The altitude from one of the vertices of an acute-angled triangle ABC meets the opposite side at D. From D, perpendiculars DE and DF are drawn to the other two sides. Prove that the length EF is the same whichever vertex is chosen.

For extra credit: if a=|BC|, b=|CA|, c=|AB|, and d=|EF|, show that Area(ABC)=½√(abcd).

2002 British Mathematical Olympiad, Round 2, Problem 1.

Just to get the ball started, if we pick a right triangle, and let B be the right angle, EF is the height of the triangle with respect to AC, so AC.EF/2= area(ABC). [It's easy to see that if we had picked A to be the right angle, EF also equals the same height.]

So... AB.BC/2= AC.EF/2= area(ABC), then AB.BC.AC.EF/4= area^2, and the result follows.

Posted by Federico Kereki on 2005-12-08 19:58:14