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 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.)
 re(3): Independent solution? Comment 12 of 12 |
(In reply to re(2): Independent solution? by McWorter)

` `
`Routh's Theorem`
`See figure www.geocities.com/bractals/rt.jpg`
`  Area(PQR)     | <PQ> x <PR> | ----------- = -----------------  Area(ABC)     | <AB> x <AC> |`
`  <XY> denotes vector from point X to point Y.`
`------------------------------------------------------------`
`Let <AN> = x<NB>, <BL> = y<LC>, and <CM> = z<MA>.`
`             x                 y                     z    <AN> = -----<AB>, <BL> = -----<BC>, and <CM> = -----<CA>            1+x               1+y                   1+z`
`                                          h <AP> = (1-h)<AB> + h<AM> = (1-h)<AB> + -----<AC>                                         1+z`
`                             kx       = (1-k)<AC> + k<AN> = -----<AB> + (1-k)<AC>                             1+x`
`             kx           h ==>  1-h = -----  and  ----- = 1-k             1+x         1+z`
`             1+z ==>  h = --------           1+z+zx`
`             zx<AB> + <AC> ==> <AP> = ---------------                 1+z+zx`
`A similar argument gives`
`             xy<BC> + <BA>     <BQ> = ---------------                 1+x+xy`
`             yz<CA> + <CB>     <CR> = ---------------                 1+y+yz`
`Therefore,`
`     <PQ> = <PA> + <AB> + <BQ> `
`               zx<AB> + <AC>            xy(<AC> - <AB>) - <AB>          = - --------------- + <AB> + ------------------------                   1+z+zx                       1+x+xy`
`                  1-xyz          = ------------------[x<AB> - (1+x)<AC>]             (1+z+zx)(1+x+xy)`
`     <PR> = <PA> + <AC> + <CR> `
`               zx<AB> + <AC>            -yz<AC> + (<AB> - <AC>)          = - --------------- + <AC> + -------------------------                   1+z+zx                       1+y+yz`
`                  1-xyz          = ------------------[(1+z)<AB> - <AC>]             (1+z+zx)(1+y+yz)`
`Hence,`
`                        (1-xyz)^2     <PQ> x <PR> = --------------------------( <AB> x <AC> )                    (1+x+xy)(1+y+yz)(1+z+zx)    `
` `

 Posted by Bractals on 2005-08-25 08:12:01

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