10,11
Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. On the rays $C_{1} C, C_{1} B_{1}$, and $C_{1} D_{1}$, segments $C_{1} M, C_{1} N$, and $C_{1} K$ are laid out, equal to
$5 / 2 C C_{1}, 5 / 2 C_{1} B_{1}$,
$5 / 2 C_{1} D_{1}$, respectively. In what ratio does the plane passing through points $M, N, K$ divide the volume of the parallelepiped
$A B C D A_{1} B_{1} C_{1} D_{1}$.
Lines $M K$ and $C D$ lie in the plane of the face $D C C_{1} D_{1}$. The point $F$ of their intersection lies on the line of intersection of the cutting plane with the plane of the face $A B C D$. Similarly, the point $E$ of intersection of the lines $B C$ and $M N$ is located on the same line. From the similarity of triangles $M C F$ and $M C_{1} K$ it follows that $C F: C_{1} K = M C: M C_{1} = 3: 5, C F = 3 / 5$ $C_{1} K = 3 / 5 \cdot 5 / 2 C_{1} D_{1} = 3 / 2 C_{1} D_{1} = 3 / 2 C D, D F = C F - C D = 1 / 2 C D$.

Similarly, $C E = 3 / 2 B C, B E = 1 / 2 B C$. Therefore, $E F \| B D$. Let $P$ and $Q$ be the points of intersection of the line $E F$ with the edges $A B$ and $A D$. Then the quadrilateral $B P F D$ is a parallelogram. Therefore, $B P = D F = 1 / 2 A B$. Similarly, $D Q = B E = 1 / 2 A D$. Therefore, $P Q$ is the midline of the triangle $A B D$. Similarly, we construct the point $R$ of intersection of the cutting plane with the edge $A A_{1}$ and prove that $R$ is the midpoint of $A A_{1}$.

Consider the triangular pyramid $A P Q R$. It is similar to the pyramid $A D B A_{1}$ with a similarity coefficient of $1 / 2$. Therefore, the volume of the pyramid $A P Q R$ is 8 times less than the volume of the pyramid $A D B A_{1}$. Let $h$ be the height of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, drawn from the vertex $A, S$ be the area of the parallelogram $A B C D, V$ be the volume of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. Then $V_{A D B A_{1}} = 1 / 3 S_{A D B} \cdot h = 1 / 3 \cdot 1 / 2 S h = 1 / 6 V$. Therefore, $V_{A P Q R} = 1 / 8 V_{A D B A_{1}} = 1 / 8 \cdot 1 / 6 V = 1 / 48 V$.

## Answer

$1: 47$.
[ Three lines intersecting at one point ]
[ Sphere touching the edges of a tetrahedron

A sphere touches all the edges of a tetrahedron. Connect the points of tangency on pairs of non-adjacent edges.

Prove that the three resulting lines intersect at one point.

## Solution

Place a mass, inversely proportional to the length of the tangent drawn from that vertex to the sphere (all three tangents from a given vertex are obviously equal), at each vertex. Then the point of tangency of an edge coincides with the center of mass of its endpoints, and, consequently, all three segments from the condition intersect at the center of mass of the resulting system of material points.

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