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

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==Solution== | ==Solution== | ||

− | We know that angle <math>BIC = 135^{\circ}</math>, as the other two angles in triangle <math>BIC</math> add to 45^{\circ}<math>. Assume that only < | + | We know that angle <math>BIC = 135^{\circ}</math>, as the other two angles in triangle <math>BIC</math> add to <math>45^{\circ}</math>. Assume that only <math>AB, BC, BI</math>, and <math>CI</math> are integers. Using the [[Law of Cosines]] on triangle BIC, |

− | < | + | <math>BC^2 = BI^2 + CI^2 - 2BI*CI*cos 135^{\circ}</math>. Observing that <math>BC^2 = AB^2 + AC^2</math> and that <math>cos 135^{\circ} = -\frac{\sqrt{2}}{2}</math>, we have |

− | < | + | <math>AB^2 + AC^2 - BI^2 - CI^2 = BI*CI*\sqrt{2}</math> |

− | < | + | <math>\sqrt{2} = \frac{AB^2 + AC^2 - BI^2 - CI^2}{BI*CI}</math> |

− | Since the right side of the equation is a rational number, the left side (i.e. < | + | Since the right side of the equation is a rational number, the left side (i.e. <math>\sqrt{2}</math>) must also be rational. Obviously since <math>\sqrt{2}</math> is irrational, this claim is false and we have a contradiction. Therefore, it is impossible for <math>AB, BC, BI</math>, and <math>CI</math> to all be integers, which invalidates the original claim that all six lengths are integers, and we are done. |

## Revision as of 21:59, 9 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.