The medians $AD$, $BE$, and $CF$ of triangle $ABC$ intersect at the centroid $G$.  The line through $G$ that is parallel to $BC$ intersects $AB$ and $AC$ at $M$ and $N$, respectively.  If the area of triangle $ABC$ is 810, then find the area of triangle $AMN$.
I know that the centroid $G$ divides each median into a 2:1 ratio, with the longer segment closer to the vertex. That means $AM = \frac{2}{3} AB$ and $AN = \frac{2}{3} AC$. So, the triangle $AMN$ is similar to the triangle $ABC$, with a similarity ratio of 2:3. That means the area of triangle $AMN$ is $\left(\frac{2}{3}\right)^2$ times the area of triangle $ABC$. The area of triangle $ABC$ is given as 810, so the area of triangle $AMN$ is $\frac{4}{9} \cdot 810 = 360$.

