0858f14c065
| 610,889
|
January 3 , 1911 was an odd date as its abbreviated representation, $1 / 3 / 1911$, can be written using only odd digits (note all four digits are written for the year). To the nearest month, how many months will have elapsed between the most recent odd date and the next odd date (today is $3 / 3 / 2001$, an even date).
|
The most recent odd date was 11/19/1999 (November has 30 days, but the assumption that it has 31 days does not change the answer), and the next odd date will be $1 / 1 / 3111$. From $11 / 19 / 1999$ to $1 / 1 / 2000$ is about 1 month. From 2000 to 3111 is 1111 years, or $12 \cdot 1111=13332$ months, so the total number of months is 13333 .
|
13333
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
January 3 , 1911 was an odd date as its abbreviated representation, $1 / 3 / 1911$, can be written using only odd digits (note all four digits are written for the year). To the nearest month, how many months will have elapsed between the most recent odd date and the next odd date (today is $3 / 3 / 2001$, an even date).
|
The most recent odd date was 11/19/1999 (November has 30 days, but the assumption that it has 31 days does not change the answer), and the next odd date will be $1 / 1 / 3111$. From $11 / 19 / 1999$ to $1 / 1 / 2000$ is about 1 month. From 2000 to 3111 is 1111 years, or $12 \cdot 1111=13332$ months, so the total number of months is 13333 .
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n1. [5]",
"solution_match": "\nSolution: "
}
|
42c9d8cb-ddc5-53b1-83fe-207f6a0acbb1
| 610,890
|
Ken is the best sugar cube retailer in the nation. Trevor, who loves sugar, is coming over to make an order. Ken knows Trevor cannot afford more than 127 sugar cubes, but might ask for any number of cubes less than or equal to that. Ken prepares seven cups of cubes, with which he can satisfy any order Trevor might make. How many cubes are in the cup with the most sugar?
|
The only way to fill seven cups to satisfy the above condition is to use a binary scheme, so the cups must contain $1,2,4,8,16,32$, and 64 cubes of sugar.
|
64
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Ken is the best sugar cube retailer in the nation. Trevor, who loves sugar, is coming over to make an order. Ken knows Trevor cannot afford more than 127 sugar cubes, but might ask for any number of cubes less than or equal to that. Ken prepares seven cups of cubes, with which he can satisfy any order Trevor might make. How many cubes are in the cup with the most sugar?
|
The only way to fill seven cups to satisfy the above condition is to use a binary scheme, so the cups must contain $1,2,4,8,16,32$, and 64 cubes of sugar.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n2. [4]",
"solution_match": "\nSolution: "
}
|
67159a0d-a7d7-5cd9-8077-39e426e2d704
| 610,891
|
Find the number of triangulations of a general convex 7 -gon into 5 triangles by 4 diagonals that do not intersect in their interiors.
|
Define the Catalan numbers by $C(n)=\frac{1}{n+1}\binom{2 n}{n}$. The current solution is the $C($ number of triangles $)=C(5)=42$.
|
42
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Find the number of triangulations of a general convex 7 -gon into 5 triangles by 4 diagonals that do not intersect in their interiors.
|
Define the Catalan numbers by $C(n)=\frac{1}{n+1}\binom{2 n}{n}$. The current solution is the $C($ number of triangles $)=C(5)=42$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n3. [7]",
"solution_match": "\nSolution: "
}
|
505dda04-e474-5dd1-b24d-982d4d10ffe9
| 610,892
|
Find $\prod_{n=2}^{\infty}\left(1-\frac{1}{n^{2}}\right)$.
|
$\prod_{n=2}^{\infty}\left(1-\frac{1}{n^{2}}\right)=\prod_{n=2}^{\infty} \frac{n^{2}-1}{n^{2}}=\prod_{n=2}^{\infty} \frac{(n-1)(n+1)}{n \cdot n}=\frac{1 \cdot 3}{2 \cdot 2} \frac{2 \cdot 4}{3 \cdot 3} \frac{3 \cdot 5}{4 \cdot 4} \frac{4 \cdot 6}{5 \cdot 5} \frac{5 \cdot 7}{6 \cdot 6} \cdots=\frac{1 \cdot 2 \cdot 3 \cdot 3 \cdot \cdot \cdot \cdot \cdot \cdot \cdot 5 \cdot 5 \cdots}{2 \cdot 2 \cdot 3 \cdot 3 \cdot 4 \cdot 4 \cdot 5 \cdot 5}=$ $\frac{1}{2}$.
|
\frac{1}{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find $\prod_{n=2}^{\infty}\left(1-\frac{1}{n^{2}}\right)$.
|
$\prod_{n=2}^{\infty}\left(1-\frac{1}{n^{2}}\right)=\prod_{n=2}^{\infty} \frac{n^{2}-1}{n^{2}}=\prod_{n=2}^{\infty} \frac{(n-1)(n+1)}{n \cdot n}=\frac{1 \cdot 3}{2 \cdot 2} \frac{2 \cdot 4}{3 \cdot 3} \frac{3 \cdot 5}{4 \cdot 4} \frac{4 \cdot 6}{5 \cdot 5} \frac{5 \cdot 7}{6 \cdot 6} \cdots=\frac{1 \cdot 2 \cdot 3 \cdot 3 \cdot \cdot \cdot \cdot \cdot \cdot \cdot 5 \cdot 5 \cdots}{2 \cdot 2 \cdot 3 \cdot 3 \cdot 4 \cdot 4 \cdot 5 \cdot 5}=$ $\frac{1}{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n4. [7]",
"solution_match": "\nSolution: "
}
|
18a45ef7-844b-5f50-b274-45fdeda89361
| 610,893
|
Let $A B C$ be a triangle with incenter $I$ and circumcenter $O$. Let the circumradius be $R$. What is the least upper bound of all possible values of $I O$ ?
|
$I$ always lies inside the convex hull of $A B C$, which in turn always lies in the circumcircle of $A B C$, so $I O<R$. On the other hand, if we first draw the circle $\Omega$ of radius $R$ about $O$ and then pick $A, B$, and $C$ very close together on it, we can force the convex hull of $A B C$ to lie outside the circle of radius $R-\epsilon$ about $O$ for any $\epsilon$. Thus the answer is $R$.
|
R
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $A B C$ be a triangle with incenter $I$ and circumcenter $O$. Let the circumradius be $R$. What is the least upper bound of all possible values of $I O$ ?
|
$I$ always lies inside the convex hull of $A B C$, which in turn always lies in the circumcircle of $A B C$, so $I O<R$. On the other hand, if we first draw the circle $\Omega$ of radius $R$ about $O$ and then pick $A, B$, and $C$ very close together on it, we can force the convex hull of $A B C$ to lie outside the circle of radius $R-\epsilon$ about $O$ for any $\epsilon$. Thus the answer is $R$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n5. $[ \\pm 6]$",
"solution_match": "\nSolution: "
}
|
b928afa5-cbc1-5c1f-b70d-fd42a9c76e13
| 610,894
|
Six students taking a test sit in a row of seats with aisles only on the two sides of the row. If they finish the test at random times, what is the probability that some student will have to pass by another student to get to an aisle?
|
The probability $p$ that no student will have to pass by another student to get to an aisle is the probability that the first student to leave is one of the students on the end, the next student to leave is on one of the ends of the remaining students, etc.: $p=\frac{2}{6} \cdot \frac{2}{5} \cdot \frac{2}{4} \cdot \frac{2}{3}$, so the desired probability is $1-p=\frac{43}{45}$.
|
\frac{43}{45}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Six students taking a test sit in a row of seats with aisles only on the two sides of the row. If they finish the test at random times, what is the probability that some student will have to pass by another student to get to an aisle?
|
The probability $p$ that no student will have to pass by another student to get to an aisle is the probability that the first student to leave is one of the students on the end, the next student to leave is on one of the ends of the remaining students, etc.: $p=\frac{2}{6} \cdot \frac{2}{5} \cdot \frac{2}{4} \cdot \frac{2}{3}$, so the desired probability is $1-p=\frac{43}{45}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n6. [8]",
"solution_match": "\nSolution: "
}
|
3e958d15-7707-5bcb-b962-bf78a8094969
| 610,895
|
Compute $1 \cdot 2+2 \cdot 3+\cdots+(n-1) n$.
|
Let $S=1 \cdot 2+2 \cdot 3+\cdots+(n-1) n$. We know $\sum_{i=1}^{n} i=\frac{n(n+1)}{2}$ and $\sum_{i=1}^{n} i^{2}=$ $\frac{n(n+1)(2 n+1)}{6}$. So $S=1(1+1)+2(2+1)+\cdots+(n-1) n=\left(1^{2}+2^{2}+\cdots+(n-1)^{2}\right)+$ $(1+2+\cdots+(n-1))=\frac{(n-1)(n)(2 n-1)}{6}+\frac{(n-1)(n)}{2}=\frac{(n-1) n(n+1)}{3}$.
We can also arrive at the solution by realizing that $\sum_{i=1}^{n} i^{2}=\sum_{i=1}^{n} i^{2}+\sum_{i=2}^{n} i^{2}+\sum_{i=3}^{n} i^{2}+\cdots+$ $\sum_{i=n}^{n} i^{2}=n \sum_{i=1}^{n} i^{2}-\left(\sum_{i=1}^{1} i^{2}+\sum_{i=1}^{2} i^{2}+\sum_{i=1}^{3} i^{2}+\cdots \sum_{n=1}^{n-1} i^{2}\right)=n \frac{n(n+1)}{2}-\left(\frac{1 \cdot 2}{2}+\frac{2 \cdot 3}{2}+\cdots+\frac{(n-1) n}{2}\right)=$ $n \frac{n(n+1)}{2}-\frac{1}{2} S=\frac{n(n+1)(2 n+1)}{6}$, so $S=\frac{(n-1) n(n+1)}{3}$.
|
\frac{(n-1) n(n+1)}{3}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Compute $1 \cdot 2+2 \cdot 3+\cdots+(n-1) n$.
|
Let $S=1 \cdot 2+2 \cdot 3+\cdots+(n-1) n$. We know $\sum_{i=1}^{n} i=\frac{n(n+1)}{2}$ and $\sum_{i=1}^{n} i^{2}=$ $\frac{n(n+1)(2 n+1)}{6}$. So $S=1(1+1)+2(2+1)+\cdots+(n-1) n=\left(1^{2}+2^{2}+\cdots+(n-1)^{2}\right)+$ $(1+2+\cdots+(n-1))=\frac{(n-1)(n)(2 n-1)}{6}+\frac{(n-1)(n)}{2}=\frac{(n-1) n(n+1)}{3}$.
We can also arrive at the solution by realizing that $\sum_{i=1}^{n} i^{2}=\sum_{i=1}^{n} i^{2}+\sum_{i=2}^{n} i^{2}+\sum_{i=3}^{n} i^{2}+\cdots+$ $\sum_{i=n}^{n} i^{2}=n \sum_{i=1}^{n} i^{2}-\left(\sum_{i=1}^{1} i^{2}+\sum_{i=1}^{2} i^{2}+\sum_{i=1}^{3} i^{2}+\cdots \sum_{n=1}^{n-1} i^{2}\right)=n \frac{n(n+1)}{2}-\left(\frac{1 \cdot 2}{2}+\frac{2 \cdot 3}{2}+\cdots+\frac{(n-1) n}{2}\right)=$ $n \frac{n(n+1)}{2}-\frac{1}{2} S=\frac{n(n+1)(2 n+1)}{6}$, so $S=\frac{(n-1) n(n+1)}{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n8. [10]",
"solution_match": "\nSolution: "
}
|
5171d1ad-d080-545d-b15b-90e88f57276d
| 610,897
|
Suppose $x$ satisfies $x^{3}+x^{2}+x+1=0$. What are all possible values of $x^{4}+2 x^{3}+$ $2 x^{2}+2 x+1$ ?
|
$x^{4}+2 x^{3}+2 x^{2}+2 x+1=(x+1)\left(x^{3}+x^{2}+x+1\right)=0$ is the only possible solution.
|
0
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Suppose $x$ satisfies $x^{3}+x^{2}+x+1=0$. What are all possible values of $x^{4}+2 x^{3}+$ $2 x^{2}+2 x+1$ ?
|
$x^{4}+2 x^{3}+2 x^{2}+2 x+1=(x+1)\left(x^{3}+x^{2}+x+1\right)=0$ is the only possible solution.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n9. [5]",
"solution_match": "\nSolution: "
}
|
bb256ad6-d037-5cc8-86dd-60be2dd16dac
| 610,898
|
Two concentric circles have radii $r$ and $R>r$. Three new circles are drawn so that they are each tangent to the big two circles and tangent to the other two new circles. Find $\frac{R}{r}$.
|
The centers of the three new circles form a triangle. The diameter of the new circles is $R-r$, so the side length of the triangle is $R-r$. Call the center of the concentric circle $O$, two vertices of the triangle $A$ and $B$, and $A B$ 's midpoint $D$. $O A$ is the average $R$ and $r$, namely $\frac{R+r}{2}$. Using the law of sines on triangle $D A O$, we get $\frac{\sin (30)}{A D}=\frac{\sin (90)}{A O} \Rightarrow R=3 r$, so $\frac{R}{r}=3$.
|
3
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Two concentric circles have radii $r$ and $R>r$. Three new circles are drawn so that they are each tangent to the big two circles and tangent to the other two new circles. Find $\frac{R}{r}$.
|
The centers of the three new circles form a triangle. The diameter of the new circles is $R-r$, so the side length of the triangle is $R-r$. Call the center of the concentric circle $O$, two vertices of the triangle $A$ and $B$, and $A B$ 's midpoint $D$. $O A$ is the average $R$ and $r$, namely $\frac{R+r}{2}$. Using the law of sines on triangle $D A O$, we get $\frac{\sin (30)}{A D}=\frac{\sin (90)}{A O} \Rightarrow R=3 r$, so $\frac{R}{r}=3$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n10. [8]",
"solution_match": "\nSolution: "
}
|
97351a09-392c-50ce-84d4-8ae2470da2a9
| 610,899
|
12 points are placed around the circumference of a circle. How many ways are there to draw 6 non-intersecting chords joining these points in pairs?
|
$C$ (number of chords) $=C(6)=132$.
|
132
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
12 points are placed around the circumference of a circle. How many ways are there to draw 6 non-intersecting chords joining these points in pairs?
|
$C$ (number of chords) $=C(6)=132$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n11. [8]",
"solution_match": "\nSolution: "
}
|
71ea1653-b37a-51c3-99c2-aca93cdae649
| 610,900
|
How many distinct sets of 8 positive odd integers sum to 20 ?
|
This is the same as the number of ways 8 nonnegative even integers sum to 12 (we subtract 1 from each integer in the above sum). All 11 possibilities are (leaving out 0s): $12,10+2,8+4,8+2+2,6+6,6+4+2,6+2+2+2+2,4+4+4,4+4+2+2$, $4+2+2+2+2,2+2+2+2+2+2$.
|
11
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many distinct sets of 8 positive odd integers sum to 20 ?
|
This is the same as the number of ways 8 nonnegative even integers sum to 12 (we subtract 1 from each integer in the above sum). All 11 possibilities are (leaving out 0s): $12,10+2,8+4,8+2+2,6+6,6+4+2,6+2+2+2+2,4+4+4,4+4+2+2$, $4+2+2+2+2,2+2+2+2+2+2$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n12. $[ \\pm 6]$",
"solution_match": "\nSolution: "
}
|
c265b892-0f38-5347-b908-4affdfff8e0d
| 610,901
|
Find the number of real zeros of $x^{3}-x^{2}-x+2$.
