12 O'clock show down

[ruhollah]

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.

[Cinjin]

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.

[ruhollah]

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.