# Difference between revisions of "2015 AIME I Problems/Problem 7"

## Problem

7. In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$, and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$.

$[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,dir(90)); label("F",F,NE); label("G",G,NE); label("H",H,W); label("J",J,S); label("K",K,SE); label("L",L,SE); label("M",M,dir(90)); label("N",N,dir(180)); [/asy]$

## Solution

We begin by denoting the length $ED$ $a$, giving us $DC = 2a$ and $EC = a\sqrt5$. Since angles $\angle DCE$ and $\angle FCJ$ are complimentary, we have that $\triangle CDE \sim \triangle JFC$ (and similarly the rest of the triangles are $1-2-\sqrt5$ triangles). We let the sidelength of $FGHJ$ be $b$, giving us:

$JC = \sqrt5 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}$ and $BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}$.

Since $BC = CJ + JC$,

$2a = \frac{b\sqrt 5}{2} + \frac{b}{\sqrt5}$,

Solving for $b$ in terms of $a$ yields $b = \frac{4a\sqrt5}{7}$.

We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\sqrt{11}$. We also draw the perpendicular from $E$ to $ML$ and label the point of intersection $P$:

$[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L); draw(P--E); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,dir(90)); label("F",F,NE); label("G",G,NE); label("H",H,W); label("J",J,S); label("K",K,SE); label("L",L,SE); label("M",M,dir(90)); label("N",N,dir(180)); label("P",P,dir(235)); [/asy]$

This gives that $AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}$ and $ME = \sqrt5 \cdot MP = \sqrt5 \cdot \frac{EP}{2} = \sqrt5 \cdot \frac{LG}{2} = \sqrt5 \cdot \frac{HG - HK - KL}{2} = \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2}$

Since $AE$ = $AM + ME$, we get

$2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2} = a$

$\Rightarrow 12\sqrt{11} + 5(\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}) = 2\sqrt5a$

$\Rightarrow \frac{-21}{2}\sqrt{11} + \frac{20a\sqrt5}{7} = 2\sqrt5a$

$\Rightarrow -21\sqrt{11} = 2\sqrt5a\frac{14 - 20}{7}$

$\Rightarrow \frac{49\sqrt{11}}{4} = \sqrt5a$

$\Rightarrow 7\sqrt{11} = \frac{4a\sqrt{5}}{7}$

So our final answer is $(7\sqrt{11})^2 = \boxed{539}$