4 [ Sphere inscribed in a dihedral angle ]

Point $O$ is located in the section $A A^{\prime} C^{\prime} C$ of a rectangular parallelepiped $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with dimensions $2 \times 6 \times 9$ such that $\angle O A B + \angle O A D + \angle O A A^{\prime} = 180^{\circ}$. A sphere with center at point $O$ touches the planes $A^{\prime} B^{\prime} C^{\prime}, A A^{\prime} B$ and does not intersect the plane $A A^{\prime} D$. Find the distance from point $O$ to this plane.
Let $\angle O A B=\alpha, \angle O A D=\beta, \angle O A A^{\prime}=\gamma$. According to the problem, $\alpha+\beta+\gamma=180^{\circ}$.

We mark a segment $A E=1$ on the ray $A O$ (Fig.1). Then the projections of point $E$ onto the lines $A B, A D$, and $A A^{\prime}$ are $\cos \alpha, \cos \beta$, and $\cos \gamma$ respectively. By the theorem of the square of the diagonal of a rectangular parallelepiped, $\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=1$. Since $\gamma=180^{\circ}-\alpha-\beta$, we have

\[
\begin{gathered}
\cos 2 \alpha+\cos 2 \beta+\cos 2\left(180^{\circ}-\alpha-\beta\right)=1, \cos 2 \alpha+\cos 2 \beta+\cos 2(\alpha+\beta)=1 \\
2 \cos 2 \alpha+2 \cos 2 \beta+2 \cos 2(\alpha+\beta)=2, 1+\cos 2 \alpha+1+\cos 2 \beta+2 \cos 2(\alpha+\beta)=2 \\
\cos 2 \alpha+\cos 2 \beta+2 \cos 2(\alpha+\beta)=0, 2 \cos (\alpha+\beta) \cos (\alpha-\beta)+2 \cos 2(\alpha+\beta)=0 \\
\cos (\alpha+\beta)(\cos (\alpha-\beta)+\cos (\alpha+\beta))=0, \cos (\alpha+\beta) \cos \alpha \cos \beta=0, \cos \gamma \cos \alpha \cos \beta=0
\end{gathered}
\]

Therefore, either $\gamma=90^{\circ}$, or $\beta=90^{\circ}$, or $\cos \alpha=90^{\circ}$.

If $\alpha=90^{\circ}$, then point $O$ lies in the plane $A A^{\prime} D$, which is impossible because the sphere with center $O$ does not have common points with this plane. If $\beta=90^{\circ}$, then point $O$ lies in the plane $A A^{\prime} B$, which is also impossible because the sphere with center $O$ touches this plane. Therefore, $\gamma=90^{\circ}$, so point $O$ lies in the plane $A B C D$, and thus on the line $A C$ at the intersection of the planes $A A^{\prime} C^{\prime} C$ and $A B C D$.

According to the problem, point $O$ lies in the section $A A^{\prime} C^{\prime} C$, so it is located on the segment $A C$. The sphere with center $O$, lying in the plane $A B C$, touches the planes $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ and $A A^{\prime} B$, so its radius is equal to the length of the edge $A A^{\prime}$ and is less than the length of the edge $A D$. Since the sphere does not have common points with the face $A A^{\prime} D$, its radius $r$ is less than the length of the edge $A B$. Therefore, $A A^{\prime}$ is the smallest edge of the parallelepiped, i.e., $r=A A^{\prime}=2$.

The distance from the center of the sphere to the plane $A A^{\prime} D$ (Fig.2) is equal to the distance from point $O$, lying on the diagonal $A C$ of the rectangle $A B C D$, to the side $A D$, so $A B>A D$. Therefore, $A B=9$ and $A D=6$. Let $H$ be the projection of point $O$ onto the side $A B$. Then the desired distance is the length of the segment $A H$. From the similarity of the right triangles $A O H$ and $A C B$, we find that

\[
A H=A B \cdot \frac{O H}{B C}=9 \cdot \frac{2}{6}=3
\]