1st APMO 1989 Problem 3 ABC is a triangle. X lies on the segment AB so that AX/AB = 1/4. CX intersects the median from A at A' and the median from B at B''. Points B', C', A'', C'' are defined similarly. Find the area of the triangle A''B''C'' divided by the area of the triangle A'B'C'. Solution
: 25/49. Let M be the midpoint of AB. We use vectors center G. Take GA = A , GB = B , GC = C . Then GM = A /2 + B /2 and GX = 3/4 A + 1/4 C . Hence GA' = 2/5 A (showing it lies on GA) = 4/5 (3/4 A + 1/4 B ) + 1/5 C , since A + B + C = 0 (which shows it lies on CX). Similarly, GB'' = 4/7 (1/2 A + 1/2 C ) (showing it lies on the median through B) = 2/7 A + 2/7 C = 5/7 (2/5 A ) + 2/7 C (showing it lies on CA' and hence on CX). Hence GB'' = -2/7 B . So we have shown that GB'' is parallel to GB' and 5/7 the length. The same applies to the distances from the centroid to the other vertices. Hence triangle A''B''C'' is similar to triangle A'B'C' and its area is 25/49 times the area of A'B'C'. 1st APMO 1989 © John Scholes jscholes@kalva.demon.co.uk 11 Apr 2002