|
Let $f(x)=x^{3}-x^{2}-x+2$, so $f^{\prime}(x)=3 x^{2}-2 x-1$. The slope is zero when $3 x^{2}-2 x-1=0$, where $x=-\frac{1}{3}$ and $x=1$. Now $f\left(\frac{1}{3}\right)>0$ and $f(1)>0$, so there are no zeros between $x=-\frac{1}{3}$ and $x=1$. Since $\lim _{x \rightarrow+\infty} f(x)>0$, there are no zeros for $x>1$. Since $\lim _{x \rightarrow-\infty} f(x)<0$, there is one zero for $x<-\frac{1}{3}$, for a total of 1 zero.
|
1
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find the number of real zeros of $x^{3}-x^{2}-x+2$.
|
Let $f(x)=x^{3}-x^{2}-x+2$, so $f^{\prime}(x)=3 x^{2}-2 x-1$. The slope is zero when $3 x^{2}-2 x-1=0$, where $x=-\frac{1}{3}$ and $x=1$. Now $f\left(\frac{1}{3}\right)>0$ and $f(1)>0$, so there are no zeros between $x=-\frac{1}{3}$ and $x=1$. Since $\lim _{x \rightarrow+\infty} f(x)>0$, there are no zeros for $x>1$. Since $\lim _{x \rightarrow-\infty} f(x)<0$, there is one zero for $x<-\frac{1}{3}$, for a total of 1 zero.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n13. [5]",
"solution_match": "\nSolution: "
}
|
85b0e791-5ac1-5c6a-af06-76f9776dd33f
| 610,902
|
Find the exact value of $1+\frac{1}{1+\frac{2}{1+\frac{1}{1++\frac{2}{1+\ldots}}}}$.
|
Let $x$ be what we are trying to find. $x-1=\frac{1}{1+\frac{2}{1+\frac{1}{1+\frac{2}{1+\ldots}}}} \Rightarrow \frac{1}{x-1}-1=$ $\frac{2}{1+\frac{1}{1+\frac{2}{1+\cdots}}} \Rightarrow \frac{2}{\frac{1}{x-1}-1}=x \Rightarrow x^{2}-2=0$, so $x=\sqrt{2}$ since $x>0$.
|
\sqrt{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find the exact value of $1+\frac{1}{1+\frac{2}{1+\frac{1}{1++\frac{2}{1+\ldots}}}}$.
|
Let $x$ be what we are trying to find. $x-1=\frac{1}{1+\frac{2}{1+\frac{1}{1+\frac{2}{1+\ldots}}}} \Rightarrow \frac{1}{x-1}-1=$ $\frac{2}{1+\frac{1}{1+\frac{2}{1+\cdots}}} \Rightarrow \frac{2}{\frac{1}{x-1}-1}=x \Rightarrow x^{2}-2=0$, so $x=\sqrt{2}$ since $x>0$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n14. [8]",
"solution_match": "\nSolution: "
}
|
44cc5ac4-c1e2-59e3-9d17-c738e7e002ac
| 610,903
|
Calculate $\sum_{n=1}^{2001} n^{3}$.
|
$\sum_{n=1}^{2001} n^{3}=\left(\sum_{n=1}^{2001} n\right)^{2}=\left(\frac{2001 \cdot 2002}{2}\right)^{2}=4012013006001$.
|
4012013006001
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Calculate $\sum_{n=1}^{2001} n^{3}$.
|
$\sum_{n=1}^{2001} n^{3}=\left(\sum_{n=1}^{2001} n\right)^{2}=\left(\frac{2001 \cdot 2002}{2}\right)^{2}=4012013006001$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n19. [9]",
"solution_match": "\nSolution: "
}
|
a9da7dc7-e0a4-5eb2-8cf4-d00aec1317ab
| 610,908
|
Karen has seven envelopes and seven letters of congratulations to various HMMT coaches. If she places the letters in the envelopes at random with each possible configuration having an equal probability, what is the probability that exactly six of the letters are in the correct envelopes?
|
0 , since if six letters are in their correct envelopes the seventh is as well.
|
0
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Karen has seven envelopes and seven letters of congratulations to various HMMT coaches. If she places the letters in the envelopes at random with each possible configuration having an equal probability, what is the probability that exactly six of the letters are in the correct envelopes?
|
0 , since if six letters are in their correct envelopes the seventh is as well.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n20. [ $\\pm 4]$",
"solution_match": "\nSolution: "
}
|
f2ca2694-5a3b-568b-a112-3199259e9b24
| 610,909
|
Evaluate $\sum_{i=1}^{\infty} \frac{(i+1)(i+2)(i+3)}{(-2)^{i}}$.
|
This is the power series of $\frac{6}{(1+x)^{4}}$ expanded about $x=0$ and evaluated at $x=-\frac{1}{2}$, so the solution is 96 .
|
96
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Evaluate $\sum_{i=1}^{\infty} \frac{(i+1)(i+2)(i+3)}{(-2)^{i}}$.
|
This is the power series of $\frac{6}{(1+x)^{4}}$ expanded about $x=0$ and evaluated at $x=-\frac{1}{2}$, so the solution is 96 .
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n21. [10]",
"solution_match": "\nSolution: "
}
|
01d909d2-bd54-5c1e-bb52-fb35499fb1b2
| 610,910
|
A man is standing on a platform and sees his train move such that after $t$ seconds it is $2 t^{2}+d_{0}$ feet from his original position, where $d_{0}$ is some number. Call the smallest (constant) speed at which the man have to run so that he catches the train $v$. In terms of $n$, find the $n$th smallest value of $d_{0}$ that makes $v$ a perfect square.
|
The train's distance from the man's original position is $t^{2}+d_{0}$, and the man's distance from his original position if he runs at speed $v$ is $v t$ at time $t$. We need to find where $t^{2}+d_{0}=v t$ has a solution. Note that this is a quadratic equation with discriminant $D=\sqrt{v^{2}-4 d_{0}}$, so it has solutions for real $D$, i.e. where $v \geq \sqrt{4 d_{0}}$, so $4 d_{0}$ must be a perfect square. This happens when $4 d_{0}$ is an even power of 2 : the smallest value is $2^{0}$, the second smallest is $2^{2}$, the third smallest is $2^{4}$, and in general the $n$th smallest is $2^{2(n-1)}$, or $4^{n-1}$.
|
4^{n-1}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
A man is standing on a platform and sees his train move such that after $t$ seconds it is $2 t^{2}+d_{0}$ feet from his original position, where $d_{0}$ is some number. Call the smallest (constant) speed at which the man have to run so that he catches the train $v$. In terms of $n$, find the $n$th smallest value of $d_{0}$ that makes $v$ a perfect square.
|
The train's distance from the man's original position is $t^{2}+d_{0}$, and the man's distance from his original position if he runs at speed $v$ is $v t$ at time $t$. We need to find where $t^{2}+d_{0}=v t$ has a solution. Note that this is a quadratic equation with discriminant $D=\sqrt{v^{2}-4 d_{0}}$, so it has solutions for real $D$, i.e. where $v \geq \sqrt{4 d_{0}}$, so $4 d_{0}$ must be a perfect square. This happens when $4 d_{0}$ is an even power of 2 : the smallest value is $2^{0}$, the second smallest is $2^{2}$, the third smallest is $2^{4}$, and in general the $n$th smallest is $2^{2(n-1)}$, or $4^{n-1}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n22. [6]",
"solution_match": "\nSolution: "
}
|
ac8e620b-010b-5e6c-b28e-43da235aeb90
| 610,911
|
Alice, Bob, and Charlie each pick a 2-digit number at random. What is the probability that all of their numbers' tens' digits are different from each others' tens' digits and all of their numbers' ones digits are different from each others' ones' digits?
|
$\frac{9}{10} \frac{8}{10} \frac{8}{9} \frac{7}{9}=\frac{112}{225}$.
|
\frac{112}{225}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Alice, Bob, and Charlie each pick a 2-digit number at random. What is the probability that all of their numbers' tens' digits are different from each others' tens' digits and all of their numbers' ones digits are different from each others' ones' digits?
|
$\frac{9}{10} \frac{8}{10} \frac{8}{9} \frac{7}{9}=\frac{112}{225}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n23. [5]",
"solution_match": "\nSolution: "
}
|
b119c9ea-1026-579d-ac15-5fd43616285d
| 610,912
|
Square $A B C D$ has side length 1. A dilation is performed about point $A$, creating square $A B^{\prime} C^{\prime} D^{\prime}$. If $B C^{\prime}=29$, determine the area of triangle $B D C^{\prime}$.
|
$29^{2}-2 \cdot \frac{1}{2}(29)\left(\frac{29}{2}\right)-\frac{1}{2}=420$.
|
420
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Square $A B C D$ has side length 1. A dilation is performed about point $A$, creating square $A B^{\prime} C^{\prime} D^{\prime}$. If $B C^{\prime}=29$, determine the area of triangle $B D C^{\prime}$.
|
$29^{2}-2 \cdot \frac{1}{2}(29)\left(\frac{29}{2}\right)-\frac{1}{2}=420$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n24. [6]",
"solution_match": "\nSolution: "
}
|
931d6657-cc4d-545b-ad65-3e9d7acb89c4
| 610,913
|
What is the remainder when 100 ! is divided by 101 ?
|
Wilson's theorem says that for $p$ a prime, $(p-1)!\equiv-1(\mathrm{p})$, so the remainder is 100 .
|
100
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
What is the remainder when 100 ! is divided by 101 ?
|
Wilson's theorem says that for $p$ a prime, $(p-1)!\equiv-1(\mathrm{p})$, so the remainder is 100 .
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n25. [ $\\pm \\mathbf{1 0}$ ]",
"solution_match": "\nSolution: "
}
|
b00004fc-aff4-5879-a7ac-9303e8a26475
| 610,914
|
Count the number of permutations $a_{1} a_{2} \ldots a_{7}$ of 1234567 with longest decreasing subsequence of length at most two (i.e. there does not exist $i<j<k$ such that $a_{i}>a_{j}>a_{k}$ ).
|
$C(7)=429$.
|
429
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Count the number of permutations $a_{1} a_{2} \ldots a_{7}$ of 1234567 with longest decreasing subsequence of length at most two (i.e. there does not exist $i<j<k$ such that $a_{i}>a_{j}>a_{k}$ ).
|
$C(7)=429$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n32. [10]",
"solution_match": "\nSolution: "
}
|
4c96adfe-ed5b-5ef1-957a-2e4682bb10c1
| 610,921
|
Through a point in the interior of a triangle $A B C$, three lines are drawn, one parallel to each side. These lines divide the sides of the triangle into three regions each. Let $a, b$, and $c$ be the lengths of the sides opposite $\angle A, \angle B$, and $\angle C$, respectively, and let $a^{\prime}, b^{\prime}$, and $c^{\prime}$ be the lengths of the middle regions of the sides opposite $\angle A, \angle B$, and $\angle C$, respectively. Find the numerical value of $a^{\prime} / a+b^{\prime} / b+c^{\prime} / c$.
|
1 .
|
1
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Through a point in the interior of a triangle $A B C$, three lines are drawn, one parallel to each side. These lines divide the sides of the triangle into three regions each. Let $a, b$, and $c$ be the lengths of the sides opposite $\angle A, \angle B$, and $\angle C$, respectively, and let $a^{\prime}, b^{\prime}$, and $c^{\prime}$ be the lengths of the middle regions of the sides opposite $\angle A, \angle B$, and $\angle C$, respectively. Find the numerical value of $a^{\prime} / a+b^{\prime} / b+c^{\prime} / c$.
|
1 .
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n44. [7]",
"solution_match": "\nSolution: "
}
|
4a93e42e-25d7-575e-9597-b92013efbab3
| 610,933
|
Find the sum of the infinite series $\frac{1}{3^{2}-1^{2}}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)+\frac{1}{5^{2}-3^{2}}\left(\frac{1}{3^{2}}-\frac{1}{5^{2}}\right)+\frac{1}{7^{2}-5^{2}}\left(\frac{1}{5^{2}}-\frac{1}{7^{2}}\right)+$
|
$1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\cdots=1$.
|
1
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find the sum of the infinite series $\frac{1}{3^{2}-1^{2}}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)+\frac{1}{5^{2}-3^{2}}\left(\frac{1}{3^{2}}-\frac{1}{5^{2}}\right)+\frac{1}{7^{2}-5^{2}}\left(\frac{1}{5^{2}}-\frac{1}{7^{2}}\right)+$
|
$1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\cdots=1$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n53. [7]",
"solution_match": "\nSolution: "
}
|
90f3bcc8-2215-56d4-b400-66485986e2e0
| 610,942
|
How many multiples of 7 between $10^{6}$ and $10^{9}$ are perfect squares?
|
$\left[\sqrt{\frac{10^{9}}{7^{2}}}\right]-\left[\sqrt{\frac{10^{6}}{7^{2}}}\right]=4517-142=4375$.
|
4375
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
How many multiples of 7 between $10^{6}$ and $10^{9}$ are perfect squares?
|
$\left[\sqrt{\frac{10^{9}}{7^{2}}}\right]-\left[\sqrt{\frac{10^{6}}{7^{2}}}\right]=4517-142=4375$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-guts-solutions.jsonl",
"problem_match": "\n55. [7]",
"solution_match": "\nSolution: "
}
|
bd3a2789-d1ec-5b87-ba72-306ba42b5a4a
| 610,944
|
What is the 18 th digit after the decimal point of $\frac{10000}{9899}$ ?
|
$\frac{10000}{9899}$ satisfies $100(x-1)=1.01 x$, so each pair of adjacent digits is generated by adding the previous two pairs of digits. So the decimal is $1.01020305081321345590 \ldots$, and the 18 th digit is 5 .
|
5
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
What is the 18 th digit after the decimal point of $\frac{10000}{9899}$ ?
|
$\frac{10000}{9899}$ satisfies $100(x-1)=1.01 x$, so each pair of adjacent digits is generated by adding the previous two pairs of digits. So the decimal is $1.01020305081321345590 \ldots$, and the 18 th digit is 5 .
