Problem 820

Answer: 10.

Solution. Suppose we have found the position of point $P$ at which the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$, respectively. Note that $V_{M N K P}=\frac{1}{3} V_{M N K M_{1} N_{1} K_{1}}$. Draw planes $\beta$ and $\beta_{1}$ through points $M$ and $M_{1}$, parallel to the plane $A B C$, and call $R$ and $R_{1}$ the points of intersection with edge $B B_{1}$, and $S$ and $S_{1}$ with edge $C C_{1}$. Note that the figures $M N K R S$ and $M_{1} N_{1} K_{1} R_{1} S_{1}$ are obtained from each other by a parallel translation, and therefore are equal, and their volumes are also equal. Then the volumes of prisms $M N K M_{1} N_{1} K_{1}$ and $M R S M_{1} R_{1} S_{1}$ are also equal. But $V_{M R S M_{1} R_{1} S_{1}}=\frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}$, from which we get that $V_{M N K P}=\frac{1}{3} \frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}$.

We need to find the position of plane $\alpha$ such that $M M_{1}$ is maximized. Note that at least one of the points $M_{1}, N_{1}, K_{1}$ lies within the original prism, from which $M M_{1}=N N_{1}=K K_{1} \leq \max \left\{A M, A_{1} M, B N, B_{1} N, C K, C_{1} K\right\}$. Substituting the given ratios in the problem, we finally get that $M M_{1}=N N_{1}=K K_{1}=B N=\frac{6}{7} B B_{1}$, from which $V_{M N K P}=\frac{1}{3} \frac{N N_{1}}{B B_{1}} V_{A B C A_{1} B_{1} C_{1}}=\frac{16}{3} \frac{6}{7} 35=10$.
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[^0]: ${ }^{1}$ see https://ru.wikipedia.org/wiki/Малая_теорема_Ферма\#Альтернативная_формулировка

[^1]: ${ }^{2}$ see https://ru.wikipedia.org/wiki/Малая_теорема_Ферма\#Альтернативная_формулировка