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 $a + b + c$ .

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

2025 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2025 AIME I Problems/Problem 9

Problem

Solution 1

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