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Difference between revisions of "2015 AIME I Problems/Problem 7"
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==Solution==
 
==Solution==
  
−
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 complimentary, we have that <math>\triangle CDE \similar \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 complimentary, we have that <math>\triangle CDE \~ \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:  
  
 
<math>JC = \sqrt5 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}</math> and <math>BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}</math>.
 
<math>JC = \sqrt5 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}</math> and <math>BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}</math>.
Line 74: Line 74:
 
L=foot(M,H,G);
 
L=foot(M,H,G);
 
draw(K--N--M--L);
 
draw(K--N--M--L);
−
P=foot(E,M,L)
+
P=foot(E,M,L);
−
label("P",P,E)
+
label("$P$",P,E);
 
label("$A$",A,NW);
 
label("$A$",A,NW);
 
label("$B$",B,SW);
 
label("$B$",B,SW);

Revision as of 19:01, 21 March 2015

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

We begin by denoting the length $ED$ $a$, giving us $DC = 2a$ and $EC = a\sqrt5$. Since angles $\angle DCE$ and $\angle FCJ$ are complimentary, we have that $\triangle CDE \~ \triangle JFC$ (Error compiling LaTeX. Unknown error_msg) (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 $BC = CJ + JC$,

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

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; 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); label("$P$",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)); [/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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