jmc

Let $Q$ be the foot of the altitude from $A$ to $BC$. Then triangles $AQD$ and $GPD$ are similar. Furthermore, $$\frac{GP}{AQ}=\frac{GD}{AD}=\frac{1}{3},$$so to find $GP$, we can find $AQ$.



The semi-perimeter of the triangle is $(11+13+20)/2=22$, so by Heron's formula, the area of triangle $ABC$ is $$\sqrt{22(22-11)(22-13)(22-20)}=66.$$Hence, the height of triangle $ABC$ with respect to base $BC$ is $AQ=2\cdot 66/BC=2\cdot 66/20=33/5$. Therefore, $GP=AQ/3=(33/5)/3=\boxed{\frac{11}{5}}$.