# 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 center of the square.

Draw $\overline{IJ} \parallel \overline{HF}$ with $I$ on $\overline{AD}$, $J$ on $\overline{BC}$ such that $\overline{IJ}$ and $\overline{EG}$ intersects at the center of the square which I'll label as $O$.

Let the area of the square be $1360a$. Then the area of $HPOI=71a$ and the area of $FPOJ=65a$. This is because $\overline{HF}$ is perpendicular to $\overline{EG}$ (given in the problem), so $\overline{IJ}$ is also perpendicular to $\overline{EG}$. These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals.

Let the side length of the square 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 $PFJO$. $$\frac12(PF+OJ)(PO)=65a$$ $$\left(17-\frac{3a}{h}\right)h=65a$$ $$h=4a$$

Thus, $KP=1.5$.

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

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

## Lazy Solution

$269+275+405+411=1360$, 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