Three circles of equal radius have a common point O and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle are collinear with the point O.
Solution
Let the triangle be ABC. Let the center of the circle touching AB and AC be D, the center of the circle touching AB and BC be E, and the center of the circle touching AC and BC be F. Because the circles center D and E have the same radius the perpendiculars from D and E to AB have the same length, so DE is parallel to AB. Similarly EF is parallel to BC and FD is parallel to CA. Hence DEF is similar and similarly oriented to ABC. Moreover D must lie on the angle bisector of A since the circle center D touches AB and AC. Similarly E lies on the angle bisector of B and F lies on the angle bisector of C. Hence the incenter I of ABC is also the incenter of DEF and acts as a center of symmetry so that corresponding points P of ABC and P' of DEF lie on a line through I with PI/P'I having a fixed ratio. But OD = OE = OF since the three circles have equal radii, so O is the circumcenter of DEF. Hence it lies on a line with I and the circumcenter of ABC.
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
14 Oct 1998