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

Latest revision as of 11:32, 16 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 $\frac{1}{2}(2x)(24-y)=x(24-y)$ .

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 $PQ = 6-\frac{b}{2}$ . 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

~ Additional edits by fermat_sLastAMC

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

Solution 4 (Almost no Algebra)

If we draw segments connecting $A$ to $F, G,$ and $H,$ we can easily verify that all of the right triangles created are congruent. Thus, triangles $AGF$ and $AGH$ have equal areas, which means, by the given conditions, that kites $AGCH$ and $AGDF$ have equal areas. Thus, by the area formula for kites,

$\frac{1}{2}(AD)(GF)=\frac{1}{2}(AC)(BH),$ or $(AD)(GF)=(24)(GH).$

Also, if we extend $AD$ to the other side of the large circle, the chord length formula gives

$(AD)(AD)=(AC)(AB),$ so $(AD)(AD)=(24)(6).$

Squaring the second equation above gives $(AD)(AD)(GF)(GF)=(24)(24)(GH)(GH).$ Dividing by the fourth equation gives $(GF)^2=4(GH)^2.$ Since we know, by the Pythagorean Theorem, that $(GF)^2+(GH)^2=144,$ we can substitute to determine that $5(GH)^2=144.$

The desired area is $(GF)(GH)=2(GH)^2=\frac{2}{5}(144),$ which is $\frac{288}{5},$ so the answer is $\boxed{293}.$

2025 AIME II (Problems • Answer Key • Resources)
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