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

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− | == | + | ==Solution 1: Pythagorean Bash== |

− | 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 ~ \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 | + | <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>. | ||

+ | |||

+ | 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 16:57, 20 March 2015

## Problem

7. In the diagram below, is a square. Point is the midpoint of . Points and lie on , and and lie on and , respectively, so that is a square. Points and lie on , and and lie on and , respectively, so that is a square. The area of is 99. Find the area of .

## Solution 1: Pythagorean Bash

We begin by denoting the length , giving us and . Since angles and are complimentary, we have that (and similarly the rest of the triangles are triangles). We let the sidelength of be , giving us:

and .

Since ,

,

Solving for in terms of yields .

We now use the given that , implying that . We also draw the perpendicular from E to ML and label the point of intersection P.

This gives that and