HMMT-Nov 2011 guts, Problem 18

Consider a cube $A B C D E F G H$, where $A B C D$ and $E F G H$ are faces, and segments $A E, B F, C G, D H$ are edges of the cube. Let $P$ be the center of face $E F G H$, and let $O$ be the center of the cube. Given that $A G=1$, determine the area of triangle $A O P$.

Solution: From $A G=1$, we get that $A E=\frac{1}{\sqrt{3}}$ and $A C=\frac{\sqrt{2}}{\sqrt{3}}$. We note that triangle $A O P$ is located in the plane of rectangle $A C G E$. Since $O P \| C G$ and $O$ is halway between $A C$ and $E G$, we get that $[A O P]=\frac{1}{8}[A C G E]$. Hence, $[A O P]=\frac{1}{8}\left(\frac{1}{\sqrt{3}}\right)\left(\frac{\sqrt{2}}{\sqrt{3}}\right)=\frac{\sqrt{2}}{24}$. $\fbox{\frac{\sqrt{2}}{24}}$.

Answer: \frac{\sqrt{2}}{24}