All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Variation on a Classic (Posted on 2005-08-23)
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?

 See The Solution Submitted by McWorter No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution | Comment 8 of 12 |
` `
`If x = |AD|/|DB| = 2, y = |BE|/|EC| = 1, and z = |CF|/|FA|= 3; thenthe ratio of the area of the central triangle to the area of thetriangle 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 2005-08-24 04:26:42

 Search: Search body:
Forums (0)