In a regular triangular pyramid $S A B C$ ( $S$ - vertex) points $D$ and $E$ are the midpoints of edges $A C$ and $B C$ respectively. Through point $E$, a plane $\beta$ is drawn, intersecting edges $A B$ and $S B$ and being equidistant from points $D$ and $B$ by a distance of $\frac{1}{2}$. Find the lengths of the segments into which the plane divides edge $S B$, if $B C=4$

$$
\text { , } S C=3 \text {. }
$$
Since the plane $\beta$ intersects the edge $A B$, points $A$ and $B$ are located on opposite sides of it. Similarly for points $B$ and $C$. Therefore, points $A$ and $C$ (and thus all points on the edge $A C$) lie on one side of $\beta$, while point $B$ lies on the other. Thus, the segment $B D$ intersects the plane $\beta$ at some point $M$ (Fig.1). Let $B 1$ and $D 1$ be the orthogonal projections of points $B$ and $D$ onto the plane $\beta$ (Fig.2), and let the plane $\beta$ intersect the edge

$A B$ at point $N$. Then point $M$ belongs to the projection $B 1 D 1$ of the segment $B D$ onto the plane $\beta$. From the equality of right triangles $B M B 1$ and $D M D 1$ (by the leg and the adjacent acute angle), it follows that $M$ is the midpoint of $B D$, and thus the line $E M$ intersects the edge $A B$ at point $N$ and $E M \| A C$, so $N$ is the midpoint of the edge $A B$. Let the plane $\beta$ intersect the edge $S B$ at point $K$. Then $K M$ is the median and altitude of the isosceles triangle $E K N$, and since $N B 1 = E B 1$ (as projections of equal obliques $B N$ and $B E$ onto the plane $\beta$), point $B 1$ lies on $K M$. Let $S H$

- be the height of the pyramid. Denote $\angle S B H = \alpha, \angle K M B = \gamma$. From right triangles $S B H$ and $B B 1 M$, we find that

$$
\begin{gathered}
\cos \alpha = \frac{B H}{B S} = \frac{\frac{3}{B D}}{B S} = \frac{4 \sqrt{3}}{9}, \sin \gamma = \frac{B B_{1}}{B M} = \frac{B B_{1}}{\frac{1}{2} B D} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{1}{2 \sqrt{3}} \\
\sin \alpha = \sqrt{1 - \left(\frac{4 \sqrt{3}}{9}\right)^{2}} = \sqrt{1 - \frac{43}{B 1}} = \frac{\sqrt{33}}{9}, \cos \gamma = \sqrt{1 - \left(\frac{1}{2 \sqrt{3}}\right)^{2}} = \sqrt{1 - \frac{1}{12}} = \frac{\sqrt{11}}{2 \sqrt{3}} \\
\sin (\alpha + \gamma) = \sin \alpha \cos \gamma + \cos \alpha \sin \gamma = \frac{\sqrt{33}}{9} \cdot \frac{\sqrt{11}}{2 \sqrt{3}} + \frac{4 \sqrt{3}}{9} \cdot \frac{1}{2 \sqrt{3}} = \frac{5}{6}
\end{gathered}
$$

From triangle $B K M$ by the sine theorem, we find that

$$
B K = \frac{B M \sin \gamma}{\sin (\alpha + \gamma)} = \frac{\sqrt{3} \cdot \frac{1}{2 \sqrt{3}}}{\frac{1}{2}} = \frac{3}{5}.
$$

Therefore,

$$
S K = S B - B K = 3 - \frac{3}{5} = \frac{12}{5}
$$

## Answer

$\frac{12}{5}, \frac{3}{5}$.