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.