ERic wroter:
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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?
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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