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

Revision as of 20:28, 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

Video Solution by grogg007

https://www.youtube.com/watch?v=ZHGAmInyhZc

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)

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 point's angle become? 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 $\alpha -2\theta.$

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 counterclockwise rotation by $2\theta^\circ.$

So a counterclockwise rotation of $60^\circ$ is the same thing as a reflection over the line forming a $30$ degree angle with the $x$ axis, followed by a reflection over the $x$ axis. The equation of this line is $y = \tan(30^\circ)x = \sqrt{3}x.$ 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 $y = (\frac{-\sqrt{3} + \sqrt{19}}{2})^2 - 4 = \frac{3 - \sqrt{57}}{2}.$ Ans = $57 + 3 + 2 = \boxed{62}.$

~grogg007

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 7: / Line 7: @@
 Link: https://www.desmos.com/calculator/ci3vodl4vs
-==Video solution by [[User:grogg007|grogg007]]==
+==Video Solution by [[User:grogg007|grogg007]]==
-https://youtu.be/wib5vos7Sd4?t=639
+https://www.youtube.com/watch?v=ZHGAmInyhZc
 ==Solution 1==
@@ Line 15: / Line 15: @@
 ~[[User:Mathkiddus|mathkiddus]]
-==Solution 2==
+==Solution 2 (Polar Coordinates)==
-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 the negative value for <math>a</math> to find the negative value of <math>y,</math> so <math>a = \frac{\sqrt{3} - \sqrt{19}}{2}.</math> Then the y-coordinate is <math>(\frac{\sqrt{3} - \sqrt{19}}{2})^2 - 4 = \frac{3 - \sqrt{57}}{2}.</math> <math>57 + 2 + 3 = \boxed{62}.</math>
+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.
-~[[User:grogg007|grogg007]], [[User:Bloggish|Bloggish]]
+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>
-IMPORTANT NOTE: The reason we can use a negative <math>a</math> coordinate to still get an intersection in the 4th quadrant is because <math>y = x^2 - 4</math> is symmetric about the y axis, meaning the actual <math>x</math> coordinate of the intersection point is -1 times the <math>a</math> value we used in the solution, so the point is <math>(\frac{\sqrt{19}-\sqrt{3}}{2},\frac{3-\sqrt{57}}{2})</math>.
+Then if we reflected the point over the <math>x</math> axis again the angle becomes <math>\alpha -2\theta.</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 counterclockwise rotation by <math>2\theta^\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.
+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> Ans = <math>57 + 3 + 2 = \boxed{62}.</math>
+~[[User:grogg007|grogg007]]
 ==Solution 3 (Similar to Solution 1)==

2025 AIME I (Problems • Answer Key • Resources)
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