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Difference between revisions of "2015 AIME I Problems/Problem 7"

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==Solution==
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==Solution 1==
 +
 
 +
Let us find the proportion of the side length of <math>KLMN</math> and <math>FJGH</math>. Let the side length of <math>KLMN=y</math> and the side length of <math>FJGH=x</math>.
 +
 
 +
<asy>
 +
pair A,B,C,D,E,F,G,H,J,K,L,M,N,P;
 +
B=(0,0);
 +
real m=7*sqrt(55)/5;
 +
J=(m,0);
 +
C=(7*m/2,0);
 +
A=(0,7*m/2);
 +
D=(7*m/2,7*m/2);
 +
E=(A+D)/2;
 +
H=(0,2m);
 +
N=(0,2m+3*sqrt(55)/2);
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G=foot(H,E,C);
 +
F=foot(J,E,C);
 +
draw(A--B--C--D--cycle);
 +
draw(C--E);
 +
draw(G--H--J--F);
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pair X=foot(N,E,C);
 +
M=extension(N,X,A,D);
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K=foot(N,H,G);
 +
L=foot(M,H,G);
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draw(K--N--M--L);
 +
P=foot(E,M,L);
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draw(P--E);
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label("$A$",A,NW);
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label("$B$",B,SW);
 +
label("$C$",C,SE);
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label("$D$",D,NE);
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label("$E$",E,dir(90));
 +
label("$F$",F,NE);
 +
label("$G$",G,NE);
 +
label("$H$",H,W);
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label("$J$",J,S);
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label("$K$",K,SE);
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label("$L$",L,SE);
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label("$M$",M,dir(90));
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label("$N$",N,dir(180));
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label("$P$",P,dir(235)); </asy>
 +
 
 +
 
 +
Now, examine <math>BC</math>. We know <math>BC=BJ+JC</math>, and triangles <math>\Delta BHJ</math> and <math>\Delta JFC</math> are similar to <math>\Delta EDC</math> since they are <math>1-2-\sqrt{5}</math> triangles. Thus, we can rewrite <math>BC</math> in terms of the side length of <math>FJGH</math>.
 +
<cmath>BJ=\frac{1}{\sqrt{5}}HJ=\frac{x}{\sqrt{5}}=\frac{x\sqrt{5}}{5}, JC=\frac{\sqrt{5}}{2}JF=\frac{x\sqrt{5}}{2}\Rightarrow BC=\frac{7x\sqrt{5}}{10}</cmath>
 +
 
 +
Now examine <math>AB</math>. We can express this length in terms of <math>x,y</math> since <math>AB=AN+NH+HB</math>. By using similar triangles as in the first part, we have
 +
<cmath>AB=\frac{1}{\sqrt{5}}y+\frac{\sqrt{5}}{2}y+\frac{2}{\sqrt{5}}x</cmath>
 +
<cmath>AB=BC\Rightarrow \frac{7y\sqrt{5}}{10}+\frac{2x\sqrt{5}}{5}=\frac{7x\sqrt{5}}{10}\Rightarrow \frac{7y\sqrt{5}}{10}=\frac{3x\sqrt{5}}{10}\Rightarrow 7y=3x</cmath>
 +
 
 +
Now, it is trivial to see that <math>[FJGH]=\left(\frac{x}{y}\right)^2[KLMN]=\left(\frac{7}{3}\right)^2[KLMN]=\boxed{539}.\Box</math>
 +
 
 +
==Solution 2==
  
 
We begin by denoting the length <math>ED</math> <math>a</math>, giving us <math>DC = 2a</math> and <math>EC = a\sqrt5</math>. Since angles <math>\angle DCE</math> and <math>\angle FCJ</math> are complementary, we have that <math>\triangle CDE \sim \triangle JFC</math> (and similarly the rest of the triangles are <math>1-2-\sqrt5</math> triangles). We let the sidelength of <math>FGHJ</math> be <math>b</math>, giving us:  
 
