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

Revision as of 23:23, 15 August 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

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

Video Solution by grogg007

https://www.youtube.com/watch?v=ZHGAmInyhZc

Solution 1

We need to relate the rotation to something simpler because substituting the equation of the rotated parabola into the original will give us a quartic.

Consider a point $(r\cos\alpha, r\sin\alpha)$ in polar coordinates. If we take this point and reflect it over a line that makes an angle $\theta$ with the $x$ axis, what will the new point's angle be? The angle between the point and the reflection line is $\alpha - \theta.$ Reflecting this point over the line will mean the point is at the same angular distance from the line but on the opposite side. So the reflected point's angle becomes $2(\theta - \alpha) + \alpha = 2\theta - \alpha.$

Then if we reflected the point over the $x$ axis again the angle becomes $\alpha -2\theta.$

$[asy] size(230); real R = 2.8; real alpha = 65; real theta = 45; real tma = 2*theta - alpha; // 2θ - α real amt = alpha - 2*theta; // α - 2θ // Normalize angles into [0,360) tma = (tma + 360) % 360; amt = (amt + 360) % 360; pair O = (0,0); // Axes draw((-R,0)--(R,0), gray(0.6)); draw((0,-R)--(0,R), gray(0.6)); label("$x$", (R,0), E); label("$y$", (0,R), N); // Circle draw(Circle(O,R), gray(0.8)); // Reflection line pair Ldir = dir(theta); draw((-R)*Ldir--(R)*Ldir, heavygray); label("line at $\theta$", 1.05*R*Ldir, dir(theta+15)); // Points pair P = R*dir(alpha); pair P1 = R*dir(tma); pair P2 = R*dir(amt); draw(O--P, blue+1bp, Arrow); draw(O--P1, red+1bp, Arrow); draw(O--P2, green+1bp, Arrow); dot(P,blue); label("$P\ (\alpha)$", P, dir(alpha+20)); dot(P1,red); label("$P'\ (2\theta-\alpha)$", P1, dir(tma-20)); dot(P2,green); label("$P''\ (\alpha-2\theta)$", P2, dir(amt-20)); // Angle marking: alpha - theta real rarc = 0.5*R; draw(arc(O, rarc, theta, alpha), dashed+blue); label("$\alpha - \theta$", 1.2*rarc*dir((alpha+theta)/2), dir((alpha+theta)/2)); [/asy]$

This essentially means that a reflection over the line forming an angle $\theta$ with the $x$ axis followed by a reflection over the $x$ axis is the same thing as a counterclockwise rotation by $2\theta^\circ.$

So a counterclockwise rotation of $60^\circ$ is the same as a reflection over the line forming a $30$ degree angle with the $x$ axis, followed by a reflection over the $x$ axis. The equation of this line is $y = \tan30^\circ \cdot x = \sqrt{3}x.$ The reflection over the $x$ axis makes it $y = -\sqrt{3}x.$ This means that the point lying on this line and the original parabola must be the intersection of the new and original images since it won't change position with the rotation.

We substitute $y = x^2 - 4:$ $x^2 - 4 = -\sqrt{3}x$ $x = \frac{-\sqrt{3} + \sqrt{19}}{2}.$

Then $y = (\frac{-\sqrt{3} + \sqrt{19}}{2})^2 - 4 = \frac{3 - \sqrt{57}}{2}.$ $57 + 3 + 2 = \boxed{62}.$

~grogg007, ~mathkiddus

Solution 2 (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

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2025 AIME I (Problems • Answer Key • Resources)
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