China Mathematical Competition

Question

Given ABC and AEF such that B is the midpoint of EF. Also, AB = EF = 1, BC = 6, CA = 33, and AB AE + AC AF = 2. The cosine of the angle between EF and BC is ______.

Accepted Answer

We have aligned 2 &= AB AE + AC AF &= AB (AB + BE) + AC (AB + BF), aligned i.e. AB^2 + AB BE + AC AB + AC BF = 2. As AB^2 = 1, AC AB = 33 1 33+1-362 33 1 = -1 and BE = -BF, we get 1 + BF (AC - AB) - 1 = 2, i.e. BF BC = 2. Defining as the angle between EF and BC, we get |BF| |BC| = 2 or 3 = 2. So = 23.