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Difference between revisions of "2015 AIME I Problems/Problem 7"
Line 28: Line 28:
 
<math>\Rightarrow \frac{-21}{2}\sqrt{11} + \frac{20a\sqrt5}{7} = 2\sqrt5a</math>
 
<math>\Rightarrow \frac{-21}{2}\sqrt{11} + \frac{20a\sqrt5}{7} = 2\sqrt5a</math>
  
−
<math>\Rightarrow -21\sqrt{11} = \sqrt5a\frac{14 - 20}{7}</math>
+
<math>\Rightarrow -21\sqrt{11} = 2\sqrt5a\frac{14 - 20}{7}</math>
  
−
<math>\Rightarrow \frac{49\sqrt{11}}{2} = \sqrt5a</math>
+
<math>\Rightarrow \frac{49\sqrt{11}}{4} = \sqrt5a</math>
  
−
<math>\Rightarrow \frac{7\sqrt{11}} = \frac{4a\sqrt{5}}{7}</math>
+
<math>\Rightarrow 7\sqrt{11} = \frac{4a\sqrt{5}}{7}</math>
 +
 
 +
So our final answer is <math>(7\sqrt{11})^2 = \boxed{539}</math>
  
  

Revision as of 17:23, 20 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 $FGHJ$.

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 $\triangle CDE ~ \triangle JFC$ (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.

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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