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 2 (Polar Coordinates, Intuition/Coincidence))
(→‎Solution 1)
 
(39 intermediate revisions by one other user not shown)
Line 3: Line 3:
  
 
==Graph==
 
==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.  
 +
<asy>
 +
size(300);
 +
import graph;
 +
 
 +
// View window
 +
real L = 6;
  
−
==Video solution by [[User:grogg007|grogg007]]==
+
// Original parabola y = x^2 - 4
−
https://youtu.be/wib5vos7Sd4?t=639
+
real f(real x){ return x^2 - 4; }
  
−
==Solution 1==
+
// Rotation by +60 degrees about the origin
−
To begin with notice, a <math>60^{\circ}</math> rotation counterclockwise about the origin on the <math>y-</math>axis is the same as a reflection over the line <math>y=-x\sqrt{3}.</math> Since the parabola <math>y=x^2-4</math> is symmetric about the <math>y-</math>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, <cmath>-x\sqrt{3}=x^2-4.</cmath> <cmath>x^2+x\sqrt{3}-4=0.</cmath><cmath>x=\frac{-\sqrt{3} \pm \sqrt{19}}{2}.</cmath> Since we want the point in the fourth quadrant we only care about the positive case, giving <cmath>y=x^2-4=\left(\frac{-\sqrt{3} + \sqrt{19}}{2}\right)^2-4=\frac{3-\sqrt{57}}{2}\implies \boxed{062}.</cmath>
+
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));
 +
}
  
−
~[[User:Mathkiddus|mathkiddus]]
+
// Axes
 +
draw((-L,0)--(L,0), gray(0.65), Arrows(4));
 +
draw((0,-L)--(0,L), gray(0.65), Arrows(4));
  
−
==Solution 2 (Polar Coordinates, Intuition/Coincidence)==
+
// Plot original parabola
 +
real xmin=-3.5, xmax=3.5;
 +
path parab = graph(f, xmin, xmax);
 +
draw(parab, blue+1.2bp);
  
−
We know that in polar coordinates, <math>r\sin\theta = y</math> and <math>r\cos\theta = x.</math> So, if we rotate any point <math>60^\circ</math> CCW we will get <cmath>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}.</cmath>
+
// 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);
  
−
<cmath>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}.</cmath>
+
// Line y = -sqrt(3) x
 +
draw((-L,  sqrt(3)*L)--(L, -sqrt(3)*L), rgb(0,0.45,0)+1.2bp);
  
−
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.
+
// Vertices
 +
pair v1 = (0,-4);              // original vertex
 +
pair v2 = rot60(v1);          // rotated vertex
 +
dot(v1, blue+4bp);
 +
dot(v2, red+4bp);
 +
</asy>
  
−
Let the intersection of the original parabola and its rotated image be <math>(a, b).</math> <math>b</math> can be written as <math>a^2 - 4</math> from the original parabola's equation. To simplify things, for now let's just say <math>y' = y = a^2 - 4</math> and <math>x' = x = a</math>. Then we get the equation <math>a^2 - 4 = \frac{(a^2 - 4) + a\sqrt{3}}{2}.</math> This equation is equivalent to <math>b = a^2 - 4 = a\sqrt{3},</math> 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 <math>a = \frac{\sqrt{3} \pm \sqrt{19}}{2}.</math>
+
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 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 reflect <math>a</math> over the <math>y</math> axis while still keeping it positive to get <math>a = -(\frac{\sqrt{3} - \sqrt{19}}{2}) = \frac{\sqrt{19} - \sqrt{3}}{2}.</math> Then the y-coordinate is <math>(\frac{\sqrt{19} - \sqrt{3}}{2})^2 - 4 = \frac{3 - \sqrt{57}}{2}.</math> <math>57 + 2 + 3 = \boxed{62}.</math>
 
  
−
~[[User:grogg007|grogg007]], [[User:Bloggish|Bloggish]]
+
We substitute <math>y = x^2 - 4:</math>
 +
<cmath>x^2 - 4 = -\sqrt{3}x</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> <math>57 + 3 + 2 = \boxed{62}.</math>
  
−
Keep in mind that this approach is not valid for other rotation problems, it just happened to work here.
+
~[[User:grogg007|grogg007]], ~[[User:Mathkiddus|mathkiddus]], ~[[User:athreyay|athreyay]]
  
−
==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 66: Line 101:
 
~[[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 08:59, 24 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 $y = -\tan60^\circ \cdot y = -\sqrt{3}x.$ 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, ~athreyay

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=258737"