A regular pentagonal pyramid has unit-length base edges and side edges such that when the side faces are rotated outward into the plane of the base, a regular star pentagon is formed.

What is the radius of the sphere inscribed in the pyramid, and what is the radius of the circumscribed sphere?
I. Solution. Let the vertices of the base (regular pentagon) be denoted by $A, B, C, D, E$, and the sixth vertex of the pyramid by $F$. Since the lateral faces form a regular star pentagon when rotated, the rotated position of $F$ around the $A E$ edge will be at the intersection of the lines $A B$ and $E D$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_a2472d86c5673be2810cg-1.jpg?height=1138&width=1808&top_left_y=323&top_left_x=148)

Figure 1

According to this, the lateral faces are isosceles triangles where the base angles are $72^{\circ}$, and the length of the lateral edges is

$$
A F=\frac{A E / 2}{\cos \angle F A E}=\frac{1}{2 \cos 72^{\circ}}=\frac{1}{2 \sin 18^{\circ}}=\frac{2}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{2}(=1.618 \text { units }).
$$

Let the center of the base be $O$ - which is the foot of the height - and the midpoint of the $A E$ edge be $G$. From the regularity of the pyramid, it follows that the center of both spheres will lie on the $F O$ height line (axis of rotational symmetry), and the inscribed sphere touches the base at $O$ and the $A E F$ face at a point on the altitude $G F$.

Let the center of the inscribed sphere be $K_{b}$, and its radius be $\varrho$. Then $K G_{b}$ bisects the angle $\angle F G O = \gamma$, and

$$
\varrho=O K_{b}=G O \operatorname{tg} \frac{\gamma}{2}=G O \sqrt{\frac{1-\cos \gamma}{1+\cos \gamma}}
$$

Using known trigonometric identities and the fact that the trigonometric functions of $18^{\circ}$ and its multiples can be calculated from the regular pentagon, we have, on the one hand,

$$
G O=A G \operatorname{tg} 54^{\circ}=\frac{1}{2} \sqrt{\frac{5+2 \sqrt{5}}{5}}=\frac{1}{2} \sqrt{1+\frac{2}{\sqrt{5}}}
$$

On the other hand, $\cos \gamma = G O / G F$, so the expression under the square root is:

$$
\begin{gathered}
\frac{1-\cos \gamma}{1+\cos \gamma}=\frac{G F-G O}{G F+G O}=\frac{G A\left(\operatorname{tg} 72^{\circ}-\operatorname{tg} 54^{\circ}\right)}{G A\left(\operatorname{tg} 72^{\circ}+\operatorname{tg} 54^{\circ}\right)}=\frac{\sin 18^{\circ}}{\sin 3 \cdot 18^{\circ}}=\frac{1}{3-4 \sin ^{2} 18^{\circ}} \\
=\frac{2}{3+\sqrt{5}}=\frac{3-\sqrt{5}}{2}
\end{gathered}
$$

Thus,

$$
\varrho=\frac{1}{2} \sqrt{\frac{5+2 \sqrt{5}}{5}} \cdot \frac{3-\sqrt{5}}{2}=\sqrt{\frac{5+\sqrt{5}}{40}}(=0.4253 \text { units }).
$$

To calculate the radius $R$ of the circumscribed sphere, we use the triangle $C O F$. Let the center of the sphere be $K_{k}$, and the midpoint of $F C$ be $H$. From the similarity of the right triangles $F K_{k} H$ and $F C O$,

$$
\frac{F H}{R}=\frac{F O}{F C}, \quad R=\frac{F C^{2}}{2 \cdot F O}=\frac{F C^{2}}{2 \sqrt{F C^{2}-O C^{2}}}
$$

Here,

$$
O C=O A=\frac{A G}{\cos 54^{\circ}}=\frac{1}{2 \sin 36^{\circ}}=\frac{2}{\sqrt{10-2 \sqrt{5}}}=\sqrt{\frac{5+\sqrt{5}}{10}}
$$

Under the square root in the denominator,

$$
F C^{2}-O C^{2}=\left(\frac{\sqrt{5}+1}{2}\right)^{2}-\frac{5+\sqrt{5}}{10}=\frac{1}{5-2 \sqrt{5}}
$$

Thus, the final result is

$$
R=\frac{1}{4} \sqrt{10+2 \sqrt{5}}(=0.9511 \text { units }).
$$

Remark. Based on our calculations, we make two observations. First, $R / \varrho=\sqrt{5}$, and second,

$$
F O=\frac{F C^{2}}{2 R}=\sqrt{1+\frac{2}{\sqrt{5}}}=R+\varrho
$$

so $K_{b}$ and $K_{k}$ coincide.

II. Solution (outline). Start from the Euclidean construction of a regular pentagon (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_a2472d86c5673be2810cg-2.jpg?height=921&width=1092&top_left_y=1255&top_left_x=495)

Let $O A$ and $O B$ be perpendicular radii of the circle $k$, $C$ the midpoint of the segment $O A$, and denote the circles centered at $C$ passing through $A$ and $B$ by $k_{0}$ and $k_{1}$, respectively. Let the points of intersection of $k_{1}$ with the line $O A$ be $D$ and $E$ (with $D$ being the closer to $O$). In the right triangle $E B D$, $B O$ is the geometric mean of the segments $D O$ and $O E = O A + A E = O B + D O$, so

$$
D O: O B=O B:(O B+D O)
$$

Thus, the ratio of $D O$ to $O B$ is the same as the ratio of the side of a regular decagon inscribed in $k$ to the radius of the circumscribed circle. Therefore, $D O$ is indeed the side of a regular decagon inscribed in $k$, and

$$
B E^{2}=D E \cdot O E
$$

so $B E$ is the diagonal of a regular pentagon inscribed in $k$, and $D B: B E=D O: O B$ implies that $D B$ is the side of a regular pentagon inscribed in $k$.

Rotate this figure around the line $D E$, and denote the spheres formed by the rotation of $k_{0}$ and $k_{1}$ by $G_{0}$ and $G_{1}$, respectively. The plane formed by the rotation of $B O$ intersects $G_{1}$ in a circle congruent to $k$, so the sides of the pentagon inscribed in this circle are equal to $B D$, and the diagonals are equal to $B E$. Thus, this pentagon, together with $E$, defines the pyramid in the problem, if $B D$ is taken to be 1 unit.

The sphere $G_{1}$ is naturally the circumscribed sphere of the pyramid. We show that $G_{0}$ is the inscribed sphere. By the construction, $G_{0}$ touches the base, which, for example, contains the diagonals from vertex $B$ that form triangles congruent to the lateral faces. Since $B C = E C$, the tangent planes from $E$ to $G_{0}$ also intersect $G_{1}$ in circles congruent to $k$, so the lateral faces of the pyramid indeed touch $G_{0}$.

If $O C = \varrho$, then $B C = \varrho \sqrt{5}$, $D O = \varrho(\sqrt{5}-1)$, and $D B^{2} = \varrho^{2}(10-2 \sqrt{5})$. If $D B = 1$, then

$$
\varrho^{2}=\frac{1}{2} \frac{1}{5-\sqrt{5}}=\frac{5+\sqrt{5}}{40}
$$

Thus, the radius of the inscribed sphere of the pyramid is $\varrho=\sqrt{\frac{5+\sqrt{5}}{40}}$ and the radius of the circumscribed sphere is $B C = \varrho \sqrt{5} = \sqrt{\frac{5+\sqrt{5}}{8}}$.