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

m (→‎Solution 2 (Polar Coordinates))
m (→‎Graph)
 
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==Graph==
 
==Graph==
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[[File:2025-AIMEI-P9-Graph.png|300px]]
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https://www.desmos.com/calculator/ci3vodl4vs
  
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Link: https://www.desmos.com/calculator/ci3vodl4vs
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==Solution 1==
  
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==Video solution by [[User:grogg007|grogg007]]==
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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.
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https://youtu.be/wib5vos7Sd4?t=639
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<asy>
 +
size(300);
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import graph;
  
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==Solution 1==
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// View window
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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>
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real L = 6;
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// Original parabola y = x^2 - 4
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real f(real x){ return x^2 - 4; }
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// Rotation by +60 degrees about the origin
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pair rot60(pair P){
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  real ang = pi/3;
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  return (P.x*cos(ang) - P.y*sin(ang), P.x*sin(ang) + P.y*cos(ang));
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}
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// Axes
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draw((-L,0)--(L,0), gray(0.65), Arrows(4));
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draw((0,-L)--(0,L), gray(0.65), Arrows(4));
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// Plot original parabola
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real xmin=-3.5, xmax=3.5;
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path parab = graph(f, xmin, xmax);
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draw(parab, blue+1.2bp);
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// Build rotated parabola by sampling, then connecting smoothly
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int N=220;
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pair[] pts;
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for(int i=0; i<=N; ++i){
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  real x = xmin + (xmax - xmin)*i/N;
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  pts.push(rot60((x, f(x))));
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}
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path rpar = pts[0];
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for(int i=1; i<=N; ++i) rpar = rpar..pts[i];
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draw(rpar, red+1.2bp);
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// Line y = -sqrt(3) x
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draw((-L,  sqrt(3)*L)--(L, -sqrt(3)*L), rgb(0,0.45,0)+1.2bp);
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// Vertices
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pair v1 = (0,-4);              // original vertex
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pair v2 = rot60(v1);          // rotated vertex
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dot(v1, blue+4bp);
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dot(v2, red+4bp);
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</asy>
  
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~[[User:Mathkiddus|mathkiddus]]
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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.
  
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==Solution 2 (Polar Coordinates)==
 
  
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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> Let the intersection of the original parabola and its rotated image be <math>(a, b).</math> We know since this is the intersection, <math>b</math> can be written as <math>a^2 - 4</math> from the original parabola's equation. We set <math>y' = y = a^2 - 4</math> and <math>x' = x = a.</math> Remember, since <math>(a,b)</math> is the point of intersection, it must satisfy both the equation of the original parabola and the equation of the rotated image. Therefore, when we apply the rotation formula to a point <math>(a,b)</math> that lies on the original parabola, the resulting rotated coordinates <math>(x',y')</math> must be equal to <math>(a,b)</math> itself for it to be an intersection point. Then we get the equation <math>a^2 - 4 = \frac{(a^2 - 4) + a\sqrt{3}}{2} \implies a = \frac{\sqrt{3} \pm \sqrt{19}}{2}.</math> Since the problem asks for the intersection in the fourth quadrant, we want a minus sign somewhere in <math>a</math> to find the negative <math>y</math> coordinate, but <math>a</math> still has to be positive for the intersection to be in the fourth quadrant, so we can say <math>a = -(\frac{\sqrt{3} - \sqrt{19}}{2}) = \frac{\sqrt{19} - \sqrt{3}}{2}</math> (we can do this because the parabola is symmetric about the <math>y</math> axis, so multiplying the <math>x</math> coordinate by negative <math>1</math> doesn't affect the <math>y</math> coordinate). 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>
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We substitute <math>y = x^2 - 4:</math>  
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<cmath>x^2 - 4 = -\sqrt{3}x</cmath>  
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<cmath>x = \frac{-\sqrt{3} + \sqrt{19}}{2}.</cmath>  
  
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~[[User:grogg007|grogg007]], [[User:Bloggish|Bloggish]]
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Then <math>y = (\frac{-\sqrt{3} + \sqrt{19}}{2})^2 - 4 = \frac{3 - \sqrt{57}}{2}.</math> <math>57 + 3 + 2 = \boxed{62}.</math>
  
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Remark: The step where we took <math>\frac{\sqrt{3} - \sqrt{19}}{2}</math> and multiplied it by negative 1 wasn't actually necessary because the <math>y</math> coordinate would have been the same anyway. This was done to find the correct x-coordinate of the intersection point even though the problem didn’t ask for it.
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~[[User:grogg007|grogg007]], ~[[User:Mathkiddus|mathkiddus]]
  
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==Solution 3 (Similar to Solution 1)==
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==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 57: 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 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

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

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