Russia — Final Round

Let us denote the given points by $A$, $B$, $C$, $D$, and $E$. Choose among them two points that are the farthest apart; let these be $A$ and $B$. We will show that it is possible to move them as required.

Draw the perpendicular bisector $\alpha$ to the segment $CD$. If $E$ lies on $\alpha$, then it is sufficient to move points $A$ and $B$ onto the line $\alpha$. Otherwise, let $E'$ be the point symmetric to $E$ with respect to $\alpha$. Note that the distance from $E$ to $\alpha$ is less than the length of one of the segments $EC$ or $ED$, that is, less than $AB$.

Thus, we can move point $A$ to point $E'$, and point $B$ to a point lying on $\alpha$ and at a distance $AB$ from $E'$. It is easy to see that $\alpha$ is the axis of symmetry of the new set of points, as required.