Try the solution of problem from RMO (Regional Mathematical Olympiad) 2015 Problem 1 based on Triangle.

**Problem: Triangle Problem**

Two circles and , with centers O and O’, respectively, are such that O’ lies on . Let A be a point on , and let M be the midpoint of AO’. Let B be another point on , such that . Then prove that the midpoint of AB lies on .

**Discussion:Â **

Suppose AB intersects at C. Join O’C. Suppose it intersects OM at D. Clearly in M is the midpoint of AO’ and DM is parallel to AC. Then D is the midpoint of O’C.

Now O’C is a chord of and we have proved that D is the midpoint of it. Therefore we can say that OD is perpendicular to O’C.

Since AB is parallel to OD (OM), therefore as O’C is perpendicular to OD, therefore O’C is also perpendicular to AB. Since AB is a chord of circle and O’C is a line from center perpendicular to the chord, hence it bisects are chord implying that C is the midpoint of AB.

## One reply on “Triangle Problem | RMO 2015 Solutions Problem 1”

[…] circles and with centers and respectively, are such that lies on Let be a point on and let be the midpoint of Let be another point on such that Then prove that the midpoint of lies on SOLUTION: Here […]

Google