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

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

Video solution by grogg007

https://youtu.be/wib5vos7Sd4?t=639

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 positive case, giving $y=x^2-4=\left(\frac{-\sqrt{3} + \sqrt{19}}{2}\right)^2-4=\frac{3-\sqrt{57}}{2}\implies \boxed{062}.$

~mathkiddus

Solution 2 (Polar Coordinates)

We know that in polar coordinates, $r\sin\theta = y$ and $r\cos\theta = x.$ So, if we rotate any point $60^\circ$ CCW we will get $y' = r\sin(\theta + 60^\circ) = r\sin\theta\cos{60^\circ} + r\sin{60^\circ}\cos\theta = y\cos{60^\circ} + x\sin{60^\circ} = \frac{y + x\sqrt{3}}{2}.$ Let the intersection of the original parabola and its rotated image be $(a, b).$ We know since this is the intersection, $b$ can be written as $a^2 - 4$ from the original parabola's equation. We set $y' = y = a^2 - 4$ and $x' = x = a.$ Remember, since $(a,b)$ is the point of intersection, it must satisfy both the equation of the original parabola and the equation of the rotated image. Therefore, when we apply the rotation formula to a point $(a,b)$ that lies on the original parabola, the resulting rotated coordinates $(x',y')$ must be equal to $(a,b)$ itself for it to be an intersection point. Then we get the equation $a^2 - 4 = \frac{(a^2 - 4) + a\sqrt{3}}{2} \implies a = \frac{\sqrt{3} \pm \sqrt{19}}{2}.$ Since the problem asks for the intersection in the fourth quadrant, we want a minus sign somewhere in $a$ to find the negative $y$ coordinate, but $a$ still has to be positive for the intersection to be in the fourth quadrant, so we can say $a = -(\frac{\sqrt{3} - \sqrt{19}}{2}) = \frac{\sqrt{19} - \sqrt{3}}{2}$ (we can do this because the parabola is symmetric about the $y$ axis, so multiplying the $x$ coordinate by negative $1$ doesn't affect the $y$ coordinate). Then the y-coordinate is $(\frac{\sqrt{19} - \sqrt{3}}{2})^2 - 4 = \frac{3 - \sqrt{57}}{2}.$ $57 + 2 + 3 = \boxed{62}.$

~grogg007, Bloggish

Remark: The step where we took $\frac{\sqrt{3} - \sqrt{19}}{2}$ and multiplied it by negative 1 wasn't actually necessary because the $y$ coordinate would have been the same anyway. This was done to find the correct x-coordinate of the intersection point, even though the problem didn’t ask for it.

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:

$\|OA\|$ = $\|OB\|$

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:

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

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 $\overline{OA}$ and $\overline{OB}$ is $60^\circ$ and $A$ and $B$ is symmetrical, the y axis should bisect the angle \angle AOB, i.e., the angle between $\overline{OB}$ 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

MathProblemSolvingSkills.com

2025 AIME I (Problems • Answer Key • Resources)
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