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 $\overline{BC}$ 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 $FPOJ=65a$ . 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 $PFJO$ . $\frac12(PF+OJ)(PO)=65a$ $(17-\frac{3a}{h})h=65a$ $h=4a$

Thus, $KP=1.5$ .

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

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

Strategy

$269+275+405+411=1360$ , a multiple of $17$ . Therefore the $\text{square}$ of the "hypotenuse" of a triangle with slope $m$ must be a multiple of $17$ . All of these triples are primitive:

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

The sides of the square could either be the longer leg or the shorter leg. Substituting $EG=FH=34$ : $\sqrt{17}\rightarrow 34\implies \{2\sqrt{17},4\sqrt{17}\}\implies 272,\textcolor{red}{1088}$ $\sqrt{34}\rightarrow 34\implies \{3\sqrt{34},5\sqrt{34}\}\implies 306,850$ $\sqrt{85}\rightarrow 34\implies \{4\sqrt{85}/5,18\sqrt{85}/5,12\sqrt{85}/5,14\sqrt{85}/5\}\implies\textcolor{red}{54.4,1101.6,489.6,666.4}$

The three viable answers here are $272,306,850$ . Since the areas add up to $1360$ parts, probably the answer should be $\boxed{850}$ . And perhaps MAA wants a high answer at the end of the AIME :).

2014 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2014 AIME I Problems/Problem 13

Contents

Problem 13

Solution

Strategy

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