$[\quad$ Prism (other). ] $[\underline{\text { Round bodies (other) })]}$

The base of the prism $A B C A 1 B 1 C 1$ is an equilateral triangle $A B C$ with side $a$. The orthogonal projection of vertex $A 1$ coincides with the center of the base $A B C$, and the lateral edge forms an angle of $60^{\circ}$ with the plane of the base. Find the lateral surface area of the prism.
Let $O$ be the center of the base $ABC$ of the given prism $ABC A1B1C1$, $M$ and $K$ be the midpoints of the edges $BC$ and $AB$ respectively, and $S$ be the lateral surface of the prism. Since $O$ is the orthogonal projection of vertex $A1$ onto the plane of the base $ABC$, $OA$ is the orthogonal projection of the lateral edge $AA1$ onto this plane. Therefore, $\angle A1AM$ is the angle between the lateral edge $AA1$ and the plane of the base $ABC$. According to the problem, $\angle A1AM = 60^\circ$. From the right triangle $A1AM$, we find that

$$
\begin{gathered}
AA1 = \frac{OA}{\cos \angle A1AM} = \frac{2}{3} \cdot \frac{AM}{\cos 60^\circ} = \frac{2}{3} \cdot \frac{1}{2} a \sqrt{3} \cdot 2 = \frac{2}{3} a \sqrt{3} \\
OA1 = OA \tan \angle A1AM = \frac{1}{3} a \sqrt{3} \tan 60^\circ = \frac{1}{3} a \sqrt{3} \cdot \sqrt{3} = a
\end{gathered}
$$

Since $AM \perp BC$, by the theorem of three perpendiculars, $AA1 \perp BC$. Since $BB1 \parallel AA1$, $BB1 \perp BC$. Therefore, $BB1C1C$ is a rectangle. Thus,

$$
S_{BB1C1C} = BC \cdot BB1 = a \cdot \frac{2}{3} a \sqrt{3} = \frac{2}{3} a^2 \sqrt{3}
$$

Since $OK$ is the orthogonal projection of segment $A1K$, by the theorem of three perpendiculars, $A1K \perp AB$. Therefore, $A1K$ is the height of the parallelogram $AA1B1B$. From the right triangle $A1KO$, we find that

$$
A1K = \sqrt{OK^2 + OA1^2} = \sqrt{\left(\frac{1}{6} a \sqrt{3}\right)^2 + a^2} = \frac{a \sqrt{13}}{\sqrt{12}}
$$

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_41a6f93ed2bb2ffd440eg-12.jpg?height=132&width=698&top_left_y=2295&top_left_x=680)

Similarly, we find that $S_{AA1C1C} = \frac{\frac{a}{2}^2 \sqrt{13}}{2 \sqrt{3}}$. Therefore,

$$
\begin{gathered}
S = S_{BB1C1C} + S_{AA1B1B} + S_{AA1C1C} = \frac{2}{3} a^2 \sqrt{3} + 2 \cdot \frac{a^2 \sqrt{13}}{2 \sqrt{3}} = \\
= \frac{2}{3} a^2 \sqrt{3} + \frac{a^2 \sqrt{13}}{\sqrt{3}} = \frac{2 a^2}{\sqrt{3}} + \frac{a^2 \sqrt{13}}{\sqrt{3}} = \frac{a^2 (2 + \sqrt{13})}{\sqrt{3}}
\end{gathered}
$$

Prove that segments of parallel lines enclosed between two parallel planes are equal.

## Solution

Let planes $\alpha$ and $\beta$ be parallel, line $a$ intersects them at points $A$ and $B$ respectively, and parallel line $a1$ intersects them at points $A1$ and $B1$. Consider plane $\gamma$ passing through the parallel lines $a$ and $a1$. It intersects the parallel planes $\alpha$ and $\beta$ along lines $AA1$ and $BB1$. Therefore, $AA1 \parallel BB1$. The opposite sides of quadrilateral $AA1B1B$ are parallel, so $AA1B1B$ is a parallelogram. Consequently, $AB = A1B1$