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-team-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution: "
}
|
771baf7b-029f-5a95-a528-f1c50cc9742c
| 610,952
|
P is a polynomial. When P is divided by $x-1$, the remainder is -4 . When P is divided by $x-2$, the remainder is -1 . When $P$ is divided by $x-3$, the remainder is 4 . Determine the remainder when P is divided by $x^{3}-6 x^{2}+11 x-6$.
|
The remainder polynomial is simply the order two polynomial that goes through the points $(1,-4),(2,-1)$, and $(3,4): x^{2}-5$.
|
x^{2}-5
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
P is a polynomial. When P is divided by $x-1$, the remainder is -4 . When P is divided by $x-2$, the remainder is -1 . When $P$ is divided by $x-3$, the remainder is 4 . Determine the remainder when P is divided by $x^{3}-6 x^{2}+11 x-6$.
|
The remainder polynomial is simply the order two polynomial that goes through the points $(1,-4),(2,-1)$, and $(3,4): x^{2}-5$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-team-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution: "
}
|
5b906612-6dd5-59a3-80fd-05f9ba22aefb
| 610,953
|
Express $\frac{\sin 10+\sin 20+\sin 30+\sin 40+\sin 50+\sin 60+\sin 70+\sin 80}{\cos 5 \cos 10 \cos 20}$ without using trigonometric functions.
|
We will use the identities $\cos a+\cos b=2 \cos \frac{a+b}{2} \cos \frac{a-b}{2}$ and $\sin a+\sin b=$ $2 \sin \frac{a+b}{2} \cos \frac{a+b}{2}$. The numerator is $(\sin 10+\sin 80)+(\sin 20+\sin 70)+(\sin 30+\sin 60)+(\sin 40+$ $\sin 60)=2 \sin 45(\cos 35+\cos 25+\cos 15+\cos 35)=2 \sin 45((\cos 35+\cos 5)+(\cos 25+\cos 15))=$ $4 \sin 45 \cos 20(\cos 15+\cos 5)=8 \sin 45 \cos 20 \cos 10 \cos 5$, so the fraction equals $8 \sin 45=4 \sqrt{2}$.
|
4 \sqrt{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Express $\frac{\sin 10+\sin 20+\sin 30+\sin 40+\sin 50+\sin 60+\sin 70+\sin 80}{\cos 5 \cos 10 \cos 20}$ without using trigonometric functions.
|
We will use the identities $\cos a+\cos b=2 \cos \frac{a+b}{2} \cos \frac{a-b}{2}$ and $\sin a+\sin b=$ $2 \sin \frac{a+b}{2} \cos \frac{a+b}{2}$. The numerator is $(\sin 10+\sin 80)+(\sin 20+\sin 70)+(\sin 30+\sin 60)+(\sin 40+$ $\sin 60)=2 \sin 45(\cos 35+\cos 25+\cos 15+\cos 35)=2 \sin 45((\cos 35+\cos 5)+(\cos 25+\cos 15))=$ $4 \sin 45 \cos 20(\cos 15+\cos 5)=8 \sin 45 \cos 20 \cos 10 \cos 5$, so the fraction equals $8 \sin 45=4 \sqrt{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-team-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nSolution: "
}
|
597b366c-85d0-5d92-8de5-54757e5fafb4
| 610,957
|
Compute $\sum_{i=1}^{\infty} \frac{a i}{a^{i}}$ for $a>1$.
|
The sum $S=a+a x+a x^{2}+a x^{3}+\cdots$ for $x<1$ can be determined by realizing that $x S=a x+a x^{2}+a x^{3}+\cdots$ and $(1-x) S=a$, so $S=\frac{a}{1-x}$. Using this, we have $\sum_{i=1}^{\infty} \frac{a i}{a^{i}}=$ $a \sum_{i=1}^{\infty} \frac{i}{a^{i}}=a\left[\frac{1}{a}+\frac{2}{a^{2}}+\frac{3}{a^{3}}+\cdots\right]=a\left[\left(\frac{1}{a}+\frac{1}{a^{2}}+\frac{1}{a^{3}}+\cdots\right)+\left(\frac{1}{a^{2}}+\frac{1}{a^{3}}+\frac{1}{a^{4}}+\cdots\right)+\cdots\right]=$ $a\left[\frac{1}{1-a}+\frac{1}{a} \frac{1}{1-a}+\frac{1}{a^{2}} \frac{1}{1-a}+\cdots\right]=\frac{a}{1-a}\left[1+\frac{1}{a}+\frac{1}{a^{2}}+\cdots\right]=\left(\frac{a}{1-a}\right)^{2}$.
|
\left(\frac{a}{1-a}\right)^{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Compute $\sum_{i=1}^{\infty} \frac{a i}{a^{i}}$ for $a>1$.
|
The sum $S=a+a x+a x^{2}+a x^{3}+\cdots$ for $x<1$ can be determined by realizing that $x S=a x+a x^{2}+a x^{3}+\cdots$ and $(1-x) S=a$, so $S=\frac{a}{1-x}$. Using this, we have $\sum_{i=1}^{\infty} \frac{a i}{a^{i}}=$ $a \sum_{i=1}^{\infty} \frac{i}{a^{i}}=a\left[\frac{1}{a}+\frac{2}{a^{2}}+\frac{3}{a^{3}}+\cdots\right]=a\left[\left(\frac{1}{a}+\frac{1}{a^{2}}+\frac{1}{a^{3}}+\cdots\right)+\left(\frac{1}{a^{2}}+\frac{1}{a^{3}}+\frac{1}{a^{4}}+\cdots\right)+\cdots\right]=$ $a\left[\frac{1}{1-a}+\frac{1}{a} \frac{1}{1-a}+\frac{1}{a^{2}} \frac{1}{1-a}+\cdots\right]=\frac{a}{1-a}\left[1+\frac{1}{a}+\frac{1}{a^{2}}+\cdots\right]=\left(\frac{a}{1-a}\right)^{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-team-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nSolution: "
}
|
ce6ccc4d-1768-5271-85d7-bfe910dcb73a
| 610,958
|
Define a monic irreducible polynomial with integral coefficients to be a polynomial with leading coefficient 1 that cannot be factored, and the prime factorization of a polynomial with leading coefficient 1 as the factorization into monic irreducible polynomials. How many not necessarily distinct monic irreducible polynomials are there in the prime factorization of $\left(x^{8}+x^{4}+1\right)\left(x^{8}+x+1\right)$ (for instance, $(x+1)^{2}$ has two prime factors)?
|
$x^{8}+x^{4}+1=\left(x^{8}+2 x^{4}+1\right)-x^{4}=\left(x^{4}+1\right)^{2}-\left(x^{2}\right)^{2}=\left(x^{4}-x^{2}+1\right)\left(x^{4}+x^{2}+1\right)=$ $\left(x^{4}-x^{2}+1\right)\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)$, and $x^{8}+x+1=\left(x^{2}+x+1\right)\left(x^{6}-x^{5}+x^{3}-x^{2}+1\right)$. If an integer polynomial $f(x)=a_{n} x^{n}+\cdots+a_{0}(\bmod p)$, where $p$ does not divide $a_{n}$, has no zeros, then $f$
has no rational roots. Taking $p=2$, we find $x^{6}-x^{5}+x^{3}-x^{2}+1$ is irreducible. The prime factorization of our polynomial is thus $\left(x^{4}-x^{2}+1\right)\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)^{2}\left(x^{6}-x^{5}+x^{3}-x^{2}+1\right)$, so the answer is 5 .
|
5
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Define a monic irreducible polynomial with integral coefficients to be a polynomial with leading coefficient 1 that cannot be factored, and the prime factorization of a polynomial with leading coefficient 1 as the factorization into monic irreducible polynomials. How many not necessarily distinct monic irreducible polynomials are there in the prime factorization of $\left(x^{8}+x^{4}+1\right)\left(x^{8}+x+1\right)$ (for instance, $(x+1)^{2}$ has two prime factors)?
|
$x^{8}+x^{4}+1=\left(x^{8}+2 x^{4}+1\right)-x^{4}=\left(x^{4}+1\right)^{2}-\left(x^{2}\right)^{2}=\left(x^{4}-x^{2}+1\right)\left(x^{4}+x^{2}+1\right)=$ $\left(x^{4}-x^{2}+1\right)\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)$, and $x^{8}+x+1=\left(x^{2}+x+1\right)\left(x^{6}-x^{5}+x^{3}-x^{2}+1\right)$. If an integer polynomial $f(x)=a_{n} x^{n}+\cdots+a_{0}(\bmod p)$, where $p$ does not divide $a_{n}$, has no zeros, then $f$
has no rational roots. Taking $p=2$, we find $x^{6}-x^{5}+x^{3}-x^{2}+1$ is irreducible. The prime factorization of our polynomial is thus $\left(x^{4}-x^{2}+1\right)\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)^{2}\left(x^{6}-x^{5}+x^{3}-x^{2}+1\right)$, so the answer is 5 .
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-team-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nSolution: "
}
|
ff679696-60ae-5c92-a9a4-5d7570ab7f73
| 610,959
|
Define $a$ ? $=(a-1) /(a+1)$ for $a \neq-1$. Determine all real values $N$ for which $(N ?) ?=\tan 15$.
|
Let $x=N$ ?. Then $(x-1) \cos 15=(x+1) \sin 15$. Squaring and rearranging terms, and using the fact that $\cos ^{2} 15-\sin ^{2} 15=\cos 30=\frac{\sqrt{3}}{2}$, we have $3 x^{2}-4 \sqrt{3} x+3=0$. Solving, we find that $x=\sqrt{3}$ or $\frac{\sqrt{3}}{3}$. However, we may reject the second root because it yields a negative value for ( $N$ ?)?. Therefore $x=\sqrt{3}$ and $N=\frac{1+x}{1-x}=\sqrt{\frac{1+\sqrt{3}}{1-\sqrt{3}}}=-2-\sqrt{3}$.
|
-2-\sqrt{3}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Define $a$ ? $=(a-1) /(a+1)$ for $a \neq-1$. Determine all real values $N$ for which $(N ?) ?=\tan 15$.
|
Let $x=N$ ?. Then $(x-1) \cos 15=(x+1) \sin 15$. Squaring and rearranging terms, and using the fact that $\cos ^{2} 15-\sin ^{2} 15=\cos 30=\frac{\sqrt{3}}{2}$, we have $3 x^{2}-4 \sqrt{3} x+3=0$. Solving, we find that $x=\sqrt{3}$ or $\frac{\sqrt{3}}{3}$. However, we may reject the second root because it yields a negative value for ( $N$ ?)?. Therefore $x=\sqrt{3}$ and $N=\frac{1+x}{1-x}=\sqrt{\frac{1+\sqrt{3}}{1-\sqrt{3}}}=-2-\sqrt{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-team-solutions.jsonl",
"problem_match": "\n11. ",
"solution_match": "\nSolution:"
}
|
25e574da-badd-56a8-8c26-bb0b742e85a7
| 610,960
|
All subscripts in this problem are to be considered modulo 6 , that means for example that $\omega_{7}$ is the same as $\omega_{1}$. Let $\omega_{1}, \ldots \omega_{6}$ be circles of radius $r$, whose centers lie on a regular hexagon of side length 1 . Let $P_{i}$ be the intersection of $\omega_{i}$ and $\omega_{i+1}$ that lies further from the center of the hexagon, for $i=1, \ldots 6$. Let $Q_{i}, i=1 \ldots 6$, lie on $\omega_{i}$ such that $Q_{i}, P_{i}, Q_{i+1}$ are colinear. Find the number of possible values of $r$.
|
Consider two consecutive circles $\omega_{i}$ and $\omega_{i+1}$. Let $Q_{i}, Q_{i}^{\prime}$ be two points on $\omega_{i}$ and $Q_{i+1}, Q_{i+1}^{\prime}$ on $\omega_{i+1}$ such that $Q_{i}, P_{i}$ and $Q_{i+1}$ are colinear and also $Q_{i}^{\prime}, P_{i}$ and $Q_{i+1}^{\prime}$. Then $Q_{i} Q_{i}^{\prime}=2 \angle Q_{i} P_{i} Q_{i}^{\prime}=2 \angle Q_{i+1} P_{i} Q_{i+1}^{\prime}=\angle Q_{i+1} Q_{i+1}^{\prime}$. Refer to the center of $\omega_{i}$ as $O_{i}$. The previous result shows that the lines $O_{i} Q_{i}$ and $O_{i+1} Q_{i+1}$ meet at the same angle as the lines $O_{i} Q_{i}^{\prime}$ and $O_{i+1} Q_{i+1}^{\prime}$, call this angle $\psi_{i}$. $\psi_{i}$ is a function solely of the circles $\omega_{i}$ and $\omega_{i+1}$ and the distance between them (we have just showed that any two points $Q_{i}$ and $Q_{i}^{\prime}$ on $\omega_{i}$ give the same value of $\psi_{i}$, so $\psi_{i}$ can't depend on this.) Now, the geometry of $\omega_{i}$ and $\omega_{i+1}$ is the same for every $i$, so $\psi_{i}$ is simply a constant $\psi$ which depends only on $r$. We know $6 \psi=0 \bmod 2 \pi$ because $Q_{7}=Q_{1}$.
We now compute $\psi$. It suffices to do the computaiton for some specific choice of $Q_{i}$. Take $Q_{i}$ to be the intersection of $O_{i} O_{i+1}$ and $\omega_{i}$ which is further from $O_{i+1}$. We are to compute the angle between $O_{i} Q_{i}$ and $O_{i+1} Q_{i+1}$ which is the same as $\angle O_{i} O_{i+1} Q_{i+1}$. Note the triangle $\triangle O_{i} P_{i} O_{i+1}$ is isosceles, call the base angle $\xi$. We have $\angle O_{i} O_{i+1} Q_{i+1}=\angle O_{i} O_{i+1} P_{i}+$ $\angle P_{i} O_{i+1} Q_{i+1}=\xi+\left(\pi-2 \angle O_{i+1} P_{i} Q_{i+1}\right)=\xi+\left(\pi-2\left(\pi-\angle Q_{i} O_{i+1} P_{i}-\angle P_{i} Q_{i} O_{i+1}\right)\right)=$ $\xi-\pi+2\left(\xi+(1 / 2) \angle P_{i} O_{i} O_{i+1}\right)=\xi-\pi+2(\xi+(1 / 2) \xi)=4 \xi-\pi$.
So we get $6(4 \xi-\pi)=0 \bmod 2 \pi$. Noting that $\xi$ must be acute, $\xi=\pi / 12, \pi / 6, \pi / 4, \pi / 3$ or $5 \pi / 12$. $r$ is uniquely determined as $(1 / 2) \sec \xi$ so there are 5 possible values of $r$.
|
5
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
All subscripts in this problem are to be considered modulo 6 , that means for example that $\omega_{7}$ is the same as $\omega_{1}$. Let $\omega_{1}, \ldots \omega_{6}$ be circles of radius $r$, whose centers lie on a regular hexagon of side length 1 . Let $P_{i}$ be the intersection of $\omega_{i}$ and $\omega_{i+1}$ that lies further from the center of the hexagon, for $i=1, \ldots 6$. Let $Q_{i}, i=1 \ldots 6$, lie on $\omega_{i}$ such that $Q_{i}, P_{i}, Q_{i+1}$ are colinear. Find the number of possible values of $r$.
|
Consider two consecutive circles $\omega_{i}$ and $\omega_{i+1}$. Let $Q_{i}, Q_{i}^{\prime}$ be two points on $\omega_{i}$ and $Q_{i+1}, Q_{i+1}^{\prime}$ on $\omega_{i+1}$ such that $Q_{i}, P_{i}$ and $Q_{i+1}$ are colinear and also $Q_{i}^{\prime}, P_{i}$ and $Q_{i+1}^{\prime}$. Then $Q_{i} Q_{i}^{\prime}=2 \angle Q_{i} P_{i} Q_{i}^{\prime}=2 \angle Q_{i+1} P_{i} Q_{i+1}^{\prime}=\angle Q_{i+1} Q_{i+1}^{\prime}$. Refer to the center of $\omega_{i}$ as $O_{i}$. The previous result shows that the lines $O_{i} Q_{i}$ and $O_{i+1} Q_{i+1}$ meet at the same angle as the lines $O_{i} Q_{i}^{\prime}$ and $O_{i+1} Q_{i+1}^{\prime}$, call this angle $\psi_{i}$. $\psi_{i}$ is a function solely of the circles $\omega_{i}$ and $\omega_{i+1}$ and the distance between them (we have just showed that any two points $Q_{i}$ and $Q_{i}^{\prime}$ on $\omega_{i}$ give the same value of $\psi_{i}$, so $\psi_{i}$ can't depend on this.) Now, the geometry of $\omega_{i}$ and $\omega_{i+1}$ is the same for every $i$, so $\psi_{i}$ is simply a constant $\psi$ which depends only on $r$. We know $6 \psi=0 \bmod 2 \pi$ because $Q_{7}=Q_{1}$.
