Rider loops a marching column

A column of soldiers $50$ miles long marches forward $50$ miles at a steady pace. In that time a rider at constant speed travels from the back of the column to the front and straight back to the rear, arriving just as the march ends. The rider's total distance is $a(b+\sqrt c)$ miles with integer $a,b,c$ and $a$ as large as possible. Find $abc$.

100. Take marching speed $1$ and rider speed $s$. The forward leg takes $\tfrac{50}{s-1}$ and the return $\tfrac{50}{s+1}$, summing to $50$, so $s^2-2s-1=0$ and $s=1+\sqrt2$. The rider covers $50(1+\sqrt2)$ miles, so $abc=50\cdot1\cdot2=100$.