8,9}

In a triangular pyramid $A B C D$, it is known that $D C=9, D B=A D$, and the edge $A C$ is perpendicular to the face $A B D$. A sphere of radius 2 touches the face $A B C$, the edge $D C$, and the face $D A B$ at the point of intersection of its medians. Find the volume of the pyramid.
Let $C$ be the vertex, and $A B D$ be the base of the triangular pyramid $C A B D$ (Fig.1); the given sphere with center $O$ touches the lateral edge $C D$ at point $P$, the base $A B D$ at point $M$ (the intersection of the medians of the isosceles triangle $A B C$), and the lateral face $A B C$ at point $Q$. The median $D K$ of the isosceles triangle $A B D$ is its height, so $D K \perp A B$. Since $A C$ is perpendicular to the plane $A B D$, then $D K \perp A C$. Therefore, the line $D K$ is perpendicular to two intersecting lines of the plane $A B C$. Consequently, $D K$ is perpendicular to the plane $A B C$. Since point $M$ lies on $D K$, then $M K$ is also perpendicular to the plane $A B C$. The segment $O Q$ is the radius of the sphere, drawn to the point of tangency with the plane $A B C$, so $O Q$ is perpendicular to this plane. Thus, the lines $O Q$ and $M K$ are parallel as perpendiculars to the same plane. Therefore, the quadrilateral $O M K Q$ lies in one plane, has three right angles, and two adjacent sides are equal as radii of the same sphere. Hence, $O M K Q$ is a square. Therefore,

$$
\begin{gathered}
K M=O M=2, D M=2 K M=4, D P=D M=4 \\
C Q=C P=C D-D P=9-4=5
\end{gathered}
$$

The line $D K$ is perpendicular to the plane $A B C$, so $D K \perp C K$. By the Pythagorean theorem,

$$
C K=\sqrt{C D^{2}-D K^{2}}=\sqrt{81-36}=\sqrt{45}=3 \sqrt{5}
$$

Consider the right triangle $A B C$ (Fig.2). Let $A K=K B=x$. Let $F$ be the orthogonal projection of point $Q$ onto $A C$. Then

$$
\begin{gathered}
C F=\sqrt{C Q^{2}-Q F^{2}}=\sqrt{C Q^{2}-A K^{2}}=\sqrt{25-x^{2}}, \\
A C=\sqrt{C K^{2}-A K^{2}}=\sqrt{45-x^{2}} \\
A C=A F+C F=K Q+C F, \text { or } \sqrt{45-x^{2}}=2+\sqrt{25-x^{2}}
\end{gathered}
$$

From this equation, we find that $x=3$. Therefore, $A B=6$ and $A C=6$. Consequently,

$$
V_{C A B D}=\frac{1}{3} S_{\triangle A B D} \cdot A C=\frac{1}{3} \cdot \frac{1}{2} A B \cdot D K \cdot A C=\frac{1}{3} \cdot \frac{1}{2} \cdot 6 \cdot 6 \cdot 6=36
$$

## Answer

36.00