In the classic problem you are given a triangle ABC with points D on AB, E on BC, and F on AC such that |AD|=2|DB|, |BE|=2|EC|, and |CF|=2|FA|. The lines AE, BF, and CD enclose a triangle inside triangle ABC. You are to find the area of this enclosed triangle relative to that of ABC. The answer is 1/7.
What if everything is the same except |BE|=|EC| and |CF|=3|FA|. What is the area of the enclosed triangle relative to that of ABC?
If x = |AD|/|DB| = 2, y = |BE|/|EC| = 1, and z = |CF|/|FA|= 3; then
the ratio of the area of the central triangle to the area of the
triangle is given by Routh's Theorem,
(xyz - 1)^2/[(xy + x + 1)(yz + y + 1)(zx + z + 1)]
= (2*1*3 - 1)^2/[(2*1 + 2 + 1)(1*3 + 1 + 1)(3*2 + 3 + 1)
= 1/10
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Posted by Bractals
on 2005-08-24 04:26:42 |
Solution
