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Variation on a Classic (Posted on 2005-08-23) Difficulty: 3 of 5
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    
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Hints/Tips 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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