$\left[\begin{array}{l}\text { Sections, developments, and skeletons (other) })] \\ {[\underline{\text { Volume of a tetrahedron and pyramid }}]}\end{array}\right.$

On the edges $B C$ and $D C$ of the triangular pyramid $A B C D$, points $N$ and $K$ are taken respectively, such that $C N=2 B N, D K: K C$ $=3: 2$. It is known that $M$ is the point of intersection of the medians of triangle $A B D$. In what ratio does the plane passing through points $M, N, K$ divide the volume of the pyramid $A B C D$?
Let $BD = a$. Consider the plane of the face $BCD$ (Fig.2). Through point $C$, draw a line $l$ parallel to $BD$. Let the extension of segment $KN$ intersect line $l$ at point $T$, and line $BD$ at point $E$. Denote $BE = x$. From the similarity of triangles $TKC$ and $EKD$, we find that

$$
CT = DE \cdot \frac{CK}{KD} = \frac{2}{3}(a + x)
$$

From the similarity of triangles $TNC$ and $ENB$, it follows that

$$
CT = BE \cdot \frac{CN}{NB} = 2x
$$

From the equation $\frac{2}{3}(a + x) = 2x$, we find that

$$
BE = x = \frac{1}{2}a, \quad DE = DB + BE = a + \frac{1}{2}a = \frac{3}{2}a.
$$

Consider the plane of the face $ABD$ (Fig.3). Through point $A$, draw a line $m$ parallel to $BD$. Let line $EM$ intersect line $m$ at point $G$, edge $AB$ at point $Q$, edge $AD$ at point $F$, and the extension of the median $DP$ of triangle $ABD$ intersect line $m$ at point $H$. From the equality of triangles $BPD$ and $APH$, we find that $AH = DB = a$, and since $M$ is the point of intersection of the medians of triangle $ABD$, then

$$
\frac{DM}{MH} = \frac{DM}{MP + PH} = \frac{DM}{\frac{1}{2}DM + \frac{1}{2}DM} = \frac{1}{2}
$$

From the similarity of triangles $GMH$ and $EMD$, it follows that

$$
GH = DE \cdot \frac{MH}{DM} = 2 \cdot \frac{3}{2}a = 3a
$$

therefore, $AG = GH - AH = 3a - a = 2a$. From the similarity of triangles $AFG$ and $DFE$, it follows that

$$
\frac{AF}{FD} = \frac{AG}{DE} = \frac{2a}{\frac{3}{2}a} = \frac{4}{3}
$$

and from the similarity of triangles $BQE$ and $AQG$,

$$
\frac{BQ}{QA} = \frac{BE}{AG} = \frac{\frac{1}{2}a}{2a} = \frac{1}{4}
$$

The plane $ADN$ (Fig.1) divides the volume $V$ of the given pyramid in the ratio $1:2$, so $V_{DABN} = \frac{1}{3}V$ and $V_{DACN} = \frac{2}{3}V$. The cutting plane intersects the lateral edges $AD$, $AB$, and $AN$ of the triangular pyramid $DABN$ at points $F$, $Q$, and $N$, respectively, so

$$
V_{AFQN} = \frac{AF}{AD} \cdot \frac{AQ}{AB} \cdot \frac{AN}{AN} \cdot V_{DABN} = \frac{4}{7} \cdot \frac{4}{5} \cdot \frac{1}{1} \cdot \frac{1}{3}V = \frac{16}{105}V
$$

Then the volume of the polyhedron $DBNQF$ is $\frac{1}{3}V - \frac{16}{105}V = \frac{19}{105}V$. The cutting plane intersects the lateral edges $DA$, $DC$, and $DN$ of the triangular pyramid $DACN$ at points $F$, $K$, and $N$, respectively, so

$$
V_{DFKN} = \frac{DF}{DA} \cdot \frac{DK}{DC} \cdot \frac{DN}{DN} \cdot V_{DACN} = \frac{3}{7} \cdot \frac{3}{5} \cdot \frac{1}{1} \cdot \frac{2}{3}V = \frac{6}{35}V
$$

Thus, the volume of the polyhedron $DFQNK$ is

$$
\frac{19}{105}V + \frac{6}{35}V = \frac{37}{105}V
$$

Then the volume of the remaining part of the pyramid $ABCD$ is

$$
V - \frac{37}{105}V = \frac{68}{105}
$$

Therefore, the desired ratio is $\frac{37}{68}$.

## Answer

$37: 68$.