[ Maximum/Minimum Distance ]

The side of the base $ABCD$ of a regular prism $ABCD A1B1C1D1$ is $2a$, and the lateral edge is $a$. Segments are considered with endpoints on the diagonal $AD1$ of the face $AA1D1D$ and the diagonal $DB1$ of the prism, parallel to the plane $AA1B1B$. a) One of these segments passes through the point $M$ on the diagonal $AD1$, for which $AM: AD1 = 2: 3$. Find its length. b) Find the minimum length of all such segments.
Through point $M$, we draw a plane parallel to the plane $A A 1 B 1 B$. Let the intersecting plane intersect the edges $A D, B C, B 1 C 1, A 1 D 1$ of the given parallelepiped at points $P, Q, R, S$, respectively, and the diagonals $B D$ and $B 1 D 1$ of the faces $A B C D$ and $A 1 B 1 C 1 D 1$ at points $E$ and $F$, respectively. Then the planes $P Q R S$ and $B D D 1 B 1$ intersect along the line $E F$. If $N$ is the point of intersection of the lines $E F$ and $D B 1$, then the segment $M N$ is parallel to the plane $A A 1 B 1 B$, and its ends lie on the lines $A D 1$ and $D B 1$. Given that $\frac{A M}{A \nu_{1}}=\frac{2}{3}$. Then

$$
\frac{P P}{B Q}=\frac{P D}{A P^{2}}=\frac{M D_{1}}{A M}=\frac{1}{2}
$$

therefore

$$
\frac{P N}{N R}=\frac{E N}{N F^{\prime}}=\frac{D E}{B_{1} F^{\prime}}=\frac{D E}{B E}=\frac{D P^{\prime}}{B Q}=\frac{1}{2}
$$

which means,

$$
\begin{gathered}
P M=\frac{2}{3} P S=\frac{2}{3} A A 1=\frac{2}{3} a \\
P N=\frac{1}{3} P R=\frac{1}{3} A B 1=\frac{1}{3} \sqrt{A A_{1}^{2}+A_{1} B_{1}^{2}}=\frac{1}{3} \sqrt{a^{2}+4 a^{2}}=\frac{1}{3} a \sqrt{5} .
\end{gathered}
$$

Let $\angle S P R=\angle A 1 A B 1=\alpha$. Then $\operatorname{tg} \alpha=\frac{A_{1} B_{1}}{A A_{1}}=2$. Therefore

$$
\cos \alpha=\frac{1}{\sqrt{1+l^{2} r_{2}}}=\frac{1}{\sqrt{5}}
$$

By the cosine rule in triangle $M P N$, we find that

$$
\begin{aligned}
& M N=\sqrt{P M^{2}+P N^{2}-2 P M \cdot P N \cos \alpha}= \\
& =\sqrt{\frac{4 \alpha^{2}}{9}+\frac{5 a^{2}}{9}-2 \cdot \frac{2}{3} a \cdot \frac{a \sqrt{5}}{3} \cdot \frac{1}{\sqrt{5}}}=\frac{1}{3} a \sqrt{5}
\end{aligned}
$$

Now suppose point $M$ moves along the segment $A D 1$. For each of its positions, we construct the plane $P Q R S$, parallel to the plane $A A 1 B 1 B$, and the point $N$ on $D B 1$. Let $\frac{A M}{A D_{1}}=x(0 \leqslant x \leqslant 1)$. Using the method described above, we find that

Then

$$
\begin{gathered}
M N 2=P M 2+P N 2-2 P M \cdot P N \cos \alpha= \\
=x 2 a 2+5(1-x) 2 a 2-2 \cdot x a \cdot(1-x) a \sqrt{5} \cdot \frac{1}{\sqrt{5}}= \\
=a 2(x 2+5-10 x+5 x 2-2 x+2 x 2)= \\
=a 2(8 x 2-12 x+5)=a 2\left(2\left(4 x 2-2 \cdot 2 x \frac{3}{2}+\frac{9}{4}\right)-\frac{9}{2}+5\right)= \\
=a 2\left(2\left(2 x-\frac{3}{2}\right) 2+\frac{1}{2}\right) \geqslant \frac{1}{2} a 2
\end{gathered}
$$

with equality achieved at $x=\frac{\mathbf{3}}{\mathbf{4}}$. Therefore, the minimum length of the segment $M N$ is $\frac{\mathbf{a}}{\sqrt{2}}$.

## Answer

$\frac{a \sqrt{5}}{3}, \frac{a}{\sqrt{2}}$.