Ιt is an extension of a well-known problem in Russia 1968 . In this problem was proved that all the opposite sides of the octagonon are equal using the fact that the side lengths are rational numbers .So now let $A_1....A_8$ th octagonon .From above we have that $A_2A_3A_6A_7$ is a parallellogram whose center is O .Now is the same way $A_3A_4A_7A_8$ is also a parallellogram and its center is also O .Therefore the point O is the center of symmetry for the octagonon and we are done. BTW , if someone wants the proof of the fact that all the opposite sides of the octagonon are equal tell.

Silouan, actually it is not necessary all of the sides of the octogon to be equal. Just the opposite ones.

This problem is well-known I think: http://www.mathlinks.ro/Forum/viewtopic.php?p=235765#p235765 . How can they give it in an olympiad? We did it in class some time ago, so I guess it's from some past olympiad or something. And indeed, all sides needn't be equal.

Indeed the opposite sides, I wrote in hurry and I make this mistake, sorry. But the problem is well known