O is a point inside the convex quadrilateral ABCD. The line through O parallel to AB meets the side BC at L and the line through O parallel to BC meets the side AB at K. The line through O parallel to AD meets the side CD at M and the line through O parallel to CD meets the side DA at N. The area of ABCD is k the area of AKON is k1 and the area of LOMC is k2. Show that k1/2 ≥ k11/2 + k21/2.
Solution
If O lies on AC, then ABCD, AKON, OLCM are similar, so the square roots of their areas are proportional to AC, AO and OC. But AC = AO + OC. Hence k1/2 = k11/2 + k21/2. So we may assume O does not lie on AC. Without loss of generality it lies on the same side of AC as D. So C and D are on opposite sides of the line AO. Rotate this line about O until it passes through C. Suppose that in the general position (during the rotation) it intersects the ray BA at W, AD at X, CD at Y and the ray BC at Z.
In the starting position W = X = A, so OW/OX = 1, but OZ/OY > 1. In the final position, OW/OX > 1 and OZ/OY = 1. So for some intermediate position OW/OX = OZ/OY. Take this position. ABCD is partitioned into 4 quadrilaterals: AKON, LOMC, BKOL and ONDM. Let area BKOL = T1, area ONDM = T2. AKON is divided into AKOX and XON. Let area AKOX = P1, area XON = Q1. LOMC is divided into LOYC and OYM. Let area LOYC = P2, area OYM = Q2. The required inequality is equivalent to k ≥ k1 + k2 + 2k11/2k21/2 and hence to T1 + T2 ≥ 2k11/2k21/2.
WBZ, WKO and OLZ are similar, so P11/2 = (WO/WZ) (P1 + P2 + T1) and P21/2 = (OZ/WZ) (P1 + P2 + T1). Hence P11/2 + P21/2 = (P1 + P2 + T1). Squaring: T1 = 2 (P1P2)1/2. Similarly, T2 = 2 (Q1Q2)1/2. Now P1/P2 = OW2/OZ2 = OX2/OY2 (by the assumption about the position of the line) = Q1/Q2. So put k = Q1/P1 = Q2/P2. Then T1 + T2 = 2 (P1P2)1/2 (1 + k) = 2( (1+k)P1(1+k)P2)1/2 = 2( (P1 + Q1)(P2 + Q2) )1/2 >= 2( k1k2)1/2.
© John Scholes
jscholes@kalva.demon.co.uk
9 Oct 2002