A plane intersects the $A B C D$ regular tetrahedron at point $K$ on edge $A B$, at point $L$ on edge $B C$, and at point $M$ on edge $C D$, such that $A K=2 K B, B L=C L$, and $C M=2 M D$. In what ratio does the plane divide the surface area of the tetrahedron?
The plane $S$ is not parallel to the base edge $AC$, since the intersection line $KL$ with the base plane $ABC$ is not parallel to $AC$. Thus, we can calculate the distance of the intersection point $P$ of the extensions of $AC$ and $KL$ from $C$ (or we can also construct it on the base plane $ACB$), and then the intersection line of $S$ and the plane $ACD$ is $PM$, and this intersects the plane at point $N$ on the edge $AD$.
![](https://cdn.mathpix.com/cropped/2024_05_02_a98342e6a6690d93b9b6g-1.jpg?height=772&width=1376&top_left_y=351&top_left_x=363)

Draw the parallel to $AC$ through $K$ and $N$, and let it intersect $CB$ at $K_1$ and $CD$ at $N_1$. Then $LK_1 = LB - K_1B = BC\left(\frac{1}{2} - \frac{1}{3}\right) = \frac{1}{6}BC = \frac{1}{3}LC$, and from similar triangles, $CP = K_1K \cdot \frac{LC}{LK_1} = BC = AC$, so $P$ is the reflection of $A$ over $C$. Continuously,

$$
\frac{N_1N}{N_1M} = \frac{CP}{CM} = \frac{3}{2} = \frac{DN_1}{N_1M} = \frac{DM - N_1M}{N_1M} = \frac{DM}{N_1M} - 1
$$

thus $N_1M = \frac{2}{5}DM$, $DN_1 = NN_1 = \frac{3}{5}DM = \frac{1}{5}DC$, that is, $N$ is the one-fifth point on the edge $DA$: $AN = 4ND$. The plane $S$ intersects the tetrahedron in the quadrilateral $KLMN$.

Hold the tetrahedron so that the edge $BD$ is closer to you than $AC$. Then the front part of the surface is given by the triangles $BKL$ and $DMN$ and the quadrilaterals $BDNK$, $BDML$. The latter are formed from the planes $BDA$, $BDC$ by removing the triangles $KNA$, $MLC$ that fall behind $S$. The ratio of the area of the 4 partial triangles to the area of the original plane is easy to determine, as one of the angles of each is $60^\circ$. Indeed, in the known formula $t = \frac{ab}{2} \sin \gamma$, the factor $\frac{\sin \gamma}{2}$ is common, so the areas of the triangles are proportional to the product of the sides enclosing the $60^\circ$ angle. The ratio of the original planes is $1 \cdot 1 = 1$, and for the parts,

$$
\begin{array}{ll}
AK \cdot AN = \frac{2}{3} \cdot \frac{4}{5} = \frac{8}{15}, & BK \cdot BL = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} \\
CL \cdot CM = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}, & DM \cdot DN = \frac{1}{3} \cdot \frac{1}{5} = \frac{1}{15}
\end{array}
$$

and therefore the ratio for the quadrilateral parts of the planes is

$$
KND B \text{ is } 1 - \frac{8}{15} = \frac{7}{15}, \quad KLC A \text{ is } \frac{5}{6}, \quad LMD B \text{ is } \frac{2}{3}, \quad MNA C \text{ is } \frac{14}{15}
$$

Thus, the sum of the ratios of the front and back surface parts is

$$
\frac{1}{6} + \frac{7}{15} + \frac{2}{3} + \frac{1}{15} = \frac{41}{30}, \quad \text{ and } \quad \frac{5}{6} + \frac{8}{15} + \frac{1}{3} + \frac{14}{15} = \frac{79}{30}
$$

and from this, the desired ratio is 41:79, the ratio for the front part containing the original edge $BD$, and the back part containing the edge $AC$.

Remark. We can also determine the position of $N$ from the intersection of $S$ and the line of the edge $BD$, which is the reflection of $B$ over $D$.