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Math Magazine Problem 1564 Solution
by Peter Y. Woo, Biola University
4/18/99
Problem. (by Wu Wei Chao, Guang Zhou Normal Univ., China) Let a convex quadrilateral ABCD satisfy ÐBAC + ÐBDC = 180°, and A is closer to line BC than D is. Let AC intersect BD at P. Prove that (AC/BD)2 > (AP . CD) / (DP . AB) .
Proof. Let [K] denote the area of any polygon K.
Let a, d, p, q represent the areas [PAB], [PCD], [PBC], and [PDA] respectively. We shall prove a < d and q < p.
Observe ÐBAD + ÐADC > ÐBAC + ÐBDC = p. Hence lines BA and CD will intersect at some point Q on the same side of line BC as A and D.
\A is closer to line CD than B is. \ [ACD] < [BCD]. \q + d < p + d. \ q < p.
Then PA × PD < PB × PC . \ PA / PC < PB / PD.
\ a / p < p / d, or p2 - a d > 0.
Now A is closer to line BC than D is.
\ [ABC] < [DBC]. \ a + p < d + p, or a < d.
\ (d - a) (p2 - a d) > 0, which is equivalent to d (p + a)2 > a (p + d)2.
\ d / (p + d) > (a / (p + a)) × (p + d) / (p + a).
\ DP / BD > (AP / AC) ([BCD] / [BCA])
= (AP / AC) (BD . CD sin ÐBDC) / (AC . AB sin ÐBAC)
= (AP . BD . CD) / (AC . AC . AB) because sin ÐBDC = sin ÐBAC.
\ (AC / BD)2 > (AP . CD) / (DP . AB). QED
