Let's calculate the edge angle at the apex of a regular square pyramid, given that the centers of the inscribed and circumscribed spheres coincide.

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I. solution: Let the common center of the two spheres be $O$. The points of tangency of the inscribed sphere are: $K$ (the center of the base square) and $O_{1}$ (on the $A S B$ pyramid face - see 1st figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_cd2f34b460290f287615g-1.jpg?height=428&width=466&top_left_y=255&top_left_x=794)

The plane of the $A S B$ pyramid face cuts the circumscribed sphere, which also has center $O$, to form the circumcircle of $\triangle A S B$, whose center is $O_{1}$, since $O O_{1}$ is perpendicular to the plane of $\triangle A B S$ by the given condition. Therefore, $O_{1}$ lies on the altitude $F S$ of the isosceles triangle $A S B$, where $F$ is the midpoint of the base edge $B C = a$.

$F K = F O_{1}$, as the two tangents drawn from point $F$ to the inscribed sphere. But $F K = \frac{a}{2} = F A = F B$, and thus $F A = F B = F O_{1}$. Since $F O_{1} \perp A B$, $\triangle A O_{1} B$ is isosceles.

Therefore, the central angle $\angle A O_{1} B$ corresponding to the chord $A B$ of the circumcircle of $\triangle A B S$ is $90^{\circ}$, and thus

$$
\angle A S B = 45^{\circ}
$$

as the inscribed angle corresponding to the chord $A B$.

Zoltán Daróczy (Debrecen Ref. g. III. o. t.)

II. solution: The faces of the pyramid are tangent to the inscribed sphere, so they are at equal distances from the center $O$, and thus they cut out congruent circles from the circumscribed sphere centered at $O$. Therefore, the circumcircle of the square $A B C D$ is congruent to the circumcircle of $\triangle A B S$. The common chord $A B$ of these two congruent circles is seen from any point on either circle at the same inscribed angle, so

$$
\angle A S B = \angle A D B = 45^{\circ}
$$

Györgyi Lackner (Bp. V., Textilip. techn. IV. o. t.)

III. solution: We can generalize our problem to a regular pyramid with $n$ sides $(n \geq 3)$. The right triangles $\triangle A F K$ and $\triangle A F O_{1}$ are congruent because the leg $A F$ is common, and the other two legs, being the two tangents drawn from $F$ to the inscribed sphere, are equal, i.e., $F K = F O_{1}$. As is known, $\angle A K F = \frac{180^{\circ}}{n}$, and thus, since the central angle is twice the inscribed angle,

$$
\angle A O_{1} F = \angle A S B = \frac{180^{\circ}}{n}.
$$

In our problem, $n = 4$, so the desired angle is $\frac{180^{\circ}}{4} = 45^{\circ}$.

Gergely Krammer (Bp. II., Rákóczi g. IV. o. t.)

Remark: The general solution highlights that for regular pyramids with a common center for the inscribed and circumscribed spheres, the sum of the dihedral angles at the apex is $180^{\circ}$. For a right circular cone, the lateral surface when unrolled is a semicircle. (For an equilateral cone.)