We now compute $\psi$. It suffices to do the computaiton for some specific choice of $Q_{i}$. Take $Q_{i}$ to be the intersection of $O_{i} O_{i+1}$ and $\omega_{i}$ which is further from $O_{i+1}$. We are to compute the angle between $O_{i} Q_{i}$ and $O_{i+1} Q_{i+1}$ which is the same as $\angle O_{i} O_{i+1} Q_{i+1}$. Note the triangle $\triangle O_{i} P_{i} O_{i+1}$ is isosceles, call the base angle $\xi$. We have $\angle O_{i} O_{i+1} Q_{i+1}=\angle O_{i} O_{i+1} P_{i}+$ $\angle P_{i} O_{i+1} Q_{i+1}=\xi+\left(\pi-2 \angle O_{i+1} P_{i} Q_{i+1}\right)=\xi+\left(\pi-2\left(\pi-\angle Q_{i} O_{i+1} P_{i}-\angle P_{i} Q_{i} O_{i+1}\right)\right)=$ $\xi-\pi+2\left(\xi+(1 / 2) \angle P_{i} O_{i} O_{i+1}\right)=\xi-\pi+2(\xi+(1 / 2) \xi)=4 \xi-\pi$.
So we get $6(4 \xi-\pi)=0 \bmod 2 \pi$. Noting that $\xi$ must be acute, $\xi=\pi / 12, \pi / 6, \pi / 4, \pi / 3$ or $5 \pi / 12$. $r$ is uniquely determined as $(1 / 2) \sec \xi$ so there are 5 possible values of $r$.
|
{
"resource_path": "HarvardMIT/segmented/en-42-2001-feb-team-solutions.jsonl",
"problem_match": "\n12. ",
"solution_match": "\nSolution: "
}
|
d687337b-a732-5107-b6c5-8f4931051683
| 610,961
|
Eight knights are randomly placed on a chessboard (not necessarily on distinct squares). A knight on a given square attacks all the squares that can be reached by moving either (1) two squares up or down followed by one squares left or right, or (2) two squares left or right followed by one square up or down. Find the probability that every square, occupied or not, is attacked by some knight.
|
0. Since every knight attacks at most eight squares, the event can only occur if every knight attacks exactly eight squares. However, each corner square must be attacked, and some experimentation readily finds that it is impossible to place a knight so as to attack a corner and seven other squares as well.
|
0
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Eight knights are randomly placed on a chessboard (not necessarily on distinct squares). A knight on a given square attacks all the squares that can be reached by moving either (1) two squares up or down followed by one squares left or right, or (2) two squares left or right followed by one square up or down. Find the probability that every square, occupied or not, is attacked by some knight.
|
0. Since every knight attacks at most eight squares, the event can only occur if every knight attacks exactly eight squares. However, each corner square must be attacked, and some experimentation readily finds that it is impossible to place a knight so as to attack a corner and seven other squares as well.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-adv-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution: "
}
|
be9731db-7b92-5861-bdbd-22a22ab280fc
| 610,962
|
A certain cafeteria serves ham and cheese sandwiches, ham and tomato sandwiches, and tomato and cheese sandwiches. It is common for one meal to include multiple types of sandwiches. On a certain day, it was found that 80 customers had meals which contained both ham and cheese; 90 had meals containing both ham and tomatoes; 100 had meals containing both tomatoes and cheese. 20 customers' meals included all three ingredients. How many customers were there?
|
230. Everyone who ate just one sandwich is included in exactly one of the first three counts, while everyone who ate more than one sandwich is included in all four counts. Thus, to count each customer exactly once, we must add the first three figures and subtract the fourth twice: $80+90+100-2 \cdot 20=230$.
|
230
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A certain cafeteria serves ham and cheese sandwiches, ham and tomato sandwiches, and tomato and cheese sandwiches. It is common for one meal to include multiple types of sandwiches. On a certain day, it was found that 80 customers had meals which contained both ham and cheese; 90 had meals containing both ham and tomatoes; 100 had meals containing both tomatoes and cheese. 20 customers' meals included all three ingredients. How many customers were there?
|
230. Everyone who ate just one sandwich is included in exactly one of the first three counts, while everyone who ate more than one sandwich is included in all four counts. Thus, to count each customer exactly once, we must add the first three figures and subtract the fourth twice: $80+90+100-2 \cdot 20=230$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-adv-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution: "
}
|
b080738e-6166-598c-ab5b-f7881acdb737
| 610,963
|
How many four-digit numbers are there in which at least one digit occurs more than once?
|
4464 . There are 9000 four-digit numbers altogether. If we consider how many four-digit numbers have all their digits distinct, there are 9 choices for the first digit (since we exclude leading zeroes), and then 9 remaining choices for the second digit, then 8 for the third, and 7 for the fourth, for a total of $9 \cdot 9 \cdot 8 \cdot 7=4536$. Thus the remaining $9000-4536=4464$ numbers have a repeated digit.
|
4464
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many four-digit numbers are there in which at least one digit occurs more than once?
|
4464 . There are 9000 four-digit numbers altogether. If we consider how many four-digit numbers have all their digits distinct, there are 9 choices for the first digit (since we exclude leading zeroes), and then 9 remaining choices for the second digit, then 8 for the third, and 7 for the fourth, for a total of $9 \cdot 9 \cdot 8 \cdot 7=4536$. Thus the remaining $9000-4536=4464$ numbers have a repeated digit.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-adv-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution: "
}
|
845b55e4-83de-5852-9857-78e72590c05f
| 80,302
|
Two fair coins are simultaneously flipped. This is done repeatedly until at least one of the coins comes up heads, at which point the process stops. What is the probability that the other coin also came up heads on this last flip?
|
$1 / 3$. Let the desired probability be $p$. There is a $1 / 4$ chance that both coins will come up heads on the first toss. Otherwise, both can come up heads simultaneously only if both are tails on the first toss, and then the process restarts as if from the beginning; thus this situation occurs with probability $p / 4$. Thus $p=1 / 4+p / 4$; solving, $p=1 / 3$. Alternate Solution: The desired event is equivalent to both coins coming up tails for $n$ successive turns (for some $n \geq 0$ ), then both coins coming up heads. For any fixed value of
$n$, the probability of this occurring is $1 / 4^{n+1}$. Since all these events are disjoint, the total probability is $1 / 4+1 / 4^{2}+1 / 4^{3}+\cdots=1 / 3$.
|
\frac{1}{3}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Two fair coins are simultaneously flipped. This is done repeatedly until at least one of the coins comes up heads, at which point the process stops. What is the probability that the other coin also came up heads on this last flip?
|
$1 / 3$. Let the desired probability be $p$. There is a $1 / 4$ chance that both coins will come up heads on the first toss. Otherwise, both can come up heads simultaneously only if both are tails on the first toss, and then the process restarts as if from the beginning; thus this situation occurs with probability $p / 4$. Thus $p=1 / 4+p / 4$; solving, $p=1 / 3$. Alternate Solution: The desired event is equivalent to both coins coming up tails for $n$ successive turns (for some $n \geq 0$ ), then both coins coming up heads. For any fixed value of
$n$, the probability of this occurring is $1 / 4^{n+1}$. Since all these events are disjoint, the total probability is $1 / 4+1 / 4^{2}+1 / 4^{3}+\cdots=1 / 3$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-adv-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution: "
}
|
3f8b3f21-3027-5934-8add-2a0a09973985
| 80,314
|
Determine the number of subsets $S$ of $\{1,2,3, \ldots, 10\}$ with the following property: there exist integers $a<b<c$ with $a \in S, b \notin S, c \in S$.
|
968 There are $2^{10}=1024$ subsets of $\{1,2, \ldots, 10\}$ altogether. Any subset without the specified property must be either the empty set or a block of consecutive integers. To specify a block of consecutive integers, we either have just one element ( 10 choices) or a pair of distinct endpoints $\left(\binom{10}{2}=45\right.$ choices). So the number of sets with our property is $1024-(1+10+45)=968$.
|
968
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Determine the number of subsets $S$ of $\{1,2,3, \ldots, 10\}$ with the following property: there exist integers $a<b<c$ with $a \in S, b \notin S, c \in S$.
|
968 There are $2^{10}=1024$ subsets of $\{1,2, \ldots, 10\}$ altogether. Any subset without the specified property must be either the empty set or a block of consecutive integers. To specify a block of consecutive integers, we either have just one element ( 10 choices) or a pair of distinct endpoints $\left(\binom{10}{2}=45\right.$ choices). So the number of sets with our property is $1024-(1+10+45)=968$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-adv-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nSolution: "
}
|
718c7d1b-d523-5959-899f-1ae3ab052849
| 610,964
|
In how many ways can the numbers $1,2, \ldots, 2002$ be placed at the vertices of a regular 2002-gon so that no two adjacent numbers differ by more than 2 ? (Rotations and reflections are considered distinct.)
|
4004. There are 2002 possible positions for the 1 . The two numbers adjacent to the 1 must be 2 and 3 ; there are two possible ways of placing these. The positions of these numbers uniquely determine the rest: for example, if 3 lies clockwise from 1 , then the number lying counterclockwise from 2 must be 4 ; the number lying clockwise from 3 must be 5; the number lying counterclockwise from 4 must now be 6 ; and so forth. Eventually, 2002 is placed adjacent to 2000 and 2001, so we do get a valid configuration. Thus there are $2002 \cdot 2$ possible arrangements.
|
4004
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In how many ways can the numbers $1,2, \ldots, 2002$ be placed at the vertices of a regular 2002-gon so that no two adjacent numbers differ by more than 2 ? (Rotations and reflections are considered distinct.)
|
4004. There are 2002 possible positions for the 1 . The two numbers adjacent to the 1 must be 2 and 3 ; there are two possible ways of placing these. The positions of these numbers uniquely determine the rest: for example, if 3 lies clockwise from 1 , then the number lying counterclockwise from 2 must be 4 ; the number lying clockwise from 3 must be 5; the number lying counterclockwise from 4 must now be 6 ; and so forth. Eventually, 2002 is placed adjacent to 2000 and 2001, so we do get a valid configuration. Thus there are $2002 \cdot 2$ possible arrangements.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-adv-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nSolution: "
}
|
4c67a37b-fe4f-5f92-9553-5a6eb0292786
| 610,965
|
Given a $9 \times 9$ chess board, we consider all the rectangles whose edges lie along grid lines (the board consists of 81 unit squares, and the grid lines lie on the borders of the unit squares). For each such rectangle, we put a mark in every one of the unit squares inside it. When this process is completed, how many unit squares will contain an even number of marks?
|
56. Consider the rectangles which contain the square in the $i$ th row and $j$ th column. There are $i$ possible positions for the upper edge of such a rectangle, $10-i$ for the lower edge, $j$ for the left edge, and $10-j$ for the right edge; thus we have $i(10-i) j(10-j)$ rectangles altogether, which is odd iff $i, j$ are both odd, i.e. iff $i, j \in\{1,3,5,7,9\}$. There are thus 25 unit squares which lie in an odd number of rectangles, so the answer is $81-25=56$.
|
56
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Given a $9 \times 9$ chess board, we consider all the rectangles whose edges lie along grid lines (the board consists of 81 unit squares, and the grid lines lie on the borders of the unit squares). For each such rectangle, we put a mark in every one of the unit squares inside it. When this process is completed, how many unit squares will contain an even number of marks?
|
56. Consider the rectangles which contain the square in the $i$ th row and $j$ th column. There are $i$ possible positions for the upper edge of such a rectangle, $10-i$ for the lower edge, $j$ for the left edge, and $10-j$ for the right edge; thus we have $i(10-i) j(10-j)$ rectangles altogether, which is odd iff $i, j$ are both odd, i.e. iff $i, j \in\{1,3,5,7,9\}$. There are thus 25 unit squares which lie in an odd number of rectangles, so the answer is $81-25=56$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-adv-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nSolution: "
}
|
593b93b5-b608-537c-a286-44b01d5cc85b
| 610,967
|
Given that $a, b, c$ are positive real numbers and $\log _{a} b+\log _{b} c+\log _{c} a=0$, find the value of $\left(\log _{a} b\right)^{3}+\left(\log _{b} c\right)^{3}+\left(\log _{c} a\right)^{3}$.
|
3. Let $x=\log _{a} b$ and $y=\log _{b} c$; then $\log _{c} a=-(x+y)$. Thus we want to compute the value of $x^{3}+y^{3}-(x+y)^{3}=-3 x^{2} y-3 x y^{2}=-3 x y(x+y)$. On the other hand, $-x y(x+y)=\left(\log _{a} b\right)\left(\log _{b} c\right)\left(\log _{c} a\right)=1$, so the answer is 3 .
|
3
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Given that $a, b, c$ are positive real numbers and $\log _{a} b+\log _{b} c+\log _{c} a=0$, find the value of $\left(\log _{a} b\right)^{3}+\left(\log _{b} c\right)^{3}+\left(\log _{c} a\right)^{3}$.
|
3. Let $x=\log _{a} b$ and $y=\log _{b} c$; then $\log _{c} a=-(x+y)$. Thus we want to compute the value of $x^{3}+y^{3}-(x+y)^{3}=-3 x^{2} y-3 x y^{2}=-3 x y(x+y)$. On the other hand, $-x y(x+y)=\left(\log _{a} b\right)\left(\log _{b} c\right)\left(\log _{c} a\right)=1$, so the answer is 3 .
