2010 USAMO Problems/Problem 4
Revision as of 23:51, 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, …')
Solution
We know that angle 
, as the other two angles in triangle 
add to \frac{\pi}{4}![$. Assume that only AB, BC, BI, and CI are integers. Using the [[Law of Cosines]] on triangle BIC,$](http://latex.artofproblemsolving.com/b/8/5/b85a23c36b0637bb5a384110dd0081fe6b0fb068.png)
BC^2 = BI^2 + CI^2 - 2BI*CI*cos \frac{3\pi}{4}
BC^2 = AB^2 + AC^2
cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}
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}
\sqrt{2}
\sqrt{2}
AB, BC, BI, and CI$ to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.
