10,11 On the edge $AC$ of a regular triangular prism $ABC A1B1C1$, a point $K$ is taken such that $AK=\frac{1}{4}, CK=\frac{3}{4}$. A plane is drawn through point $K$, forming an angle $\operatorname{arctg}^{\frac{7}{6}}$ with the plane $ABC$ and dividing the prism into two polyhedra with equal surface areas. Find the volume of the prism, given that a sphere can be circumscribed around one of these polyhedra, but not around the other.
Let $\phi 1$ and $\phi 2$ be polyhedra into which a plane $\alpha$ cuts a prism, such that a sphere can be circumscribed around $\phi 1$, but not around $\phi 2$. Let $S 1$ and $S 2$ be the areas of their surfaces, respectively. Each face of a polyhedron inscribed in a sphere is an inscribed polygon, since the intersection of the sphere with the plane of this face is a circle on which the vertices of this face lie. Suppose that the plane $\alpha$ intersects the edge $A 1 C 1$ of the prism at some point $P$. If $P K \| A A 1$, then the plane $\alpha$ is perpendicular to the plane $A B C$, which is impossible since the angle between these planes is $\operatorname{arctg} \frac{7}{6}$. If the line $P K$ is not parallel to

$C C 1$, then it divides the rectangle into two right trapezoids, which is also impossible since a circle cannot be circumscribed around a right trapezoid. Therefore, the plane $\alpha$ must intersect either the edge $C C 1$ or the edge $A A 1$. 1. Suppose the plane $\alpha$ intersects the edge $C C 1$ at some point $N$. Then the face of the polyhedron $\phi 1$ can only be the triangle $K C N$, since the circle passing through the points $A, A 1$, and $C 1$ is the circle circumscribed around the rectangle $A A 1 C 1 C$, and it cannot pass through the points $N$ and $K$. If in this case $\alpha$ intersects the edge $B C$ at some point $Q$, then the polyhedron $\phi 1$ is a triangular pyramid $C K Q N$, and the area $S 1$ of its surface is obviously less than the area $S 2$. If, however, $\alpha$ intersects the edge $B 1 C 1$ at some point $H$, then the point $D$ of intersection of the lines $N H$ and $B B 1$ would lie on the extension of the edge $B B 1$ beyond the point $B 1$, and the point $F$ of intersection of the lines $N K$ and $A A 1$ - on the extension of the edge $A A 1$ beyond the point $A$, so the line $D F$ (and therefore the plane $\alpha$) would divide the rectangle into two right trapezoids. Similarly, the plane $\alpha$ cannot intersect the edges $A B$ and $A 1 B 1$. Thus, the only possibility left is that the plane $\alpha$ intersects the edge $B B 1$ at some point $M$. In this case, $M N \| B C$

, since otherwise the line $M N$ would divide the rectangle $B B 1 C 1 C$ into two right trapezoids.

Therefore, the plane $\alpha$ intersects the base $A B C$ along a line passing through the point $K$ parallel to $B C$.

Let this line intersect the edge $A B$ at the point $L$. Then

$$
\frac{A L}{L B}=\frac{A K}{K C}=\frac{1}{3}, A L=A K=\frac{1}{4}, B L=C K=\frac{3}{4},
$$

2. Suppose the plane $\alpha$ intersects the edge $C C 1$ at some point $N'$. Reasoning similarly, we prove that $\alpha$ intersects the base $A B C$ along some segment $K L'$, where $K L' \| A B$ and

$$
C L'=C K=\frac{3}{4}, B L'=A K=\frac{1}{4}.
$$

In this case,

$$
S 1=S_{A B L' K}+2 S_{\triangle K A N'}+S_{\text{sec.}}<\frac{7}{16} S_{\triangle A B C}+2 \cdot \frac{1}{5 S_{A C C}} 1 A 1+S_{\text{sec.}}<S 2
$$

