# Difference between revisions of "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

Notice that $269+411=275+405$. This means $\overline{EG}$ passes through the centre of the square.

Draw $\overline{IJ} \parallel \overline{HF}$ with $I$ on $\overline{AD}$, $J$ on $\overline{EC}$ such that $\overline{IJ}$ and $\overline{EG}$ intersects at the centre of the square $O$.

Let the area of the square be $1360a$. Then the area of $HPOI=71a$ and the area of $FPOJ=65a$.

Let the side side length be $d=\sqrt{1360a}$.

Draw $\overline{OK}\parallel \overline{HI}$ and intersects $\overline{HF}$ at $K$. $OK=d\cdot\frac{HFJI}{ABCD}=\frac{d}{10}$.

The area of $HKOI=\frac12\cdot HFJI=68a$, so the area of $POK=3a$.

Let $\overline{PO}=h$. Then $KP=\frac{6a}{h}$

Consider the area of $PFJO$. $$\frac12(PF+OJ)(PO)=65a$$ $$(17-\frac{3a}{h})h=65a$$ $$h=4a$$

Thus, $KP=1.5$.

Solving $(4a)^2+1.5^2=(\frac{d}{10})^2=13.6a$, we get $a=\frac58$.

Therefore, the area of $ABCD=1360a=\boxed{850}$