MathDB

4

Part of 1969 Canada National Olympiad

Problems(1)

Perpendiculars

Source: Canada 1969, Problem 4

5/14/2006
Let ABCABC be an equilateral triangle, and PP be an arbitrary point within the triangle. Perpendiculars PD,PE,PFPD,PE,PF are drawn to the three sides of the triangle. Show that, no matter where PP is chosen, PD+PE+PFAB+BC+CA=123. \frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}.
ratiogeometrytrigonometryarea of a triangleHeron's formula