## Task B-4.7.

The length of the side of the base of a regular right quadrilateral prism is 1. The plane $\pi$ contains one vertex of the upper base and one diagonal of the lower base, and divides the prism into two parts of different volumes. The lines in which the plane $\pi$ intersects the two lateral sides of the prism form an angle $\varphi$ such that $\cos \varphi=\frac{1}{3}$. Calculate the surface area and volume of the smaller of the resulting geometric bodies.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_63d9d2269da4cabd8f72g-26.jpg?height=734&width=762&top_left_y=1021&top_left_x=567)

According to the conditions of the problem, we have $|C D|=1$ and $\varangle B C_{1} D=\varphi$. We easily obtain $|B D|=\sqrt{2}$, $|O D|=\frac{\sqrt{2}}{2}$.

From $\triangle O C_{1} D$, it follows that $\left|C_{1} D\right|=\frac{|O D|}{\sin \frac{\varphi}{2}}=\frac{\sqrt{2}}{2 \sin \frac{\varphi}{2}}$.

From $\triangle C_{1} C D$, it follows that $\left|C C_{1}\right|^{2}=\left|C_{1} D\right|^{2}-|C D|^{2}=\frac{1}{2 \sin ^{2} \frac{\varphi}{2}}-1$.

Since $\cos \varphi=\frac{1}{3}$, it follows that $\sin ^{2} \frac{\varphi}{2}=\frac{1-\cos \varphi}{2}=\frac{1}{3}$, so $\left|C C_{1}\right|=\frac{\sqrt{2}}{2}$.

The volume of the pyramid $B D C C_{1}$ is

$$
V=\frac{1}{3} P_{B D C} \cdot\left|C C_{1}\right|=\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{12}
$$

Notice that $|O C|=\left|C C_{1}\right|=\frac{\sqrt{2}}{2}$, so $\left|O C_{1}\right|=1$.

$$
\begin{aligned}
P_{\triangle B D C} & =\frac{1}{2} \\
P_{\triangle D C C_{1}} & =\frac{|D C| \cdot\left|C C_{1}\right|}{2}=\frac{1}{2} \cdot 1 \cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{4} \\
P_{\triangle B D C_{1}} & =\frac{|B D| \cdot\left|O C_{1}\right|}{2}=\frac{1}{2} \cdot \sqrt{2} \cdot 1=\frac{\sqrt{2}}{2}
\end{aligned}
$$

The surface area of the pyramid is:

$$
O=P_{\triangle B D C}+2 \cdot P_{\triangle D C C_{1}}+P_{\triangle B D C_{1}}=\frac{1}{2}+2 \cdot \frac{\sqrt{2}}{4}+\frac{\sqrt{2}}{2}=\frac{1}{2}+\sqrt{2}
$$