[ Properties of sections [Theorems of Ceva and Menelaus ]

In tetrahedron $A B C D$, a plane is drawn through the midpoint $M$ of edge $A D$, vertex $C$, and point $N$ on edge $B D$ such that $B N: N D=2: 1$. In what ratio does this plane divide the segment $K P$, where $K$ and $P$ are the midpoints of edges $A B$ and $C D$ respectively?
Let's conduct a plane through edge $AB$ and point $P$ (Fig.1). Planes $ABP$ and $CMN$ intersect along line $QL$, where $Q$ is the intersection point of lines $CM$ and $AP$, and $L$ is the intersection of lines $BP$ and $CN$. Consider the plane of triangle $BCD$ (Fig.2). Through point $B$, draw a line parallel to $CD$, and extend segments $CN$ to intersect this line at point $T$. Denote $CP=DP=a$. From the similarity of triangles $BNT$ and $DNC$, we find that

$$
BT=CD \cdot \frac{BN}{ND}=2a \cdot 2=4a
$$

and from the similarity of triangles $CLP$ and $TLB$,

$$
\frac{PL}{LB}=\frac{CP}{B'L'}=\frac{\mathbf{a}}{4\mathbf{a}}=\frac{\mathbf{1}}{4}
$$

Since $Q$ is the intersection point of the medians of triangle $ACD$,

$$
\frac{PQ}{QA}=\frac{1}{2}
$$

It remains to find in what ratio segment $QL$ divides the median $PK$ of triangle $APB$. Let $O$ be the intersection point of these segments (Fig.3). Through point $Q$, draw a line parallel to $AB$ until it intersects side $BP$ at point $H$. Then

$$
\frac{PPH}{PPB}=\frac{PPQ}{PPA}=\frac{1}{3}
$$

therefore

$$
\frac{PPL}{PPH}=\frac{1PPB}{\frac{1}{3}PPB}=\frac{3}{5}, \frac{PPL}{LH}=\frac{3}{2}
$$

Let $G$ be the intersection of $PK$ and $QH$. Then $G$ is the midpoint of $QH$. Consider triangle $PQH$. Through vertex $P$, draw a line parallel to $QH$. Extend $QL$ to intersect this line at point $S$. Denote $QG$ $=GH=b$. From the similarity of triangles $PLS$ and $HLQ$, we find that

$$
PS=QH \cdot \frac{PL}{LH}=2b \cdot \frac{3}{2}=3b
$$

and from the similarity of triangles $POS$ and $GOQ$,

$$
\frac{PO}{OG}=\frac{PS}{QG}=\frac{3b}{b}=3
$$

Denote $OG=c$. Then

$$
PO=3c, PG=4c, KG=2PG=8c, OK=KG+OG=9c
$$

Therefore,

$$
\frac{PO}{OK}=\frac{3d}{3:}=\frac{1}{3}
$$

## Answer

$1: 3$