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-adv-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nSolution: "
}
|
3815e905-e7dc-51bd-897f-75ba51ea1bc4
| 610,968
|
One fair die is rolled; let $a$ denote the number that comes up. We then roll $a$ dice; let the sum of the resulting $a$ numbers be $b$. Finally, we roll $b$ dice, and let $c$ be the sum of the resulting $b$ numbers. Find the expected (average) value of $c$.
|
$343 / 8$ The expected result of an individual die roll is $(1+2+3+4+5+6) / 6=$ $7 / 2$. For any particular value of $b$, if $b$ dice are rolled independently, then the expected sum is $(7 / 2) b$. Likewise, when we roll $a$ dice, the expected value of their sum $b$ is $(7 / 2) a$, so the expected value of $c$ is $(7 / 2)^{2} a$. Similar reasoning again shows us that the expected value of $a$ is $7 / 2$ and so the expected value of $c$ overall is $(7 / 2)^{3}=343 / 8$.
|
\frac{343}{8}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
One fair die is rolled; let $a$ denote the number that comes up. We then roll $a$ dice; let the sum of the resulting $a$ numbers be $b$. Finally, we roll $b$ dice, and let $c$ be the sum of the resulting $b$ numbers. Find the expected (average) value of $c$.
|
$343 / 8$ The expected result of an individual die roll is $(1+2+3+4+5+6) / 6=$ $7 / 2$. For any particular value of $b$, if $b$ dice are rolled independently, then the expected sum is $(7 / 2) b$. Likewise, when we roll $a$ dice, the expected value of their sum $b$ is $(7 / 2) a$, so the expected value of $c$ is $(7 / 2)^{2} a$. Similar reasoning again shows us that the expected value of $a$ is $7 / 2$ and so the expected value of $c$ overall is $(7 / 2)^{3}=343 / 8$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-adv-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nSolution: "
}
|
b252682d-cf58-56c9-ac1e-05b192964e9b
| 610,969
|
Nine nonnegative numbers have average 10 . What is the greatest possible value for their median?
|
18 If the median is $m$, then the five highest numbers are all at least $m$, so the sum of all the numbers is at least $5 m$. Thus $90 \geq 5 m \Rightarrow m \leq 18$. Conversely, we can achieve $m=18$ by taking four 0 's and five 18's.
|
18
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Nine nonnegative numbers have average 10 . What is the greatest possible value for their median?
|
18 If the median is $m$, then the five highest numbers are all at least $m$, so the sum of all the numbers is at least $5 m$. Thus $90 \geq 5 m \Rightarrow m \leq 18$. Conversely, we can achieve $m=18$ by taking four 0 's and five 18's.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-alg-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution: "
}
|
83fb36ec-e522-58ca-8430-bb8a65807787
| 610,970
|
$p$ and $q$ are primes such that the numbers $p+q$ and $p+7 q$ are both squares. Find the value of $p$.
|
2. Writing $x^{2}=p+q, y^{2}=p+7 q$, we have $6 q=y^{2}-x^{2}=(y-x)(y+x)$. Since $6 q$ is even, one of the factors $y-x, y+x$ is even, and then the other is as well; thus $6 q$ is divisible by $4 \Rightarrow q$ is even $\Rightarrow q=2$ and $6 q=12$. We may assume $x, y$ are both taken to be positive; then we must have $y-x=2, y+x=6 \Rightarrow x=2$, so $p+2=2^{2}=4 \Rightarrow p=2$ also.
|
2
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
$p$ and $q$ are primes such that the numbers $p+q$ and $p+7 q$ are both squares. Find the value of $p$.
|
2. Writing $x^{2}=p+q, y^{2}=p+7 q$, we have $6 q=y^{2}-x^{2}=(y-x)(y+x)$. Since $6 q$ is even, one of the factors $y-x, y+x$ is even, and then the other is as well; thus $6 q$ is divisible by $4 \Rightarrow q$ is even $\Rightarrow q=2$ and $6 q=12$. We may assume $x, y$ are both taken to be positive; then we must have $y-x=2, y+x=6 \Rightarrow x=2$, so $p+2=2^{2}=4 \Rightarrow p=2$ also.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-alg-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution: "
}
|
de1b1d2d-d2a8-5512-8686-f8bc8d21a533
| 610,971
|
Real numbers $a, b, c$ satisfy the equations $a+b+c=26,1 / a+1 / b+1 / c=28$. Find the value of
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{c}+\frac{c}{b}+\frac{b}{a} .
$$
|
725 Multiplying the two given equations gives
$$
\frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{c}{c}=26 \cdot 28=728,
$$
and subtracting 3 from both sides gives the answer, 725 .
|
725
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Real numbers $a, b, c$ satisfy the equations $a+b+c=26,1 / a+1 / b+1 / c=28$. Find the value of
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{c}+\frac{c}{b}+\frac{b}{a} .
$$
|
725 Multiplying the two given equations gives
$$
\frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{c}{c}=26 \cdot 28=728,
$$
and subtracting 3 from both sides gives the answer, 725 .
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-alg-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution: "
}
|
07f7baf1-9690-5960-90c7-1dd0fa49c36b
| 610,972
|
Find the greatest common divisor of the numbers $2002+2,2002^{2}+2,2002^{3}+2, \ldots$.
|
6. Notice that $2002+2$ divides $2002^{2}-2^{2}$, so any common divisor of $2002+2$ and $2002^{2}+2$ must divide $\left(2002^{2}+2\right)-\left(2002^{2}-2^{2}\right)=6$. On the other hand, every number in the sequence is even, and the $n$th number is always congruent to $1^{n}+2 \equiv 0$ modulo 3 . Thus, 6 divides every number in the sequence.
|
6
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the greatest common divisor of the numbers $2002+2,2002^{2}+2,2002^{3}+2, \ldots$.
|
6. Notice that $2002+2$ divides $2002^{2}-2^{2}$, so any common divisor of $2002+2$ and $2002^{2}+2$ must divide $\left(2002^{2}+2\right)-\left(2002^{2}-2^{2}\right)=6$. On the other hand, every number in the sequence is even, and the $n$th number is always congruent to $1^{n}+2 \equiv 0$ modulo 3 . Thus, 6 divides every number in the sequence.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-alg-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nSolution: "
}
|
490fccaa-aa5a-5846-9229-b765ab90c9bc
| 610,974
|
Find the sum of the even positive divisors of 1000 .
|
2184 . Notice that $2 k$ is a divisor of 1000 iff $k$ is a divisor of 500 , so we need only find the sum of the divisors of 500 and multiply by 2 . This can be done by enumerating the divisors individually, or simply by using the formula: $\sigma\left(2^{2} \cdot 5^{3}\right)=\left(1+2+2^{2}\right)(1+5+$ $\left.5^{2}+5^{3}\right)=1092$, and then doubling gives 2184. Alternate Solution: The sum of all the divisors of 1000 is $\left(1+2+2^{2}+2^{3}\right)\left(1+5+5^{2}+5^{3}\right)=2340$. The odd divisors of 1000 are simply the divisors of 125 , whose sum is $1+5+5^{2}+5^{3}=156$; subtracting this from 2340 , we are left with the sum of the even divisors of 1000 , which is 2184 .
|
2184
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find the sum of the even positive divisors of 1000 .
|
2184 . Notice that $2 k$ is a divisor of 1000 iff $k$ is a divisor of 500 , so we need only find the sum of the divisors of 500 and multiply by 2 . This can be done by enumerating the divisors individually, or simply by using the formula: $\sigma\left(2^{2} \cdot 5^{3}\right)=\left(1+2+2^{2}\right)(1+5+$ $\left.5^{2}+5^{3}\right)=1092$, and then doubling gives 2184. Alternate Solution: The sum of all the divisors of 1000 is $\left(1+2+2^{2}+2^{3}\right)\left(1+5+5^{2}+5^{3}\right)=2340$. The odd divisors of 1000 are simply the divisors of 125 , whose sum is $1+5+5^{2}+5^{3}=156$; subtracting this from 2340 , we are left with the sum of the even divisors of 1000 , which is 2184 .
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-alg-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nSolution: "
}
|
18bf7ee8-b16f-53f4-9434-30b9b46f62d1
| 610,975
|
The real numbers $x, y, z, w$ satisfy
$$
\begin{aligned}
2 x+y+z+w & =1 \\
x+3 y+z+w & =2 \\
x+y+4 z+w & =3 \\
x+y+z+5 w & =25 .
\end{aligned}
$$
Find the value of $w$.
|
$11 / 2$. Multiplying the four equations by $12,6,4,3$ respectively, we get
$$
\begin{aligned}
24 x+12 y+12 z+12 w & =12 \\
6 x+18 y+6 z+6 w & =12 \\
4 x+4 y+16 z+4 w & =12 \\
3 x+3 y+3 z+15 w & =75
\end{aligned}
$$
Adding these yields $37 x+37 y+37 z+37 w=111$, or $x+y+z+w=3$. Subtract this from the fourth given equation to obtain $4 w=22$, or $w=11 / 2$.
|
\frac{11}{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
The real numbers $x, y, z, w$ satisfy
$$
\begin{aligned}
2 x+y+z+w & =1 \\
x+3 y+z+w & =2 \\
x+y+4 z+w & =3 \\
x+y+z+5 w & =25 .
\end{aligned}
$$
Find the value of $w$.
|
$11 / 2$. Multiplying the four equations by $12,6,4,3$ respectively, we get
$$
\begin{aligned}
24 x+12 y+12 z+12 w & =12 \\
6 x+18 y+6 z+6 w & =12 \\
4 x+4 y+16 z+4 w & =12 \\
3 x+3 y+3 z+15 w & =75
\end{aligned}
$$
Adding these yields $37 x+37 y+37 z+37 w=111$, or $x+y+z+w=3$. Subtract this from the fourth given equation to obtain $4 w=22$, or $w=11 / 2$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-alg-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nSolution: "
}
|
6c901385-8e66-598f-ae6b-717f5a430db5
| 610,976
|
Determine the value of the sum
$$
\frac{3}{1^{2} \cdot 2^{2}}+\frac{5}{2^{2} \cdot 3^{2}}+\frac{7}{3^{2} \cdot 4^{2}}+\cdots+\frac{29}{14^{2} \cdot 15^{2}} .
$$
|
224/225 The sum telescopes as
$$
\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)+\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)+\cdots+\left(\frac{1}{14^{2}}-\frac{1}{15^{2}}\right)=\frac{1}{1^{2}}-\frac{1}{15^{2}}=\frac{224}{225 .}
$$
|
\frac{224}{225}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Determine the value of the sum
$$
\frac{3}{1^{2} \cdot 2^{2}}+\frac{5}{2^{2} \cdot 3^{2}}+\frac{7}{3^{2} \cdot 4^{2}}+\cdots+\frac{29}{14^{2} \cdot 15^{2}} .
$$
|
224/225 The sum telescopes as
$$
\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)+\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)+\cdots+\left(\frac{1}{14^{2}}-\frac{1}{15^{2}}\right)=\frac{1}{1^{2}}-\frac{1}{15^{2}}=\frac{224}{225 .}
$$
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-alg-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nSolution: "
}
|
d4dcae4d-5f8f-5fec-9802-1ab162bb516f
| 610,977
|
For any positive integer $n$, let $f(n)$ denote the number of 1 's in the base- 2 representation of $n$. For how many values of $n$ with $1 \leq n \leq 2002$ do we have $f(n)=f(n+1)$ ?
|
501. If $n$ is even, then $n+1$ is obtained from $n$ in binary by changing the final 0 to a 1 ; thus $f(n+1)=f(n)+1$. If $n$ is odd, then $n+1$ is obtained by changing the last 0 to a 1 , the ensuing string of 1 s to 0 s , and then changing the next rightmost 0 to a 1. This produces no net change in the number of 1 's iff $n$ ends in 01 in base 2. Thus,
$f(n+1)=f(n)$ if and only if $n$ is congruent to $1 \bmod 4$, and there are 501 such numbers in the specified range.
|
501
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
For any positive integer $n$, let $f(n)$ denote the number of 1 's in the base- 2 representation of $n$. For how many values of $n$ with $1 \leq n \leq 2002$ do we have $f(n)=f(n+1)$ ?
|
501. If $n$ is even, then $n+1$ is obtained from $n$ in binary by changing the final 0 to a 1 ; thus $f(n+1)=f(n)+1$. If $n$ is odd, then $n+1$ is obtained by changing the last 0 to a 1 , the ensuing string of 1 s to 0 s , and then changing the next rightmost 0 to a 1. This produces no net change in the number of 1 's iff $n$ ends in 01 in base 2. Thus,
$f(n+1)=f(n)$ if and only if $n$ is congruent to $1 \bmod 4$, and there are 501 such numbers in the specified range.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-alg-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nSolution: "
}
|
3a2e8035-1284-5206-85f9-3ed227171235
| 610,978
|
Determine the value of
$$
2002+\frac{1}{2}\left(2001+\frac{1}{2}\left(2000+\cdots+\frac{1}{2}\left(3+\frac{1}{2} \cdot 2\right)\right) \cdots\right) .
$$
|
4002. We can show by induction that $n+\frac{1}{2}\left([n-1]+\frac{1}{2}\left(\cdots+\frac{1}{2} \cdot 2\right) \cdots\right)=2(n-1)$. For $n=3$ we have $3+\frac{1}{2} \cdot 2=4$, giving the base case, and if the result holds for $n$, then $(n+1)+\frac{1}{2} 2(n-1)=2 n=2(n+1)-2$. Thus the claim holds, and now plug in $n=2002$. Alternate Solution: Expand the given expression as $2002+2001 / 2+2000 / 2^{2}+\cdots+2 / 2^{2000}$. Letting $S$ denote this sum, we have $S / 2=2002 / 2+2001 / 2^{2}+\cdots+2 / 2^{2001}$, so $S-S / 2=$ $2002-\left(1 / 2+1 / 4+\cdots+1 / 2^{2000}\right)-2 / 2^{2001}=2002-\left(1-1 / 2^{2000}\right)-1 / 2^{2000}=2001$, so $S=4002$.
|
4002
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Determine the value of
$$
2002+\frac{1}{2}\left(2001+\frac{1}{2}\left(2000+\cdots+\frac{1}{2}\left(3+\frac{1}{2} \cdot 2\right)\right) \cdots\right) .
$$
|
4002. We can show by induction that $n+\frac{1}{2}\left([n-1]+\frac{1}{2}\left(\cdots+\frac{1}{2} \cdot 2\right) \cdots\right)=2(n-1)$. For $n=3$ we have $3+\frac{1}{2} \cdot 2=4$, giving the base case, and if the result holds for $n$, then $(n+1)+\frac{1}{2} 2(n-1)=2 n=2(n+1)-2$. Thus the claim holds, and now plug in $n=2002$. Alternate Solution: Expand the given expression as $2002+2001 / 2+2000 / 2^{2}+\cdots+2 / 2^{2000}$. Letting $S$ denote this sum, we have $S / 2=2002 / 2+2001 / 2^{2}+\cdots+2 / 2^{2001}$, so $S-S / 2=$ $2002-\left(1 / 2+1 / 4+\cdots+1 / 2^{2000}\right)-2 / 2^{2001}=2002-\left(1-1 / 2^{2000}\right)-1 / 2^{2000}=2001$, so $S=4002$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-alg-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nSolution: "
}
|
8dbd80e2-11c2-57ad-b28f-189abcc10cfe
| 610,979
|
Two circles have centers that are $d$ units apart, and each has diameter $\sqrt{d}$. For any $d$, let $A(d)$ be the area of the smallest circle that contains both of these circles. Find $\lim _{d \rightarrow \infty} \frac{A(d)}{d^{2}}$.
|
This equals $\lim _{d \rightarrow \infty} \frac{\pi\left(\frac{d+\sqrt{d}}{2}\right)^{2}}{d^{2}}=\frac{\pi}{4}$.
|
\frac{\pi}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Two circles have centers that are $d$ units apart, and each has diameter $\sqrt{d}$. For any $d$, let $A(d)$ be the area of the smallest circle that contains both of these circles. Find $\lim _{d \rightarrow \infty} \frac{A(d)}{d^{2}}$.
|
This equals $\lim _{d \rightarrow \infty} \frac{\pi\left(\frac{d+\sqrt{d}}{2}\right)^{2}}{d^{2}}=\frac{\pi}{4}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-calc-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution: "
}
|
b9ce09e9-d920-5ae9-923a-527b63882335
| 610,980
|
Find $\lim _{h \rightarrow 0} \frac{x^{2}-(x+h)^{2}}{h}$.
|
This equals $\lim _{h \rightarrow 0} \frac{x^{2}-x^{2}-h^{2}-2 h x}{h}=\lim _{h \rightarrow 0}-h-2 x=-2 x$. Alternate Solution: This is the definition of the derivative of $-x^{2}$ with respect to $x$, which is $-2 x$.
|
-2x
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Find $\lim _{h \rightarrow 0} \frac{x^{2}-(x+h)^{2}}{h}$.
