question stringlengths 39 3.34k | thinking stringlengths 3.22k 13.5k | answer stringlengths 1 186 | id stringlengths 4 7 | content stringlengths 3.22k 13.5k | original_id int64 41 93.7k |
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The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$. Suppose that the radius of the circle is $5$, that $BC=6$, and that $AD$ is bisected by $BC$. Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$. It follows that the sine of the central angle o... | <think>Solution 1
As with some of the other solutions, we analyze this with a locus—but a different one. We'll consider: given a point $P$ and a line $\ell,$ what is the set of points $X$ such that the midpoint of $PX$ lies on line $\ell$? The answer to this question is: a line $m$ parallel to $\ell$, such that $m$ a... | 175 | 1_43738 | <think>Solution 1
As with some of the other solutions, we analyze this with a locus—but a different one. We'll consider: given a point $P$ and a line $\ell,$ what is the set of points $X$ such that the midpoint of $PX$ lies on line $\ell$? The answer to this question is: a line $m$ parallel to $\ell$, such that $m$ a... | 43,738 |
A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are... | <think>Solution 1 (Easiest)
Let $P_n$ represent the probability that the bug is at its starting vertex after $n$ moves. If the bug is on its starting vertex after $n$ moves, then it must be not on its starting vertex after $n-1$ moves. At this point it has $\frac{1}{2}$ chance of reaching the starting vertex in the nex... | 683 | 1_49969 | <think>Solution 1 (Easiest)
Let $P_n$ represent the probability that the bug is at its starting vertex after $n$ moves. If the bug is on its starting vertex after $n$ moves, then it must be not on its starting vertex after $n-1$ moves. At this point it has $\frac{1}{2}$ chance of reaching the starting vertex in the nex... | 49,969 |
2. Let $a$ and $b$ be distinct real numbers such that the equations $x^{2}+a x+b=0$ and $x^{2}+b x+a=0$ have a common solution. What is $a+b$? | <think>II/2. We equate the quadratic equations and get $x^{2}+a x+b=x^{2}+b x+a$. This leads to $a x+b=b x+a$ and $(x-1)(a-b)=0$. Since $a \neq b$, it follows that $x=1$. Substituting $x=1$ into one of the equations, we get $a+b=-1$.
(a-b)=0$. Since $a \neq b$, it follows that $x=1$. Substituting $x=1$ into one of the equations, we get $a+b=-1$.
\left(p^{2}-4\right)+9$. What is the smallest possible sum of the digits of the number $n$, if $p$ is a prime number? For which prime numbers $p$ is this sum achieved? | <think>IV/3. Let's calculate the first few numbers $n$. For $p=2$, we get $n=9$, for $p=3$, $n=49$, and for $p=5$, $n=513$. Now let $p>5$. Write $n$ as $n=(p-2)(p-1)(p+1)(p+2)+9$. Since $(p-2),(p-1), p,(p+1),(p+2)$ are five consecutive numbers, at least one is divisible by 5. If $p>5$, the number $p$ is not divisible b... | 9 | 1_11174 | <think>IV/3. Let's calculate the first few numbers $n$. For $p=2$, we get $n=9$, for $p=3$, $n=49$, and for $p=5$, $n=513$. Now let $p>5$. Write $n$ as $n=(p-2)(p-1)(p+1)(p+2)+9$. Since $(p-2),(p-1), p,(p+1),(p+2)$ are five consecutive numbers, at least one is divisible by 5. If $p>5$, the number $p$ is not divisible b... | 11,174 |
2. Find all real numbers $x$ and $y$ for which $x+y^{2}=x y+1$ and $x y=4+y$. | <think>I/2. From the first equation, it follows that $x(1-y)=1-y^{2}$ or $(1-y)(x-1-y)=0$. If $y=1$, this equation holds, and from the second, we get $x=5$. In the case $y \neq 1$, we get $x=1+y$. Together with the second equation, we then have $(1+y) y=4+y$ or $y^{2}=4$. From this, it follows that $y=2$ or $y=-2$.
Th... | 5,1;3,2;-1,-2 | 1_48785 | <think>I/2. From the first equation, it follows that $x(1-y)=1-y^{2}$ or $(1-y)(x-1-y)=0$. If $y=1$, this equation holds, and from the second, we get $x=5$. In the case $y \neq 1$, we get $x=1+y$. Together with the second equation, we then have $(1+y) y=4+y$ or $y^{2}=4$. From this, it follows that $y=2$ or $y=-2$.
