All vertices of the regular pyramid $P A B C D$ lie on the lateral surface of a cylinder, the axis of which is perpendicular to the plane $P A B$. Find the radius of the base of the cylinder, if $A B=a$.

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Since the axis of the cylinder is perpendicular to the plane of face $ABP$, the orthogonal projection of the cylinder onto the plane of face $ABP$ is a circle, the radius of which is equal to the radius of the base of the cylinder. On this circle lie the points $A, B$, and $P$, as well as the orthogonal projections $C_1$ and $D_1$ of the vertices $C$ and $D$ of the pyramid $PABCD$. Since $CD$ is parallel to the plane $ABP$, $C_1D_1 = CD = AB$ and $C_1D_1 \| CD \| AB$, so $AB C_1 D_1$ is a rectangle. Its center $O_1$ coincides with the center of the circle. Therefore, $O_1A = O_1B = O_1P$. On the other hand, point $O_1$ is the orthogonal projection of the center $O$ of the square $ABCD$, so $O_1A, O_1B$, and $O_1P$ are the orthogonal projections of segments $OA, OB$, and $OP$ onto the plane $ABP$, which means $OP = OA = OB = \frac{\mathbf{a}}{\sqrt{2}}$. Consequently, $AP = BP = AB = a$, and the radius of the circle circumscribed around an equilateral triangle with side $a$ is $\frac{\frac{a}{\sqrt{3}}}{\sqrt{3}}$.

## Answer

$\frac{a}{\sqrt{3}}$.