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Difference between revisions of "2015 AIME I Problems/Problem 7"
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==Problem==
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==Solution 1: Pythagorean Bash==
  
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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  
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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:
  
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<math>JC = \sqrt5 * FC = \sqrt5 * FJ/2 = \frac{b\sqrt 5}{2}</math> and <math>BJ = \frac{1}{\sqrt5} * HJ = \frac{b}{\sqrt5}</math>.
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<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>.
  
 
Since <math>BC = CJ + JC</math>,  
 
Since <math>BC = CJ + JC</math>,  
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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>.
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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.
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This gives that <math>AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}</math> and <math>ME = \sqrt5 \cdot MP = \sqrt5 \cdot (ML - EG) = \sqrt5 \cdot (ML - (EC - (GF + FC))</math>

Revision as of 17:57, 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 1: Pythagorean Bash

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 $ME = \sqrt5 \cdot MP = \sqrt5 \cdot (ML - EG) = \sqrt5 \cdot (ML - (EC - (GF + FC))$

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