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

(→‎Problem)
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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 ~ \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 \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:  
  
 
<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 51: Line 51:
 
Solving for <math>b</math> in terms of <math>a</math> yields <math>b = \frac{4a\sqrt5}{7}</math>.
 
Solving for <math>b</math> in terms of <math>a</math> yields <math>b = \frac{4a\sqrt5}{7}</math>.
  
−
We now use the given that <math>[KLMN] = 99</math>, implying that <math>KL = LM = MN = NK = 3\sqrt{11}</math>. We also draw the perpendicular from E to ML and label the point of intersection P.
+
We now use the given that <math>[KLMN] = 99</math>, implying that <math>KL = LM = MN = NK = 3\sqrt{11}</math>. 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 <math>AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}</math>  
 
This gives that <math>AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}</math>  

Revision as of 19:00, 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 $\triangle CDE \similar \triangle JFC$ (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: 

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));  (Error making remote request. Unknown error_msg)

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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