2010 USAMO Problems/Problem 4

Revision as of 22:51, 5 May 2010 by Aimesolver (talk | contribs) (Created page with '==Solution== We know that angle <math>BIC = \frac{3\pi}{4}</math>, as the other two angles in triangle <math>BIC</math> add to \frac{\pi}{4}<math>. Assume that only AB, BC, BI, …')
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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}$$ (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.

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