MathDB
Perpendiculars

Source: Canada 1969, Problem 4

May 14, 2006
ratiogeometrytrigonometryarea of a triangleHeron's formula

Problem Statement

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}}.
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