Pointed Pointy Triangle (My bad sense of naming things)

Let the unit triangle, $\Delta ABC$ , be centered on the origin, $O$ . Let it be rotated such that point $A$ is located at $(\frac{\sqrt{3}}{6},\frac{1}{2})$ , point $B$ is located at $(\frac{\sqrt{3}}{6}, -\frac{1}{2})$ , and point $C$ is located at $(-\frac{\sqrt{3}}{3},0)$ . The right side of the triangle will be the line $x=\frac{\sqrt{3}}{6}$ . In polar form, this is $r=\frac{\sqrt{3}}{6}\sec{\theta}$ .

Now consider $\Delta OAB$ bound by the lines $\theta=-\frac{\pi}{3}$ , $\theta=\frac{\pi}{3}$ , and $r=\frac{\sqrt{3}}{6}\sec{\theta}$ . The area of this triangle is $B=\frac{\sqrt{3}}{12}$ .

The distance from the origin to any point in $\Delta OAB$ 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 $\Delta OAB$ is the same as the average distance from the origin to any point in the unit triangle, $\Delta ABC$ . We can calculate this average distance using the double integral:

$\frac{1}{B}\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\int\limits_{0}^{\frac{\sqrt{3}}{6}\sec{\theta}}rdA$

$4\sqrt{3}\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\int\limits_{0}^{\frac{\sqrt{3}}{6}\sec{\theta}}r^2drd\theta$

$\frac{1}{18}\int\limits_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\sec^3{\theta}d\theta$

This evaluates to $\frac{\ln(2+\sqrt{3})+2\sqrt{3}}{18}$ .

$a=2$ , $b=3$ , and $c=18$ . Therefore, $a+b+c=2+3+18=\boxed{23}$ .

Let the triangle be ABC and O be the center. Let D be the midpoint of AB. The average distance of Points of traingle OAD from O would be equal to the average distance of points of ABC from O due to symmetry. OA = 1/root(3) Let's call it x. OD = 1/(2 (root(3))) Let's call it y. AD = 1/2 Let's call it z. Let u = (x+y)/z and v = (x-y)/c Formula: Average distance of all points of Triangle OAD from vertex O = (z/6) ( (u (1+(v^2))) + ((1/2) (1 - u^2) (1 - v^2) ln ( (u-1)/(u+1) ))) [Proof : http://www.mathpages.com/home/kmath283/kmath283.htm]

Therefore a = 2, b =3, c =18. My first solution. Yay!!