Given the tetrahedron $DABC = T$. The rays $DA, DB, DC$ are intersected by a plane at points $A_1, B_1$, and $C_1$ respectively. Show that the ratio of the volumes of $T$ and the tetrahedron $DA_1B_1C_1 = T_1$ is

$$
\frac{DA \cdot DB \cdot DC}{DA_1 \cdot DB_1 \cdot DC_1}
$$

A plane bisects the edges $KL$ and $LM$ of the tetrahedron $KLMN$, and intersects the edge $MN$ at point $P$ such that $MP = 2PN$. Determine the ratio of the volumes of the two parts of the tetrahedron that are cut by the plane.
a) Let $\angle ADC = \alpha$, and the angle between the ray $DB$ and the plane $ACD$ be $\beta$. Then the area of triangle $ACD$ is

$$
\frac{DA \cdot DC \cdot \sin \alpha}{2}
$$

and the height of the tetrahedron $ABCD$ corresponding to the base triangle $ACD$ is $DB \cdot \sin \beta$, so the volume of the tetrahedron $ABCD$ is

$$
\frac{DA \cdot DB \cdot DC \cdot \sin \alpha \cdot \sin \beta}{6}
$$

$1985-03-107-1 . \mathrm{eps}$

Similarly, the volume of the tetrahedron $A_1B_1C_1D_1$ is

$$
\frac{DA_1 \cdot DB_1 \cdot DC_1 \cdot \sin \alpha \cdot \sin \beta}{6}
$$

thus the ratio of the two volumes is indeed

$$
\frac{DA \cdot DB \cdot DC}{DA_1 \cdot DB_1 \cdot DC_1}
$$

provided that none of $A_1, B_1, C_1$ coincide with $D$.

The statement is also valid if the points $A_1, B_1, C_1$ - possibly all three - are taken on the extensions of the edges $DA, DB, DC$ beyond $D$, since in that case $\alpha$ and $\beta$ are replaced by $180^\circ - \alpha$ and $180^\circ - \beta$, respectively, but their sines remain the same.
b)

# $1985-03-107-2 . \mathrm{eps}$ 

Let the midpoints of $LM$ and $LK$ be $Q$ and $R$, respectively. We claim that the plane $\Sigma = PQR$ intersects the line $LN$ at a point $X$ which is the reflection of $L$ over $N$, i.e., $XN = NL$. Indeed, choosing $X$ this way, $XQ$ and $MN$ are medians of triangle $LMX$, and their intersection is the centroid, which trisects $MN$, thus coinciding with $P$. Conversely: the line $QP$ and hence the plane $\Sigma$ intersects $LN$ at $X$.

Similarly, we find that the line $XR$ - and hence the plane - intersects the edge $KN$ at point $S$ such that $KS = 2 \cdot SN$.

We already knew that $\Sigma$ does not intersect the edge $KM$ but is parallel to it, since the line $QR$ is the midline of triangle $LKM$.

Thus, $\Sigma$ intersects the faces of the tetrahedron $KLMN = T$ along the sides of the quadrilateral $PQRS$, and the parts of $T$ are pentahedra, with additional faces being 2 quadrilaterals and 2 triangles. The part that contains the entire edge $LN$ is obtained as the difference of the tetrahedra $LQRX$ and $NPSX$. For these - paired with $T$ - the theorem proved in part $a)$, or its extension, can be applied. The first pair is $LRQX$ and $LKMN$, the second pair is $NSPX$ and $NKM L$. The volume ratios are products of 3 factors, each factor being the ratio of two segments on the same line, one of whose endpoints is common, at $L$ or $N$:

$$
\begin{aligned}
& \frac{LR}{LK} \cdot \frac{LQ}{LM} \cdot \frac{LX}{LN} = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{1} = \frac{1}{2} \\
& \frac{NS}{NK} \cdot \frac{NP}{NM} \cdot \frac{NX}{NL} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{1} = \frac{1}{9}
\end{aligned}
$$

Thus, the volume of the part of the bisected tetrahedron that contains $LN$ is $\frac{1}{2} - \frac{1}{9} = \frac{7}{18}$ of the original. Therefore, the volume of the other part, which contains the edge $KM$, is $\frac{11}{18}$ of the original, and the ratio of the volumes of the two parts is $7:11$.