10,11 |
| :---: | :---: | :---: |
|  | Volume Ratios |  |
|  | Volume of a Parallelepiped |  |
|  | Two Pairs of Similar Triangles |  |
|  | Midline of a Triangle |  |
|  | The Volume of a Body is the Sum of the Volumes of its Parts |  |

Points \( M, N, K \) are the midpoints of the edges \( AB, BC, DD_1 \) of the parallelepiped \( ABCD A_1 B_1 C_1 D_1 \).

a) Construct the section of the parallelepiped by the plane passing through points \( M, N, K \).

b) In what ratio does this plane divide the edge \( CC_1 \) and the diagonal \( DB_1 \)?

c) In what ratio does this plane divide the volume of the parallelepiped?
a) Let's extend segments $M N$ and $D C$ until they intersect at point $Q$. The line $Q K$ is the line of intersection of the cutting plane with the plane of the face $C D D_{1} C_{1}$. Denote by $E$ the point of intersection of this line with the edge $C C_{1}$.

Similarly, construct the point $P$ of intersection of the line $M N$ with the plane of the face $A D D_{1} A_{1}$ and the point $F$ of intersection of the cutting plane with the edge $A A_{1}$. Thus, the desired section is the pentagon MNEKF.

b) Since $M N$ is the midline of triangle $A B C$, $M N \| A C$. Therefore, the quadrilateral $A M Q C$ is a parallelogram. Consequently,

$Q C = A M = 1/2 A B = 1/2 C D$. From the similarity of triangles $Q C E$ and $Q D K$, it follows that $C E: D K = Q C: Q D = 1: 3$.

Then $C E = 1/3 D K = 1/6 D D_{1} = 1/6 C C_{1}$.

Therefore, $C E: E C_{1} = 1: 5$. Let $H$ be the point of intersection of the diagonals of the parallelogram $A B C D$, and $L$ be the point of intersection of the segments $M N$ and $B D$. Then $L$ is the midpoint of $B H$, $D L = D H + H L = 1/2 B D + 1/4 B D = 3/4 B D$.

The cutting plane intersects the plane of the parallelogram $B D D_{1} B_{1}$ along the line $L K$, so the point $O$ of intersection of the diagonal $D B_{1}$ of the parallelepiped with the segment $L K$ is the point of intersection of the diagonal $D B_{1}$ with the cutting plane. Let $R$ be the point of intersection of the lines $L K$ and $B_{1} D_{1}$, lying in the plane of the parallelogram $B D D_{1} B_{1}$. From the equality of triangles $D_{1} K R$ and $D K L$, it follows that $D_{1} R = D L = 3/4 B D = 3/4 B_{1} D_{1}$. Therefore, $B_{1} R = B_{1} D_{1} + D_{1} R = 7/4 B_{1} D_{1}$.

From the similarity of triangles $L O D$ and $R O B_{1}$, we find that $D O: O B_{1} = D L: B_{1} R = 3: 7$.

c) Let $S$ be the area of the parallelogram $A B C D$, $h$ be the height of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (the distance between the parallel planes of the faces $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$), and $V$ be the volume of the given parallelepiped. Since $M N$ is the midline of triangle $A B C$, $S_{B M N} = 1/4 S_{A B C} = S / 8$,

$S_{A M P} = S_{C N Q} = S_{B M N} = S / 8$, $S_{P Q D} = S - S_{B M N} + S_{A M P} + S_{C N Q} = 9 S / 8$, $V_{K P Q D} = 1/3 S_{P D Q} \cdot h / 2 = 1/3 \cdot 9/8 S \cdot h / 2 = 3 V / 16$.

If $h_{1}$ is the height of the triangular pyramid $E C N Q$, drawn from the vertex $E$, then $h_{1}: h = E C: E C_{1} = 1: 6$.

Thus,

$V_{E C N Q} = 1/3 \cdot 1/8 S \cdot h / 6 = 1/144 S h = V / 144$.

Similarly, $V_{F A M P} = V / 144$. Therefore, the cutting plane cuts off from the given parallelepiped a polyhedron with a volume equal to

$V_{K P D Q} - V_{E C N Q} - V_{F A M P} = 3 V / 16 - V / 72 = 25 V / 144$.

## Answer

b) $1: 5, 3: 7$; c) $25: 119$