In a regular pyramid $P A B C$, the side of the base $A B C$ is $a$, and the lateral edge is $2a$. Points $P, B$, and $C$ lie on the lateral surface of a cone with its vertex at point $A$. Find the angle at the vertex of the axial section of the cone.
On the extensions of edges $A B$ and $A C$ beyond points $B$ and $C$, we lay off segments $B M$ and $C N$, each equal to $a$. Then $A P = A M = A N = 2 a$, so points $P, M$, and $N$ lie on the circumference of the base of a cone with vertex $A$ and slant height $2 a$. The circle at the base of this cone is circumscribed around triangle $M N P$. Let $r$ be its radius. Denote $P M = P N = x$. In triangle $A P M$, using the formula for the median,

$$
4 P B^2 = 2 P A^2 + 2 P M^2 - \text{AM}^2, \text{ or } 16 a^2 = 8 a^2 + 2 \times 2 - 4 a^2,
$$

from which $x = a \sqrt{6}$. Since $B C$ is the midline of the triangle, $M N = 2 B C = 2 a$. If $P K$ is the height of the isosceles triangle $M P N$, then

$$
\begin{gathered}
P K = \sqrt{P M^2 - M K^2} = \sqrt{6 a^2 - a^2} = a \sqrt{5}, \\
\sin \angle P M N = \frac{P K}{P M} = \frac{\sqrt{5}}{\sqrt{6}}.
\end{gathered}
$$

Therefore,

$$
r = \frac{P N}{2 \sin \angle P M N} = \frac{a \sqrt{6}}{\frac{2 \sqrt{6}}{\sqrt{6}}} = \frac{3 a}{\sqrt{5}}
$$

Let $A O$ be the height of the cone, $\alpha$ be the angle at the vertex of the axial section. Then

$$
\sin \frac{\alpha}{2} = \frac{O M}{A M} = \frac{r}{2 a} = \frac{\sqrt{5}}{2 a} = \frac{3}{2 \sqrt{5}}
$$

## Answer

$2 \arcsin \frac{3}{2 \sqrt{5}}$