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

Latest revision as of 23:00, 17 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

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

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. $[asy] size(300); import graph; // View window real L = 6; // Original parabola y = x^2 - 4 real f(real x){ return x^2 - 4; } // Rotation by +60 degrees about the origin pair rot60(pair P){ real ang = pi/3; return (P.x*cos(ang) - P.y*sin(ang), P.x*sin(ang) + P.y*cos(ang)); } // Axes draw((-L,0)--(L,0), gray(0.65), Arrows(4)); draw((0,-L)--(0,L), gray(0.65), Arrows(4)); // Plot original parabola real xmin=-3.5, xmax=3.5; path parab = graph(f, xmin, xmax); draw(parab, blue+1.2bp); // Build rotated parabola by sampling, then connecting smoothly int N=220; pair[] pts; for(int i=0; i<=N; ++i){ real x = xmin + (xmax - xmin)*i/N; pts.push(rot60((x, f(x)))); } path rpar = pts[0]; for(int i=1; i<=N; ++i) rpar = rpar..pts[i]; draw(rpar, red+1.2bp); // Line y = -sqrt(3) x draw((-L, sqrt(3)*L)--(L, -sqrt(3)*L), rgb(0,0.45,0)+1.2bp); // Vertices pair v1 = (0,-4); // original vertex pair v2 = rot60(v1); // rotated vertex dot(v1, blue+4bp); dot(v2, red+4bp); [/asy]$

Notice that the vertices of the original parabola and its image are symmetrical about the angle bisector of the 60 degree rotation (shown in green).

The equation of this line is $y = -\tan60^\circ \cdot x = -\sqrt{3}x.$ 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

MathProblemSolvingSkills.com

@@ Line 3: / Line 3: @@
 ==Graph==
-[[File:2025-AIMEI-P9-Graph.png|300px]]
+https://www.desmos.com/calculator/ci3vodl4vs
-Link: https://www.desmos.com/calculator/ci3vodl4vs
 ==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 <math>(r\cos\alpha, r\sin\alpha)</math> in polar coordinates.  If we take this point and reflect it over a line that makes an angle <math>\theta</math> with the <math>x</math> axis, what will the new point's angle be? The angle between the point and the reflection line is <math>\alpha - \theta.</math> 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 <math>2(\theta - \alpha) + \alpha = 2\theta - \alpha.</math>
-Then if we reflected the point over the <math>x</math> axis again the angle becomes <math>\alpha -2\theta.</math>
 <asy>
-size(230);
+size(300);
+import graph;
-real R = 2.8;
-real alpha = 65;
-real theta = 45;
-real tma = 2*theta - alpha;
-real amt = alpha - 2*theta;
-tma = (tma + 360) % 360;
-amt = (amt + 360) % 360;
-pair O = (0,0);
-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);
-draw(Circle(O,R), gray(0.8));
+// View window
+real L = 6;
+// Original parabola y = x^2 - 4
+real f(real x){ return x^2 - 4; }
-pair Ldir = dir(theta);
+// Rotation by +60 degrees about the origin
-draw((-R)*Ldir--(R)*Ldir, heavygray);
+pair rot60(pair P){
-label("line at $\theta$", 1.05*R*Ldir, dir(theta+15));
+  real ang = pi/3;
+  return (P.x*cos(ang) - P.y*sin(ang), P.x*sin(ang) + P.y*cos(ang));
+}
+// Axes
+draw((-L,0)--(L,0), gray(0.65), Arrows(4));
+draw((0,-L)--(0,L), gray(0.65), Arrows(4));
-pair P  = R*dir(alpha);
+// Plot original parabola
-pair P1 = R*dir(tma);
+real xmin=-3.5, xmax=3.5;
-pair P2 = R*dir(amt);
+path parab = graph(f, xmin, xmax);
+draw(parab, blue+1.2bp);
-draw(O--P, blue+1bp, Arrow);
+// Build rotated parabola by sampling, then connecting smoothly
-draw(O--P1, red+1bp, Arrow);
+int N=220;
-draw(O--P2, green+1bp, Arrow);
+pair[] pts;
+for(int i=0; i<=N; ++i){
+  real x = xmin + (xmax - xmin)*i/N;
+  pts.push(rot60((x, f(x))));
+}
+path rpar = pts[0];
+for(int i=1; i<=N; ++i) rpar = rpar..pts[i];
+draw(rpar, red+1.2bp);
-dot(P,blue);   label("$P\ (\alpha)$", P, dir(alpha+20));
+// Line y = -sqrt(3) x
-dot(P1,red);   label("$P'\ (2\theta-\alpha)$", P1, dir(tma-20));
+draw((-L,  sqrt(3)*L)--(L, -sqrt(3)*L), rgb(0,0.45,0)+1.2bp);
-dot(P2,green); label("$P''\ (\alpha-2\theta)$", P2, dir(amt-20));
-real rarc = 0.5*R;
+// Vertices
-draw(arc(O, rarc, theta, alpha), dashed+blue);
+pair v1 = (0,-4);              // original vertex
-label("$\alpha - \theta$", 1.2*rarc*dir((alpha+theta)/2), dir((alpha+theta)/2));
+pair v2 = rot60(v1);           // rotated vertex
+dot(v1, blue+4bp);
+dot(v2, red+4bp);
 </asy>
+Notice that the vertices of the original parabola and its image are symmetrical about the angle bisector of the 60 degree rotation (shown in green).
+The equation of this line is <math>y = -\tan60^\circ \cdot x = -\sqrt{3}x.</math> 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.
-This essentially means that a reflection over the line forming an angle <math>\theta</math> with the <math>x</math> axis followed by a reflection over the <math>x</math> axis is the same thing as a clockwise rotation by <math>2\theta^\circ.</math>
-A counterclockwise rotation of <math>60^\circ</math> is the same thing as a clockwise rotation of <math>300^\circ.</math>
-So a counterclockwise rotation of <math>60^\circ</math> is the same as a reflection over the line forming a <math>150 - 90 = 60</math> degree angle with the <math>x</math> axis, followed by a reflection over the <math>x</math> axis. The equation of this line is <math>y = \tan60^\circ \cdot x = \sqrt{3}x.</math> The reflection over the <math>x</math> axis makes it <math>y = -\sqrt{3}x.</math> 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.

2025 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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