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. Unknown error_msg)\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.

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2010 USAMO Problems/Problem 4

Problem

Solution