Ten gangsters are standing on a flat surface, and the distances between them are all distinct. Suddenly, each of them fatally shoots the one among the other nine gangsters that is the nearest. What is the maximum possible number of surviving gangsters?
I guess that it is possible for 7 to survive.
First, 6 survivors is easy.
One way: 8 men stand at the corner of two almost-squares, widely separated. one man stands in the center of each almost-square. All corners shoot at the center men, and the center men each shoot at a corner. 6 survive.
Alternatively, 5 men stand at the corner of an almost pentagon and 3 stand at the corners of an almost-equilateral-triangle, with the two widely separated. One man stands in the center of each.
7 survivors is, I suspect, difficult but possible. I would try placing 7 men at the corners of an almost-heptagon. I guess that three men can be placed in the center such that the 7 corners all survive and the 3 center men all die. one center man is placed so that it is closed to three corners, another so that it is closest to the next three corners, and the last so that it closest to the last corner. The three center men need to be close enough so that they all shoot each other. I have not actually done this.
I do not believe 8 survivors is possible. I have tried using an almost-hexagon with two men in the center, but this seems to fail.
A guess (spoiler?)