|
This equals $\lim _{h \rightarrow 0} \frac{x^{2}-x^{2}-h^{2}-2 h x}{h}=\lim _{h \rightarrow 0}-h-2 x=-2 x$. Alternate Solution: This is the definition of the derivative of $-x^{2}$ with respect to $x$, which is $-2 x$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-calc-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution: "
}
|
fd9badd9-74aa-5f36-97ba-b3d17be45ffe
| 610,981
|
We are given the values of the differentiable real functions $f, g, h$, as well as the derivatives of their pairwise products, at $x=0$ :
$$
f(0)=1 ; g(0)=2 ; h(0)=3 ; \quad(g h)^{\prime}(0)=4 ; \quad(h f)^{\prime}(0)=5 ; \quad(f g)^{\prime}(0)=6 .
$$
Find the value of $(f g h)^{\prime}(0)$.
|
16 By the product rule, $(f g h)^{\prime}=f^{\prime} g h+f g^{\prime} h+f g h^{\prime}=\left((f g)^{\prime} h+(g h)^{\prime} f+\right.$ $\left.(h f)^{\prime} g\right) / 2$. Evaluated at 0 , this gives 16.
|
16
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
We are given the values of the differentiable real functions $f, g, h$, as well as the derivatives of their pairwise products, at $x=0$ :
$$
f(0)=1 ; g(0)=2 ; h(0)=3 ; \quad(g h)^{\prime}(0)=4 ; \quad(h f)^{\prime}(0)=5 ; \quad(f g)^{\prime}(0)=6 .
$$
Find the value of $(f g h)^{\prime}(0)$.
|
16 By the product rule, $(f g h)^{\prime}=f^{\prime} g h+f g^{\prime} h+f g h^{\prime}=\left((f g)^{\prime} h+(g h)^{\prime} f+\right.$ $\left.(h f)^{\prime} g\right) / 2$. Evaluated at 0 , this gives 16.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-calc-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution: "
}
|
06fcc9a3-b6cc-5e46-9c77-99e3a9ff1d58
| 610,982
|
Find the area of the region in the first quadrant $x>0, y>0$ bounded above the graph of $y=\arcsin (x)$ and below the graph of the $y=\arccos (x)$.
|
We can integrate over $y$ rather than $x$. In particular, the solution is $\int_{0}^{\pi / 4} \sin y \mathrm{~d} y+$ $\int_{\pi / 4}^{\pi / 2} \cos y \mathrm{~d} y=\left(1-\frac{1}{\sqrt{2}}\right) 2=2-\sqrt{2}$.
|
2-\sqrt{2}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Find the area of the region in the first quadrant $x>0, y>0$ bounded above the graph of $y=\arcsin (x)$ and below the graph of the $y=\arccos (x)$.
|
We can integrate over $y$ rather than $x$. In particular, the solution is $\int_{0}^{\pi / 4} \sin y \mathrm{~d} y+$ $\int_{\pi / 4}^{\pi / 2} \cos y \mathrm{~d} y=\left(1-\frac{1}{\sqrt{2}}\right) 2=2-\sqrt{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-calc-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution: "
}
|
c020d4cb-5808-5e51-a305-778a77e11174
| 610,983
|
What is the minimum vertical distance between the graphs of $2+\sin (x)$ and $\cos (x)$ ?
|
The derivative of $2+\sin (x)-\cos (x)$ is $\cos x+\sin x$, which in the interval $0 \leq x<2 \pi$ is zero at $x=\frac{3 \pi}{4}, \frac{7 \pi}{4}$. At $\frac{7 \pi}{4}$, when $\sin (x)$ is negative and $\cos (x)$ is positive, the distance reaches its minimal value of $2-\sqrt{2}$.
|
2-\sqrt{2}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
What is the minimum vertical distance between the graphs of $2+\sin (x)$ and $\cos (x)$ ?
|
The derivative of $2+\sin (x)-\cos (x)$ is $\cos x+\sin x$, which in the interval $0 \leq x<2 \pi$ is zero at $x=\frac{3 \pi}{4}, \frac{7 \pi}{4}$. At $\frac{7 \pi}{4}$, when $\sin (x)$ is negative and $\cos (x)$ is positive, the distance reaches its minimal value of $2-\sqrt{2}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-calc-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nSolution: "
}
|
bfdd51b8-bb21-5549-b92c-1e00aabeaa94
| 610,984
|
Determine the positive value of $a$ such that the parabola $y=x^{2}+1$ bisects the area of the rectangle with vertices $(0,0),(a, 0),\left(0, a^{2}+1\right)$, and $\left(a, a^{2}+1\right)$.
|
$\sqrt{3}$ The area of the rectangle is $a^{3}+a$. The portion under the parabola has area $\int_{0}^{a} x^{2}+1 d x=a^{3} / 3+a$. Thus we wish to solve the equation $a^{3}+a=2\left(a^{3} / 3+a\right)$; dividing by $a$ and rearranging gives $a^{2} / 3=1$, so $a=\sqrt{3}$.
|
\sqrt{3}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Determine the positive value of $a$ such that the parabola $y=x^{2}+1$ bisects the area of the rectangle with vertices $(0,0),(a, 0),\left(0, a^{2}+1\right)$, and $\left(a, a^{2}+1\right)$.
|
$\sqrt{3}$ The area of the rectangle is $a^{3}+a$. The portion under the parabola has area $\int_{0}^{a} x^{2}+1 d x=a^{3} / 3+a$. Thus we wish to solve the equation $a^{3}+a=2\left(a^{3} / 3+a\right)$; dividing by $a$ and rearranging gives $a^{2} / 3=1$, so $a=\sqrt{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-calc-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nSolution: "
}
|
5db4bddd-91d1-5fb5-be9a-1f0437f17385
| 610,985
|
Denote by $\langle x\rangle$ the fractional part of the real number $x$ (for instance, $\langle 3.2\rangle=0.2$ ). A positive integer $N$ is selected randomly from the set $\{1,2,3, \ldots, M\}$, with each integer having the same probability of being picked, and $\left\langle\frac{87}{303} N\right\rangle$ is calculated. This procedure is repeated $M$ times and the average value $A(M)$ is obtained. What is $\lim _{M \rightarrow \infty} A(M)$ ?
|
This method of picking $N$ is equivalent to uniformly randomly selecting a positive integer. Call this the average value of $\left\langle\frac{87}{303} N\right\rangle$ for $N$ a positive integer. In lowest terms, $\frac{87}{303}=\frac{29}{101}$, so the answer is the same as the average value of $\frac{0}{101}, \frac{1}{101}, \ldots, \frac{100}{101}$, which is $\frac{1+2+\cdots+100}{101 \cdot 101}=\frac{100 \cdot 101 / 2}{101 \cdot 101}=\frac{50}{101}$.
|
\frac{50}{101}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Denote by $\langle x\rangle$ the fractional part of the real number $x$ (for instance, $\langle 3.2\rangle=0.2$ ). A positive integer $N$ is selected randomly from the set $\{1,2,3, \ldots, M\}$, with each integer having the same probability of being picked, and $\left\langle\frac{87}{303} N\right\rangle$ is calculated. This procedure is repeated $M$ times and the average value $A(M)$ is obtained. What is $\lim _{M \rightarrow \infty} A(M)$ ?
|
This method of picking $N$ is equivalent to uniformly randomly selecting a positive integer. Call this the average value of $\left\langle\frac{87}{303} N\right\rangle$ for $N$ a positive integer. In lowest terms, $\frac{87}{303}=\frac{29}{101}$, so the answer is the same as the average value of $\frac{0}{101}, \frac{1}{101}, \ldots, \frac{100}{101}$, which is $\frac{1+2+\cdots+100}{101 \cdot 101}=\frac{100 \cdot 101 / 2}{101 \cdot 101}=\frac{50}{101}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-calc-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nSolution: "
}
|
8a0de652-6c1d-5364-9631-a7a3e15735d9
| 610,986
|
Evaluate $\int_{0}^{(\sqrt{2}-1) / 2} \frac{\mathrm{~d} x}{(2 x+1) \sqrt{x^{2}+x}}$.
|
Let $u=\sqrt{x^{2}+x}$. Then $d u=\frac{2 x+1}{2 \sqrt{x^{2}+x}} d x$. So the integral becomes $2 \int \frac{d u}{\left(4 x^{2}+4 x+1\right)}$, or $2 \int \frac{d u}{4 u^{2}+1}$. This is $\tan ^{-1}(2 u)$, yielding a final answer of $\tan ^{-1}\left(2 \sqrt{x^{2}+x}\right)+C$ for the indefinite integral. The definite integral becomes $\tan ^{-1}(1)-\tan ^{-1}(0)=\frac{\pi}{4}$.
|
\frac{\pi}{4}
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Evaluate $\int_{0}^{(\sqrt{2}-1) / 2} \frac{\mathrm{~d} x}{(2 x+1) \sqrt{x^{2}+x}}$.
|
Let $u=\sqrt{x^{2}+x}$. Then $d u=\frac{2 x+1}{2 \sqrt{x^{2}+x}} d x$. So the integral becomes $2 \int \frac{d u}{\left(4 x^{2}+4 x+1\right)}$, or $2 \int \frac{d u}{4 u^{2}+1}$. This is $\tan ^{-1}(2 u)$, yielding a final answer of $\tan ^{-1}\left(2 \sqrt{x^{2}+x}\right)+C$ for the indefinite integral. The definite integral becomes $\tan ^{-1}(1)-\tan ^{-1}(0)=\frac{\pi}{4}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-calc-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nSolution: "
}
|
2a2c20d3-017e-5b26-bae7-bad1a3aa12d1
| 610,987
|
Suppose $f$ is a differentiable real function such that $f(x)+f^{\prime}(x) \leq 1$ for all $x$, and $f(0)=0$. What is the largest possible value of $f(1)$ ? (Hint: consider the function $e^{x} f(x)$.)
|
$1-1 / e$ Let $g(x)=e^{x} f(x)$; then $g^{\prime}(x)=e^{x}\left(f(x)+f^{\prime}(x)\right) \leq e^{x}$. Integrating from 0 to 1 , we have $g(1)-g(0)=\int_{0}^{1} g^{\prime}(x) d x \leq \int_{0}^{1} e^{x} d x=e-1$. But $g(1)-g(0)=e \cdot f(1)$, so we get $f(1) \leq(e-1) / e$. This maximum is attained if we actually have $g^{\prime}(x)=e^{x}$ everywhere; this entails the requirement $f(x)+f^{\prime}(x)=1$, which is met by $f(x)=1-e^{-x}$.
|
1-1/e
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
Suppose $f$ is a differentiable real function such that $f(x)+f^{\prime}(x) \leq 1$ for all $x$, and $f(0)=0$. What is the largest possible value of $f(1)$ ? (Hint: consider the function $e^{x} f(x)$.)
|
$1-1 / e$ Let $g(x)=e^{x} f(x)$; then $g^{\prime}(x)=e^{x}\left(f(x)+f^{\prime}(x)\right) \leq e^{x}$. Integrating from 0 to 1 , we have $g(1)-g(0)=\int_{0}^{1} g^{\prime}(x) d x \leq \int_{0}^{1} e^{x} d x=e-1$. But $g(1)-g(0)=e \cdot f(1)$, so we get $f(1) \leq(e-1) / e$. This maximum is attained if we actually have $g^{\prime}(x)=e^{x}$ everywhere; this entails the requirement $f(x)+f^{\prime}(x)=1$, which is met by $f(x)=1-e^{-x}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-calc-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nSolution: "
}
|
f359216f-ebbb-5a5b-9053-78c14bc247d4
| 610,988
|
A continuous real function $f$ satisfies the identity $f(2 x)=3 f(x)$ for all $x$. If $\int_{0}^{1} f(x) d x=1$, what is $\int_{1}^{2} f(x) d x$ ?
|
5 Let $S=\int_{1}^{2} f(x) d x$. By setting $u=2 x$, we see that $\int_{1 / 2}^{1} f(x) d x=$ $\int_{1 / 2}^{1} f(2 x) / 3 d x=\int_{1}^{2} f(u) / 6 d u=S / 6$. Similarly, $\int_{1 / 4}^{1 / 2} f(x) d x=S / 36$, and in general $\int_{1 / 2^{n}}^{1 / 2^{n-1}} f(x) d x=S / 6^{n}$. Adding finitely many of these, we have $\int_{1 / 2^{n}}^{1} f(x) d x=S / 6+S / 36+$ $\cdots+S / 6^{n}=S \cdot\left(1-1 / 6^{n}\right) / 5$. Taking the limit as $n \rightarrow \infty$, we have $\int_{0}^{1} f(x) d x=S / 5$. Thus $S=5$, the answer.
|
5
|
Yes
|
Yes
|
math-word-problem
|
Calculus
|
A continuous real function $f$ satisfies the identity $f(2 x)=3 f(x)$ for all $x$. If $\int_{0}^{1} f(x) d x=1$, what is $\int_{1}^{2} f(x) d x$ ?
|
5 Let $S=\int_{1}^{2} f(x) d x$. By setting $u=2 x$, we see that $\int_{1 / 2}^{1} f(x) d x=$ $\int_{1 / 2}^{1} f(2 x) / 3 d x=\int_{1}^{2} f(u) / 6 d u=S / 6$. Similarly, $\int_{1 / 4}^{1 / 2} f(x) d x=S / 36$, and in general $\int_{1 / 2^{n}}^{1 / 2^{n-1}} f(x) d x=S / 6^{n}$. Adding finitely many of these, we have $\int_{1 / 2^{n}}^{1} f(x) d x=S / 6+S / 36+$ $\cdots+S / 6^{n}=S \cdot\left(1-1 / 6^{n}\right) / 5$. Taking the limit as $n \rightarrow \infty$, we have $\int_{0}^{1} f(x) d x=S / 5$. Thus $S=5$, the answer.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-calc-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nSolution: "
}
|
5430d9f8-b080-5860-a4a4-082cf234a6ea
| 610,989
|
What is the maximum number of lattice points (i.e. points with integer coordinates) in the plane that can be contained strictly inside a circle of radius 1 ?
|
4. The circle centered at $(1 / 2,1 / 2)$ shows that 4 is achievable. On the other hand, no two points within the circle can be at a mutual distance of 2 or greater. If there are more than four lattice points, classify all such points by the parity of their coordinates: (even, even), (even, odd), (odd, even), or (odd, odd). Then some two points lie in the same class. Since they are distinct, this means either their first or second coordinates must differ by at least 2 , so their distance is at least 2 , a contradiction.
|
4
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
What is the maximum number of lattice points (i.e. points with integer coordinates) in the plane that can be contained strictly inside a circle of radius 1 ?
|
4. The circle centered at $(1 / 2,1 / 2)$ shows that 4 is achievable. On the other hand, no two points within the circle can be at a mutual distance of 2 or greater. If there are more than four lattice points, classify all such points by the parity of their coordinates: (even, even), (even, odd), (odd, even), or (odd, odd). Then some two points lie in the same class. Since they are distinct, this means either their first or second coordinates must differ by at least 2 , so their distance is at least 2 , a contradiction.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-gen1-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution: "
}
|
f096121e-a3f4-5372-95ca-75e56b5f6414
| 610,990
|
How many triples $(A, B, C)$ of positive integers (positive integers are the numbers $1,2,3,4, \ldots$ ) are there such that $A+B+C=10$, where order does not matter (for instance the triples $(2,3,5)$ and $(3,2,5)$ are considered to be the same triple) and where two of the integers in a triple could be the same (for instance $(3,3,4)$ is a valid triple).
