# Solution.

This is a segment of length $2 a$. Let $A B C$ be the medial triangle of $A_{0} B_{0} C_{0}$, then $D$ runs along the segment $A C_{0}$, and $E$ runs along $B A_{0}$. The quadrilateral $D E F^{\prime} C_{0}$ is cyclic, with arcs $D E$ and $E F^{\prime}$ each 60 degrees. Therefore, the angle $F^{\prime} C_{0} B$ is 60 degrees, meaning $F^{\prime}$ runs along the ray from $C_{0}$ parallel to $A B$. However, since the angle $F^{\prime} E C_{0}$ is definitely less than 60 degrees, it runs not along the ray, but along a segment of length $2 a$. It is also necessary to note that all points are reached, and that $F^{\prime}$ cannot be in the other half-plane relative to $A C_{0}$.
1.3. Let the initial triangle $A B C$ be fixed and its side equal to $a$, while the other points are not fixed. $F^{\prime}$ is a point different from $F$ such that triangle $D E F^{\prime}$ is equilateral. Find the geometric locus of points $F^{\prime}$ and its length.