[ Right Pyramid ]  [ Unfolding helps solve the problem. ]

The lateral edge of a regular quadrilateral pyramid is $b$, and the plane angle at the vertex is $\alpha$. Find the length of the shortest closed path on the surface of the pyramid, starting and ending at the vertex of the base and intersecting all lateral edges of the pyramid.
Let $A$ be the vertex of the base $ABCD$ of a regular quadrilateral pyramid $PABCD$ (Fig.1). Consider the development of the lateral surface of the pyramid, consisting of four equal isosceles triangles $A'P'B'$, $B'P'C'$, $C'P'B'$, and $D'P'A''$ with an angle $\alpha$ at their common vertex $P'$. If $\alpha \leqslant 45^{\circ}$, then $\angle A'P'A'' \leqslant 180^{\circ}$, so the specified path from $A$ to $A$ along the surface of the pyramid is the sum of the segments $A'M$, $MN$, $NK$, and $KA''$, where points $M$, $N$, and $K$ lie on segments $B'P'$, $C'P'$, and $D'P'$ respectively (Fig.2). In this case,

$$
A'M + MN + NK + KA'' \geqslant A'A'' = 2A'P' \sin 2\alpha = 2b \sin 2\alpha
$$

and the shortest path on the considered development is the segment $A'A''$. If, however, $\alpha > 45^{\circ}$ (Fig.3), then $\angle A'P'N = \angle NP'A'' = 2\alpha > 90^{\circ}$, so

$$
\begin{gathered}
A'M + MN + NK + KA'' = (A'M + MN) + (NK + KA'') \geqslant \\
\geqslant A'N + NA'' \geqslant A'P' + A''P' = 2b
\end{gathered}
$$

(in triangles $A'P'N$ and $A''P'N$, sides $A'N$ and $A''N$ lie opposite obtuse angles). In this case, the shortest path on the considered development consists of segments $A'P'$ and $P'A''$.

## Answer

$2b \sin 2\alpha$, if $\alpha \leqslant 45^{\circ} ; 2b$, if $45^{\circ} < \alpha < 90^{\circ}$.