8. [6] Let $O$ be the point $(0,0)$. Let $A, B, C$ be three points in the plane such that $A O=15, B O=15$, and $C O=7$, and such that the area of triangle $A B C$ is maximal. What is the length of the shortest side of $A B C$ ?
Answer: 20 We claim that $O$ should be the orthocenter of the triangle $A B C$. If $O$ is not on an altitude of $\triangle A B C$, suppose (without loss of generality) that $\overline{A O}$ is not perpendicular to $\overline{B C}$. We can
rotate $A$ around $O$, leaving $B$ and $C$ fixed, to make $\overline{A O}$ perpendicular to $\overline{B C}$, which strictly increases the area. Therefore, if $[A B C]$ is maximal then $\triangle A B C$ is an isosceles triangle with orthocenter $O$ and base $\overline{A B}$.

Let $F$ be the foot of the perpendicular from $C$ to $\overline{A B}$. Since $\angle F O A$ and $\angle C O E$ are vertical, $\angle F A O=$ $\angle O C E$. Then $\triangle F A O$ is similar to $\triangle F C B$, so we have $\frac{A F}{O F}=\frac{C F}{B F}=\frac{O F+7}{A F}$, so $A F^{2}=O F^{2}+7 \cdot O F$. Since $A F^{2}=225-O F^{2}, 2 \cdot O F^{2}+7 \cdot O F-225=0$, so $O F \stackrel{=}{=}$. Then $A F=12$, so $A B=24$ and $B C=20$. Thus, the length of the shortest side of $\triangle A B C$ is 20 .