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?
(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)
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Posted by Bractals
on 2005-08-25 08:12:01 |
re(3): Independent solution?
