Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
m (→‎Video Solution by grogg007)
 
(10 intermediate revisions by the same user not shown)
Line 6: Line 6:
  
 
Link: https://www.desmos.com/calculator/ci3vodl4vs
 
Link: https://www.desmos.com/calculator/ci3vodl4vs
−
 
−
==Video Solution by [[User:grogg007|grogg007]]==
 
−
https://www.youtube.com/watch?v=ZHGAmInyhZc
 
  
 
==Solution 1==
 
==Solution 1==
Line 15: Line 12:
  
  
−
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 point's angle become? 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>  
+
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>  
 
Then if we reflected the point over the <math>x</math> axis again the angle becomes <math>\alpha -2\theta.</math>  
 +
 +
<asy>
 +
size(230);
 +
 +
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));
 +
 +
 +
pair Ldir = dir(theta);
 +
draw((-R)*Ldir--(R)*Ldir, heavygray);
 +
label("line at $\theta$", 1.05*R*Ldir, dir(theta+15));
 +
 +
 +
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));
 +
 +
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 <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 counterclockwise rotation by <math>2\theta^\circ.</math>  
+
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 thing as a reflection over the line forming a <math>30</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 = \tan(30^\circ)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.  
+
So a counterclockwise rotation of <math>60^\circ</math> is the same as a reflection over the line forming a <math>150</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.  
  
  
 
We substitute <math>y = x^2 - 4:</math>  
 
We substitute <math>y = x^2 - 4:</math>  
 
<cmath>x^2 - 4 = -\sqrt{3}x</cmath>  
 
<cmath>x^2 - 4 = -\sqrt{3}x</cmath>  
−
<cmath>x = \frac{-\sqrt{3} + \sqrt{19}}{2}.</cmath>
+
<cmath>x = \frac{-\sqrt{3} + \sqrt{19}}{2}.</cmath>  
  
−
Then <math>y = (\frac{-\sqrt{3} + \sqrt{19}}{2})^2 - 4 = \frac{3 - \sqrt{57}}{2}.</math> Ans = <math>57 + 3 + 2 = \boxed{62}.</math>
+
Then <math>y = (\frac{-\sqrt{3} + \sqrt{19}}{2})^2 - 4 = \frac{3 - \sqrt{57}}{2}.</math> <math>57 + 3 + 2 = \boxed{62}.</math>
  
 
~[[User:grogg007|grogg007]], ~[[User:Mathkiddus|mathkiddus]]
 
~[[User:grogg007|grogg007]], ~[[User:Mathkiddus|mathkiddus]]
  
−
==Solution 3 (Similar to Solution 1)==
+
==Solution 2 (Similar to Solution 1)==
  
 
Note that this question is equivalent to finding a point <math>B</math> in the fourth quadrant, such that when a point <math>A</math> on the graph of <math>y = x^2 - 4</math> is rotated <math>60^\circ</math> counterclockwise around the origin, it lands on <math>B</math>, which is also on the graph.
 
Note that this question is equivalent to finding a point <math>B</math> in the fourth quadrant, such that when a point <math>A</math> on the graph of <math>y = x^2 - 4</math> is rotated <math>60^\circ</math> counterclockwise around the origin, it lands on <math>B</math>, which is also on the graph.
Line 69: Line 114:
 
~[[User:IDKHowtoaddsolution|IDKHowtoaddsolution]]
 
~[[User:IDKHowtoaddsolution|IDKHowtoaddsolution]]
  
−
The last part of this solution is essentially Solution 1.  
+
The last part of this solution is essentially Solution 1.
  
 
== Video Solution ==
 
== Video Solution ==

Latest revision as of 13:42, 16 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

2025-AIMEI-P9-Graph.png

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 $(r\cos\alpha, r\sin\alpha)$ 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; 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)); pair Ldir = dir(theta); draw((-R)*Ldir--(R)*Ldir, heavygray); label("line at $\theta$", 1.05*R*Ldir, dir(theta+15)); 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)); 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 clockwise rotation by $2\theta^\circ.$

A counterclockwise rotation of $60^\circ$ is the same thing as a clockwise rotation of $300^\circ.$

So a counterclockwise rotation of $60^\circ$ is the same as a reflection over the line forming a $150$ degree angle with the $x$ axis, followed by a reflection over the $x$ axis. The equation of this line is $y = \tan60^\circ \cdot x = \sqrt{3}x.$ 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

MathProblemSolvingSkills.com

See also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_AIME_I_Problems/Problem_9&oldid=257822"