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

Revision as of 17:01, 16 April 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

NOTE: The rotation matrix used here is actually incorrect (it is for clockwise rotation), it should be:

\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}

Due to symmetry, the answer of the intersection point in the fourth quadrant turned out to be the same. However, this is just coincidence; the solution should be edited to display the proper algebraic steps with the correct rotation matrix.

The new matrix causes some signs to be flipped, so the latter method of plugging in $p$ and $p^2-4$ does not work.

@@ Line 74: / Line 74: @@
 Due to symmetry, the answer of the intersection point in the fourth quadrant turned out to be the same. However, this is just coincidence; the solution should be edited to display the proper algebraic steps with the correct rotation matrix.
+The new matrix causes some signs to be flipped, so the latter method of plugging in <math>p</math> and <math>p^2-4</math> does not work.
 ==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
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