The base of the pyramid is the convex regular pentagon $ABCDE$, and the edges leading to the opposite vertex $M$ are of unit length. The edge $MA$ is perpendicular to the side face $MCD$.

What is the volume and surface area of the pyramid?
Since $M$ is at an equal distance from each vertex of the base, the same is true for the orthogonal projection $O$ of $M$ onto the base. Thus, a circle can be circumscribed around the base pentagon, which means it is a regular pentagon because the inscribed equilateral (convex) figure in a circle is regular.

![](https://cdn.mathpix.com/cropped/2024_05_02_ef4b67911bd82031f363g-1.jpg?height=585&width=857&top_left_y=301&top_left_x=623)

Since $M A$ is perpendicular to the plane $M C D$, it is also perpendicular to the symmetry axis $M F$ of this plane, where $F$ is the midpoint of the edge $C D$.

![](https://cdn.mathpix.com/cropped/2024_05_02_ef4b67911bd82031f363g-1.jpg?height=483&width=851&top_left_y=1038&top_left_x=623)

From the right triangle $M A F$, we can determine the necessary dimensions $O M = m$, $O F = \varrho$, $M F = b$, with $O A = r$ and $C D = a$ (the hypotenuse $A F$ passes through $O$, since the pentagon is regular). $\varrho = O C \cos 36^{\circ} = r \cos 36^{\circ}$,

$$
A F = \frac{A M^2}{A O} = \frac{1}{r} = r + \varrho = r \left(1 + \cos 36^{\circ}\right)
$$

from which

$$
\begin{gathered}
r = \frac{1}{\sqrt{1 + \cos 36^{\circ}}}, \quad \varrho = \frac{\cos 36^{\circ}}{\sqrt{1 + \cos 36^{\circ}}} = \frac{\sqrt{5} + 1}{2 \sqrt{5 + \sqrt{5}}} = \frac{1}{2} \sqrt{1 + \frac{\sqrt{5}}{5}} \\
m = O M = \sqrt{O F \cdot O A} = \sqrt{\varrho r} = r \sqrt{\cos 36^{\circ}} = \sqrt{\frac{\cos 36^{\circ}}{1 + \cos 36^{\circ}}} = \sqrt{\frac{\sqrt{5} + 1}{5 + \sqrt{5}}} = \sqrt{\frac{1}{\sqrt{5}}}
\end{gathered}
$$

since $\cos 36^{\circ}$ is half the ratio of the diagonal to the side of a regular pentagon, which is $\frac{\sqrt{5} + 1}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_ef4b67911bd82031f363g-2.jpg?height=872&width=849&top_left_y=125&top_left_x=627)

Furthermore,

$$
\begin{gathered}
b = \sqrt{m^2 + \varrho^2} = r \sqrt{\cos 36^{\circ} (1 + \cos 36^{\circ})} = \sqrt{\cos 36^{\circ}} = \frac{1}{2} \sqrt{1 + \sqrt{5}} \\
a = 2 r \sin 36^{\circ} = \frac{2 \sqrt{1 - \cos^2 36^{\circ}}}{\sqrt{1 + \cos 36^{\circ}}} = 2 \sqrt{1 - \cos 36^{\circ}} = \sqrt{3 - \sqrt{5}}
\end{gathered}
$$

The area of the base and the surface area of the pyramid are:

$$
t = \frac{5}{2} a \varrho = \frac{1}{4} \sqrt{50 - 10 \sqrt{5}} = 1.31 \text{ area units}
$$

and

$$
F = t + \frac{5}{2} a b = \frac{5}{4} \sqrt{2 (\sqrt{5} - 1)} = 3.28 \text{ area units}
$$

The volume is

$$
V = \frac{t m}{3} = \frac{1}{12} \sqrt{10 (\sqrt{5} - 1)} = 0.293 \text{ volume units}
$$

Remark. Many solutions used the fact that the base is a regular pentagon without proof. These are not complete solutions. The editorial board deliberately omitted stating the regularity to avoid giving unnecessary information and to make proving it part of the task for the solvers.