|
The triples can merely be enumerated: $(1,1,8),(1,2,7),(1,3,6),(1,4,5)$, $(2,2,6),(2,3,5),(2,4,4)$, and $(3,3,4)$. There are 8 elements.
|
8
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many triples $(A, B, C)$ of positive integers (positive integers are the numbers $1,2,3,4, \ldots$ ) are there such that $A+B+C=10$, where order does not matter (for instance the triples $(2,3,5)$ and $(3,2,5)$ are considered to be the same triple) and where two of the integers in a triple could be the same (for instance $(3,3,4)$ is a valid triple).
|
The triples can merely be enumerated: $(1,1,8),(1,2,7),(1,3,6),(1,4,5)$, $(2,2,6),(2,3,5),(2,4,4)$, and $(3,3,4)$. There are 8 elements.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-gen1-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution: "
}
|
71eef604-5463-505c-93b6-294a9a6fd242
| 610,991
|
We call a set of professors and committees on which they serve a university if
(1) given two distinct professors there is one and only one committee on which they both serve,
(2) given any committee, $C$, and any professor, $P$, not on that committee, there is exactly one committee on which $P$ serves and no professors on committee $C$ serve, and
(3) there are at least two professors on each committee; there are at least two committees. What is the smallest number of committees a university can have?
|
Let $C$ be any committee. Then there exists a professor $P$ not on $C$ (or else there would be no other committees). By axiom 2, $P$ serves on a committee $D$ having no common members with $C$. Each of these committees has at least two members, and for each $Q \in C, R \in D$, there exists (by axiom 1) a committee containing $Q$ and $R$, which (again by axiom 1) has no other common members with $C$ or $D$. Thus we have at least
$2+2 \cdot 2=6$ committees. This minimum is attainable - just take four professors and let any two professors form a committee.
|
6
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
We call a set of professors and committees on which they serve a university if
(1) given two distinct professors there is one and only one committee on which they both serve,
(2) given any committee, $C$, and any professor, $P$, not on that committee, there is exactly one committee on which $P$ serves and no professors on committee $C$ serve, and
(3) there are at least two professors on each committee; there are at least two committees. What is the smallest number of committees a university can have?
|
Let $C$ be any committee. Then there exists a professor $P$ not on $C$ (or else there would be no other committees). By axiom 2, $P$ serves on a committee $D$ having no common members with $C$. Each of these committees has at least two members, and for each $Q \in C, R \in D$, there exists (by axiom 1) a committee containing $Q$ and $R$, which (again by axiom 1) has no other common members with $C$ or $D$. Thus we have at least
$2+2 \cdot 2=6$ committees. This minimum is attainable - just take four professors and let any two professors form a committee.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-gen1-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution: "
}
|
f97d76f0-8a13-5635-b4b6-187db28a6b07
| 610,992
|
A circle is inscribed in a square dartboard. If a dart is thrown at the dartboard and hits the dartboard in a random location, with all locations having the same probability of being hit, what is the probability that it lands within the circle?
|
The answer is the area of the circle over the area of the square, which is $\frac{\pi}{4}$.
|
\frac{\pi}{4}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
A circle is inscribed in a square dartboard. If a dart is thrown at the dartboard and hits the dartboard in a random location, with all locations having the same probability of being hit, what is the probability that it lands within the circle?
|
The answer is the area of the circle over the area of the square, which is $\frac{\pi}{4}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-gen1-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nSolution: "
}
|
e1b52b18-2d93-5112-b235-ea886aa845b0
| 610,995
|
Count the number of triangles with positive area whose vertices are points whose $(x, y)$-coordinates lie in the set $\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\}$.
|
There are $\binom{9}{3}=84$ triples of points. 8 of them form degenerate triangles (the ones that lie on a line), so there are $84-8=76$ nondegenerate triangles.
|
76
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Count the number of triangles with positive area whose vertices are points whose $(x, y)$-coordinates lie in the set $\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\}$.
|
There are $\binom{9}{3}=84$ triples of points. 8 of them form degenerate triangles (the ones that lie on a line), so there are $84-8=76$ nondegenerate triangles.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-gen1-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nSolution: "
}
|
ccfa9e0b-545f-51e7-bce5-9733a6c91979
| 610,996
|
The squares of a chessboard are numbered from left to right and top to bottom (so that the first row reads $1,2, \ldots, 8$, the second reads $9,10, \ldots, 16$, and so forth). The number 1 is on a black square. How many black squares contain odd numbers?
|
16. The black squares in the $n$th row contain odd numbers when $n$ is odd and even numbers when $n$ is even; thus there are four rows where the black squares contain odd numbers, and each such row contributes four black squares.
|
16
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
The squares of a chessboard are numbered from left to right and top to bottom (so that the first row reads $1,2, \ldots, 8$, the second reads $9,10, \ldots, 16$, and so forth). The number 1 is on a black square. How many black squares contain odd numbers?
|
16. The black squares in the $n$th row contain odd numbers when $n$ is odd and even numbers when $n$ is even; thus there are four rows where the black squares contain odd numbers, and each such row contributes four black squares.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-gen2-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution: "
}
|
71eb4c11-7ff2-52bc-8f4b-2a51b89ffa1f
| 610,997
|
You are in a completely dark room with a drawer containing 10 red, 20 blue, 30 green, and 40 khaki socks. What is the smallest number of socks you must randomly pull out in order to be sure of having at least one of each color?
|
91 . The maximum number of socks that can be pulled out without representing every color is 20 blue +30 green +40 khaki $=90$, so 91 is the minimum needed to ensure that this doesn't happen.
|
91
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
You are in a completely dark room with a drawer containing 10 red, 20 blue, 30 green, and 40 khaki socks. What is the smallest number of socks you must randomly pull out in order to be sure of having at least one of each color?
|
91 . The maximum number of socks that can be pulled out without representing every color is 20 blue +30 green +40 khaki $=90$, so 91 is the minimum needed to ensure that this doesn't happen.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-gen2-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution: "
}
|
b31bc265-a4a0-5cea-bd16-ec4f03686fe1
| 610,998
|
Solve for $x$ in $3=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$.
|
Squaring both sides and subtracting $x$ from both sides, we get $9-x=3$, or $x=6$.
|
6
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Solve for $x$ in $3=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$.
|
Squaring both sides and subtracting $x$ from both sides, we get $9-x=3$, or $x=6$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-gen2-solutions.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution: "
}
|
ac9cef03-859b-5ccc-a680-e4dc09f0b70f
| 610,999
|
Dan is holding one end of a 26 inch long piece of light string that has a heavy bead on it with each hand (so that the string lies along two straight lines). If he starts with his hands together at the start and leaves his hands at the same height, how far does he need to pull his hands apart so that the bead moves upward by 8 inches?
|
After he pulls the bead is 5 inches below his hands, and it is 13 inches from each hand. Using the Pythagorean theorem, his hands must be $2 \cdot 12=24$ inches apart.
|
24
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Dan is holding one end of a 26 inch long piece of light string that has a heavy bead on it with each hand (so that the string lies along two straight lines). If he starts with his hands together at the start and leaves his hands at the same height, how far does he need to pull his hands apart so that the bead moves upward by 8 inches?
|
After he pulls the bead is 5 inches below his hands, and it is 13 inches from each hand. Using the Pythagorean theorem, his hands must be $2 \cdot 12=24$ inches apart.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-gen2-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution: "
}
|
9b8cbff6-1926-5767-bb36-31b61f3eb401
| 611,000
|
$A$ and $B$ are two points on a circle with center $O$, and $C$ lies outside the circle, on ray $A B$. Given that $A B=24, B C=28, O A=15$, find $O C$.
|
41. Let $M$ be the midpoint of $A B$; then $\triangle O M B$ is a right triangle with $O B=15, M B=12$, so $O M=9$. Now $\triangle O M C$ is a right triangle with $O M=9, M C=40$, so $O C=41$.
|
41
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
$A$ and $B$ are two points on a circle with center $O$, and $C$ lies outside the circle, on ray $A B$. Given that $A B=24, B C=28, O A=15$, find $O C$.
|
41. Let $M$ be the midpoint of $A B$; then $\triangle O M B$ is a right triangle with $O B=15, M B=12$, so $O M=9$. Now $\triangle O M C$ is a right triangle with $O M=9, M C=40$, so $O C=41$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-gen2-solutions.jsonl",
"problem_match": "\n9. ",
"solution_match": "\nSolution: "
}
|
ca372a37-8def-5f48-a5fe-a5cf0b47b1e4
| 611,001
|
Dan is holding one end of a 26 inch long piece of light string that has a heavy bead on it with each hand (so that the string lies along straight lines). If he starts with his hands together at the start and leaves his hands at the same height, how far does he need to pull his hands apart so that the bead moves upward by 8 inches?
|
After he pulls the bead is 5 inches below his hands, and it is 13 inches from each hand. Using the Pythagorean theorem, his hands must be $2 \cdot 12=24$ inches apart.
|
24
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Dan is holding one end of a 26 inch long piece of light string that has a heavy bead on it with each hand (so that the string lies along straight lines). If he starts with his hands together at the start and leaves his hands at the same height, how far does he need to pull his hands apart so that the bead moves upward by 8 inches?
|
After he pulls the bead is 5 inches below his hands, and it is 13 inches from each hand. Using the Pythagorean theorem, his hands must be $2 \cdot 12=24$ inches apart.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-geo-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution: "
}
|
fdb4270e-cfc0-5eb0-967f-c81596e28026
| 611,002
|
We call a set of professors and committees on which they serve a university if
(1) given two distinct professors there is one and only one committee on which they both serve,
(2) given any committee, $C$, and any professor, $P$, not on that committee, there is exactly one committee on which $P$ serves and no professors on committee $C$ serve, and
(3) there are at least two professors on each committee; there are at least two committees.
What is the smallest number of committees a university can have?
|
Let $C$ be any committee. Then there exists a professor $P$ not on $C$ (or else there would be no other committees). By axiom $2, P$ serves on a committee $D$ having no common members with $C$. Each of these committees has at least two members, and for each $Q \in C, R \in D$, there exists (by axiom 1) a committee containing $Q$ and $R$, which (again by axiom 1) has no other common members with $C$ or $D$. Thus we have at least $2+2 \cdot 2=6$ committees. This minimum is attainable - just take four professors and let any two professors form a committee.
|
6
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
We call a set of professors and committees on which they serve a university if
(1) given two distinct professors there is one and only one committee on which they both serve,
(2) given any committee, $C$, and any professor, $P$, not on that committee, there is exactly one committee on which $P$ serves and no professors on committee $C$ serve, and
(3) there are at least two professors on each committee; there are at least two committees.
What is the smallest number of committees a university can have?
|
Let $C$ be any committee. Then there exists a professor $P$ not on $C$ (or else there would be no other committees). By axiom $2, P$ serves on a committee $D$ having no common members with $C$. Each of these committees has at least two members, and for each $Q \in C, R \in D$, there exists (by axiom 1) a committee containing $Q$ and $R$, which (again by axiom 1) has no other common members with $C$ or $D$. Thus we have at least $2+2 \cdot 2=6$ committees. This minimum is attainable - just take four professors and let any two professors form a committee.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-geo-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution: "
}
|
4b03fc83-0152-5128-b201-7332576a1982
| 611,003
|
Consider a square of side length 1 . Draw four lines that each connect a midpoint of a side with a corner not on that side, such that each midpoint and each corner is touched by only one line. Find the area of the region completely bounded by these lines.
|
In unit square $A B C D$, denote by $E, F, G, H$ the respective midpoints of sides $A B, B C, C D, D A$. Let $I$ be the intersection of $A F$ and $D E$, let $J$ be the intersection of $B G$ and $A F$, let $K$ be the intersection of $C H$ and $B G$, and let $L$ be the intersection of $D E$ and $C H$. We want to find the area of square $I J K L$. The area of $A B F$ is $\frac{1}{4}$, which is equal to $\frac{1}{2} A F \cdot B J=\frac{\sqrt{5}}{4} B J$, so $B J=\frac{1}{\sqrt{5}}$. Using similar triangles, $G K=J F=\frac{1}{2} B J$. Thus the length of a side of $I J K L$ is $J K=\frac{\sqrt{5}}{2}-\frac{1}{\sqrt{5}}-\frac{1}{2} \frac{1}{\sqrt{5}}=\frac{1}{\sqrt{5}}$, and the area of $I J K L$ is $\frac{1}{5}$.
|
\frac{1}{5}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Consider a square of side length 1 . Draw four lines that each connect a midpoint of a side with a corner not on that side, such that each midpoint and each corner is touched by only one line. Find the area of the region completely bounded by these lines.
|
In unit square $A B C D$, denote by $E, F, G, H$ the respective midpoints of sides $A B, B C, C D, D A$. Let $I$ be the intersection of $A F$ and $D E$, let $J$ be the intersection of $B G$ and $A F$, let $K$ be the intersection of $C H$ and $B G$, and let $L$ be the intersection of $D E$ and $C H$. We want to find the area of square $I J K L$. The area of $A B F$ is $\frac{1}{4}$, which is equal to $\frac{1}{2} A F \cdot B J=\frac{\sqrt{5}}{4} B J$, so $B J=\frac{1}{\sqrt{5}}$. Using similar triangles, $G K=J F=\frac{1}{2} B J$. Thus the length of a side of $I J K L$ is $J K=\frac{\sqrt{5}}{2}-\frac{1}{\sqrt{5}}-\frac{1}{2} \frac{1}{\sqrt{5}}=\frac{1}{\sqrt{5}}$, and the area of $I J K L$ is $\frac{1}{5}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-geo-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nSolution: "
}
|
ad356761-6c64-528b-bf76-dd8e7aac006c
| 611,004
|
If we pick (uniformly) a random square of area 1 with sides parallel to the $x$ - and $y$-axes that lies entirely within the 5 -by- 5 square bounded by the lines $x=0, x=5, y=$ $0, y=5$ (the corners of the square need not have integer coordinates), what is the probability that the point $(x, y)=(4.5,0.5)$ lies within the square of area 1 ?
|
The upper-left corner of the unit square is picked uniformly from the square $0 \leq x \leq 4 ; 1 \leq y \leq 5$, and for it to contain the desired point it must lie in the square $3.5 \leq x \leq 4 ; 1 \leq y \leq 1.5$. The answer is the ratio of the squares' areas, $\frac{1}{4} / 16=\frac{1}{64}$.
|
\frac{1}{64}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
If we pick (uniformly) a random square of area 1 with sides parallel to the $x$ - and $y$-axes that lies entirely within the 5 -by- 5 square bounded by the lines $x=0, x=5, y=$ $0, y=5$ (the corners of the square need not have integer coordinates), what is the probability that the point $(x, y)=(4.5,0.5)$ lies within the square of area 1 ?
|
The upper-left corner of the unit square is picked uniformly from the square $0 \leq x \leq 4 ; 1 \leq y \leq 5$, and for it to contain the desired point it must lie in the square $3.5 \leq x \leq 4 ; 1 \leq y \leq 1.5$. The answer is the ratio of the squares' areas, $\frac{1}{4} / 16=\frac{1}{64}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-geo-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nSolution: "
}
|
57de09e9-a2b1-54b4-a195-b7d1f8e62413
| 611,005
|
Equilateral triangle $A B C$ of side length 2 is drawn. Three squares external to the triangle, $A B D E, B C F G$, and $C A H I$, are drawn. What is the area of the smallest triangle that contains these squares?
