Let $O$ and $H$ be the circumcenter and orthocenter of triangle $ABC$, respectively.  Let $a$, $b$, and $c$ denote the side lengths, and let $R$ denote the circumradius.  Find $OH^2$ if $R = 7$ and $a^2 + b^2 + c^2 = 29$.
I recall a formula that relates $OH^2$ to $R$ and the side lengths of the triangle. The formula is $OH^2 = 9R^2 - (a^2 + b^2 + c^2)$. I can plug in the given values of $R$ and $a^2 + b^2 + c^2$ and simplify. $OH^2 = 9 \cdot 7^2 - 29 = 9 \cdot 49 - 29 = 441 - 29 = 412$. 