See the Wednesday Problem’s __vague rules here__.

Some games again. According to __Sam__, from whom I heard this riddle, it divides people into cathegories i) those who realize the answer immidiately and ii) those who think quite long about it.

The game board is a round table of diameter d>0. The two players place one after another coins of diameter s>0 on the table. The coins cannot overlap. If there is no place left on the table to place the next coin, the player whose turn it is, loses. Which one of the players has a winning strategy?

[tex]2\frac{1}{2}/5[/tex]

The second player to move has a winning strategy.

If the coins have the same size as the table (d=s), then the beginner wins, since the second player cannot place the next coin. The interesting part is when the coins have the size (much) smaller than the table (3s<d). You claim that in this case second player to move wins? How?

I was wrong about it, the first always wins (forgot that the someone has to put one coin in the middle):

The first starts off by putting one coin in the middle of the table.

The second player then makes a move, and then the first player copies that move by placing the coin on the ray through the origin (centre of the table), at the same distance (look at the origin as a mirror point). Because the first player will only

run out of moves when the second one runs out first, he wins.

That’s correct.

Inside every cynical person, there is a depressed idealist

Sent via Blackberry

but of course, you may like the algebraic proof.

There’s no accounting for taste.