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

Latest revision as of 20:07, 13 October 2023

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 $\overline{AD}$ . $[asy] size(150); defaultpen(fontsize(10pt)); pair A,B,C,D,E,F,Fp,G,Gp,H,O,I,J,R,S,T; A=dir(45*3); B=dir(-45*3); C=dir(-45); D=dir(45); O = origin; real theta=15; E=extension(A,B,O,dir(180+theta)); G=extension(C,D,O,dir(theta)); I=extension(A,D,O,dir(90+theta)); J=extension(B,C,O,dir(-90+theta)); H=(A+I)/2; F=H+(J-I); R=midpoint(H--F); S=midpoint(C--D); T=(R.x, A.y); draw(A--B--C--D--cycle^^E--G^^F--H, black+0.8); draw(S--R--T, gray+0.4); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); dot("$Q$",O,dir(-90)); dot("$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 $[RPQ]= \tfrac 12 \cdot \tfrac 1{10}s\sin\theta \cdot \tfrac 1{10}s\cos\theta = \tfrac 1{400}s^2\sin 2\theta.$ Therefore the area of $[DHPG]$ is $s^2(\tfrac 3{10}+\tfrac 1{400}\sin 2\theta)$ . Because the area of trapezoid $CDHF$ is $\tfrac 35 s^2$ , the area of $CGPF$ is $s^2(\tfrac 3{10}-\tfrac 1{400}\sin 2\theta)$ .

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 $\theta =\angle RHT > \angle QAT = 45^\circ$ , so $\cos 2\theta = -\sqrt{1-\sin^2 2\theta}= -\tfrac 8{17}$ and $\sin^2\theta = \tfrac{1}{2}(1-\cos 2\theta) = \tfrac{25}{34}$ . Then $[ABCD]=s^2 = EG^2\sin^2\theta = 34^2 \cdot \tfrac {25}{34} = 850.$

Solution 1

Let $s$ be the side length of $ABCD$ , let $[ABCD]=1360a$ . Let $Q$ and $R$ be the midpoints of $\overline{EG}$ and $\overline{FH}$ , respectively; because $269+411=275+405$ , $Q$ is also the center of the square. Draw $\overline{IJ} \parallel \overline{HF}$ through $Q$ , with $I$ on $\overline{AD}$ , $J$ on $\overline{BC}$ . $[asy] size(150); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,G,H,I,J,L,P,Q,R,S; Q=MP("Q",origin,down); A=MP("A",(-1,1),dir(135)); B=MP("B",(-1,-1),dir(225)); C=MP("C",(1,-1),dir(-45)); D=MP("D",(1,1),dir(45)); real theta = 20; real shift=0.4; E=MP("E",extension(A,B,Q,dir(theta)),left); J=MP("J",extension(B,C,Q,dir(90+theta)),down); F=MP("F",J+(shift*left),down); G=MP("G",extension(C,D,Q,dir(theta)),right); I=MP("I",extension(A,D,Q,dir(90+theta)),up); H=MP("H",I+(shift*left),up); P=MP("P",extension(E,G,F,H),2*dir(-110)); R=MP("R",extension(F,H,Q,left),left); draw(A--B--C--D--cycle^^E--G^^F--H, black+1); draw(R--Q^^I--J, gray); [/asy]$ Segments $\overline{EG}$ and $\overline{IJ}$ divide the square into four congruent quadrilaterals, each of area $\tfrac 14 [ABCD]=340a$ . Then $[HFJI]=[ABJI]-[ABFH]=136a.$ The fraction of the total area occupied by parallelogram $HFJI$ is $\tfrac 1{10}$ , so $RQ=\tfrac{s}{10}$ .

Because $[HFJI]= HF\cdot PQ$ , with $HF=34$ , we get $PQ=4a$ . Now $[PQR]=[HPQI]-[HRQI]= ([AEQI]-[AEPH])-\tfrac 12[IJFH] = 3a,$ and because $[PQR]=\tfrac 12 \cdot PQ\cdot PR$ , with $PQ=4a$ , we get $PR=\tfrac 32$ . By Pythagoras' Theorem on $\triangle PQR$ , we get $16a^2+\frac 94 =\tfrac{68}{5}a,\quad \text{i.e.}\quad 320a^2-272a+45=0,$ with roots $a=\tfrac 9{40}$ 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 $[ABCD]=850$ .

Solution 2 (Fakesolve)

