Given the tetrahedron $ABCD$, the length of edge $AB$ is $a$, and the length of edge $CD$ is $b$. The distance between the lines containing edges $AB$ and $CD$ is $d$, and the (one of the) angles between them is $\omega$. The tetrahedron is divided into two parts by a plane $\varepsilon$ that is parallel to edges $AB$ and $CD$. What is the ratio of the volumes of these two parts, given that the distance from line $AB$ to plane $\varepsilon$ is $k$ times the distance from line $CD$ to plane $\varepsilon$?

(International Mathematical Olympiad problem, 1965. CMO6513)
Let the part of the cut tetrahedron containing $AB$ and $CD$ be denoted by $T_{AB}$ and $T_{CD}$, respectively, and their volumes by $V_{AB}$ and $V_{CD}$. Let $\varepsilon$ intersect the face $ABC$ at $EF$ and the face $ABD$ at $HG$, so $EFGH$ forms a parallelogram because $EF \parallel AB \parallel HG$ and $EH \parallel CD \parallel FG$. Thus, a plane parallel to $ABD$ can be drawn through $EF$, intersecting $CD$ at $J$. The triangle $EFJ$ divides $T_{CD}$ into the prism $DGHJFE = P$ and the tetrahedron $EFJC = T'$. Let the volumes of these be $V_P$ and $V_J$, and the volume of the entire tetrahedron be $V$.

![](https://cdn.mathpix.com/cropped/2024_05_02_0da6849bff83cda6183fg-1.jpg?height=529&width=675&top_left_y=405&top_left_x=711)

The desired ratio is

$$
q=\frac{V_{CD}}{V-V_{CD}}=\frac{V_P+V_J}{V-V_P-V_J}
$$

According to the assumption, the ratio $AE: CE$ is $k$, and thus $CE = CA / (k+1)$. This is because if we denote the projections of $A$ and $C$ onto $\varepsilon$ as $A'$ and $C'$, respectively, the right triangles $AA'E$ and $CC'E$ are similar, and $AE: CE = AA': CC' = k$. Therefore, $EF = AB / (k+1)$, and the ratio of the areas of the similar triangles $EFJ$ and $ABD$ is $1 / (k+1)^2$. The ratio of the heights from $C$ in the tetrahedrons $T'$ and $ABCD$ is $CK': CK = CE: CA = 1 / (k+1)$, so $V_J = V / (k+1)^3$. On the other hand, the ratio of the heights of $P$ and $T'$ is $K'K: CK' = EA: CE = k$, so $V_P: V_J = 3k$, and $V_P = 3kV / (k+1)^3$, and

$$
q=\frac{V(3k+1) / (k+1)^3}{V[1-(3k+1) / (k+1)^3]}=\frac{3k+1}{k^3+3k^2}
$$

To determine the ratio, no need was found for the data $a, b, d, \omega$, as could be foreseen in advance.

András Gáspár (Budapest, Vasútgépészeti t. III. o. t.)