[ Cross-sectional Area ]

In a regular quadrilateral prism, two parallel sections are made: one passes through the midpoints of two adjacent sides of the base and the midpoint of the axis, the other divides the axis in the ratio $1: 3$. Knowing that the area of the first section is 12, find the area of the second.
Let the plane passing through the midpoint $Q$ of the axis $O O_{1}$ of a regular quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ and the midpoints $M$ and $N$ of the sides $A D$ and $C D$ of the base $A B C D$ intersect the diagonal $B D$ of the base $A B C D$ at point $K$, and the diagonal $B_{1} D_{1}$ of the base $A_{1} B_{1} C_{1} D_{1}$ at point $L$. Since point $K$ lies on the midline $M N$ of triangle $A D C$, $K$ is the midpoint of $O D$, and since triangle $Q O_{1} L$ is equal to triangle $Q O K$, $L$ is the midpoint of $O_{1} B_{1}$. Let the line $M N$ intersect the lines $A B$ and $B C$ at points $E$ and $F$, respectively, and the line $K L$ intersect the line $B B_{1}$ at point $G$. Denote $M N=a$. Then

$$
A C=2 a, M E=N F=M N=a, E F=3 a
$$

Let the line $E G$ intersect $A A_{1}$ and $A_{1} B_{1}$ at points $H$ and $M_{1}$, respectively, and the line $F G$ intersect $C C_{1}$ and $B_{1} C_{1}$ at points $T$ and $N_{1}$, respectively. Then the hexagon $M N T N_{1} M_{1} H$ is the first section. It is clear that $M H\|F G, N T\| E G$ and $M_{1} N_{1} \| E F$, and $M_{1} N_{1}=M N=a$. Each of the triangles $E H M, N T F$ and $M_{1} G N_{1}$ is similar to triangle $E G F$ with a ratio of

$$
E M / E F=F N / E F=M_{1} N_{1} / E F=a /(3 a)=1 / 3 .
$$

Thus, the area $x$ of each of the triangles $E H M, N T F$ and $M_{1} G N_{1}$ is one-ninth of the area $S$ of triangle $E G F$. Therefore,

$$
S-3 \cdot S / 9=12
$$

from which we find that $S=18$.

The second cutting plane passes through the midpoint $R$ of the segment $O Q$. Let this plane intersect the diagonal $B D$ of the base $A B C D$ at point $U$. Then $R U \| Q K$. Therefore, $U$ is the midpoint of $K O$. Thus, $U K=O K / 2=B_{1} L$.

Hence, the line $Q R$ passes through the vertex $B_{1}$.

Let the line passing through point $U$ parallel to $E F$ intersect the lines $B E, A D, C D$ and $B F$ at points $E_{1}, M_{2}, N_{2}$ and $F_{1}$, respectively, and the lines $E_{1} B_{1}$ and $F_{1} B_{1}$ intersect the edges $A A_{1}$ and $C C_{1}$ at points $H_{1}$ and $T_{1}$, respectively.

Then the pentagon $B_{1} H_{1} M_{2} N T_{1}$ is the second section.

The points $M_{2}$ and $N_{2}$ are the midpoints of $A M$ and $C N$, respectively, so

$$
\begin{gathered}
N_{2} F_{1}=\frac{1}{2} N F=a / 2, M_{2} E_{1}=\frac{1}{2} E M=a / 2 \\
M_{2} N_{2}=\frac{1}{2}(A C+M N)=\frac{1}{2}(2 a+a)=3 a / 2 \\
E_{1} F_{1}=\frac{1}{2}(A C+E F)=\frac{1}{2}(2 a+3 a)=5 a / 2
\end{gathered}
$$

The sides of triangle $E_{1} B_{1} F_{1}$ are respectively parallel to the sides of triangle $E G F$. Therefore, triangle $E_{1} B_{1} F_{1}$ is similar to triangle $E G F$, with a similarity ratio of $E_{1} F_{1} / E F=(5 a / 2) /(3 a)=5 / 6$. Let $S_{1}$ be the area of triangle $E_{1} B_{1} F_{1}$. Then

$$
S_{1}=(5 / 6)^{2} \cdot S=(25 / 36) \cdot 18=25 / 2
$$

Each of the triangles $E_{1} M_{2} H_{1}$ and $N_{2} F_{1} T_{1}$ is similar to triangle $E_{1} F_{1} B_{1}$, since $M_{2} H_{1} \| B_{1} F_{1}$ and $N_{2} T_{1} \| B_{1} H_{1}$, with a similarity ratio of $E_{1} M_{2} / E_{1} F_{1}=F_{1} N_{2} / E_{1} F_{1}=1 / 5$. Therefore, the area of each of these triangles is $S_{1} / 25=1 / 2$. Thus, the area of the second section is $25 / 2$ - $1=23 / 2$.

## [ Systems of points and segments (other). Extreme principle (other) $\quad]$

## Author: Razaroov A.

The numbers from 1 to 1000 are arranged in a circle.

Prove that they can be connected by 500 non-intersecting segments, the difference of the numbers at the ends of which (in absolute value) is no more than 749.

## Solution

Color the points numbered from 251 to 750 inclusive in red, and the rest in blue. Note that the maximum difference between the numbers of blue and red points is $1000-251=750-1=749$.

Connect neighboring points of different colors with a segment (such points obviously exist). Temporarily remove them and again connect neighboring points of different colors. Continue this process until all points are connected. By restoring the removed points and segments, we obtain a system of non-intersecting segments that satisfy the conditions of the problem.