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=2DB, BE=2EC, and CF=2FA. 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=3FA. 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

Posted by Bractals
on 20050824 04:26:42 