# 2014 AIME I Problems/Problem 13

## Problem 13

On square $ABCD$, points $E,F,G$, and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$. Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$, and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$.

$[asy] pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("A",A,dir(135)); dot("B",B,dir(215)); dot("C",C,dir(305)); dot("D",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("H",H,dir(90)); dot("F",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("E",E,dir(180)); dot("G",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("P",P,dir(60)); label("w", intersectionpoint( A--P, E--H )); label("x", intersectionpoint( B--P, E--F )); label("y", intersectionpoint( C--P, G--F )); label("z", intersectionpoint( D--P, G--H ));[/asy]$

## Solution (Official Solution, MAA)

Let $s$ be the side length of $ABCD$, let $Q$, and $R$ be the midpoints of $\overline{EG}$ and $\overline{FH}$, respectively, let $S$ be the foot of the perpendicular from $Q$ to $\overline{CD}$, let $T$ be the foot of the perpendicular from $R$ to $\overline{AD}$. $[asy] size(150); defaultpen(fontsize(10pt)); pair A,B,C,D,E,F,Fp,G,Gp,H,O,I,J,R,S,T; A=dir(45*3); B=dir(-45*3); C=dir(-45); D=dir(45); O = origin; real theta=15; E=extension(A,B,O,dir(180+theta)); G=extension(C,D,O,dir(theta)); I=extension(A,D,O,dir(90+theta)); J=extension(B,C,O,dir(-90+theta)); H=(A+I)/2; F=H+(J-I); R=midpoint(H--F); S=midpoint(C--D); T=(R.x, A.y); draw(A--B--C--D--cycle^^E--G^^F--H, black+0.8); draw(S--R--T, gray+0.4); dotfactor = 3; dot("A",A,dir(135)); dot("B",B,dir(215)); dot("C",C,dir(305)); dot("D",D,dir(45)); dot("H",H,dir(90)); dot("F",F,dir(270)); dot("E",E,dir(180)); dot("G",G,dir(0)); dot("Q",O,dir(-90)); dot("R",R,dir(-180)); dot("S",S,dir(0)); dot("T",T,dir(90)); pair P = extension(F,H,E,G); dot("P",P,dir(180+60)); [/asy]$ The fraction of the area of the square $ABCD$ which is occupied by trapezoid $BCGE$ is $$\frac{275+405}{269+275+405+411}=\frac 12,$$so $Q$ is the center of $ABCD$. Thus $R$, $Q$, $S$ are collinear, and $RT=QS=\tfrac 12 s$. Similarly, the fraction of the area occupied by trapezoid $CDHF$ is $\tfrac 35$, so $RS=\tfrac 35s$ and $RQ=\tfrac{1}{10}s$.

Because $\triangle QSG \cong \triangle RTH$, the area of $DHPG$ is the sum $$[DHPG]=[DTRS]+[RPQ].$$ Rectangle $DTRS$ has area $RS\cdot RT = \tfrac 35s\cdot \tfrac 12 s = \tfrac{3}{10}s^2$. If $\angle QRP = \theta$ , then $\triangle RPQ$ has area $$[RPQ]= \tfrac 12 \cdot \tfrac 1{10}s\sin\theta \cdot \tfrac 1{10}s\cos\theta = \tfrac 1{400}s^2\sin 2\theta.$$Therefore the area of $[DHPG]$ is $s^2(\tfrac 3{10}+\tfrac 1{400}\sin 2\theta)$. Because the area of trapezoid $CDHF$ is $\tfrac 35 s^2$, the area of $CGPF$ is $s^2(\tfrac 3{10}-\tfrac 1{400}\sin 2\theta)$.

Because these areas are in the ratio $411:405=(408+3):(408-3)$, it follows that $$\frac{\frac 1{400}\sin 2\theta}{\frac 3{10}}=\frac 3{408},$$from which we get $\sin 2\theta = \tfrac {15}{17}$. Note that $\theta =\angle RHT > \angle QAT = 45^\circ$, so $\cos 2\theta = -\sqrt{1-\sin^2 2\theta}= -\tfrac 8{17}$ and $\sin^2\theta = \tfrac{1}{2}(1-\cos 2\theta) = \tfrac{25}{34}$. Then $$[ABCD]=s^2 = EG^2\sin^2\theta = 34^2 \cdot \tfrac {25}{34} = 850.$$

