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Difference between revisions of "2025 AIME II Problems/Problem 6"

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== Solution ==
 
== Solution ==
  
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Denote the intersection of <math>BC</math> and <math>w_1</math> as <math>P</math>, the intersection of <math>BC</math> and <math>GH</math> be <math>Q</math>, and the center of <math>w_2</math> to be <math>O</math>. Additionally, let <math>EF = GH = a, FG = EH = b</math>. We have that <math>CP = 18</math> and <math>PQ = \frac{6-b}{2}</math>. Considering right triangle <math>OAD</math>, <math>AD = 12</math>. Letting <math>R</math> be the intersection of <math>AD</math> and <math>FG</math>, <math>DR = 12 - \frac{b}{2}</math>
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Denote the intersection of <math>BC</math> and <math>w_1</math> as <math>P</math>, the intersection of <math>BC</math> and <math>GH</math> be <math>Q</math>, and the center of <math>w_2</math> to be <math>O</math>. Additionally, let <math>EF = GH = a, FG = EH = b</math>. We have that <math>CP = 18</math> and <math>PQ = \frac{6-b}{2}</math>. Considering right triangle <math>OAD</math>, <math>AD = 12</math>. Letting <math>R</math> be the intersection of <math>AD</math> and <math>FG</math>, <math>DR = 12 - \frac{b}{2}</math>. Using the equivalent area ratios: <cmath> \frac{a(24-\frac{b}{2})}{2} = \frac{(12-\frac{a}{2})b}{2} </cmath>
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This equation gives <math>b=2a</math>. Using pythagorean theorem on triangle <math>GHE</math> gives that <math>a^2+b^2 = 144</math>. Plugging the reuslt <math>b=2a</math> into this equation gives that the area of the triangle is <math>\frac{288}{5} \to \boxed{293}</math>

Revision as of 02:34, 14 February 2025

Problem

Circle $\omega_1$ with radius $6$ centered at point $A$ is internally tangent at point $B$ to circle $\omega_2$ with radius $15$. Points $C$ and $D$ lie on $\omega_2$ such that $\overline{BC}$ is a diameter of $\omega_2$ and $\overline{BC} \perp \overline{AD}$. The rectangle $EFGH$ is inscribed in $\omega_1$ such that $\overline{EF} \perp \overline{BC}$, $C$ is closer to $\overline{GH}$ than to $\overline{EF}$, and $D$ is closer to $\overline{FG}$ than to $\overline{EH}$, as shown. Triangles $\triangle DGF$ and $\triangle CHG$ have equal areas. The area of rectangle $EFGH$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[asy] size(5cm); defaultpen(fontsize(10pt)); pair A = (9, 0), B = (15, 0), C = (-15, 0), D = (9, 12), E = (9+12/sqrt(5), -6/sqrt(5)), F = (9+12/sqrt(5), 6/sqrt(5)), G = (9-12/sqrt(5), 6/sqrt(5)), H = (9-12/sqrt(5), -6/sqrt(5)); filldraw(G--H--C--cycle, lightgray); filldraw(D--G--F--cycle, lightgray); draw(B--C); draw(A--D); draw(E--F--G--H--cycle); draw(circle(origin, 15)); draw(circle(A, 6)); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); label("$A$", A, (.8, -.8)); label("$B$", B, (.8, 0)); label("$C$", C, (-.8, 0)); label("$D$", D, (.4, .8)); label("$E$", E, (.8, -.8)); label("$F$", F, (.8, .8)); label("$G$", G, (-.8, .8)); label("$H$", H, (-.8, -.8)); label("$\omega_1$", (9, -5)); label("$\omega_2$", (-1, -13.5)); [/asy]

Solution

Denote the intersection of $BC$ and $w_1$ as $P$, the intersection of $BC$ and $GH$ be $Q$, and the center of $w_2$ to be $O$. Additionally, let $EF = GH = a, FG = EH = b$. We have that $CP = 18$ and $PQ = \frac{6-b}{2}$. Considering right triangle $OAD$, $AD = 12$. Letting $R$ be the intersection of $AD$ and $FG$, $DR = 12 - \frac{b}{2}$. Using the equivalent area ratios: \[\frac{a(24-\frac{b}{2})}{2} = \frac{(12-\frac{a}{2})b}{2}\]

This equation gives $b=2a$. Using pythagorean theorem on triangle $GHE$ gives that $a^2+b^2 = 144$. Plugging the reuslt $b=2a$ into this equation gives that the area of the triangle is $\frac{288}{5} \to \boxed{293}$

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