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

Revision as of 18:42, 5 January 2022

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 (Official Solution, MAA)

Let $s$ be the side length of $ABCD$ , let $Q$ , and $R$ be the midpoints of $\overline{EG}$ and $\overline{FH}$ , respectively, let $S$ be the foot of the perpendicular from $Q$ to $\overline{CD}$ , let $T$ be the foot of the perpendicular from $R$ to $\overline{AD}$ . $[asy] size(150); defaultpen(fontsize(10pt)); pair A,B,C,D,E,F,Fp,G,Gp,H,O,I,J,R,S,T; A=dir(45*3); B=dir(-45*3); C=dir(-45); D=dir(45); O = origin; real theta=15; E=extension(A,B,O,dir(180+theta)); G=extension(C,D,O,dir(theta)); I=extension(A,D,O,dir(90+theta)); J=extension(B,C,O,dir(-90+theta)); H=(A+I)/2; F=H+(J-I); R=midpoint(H--F); S=midpoint(C--D); T=(R.x, A.y); draw(A--B--C--D--cycle^^E--G^^F--H, black+0.8); draw(I--J^^S--R--T, gray+0.4); dotfactor = 3; dot("$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$",G,dir(0)); dot("$Q$",O,dir(60)); dot("$R$",R,dir(-180)); dot("$S$",S,dir(0)); dot("$T$",T,dir(90)); pair P = extension(F,H,E,G); dot("$P$",P,dir(180+60)); [/asy]$ The fraction of the area of the square $ABCD$ which is occupied by trapezoid $BCGE$ is $\frac{275+405}{269+275+405+411}=\frac 12,$ so $Q$ is the center of $ABCD$ . Thus $R$ , $Q$ , $S$ are collinear, and $RT=QS=\tfrac 12 s$ . Similarly, the fraction of the area occupied by trapezoid $CDHF$ is $\tfrac 35$ , so $RS=\tfrac 35s$ and $RQ=\tfrac{1}{10}s$ .

Because $\triangle QSG \cong \triangle RTH$ , the area of $DHPG$ is the sum $[DHPG]=[DTRS]+[RPQ].$ Rectangle $DTRS$ has area $RS\cdot RT = \tfrac 35s\cdot \tfrac 12 s = \tfrac{3}{10}s^2$ . If $\angle QRP = \theta$ , then $\triangle RPQ$ has area $[RPQ]= \tfrac 12 \cdot \tfrac 1{10}s\cos\theta = \tfrac 1{400}s^2\sin 2\theta.$ Therefore $[DHPG]=s^2(\tfrac 3{10}-\tfrac 1{400}\sin 2\theta)$ .

Because these areas are in the ratio $411:405=(408+3):(408-3)$ , it follows that $\frac{\frac 1{400}\sin 2\theta}{\frac 3{10}}=\frac 3{408},$ from which we get $\sin 2\theta = \tfrac {15}{17}$ . Note that $\theta > 45^\circ$ , so $\cos 2\theta = -\tfrac 8{17}$ and $\sin^2\theta = \tfrac{25}{34}$ . Then $[ABCD]=s^2 = EG^2\sin^2\theta = 850.$

Solution 1

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 $O$ . $[asy] size(150); defaultpen(fontsize(10pt)); pair A,B,C,D,E,F,Fp,G,Gp,H,O,I,J,K; A=dir(45*3); B=dir(-45*3); C=dir(-45); D=dir(45); O = origin; real theta=15; E=extension(A,B,O,dir(180+theta)); G=extension(C,D,O,dir(theta)); I=extension(A,D,O,dir(90+theta)); J=extension(B,C,O,dir(-90+theta)); H=(A+I)/2; F=H+(J-I); K=H-I; draw(A--B--C--D--cycle^^E--G^^F--H, black+0.8); draw(I--J^^O--K, gray+0.4); dotfactor = 3; dot("$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$",O,dir(60)); dot("$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 $ABCD=1360a=\boxed{850}$

Solution 2 (Lazy)

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

@@ Line 29: / Line 29: @@
 label("$y$", intersectionpoint( C--P, G--F ));
 label("$z$", intersectionpoint( D--P, G--H ));</asy>
+== Solution (Official Solution, MAA)==
+Let <math>s</math> be the side length of <math>ABCD</math>, let <math>Q</math>, and <math>R</math> be the midpoints of <math>\overline{EG}</math> and <math>\overline{FH}</math>, respectively, let <math>S</math> be the foot of the perpendicular from <math>Q</math> to <math>\overline{CD}</math>, let <math>T</math> be the foot of the perpendicular from <math>R</math> to <math>\overline{AD}</math>.
+<asy> size(150); defaultpen(fontsize(10pt)); pair A,B,C,D,E,F,Fp,G,Gp,H,O,I,J,R,S,T;  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); R=midpoint(H--F); S=midpoint(C--D); T=(R.x, A.y); draw(A--B--C--D--cycle^^E--G^^F--H, black+0.8);  draw(I--J^^S--R--T, 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("$Q$",O,dir(60)); dot("$R$",R,dir(-180)); dot("$S$",S,dir(0)); dot("$T$",T,dir(90));  pair P = extension(F,H,E,G); dot("$P$",P,dir(180+60));  </asy>
+The fraction of the area of the square <math>ABCD</math> which is occupied by trapezoid <math>BCGE</math> is <cmath>\frac{275+405}{269+275+405+411}=\frac 12,</cmath>so <math>Q</math> is the center of <math>ABCD</math>. Thus <math>R</math>, <math>Q</math>, <math>S</math> are collinear, and <math>RT=QS=\tfrac 12 s</math>. Similarly, the fraction of the area occupied by trapezoid <math>CDHF</math> is <math>\tfrac 35</math>, so <math>RS=\tfrac 35s</math> and <math>RQ=\tfrac{1}{10}s</math>.
+Because <math>\triangle QSG \cong \triangle RTH</math>, the area of <math>DHPG</math> is the sum <cmath>[DHPG]=[DTRS]+[RPQ].</cmath> Rectangle <math>DTRS</math> has area <math>RS\cdot RT = \tfrac 35s\cdot \tfrac 12 s = \tfrac{3}{10}s^2</math>. If <math>\angle QRP = \theta</math> , then <math>\triangle RPQ</math> has area <cmath>[RPQ]= \tfrac 12 \cdot \tfrac 1{10}s\cos\theta = \tfrac 1{400}s^2\sin 2\theta.</cmath>Therefore <math>[DHPG]=s^2(\tfrac 3{10}-\tfrac 1{400}\sin 2\theta)</math>.
+Because these areas are in the ratio <math>411:405=(408+3):(408-3)</math>, it follows that <cmath>\frac{\frac 1{400}\sin 2\theta}{\frac 3{10}}=\frac 3{408},</cmath>from which we get <math>\sin 2\theta = \tfrac {15}{17}</math>. Note that <math>\theta > 45^\circ</math>, so <math>\cos 2\theta = -\tfrac 8{17}</math> and <math>\sin^2\theta = \tfrac{25}{34}</math>. Then <cmath>[ABCD]=s^2 = EG^2\sin^2\theta = 850.</cmath>
 == Solution 1==

2014 AIME I (Problems • Answer Key • Resources)
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