14th Balkan 1997

ABCD is a convex quadrilateral. X is a point inside it. XA² + XB² + XC² + XD² is twice the area of the quadrilateral. Show that it is a square and that X is its center.

Solution

XA² + XB² ≥ 2·XA·XB with equality iff XA = XB. But XA·XB ≥ 2 area XAB with equality iff ∠AXB = 90^o. Hence (XA² + XB²) + (XB² + XC²) + (XC² + XD²) + (XD² + XA²) ≥ 4 (area XAB + area XBC + area XCD + area XDA) with equality iff ∠AXB = ∠BXC = ∠CXD = ∠DXA = 90^o and XA = XB = XC = XD. But we are given that equality holds. Hence AXC and BXD are straight line segments with X as their midpoint. Hence ABCD is a square center X.

14th Balkan 1997

© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002