Let cevians BE and CF of ΔABC intersect at point D.

If Area(ΔABC) = 1, then find the maximum Area(ΔDEF) as points E and F vary over sides AC and AB respectively. Prove your result.

Let E and F divide AC and AB in the ratios x:(1-x) and y:(1-y) respectively.
Using square brackets to denote areas, with [ABC] = 1, and [DEF] = U,

[CDE]    = [ACF] - [AEF] - [DEF]               =  y - xy - U
[BDF]    = [ABE] - [AEF] - [DEF]                          =  x - xy - U   
[BCD]   = [ABC] - [ABE] - [ACF] + [AEF] + [DEF] =  1 - x - y + xy + U

Also,                     [DEF]/[BDF]  =  [CDE]/[BCD]
so:                        [DEF][BCD]   =  [BDF][CDE]

which gives        U(1 - x - y + xy + U)  =  (x - xy - U)(y - xy - U)
                        U(1 - x - y + xy +x - xy +y - xy)  =  (x - xy)(y - xy)
                        U  =  xy(1 - x)(1 - y)/(1 - xy)                                          (1)

For maximum U, putting the partial derivative of U wrt x equal to zero:

(1 - xy)(1 - 2x) = x(1 - x)(-y)     which gives        x²y - 2x + 1 = 0             (2)

Similarly for the partial derivative wrt y:              y²x - 2y + 1 = 0             (3)

(2) - (3) gives                (x - y)(xy - 2) = 0

Since 0 < x,y < 1, (xy - 2) < 0, so the only solution is x = y ( = x_m say).

Equation (2) now gives:              x_m³ - 2x_m + 1 = 0          

                                    (x_m - 1)(x_m² + x_m - 1) = 0

Since x < 1, the only solution is: x_m = (sqrt(5) - 1)/2   ~ 0.61803..

and (1) gives     U_max      =  x_m²(1 - x_m)²/(1 - x_m²)

                                    =  x_m²(1 - x_m)/(1 + x_m)

                                    =  0.25(sqrt(5) - 1)²(3 - sqrt(5))/(sqrt(5) + 1)

   Maximum area            =  (5sqrt(5) - 11)/2

                                    ~  0.09017

Posted by Harry on 2011-06-08 01:10:52