Asymmetric Optimization

This shouldn't be taken as a full proof but it may point the way.

Let $A=(0,0)$ , $B=(B_{x},B_{y})$ , and $C=(1,0)$ . The perimeter of the triangle $p=\sqrt{(B_{x}-1)^{2}+B_{y}^{2}}+1+\sqrt{B_{x}^{2}+B_{y}^{2}}$ .

The incenter coordinates are $I=(\frac{B_{x}+\sqrt{B_{x}^{2}+B_{y}^{2}}}{p},\frac{B_{y}}{p})$

I'm going to leave out finding the coordinates of $A'$ , $B'$ , and $C'$ as this part is messy and things cancel when the ratios are found.

Let $B_{x}=x$ and $B_{y}=y$ for simplicity

$\frac{AI}{AA'}=\frac{\sqrt{x^{2}+y^{2}}+1}{p}$ , $\frac{BI}{BB'}=\frac{\sqrt{(x+1)^{2}+y^{2}}+\sqrt{x^{2}+y^{2}}}{p}$ , $\frac{CI}{CC'}=\frac{\sqrt{(x+1)^{2}+y^{2}}+1}{p}$

Now the function to be maximized is $f(x,y)=\large \frac{(\sqrt{x^{2}+y^{2}}+1)^{3}(\sqrt{(x+1)^{2}+y^{2}}+\sqrt{x^{2}+y^{2}})^{2}(\sqrt{(x+1)^{2}+y^{2}}+1)}{p^{6}}$

A little playing around convinced me that the maximum actually occurs as $y \rightarrow 0$ , so let the triangle degenerate by setting $y=0$ . Enough cancels to let this formula work.

$f(x,0)=f(x)=\frac{(x+1)^3(2x-1)^2}{64x^{5}}$

Which has maximum $f(\frac{5}{4})=\frac{6561}{50000}=0.13122=P$

So $\lfloor 1000\cdot P \rfloor = \boxed{131}$

It's rather nice that $P$ is a rational number with a terminating decimal.