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Difference between revisions of "2025 AIME I Problems/Problem 9"

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==Problem==
 
==Problem==
 
The parabola with equation <math>y = x^2 - 4</math> is rotated <math>60^\circ</math> counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has <math>y</math>-coordinate <math>\frac{a - \sqrt{b}}{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers, and <math>a</math> and <math>c</math> are relatively prime. Find <math>a + b + c</math>.
 
The parabola with equation <math>y = x^2 - 4</math> is rotated <math>60^\circ</math> counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has <math>y</math>-coordinate <math>\frac{a - \sqrt{b}}{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers, and <math>a</math> and <math>c</math> are relatively prime. Find <math>a + b + c</math>.
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 +
==Graph==
 +
https://www.desmos.com/calculator/ci3vodl4vs
 +
 
==Solution 1==
 
==Solution 1==
 
To begin with notice, a <math>60^{\circ}</math> rotation counterclockwise about the origin  on the <math>y-</math>axis is the same as a reflection over the line <math>y=-x\sqrt{3}.</math> Since the parabola <math>y=x^2-4</math> is symmetric about the <math>y-</math>axis as well, we can simply reflect it over the line. In addition any point of intersection between the line and parabola will also be on the rotated parabola. So we solve for the intersection, <cmath>-x\sqrt{3}=x^2-4.</cmath> <cmath>x^2+x\sqrt{3}-4=0.</cmath><cmath>x=\frac{-\sqrt{3} \pm \sqrt{19}}{2}.</cmath> Since we want the point in the fourth quadrant we only care about the negative case giving us, <cmath>y=x^2-4=\left(\frac{-\sqrt{3} - \sqrt{19}}{2}\right)^2-4=\frac{3-\sqrt{57}}{2}\implies \boxed{062}.</cmath>
 
To begin with notice, a <math>60^{\circ}</math> rotation counterclockwise about the origin  on the <math>y-</math>axis is the same as a reflection over the line <math>y=-x\sqrt{3}.</math> Since the parabola <math>y=x^2-4</math> is symmetric about the <math>y-</math>axis as well, we can simply reflect it over the line. In addition any point of intersection between the line and parabola will also be on the rotated parabola. So we solve for the intersection, <cmath>-x\sqrt{3}=x^2-4.</cmath> <cmath>x^2+x\sqrt{3}-4=0.</cmath><cmath>x=\frac{-\sqrt{3} \pm \sqrt{19}}{2}.</cmath> Since we want the point in the fourth quadrant we only care about the negative case giving us, <cmath>y=x^2-4=\left(\frac{-\sqrt{3} - \sqrt{19}}{2}\right)^2-4=\frac{3-\sqrt{57}}{2}\implies \boxed{062}.</cmath>
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 +
~[[User:Mathkiddus|mathkiddus]]
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 +
==Solution 2==
 +
To rotate the curve <math>y=x^2-4</math> counterclockwise by an angle of <math>60^\circ</math> about the origin, we will use the rotation matrix as follows:
 +
 +
\begin{gather}
 +
\begin{bmatrix} x' \\ y' \end{bmatrix}
 +
=
 +
\begin{bmatrix}
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\cos{\theta} & \sin{\theta} \\
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-\sin{\theta} & \cos{\theta}
 +
\end{bmatrix}
 +
\begin{bmatrix}
 +
x \\ y
 +
\end{bmatrix}
 +
\end{gather}
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 +
Carrying in <math>\theta=\frac{\pi}{3}</math>, the rotation matrix becomes
 +
 +
\begin{gather}
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\begin{bmatrix} x' \\ y' \end{bmatrix}
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=
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\begin{bmatrix}
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\frac{1}{2} & \frac{\sqrt{3}}{2} \\
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-\frac{\sqrt{3}}{2} & \frac{1}{2}
 +
\end{bmatrix}
 +
\begin{bmatrix}
 +
x \\ y
 +
\end{bmatrix}
 +
\end{gather}
 +
 +
which leads to the following equations:
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<cmath>x'=\frac{1}{2}x+\frac{\sqrt{3}}{2}y</cmath>
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<cmath>y'=-\frac{\sqrt{3}}{2}x+\frac{1}{2}y</cmath>
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 +
Substituting <math>y</math> with <math>x^2-4</math> yields
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<cmath>x'=\frac{\sqrt{3}}{2}x^2+\frac{1}{2}x-2\sqrt{3}</cmath>
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<cmath>y'=\frac{1}{2}x^2-\frac{\sqrt{3}}{2}x-2</cmath>
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 +
We wish to find the coordinates of the intersection point. Let the point of intersection be <math>(p, p^2-4)</math>, then
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 +
<cmath>p^2-4=\frac{1}{2}p^2-\frac{\sqrt{3}}{2}p-2</cmath>
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 +
Solving this quadratic equation yields
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 +
<cmath>p_1=\frac{-\sqrt{3}+\sqrt{19}}{2}, \, p_2=\frac{-\sqrt{3}-\sqrt{19}}{2}</cmath>
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 +
Since the problem asks for the intersection point in the fourth quadrant, <math>p=\frac{-\sqrt{3}+\sqrt{19}}{2}</math>. Therefore, the point of intersection has <math>y</math>-coordinate <math>\frac{3-\sqrt{57}}{2}</math>, with final answer <math>3+57+2=\boxed{062}</math>
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~[[User:Bloggish|Bloggish]]
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 +
==See also==
 +
{{AIME box|year=2025|num-b=8|num-a=10|n=I}}
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 +
{{MAA Notice}}

