10.268. Point $C$ moves along the segment $A B$ of length $l$. On the segments $A C$ and $C B$, as bases, regular triangles are constructed on one side of $A B$. Where should point $C$ be taken so that the distance between the vertices of the triangles is the smallest?
Solution.

Let $A C=x$; then $C B=l-x$ (Fig. 10.68). Draw $D L \| A B$ and drop perpendiculars from points $D$ and $E$ to $A B$. Since triangles $A D C$ and $C E B$ are equilateral, $D L=F M=F C+C M=\frac{x}{2}+\frac{l-x}{2}=\frac{l}{2}$, $E L=\frac{(l-x) \sqrt{3}}{2}-\frac{x \sqrt{3}}{2}=\frac{(l-2 x) \sqrt{3}}{2}$. Therefore, $D E^{2}=\frac{l^{2}}{4}+\frac{3(l-2 x)^{2}}{4}$.

The distance $D E$ is minimized when $l-2 x=0$, from which $x=l / 2$, i.e., point $C$ should be taken at the midpoint of segment $A B$.

Answer: at the midpoint of segment $A B$.