Task B-3.5. A cube is determined by its vertices $A B C D A_{1} B_{1} C_{1} D_{1}$. Determine the ratio of the volume of the tetrahedron $A C B_{1} D_{1}$ to the volume of the given cube. In what ratio are their surface areas?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8c954f607b7f1570b01bg-13.jpg?height=545&width=615&top_left_y=687&top_left_x=728)

(1 point)

The volume of the given tetrahedron is calculated by subtracting the volume of 4 triangular pyramids, each with a base area of $\frac{a^{2}}{2}$ and height $a$, from the volume of the cube, where $a$ is the length of the edge of the cube.

$$
\begin{aligned}
V_{\text {tetrahedron }} & =V_{\text {cube }}-4 V_{\text {pyramid }}= \\
& =a^{3}-4 \frac{1}{3} \cdot \frac{a^{2}}{2} \cdot a= \\
& =\frac{1}{3} a^{3}=\frac{1}{3} V_{\text {cube }}
\end{aligned}
$$

Then,

$$
V_{\text {tetrahedron }}: V_{\text {cube }}=1: 3
$$

The ratio of the surface area of the tetrahedron to the surface area of the cube is

$$
O_{\text {tetrahedron }}: O_{\text{cube}}=\frac{4 \cdot \frac{(a \sqrt{2})^{2} \sqrt{3}}{4}}{6 a^{2}}=\frac{2 a^{2} \sqrt{3}}{6 a^{2}}=\sqrt{3}: 3
$$