# 2010 USA Mathematical Olympiad, Problem 4

Let ABC be a triangle with ∠A = 90°. Points D and E lie on sides AC and AB, respectively, such that ∠ABD = ∠DBC and ∠ACE = ∠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 by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Jim Krause) Applying the law of the cosines, we have

BC² = BI² + CI² - 2 BI × CI cos∠BIC

But ∠BIC = 180° - ½ (180° - ∠A) = 135° and cos∠BIC = - ½ \sqrt{2}

The above equation becomes BC² = BI² + CI² + \sqrt{2} BI × CI

Or \sqrt{2} BI × CI = BC² - BI² - CI²

Now assume that it is possible for segments AB, AC, BI, ID, CI, IE to all have integer lengths. BC² is then also an integer because BC² = AB² + AC² which, in turn, requires \sqrt{2} BI × CI to be an integer.

But since \sqrt{2} is an irrational number, the product of \sqrt{2} with an integer is not an integer. Therefore, our assumption is not possible, and its not possible for segments AB, AC, BI, ID, CI, IE to all have integer lengths.