IMO 1982 SL 13

Problem

C1 (NET 1)IMO2 A scalene triangle A1A2A3 is given with sides a1, a2, a3 (ai is the side opposite to Ai). For all i = 1, 2, 3, Mi is the midpoint of side ai, Ti is the point where the incircle touches side ai, and the reflection of Ti in the interior bisector of Ai yields the point Si. Prove that the lines M1S1, M2S2, and M3S3 are concurrent.

Solution

Points S1, S2, S3 clearly lie on the inscribed circle. Let : XY denote the oriented arc XY . The arcs  T2S1 and  T1T3 are equal, since they are symmetric with respect to the bisector of \angleA1. Similarly,  T3T2 =  S2T1. Therefore  T3S1 =  T3T2 +  T2S1

 S2T1 +  T1T3

 S2T3. It follows that S1S2 is parallel to A1A2, and consequently S1S2 \parallel M1M2. Analogously S1S3 \parallelM1M3 A1 A2 A3 S1 S2 S3 T1 T2 T3 and S2S3 \parallelM2M3. Since the circumcircles of \triangleM1M2M3 and \triangleS1S2S3 are not equal, these triangles are not congruent and hence they must be homothetic. Then all the lines MiSi pass through the center of homothety. Second solution. Set the complex plane so that the incenter of \triangleA1A2A3 is the unit circle centered at the origin. Let ti, si respectively denote the complex numbers of modulus 1 corresponding to Ti, Si. Clearly t1t1 =

t2t2 = t3t3 = 1. Since T2T3 and T1S1 are parallel, we obtain t2t3 = t1s1, or s1 = t2t3t1. Similarly s2 = t1t3t2, s3 = t1t2t3, from which it follows that s2 −s3 = t1(t3t2 −t2t3). Since the number in parentheses is strictly imaginary, we conclude that OT1 \perpS2S3 and consequently S2S3 \parallelA2A3. We proceed as in the first solution.