Many years ago I solved the even more general problem, and discovered that when stated slightly differently the formula comes out beautifully. Given triangle ABC, A' divides side BC in the ratio of m:1, B' divides CA in the ratio n:1, and C' divides AB in the ratio p:1. Now draw cevians AA', BB', and CC'. The ratio of the area of the center triangle to the area of the original triangle is: f(m,n,p) = (mnp-1)^2/((mn+n+1)(mp+m+1)(np+p+1)) Checking, well known cases: f(2,2,2) = 1/7, and f(p,p,p) = (p-1)^3/(p^3-1) (equivalent to but nicer than the formula in the Feynman's Triangle reference). Also the case where mnp=1 (area=0) is precisely Ceva's Theorem. I'm curious if anyone has seen this or similar result before. Nick Fred lunnon wrote:
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
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