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.

Start the construction at the vertex A with angles Alpha1 and Alpha2 (Alpha1+Alpha2=Alpha) created by the altitude AD. Let Beta and Gamma be the other two acute angles. The law of cosines aplied to the inscribed triangle gives
|EF|^2 = |DE|^2+|DF|^2-2|DE||DF|Cos(Beta+Gamma).
Substituting Beta+Game=Pi-Alpha1-Alpha2, |AD|Sin(Alpha1)=|DE| and |AD|Sin(Alpha2)=|DF| and simplifying (Ha Ha) gives
|EF|^2 = (|AD|Sin(Alpha))^2
|EF|^2 = (|AD||BC|)^2 (Sin(Alpha)/|BA|)^2
|EF|^2 = (2*Area)^2 (Sin(Alpha)/|BA|)^2

Note that the right hand side is constant; by the law of sines, regardless of the beginning vertex the second factor is constant.

Posted by owl on 2005-12-09 10:40:05