On 6/9/07, Dan Asimov <dasimov@earthlink.net> wrote:
A slightly more general problem, and fun to solve, is:
Suppose 0 <= p <= 1. Given a triangle ABC, mark the 3 points that are the fraction p of the way from A to B; from B to C; and from C to A. Now draw the segment from each of A,B,C to the marked point on the opposite side.
Express the area of the interior triangle thus created, as a fraction F(p) of the area of ABC ? (As a check, F(1/3) = 1/7.)
And a still more general problem: draw three lines which cut sides AB, BC, CA in (directed) ratios p,q,r; q,r,p; r,p,q respectively. Then the ratio of the areas of secondary to primary triangle depends only on p,q,r [where by Menelaus' theorem, p q r + (1 - p)(1 - q)(1 - r) = 0]. Dan Asimov mentions the special case q = 0, r = 1, where the lines meet the vertices A,B,C; deVilliers also mentions the case q = p, r = (1-p)^2/(1-2p+2p^2), where the secondary triangle touches the sides of the primary; also a pair of analogues for parallelograms. But he doesn't actually mention the general case --- I don't know whether he was aware of it, nor an expression for the general ratio! Fred Lunnon