0
$\begingroup$

Two players place coins of identical size (say quarters) on a round table. Each player has to place exactly one coin on the table without overlap with the coins already on the table. The first player who cannot put a coin on the table loses.

Prove that the first player has a winning strategy.

$\endgroup$
1
  • 1
    $\begingroup$ HINT: Cut a coin-sized hole out of the centre of the table. Now show that on this table the second player has a winning strategy. $\endgroup$ Dec 8, 2015 at 17:57

1 Answer 1

1
$\begingroup$

The first player puts a coin at the center of the table and after that, whatever position of the table the second player puts a coin, the first player keeps his coin at a position which is the reflection of the previous coin through the center coin i.e. rotated through $180$ degrees. Or in other words, as fleablood puts it; the position is "collinear to the center and the previous coin at an equal distance from the center as the previous coin".

$\endgroup$
6
  • $\begingroup$ Not mirror. That'd allow second player to place coins on the axis of symmetry which can not be matched. 180 degree rotated. $\endgroup$
    – fleablood
    Dec 8, 2015 at 18:05
  • $\begingroup$ @fleablood I am reflecting through a point. There is no axis of symmetry. If I got you right. $\endgroup$ Dec 8, 2015 at 18:08
  • 1
    $\begingroup$ Yes, but you used the term "mirror image". A mirror image does have an axis of symmetry. The "mirror image" of (x, y) is (x, -y) and so any point (x, 0) will be unmatchable. I think what you meant was "reflected through the central point" or in other words "rotated 180 degrees". The reflection of (x, y) is (-x,-y). No point in this strategy is matchable. $\endgroup$
    – fleablood
    Dec 8, 2015 at 18:31
  • $\begingroup$ Yes, I meant that only. $\endgroup$ Dec 8, 2015 at 18:35
  • $\begingroup$ I suppose another way of putting it, although not necessarily clearer, is "colinear to the center and the previous coin at an equidistance for the center as the previous coin". $\endgroup$
    – fleablood
    Dec 8, 2015 at 18:35

Not the answer you're looking for? Browse other questions tagged .