[ $\quad \underline{\text { Tangents to Spheres }}$]

In a regular hexagonal pyramid $\operatorname{SABCDEF}$ ( $S$ - apex) the side of the base is $2 \sqrt{3}$, and the height of the pyramid $S H$ is 6. A plane passes through point $E$ perpendicular to line $A S$ and intersects segment $S H$ at point $O$. Points $P$ and $Q$ are located on lines $A S$ and $C E$ respectively, such that line $P Q$ is tangent to a sphere of radius $\frac{\sqrt{5}}{2}$ centered at point $O$. Find the minimum length of segment $P Q$.
Let the diagonals $A D$ and $C E$ of the regular hexagon $A B C D E F$ intersect at point $N$. From the properties of a regular hexagon, it follows that $N$ is the midpoint of segments $D H$ and $C E$, and $A D \perp C E$. Then, by the theorem of three perpendiculars, $C E \perp S A$, which means that the line $C E$ lies in the plane passing through point $E$ and perpendicular to the line $S A$. Let this plane intersect the line $S A$ at point $K$. Then

\[
\begin{gathered}
A H=A B=2 \sqrt{3}, N H=\frac{1}{2} D H=\sqrt{3}, A N=N H+A H=3 \sqrt{3}, \\
\operatorname{tg} \angle S A H=\frac{S_{H}}{A H}=\frac{6}{2 \sqrt{3}}=\sqrt{3}, \angle S A H=60^{\circ}, N K=A N \sin 60^{\circ}=3 \sqrt{3} \cdot \frac{\sqrt{3}}{2}=\frac{9}{2}, \\
O N=\frac{O N}{\text { con } N 30^{\circ}}=\frac{\sqrt{3}}{\sqrt{3}}=2, O K=N K-O N=\frac{\frac{9}{2}}{2}-2=\frac{5}{2}.
\end{gathered}
\]

Let the line passing through points $P$ and $Q$ be tangent to the sphere at point $M$. Then $O M \perp P Q$. Denote $P K=x$, $N Q=y$. From the right triangles $N P K, K O P, M O P, Q N O$, and $N P Q$, we find that

\[
\begin{gathered}
O P 2=P K 2+O K 2=x 2+\frac{25}{4}, N P 2=P K 2+N K 2=x 2+\frac{81}{4} \\
M P 2=O P 2-O M 2=\left(x 2+\frac{25}{4}\right)-\frac{5}{4}=x 2+5, \\
Q O 2=N Q 2+N O 2=y 2+4, Q M 2=Q O 2-O M 2=(y 2+4)-\frac{5}{4}=y 2+\frac{\mathbf{1 1}}{4} \\
P Q 2=N P 2+N Q 2=\left(x 2+\frac{81}{4}\right)+y 2=x 2+y 2+\frac{81}{4}
\end{gathered}
\]

Since $P Q=P M+M Q$, we get the equation

\[
\sqrt{x^{2}+5}+\sqrt{y^{2}+\frac{11}{4}}=\sqrt{x^{2}+y^{2}+\frac{81}{4}}
\]

from which we find that $y 2=\frac{625}{16\left(x^{2}+5\right)}-\frac{11}{4}$. Then

\[
\begin{gathered}
P Q 2=x 2+y 2+\frac{81}{4}=x 2+\frac{625}{16\left(x^{2}+5\right)-4}+\frac{11}{4}+x 1 \\
=x 2+5+\frac{625}{16\left(x^{2}+5\right)}+\frac{81}{4}-5= \\
=x 2+5+\frac{625}{16\left(x^{2}+5\right)}+\frac{35}{2}-5 \geqslant 2 \sqrt{\left(x^{2}+5\right) \cdot \frac{625}{16\left(x^{2}+5\right)}}+\frac{25}{2}=\frac{25}{2}+\frac{25}{2}=25
\end{gathered}
\]

with equality achieved when

\[
x 2+5=\frac{625}{16\left(x^{2}+5\right)} \Leftrightarrow(x 2+5) 2=\frac{625}{16} \Leftrightarrow x 2=\frac{5}{4}.
\]

Therefore, the minimum length of the segment $P Q$ is 5.

## Answer

$P Q=5$.