Points $M$ and $N$ are the midpoints of edges $A A_1$ and $C C_1$ of the parallelepiped $A B C D A_1 B_1 C_1 D_1$. The lines $A_1 C, B_1 M$, and $B N$ are pairwise perpendicular. Find the volume of the parallelepiped, given that $A_1 C=a, B_1 M=b, B N=c$.
Consider the section of the given parallelepiped by a plane passing through points $M, D 1$ and $B 1$. Since $M D 1 \| B N$ and $B N \perp B 1 M$, the triangle $M B 1 D 1$ is a right triangle. If $S$ is its area, then

$$
S=\frac{1}{2} M B 1 \cdot M D 1=\frac{1}{2} b c
$$

Let $O$ be the center of the parallelogram $A 1 B 1 C 1 D 1$. The planes $M B 1 D 1$ and $A A 1 C 1 C$ intersect along the line $M O$, so the line $A 1 C$ intersects the plane of the triangle $M B 1 D 1$ at a point $P$ lying on the line $M O$. Since $M O$ is the midline of the triangle $A A 1 C 1$,

$$
A 1 P=\frac{1}{2} \cdot \frac{1}{2} A 1 C=\frac{1}{4} A 1 C=\frac{1}{4} a
$$

The line $A 1 C$ is perpendicular to two intersecting lines $B 1 M$ and $D 1 M$ in the plane $M B 1 D 1$, so $A 1 P$ is the height of the triangular pyramid $A 1 M B 1 D 1$. Then

$$
V_{A} 1 M B 1 D 1=\frac{1}{3} S_{\triangle M B} 1 D 1 \cdot A 1 P=\frac{1}{3} \cdot \frac{1}{2} b c \cdot \frac{1}{4} a=\frac{1}{24} a b c
$$

Let $S$ be the area of the face $A 1 B 1 C 1 D 1$, and $h$ be the height of the parallelepiped dropped onto this face. Then

$$
V_{A B C D A} 1 B 1 C 1 D 1=S h, V_{A} 1 M B 1 D 1=\frac{1}{3} \cdot \frac{1}{2} S \cdot \frac{1}{2} h=\frac{1}{12} S h
$$

Therefore,

$$
V_{A B C D A} 1 B 1 C 1 D 1=12 V_{A} 1 M B 1 D 1=\frac{1}{2} a b c
$$

## Answer

$\frac{1}{2} a b c$

All edges of the regular triangular prism $A B C A 1 B 1 C 1$ are equal to $a$. Segments with endpoints on the lines $A B 1$ and $B C 1$, perpendicular to the line $A C 1$, are considered. Find the smallest length of such segments.

## Solution

Let $\overrightarrow{A B}=\vec{y}, \overrightarrow{A C}=\vec{y}, \overrightarrow{A A_{1}}=\vec{z}(x=y=z=a), \frac{\frac{A M}{A B_{1}}}{}=\alpha, \frac{B N}{B C_{1}}=\beta$. Then

$$
\begin{aligned}
& \overrightarrow{A C_{1}}=\vec{y}+\vec{z}, \overrightarrow{M N}=\overrightarrow{M A}+\overrightarrow{A B}+\overrightarrow{B N}=-\alpha \overrightarrow{A B_{1}}+\overrightarrow{A B}+\beta \overrightarrow{B C_{1}}= \\
& =-\alpha(\vec{x}+\vec{z})+\vec{x}+\beta(-\vec{x}+\vec{y}+\vec{z})=-(\alpha+\beta-1) \vec{x}+\beta \vec{y}+(\beta-\alpha) \vec{z}
\end{aligned}
$$

From the problem's condition, it follows that $\overrightarrow{M N} \cdot \overrightarrow{A C_{1}}=0$. Additionally,

$$
\vec{x} \cdot \vec{y}=x y \cos 60^{\circ}=\frac{1}{2} a 2, \vec{x} \cdot \vec{z}=\vec{y} \cdot \vec{z}=0,
$$

so

$$
\begin{gathered}
(-(\alpha+\beta+1) \vec{x}+\beta \vec{y}+(\beta-\alpha) \vec{z})(\vec{y}+\vec{z})=-\frac{1}{2}(\alpha+\beta-1) a 2+\beta a 2+(\beta-\alpha) a 2= \\
\quad=\left(-\frac{1}{2}(\alpha+\beta-1)+\beta+(\beta-\alpha)\right) a 2=\frac{1}{2}(3 \beta-3 \alpha+1) a 2=\frac{3}{2}\left(\beta-\alpha+\frac{1}{3}\right) a 2=0
\end{gathered}
$$

from which $\beta=\alpha-\frac{1}{3}$. Thus,

$$
\overrightarrow{M N}=-\left(2 \alpha-\frac{4}{3}\right) \vec{x}+\left(\alpha-\frac{1}{3}\right) \vec{y}-\frac{1}{3} \vec{z},
$$

so,

$$
\begin{gathered}
M N 2=\left(-\left(2 \alpha-\frac{4}{3}\right) \vec{x}+\left(\alpha-\frac{1}{3}\right) \vec{y}-\frac{1}{3} \vec{z}\right) 2= \\
=\left(2 \alpha-\frac{\mathbf{4}}{3}\right) 2 a 2+\left(\alpha-\frac{1}{3}\right) 2 a 2+\frac{1}{3} a 2-2\left(2 \alpha-\frac{\mathbf{4}}{3}\right)\left(\alpha-\frac{1}{3}\right) \cdot \frac{1}{2} a 2= \\
=\left(4 \alpha 2-\frac{16}{3}+\frac{16}{3}+\alpha 2-\frac{2}{3} \alpha+\frac{1}{3}+\frac{1}{3}-2 \alpha 2+2 \alpha-\frac{\mathbf{4}}{3}\right) a 2= \\
=\left(3 \alpha 2-4 \alpha+\frac{\mathbf{1 4}}{3}\right) a 2=\left(3\left(\alpha 2-2 \cdot \frac{\mathbf{2}}{3} \alpha+\frac{\mathbf{4}}{3}-\frac{4}{3}\right)+\frac{\mathbf{1 4}}{9}\right) a 2= \\
=\left(3\left(\alpha-\frac{2}{3}\right) 2-\frac{\mathbf{4}}{3}+\frac{\mathbf{1 4}}{9}\right) a 2=\left(3\left(\alpha-\frac{2}{3}\right) 2-\frac{2}{3}\right) a 2 \geqslant \frac{2}{9} a 2
\end{gathered}
$$

with equality achieved at $\alpha=\frac{2}{3}$. Therefore, the smallest length is $\frac{a \sqrt{2}}{3}$.

## Answer

$\frac{\alpha \sqrt{2}}{3}$