# Difference between revisions of "2013 AIME I Problems/Problem 3"

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+ | == Problem 3 == | ||

+ | Let <math>ABCD</math> be a square, and let <math>E</math> and <math>F</math> be points on <math>\overline{AB}</math> and <math>\overline{BC},</math> respectively. The line through <math>E</math> parallel to <math>\overline{BC}</math> and the line through <math>F</math> parallel to <math>\overline{AB}</math> divide <math>ABCD</math> into two squares and two nonsquare rectangles. The sum of the areas of the two squares is <math>\frac{9}{10}</math> of the area of square <math>ABCD.</math> Find <math>\frac{AE}{EB} + \frac{EB}{AE}.</math> | ||

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+ | == Solution == | ||

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It's important to note that <math>\dfrac{AE}{EB} + \dfrac{EB}{AE}</math> is equivalent to <math>\dfrac{AE^2 + EB^2}{(AE)(EB)}</math> | It's important to note that <math>\dfrac{AE}{EB} + \dfrac{EB}{AE}</math> is equivalent to <math>\dfrac{AE^2 + EB^2}{(AE)(EB)}</math> | ||

## Revision as of 22:58, 15 March 2013

## Problem 3

Let be a square, and let and be points on and respectively. The line through parallel to and the line through parallel to divide into two squares and two nonsquare rectangles. The sum of the areas of the two squares is of the area of square Find

## Solution

It's important to note that is equivalent to

We define as the length of the side of larger inner square, which is also , as the length of the side of the smaller inner square which is also , and as the side length of . Since we are given that the sum of the areas of the two squares is of the the area of ABCD, we can represent that as . The sum of the two nonsquare rectangles can then be represented as .

Looking back at what we need to find, we can represent as . We have the numerator, and dividing by two gives us the denominator . Dividing gives us an answer of .