ERic wroter: << Draw a triangle ABC (clockwise labels) with each side devided into 3 equal parts (labels A,A1,A2,B,B1,B2,C,C1,C2): AA1=A1A2=A2B, BB1=B1B2=B2C, CC1=C1C2=C2A. Now draw AB1, BC1, CA1 -- those lines shape an "inside" triangle "I" having no common point with ABC. Question: Does an ABC triangle exist such that the surface of the inner "I" triangle is exactly 1/7th of ABC's surface?
This looks to me like putting the answer before the question! 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.) --Dan