On Saturday 09 June 2007 16:33, Dan Asimov wrote:
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.)
Working in coordinates (taking ABC to be, say, the canonical isosceles right triangle) yields the answer pretty quickly, especially if you notice at the outset that it obviously has to be a function of pq where q=1-p. But it's rather pedestrian and doesn't particularly give me the feeling of understanding *why* the answer is what it is. Is there a better solution? (Or you can appeal to something in the Ceva-Menelaus-etc. complex, but that's kinda cheating.) I suspect there's a rabbit-out-of-hat proof that finds an obvious reason why the degrees of the numerator and denominator (as polynomials in pq) have to be what they are, and then just checks a couple of obvious values -- e.g. when pq=1/4 we have to get 0, and when pq=0 we have to get 1. But I don't currently see what that obvious reason might be. -- g