10,11

The side of the base of a regular quadrilateral pyramid is equal to $a$, and the dihedral angle at the base is $60^{\circ}$. Find the radius of the sphere that touches two adjacent lateral edges, the opposite lateral face, and the base.
Let a sphere of radius $R$ with center $O$ touch the base $ABCD$ of a regular quadrilateral pyramid $PABCD$ at point $M$, the lateral face $CPD$ at point $N$, and the lateral edges $AP$ and $BP$ at points $K$ and $L$ respectively. Since $PK = PL$ and the center $O$ of the sphere is equidistant from points $K$ and $L$, the point $O$ lies in the plane passing through the midpoint of $KL$ perpendicular to $KL$, i.e., in the plane $PEF$, where $E$ and $F$ are the midpoints of the sides of the base $AB$ and $CD$ respectively. The plane $PEF$ is perpendicular to the lines $AB$ and $CD$. Therefore, $PEF$ is the linear angle of the dihedral angle of the pyramid at the edge $AB$. According to the problem, $\angle PEF = 60^\circ$. Thus, $PEF$ is an equilateral triangle with side length $a$. Consider the section of the sphere by the plane $PEF$. We obtain a circle of radius $R$ with center $O$, touching the sides $FE$ and $PF$ of the equilateral triangle $PEF$ at points $M$ and $N$ respectively. Since $FO$ is the angle bisector of $\angle PFE$ of the equilateral triangle $PFE$, the extension of segment $FO$ passes through the center $Q$ of the circle of intersection of the specified sphere with the plane of the face $PAB$. Let $PN = PL = PK = x$. From the right triangles $BEP$ and $PQL$, we find that

\[
\begin{gathered}
BP = \sqrt{BE^2 + PE^2} = \sqrt{\frac{a^2}{4} + a^2} = \frac{a \sqrt{5}}{2}, \\
\sin \angle PBE = \frac{BE'}{BP'} = \frac{\frac{a}{a \sqrt{5}}}{2} = \frac{2}{\sqrt{5}}, \sin \angle BPE = \frac{BE'}{BP'} = \frac{\frac{a}{2}}{\frac{a \sqrt{5}}{2}} = \frac{1}{\sqrt{5}},
\end{gathered}
\]

\[
PQ = \frac{PL}{\sin \angle PQL} = \frac{PL}{\sin \angle PBE} = \frac{\frac{x}{2}}{\sqrt{5}} = \frac{a \sqrt{5}}{2}
\]

Then

\[
\frac{x \sqrt{5}}{2} = PQ = \frac{1}{2} PE = \frac{a}{2}
\]

Thus,

\[
FM = FN = a - x = a - \frac{a}{\sqrt{5}}
\]

Therefore,

\[
R = OM = \frac{a - x}{\sqrt{3}} = \frac{a - \frac{a}{\sqrt{5}}}{\sqrt{3}} = \frac{a(\sqrt{5} - 1)}{\sqrt{15}}
\]

## Answer

$\frac{\mathbf{a}(\sqrt{5}-1)}{\sqrt{15}}$