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

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label("$z$", intersectionpoint( D--P, G--H ));</asy>
 
label("$z$", intersectionpoint( D--P, G--H ));</asy>
  
== Solution ==
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== 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.  
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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>.
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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>.
  
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.
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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>
  
Let the side length of the square be <math>d=\sqrt{1360a}</math>.
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== 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>.  
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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>.  
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<asy> size(150);
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defaultpen(fontsize(9pt));
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pair A,B,C,D,E,F,G,H,I,J,L,P,Q,R,S;
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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);
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draw(A--B--C--D--cycle^^E--G^^F--H, black+1); draw(R--Q^^I--J, gray); 
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</asy>
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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>.
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Because <math>[HFJI]= HF\cdot PQ</math>, with <math>HF=34</math>, we get <math>PQ=4a</math>.
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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>.
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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>
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==Solution 2 (Lazy)==
 
 
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>.
 
<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:
 
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:
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== Solution 3==
 
== Solution 3==
  
Continue in the same way as solution 1 to get that POK has area 3a, and OK = d/10. You can then find PK has length 3/2.
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Continue in the same way as solution 1 to get that <math>POK</math> has area <math>3a</math>, and <math>OK = \frac{d}{10}</math>. You can then find <math>PK</math> has length <math>\frac 32</math>.
  
Then, if we drop a perpendicular from H to BC get L, We get HLF ~ OPK.
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Then, if we drop a perpendicular from <math>H</math> to <math>BC</math> at <math>L</math>, We get <math>\triangle HLF \sim \triangle OPK</math>.
  
Thus, <math>LF = \frac{15\cdot 34}{d}</math>, and we know HL = d, and HF = 34. Thus, we can set up an equation in terms of d using the Pythagorean theorem.
+
Thus, <math>LF = \frac{15\cdot 34}{d}</math>, and we know <math>HL = d</math>, and <math>HF = 34</math>. Thus, we can set up an equation in terms of <math>d</math> using the Pythagorean theorem.
  
 
<cmath>\frac{15^2 \cdot 34^2}{d^2} + d^2 = 34^2</cmath>
 
<cmath>\frac{15^2 \cdot 34^2}{d^2} + d^2 = 34^2</cmath>
  
<cmath>d^4 - 34^2 d^2 + 15^2 \cdot 34^2</cmath>
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<cmath>d^4 - 34^2 d^2 + 15^2 \cdot 34^2 = 0</cmath>
 +
 
 +
<cmath>(d^2 - 34 \cdot 25)(d^2 - 34 \cdot 9) = 0</cmath>
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<math>d^2 = 34 \cdot 9</math> is extraneous, so <math>d^2 = 34 \cdot 25</math>. Since the area is <math>d^2</math>, we have it is equal to <math>34 \cdot 25 = \boxed{850}</math>
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 +
-Alexlikemath
 +
 
 +
==Video Solution by Punxsutawney Phil==
  
<cmath>(d^2 - 34 \cdot 25)(d^2 - 34 \cdot 9) 0</cmath>
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https://youtube.com/watch?v=wrxET2c0ZgU
  
<math>34 \cdot 9</math> is extraneous, so <math>d^2 = 34 \cdot 25</math>. Since the area is d^2, we have it is equal to <math>34 \cdot 25 = \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 03:54, 24 November 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(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 \[[RPQ]= \tfrac 12 \cdot \tfrac 1{10}s\cos\theta = \tfrac 1{400}s^2\sin 2\theta.\]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

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 (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

Video Solution by Punxsutawney Phil

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

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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