We begin by denoting the length <math>ED</math> <math>a</math>, giving us <math>DC = 2a</math> and <math>EC = a\sqrt5</math>. Since angles <math>\angle DCE</math> and <math>\angle FCJ</math> are complementary, we have that <math>\triangle CDE \sim \triangle JFC</math> (and similarly the rest of the triangles are <math>1-2-\sqrt5</math> triangles). We let the sidelength of <math>FGHJ</math> be <math>b</math>, giving us:  

Revision as of 21:48, 5 March 2019

Problem

7. In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$, and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$.

[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy]

Solution 1

Let us find the proportion of the side length of $KLMN$ and $FJGH$. Let the side length of $KLMN=y$ and the side length of $FJGH=x$.

[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L); draw(P--E); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); label("$P$",P,dir(235)); [/asy]


Now, examine $BC$. We know $BC=BJ+JC$, and triangles $\Delta BHJ$ and $\Delta JFC$ are similar to $\Delta EDC$ since they are $1-2-\sqrt{5}$ triangles. Thus, we can rewrite $BC$ in terms of the side length of $FJGH$. \[BJ=\frac{1}{\sqrt{5}}HJ=\frac{x}{\sqrt{5}}=\frac{x\sqrt{5}}{5}, JC=\frac{\sqrt{5}}{2}JF=\frac{x\sqrt{5}}{2}\Rightarrow BC=\frac{7x\sqrt{5}}{10}\]

Now examine $AB$. We can express this length in terms of $x,y$ since $AB=AN+NH+HB$. By using similar triangles as in the first part, we have \[AB=\frac{1}{\sqrt{5}}y+\frac{\sqrt{5}}{2}y+\frac{2}{\sqrt{5}}x\] \[AB=BC\Rightarrow \frac{7y\sqrt{5}}{10}+\frac{2x\sqrt{5}}{5}=\frac{7x\sqrt{5}}{10}\Rightarrow \frac{7y\sqrt{5}}{10}=\frac{3x\sqrt{5}}{10}\Rightarrow 7y=3x\]

Now, it is trivial to see that $[FJGH]=\left(\frac{x}{y}\right)^2[KLMN]=\left(\frac{7}{3}\right)^2[KLMN]=\boxed{539}.\Box$

Solution 2

We begin by denoting the length $ED$ $a$, giving us $DC = 2a$ and $EC = a\sqrt5$. Since angles $\angle DCE$ and $\angle FCJ$ are complementary, we have that $\triangle CDE \sim \triangle JFC$ (and similarly the rest of the triangles are $1-2-\sqrt5$ triangles). We let the sidelength of $FGHJ$ be $b$, giving us:

$JC = \sqrt5 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}$ and $BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}$.

Since $BC = CJ + BJ$,

$2a = \frac{b\sqrt 5}{2} + \frac{b}{\sqrt5}$,

Solving for $b$ in terms of $a$ yields $b = \frac{4a\sqrt5}{7}$.

We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\sqrt{11}$. We also draw the perpendicular from $E$ to $ML$ and label the point of intersection $P$:

[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L); draw(P--E); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); label("$P$",P,dir(235)); [/asy]

This gives that $AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}$ and $ME = \sqrt5 \cdot MP = \sqrt5 \cdot \frac{EP}{2} = \sqrt5 \cdot \frac{LG}{2} = \sqrt5 \cdot \frac{HG - HK - KL}{2} = \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2}$

Since $AE$ = $AM + ME$, we get

$2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2} = a$

$\Rightarrow 12\sqrt{11} + 5(\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}) = 2\sqrt5a$

$\Rightarrow \frac{-21}{2}\sqrt{11} + \frac{20a\sqrt5}{7} = 2\sqrt5a$

$\Rightarrow -21\sqrt{11} = 2\sqrt5a\frac{14 - 20}{7}$

$\Rightarrow \frac{49\sqrt{11}}{4} = \sqrt5a$

$\Rightarrow 7\sqrt{11} = \frac{4a\sqrt{5}}{7}$

So our final answer is $(7\sqrt{11})^2 = \boxed{539}$

See also

2015 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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