10,11

The base of an inclined parallelepiped is a rectangle $A B C D$; $A A 1, B B 1, C C 1$ and $D D 1$ are the lateral edges. A sphere with center at point $O$ touches the edges $B C, A 1 B 1$ and $D D 1$ at points $B, A 1$ and $D 1$, respectively. Find $\angle A 1 O B$, if $A D=4$, and the height of the parallelepiped is 1.
Let's first prove the following statement: if a line is tangent to a sphere at point $E$, then the plane passing through the center of the sphere and perpendicular to this line passes through point $E$. Indeed, if the plane passing through the center $O$ of the sphere and perpendicular to the tangent line $l$ intersects this line at a point $F$ different from $E$, then $O F \perp l$ and $O E \perp l$, which is impossible, since only one line can be drawn through point $O$ perpendicular to $l$. The statement is proved. Since the sphere passes through points $A 1$ and $D 1$, its center $O$ lies in the plane $\alpha$, passing through the midpoint $N 1$ of segment $A 1 D 1$ and perpendicular to $A 1 D 1$, and since $B C \| A 1 D 1$, the plane $\alpha$ is perpendicular to the line $B C$, tangent to the sphere at point $B$. From the previously proven statement, it follows that the plane $\alpha$ passes through point $B$. Since $A B C D$ is a rectangle, the line $A B$ is perpendicular to $B C$, so this line also lies in the plane $\alpha$. The sphere is tangent to the line $A 1 B 1$ at point $A 1$, so the plane $\beta$, passing through point $O$ and perpendicular to the line $A 1 B 1$, passes through point $A 1$, and since $D 1 A 1 \perp A 1 B 1$, the line $A 1 D 1$, and therefore point $N 1$, lies in the plane $\beta$. Since $O$ and $N 1$ are common points of planes $\alpha$ and $\beta$, these planes intersect along the line $O N 1$. Let the lines $A B$ and $O N 1$, lying in the plane $\alpha$, intersect at point $N$. Then $N N 1 \perp A D$ (since the line $N N 1$ lies in the plane $\alpha$, perpendicular to $A 1 D 1$, and therefore to $A D$) and $N N 1 \perp A B$ (since the line $N N 1$ lies in the plane $\beta$, perpendicular to $A 1 B 1$, and therefore to $A B$). Therefore, $N N 1$ is the height of the parallelepiped, $N N 1=1$. Let $A N=x, N O=y$, $O A 1=O B=O D 1=R$. From the right triangles $A 1 N 1 O$ and $O N B$ we find that

$$
\begin{aligned}
& R 2=O A 12=O N 12+A 1 N 12=4+(y+1) 2 \\
& R 2=O B 2=O N 2+B N 2=y 2+(x+b) 2
\end{aligned}
$$

Let $A 1 A 2$ also be the height of the parallelepiped, dropped onto the plane $A B C D$. Then $A 2 N\|A 1 D 1\| A D$, so $A 2 N \perp A B$. From the right triangles $A A 2 N$ and $A A 1 A 2$ we find that

$$
A A 22=A N 2+A 2 N 2=A N 2+A 1 N 12=x 2+4 \text {, }
$$

$$
A A 12=A 1 A 22+A A 22=1+x 2+4=5+x 2
$$

and from the right triangles $O D 1 D$ and $O N D-$

$$
O D 2=O D 12+D D 12=R 2+A A 12=R 2+5+x 2
$$

$$
O D 2=N D 2+O N 2=(A N 2+A D 2)+O N 2=(x 2+16)+y 2
$$

From the equality $R 2+5+x 2=x 2+16+y 2$ it follows that $R 2=y 2+11$. Let $\angle A 1 O B=\phi$. From the isosceles triangle $A 1 O B$ we find that $A 1 B=2 R \sin \frac{\varphi}{2}$, and from the right triangles $A 2 B N$ and $A 1 A 2 B-$

$$
A 2 B 2=B N 2+A 2 N 2=(x+b) 2+4
$$

$$
A 1 B=\sqrt{A_{1} A_{2}^{2}+A_{2} B^{2}}=\sqrt{1+(x+b)^{2}+4}=\sqrt{5+(b+x)^{2}}
$$

Therefore,

$$
\sin \frac{\varphi}{2}=\frac{A_{1} B}{2 R}=\frac{\sqrt{5+(b+x)^{2}}}{2 R}
$$

From the system

$$
\left\{\begin{array}{l}
R^{2}=4+(y+1)^{2} \\
R^{2}=y^{2}+11
\end{array}\right.
$$

we find that $y=3$ and $R=2 \sqrt{5}$. Then

$$
(b+x) 2=R 2-y 2=20-9=11
$$

Therefore,

$$
\sin \frac{\varphi}{2}=\frac{\sqrt{5+(b+x)^{2}}}{2 R}=\frac{\sqrt{5+11}}{4 \sqrt{5}}=\frac{1}{\sqrt{5}}
$$

## Answer

$2 \arcsin \frac{1}{\sqrt{5}}$