Latest revision as of 10:14, 18 February 2025

Problem

The parabola with equation $y = x^2 - 4$ is rotated $60^\circ$ counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has $y$-coordinate $\frac{a - \sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers, and $a$ and $c$ are relatively prime. Find $a + b + c$.

Graph

https://www.desmos.com/calculator/ci3vodl4vs

Solution 1

To begin with notice, a $60^{\circ}$ rotation counterclockwise about the origin on the $y-$axis is the same as a reflection over the line $y=-x\sqrt{3}.$ Since the parabola $y=x^2-4$ is symmetric about the $y-$axis as well, we can simply reflect it over the line. In addition any point of intersection between the line and parabola will also be on the rotated parabola. So we solve for the intersection, \[-x\sqrt{3}=x^2-4.\] \[x^2+x\sqrt{3}-4=0.\]\[x=\frac{-\sqrt{3} \pm \sqrt{19}}{2}.\] Since we want the point in the fourth quadrant we only care about the negative case giving us, \[y=x^2-4=\left(\frac{-\sqrt{3} - \sqrt{19}}{2}\right)^2-4=\frac{3-\sqrt{57}}{2}\implies \boxed{062}.\]

~mathkiddus

Solution 2

To rotate the curve $y=x^2-4$ counterclockwise by an angle of $60^\circ$ about the origin, we will use the rotation matrix as follows:

\begin{gather} \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \end{gather}

Carrying in $\theta=\frac{\pi}{3}$, the rotation matrix becomes

\begin{gather} \begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \end{gather}

which leads to the following equations: \[x'=\frac{1}{2}x+\frac{\sqrt{3}}{2}y\] \[y'=-\frac{\sqrt{3}}{2}x+\frac{1}{2}y\]

Substituting $y$ with $x^2-4$ yields \[x'=\frac{\sqrt{3}}{2}x^2+\frac{1}{2}x-2\sqrt{3}\] \[y'=\frac{1}{2}x^2-\frac{\sqrt{3}}{2}x-2\]

We wish to find the coordinates of the intersection point. Let the point of intersection be $(p, p^2-4)$, then

\[p^2-4=\frac{1}{2}p^2-\frac{\sqrt{3}}{2}p-2\]

Solving this quadratic equation yields

\[p_1=\frac{-\sqrt{3}+\sqrt{19}}{2}, \, p_2=\frac{-\sqrt{3}-\sqrt{19}}{2}\]

Since the problem asks for the intersection point in the fourth quadrant, $p=\frac{-\sqrt{3}+\sqrt{19}}{2}$. Therefore, the point of intersection has $y$-coordinate $\frac{3-\sqrt{57}}{2}$, with final answer $3+57+2=\boxed{062}$

~Bloggish

See also

2025 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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