This feels too easy so I'm not sure if it's correct ThE-dArK-lOrD wrote: Three diagonals of a regular $n$-gon prism intersect at an interior point $O$. Show that $O$ is the center of the prism. (The diagonal of the prism is a segment joining two vertices not lying on the same face of the prism.) Let $ABC$ and $DEF$ be the two triangles on the $n$-gon faces with $\overline{AD}, \overline{BE}, \overline{CF}$ concurrent at point $O$; let $X,Y$ be their circumcenters. Let $\lambda$ be the ratio of distance of $O$ from the planes of $\triangle ABC$ and $\triangle DEF$. Since these two planes are parallel we conclude $$\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}=\lambda$$hence $\lambda$ is the similarity coefficient between the two triangles. Thus, $\overline{XY}$ passes through $P$ and $\triangle PXA \sim \triangle PYD$. Now $AX=DY$ hence $\lambda=1$ and $PX=PY$ proving that $P$ is the center of the prism.