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2014 AIME I Problems/Problem 13

Revision as of 13:03, 31 December 2021 by Shihan (talk | contribs) (→‎Solution)

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 $O$. [asy] size(200); pair A,B,C,D,E,F,Fp,G,Gp,H,O,I,J,K; A=dir(45*3); B=dir(-45*3); C=dir(-45); D=dir(45); O = origin; real theta=15; E=extension(A,B,O,dir(180+theta)); G=extension(C,D,O,dir(theta)); I=extension(A,D,O,dir(90+theta)); J=extension(B,C,O,dir(-90+theta)); H=(A+I)/2; F=H+(J-I); K=H-I; draw(A--B--C--D--cycle^^E--G^^F--H, black+0.8); draw(I--J^^O--K, gray+0.4); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); dot("$I$",I,dir(90)); dot("$J$",J,dir(270)); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); dot("$O$",O,dir(-30)); dot("$K$",K,dir(-180)); pair P = extension(F,H,E,G); dot("$P$",P,dir(180+60)); [/asy] Let the area of the square be $1360a$. Then $[HPOI]=71a$ and $[FPOJ]=65a$. 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 $K$. Then \[OK=d\cdot\tfrac{[HFJI]}{[ABCD]}=\frac{1}{10}d.\] Then $[HKOI]=\tfrac12\cdot [HFJI]=68a$, so $[POK]=3a$. 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, $KP=1.5$. Now we solve \[(4a)^2+1.5^2=\left(\frac{d}{10}\right)^2=13.6a,\] to get $a=\tfrac 9{40}$ or $a=\tfrac58$.

The former leads to a square with diagonal less than $34$, which can't be, since $EG=FH=34$; therefore $a=\tfrac 58$ and 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

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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