Squares just got a whole lot serious

Let the unit square be centered on the origin. The right side of the square will be the line $x=\frac{1}{2}$ . In polar form, this is $r=\frac{1}{2}\sec{\theta}$ .

Now consider the triangle bound by the lines $\theta=-\frac{\pi}{4}$ , $\theta=\frac{\pi}{4}$ , and $r=\frac{1}{2}\sec{\theta}$ . The area of this triangle is $A=\frac{1}{4}$ .

The distance from the origin to any point on the triangle will be the value of $r$ for that point. Without loss of generality, we can conclude that the average distance from the origin to a point in the triangle is the same as the average distance from the origin to any point in the unit square. We can calculate this average distance using the double integral:

$\frac{1}{A}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\int\limits_{0}^{\frac{1}{2}\sec{\theta}}rdA$

$4\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\int\limits_{0}^{\frac{1}{2}\sec{\theta}}r^2drd\theta$

$\frac{1}{6}\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^3{\theta}d\theta$

This evaluates to $\frac{\ln(1+\sqrt{2})+\sqrt{2}}{6}$ .

$a=1$ , $b=2$ , and $c=6$ . Therefore, $a+b+c=1+2+6=\boxed{9}$ .