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Given a square of side length as least 2, there are two players who must in turn place a unit radius circle into the square without overlapping any previously placed circle.

The loser is the first player who can't place a circle.

For example, if the square has side length 2 then the first player wins. If the square is just big enough to fit two circles with side length 2+2√2, then again the first player wins by putting a circle into the center.

What is the perfect way to play? This will depend on the size of the square.

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  • $\begingroup$ Could you give an example where the second player wins? It seems that the first player always wins by placing in the center and then mirroring whatever the second player puts. $\endgroup$
    – WhatsUp
    Jun 26 at 17:15
  • $\begingroup$ @WhatsUp This is part of the puzzle! Can the second player ever win? $\endgroup$
    – graffe
    Jun 26 at 17:16

1 Answer 1

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OK, so let me try to write down a proof.

If the first player places a circle in the center, then whatever the second player puts, the first player can always put a symmetric circle on the other side of the center.

To show that this strategy works, it suffices to prove that any circle $C$ placed by the second player cannot cover two points that are symmetric with respect to the center $O$. It then follows that the symmetric image of that circle is an available position and restores symmetry.

Suppose the contrary, that $C$ covers two points $P, Q$ that are symmetric with respect to the center $O$. Then we have $|PO| = |OQ| = \frac 12|PQ|$.
Since $P, Q$ both lie in the circle $C$, we have $|PQ| \leq 2$, which then implies $|PO| = |OQ| \leq 1$.
However, all points that are within a radius of $1$ from $O$ have been covered by the first circle that the first player places. This is a contradiction.

(I realized a shorter proof: since a circle is convex, if $P, Q$ both lie in $C$ then their center point $O$ also lies in $C$, which is contracition.)

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