which contradicts the condition. Thus, it is established that the plane $\alpha$ can intersect the prism only along the isosceles trapezoid $K L M N$. 3. Let $T$ and $T 1$ be the midpoints of $B C$ and $B 1 C 1$, respectively, $R$ be the point of intersection of the median $A T$ of the equilateral triangle $A B C$ with the segment $K L$, and $E$ be the point of intersection of the segments $T T 1$ and $M N$. Then $E R T$ is the linear angle of the dihedral angle between the base plane of the prism and the plane $\alpha$. Denote $\angle E R T=\phi, A A 1=h, A B=B C=A C=a$. By the condition of the problem, $\operatorname{tg} \phi=\frac{7}{6}, a=\frac{1}{4}+\frac{3}{4}=1$. From the right triangle $E R T$ we find that

$$
E T=R T \operatorname{tg} \phi=\frac{3}{4} A T \operatorname{tg} \phi=\frac{3}{4} \cdot \frac{a \sqrt{3}}{2} \operatorname{tg} \phi=\frac{3}{4} \cdot \frac{\sqrt{3}}{2} \cdot \frac{7}{6}=\frac{7 \sqrt{3}}{16}
$$

Then

$$
S_{B C N M}=B C \cdot E T=1 \cdot \frac{7 \sqrt{3}}{16}=\frac{7 \sqrt{3}}{16}
$$

$$
\begin{aligned}
& S_{\triangle K C N}=S_{\triangle L B M}=\frac{1}{2} B L \cdot B M=\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{7 \sqrt{3}}{16}=\frac{21 \sqrt{3}}{128} \\
& S_{B C K L}=S_{\triangle A B C}-S_{\triangle A K L}=S_{\triangle A B C}-\frac{1}{16} S_{\triangle A B C}=\frac{15}{16} \cdot \frac{2^{2} \sqrt{3}}{4}=\frac{15 \sqrt{3}}{64}, \\
& S_{M B 1 C 1 N}=B 1 C 1 \cdot(h-B M)=h-\frac{7 \sqrt{3}}{16}, \\
& S_{A K N C 1 A 1}=S_{A L M B 1 A 1}=S_{A A 1 B 1 B}-S_{\triangle L B M}=A B \cdot h-\frac{21 \sqrt{3}}{128}=h-\frac{21 \sqrt{3}}{128}, \\
& S_{\triangle A K L}=\frac{1}{16} \cdot \frac{a^{2} \sqrt{3}}{4}=\frac{\sqrt{3}}{64} \\
& S 1=S_{B C N M}+2 S_{\triangle L B M}+S_{B C K L}+S_{\text{sec.}}= \\
& =\frac{7 \sqrt{3}}{16}+\frac{21 \sqrt{3}}{64}+\frac{15 \sqrt{3}}{64}+S_{\text{sec.}}=\sqrt{3}+S_{\text{sec.}} \\
& S 2=S_{M B 1 C 1 N}+2 S_{A K N C 1 A 1}+S_{\triangle A B C}+S_{\triangle A K L}+S_{\text{sec.}}= \\
& =h-\frac{7 \sqrt{3}}{16}+2\left(h-\frac{21 \sqrt{3}}{128}\right)+\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{64}+S_{\text{sec.}}=3 h-\frac{\sqrt{3}}{2}+S_{\text{sec.}} \\
& \text{ since } S 1=S 2 \text{, then } \\
& \sqrt{3}+S_{\text{sec.}}=3 h-\frac{\sqrt{3}}{2}+S_{\text{sec.}}
\end{aligned}
$$

from which $h=\frac{\sqrt{3}}{2}$. Therefore, if $V$ is the volume of the prism, then

$$
V=S_{\triangle A B C} \cdot h=\frac{\sqrt{3}}{4} \cdot \frac{\sqrt{3}}{2}=\frac{3}{8}
$$

## Answer

$\frac{3}{8}$.

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