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

## Problem

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 1

Let us find the proportion of the side length of $KLMN$ and $FJGH$. Let the side length of $KLMN=y$ and the side length of $FJGH=x$.

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

Now, examine $BC$. We know $BC=BJ+JC$, and triangles $\Delta BHJ$ and $\Delta JFC$ are similar to $\Delta EDC$ since they are $1-2-\sqrt{5}$ triangles. Thus, we can rewrite $BC$ in terms of the side length of $FJGH$. $$BJ=\frac{1}{\sqrt{5}}HJ=\frac{x}{\sqrt{5}}=\frac{x\sqrt{5}}{5}, JC=\frac{\sqrt{5}}{2}JF=\frac{x\sqrt{5}}{2}\Rightarrow BC=\frac{7x\sqrt{5}}{10}$$

Now examine $AB$. We can express this length in terms of $x,y$ since $AB=AN+NH+HB$. By using similar triangles as in the first part, we have $$AB=\frac{1}{\sqrt{5}}y+\frac{\sqrt{5}}{2}y+\frac{2}{\sqrt{5}}x$$ $$AB=BC\Rightarrow \frac{7y\sqrt{5}}{10}+\frac{2x\sqrt{5}}{5}=\frac{7x\sqrt{5}}{10}\Rightarrow \frac{7y\sqrt{5}}{10}=\frac{3x\sqrt{5}}{10}\Rightarrow 7y=3x$$

Now, it is trivial to see that $[FJGH]=\left(\frac{x}{y}\right)^2[KLMN]=\left(\frac{7}{3}\right)^2\cdot 99=\boxed{539}.\Box$

## Solution 2

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

We begin by denoting the length $ED$ $a$, giving us $DC = 2a$ and $EC = a\sqrt5$. Since angles $\angle DCE$ and $\angle FCJ$ are complementary, 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 $BC = CJ + BJ$,

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

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 $P$ as in the diagram at the top

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 $(7\sqrt{11})^2 = \boxed{539}$.

## Solution 3

This is a relatively quick solution but a fakesolve. We see that with a ruler, $KL = \frac{3}{2}$ cm and $HG = \frac{7}{2}$ cm. Thus if $KL$ corresponds with an area of $99$, then $HG$ ($FGHJ$'s area) would correspond with $99*(\frac{7}{3})^2 = \boxed{539}$ - aops5234