Area Enclosed by Cevians

It can shown that the intersection area is given by

$\dfrac{\text{Area}}{[ABC] } = \dfrac{ (1 - 2 r)^2}{ r^2 - r + 1 } = \dfrac{ 4 (r - \frac{1}{2} )^2 }{ (r - \frac{1}{2})^2 + \frac{3}{4} }$

And as such, the area is a function of $(r - \frac{1}{2} )^2$ , thus values that are symmetric about $r = \frac{1}{2}$ will result in the same area.

In this problem, we started with $r = 0.3$ which is $0.2$ away from $0.5$ , thus the other value is $0.5 + 0.2 = 0.7$ .

We can find the ratio of the area of "intersection" triangle to that of the large triangle $ABC$ is given by Routh's theorem . If $\dfrac {CD}{BD} = x$ , $\dfrac {AE}{CE} = y$ , and $\dfrac {BF}{AF}=z$ , then the ratio of areas is given by:

$\rho = \frac {(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}$

For this problem $x=y=z = \dfrac {0.3}{0.7} = \dfrac 37$ . If take the reciprocals of $x$ , $y$ , and $z$ , the area of both triangles do not change, therefore, $\dfrac 1x = \dfrac 1y = \dfrac 1z = \dfrac 73$ gives the same $\rho$ , therefore the other $r=\boxed{0.7}$ .

In this solution we are going to use Rouths theorem .

This theorem is only concerned of the area of the original triangle and taking the ratios in some particular orientation (clockwise or anticlockwise) And now since we can look at the same triangle other way around.

so,The answer is $\boxed{1-r}$