Intersection in a Quadrilateral (Posted on 2010-08-14)

	Submitted by Bractals
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Let ST denote the vector from point S to point T. Let OC = -cOA and OD = -dOB, where c,d > 0. Let P be the intersection of diagonal AC and line segment EF. If OP = xOA, then O in ABEF <==> x < 0.     xOA = OP = yOE + (1-y)OF                        = y(OB+BE) + (1-y)(OA+AF)                        = y(OB+½BC) + (1-y)(OA+½AD)                        = y(OB+½[OC - OB]) + (1-y)(OA+½[OD - OA])                        = ½y(OB + OC) + ½(1-y)(OA + OD)                        = ½y(OB - cOA) + ½(1-y)(OA - dOB) Thus,    2x = -yc + (1 - y)    and    0 = y - d(1 - y). Therefore,       x = ½(1 - cd)/(1 + d) Therefore, O in ABEF <==> cd > 1. Area(COD) = ½|OC×OD| = ½|(-cOA)×(-dOB)| = cd(½|OA×OB|)                     = cd Area(AOB). Therefore, Area(AOB) < Area(COD) <==> 1 < cd <==> O in ABEF.

	Subject	Author	Date
	Vector Approach (spoiler)	Harry	2010-08-18 19:27:14