, 2011

Adopting the usual notation, $2a = b + c$, hence $s = a + b + c = \frac{3}{2}a$. The area of the triangle is $K = sr = \frac{3}{2}ar = \frac{1}{2}a h_a$, where $h_a$ is the altitude from $A$ onto $BC$. Therefore $h_a = 3r$, and so the incenter is a third of the way to the vertex. The centroid is also a third of the way to the vertex; therefore the line joining the incenter and centroid is parallel to $BC$.