2025 AIME I Problems/Problem 9

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:

$\begin{matrix} [\begin{matrix} x^{'} \\ y^{'} \end{matrix}] = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] \end{matrix}$

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

$\begin{matrix} [\begin{matrix} x^{'} \\ y^{'} \end{matrix}] = [\begin{matrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ - \frac{\sqrt{3}}{2} & \frac{1}{2} \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] \end{matrix}$

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:

$\begin{matrix} [\begin{matrix} x^{'} \\ y^{'} \end{matrix}] = [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] \end{matrix}$

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.

In particular, $y'=\frac{1}{2}x^2-\frac{\sqrt{3}}{2}x-2$ becomes $y'=\frac{1}{2}x^2+\frac{\sqrt{3}}{2}x-2$ .

(Multiplied $x$ with POSITIVE $\sqrt{3}/2$ not NEGATIVE in the matrix)

This means that

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

is actually:

$p^2-4=\frac{1}{2}p^2+\frac{\sqrt{3}}{2}p-2$

Using the Quadratic formula yields:

$\frac{\sqrt3 \pm \sqrt{19}}{2}$

Squaring means that you have to assume that at the end there is a negative sign in front of the radical, since we have to choose that both the signs are positive for our point to be in the fourth quadrant.

Solution 3 (Similar to Solution 1)

Note that this question is equivalent to finding a point $B$ in the fourth quadrant, such that when a point $A$ on the graph of $y = x^2 - 4$ is rotated $60^\circ$ counterclockwise around the origin, it lands on $B$ , which is also on the graph.

The first thing to note is that point $A$ and $B$ must be equidistant to the origin. If we express the coordinates of $A$ as $(a, b)$ , and the coordinates of $B$ as $(x, y)$ , we have:

$‖ O A ‖$ = $‖ O B ‖$

Which means that:

$\sqrt{a^{2} + b^{2}} = \sqrt{x^{2} + y^{2}}$

Since $b = a^2 - 4$ and $y = x^2 - 4$ , we have $a^2 = b + 4$ and $x^2 = y + 4$ , substituting this into the previous equation and squaring both sides yields:

$2 a^{2} + 4 = 2 x^{2} + 4$

Meaning that $a^{2} = x^{2}$ , since $A$ and $B$ clearly cannot coincide, we must have $a = - x$ , since $y = x^2 - 4$ is an even function, this means that point $A$ and $B$ are just reflections of each other over the y axis. The angle between $\overset{―}{O A}$ and $\overset{―}{O B}$ is $60^\circ$ and $A$ and $B$ is symmetrical, the y axis should bisect the angle \angle AOB, i.e., the angle between $\overset{―}{O B}$ and the y axis is:

$\frac{60^\circ}{2} = 30^\circ$

Therefore the point $B$ must lie on the line $y = -\sqrt{3}x$

We have:

$\begin{cases}y = x^2 - 4 \\ y = -\sqrt{3}x \end{cases}$

$x^{2} - 4 = - \sqrt{3} x$

Using the quadratic formula and keeping in mind that the x value is positive (since $B$ is in the fourth quadrant) yields $x = \frac{\sqrt{19} - \sqrt{3}}{2}$ .

Substituting into $y = -\sqrt{3}x$ We get $y=\frac{3-\sqrt{57}}{2}\implies \boxed{062}.$

~IDKHowtoaddsolution

The last part of this solution is essentially Solution 1.

Video Solution

2025 AIME I #9

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2025 AIME I (Problems • Answer Key • Resources)
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