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Found this on the net today and lost.

On a table you have a square made of 4 coins at the corner at distance 1. So, the square is of size 1×1. In a valid move, you can choose any two coin let’s call them mirror and jumper. Now, you move the jumper in a new position which is its mirror image with respect to mirror. That is, imagine that mirror is a centre of a circle and the jumper is on the periphery. You move the jumper to a diagonally opposite point on that circle. With any number of valid moves, can you form a square of size 2×2? If yes, how? If no, why not?

I dont think its possible since if it is, then the reverse must be true i.e. say 1/2 X 1/2 scale wise, that leaves the coins in the initial position forever on the vertices.But this is well, the anti proof, how to prove mathematically that 2X2 is not a possibility.

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up vote 6 down vote accepted

No. Should not be possible. Think of the coins as positioned on the points (0,0),(0,1),(1,0),(1,1). The no. of even x-coordinates (0) is the same as the odd x-coordinates(1). Any reflection that you have defined would preserve the parity of the coordinates (adding an even no. to them). So in the end you can't get all the x-coordinates to be of the same parity (as required by a 2x2 square).

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+1, nice answer –  Slartibartfast Jul 22 '12 at 14:37
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