In the triangle ABC, AB = AC. A circle is tangent internally to the circumcircle of the triangle and also to AB, AC at P, Q respectively. Prove that the midpoint of PQ is the center of the incircle of the triangle.
Solution
It is not a good idea to get bogged down in complicated formulae for the various radii. The solution is actually simple.
By symmetry the midpoint, M, is already on the angle bisector of A, so it is sufficient to show it is on the angle bisector of B. Let the angle bisector of A meet the circumcircle again at R. AP is a tangent to the circle touching AB at P, so ∠PRQ = ∠APQ = ∠ABC. Now the quadrilateral PBRM is cyclic because the angles PBR, PMR are both 90o. Hence ∠PBM = ∠PRM = (∠PRQ)/2, so BM does indeed bisect angle B as claimed.
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
12 Oct 1998