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

Revision as of 18:56, 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

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, Intuition/Coincidence)

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 $y' = r\sin(\theta + 60^\circ) = r\sin\theta\cos{60^\circ} + r\sin{60^\circ}\cos\theta = y\cos{60^\circ} + x\sin{60^\circ} = \frac{y + x\sqrt{3}}{2}.$

$x' = r\cos(\theta + 60^\circ) = r\cos\theta\cos{60^\circ} - r\sin{60^\circ}\sin\theta = x\cos{60^\circ} - y\sin{60^\circ} = \frac{x - y\sqrt{3}}{2}.$

Now, if we created the equation of the rotated parabola and tried to find its intersection with the original parabola we would be dealing with a quartic, so we need to look for something else.

Let the intersection of the original parabola and its rotated image be $(a, b).$ $b$ can be written as $a^2 - 4$ from the original parabola's equation. To simplify things, for now let's just say $y' = y = a^2 - 4$ and $x' = x = a$ . Then we get the equation $a^2 - 4 = \frac{(a^2 - 4) + a\sqrt{3}}{2}.$ This equation is equivalent to $b = a^2 - 4 = a\sqrt{3},$ which we find is the line of symmetry of both parabolas but reflected over the y axis, so this equation is valid. We can solve and get $a = \frac{\sqrt{3} \pm \sqrt{19}}{2}.$

We reflect $a$ over the $y$ axis while still keeping it positive to get $a = -(\frac{\sqrt{3} - \sqrt{19}}{2}) = \frac{\sqrt{19} - \sqrt{3}}{2}.$ 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

Keep in mind that this approach is not valid for other rotation problems, it just happened to work here.

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

@@ Line 6: / Line 6: @@
 Link: https://www.desmos.com/calculator/ci3vodl4vs
-==Video solution by [[User:grogg007|grogg007]]==
-https://youtu.be/wib5vos7Sd4?t=639
 ==Solution 1==

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