[ Rectangular parallelepipeds ] [ Sphere inscribed in a dihedral angle ]

Point $O$ is located in the section $A C C^{\prime} A^{\prime}$ of a rectangular parallelepiped $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with dimensions $2 \times 3 \times 6$ such that $\angle O C B + \angle O C D + \angle O C C^{\prime} = 180^{\circ}$. A sphere with center at point $O$ touches the planes $A^{\prime} B^{\prime} C^{\prime}, C C^{\prime} D$ and does not intersect the plane $B B^{\prime} C$. Find the distance from point $O$ to this plane.
Let $\angle O C B=\alpha, \angle O C D=\beta, \angle O C C^{\prime}=\gamma$. According to the problem, $\alpha+\beta+\gamma=180^{\circ}$.

We mark a segment $C E=1$ on the ray $C O$. Then the projections of point $E$ on the lines $C B, C D$, and $C C^{\prime}$ 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 $\beta=90^{\circ}$, then point $O$ lies in the plane $B B^{\prime} C$, which is impossible because the sphere with center $O$ does not have common points with this plane. If $\alpha=90^{\circ}$, then point $O$ lies in the plane $C C^{\prime} D$, 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 $C C^{\prime} D$, so its radius is equal to the length of the edge $C C^{\prime}$ and is less than the length of the edge $A B$. Since the sphere does not have common points with the face $B B^{\prime} C$, its radius $r$ is less than the length of the edge $A B$. Therefore, $C C^{\prime}$ is the smallest edge of the parallelepiped, i.e., $r=C C^{\prime}=2$.

The distance from the center of the sphere to the plane $B B^{\prime} C$ is equal to the distance from point $O$, lying on the diagonal $A C$ of the rectangle $A B C D$, to the side $B C$, so $C D>B C$. Therefore, $C D=6$ and $B C=3$. Let $H$ be the projection of point $O$ on the side $C D$. Then the desired distance is the length of the segment $C H$. From the similarity of the right triangles $C O H$ and $C A D$, we find that

\[
C H=C D \cdot \frac{O H}{A D}=6 \cdot \frac{2}{3}=4 .
\]

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## Answer

4.