|
The equilateral triangle with sides lying on lines $D G, E H$, and $F I$ has minimal area. (The only other reasonable candidate is the triangle with sides along $D E, F G, H I$, but a quick sketch shows that it is larger.) Let $J, K$, and $L$ be the vertices of this triangle closest to $D, H$, and $F$, respectively. Clearly, $K I=F L=2$. Triangle $F C I$ is a $30^{\circ}-30^{\circ}-120^{\circ}$ triangle, so we can calculate the length of $F I$ as $2 \sqrt{3}$, making the side length of $\triangle J K L$ $4+2 \sqrt{3}$, and its area $12+7 \sqrt{3}$.
|
12+7 \sqrt{3}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Equilateral triangle $A B C$ of side length 2 is drawn. Three squares external to the triangle, $A B D E, B C F G$, and $C A H I$, are drawn. What is the area of the smallest triangle that contains these squares?
|
The equilateral triangle with sides lying on lines $D G, E H$, and $F I$ has minimal area. (The only other reasonable candidate is the triangle with sides along $D E, F G, H I$, but a quick sketch shows that it is larger.) Let $J, K$, and $L$ be the vertices of this triangle closest to $D, H$, and $F$, respectively. Clearly, $K I=F L=2$. Triangle $F C I$ is a $30^{\circ}-30^{\circ}-120^{\circ}$ triangle, so we can calculate the length of $F I$ as $2 \sqrt{3}$, making the side length of $\triangle J K L$ $4+2 \sqrt{3}$, and its area $12+7 \sqrt{3}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-geo-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nSolution: "
}
|
0d0d0f3c-1d4d-5073-a8fc-463e2ae3af2c
| 611,006
|
Equilateral triangle $A B C$ of side length 2 is drawn. Three squares containing the triangle, $A B D E, B C F G$, and $C A H I$, are drawn. What is the area of the smallest triangle that contains these squares?
|
The equilateral triangle with sides lying on lines $D E, F G$, and $H I$ has minimal area. Let $J, K$, and $L$ be the vertices of this triangle closest to $D, H$, and $F$, respectively. Clearly, $D E=2$. Denote by $M$ the intersection of $C I$ and $B D$. Using the $30^{\circ}-30^{\circ}-120^{\circ}$ triangle $B C M$, we get $B M=2 / \sqrt{3}$, and thus $M D=2-2 / \sqrt{3}$. By symmetry, $M J$ bisects angle $D M I$, from which we see that $\triangle J M D$ is a $30^{\circ}-60^{\circ}-90^{\circ}$. We then get $J D=2 \sqrt{3}-2$, making the side length of $J K L 4 \sqrt{3}-2$, and its area $13 \sqrt{3}-12$.
|
13 \sqrt{3}-12
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Equilateral triangle $A B C$ of side length 2 is drawn. Three squares containing the triangle, $A B D E, B C F G$, and $C A H I$, are drawn. What is the area of the smallest triangle that contains these squares?
|
The equilateral triangle with sides lying on lines $D E, F G$, and $H I$ has minimal area. Let $J, K$, and $L$ be the vertices of this triangle closest to $D, H$, and $F$, respectively. Clearly, $D E=2$. Denote by $M$ the intersection of $C I$ and $B D$. Using the $30^{\circ}-30^{\circ}-120^{\circ}$ triangle $B C M$, we get $B M=2 / \sqrt{3}$, and thus $M D=2-2 / \sqrt{3}$. By symmetry, $M J$ bisects angle $D M I$, from which we see that $\triangle J M D$ is a $30^{\circ}-60^{\circ}-90^{\circ}$. We then get $J D=2 \sqrt{3}-2$, making the side length of $J K L 4 \sqrt{3}-2$, and its area $13 \sqrt{3}-12$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-geo-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nSolution: "
}
|
1cf87fb6-a27a-5f46-8312-a78c1b01ee06
| 611,007
|
Let $\triangle A B C$ be equilateral, and let $D, E, F$ be points on sides $B C, C A, A B$ respectively, with $F A=9, A E=E C=6, C D=4$. Determine the measure (in degrees) of $\angle D E F$.
|
60. Let $H, I$ be the respective midpoints of sides $B C, A B$, and also extend $C B$ and $E F$ to intersect at $J$. By equal angles, $\triangle E I F \sim \triangle J B F$. However, $B F=12-9=$ $3=9-6=I F$, so in fact $\triangle E I F \cong \triangle J B F$, and then $J B=6$. Now let $H I$ intersect $E F$ at $K$, and notice that $\triangle E I K \sim \triangle J H K \Rightarrow I K / H K=E I / J H=6 / 12=1 / 2 \Rightarrow H K=4$, since $I K+H K=H I=6$. Now consider the $60^{\circ}$ rotation about $E$ carrying triangle $C H E$ to triangle HIE; we see that it also takes $D$ to $K$, and thus $\angle D E F=\angle D E K=60^{\circ}$.
|
60
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $\triangle A B C$ be equilateral, and let $D, E, F$ be points on sides $B C, C A, A B$ respectively, with $F A=9, A E=E C=6, C D=4$. Determine the measure (in degrees) of $\angle D E F$.
|
60. Let $H, I$ be the respective midpoints of sides $B C, A B$, and also extend $C B$ and $E F$ to intersect at $J$. By equal angles, $\triangle E I F \sim \triangle J B F$. However, $B F=12-9=$ $3=9-6=I F$, so in fact $\triangle E I F \cong \triangle J B F$, and then $J B=6$. Now let $H I$ intersect $E F$ at $K$, and notice that $\triangle E I K \sim \triangle J H K \Rightarrow I K / H K=E I / J H=6 / 12=1 / 2 \Rightarrow H K=4$, since $I K+H K=H I=6$. Now consider the $60^{\circ}$ rotation about $E$ carrying triangle $C H E$ to triangle HIE; we see that it also takes $D$ to $K$, and thus $\angle D E F=\angle D E K=60^{\circ}$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-geo-solutions.jsonl",
"problem_match": "\n10. ",
"solution_match": "\nSolution: "
}
|
ec744ea3-7bca-5294-96e2-20ed5f5a98c5
| 611,008
|
An $(l, a)$-design of a set is a collection of subsets of that set such that each subset contains exactly $l$ elements and that no two of the subsets share more than $a$ elements. How many $(2,1)$-designs are there of a set containing 8 elements?
|
There are $\binom{8}{2}=28$ 2-element subsets. Any two distinct such subsets have at most 1 common element; hence, for each subset, we can decide independently whether or not it belongs to the design, and we thus obtain $2^{28}$ designs.
|
2^{28}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
An $(l, a)$-design of a set is a collection of subsets of that set such that each subset contains exactly $l$ elements and that no two of the subsets share more than $a$ elements. How many $(2,1)$-designs are there of a set containing 8 elements?
|
There are $\binom{8}{2}=28$ 2-element subsets. Any two distinct such subsets have at most 1 common element; hence, for each subset, we can decide independently whether or not it belongs to the design, and we thus obtain $2^{28}$ designs.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution: "
}
|
c5872a3b-b764-52ba-b9e1-b599583cc2c0
| 611,009
|
A lattice point in the plane is a point of the form $(n, m)$, where $n$ and $m$ are integers. Consider a set $S$ of lattice points. We construct the transform of $S$, denoted by $S^{\prime}$, by the following rule: the pair $(n, m)$ is in $S^{\prime}$ if and only if any of $(n, m-1),(n, m+1),(n-1, m)$, $(n+1, m)$, and $(n, m)$ is in $S$. How many elements are in the set obtained by successively transforming $\{(0,0)\} 14$ times?
|
Transforming it $k \geq 1$ times yields the "diamond" of points ( $n, m$ ) such that $|n|+|m| \leq k$. The diamond contains $(k+1)^{2}+k^{2}$ lattice points (this can be seen by rotating the plane 45 degrees and noticing the lattice points in the transforms form two squares, one of which is contained in the other), so the answer is 421 .
|
421
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A lattice point in the plane is a point of the form $(n, m)$, where $n$ and $m$ are integers. Consider a set $S$ of lattice points. We construct the transform of $S$, denoted by $S^{\prime}$, by the following rule: the pair $(n, m)$ is in $S^{\prime}$ if and only if any of $(n, m-1),(n, m+1),(n-1, m)$, $(n+1, m)$, and $(n, m)$ is in $S$. How many elements are in the set obtained by successively transforming $\{(0,0)\} 14$ times?
|
Transforming it $k \geq 1$ times yields the "diamond" of points ( $n, m$ ) such that $|n|+|m| \leq k$. The diamond contains $(k+1)^{2}+k^{2}$ lattice points (this can be seen by rotating the plane 45 degrees and noticing the lattice points in the transforms form two squares, one of which is contained in the other), so the answer is 421 .
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution: "
}
|
ec0eb93c-73bf-5382-adf2-76a3cdaeca84
| 611,010
|
How many ways are there of using diagonals to divide a regular 6 -sided polygon into triangles such that at least one side of each triangle is a side of the original polygon and that each vertex of each triangle is a vertex of the original polygon?
|
The number of ways of triangulating a convex $(n+2)$-sided polygon is $\binom{2 n}{n} \frac{1}{n+1}$, which is 14 in this case. However, there are two triangulations of a hexagon which produce one triangle sharing no sides with the original polygon, so the answer is $14-2=12$.
|
12
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
How many ways are there of using diagonals to divide a regular 6 -sided polygon into triangles such that at least one side of each triangle is a side of the original polygon and that each vertex of each triangle is a vertex of the original polygon?
|
The number of ways of triangulating a convex $(n+2)$-sided polygon is $\binom{2 n}{n} \frac{1}{n+1}$, which is 14 in this case. However, there are two triangulations of a hexagon which produce one triangle sharing no sides with the original polygon, so the answer is $14-2=12$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution: "
}
|
08242c3b-91ef-5f9c-aba1-d8d78cef9a10
| 611,012
|
Two $4 \times 4$ squares are randomly placed on an $8 \times 8$ chessboard so that their sides lie along the grid lines of the board. What is the probability that the two squares overlap?
|
529/625. Each square has 5 horizontal $\cdot 5$ vertical $=25$ possible positions, so there are 625 possible placements of the squares. If they do not overlap, then either one square lies in the top four rows and the other square lies in the bottom four rows, or one square lies in the left four columns and the other lies in the right four columns. The first possibility can happen in $2 \cdot 5 \cdot 5=50$ ways (two choices of which square goes on top, and five horizontal positions for each square); likewise, so can the second. However,
this double-counts the 4 cases in which the two squares are in opposite corners, so we have $50+50-4=96$ possible non-overlapping arrangements $\Rightarrow 25^{2}-96=529$ overlapping arrangements.
|
\frac{529}{625}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Two $4 \times 4$ squares are randomly placed on an $8 \times 8$ chessboard so that their sides lie along the grid lines of the board. What is the probability that the two squares overlap?
|
529/625. Each square has 5 horizontal $\cdot 5$ vertical $=25$ possible positions, so there are 625 possible placements of the squares. If they do not overlap, then either one square lies in the top four rows and the other square lies in the bottom four rows, or one square lies in the left four columns and the other lies in the right four columns. The first possibility can happen in $2 \cdot 5 \cdot 5=50$ ways (two choices of which square goes on top, and five horizontal positions for each square); likewise, so can the second. However,
this double-counts the 4 cases in which the two squares are in opposite corners, so we have $50+50-4=96$ possible non-overlapping arrangements $\Rightarrow 25^{2}-96=529$ overlapping arrangements.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n5. ",
"solution_match": "\nSolution: "
}
|
219ac7cb-4ab2-5801-937d-0892b2337d88
| 611,013
|
Find all values of $x$ that satisfy $x=1-x+x^{2}-x^{3}+x^{4}-x^{5}+\cdots$ (be careful; this is tricky).
|
Multiplying both sides by $1+x$ gives $(1+x) x=1$, or $x=\frac{-1 \pm \sqrt{5}}{2}$. However, the series only converges for $|x|<1$, so only the answer $x=\frac{-1+\sqrt{5}}{2}$ makes sense.
|
\frac{-1+\sqrt{5}}{2}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all values of $x$ that satisfy $x=1-x+x^{2}-x^{3}+x^{4}-x^{5}+\cdots$ (be careful; this is tricky).
|
Multiplying both sides by $1+x$ gives $(1+x) x=1$, or $x=\frac{-1 \pm \sqrt{5}}{2}$. However, the series only converges for $|x|<1$, so only the answer $x=\frac{-1+\sqrt{5}}{2}$ makes sense.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n6. ",
"solution_match": "\nSolution: "
}
|
1444a947-493d-52bb-a8f6-b6106da266f3
| 611,014
|
A rubber band is 4 inches long. An ant begins at the left end. Every minute, the ant walks one inch along rightwards along the rubber band, but then the band is stretched (uniformly) by one inch. For what value of $n$ will the ant reach the right end during the $n$th minute?
|
7 The ant traverses $1 / 4$ of the band's length in the first minute, $1 / 5$ of the length in the second minute (the stretching does not affect its position as a fraction of the band's length), $1 / 6$ of the length in the third minute, and so on. Since
$$
1 / 4+1 / 5+\cdots+1 / 9<0.25+0.20+0.167+0.143+0.125+0.112=0.997<1
$$
the ant does not cover the entire band in six minutes. However,
$$
1 / 4+\cdots+1 / 10>0.25+0.20+0.16+0.14+0.12+0.11+0.10=1.08>1
$$
so seven minutes suffice.
|
7
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
A rubber band is 4 inches long. An ant begins at the left end. Every minute, the ant walks one inch along rightwards along the rubber band, but then the band is stretched (uniformly) by one inch. For what value of $n$ will the ant reach the right end during the $n$th minute?
|
7 The ant traverses $1 / 4$ of the band's length in the first minute, $1 / 5$ of the length in the second minute (the stretching does not affect its position as a fraction of the band's length), $1 / 6$ of the length in the third minute, and so on. Since
$$
1 / 4+1 / 5+\cdots+1 / 9<0.25+0.20+0.167+0.143+0.125+0.112=0.997<1
$$
the ant does not cover the entire band in six minutes. However,
$$
1 / 4+\cdots+1 / 10>0.25+0.20+0.16+0.14+0.12+0.11+0.10=1.08>1
$$
so seven minutes suffice.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n7. ",
"solution_match": "\nSolution: "
}
|
086caa26-1592-50e4-b1ca-cf90709fb0d2
| 611,015
|
Draw a square of side length 1 . Connect its sides' midpoints to form a second square. Connect the midpoints of the sides of the second square to form a third square. Connect the midpoints of the sides of the third square to form a fourth square. And so forth. What is the sum of the areas of all the squares in this infinite series?
|
The area of the first square is 1 , the area of the second is $\frac{1}{2}$, the area of the third is $\frac{1}{4}$, etc., so the answer is $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots=2$.
|
2
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Draw a square of side length 1 . Connect its sides' midpoints to form a second square. Connect the midpoints of the sides of the second square to form a third square. Connect the midpoints of the sides of the third square to form a fourth square. And so forth. What is the sum of the areas of all the squares in this infinite series?
|
The area of the first square is 1 , the area of the second is $\frac{1}{2}$, the area of the third is $\frac{1}{4}$, etc., so the answer is $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots=2$.
|
{
"resource_path": "HarvardMIT/segmented/en-52-2002-feb-guts-solutions.jsonl",
"problem_match": "\n8. ",
"solution_match": "\nSolution: "
}
|
41b8d0b6-9761-5822-a7d1-7e429c5ae77f
| 611,016
|
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