At the base of the quadrilateral pyramid $S A B C D$ lies a rhombus $A B C D$ with an acute angle at vertex $A$. The height of the rhombus is 4, and the point of intersection of its diagonals is the orthogonal projection of vertex $S$ onto the base plane. A sphere with a radius of 2 touches all the faces of the pyramid. Find the volume of the pyramid if the distance from the center of the sphere to the line $A C$ is $\frac{2 \sqrt{2}}{3} A B$.
The center of a sphere that touches two intersecting planes lies in the bisector plane of one of the dihedral angles formed by these planes. For the given pyramid, such a sphere is either inscribed in the pyramid, or touches the plane of the base and the extensions of all lateral faces, or touches one of the lateral faces and the extensions of the other faces. Let $S K$ and $S N$ be the heights of triangles $A S B$ and $D S C$ respectively, and $H$ be the center of the rhombus $A B C D$. If the sphere is inscribed in the pyramid $S A B C D$, then its center $O$ lies on the height $S H$, since the bisector planes of the dihedral angles formed by adjacent lateral faces of such a pyramid intersect along the line $S H$. By the theorem of three perpendiculars, $H K \perp A B$ and $H N \perp C D$, and since $A B \parallel C D$, the points $K, H$, and $N$ lie on the same line - the height of the rhombus $A B C D$. The plane $A S B$ passes through the line $A D$, which is perpendicular to the plane $K S N$, so the planes $A S B$ and $K S N$ are perpendicular, meaning that the perpendicular $O E$, dropped from point $O$ to the line $S K$ of intersection of the planes $A S B$ and $K S N$ (i.e., the radius of the sphere), lies in the plane $K S N$. Consider the section of the pyramid by the plane $K S N$. In the quadrilateral $O E K H$, it is known that $O E = O H = K H = 2$ and $\angle O E K = \angle O H K = 90^{\circ}$. From the equality of right triangles $O H K$ and $O E K$ (by the leg and hypotenuse), it follows that $E K = K H = 2$, so $O E K H$ is a square, which is impossible, since $\angle E K H < 90^{\circ}$. Thus, the sphere cannot be inscribed in the given pyramid. Similarly, it is proven that the sphere cannot touch the face $A B C D$ and the extensions of all lateral faces of the pyramid. Let the sphere touch the face $D S C$ and the extensions of the other faces of the pyramid. The planes $A S B$ and $C S D$ pass through parallel lines $A B$ and $C D$, so the line $l$ of their intersection is parallel to each of these lines. Let $F$ be a point on the extension of the segment $K S$ beyond point $S$. Then $F S N$ is the linear angle of the dihedral angle containing the center $O 1$ of the sphere. Then the bisector $m$ of this angle lies in the bisector plane $\alpha$ of the considered dihedral angle and is parallel to $K N$ as the bisector of the external angle at vertex $S$ of the isosceles triangle $K S N$, and since two intersecting lines $l$ and $m$ of the plane $\alpha$ are respectively parallel to the lines $A B$ and $K N$ of the plane $A B C D$, the plane $\alpha$ is parallel to the plane $A B C D$. Similarly, it is proven that the plane $\alpha$ is the bisector plane of the dihedral angle between the extensions of the faces $A S D$ and $B S C$. Thus, the radius of the sphere is equal to the distance between the parallel planes $\alpha$ and $A B C D$. Therefore, $S H = 2$, and since $N H = \frac{1}{2} K N = 2$, then $S N = 2 \sqrt{2}$. The bisector plane of the dihedral angle between the planes $A B C D$ and $C S D$, containing the center of this sphere, passes through the line $C D$, parallel to the plane $\alpha$, and intersects the plane $\alpha$, so the line of intersection is parallel to the line $C D$. Similarly, the other bisector planes of the dihedral angles adjacent to the dihedral angles of the pyramid at the edges of its base intersect the plane $\alpha$ along lines parallel to these edges. Let $O_{A} O_{B} O_{C} O_{D}$ be the quadrilateral formed by the intersection of these four lines. Its opposite sides are pairwise parallel, so it is a parallelogram. Its diagonals are respectively parallel to the diagonals of the rhombus $A B C D$, so $O_{A} O_{B} O_{C} O_{D}$ is a rhombus similar to the rhombus $A B C D$. Thus, the center of the sphere mentioned in the problem condition cannot be a point different from the points $O_{A}, O_{B}, O_{C}$, and $O_{D}$, since only these points of the plane $\alpha$ are equidistant from the planes of all faces of the pyramid.

