In the regular pyramid $S M N P Q$ ( $S$ - vertex), points $H$ and $F$ are the midpoints of edges $M N$ and $N P$ respectively, and point $E$ lies on segment $S H$, such that $S H=3, S E=\frac{9}{1}$. The distance from point $S$ to line $E F$ is $\sqrt{5}$. Find the volume of the pyramid. A sphere of radius 1 is given with its center at point $S$. Consider all possible regular tetrahedra $A B C D$ such that points $C$ and $D$ lie on line $E F$, and line $A B$ is tangent to the sphere at one of the points on segment $A B$. Find the minimum value of the edge length of the considered tetrahedra.
Let $K$ be the midpoint of $F H$ (Fig.1), $G$ be the foot of the perpendicular dropped from vertex $S$ to line $F E$, $L$ be the foot of the perpendicular dropped from point $E$ to $F H$, and $F T$ be the height of triangle $S F H$. Let $H F=2 x$. Since the pyramid $S M N P Q$ is regular, $S F=S H=3$. Therefore,

\[
\begin{gathered}
S K=\sqrt{S H^{2}-K H^{2}}=\sqrt{9-x^{2}}, F T=\frac{F H S K}{S H}=\frac{2 \pi \sqrt{9-x^{2}}}{3}, \\
E L=S K \cdot \frac{E H}{S H}=\sqrt{9-x^{2}} \cdot \frac{1}{1}=\frac{\sqrt{9-x^{2}}}{1}, \\
H L=\frac{1}{1} K H=\frac{\pi}{1}, F L=F H-H L=\frac{7 \pi}{1} \\
F E=\sqrt{E L^{2}+F L^{2}}=\sqrt{\frac{19 x^{2}}{16}+\frac{9-x^{2}}{16}}=\frac{\sqrt{19 x^{2}+9-x^{2}}}{1}=\frac{\sqrt{12 r^{2}+9}}{1},
\end{gathered}
\]

and since $F E \cdot S G=S E \cdot F T$ (twice the area of triangle $S E F$), we have the equation

\[
\frac{\sqrt{18 e^{2}+9}}{1} \cdot \sqrt{5}=\frac{9}{1} \cdot \frac{2 m \sqrt{9-x^{2}}}{3}.
\]

After obvious transformations, we get the equation

\[
36 \times 4-84 \times 2+45=0,
\]

from which $x=\sqrt{\frac{3}{2}}$ or $x=\sqrt{\frac{5}{6}}$. Let $S O$ be the height of the pyramid $S M N P Q$. Denote the side of the square $M N P Q$ by $a$. Then

\[
\begin{gathered}
2 x=F H=\frac{1}{2} M P=\frac{\alpha \sqrt{2}}{2}, a=2 x \sqrt{2} \\
S O=\sqrt{S H^{2}-O H^{2}}=\sqrt{9-\frac{\alpha^{2}}{1}}=\sqrt{9-2 x^{2}}
\end{gathered}
\]

If $x=\sqrt{\frac{3}{2}}$, then

\[
a=2 \sqrt{3}, S_{M N P Q}=a 2=12, S O=\sqrt{6}
\]

Therefore,

\[
V_{S M N P Q}=\frac{1}{2} S_{M N P Q} \cdot S O=\frac{1}{3} \cdot 12 \cdot \sqrt{6}=4 \sqrt{6}.
\]

If $x=\sqrt{\frac{5}{6}}$, then

\[
a=\frac{2 \sqrt{5}}{\sqrt{3}}, S_{M N P Q}=a 2=\frac{2 m}{3}, S O=\frac{\sqrt{22}}{\sqrt{3}}
\]

Therefore,

\[
V_{S M N P Q}=\frac{1}{3} S_{M N P Q} \cdot S O=\frac{1}{3} \cdot \frac{29}{3} \cdot \frac{\sqrt{22}}{\sqrt{3}}=\frac{20 \sqrt{22}}{9 \sqrt{3}}.
\]

Let the vertices $C$ and $D$ of the regular tetrahedron $A B C D$ lie on the line $E F$ (Fig.2), and the vertices $A$ and $B$ lie on the line tangent to the given sphere at point $X$, which lies on the segment $A B$. Since opposite edges of a regular tetrahedron are perpendicular, $C D \perp A B$. Through the line $A B$, draw a plane perpendicular to the line $E F$ (Fig.3). This plane intersects the sphere along a circle of radius $r \leqslant 1$ with center $S 1$, and the line $E F$ at some point $Y$. Let $b$ be the edge of the tetrahedron $A B C D$. Then the distance between the lines $C D$ and $A B$ is $\frac{b}{\sqrt{2}}$. This distance is equal to the length of the perpendicular $Y Z$ dropped from point $Y$ to the line $A B$. Denote $Y Z=t, \angle Y S 1 X=\alpha$. Drop the perpendicular $Y U$ from point $Y$ to the line $S 1 X$. From the right triangle $S 1 Y U$, we find that

\[
r+t=S 1 X+X U=S 1 U=S 1 Y \cos \alpha=\sqrt{5} \cos \alpha
\]

from which $t=\sqrt{5} \cos \alpha-r$. Since the cosine decreases as the angle increases, the smallest value of $t$ is achieved when $\alpha$ is the largest possible. Therefore, the perpendicular $Y Z$ is the smallest when point $X$ coincides with point $A$ or point $B$. In this case, the edge of the tetrahedron $A B C D$ will be the smallest for the given $r \leqslant 1$. Suppose the line $A B$ is tangent to the circle with center $S 1$ at point $A$ (Fig.4). Drop the perpendicular $Y V$ from point $Y$ to the line $S 1 A$. In the right triangle $S 1 V Y$, it is known that

\[
S 1 Y=\sqrt{5}, Y V=A Z=\frac{b}{2}, S 1 V=S 1 A+A V=S 1 A+Y Z=r+\frac{b}{\sqrt{2}}
\]

By the Pythagorean theorem,

\[
S 1 V 2+Y V 2=S 1 Y 2, u l i\left(r+\frac{b}{\sqrt{2}}\right) 2+\frac{b^{2}}{1}=5
\]

After obvious transformations, we get the quadratic equation

\[
3 b 2+4 b r \sqrt{2}+4 r 2-20=0
\]

Since $r \leqslant 1$, this equation has a unique positive root

\[
b=\frac{2}{3}\left(\sqrt{15-\tau^{2}}-\sqrt{2}\right).
\]

The obtained expression takes the smallest value when $r$ is the largest possible, i.e., when $r=1$. In this case,

\[
b=\frac{2}{3}(\sqrt{14}-\sqrt{2})=\frac{2}{3} \sqrt{2}(\sqrt{7}-1).
\]

## Answer

$V=4 \sqrt{6}$ or $V=\frac{20 \sqrt{22}}{9 \sqrt{3}} ; a_{\text {min }}=\frac{2 \sqrt{2}(\sqrt{7}-1)}{3}$.