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

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==Strategy==
 
==Strategy==
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<math>269+275+405+411=1360</math>, a multiple of <math>17</math>.
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<math>269+275+405+411=1360</math>, a multiple of <math>17</math>. In addition, <math>EG=FH=34</math>, which is <math>17\cdot 2</math>.
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Therefore the <math>\text{square}</math> of the "hypotenuse" of a triangle with slope <math>m</math> must be a multiple of <math>17</math>. All of these triples are primitive:
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Therefore, we suspect the <math>\text{square}</math> of the "hypotenuse" of a triangle with slope <math>m</math> must be a multiple of <math>17</math>. All of these triples are primitive:
  
 
<cmath>17=1^2+4^2</cmath>
 
<cmath>17=1^2+4^2</cmath>

Revision as of 13:08, 14 February 2016

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 $17$. In addition, $EG=FH=34$, which is $17\cdot 2$. Therefore, we suspect the $\text{square}$ of the "hypotenuse" of a triangle with slope $m$ must be a multiple of $17$. All of these triples are primitive:

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

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 $\boxed{850}$, a multiple of $5$. And perhaps MAA wants a high answer at the end of the AIME :).

See also

2014 AIME I (Problems • Answer Key • Resources)
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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