Two paths on a halfsquare

In $\triangle ACP$ ,By Pythagorean theorem , we have $x^2+y^2=(2y)^2=4y^2\implies x^2=3y^2 \implies x=y\sqrt{3}$

$\dfrac{BP}{BC}$ is given by $\dfrac{x-y}{x}=\dfrac{y\sqrt{3}-y}{y\sqrt{3}}=\dfrac{\sqrt{3}-1}{\sqrt{3}}=\dfrac{3-\sqrt{3}}{3}$

$\Rightarrow a=b=c=3$ , Hence $a+b+c=\boxed{9}$

Let B be the origin, A=(2,0), C=(1,1), and P=(p,p). The ratio sought is equivalent to the x-coordinate of P.

The green path has length $\sqrt{2}p+\sqrt{(p-2)^2+p^2}$

The purple path has length $\sqrt{2}(1-p)+\sqrt{2}$

Set the two equal and divide both sides by $\sqrt{2}$

$\sqrt{p^2-2p+2}=2-2p$

Square both sides then use the quadratic formula to get

$p=\frac{3-\sqrt{3}}{3}$

From which we have $a=3$ , $b=3$ , $c=3$ and $a+b+c=9$