10,11

Find the volume of a regular quadrilateral pyramid with a base side of $a$ and plane angles at the vertex equal to the angles of the lateral edges with the base plane.
Let $PO$ be the height of the given regular pyramid $PABCD$, and $M$ be the midpoint of $BC$. Denote $\angle BPC = \angle OCP = \alpha$. From the right triangles $CMP$ and $COP$, we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_41a6f93ed2bb2ffd440eg-29.jpg?height=94&width=620&top_left_y=1800&top_left_x=719)

From this, we obtain the equation $\cos \alpha = \sqrt{2} \sin \frac{\alpha}{2}$. Since $\alpha$ is an acute angle, $\cos \alpha > 0$ and $\sin \frac{\alpha}{2} > 0$. Squaring both sides of this equation, we get

$$
\cos 2\alpha = 2 \sin 2 \frac{\alpha}{2}, \cos 2\alpha = 1 - \cos \alpha, \cos 2\alpha + \cos \alpha - 1 = 0
$$

from which $\cos \alpha = \frac{\sqrt{\sqrt{5} - 1}}{2} = \frac{2}{\sqrt{5} + 1}$. Then

$$
\begin{gathered}
\tan \alpha = \sqrt{\frac{1}{\cos^2 \alpha} - 1} = \sqrt{\frac{(\sqrt{5} + 1)^2}{1} - 1} = \frac{\sqrt{2 + 2\sqrt{5}}}{2} \\
PO = OC \tan \alpha = \frac{a \sqrt{2 + 2\sqrt{5}}}{2 \sqrt{2}} = \frac{a \sqrt{1 + \sqrt{5}}}{2}
\end{gathered}
$$

Therefore,

$$
V_{PABCD} = \frac{1}{3} S_{ABCD} \cdot PO = \frac{1}{3} a^2 \cdot \frac{a \sqrt{1 + \sqrt{5}}}{2} = \frac{a^2 \sqrt{1 + \sqrt{5}}}{6}
$$

## Answer

$\frac{a^2 \sqrt{1 + \sqrt{5}}}{6}$

Find the distance between the midpoints of two skew edges of a cube, the total surface area of which is 36.

## Solution

Let $M$ and $N$ be the midpoints of the skew edges $AB$ and $CC1$ of the cube $ABCD A1B1C1D1$ with edge $a$, $M1$ be the orthogonal projection of point $M$ onto the plane $A1B1C1D1$, and $K$ be the orthogonal projection of point $N$ onto the line $MM1$. Then

$$
\begin{gathered}
MN^2 = NK^2 + MK^2 = C1M1^2 + MK^2 = D1C1^2 + D1M1^2 + MK^2 = \\
= a^2 + \frac{a^2}{1} + \frac{a^2}{1} = \frac{3a^2}{2}.
\end{gathered}
$$

From the problem statement, it follows that $6a^2 = 36$, hence $a^2 = 6$. Therefore,

$$
MN^2 = \frac{3a^2}{2} = 3 \cdot \frac{6}{2} = 9.
$$

Thus, $MN = 3$.

## Answer

3.00