4. Two circles with radius $r$ touch each other externally. Construct a line that intersects both circles such that it defines three segments of equal length on it.
Solution. Analysis. Let $N^{\prime}$ and $N^{\prime \prime}$ be the feet of the perpendiculars from the centers of the circles of the required segments. If we denote the length of these segments by $x$, then $A^{\prime} N^{\prime}=A^{\prime \prime} N^{\prime \prime}=\frac{x}{2}$. However, $S^{\prime} A^{\prime}=S^{\prime \prime} A^{\prime \prime}$,

![](https://cdn.mathpix.com/cropped/2024_06_05_ff57faf93329e9641894g-4.jpg?height=291&width=502&top_left_y=1037&top_left_x=960)
therefore, the right triangles $A^{\prime} N^{\prime} S^{\prime}$ and $A^{\prime \prime} N^{\prime \prime} S$ are congruent. Hence, $S^{\prime} N^{\prime}=S^{\prime \prime} N^{\prime \prime}$, and thus the quadrilateral $S^{\prime} S^{\prime \prime} N^{\prime \prime} N^{\prime}$ is a rectangle. Therefore, $N^{\prime} N^{\prime \prime}=2 r$ and since $A^{\prime} N^{\prime}=A^{\prime \prime} N^{\prime \prime}=\frac{x}{2}$ and $A^{\prime} A^{\prime \prime}=x$, we get $2 r=2 x$, or $r=x$. Therefore, triangle $S^{\prime} T A^{\prime}$ is equilateral.

Construction. We construct the equilateral triangle $S^{\prime} T A^{\prime}$. Then through $A^{\prime}$ we draw a line $p$ parallel to the line $S^{\prime} S^{\prime \prime}$, and this is the required line.

The proof and discussion follow from the analysis and construction. The details are left to the reader as an exercise.