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12 O'clock show down
#1
12 O'clock show down
2011 gangsters are spread around on a flat plane so that their mutual distances are different. At 12 O'clock every gangster shoots its nearest neighbor.

Prove that nobody is hit by more than 5 bullets. Prove that the paths of these bullets don't cross.
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#2
RE: 12 O'clock show down
Is it the year 2011 or is it the number 2,011 gangsters. Also what's to stop them from standing in a circle and everyone shooting the guy to their right/left? Than each man would only get hit with 1 bullet and the paths of the bullets wouldn't cross. Plus to insure that they were different distances apart you could simply measure each man an 1/8 of an inch apart further than the man prior.

I think you need to clarify the problem ... a lot.
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#3
RE: 12 O'clock show down
(April 22, 2011 at 1:54 am)Cinjin Cain Wrote: Is it the year 2011 or is it the number 2,011 gangsters. Also what's to stop them from standing in a circle and everyone shooting the guy to their right/left? Than each man would only get hit with 1 bullet and the paths of the bullets wouldn't cross. Plus to insure that they were different distances apart you could simply measure each man an 1/8 of an inch apart further than the man prior.

I think you need to clarify the problem ... a lot.

Its 2011 gangsters and their mutual distances are different... hence can't spread out equally around one big circle if that's what you are suggesting.
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