Problem 1. In the right rectangular parallelepiped $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime} \mathrm{with}$ $A B=12 \sqrt{3} \mathrm{~cm}$ and $A A^{\prime}=18 \mathrm{~cm}$, consider the points $P \in\left[A A^{\prime}\right]$ and $N \in\left[A^{\prime} B^{\prime}\right]$ such that $A^{\prime} N=3 B^{\prime} N$.

Determine the length of the segment $[A P]$ such that, for any point $M \in[B C]$, the triangle $M N P$ is a right triangle at $N$.

Mathematical Gazette
Solution. Since $B C \perp\left(A B B^{\prime}\right)$, it follows that $B C \perp P N \ldots \ldots \ldots \ldots$ 1p

Since $P N \perp N M$, it follows that $P N \perp(N B C)$, from which $P N \perp N B$, meaning triangle $N B P$ is right-angled at $N \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$.....................................

Denoting $A P=x$, we obtain $B P^{2}=x^{2}+432, P N^{2}=(18-x)^{2}+243$ and $B N^{2}=351$. Since $B P^{2}=P N^{2}+B N^{2}$, we get $x=13.5 \mathrm{~cm} \ldots .4 \mathbf{p}$