Since $S O_{D} \parallel H N$, the point $O_{D}$ lies in the plane $K S N$, and $N O_{D}$ is the bisector of the angle $S N G$ adjacent to the angle $S N K$. Therefore, $\angle S O_{D} N = \angle O_{D} N G = \angle S N O_{D}$, so $S O_{D} = S N$, and the similarity coefficient of the rhombuses $O_{A} O_{B} O_{C} O_{D}$ and $A B C D$ is $\frac{S O_{D}}{N H} = \frac{S N}{N H} = \sqrt{2}$. The points $O_{A}$ and $O_{C}$ do not satisfy the problem condition, since then their distance to the line $A C$ is equal to the distance between the planes $\alpha$ and $A B C D$, i.e., 2, so $\frac{2 \sqrt{2}}{3} A B = 2$, from which $A D = A B = \frac{3}{\sqrt{2}} < 4 = K N$, which is impossible, since the side of the rhombus is less than its height. Consider the point $O_{D}$. Let $A B = x$. From the similarity of the rhombuses, it follows that $D H = \frac{1}{\sqrt{2}} S O_{D}$. Since the orthogonal projection of the line $O_{D} H$ onto the plane of the base of the pyramid is the line $D H$ and $D H \perp A C$, by the theorem of three perpendiculars, $O_{D} H \perp A C$, so the length of the segment $O_{D} H$ is the distance from the point $O_{D}$ to the line $A C$, and since by the condition this distance is equal to $2 \sqrt{2} \frac{2 \sqrt{2}}{3} x$, from the right triangle $S H O_{D}$ we find that

$$
S O_{D}^2 = O_{D} H^2 - S H^2 = \left(\frac{2 \sqrt{2}}{3} x\right)^2 - 2^2 = \frac{8}{9} x^2 - 4
$$

Then

$$
D H^2 = \frac{1}{2} S O_{D}^2 = \frac{4}{9} x^2 - 2
$$

Let $Q$ be the projection of point $B$ onto the side $C D$ of the rhombus $A B C D$. Then

$$
\begin{gathered}
B Q = K N = 4, \quad C Q = \sqrt{B C^2 - B Q^2} = \sqrt{x^2 - 16}, \quad D Q = C D - C Q = x - \sqrt{x^2 - 16}, \\
B D^2 = B Q^2 + D Q^2 = 16 + \left(x - \sqrt{x^2 - 16}\right)^2, \\
D H^2 = \frac{1}{4} B D^2 = \frac{1}{4} \left(16 + \left(x - \sqrt{x^2 - 16}\right)^2\right) = \frac{1}{2} \left(x^2 - x \sqrt{x^2 - 16}\right),
\end{gathered}
$$

From the equation $\frac{4}{9} x^2 - 2 = \frac{1}{2} \left(x^2 - x \sqrt{x^2 - 16}\right)$, we find that $x = 3 \sqrt{2}$. Therefore,

$$
V_{S A B C D} = \frac{1}{3} S_{A B C D} \cdot S H = \frac{1}{3} A B \cdot K N \cdot S H = \frac{1}{3} \cdot 3 \sqrt{2} \cdot 4 \cdot 2 = 8 \sqrt{2}
$$

## Answer

$8 \sqrt{2}$.