Tracing a straight line

By symmetry, the maximum distance from center of the two intersecting lines $O$ occurs when the two sliding vertices $A$ and $B$ of the equilateral triangle are equal distances away from $O$ , so when $AO = BO$ .

This means $\triangle AOB$ is an isosceles triangle, and since $\angle AOB = 180° - 60° = 120°$ , $\angle OAB = \frac{180° - 120°}{2} = 30°$ and $\angle AOC = \frac{1}{2}120° = 60°$ . Also, since $\triangle ABC$ is a unit equilateral triangle, $AB = 1$ and $\angle BAC = 60°$ . This means $\angle OAC = 30° + 60° = 90°$ .

Solving $\triangle ACO$ with $AB = 1$ , $\angle OAC = 90°$ , and $\angle AOC = 60°$ gives $OC = \frac{2}{\sqrt{3}}$ . By symmetry the straight line is twice this length, which is $2 \cdot \frac{2}{\sqrt{3}} = \boxed{\frac{4}{\sqrt{3}}}$ .