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Show that D^2-H^2 \geq P^2/4

Source: Iran Pre-Preparation Course Examination 1997, E3, P5

March 29, 2011
geometryperimetergeometry unsolved

Problem Statement

Let OO be a point in the plane and let FF be a (not necessary convex) polygon. Let PP be the perimeter of FF, let DD be sum of the distances of the point OO from the vertices of FF, and let HH be sum of the distances of the point OO from the lines that pass through the vertices of FF. Show that D2−H2≥P24.D^2-H^2 \geq \frac{P^2}{4}.
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