A_{0}B_{0}C_{0} and A_{1}B_{1}C_{1} are acute-angled triangles. Construct the triangle ABC with the largest possible area which is circumscribed about A_{0}B_{0}C_{0} (BC contains A_{0}, CA contains B_{0}, and AB contains C_{0}) and similar to A_{1}B_{1}C_{1}.

**Solution**

Take any triangle similar to A_{1}B_{1}C_{1} and circumscribing A_{0}B_{0}C_{0}. For example, take an arbitrary line through A_{0} and then lines through B_{0} and C_{0} at the appropriate angles to the first line. Label the triangle's vertices X, Y, Z so that A_{0} lies on YZ, B_{0} on ZX, and C_{0} on XY. Now any circumscribed ABC (labeled with the same convention) must have C on the circle through A_{0}, B_{0} and Z, because it has ∠C = ∠Z = ∠C_{1}. Similarly it must have B on the circle through C_{0}, A_{0} and Y, and it must have A on the circle through B_{0}, C_{0} and X.

Consider the side AB. It passes through C_{0}. Its length is twice the projection of the line joining the centers of the two circles onto AB (because each center projects onto the midpoint of the part of AB that is a chord of its circle). But this projection is maximum when it is parallel to the line joining the two centers. The area is maximised when AB is maximised (because all the triangles are similar), so we take AB parallel to the line joining the centers. [Note, in passing, that this proves that the other sides must also be parallel to the lines joining the respective centers and hence that the three centers form a triangle similar to A_{1}B_{1}C_{1}.]

Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

© John Scholes

jscholes@kalva.demon.co.uk

29 Sep 1998

Last corrected/updated 27 Oct 2002