ABC is an arbitary triangle, and M is the midpoint of BC. How many pieces are needed to dissect AMB into triangles which can be reassembled to give AMC?
Let N be the midpoint of AB and L the midpoint of AC. Cut AMB along MN into AMN and BMN. Then AMN is congruent to MAL and BNM is congruent ot MLC, so the two pieces can be reassembled to AMC. Clearly one piece is not sufficient unless AB = AC.
43rd Putnam 1982
© John Scholes
16 Jan 2001