A non-isosceles triangle A1A2A3 has sides a1, a2, a3 with ai opposite Ai. Mi is the midpoint of side ai and Ti is the point where the incircle touches side ai. Denote by Si the reflection of Ti in the interior bisector of ∠Ai. Prove that the lines M1S1, M2S2 and M3S3 are concurrent.
Solution
Let Bi be the point of intersection of the interior angle bisector of the angle at Ai with the opposite side. The first step is to figure out which side of Bi Ti lies. Let A1 be the largest angle, followed by A2. Then T2 lies between A1 and B2, T3 lies between A1 and B3, and T1 lies between A2 and B1. For ∠OB2A1 = 180o - A1 - A2/2 = A3 + A2/2. But A3 + A2/2 < A1 + A2/2 and their sum is 180o, so A3 + A2/2 < 90o. Hence T2 lies between A1 and B2. Similarly for the others.
Let O be the center of the incircle. Then ∠T1OS2 = ∠T1OT2 - 2 ∠T2OB2 = 180o - A3 - 2(90o - ∠OB2T2) = 2(A3 + A2/2) - A3 = A2 + A3. A similar argument shows ∠T1OS3 = A2 + A3. Hence S2S3 is parallel to A2A3.
Now ∠T3OS2 = 360o - ∠T3OT1 - ∠T1OS2 = 360o - (180o - A2) - (A2 + A3) = 180o - A3 = A1 + A2. ∠T3OS1 = ∠T3OT1 + 2 ∠T1OB1 = (180o - A2) + 2(90o - ∠OB1T1) = 360o - A2 - 2(A3 + A1/2) = 2(A1 + A2 + A3) - A2 - 2A3 - A1 = A1 + A2 = ∠T3OS2. So S1S2 is parallel to A1A2. Similarly we can show that S1S3 is parallel to A1A3.
So S1S2S3 is similar to A1A2A3 and turned through 180o. But M1M2M3 is also similar to A1A2A3 and turned through 180o. So S1S2S3 and M1M2M3 are similar and similarly oriented. Hence the lines through corresponding vertices are concurrent.
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