The Light At The End Of The Tunnel

A quick overview of my solution method ......

Because of the properties of reflection, the distance the light travels before it exits the tunnel will be the same as it would be if it were to travel along its original path, not striking any mirrors, until it crosses the line $x = 10.$ Thus the distance the light will travel as a function of its initial angle $\theta$ is $L(\theta) = 10\sec(\theta).$ Using the averaging formula, the expected distance the light will travel is then

$\dfrac{1}{\dfrac{5\pi}{12} - \dfrac{\pi}{12}}\displaystyle\int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} 10\sec(\theta) d\theta = \dfrac{30}{\pi}\ln(\sec(\theta) + \tan(\theta))$

evaluated from $\theta = \dfrac{\pi}{12}$ to $\theta = \dfrac{5\pi}{12}.$ This comes to

$\dfrac{30}{\pi}\ln\left(\dfrac{\sec(\frac{5\pi}{12}) + \tan(\frac{5\pi}{12})}{\sec(\frac{\pi}{12}) + \tan(\frac{\pi}{12})}\right) =$

$\dfrac{30}{\pi}\ln(3 + 2\sqrt{2}) = \dfrac{10}{\pi}\ln(3 + 2\sqrt{2})^{3} = \dfrac{10}{\pi}\ln(99 + 70\sqrt{2}).$

Thus $a + b + c = 99 + 70 + 2 = \boxed{171}.$

(Note that to calculate the trig expression inside the ln function the facts that $\frac{5\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}$ and $\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6}$ were put to good use. I will expand further in the morning if requested.)