Th... | 48,785 |
Point $P$ is inside $\triangle ABC$. Line segments $APD$, $BPE$, and $CPF$ are drawn with $D$ on $BC$, $E$ on $AC$, and $F$ on $AB$ (see the figure below). Given that $AP=6$, $BP=9$, $PD=6$, $PE=3$, and $CF=20$, find the area of $\triangle ABC$.
[AIME 1989 Problem 15.png](https://artofproblemsolving.com/wiki/index.php... | <think>Solution 1 (Ceva's Theorem, Stewart's Theorem)
Let $[RST]$ be the area of polygon $RST$. We'll make use of the following fact: if $P$ is a point in the interior of triangle $XYZ$, and line $XP$ intersects line $YZ$ at point $L$, then $\dfrac{XP}{PL} = \frac{[XPY] + [ZPX]}{[YPZ]}.$
This is true because triangl... | 108 | 1_17629 | <think>Solution 1 (Ceva's Theorem, Stewart's Theorem)
Let $[RST]$ be the area of polygon $RST$. We'll make use of the following fact: if $P$ is a point in the interior of triangle $XYZ$, and line $XP$ intersects line $YZ$ at point $L$, then $\dfrac{XP}{PL} = \frac{[XPY] + [ZPX]}{[YPZ]}.$
This is true because triangl... | 17,629 |
[u]Round 1[/u]
[b]p1. [/b] What is the sum of the digits in the binary representation of $2023$?
[b]p2.[/b] Jack is buying fruits at the EMCCmart. Three apples and two bananas cost $\$11.00$. Five apples and four bananas cost $\$19.00$. In cents, how much more does an apple cost than a banana?
[b]p3.[/b] Define $... | <think>1. **Problem 1:**
To find the sum of the digits in the binary representation of \(2023\), we first convert \(2023\) to binary.
\[
2023_{10} = 11111100111_2
\]
Next, we sum the digits of the binary number:
\[
1 + 1 + 1 + 1 + 1 + 1 + 0 + 0 + 1 + 1 + 1 = 9
\]
Therefore, the sum of the... | 25 | 1_47661 | <think>1. **Problem 1:**
To find the sum of the digits in the binary representation of \(2023\), we first convert \(2023\) to binary.
\[
2023_{10} = 11111100111_2
\]
Next, we sum the digits of the binary number:
\[
1 + 1 + 1 + 1 + 1 + 1 + 0 + 0 + 1 + 1 + 1 = 9
\]
Therefore, the sum of the... | 47,661 |
In [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle) $ABC$ with [right angle](https://artofproblemsolving.com/wiki/index.php/Right_angle) $C$, $CA = 30$ and $CB = 16$. Its legs $CA$ and $CB$ are extended beyond $A$ and $B$. [Points](https://artofproblemsolving.com/wiki/index.php/Point) $... | <think>Solution 1
[AIME I 2007-9.png](https://artofproblemsolving.com/wiki/index.php/File:AIME_I_2007-9.png)
Label the points as in the diagram above. If we draw $\overline{O_1A}$ and $\overline{O_2B}$, we form two [right triangles](https://artofproblemsolving.com/wiki/index.php/Right_triangle). As $\overline{AF}$ and ... | 737 | 1_11659 | <think>Solution 1
[AIME I 2007-9.png](https://artofproblemsolving.com/wiki/index.php/File:AIME_I_2007-9.png)
Label the points as in the diagram above. If we draw $\overline{O_1A}$ and $\overline{O_2B}$, we form two [right triangles](https://artofproblemsolving.com/wiki/index.php/Right_triangle). As $\overline{AF}$ and ... | 11,659 |
"Real numbers $x$ and $y$ satisfy $x + y = 4$ and $x \\cdot y = -2$. What is the value of \n\\[x + \(...TRUNCATED) | "<think>Solution 1\n\\[x + \\frac{x^3}{y^2} + \\frac{y^3}{x^2} + y=x+\\frac{x^3}{y^2}+y+\\frac{y^3}{(...TRUNCATED) | 440 | 1_64863 | "<think>Solution 1\n\\[x + \\frac{x^3}{y^2} + \\frac{y^3}{x^2} + y=x+\\frac{x^3}{y^2}+y+\\frac{y^3}{(...TRUNCATED) | 64,863 |
"Find the smallest positive integer $N$ for which $N$ is divisible by $19$, and when the digits of $(...TRUNCATED) | "<think>To solve this problem, we need to find the smallest positive integer \\( N \\) such that:\n1(...TRUNCATED) | 2394 | 1_63027 | "<think>To solve this problem, we need to find the smallest positive integer \\( N \\) such that:\n1(...TRUNCATED) | 63,027 |
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