## Solution 1

Let $s$ be the side length of $ABCD$, let $[ABCD]=1360a$. Let $Q$ and $R$ be the midpoints of $\overline{EG}$ and $\overline{FH}$, respectively; because $269+411=275+405$, $Q$ is also the center of the square. Draw $\overline{IJ} \parallel \overline{HF}$ through $Q$, with $I$ on $\overline{AD}$, $J$ on $\overline{BC}$. $[asy] size(150); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,G,H,I,J,L,P,Q,R,S; Q=MP("Q",origin,down); A=MP("A",(-1,1),dir(135)); B=MP("B",(-1,-1),dir(225)); C=MP("C",(1,-1),dir(-45)); D=MP("D",(1,1),dir(45)); real theta = 20; real shift=0.4; E=MP("E",extension(A,B,Q,dir(theta)),left); J=MP("J",extension(B,C,Q,dir(90+theta)),down); F=MP("F",J+(shift*left),down); G=MP("G",extension(C,D,Q,dir(theta)),right); I=MP("I",extension(A,D,Q,dir(90+theta)),up); H=MP("H",I+(shift*left),up); P=MP("P",extension(E,G,F,H),2*dir(-110)); R=MP("R",extension(F,H,Q,left),left); draw(A--B--C--D--cycle^^E--G^^F--H, black+1); draw(R--Q^^I--J, gray); [/asy]$ Segments $\overline{EG}$ and $\overline{IJ}$ divide the square into four congruent quadrilaterals, each of area $\tfrac 14 [ABCD]=340a$. Then $$[HFJI]=[ABJI]-[ABFH]=136a.$$ The fraction of the total area occupied by parallelogram $HFJI$ is $\tfrac 1{10}$, so $RQ=\tfrac{s}{10}$.

Because $[HFJI]= HF\cdot PQ$, with $HF=34$, we get $PQ=4a$. Now $$[PQR]=[HPQI]-[HRQI]= ([AEQI]-[AEPH])-\tfrac 12[IJFH] = 3a,$$ and because $[PQR]=\tfrac 12 \cdot PQ\cdot PR$, with $PQ=4a$, we get $PR=\tfrac 32$. By Pythagoras' Theorem on $\triangle PQR$, we get $$16a^2+\frac 94 =\tfrac{68}{5}a,\quad \text{i.e.}\quad 320a^2-272a+45=0,$$ with roots $a=\tfrac 9{40}$ or $a=\tfrac58$. The former leads to a square with diagonal less than $34$, which can't be, since $EG=FH=34$; therefore $a=\tfrac 58$ and $[ABCD]=850$.

## Solution 2 (Fakesolve)

$269+275+405+411=1360$, a multiple of $17$. In addition, $EG=FH=34$, which is $17\cdot 2$. Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to $EG$ and $FH$ must be a multiple of $17$. All of these triples are primitive:

$$17=1^2+4^2$$ $$34=3^2+5^2$$ $$51=\emptyset$$ $$68=\emptyset\text{ others}$$ $$85=2^2+9^2=6^2+7^2$$ $$102=\emptyset$$ $$119=\emptyset \dots$$

The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting $EG=FH=34$: $$\sqrt{17}\rightarrow 34\implies 8\sqrt{17}\implies A=\textcolor{red}{1088}$$ $$\sqrt{34}\rightarrow 34\implies 5\sqrt{34}\implies A=850$$ $$\sqrt{85}\rightarrow 34\implies \{18\sqrt{85}/5,14\sqrt{85}/5\}\implies A=\textcolor{red}{1101.6,666.4}$$

Thus, $\boxed{850}$ is the only valid answer.

## Solution 3

Continue in the same way as solution 1 to get that $POK$ has area $3a$, and $OK = \frac{d}{10}$. You can then find $PK$ has length $\frac 32$.

Then, if we drop a perpendicular from $H$ to $BC$ at $L$, We get $\triangle HLF \sim \triangle OPK$.

Thus, $LF = \frac{15\cdot 34}{d}$, and we know $HL = d$, and $HF = 34$. Thus, we can set up an equation in terms of $d$ using the Pythagorean theorem.

$$\frac{15^2 \cdot 34^2}{d^2} + d^2 = 34^2$$

$$d^4 - 34^2 d^2 + 15^2 \cdot 34^2 = 0$$

$$(d^2 - 34 \cdot 25)(d^2 - 34 \cdot 9) = 0$$

$d^2 = 34 \cdot 9$ is extraneous, so $d^2 = 34 \cdot 25$. Since the area is $d^2$, we have it is equal to $34 \cdot 25 = \boxed{850}$

-Alexlikemath

## Video Solution

~MathProblemSolvingSkills.com

## See also

 2014 AIME I (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.