Shooting At Each Other - Part One

The maximum number of people occurs when each person is at a vertex of a regular pentagon and they are all shooting at a person at the center of the pentagon. It's up to you to prove why.

Let point A be the person being shot. Let points B,C,D,E.... Be the shooters. These points B,C,D.... Must all be equal distance from A, and from each other. Let the distance between the shooter points and A be n. Therefore, the distances between the shooter points must be greater than n, for if it were less than n, the points would be closer to its neighbour than to A. If the distance between the shooters was n, the shooters would shoot more than one person, A and their neighbouring shooter, but this is impossible for the problem does not allow for this. Therefore, the distance between them must be greater than n. If the distance were n, then any two adjacent shooters would form an equilateral triangle with A, so the two shooters would be 60° Apart from each other. Therefore, this would allow for 6 points. However, the distance between shooters must be greater than n, so all possible numbers of shooters lie in the set [0,6). Clearly must there be an integer number of shooters, and the highest integer in the set is 5.