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

Revision as of 21:25, 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 1 (Thorough)

Let $GH=2x$ and $GF=2y$ . Notice that since $\overline{BC}$ is perpendicular to $\overline{GH}$ (can be proven using basic angle chasing) and $\overline{BC}$ is an extension of a diameter of $\omega_1$ , then $\overline{CB}$ is the perpendicular bisector of $\overline{GH}$ . Similarly, since $\overline{AD}$ is perpendicular to $\overline{GF}$ (also provable using basic angle chasing) and $\overline{AD}$ is part of a diameter of $\omega_1$ , then $\overline{AD}$ is the perpendicular bisector of $\overline{GF}$ .

From the Pythagorean Theorem on $\triangle GFH$ , we have $(2x)^2+(2y)^2=12^2$ , so $x^2+y^2=36$ . To find our second equation for our system, we utilize the triangles given.

Let $I=\overline{GH}\cap\overline{CB}$ . Then we know that $GFBI$ is also a rectangle since all of its angles can be shown to be right using basic angle chasing, so $FG=IB$ . We also know that $CI+IB=2\cdot 15=30$ . $IA=y$ and $AB=6$ , so $CI=30-y-6=24-y$ . Notice that $CI$ is a height of $\triangle CHG$ , so its area is $\frac{1}{2}(2x)(30-2y)=x(24-y)$ .

Next, extend $\overline{DA}$ past $A$ to intersect $\omega_2$ again at $D'$ . Since $\overline{BC}$ is given to be a diameter of $\omega_2$ and $\overline{BC}\perp\overline{AD}$ , then $\overline{BC}$ is the perpendicular bisector of $\overline{DD'}$ ; thus $DA=D'A$ . By Power of a Point, we know that $CA\cdot AB=DA\cdot AD'$ . $CA=30-6=24$ and $AB=6$ , so $DA\cdot AD'=(DA)^2=24\cdot6=144$ and $DA=D'A=12$ .

Denote $J=\overline{DA}\cap\overline{GF}$ . We know that $DJ=DA-AJ=12-x$ (recall that $GI=IH=x$ , and it can be shown that $GIAJ$ is a rectangle). $\overline{DJ}$ is the height of $\triangle DGF$ , so its area is $\frac{1}{2}(2y)(12-x)=y(12-x)$ .

We are given that $[DGF]=[CHG]$ ( $[ABC]$ denotes the area of figure $ABC$ ). As a result, $x(24-y)=y(12-x)$ . This can be simplified to $y=2x$ . Substituting this into the Pythagorean equation yields $5x^2=36$ and $x=\frac{6}{\sqrt{5}}$ . Then $y=\frac{12}{\sqrt{5}}$ .

$[EFGH]=2x\cdot2y=2\cdot\frac{6}{\sqrt{5}}\cdot2\cdot\frac{12}{\sqrt{5}}=\frac{288}{5}$ , so the answer is $288+5=\boxed{293}$ .

~ eevee9406

~ Edited by aoum

Solution 2 (Faster)

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 the 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 $\frac{288}{5} \to \boxed{293}$ .

~ Vivdax

~ Edited by aoum

Solution 3 (Not Recommended)

You can use your ruler to check that $GF=2GH.$ Then, we have $GF^2+GH^2=36,$ and we solve the system of equations to get $\frac{288}{5},$ so the answer is $\boxed{293}.$

Note: This method is not recommended as the diagrams are not necessarily drawn to scale. However, it can be used in emergency situations or to verify the answer.

~derekwang2048

@@ Line 66: / Line 66: @@
 == Solution 3 (Not Recommended) ==
-You can use your ruler to check that <math>GF=2GH.</math> Then, we have <math>GF^2+GH^2+36,</math> and we solve the system of equations to get <math>\frac{288}{5},</math> so the answer is <math>\boxed{293}.</math>
+You can use your ruler to check that <math>GF=2GH.</math> Then, we have <math>GF^2+GH^2=36,</math> and we solve the system of equations to get <math>\frac{288}{5},</math> so the answer is <math>\boxed{293}.</math>
 Note: This method is not recommended as the diagrams are not necessarily drawn to scale. However, it can be used in emergency situations or to verify the answer.

2025 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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