The diagram shows a cube with 6 unit edges, and the points $K$ and $L$ are the trisection points of the edge $A E$. The cube is divided into parts by the planes $L H G$ and $K F G$. What is the volume of the part containing vertex $B$?

![](https://cdn.mathpix.com/cropped/2024_05_02_5cf585cbd165991a3e19g-1.jpg?height=339&width=396&top_left_y=264&top_left_x=862)
Solution. Let the points dividing the edge $D H$ into thirds be $M$ and $N$ (as shown in the figure), and the points dividing the edge $B F$ into thirds be $P$ and $Q$. The segment $K M$ is parallel to $F G$, and $K F$ is parallel to $M G$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9e876b3a716aef05c9e9g-1.jpg?height=461&width=455&top_left_y=271&top_left_x=824)

The parallelogram $K F G M$ lies in the plane $K F G$, since $K F$ and $F G$ are two intersecting lines in the plane. This plane cuts a prism out of the cube. The base of the prism is the right triangle $E K F$, and its height is $E H$. The volume of the prism is: $V_{\mathrm{h}}=\frac{6 \cdot 4}{2} \cdot 6=72$ cubic units.

The parallelism of $G Q$ and $H L$, as well as $L Q$ and $G H$, implies that $G H L Q$ is a parallelogram. The midline of the rectangle $K E F P$ is $L Q$, and its diagonal is $K F$, with their intersection point, $R$, being the midpoint of both segments. We have also cut off a pyramid with base $R Q F$ and height $F G$. The volume of the pyramid is: $V_{\mathrm{g}}=\frac{1}{3} \cdot \frac{3 \cdot 2}{2} \cdot 6=6$ cubic units.

![](https://cdn.mathpix.com/cropped/2024_05_02_9e876b3a716aef05c9e9g-1.jpg?height=482&width=455&top_left_y=1133&top_left_x=824)

The volume of the part containing vertex $B$ is obtained by subtracting the volumes of the prism and the pyramid from the volume of the cube:

$$
V=V_{\mathrm{k}}-\left(V_{\mathrm{h}}+V_{\mathrm{g}}\right)=216-78=138
$$

cubic units.