Problem 12.5. A plane intersects a tetrahedron $A B C D$ and divides the medians of the triangles $D A B, D B C$ and $D C A$ through $D$ in ratios $1: 2,1: 3$ and $1: 4$ from $D$, respectively. Find the ratio of the volumes of the two parts of the tetrahedron cut by the plane.

Oleg Mushkarov
12.5. Let the plane meet the edges $D A, D B$ and $D C$ at points $P, Q$ and $R$, respectively. Set

$$
\frac{D P}{D A}=x, \frac{D Q}{D B}=y \text { and } \frac{D R}{D C}=z
$$

Let $M$ be the midpoint of $A B$ and $L=$ $D M \cap P Q$. It follows from the condition of the

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-196.jpg?height=456&width=438&top_left_y=427&top_left_x=1137)
problem that $\frac{D L}{D M}=\frac{1}{3}$. Therefore

$$
\frac{S_{D L P}}{S_{D A M}}=\frac{D P \cdot D L}{D A \cdot D M}=\frac{x}{3} ; \quad \frac{S_{D L Q}}{S_{D M B}}=\frac{D L \cdot D Q}{D M \cdot D B}=\frac{y}{3}
$$

Since $S_{D A M}=S_{D M B}=\frac{1}{2} S_{D A B}$ we conclude that

$$
2 x y=2 \frac{D P \cdot D Q}{D A \cdot D B}=\frac{S_{D P Q}}{\frac{1}{2} S_{D A B}}=\frac{S_{D P L}}{S_{D A M}}+\frac{S_{D L Q}}{S_{D M B}}=\frac{x+y}{3}
$$

i.e. $\frac{1}{x}+\frac{1}{y}=6$. Analogously $\frac{1}{y}+\frac{1}{z}=8$ and $\frac{1}{z}+\frac{1}{x}=10$. Solving this system we obtain $x=\frac{1}{4}, y=\frac{1}{2}$ and $z=\frac{1}{6}$. Thus

$$
\frac{V_{D P Q R}}{V_{D A B C}}=\frac{D P \cdot D Q \cdot D R}{D A \cdot D B \cdot D C}=x y z=\frac{1}{48}
$$

and therefore the desired ratio equals $1: 47$.