If no one won after placing the counters, can the players always
slide a counter until someone wins?
The empty space can be one of three positions up to symmetry:
a corner, the center, or a middle of the side of the square.
Since they are sliding, there is no 3-in-a-row yet,
so the next player has the red spaces. The last player must have
the yellow space between the two reds.
This gives the final placement:
This position could not occur after one of the players slid a
counter, since then the blue player must have slid a counter out
of this open place, so he/she must have had a 3-in-a-row before the move.
However, in order for the position above to occur immediately
following the original play, the blue player must have had an
opportunity to win in the last move and did not. The blue player
had the last move and could have chosen the open place along the edge,
which would have completed a 3-in-a-row, either horizontally or
vertically (depending of where he/she had played previous counters).
Since the players can always slide a counter, someone will win.
If whoever is in the middle stays there, all the sliding counters
will move in a circle around the center. When we consider the
7 counters around the outside (and one empty space), either the player
with his/her counter in the middle will have two counters which are
separated by exactly two of the other player's counters and will
eventually win (why?) or one of the players has three counters in a
row and will win when these are all along one side. These are the
only possibilities since if the player with the counter in the
center does not have two counters separated by two of the other player's
two of this player's counters must be next to each other and the other
player must have at least 3 counters in a row.