On 6/16/07, Nick Baxter <nickb@baxterweb.com> wrote:
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.
All of which prompts the thought that the "real" theorem simply expresses the ratio of the areas of two general triangles A_1 A_2 A_3, B_1 B_2 B_3, as a function of the linear ratios p_ij in which side i of A cuts side j of B; or alternatively q_ij in which B cuts A. Re-scaling as Nick suggests might improve the aesthetics. As before, setting area ratio to 0 gives a (probably not terribly interesting) generalisation of Ceva. And while we're at it, what is the relation between the p_ij and q_ij? This is all starting to resemble barycentric coordinates ... WFL