THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

a) Let $\{O\}=A^{\prime }M\cap B^{\prime }N\cap C^{\prime }P$. The plane $(B^{\prime }N,C^{\prime }P)$ intersects the parallel planes $(ABC)$ and $(A^{\prime }{B}^{\prime }{C}^{\prime })$ along parallel straight lines, hence $NP\parallel B^{\prime }{C}^{\prime }$, so $PN\parallel BC$. In the same way $MP\parallel AC$, $NM\parallel AB$. Then $PNCM$ and $PNMB$ are parallelograms, hence $M$ is the midpoint of $[BC]$. Similarly $N$ and $P$ are the midpoints of the sides $[AC]$ and $[AB]$.

The straight line $C^{\prime }P$ is perpendicular on $A^{\prime }M$ and on $B^{\prime }N$, therefore on every straight line from their plane, in particular on $A^{\prime }{B}^{\prime }$. So $AB\perp C^{\prime }P$ and $AB\perp C{C}^{\prime }$, hence $AB\perp (PC{C}^{\prime })$, whence $AB\perp PC$. This shows that, in triangle $ABC$, $[CP]$ is a median and an altitude, hence $AC=BC$. Analogously $AB=AC$, so $△ABC$ is equilateral and the prism is regular.



b) Let $M$, $N$, $P$ be the midpoints of the sides. Then $A^{\prime }{B}^{\prime }MN$ is a trapezoid whose diagonals meet at a point $O$ such that $\frac{A^{\prime }O}{OM}=\frac{B^{\prime }O}{ON}=2$. Analogously, $A^{\prime }{C}^{\prime }MP$ is a trapezoid whose diagonals meet at a point $O^{\prime }$ such that $\frac{A^{\prime }{O}^{\prime }}{O^{\prime }M}=\frac{B^{\prime }{O}^{\prime }}{O^{\prime }N}=2$. Therefore, points $O^{\prime }$ and $O$ coincide, hence $A^{\prime }M$, $B^{\prime }N$ and $C^{\prime }P$ are concurrent.



Denoting $A{A}^{\prime }=h$, $AB=l$, we get $BN=\frac{l\sqrt{3}}{2}$, $A^{\prime }{M}^2=B^{\prime }{N}^2=h^2+\frac{3l^2}{4}$, $A^{\prime }{O}^2=B^{\prime }{O}^2=\frac{4h^2}{9}+\frac{l^2}{3}$. Then $A^{\prime }O\perp B^{\prime }O\iff A^{\prime }{O}^2+B^{\prime }{O}^2=A^{\prime }{B}^{\prime 2}\iff \frac{8h^2}{9}+\frac{2l^2}{3}=l^2\iff \frac{h}{l}=\frac{\sqrt{6}}{4}$.