Was there a rule that the jump had to be over a nearest point? Brent On 1/29/2014 6:56 AM, Michael Kleber wrote:
I think you can do effectively the same thing as Adam without needing ideas like Gaussian integer GCD. Eric's problem asked about forming a new square with size "t" > "s", and that's because it's obvious you can never make a square that's *smaller* than the unit square you started with, as all points you ever get are part of the square lattice with the original square as its unit. But since the point-replacing process is invertible, if it can't give you a smaller square, it can't give you a larger one either.
--Michael
On Wed, Jan 29, 2014 at 9:46 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
That's a very beautiful problem. I've appended a short solution to the end of this e-mail.
----- Original Message ----- From: Eric Angelini Sent: 01/29/14 10:30 AM To: math-fun Subject: [math-fun] Jumping square points
Hello Math-Fun, Four points shape a square of size "s". Every minute a point randomly jumps over another point and lands symmetrically behind it. Is it possible that, at some stage, the 4 points shape a new square having size "t" > "s"? Best, É.
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No. Operate in the ring Z[i] of Gaussian integers, and let {a,b,c,d} be the Gaussian integers corresponding to the vertices of the square. When you apply the process to obtain {a,2a-b,c,d}, the greatest common divisor (again a Gaussian integer, up to multiplication by the ring units) remains invariant (similar to Euclid's algorithm).
Hence, if they form a square at any time, it must be the same size and orientation as the original square.
Sincerely,
Adam P. Goucher
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