Difference between revisions of "2014 AIME I Problems/Problem 13"

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== Solution ==
 
== Solution ==
  
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Notice that <math>269+411=275+405</math>. This means <math>\overline{EG}</math> passes through the centre of the square.
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Draw <math>\overline{IJ} \parallel \overline{HF}</math> with <math>I</math> on <math>\overline{AD}</math>, <math>J</math> on <math>\overline{EC}</math> such that <math>\overline{IJ}</math> and <math>\overline{EG}</math> intersects at the centre of the square <math>O</math>.
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Let the area of the square be <math>1360a</math>. Then the area of <math>HPOI=71a</math> and the area of <math>FPOJ=65a</math>.
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Let the side side length be <math>d=\sqrt{1360a}</math>.
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Draw <math>\overline{OK}\parallel \overline{HI}</math> and intersects <math>\overline{HF}</math> at <math>K</math>. <math>OK=d\cdot\frac{HFJI}{ABCD}=\frac{d}{10}</math>.
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The area of <math>HKOI=\frac12\cdot HFJI=68a</math>, so the area of <math>POK=3a</math>.
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Let <math>\overline{PO}=h</math>. Then <math>KP=\frac{6a}{h}</math>
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Consider the area of <math>PFJO</math>.
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<cmath>\frac12(PF+OJ)(PO)=65a</cmath>
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<cmath>(17-\frac{3a}{h})h=65a</cmath>
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<cmath>h=4a</cmath>
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Thus, <math>KP=1.5</math>.
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Solving <math>(4a)^2+1.5^2=(\frac{d}{10})^2=13.6a</math>, we get <math>a=\frac58</math>.
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Therefore, the area of <math>ABCD=1360a=\boxed{850}</math>
 
== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=12|num-a=14}}
 
{{AIME box|year=2014|n=I|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:50, 23 March 2014

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 $ABCD=1360a=\boxed{850}$

See also

2014 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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