Let a, b, and c be the lengths of the sides opposite vertices A, B, and C respectively.
Let x = |AF| and y = |AE|. A vector will be denoted by a two bold capital letters.
We want to get all vectors in terms of two independent vectors AB and AC.
AD = (1-u)AB + uAE = (1-u)AB + u(y/b)AC (1)
and
AD = vAF + (1-v)AC = v(x/c)AB + (1-v)AC
Solving for u we get
b(c-x)
u = --------
bc-cx
Plugging u into equation (1) we get
x(b-y) y(c-x)
AD = --------AB + --------AC
bc-xy bc-xy
-x(b-y) xy(b-y)
DE = AE - AD = ---------AB + ----------AC
bc-xy b(bc-xy)
xy(c-x) -y(c-x)
DF = AF - AD = ----------AB + ---------AC
c(bc-xy) bc-xy
We will use DE and DF to get Area(ΔDEF)
Area(ΔDEF) = (1/2)|DExDF|
xy(b-y)(c-x)
= (1/2)|--------------ABxAC|
bc(bc-xy)
Since Area(ΔABC) = 1, |ABxAC| = 2. Therefore,
xy(b-y)(c-x)
Area(ΔDEF) = -------------- (2)
bc(bc-xy)
∂ Area(ΔDEF)
-------------- = 0 gives bc(c-2x) + x2y = 0 (3)
∂x
∂ Area(ΔDEF)
-------------- = 0 gives bc(b-2y) + xy2 = 0
∂y
Combining these two gives
y b
--- = --- (4)
x c
which means that the cevian triangle with the maximum area
occurs when EF and CB are parallel. Combining equations (3)
and (4) gives
x3 - 2c2 + c3 = 0
or
(x-c)[x(x+c) - c2] = 0
Solving for the correct x gives
x = c(√5 - 1)/2 (5)
This is the distance between A and F.
Combining equations (2), (4), and (5) gives
5√5 - 11
Area(ΔDEF) = ----------.
2
QED
See Harry's post for an alternate solution.
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