2010 USAMO Problems/Problem 4

Revision as of 22:51, 5 May 2010 by Aimesolver (talk | contribs) (Solution)


We know that angle $BIC = \frac{3\pi}{4}$, as the other two angles in triangle $BIC$ add to $\frac{\pi}{4}$. Assume that only AB, BC, BI, and CI are integers. Using the Law of Cosines on triangle BIC,

$BC^2 = BI^2 + CI^2 - 2BI*CI*cos \frac{3\pi}{4}$. Observing that $BC^2 = AB^2 + AC^2$ and that $cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}$, we have

$AB^2 + AC^2 - BI^2 - CI^2 = BI*CI*\sqrt{2}$

$\sqrt{2} = \frac{AB^2 + AC^2 - BI^2 - CI^2}{BI*CI}$

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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