smc

Start by considering the triangle traced by $P$ as the circle moves around the triangle. It turns out this triangle is similar to the $6-8-10$ triangle (Proof: Realize that the slope of the line made while the circle is on $AC$ is the same as line $AC$ and that it makes a right angle when the circle switches from being on $AB$ to $BC$). Then, drop the perpendiculars as shown. Since the smaller triangle is also a $6-8-10=3-4-5$ triangle, we can label the sides $EF,$ $CE,$ and $DF$ as $3x,4x,$ and $5x$ respectively. Now, it is clear that $GB=DE+1=4x+1$, so $AH=AG=8-GB=7-4x$ since $AH$ and $AG$ are both tangent to the circle P at some point. We can apply the same logic to the other side as well to get $CI=5-3x$. Finally, since we have $HI=DF=5x$, we have $AC=10=(7-4x)+(5x)+(5-3x)=12-2x$, so $x=1$ and $3x+4x+5x=\boxed{12}$