# 2010 USAMO Problems/Problem 4

## Problem

Let $ABC$ be a triangle with $\angle A = 90^{\circ}$. Points $D$ and $E$ lie on sides $AC$ and $AB$, respectively, such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$. Segments $BD$ and $CE$ meet at $I$. Determine whether or not it is possible for segments $AB, AC, BI, ID, CI, IE$ to all have integer lengths.

## Solution

We know that angle $BIC = 135^{\circ}$, as the other two angles in triangle $BIC$ add to 45^{\circ} $. Assume that only$AB, BC, BI $, and$CI $are integers. Using the [[Law of Cosines]] on triangle BIC,$BC^2 = BI^2 + CI^2 - 2BI*CI*cos 135^{\circ} $. Observing that$BC^2 = AB^2 + AC^2 $and that$cos 135^{\circ} = -\frac{\sqrt{2}}{2} $, we have$AB^2 + AC^2 - BI^2 - CI^2 = BI*CI*\sqrt{2} (Error compiling LaTeX. ! Missing $inserted.)\sqrt{2} = \frac{AB^2 + AC^2 - BI^2 - CI^2}{BI*CI} $Since the right side of the equation is a rational number, the left side (i.e.$\sqrt{2} $) must also be rational. Obviously since$\sqrt{2} $is irrational, this claim is false and we have a contradiction. Therefore, it is impossible for$AB, BC, BI $, and$CI$ to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.