# Difference between revisions of "2010 USAMO Problems/Problem 4"

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==Problem== | ==Problem== | ||

− | Let <math>ABC</math> be a triangle with <math>\angle A = 90^{\ | + | Let <math>ABC</math> be a triangle with <math>\angle A = 90^{\circ}</math>. Points <math>D</math> |

and <math>E</math> lie on sides <math>AC</math> and <math>AB</math>, respectively, such that <math>\angle | and <math>E</math> lie on sides <math>AC</math> and <math>AB</math>, respectively, such that <math>\angle | ||

ABD = \angle DBC</math> and <math>\angle ACE = \angle ECB</math>. Segments <math>BD</math> and | ABD = \angle DBC</math> and <math>\angle ACE = \angle ECB</math>. Segments <math>BD</math> and |

## Revision as of 11:48, 6 May 2010

## Problem

Let be a triangle with . Points and lie on sides and , respectively, such that and . Segments and meet at . Determine whether or not it is possible for segments to all have integer lengths.

## Solution

We know that angle , as the other two angles in triangle add to . Assume that only , and are integers. Using the Law of Cosines on triangle BIC,

. Observing that and that , we have

Since the right side of the equation is a rational number, the left side (i.e. ) must also be rational. Obviously since is irrational, this claim is false and we have a contradiction. Therefore, it is impossible for , and to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.