The side of the base $ABC$ of a regular pyramid $ABCD$ is equal to $4\sqrt{3}$, and $\angle DAB = \operatorname{arctg} \sqrt{\frac{37}{3}}$. Points $A1$, $B1$, $C1$ are the midpoints of the edges $AD$, $BD$, $CD$ respectively. Find: 1) the angle between the lines $BA1$ and $AC1$; 2) the distance between the lines $BA1$ and $AC1$; 3) the radius of the sphere that touches the plane $ABC$ and the segments $AC1$, $BA1$, and $CB1$.
1) Let K be the midpoint of AB. From the right triangle ADK, we find that

$$
A D=\frac{A K}{\cos \angle D A B}=A K \sqrt{1+\tan^2 \angle D A B}=2 \sqrt{3} \cdot \sqrt{1+\frac{37}{3}}=4 \sqrt{10}
$$

By the formula for the median of a triangle,

$$
B A 12=\frac{1}{4}(2 B D 2+2 A B 2-A D 2)=\frac{1}{4}(2 \cdot 160+2 \cdot 48-160)=64
$$

thus, $BC1 = AC1 = BA1 = 8$. Let $M$ be the midpoint of segment DC1. Then A1M is the midline of triangle ADC1, so $A 1 M=\frac{1}{2} A C 1=4$, and since $A 1 M \| A C 1$, the angle between the skew lines BA1 and AC1 is equal to the angle between the intersecting lines BA1 and A1M. From triangle DBC1, by the formula for the median, we find that

$$
B M 2=\frac{1}{4}(2 B C 12+2 B D 2-D C 12)=\frac{1}{4}(2 \cdot 64+2 \cdot 160-40)=102
$$

By the cosine rule,

$$
\cos \angle B A 1 M=\frac{A_{1} M^{2}+B A_{1}^{2}-B M^{2}}{2 A_{1} M \cdot B A_{1}}=\frac{16+64-102}{2 \cdot 4 \cdot 8}=-\frac{11}{32}
$$

Let $\phi$ be the desired angle. The angle between the lines cannot be greater than $90^{\circ}$, so

$$
\begin{gathered}
\cos \phi=\cos \left(180^{\circ}-\angle B A 1 M\right)=-\cos \angle B A 1 M=\frac{11}{32} \\
\sin \phi=\sqrt{1-\cos ^{2} \varphi}=\sqrt{1-\left(\frac{11}{32}\right)^{2}}=\frac{\sqrt{3} \sqrt{301}}{32}
\end{gathered}
$$

2) Let $D H$ be the height of the pyramid $A B C D, V$ be its volume. Then

$$
\begin{gathered}
A H=\frac{2}{3} \cdot \frac{B C \sqrt{3}}{2}=\frac{2}{3} \cdot \frac{4 \sqrt{3} \cdot \sqrt{3}}{2}=4, D H=\sqrt{A D^{2}-A H^{2}}=\sqrt{160-16}=12 . \\
V=\frac{1}{3} S_{\triangle A B C} \cdot D H=\frac{1}{3} \cdot \frac{(4 \sqrt{3})^{2} \sqrt{3}}{4} \cdot 12=48 \sqrt{3} .
\end{gathered}
$$

Consider the tetrahedron ABCD with vertex C and the tetrahedron AC1BA1 with vertex C1. The height of the tetrahedron AC1BA1, drawn from vertex $C 1$, is half the height of the tetrahedron $A B C D$, drawn from vertex $C$, and $S_{\triangle A B A} 1=$ $\frac{1}{2} S_{\triangle A B D}$, so

$$
V_{A C} 1 B A 1=\frac{1}{2} \cdot \frac{1}{2} V_{A B C D}=\frac{1}{4} V=12 \sqrt{3}
$$

On the other hand,

$$
V_{A C} 1 B A 1=\frac{1}{6} B A 1 \cdot A C 1 \cdot d \cdot \sin \phi
$$

where $d$ is the desired distance between the lines BA1 and AC1. From the equation

$$
12 \sqrt{3}=\frac{1}{6} \cdot 8 \cdot 8 \cdot d \cdot \frac{\sqrt{3} \sqrt{301}}{32}
$$

we find that $d=\frac{36}{\sqrt{301}}$. 3) It can be proven that the center O of the specified sphere of radius $r$ lies on the height $D H$ of the pyramid and the sphere touches the base plane at point $H$. If $F$ is the point of tangency of the sphere with the line $B A 1$, then $B H=B F$ as segments of tangents drawn from one point to the sphere, and since $B H=A H=4, and ~ B A 1=8$, then $B F=$ $\frac{1}{2} B A 1$, i.e., $F$ is the midpoint of BA1. Denote $\angle A D H=\alpha$. From the right triangles $A D H$ and A1OF, we find that

$$
\begin{gathered}
\cos \alpha=\frac{D H}{A D}=\frac{12}{4 \sqrt{10}}=\frac{3}{\sqrt{10}} \\
O A 12=O F 2+A 1 F 2=r 2+16
\end{gathered}
$$

and since $D O=D H-O H=12-r$, applying the cosine rule to triangle $O D A 1$, we get the equation

$$
r 2+16=4 \cdot 10+(12-r) 2-2 \cdot 2 \sqrt{10} \cdot(12-r) \cdot \frac{3}{\sqrt{10}}
$$

from which we find that $r=2$.

## Answer

$\arccos \frac{11}{32} ; \frac{36}{\sqrt{301}} ; 2$.