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

Video Solution

https://youtu.be/Kcug2ALOjkA?si=VoImhnX5rAKhprgk

~MathProblemSolvingSkills.com

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=wrxET2c0ZgU

@@ Line 30: / Line 30: @@
 label("$z$", intersectionpoint( D--P, G--H ));</asy>
-== Solution ==
+== Solution (Official Solution, MAA)==
-Notice that <math>269+411=275+405</math>. This means <math>\overline{EG}</math> passes through the center of the square.
+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(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("$H$",H,dir(90)); dot("$F$",F,dir(270));  dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); dot("$Q$",O,dir(-90)); 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>.
-Draw <math>\overline{IJ} \parallel \overline{HF}</math> with <math>I</math> on <math>\overline{AD}</math>, <math>J</math> on <math>\overline{BC}</math> such that <math>\overline{IJ}</math> and <math>\overline{EG}</math> intersects at the center of the square which I'll label as <math>O</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\sin\theta \cdot \tfrac 1{10}s\cos\theta = \tfrac 1{400}s^2\sin 2\theta.</cmath>Therefore the area of <math>[DHPG]</math> is <math>s^2(\tfrac 3{10}+\tfrac 1{400}\sin 2\theta)</math>. Because the area of trapezoid <math>CDHF</math> is <math>\tfrac 35 s^2</math>, the area of <math>CGPF</math> is <math>s^2(\tfrac 3{10}-\tfrac 1{400}\sin 2\theta)</math>.
-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>. This is because <math>\overline{HF}</math> is perpendicular to <math>\overline{EG}</math> (given in the problem), so <math>\overline{IJ}</math> is also perpendicular to <math>\overline{EG}</math>. These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals.
+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 =\angle RHT > \angle QAT = 45^\circ</math>, so <math>\cos 2\theta = -\sqrt{1-\sin^2 2\theta}= -\tfrac 8{17}</math> and <math>\sin^2\theta = \tfrac{1}{2}(1-\cos 2\theta) = \tfrac{25}{34}</math>. Then <cmath>[ABCD]=s^2 = EG^2\sin^2\theta = 34^2 \cdot \tfrac {25}{34} = 850.</cmath>
-Let the side length of the square be <math>d=\sqrt{1360a}</math>.
+== Solution 1==
-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>.
+Let <math>s</math> be the side length of <math>ABCD</math>, let <math>[ABCD]=1360a</math>. Let <math>Q</math> and <math>R</math> be the midpoints of <math>\overline{EG}</math> and <math>\overline{FH}</math>, respectively; because <math>269+411=275+405</math>, <math>Q</math> is also the center of the square. Draw <math>\overline{IJ} \parallel \overline{HF}</math> through <math>Q</math>, with <math>I</math> on <math>\overline{AD}</math>, <math>J</math> on <math>\overline{BC}</math>.
+<asy> size(150);
+defaultpen(fontsize(9pt));
+pair A,B,C,D,E,F,G,H,I,J,L,P,Q,R,S;
+Q=MP("Q",origin,down); A=MP("A",(-1,1),dir(135)); B=MP("B",(-1,-1),dir(225)); C=MP("C",(1,-1),dir(-45)); D=MP("D",(1,1),dir(45)); real theta = 20; real shift=0.4; E=MP("E",extension(A,B,Q,dir(theta)),left); J=MP("J",extension(B,C,Q,dir(90+theta)),down); F=MP("F",J+(shift*left),down); G=MP("G",extension(C,D,Q,dir(theta)),right); I=MP("I",extension(A,D,Q,dir(90+theta)),up); H=MP("H",I+(shift*left),up); P=MP("P",extension(E,G,F,H),2*dir(-110)); R=MP("R",extension(F,H,Q,left),left);
+draw(A--B--C--D--cycle^^E--G^^F--H, black+1); draw(R--Q^^I--J, gray);
+</asy>
+Segments <math>\overline{EG}</math> and <math>\overline{IJ}</math> divide the square into four congruent quadrilaterals, each of area <math>\tfrac 14 [ABCD]=340a</math>. Then <cmath>[HFJI]=[ABJI]-[ABFH]=136a.</cmath> The fraction of the total area occupied by parallelogram <math>HFJI</math> is <math>\tfrac 1{10}</math>, so <math>RQ=\tfrac{s}{10}</math>.
-The area of <math>HKOI=\frac12\cdot HFJI=68a</math>, so the area of <math>POK=3a</math>.
+Because <math>[HFJI]= HF\cdot PQ</math>, with <math>HF=34</math>, we get <math>PQ=4a</math>.
+Now <cmath>[PQR]=[HPQI]-[HRQI]= ([AEQI]-[AEPH])-\tfrac 12[IJFH] = 3a,</cmath> and because <math>[PQR]=\tfrac 12 \cdot PQ\cdot PR</math>, with <math>PQ=4a</math>, we get <math>PR=\tfrac 32</math>.
+By Pythagoras' Theorem on <math>\triangle PQR</math>, we get <cmath>16a^2+\frac 94 =\tfrac{68}{5}a,\quad \text{i.e.}\quad 320a^2-272a+45=0,</cmath> with roots <math>a=\tfrac 9{40}</math> or <math>a=\tfrac58</math>. The former leads to a square with diagonal less than <math>34</math>, which can't be, since <math>EG=FH=34</math>; therefore <math>a=\tfrac 58</math> and <math>[ABCD]=850</math>.
-Let <math>\overline{PO}=h</math>. Then <math>KP=\frac{6a}{h}</math>
+==Solution 2 (Fakesolve)==
-Consider the area of <math>PFJO</math>.
-<cmath>\frac12(PF+OJ)(PO)=65a</cmath>
-<cmath>\left(17-\frac{3a}{h}\right)h=65a</cmath>
-<cmath>h=4a</cmath>
-Thus, <math>KP=1.5</math>.
-Solving <math>(4a)^2+1.5^2=\left(\frac{d}{10}\right)^2=13.6a</math>, we get <math>a=\frac58</math>.
-Therefore, the area of <math>ABCD=1360a=\boxed{850}</math>
-==Lazy Solution==
 <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>.
 Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to <math>EG</math> and <math>FH</math>  must be a multiple of <math>17</math>. All of these triples are primitive:
@@ Line 93: / Line 91: @@
 -Alexlikemath
+==Video Solution==
+https://youtu.be/Kcug2ALOjkA?si=VoImhnX5rAKhprgk
+~MathProblemSolvingSkills.com
+==Video Solution by Punxsutawney Phil==
+https://youtube.com/watch?v=wrxET2c0ZgU
 == See also ==
 {{AIME box|year=2014|n=I|num-b=12|num-a=14}}
 {{MAA Notice}}

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

Page

Toolbox

Search