N gunmen in a field

Question

Let n be an odd integer. In some field, n gunmen are placed such that all pairwise distances between them are different. At a signal, every gunman takes out his gun and shoots the closest gunman. Prove the following:
a) At least one gunman survives.
b) No gunman is shot more than 5 times.
c) The trajectory of the bullets does not intersect.

I solved part A with the extremal principle, however, I think it lacks rigour.(I will be posting my solution shortly.) I would appreciate a rigorous one or an alternate solution.
I am clueless when it comes to part B.
Solved part C with some geometry, just thought that I should share it anyway.

EDITS: Part C is solved with the theorem: In a triangle, sides opposite to larger angles are longer. Suppose A $\rightarrow$ C and B$\rightarrow$D and their trajectories intersect, this implies four inequalities with the sides and diagonals of $ABCD$. And with the above theorem we prove that the sum of the angles is $<360$. Contradiction.

EDIT 2: The best solutions seem to be from Maciek for Part B and Micah for part A. Since Maciek has enough upvotes and will stay at the top of the feed, I have "accepted" Micah's answer.

Answer 1

Another version of a):

By the pigeonhole principle, "someone doesn't get shot" is equivalent to "someone gets shot twice" whenever $n > 1$. Consider the two gunmen who are closest together. They will shoot each other. If anyone else shoots them, we're done; if not, remove them and induct.

More formally:

We'll proceed by induction. If $n=1$, the gunman survives. Now, let $n>1$ be odd, and suppose that someone survives in every configuration involving $k<n$ gunmen with $k$. Take a configuration involving $n$ gunmen, and consider the two gunmen who are closest together. They will shoot each other. If anyone else shoots them, then one of them is being shot twice, and thus by the pigeonhole principle there must be some other gunman who isn't being shot at all. If nobody else shoots them, we can remove them from the field without affecting whom anybody else is shooting. This will give a configuration involving $n-2$ gunmen, which is an odd number; thus, by the induction hypothesis, at least one of those gunmen will survive. In either case, we have shown that there is a surviving gunman, so this completes the proof.

Answer 2

Suppose that two gunmen $(A, B)$ shoot a special individual $C$. Then if you consider the triangle $ABC$, the longest side is $|AB|$, so the $\angle ACB > 60^\circ$. And you cannot put 6 people around somebody where all the angles are greater than $60^\circ$.

Answer 3

Hints

(a)

Let each shooting be a directed edge. Then prove that any cycle must be a 2-cycle. Then prove that removing all cycles results in a nonzero number of paths.

(b)

Maciek gave a more or less full answer already so there is no hint left to give.

(c)

It is indeed a geometric fact.

Answer 4

a) All pairwise distances are different. --edit: this argument is insufficient --

One gunman must be further away from any peers than any other, and his peers will shoot each other instead of him. If there were an even number, they could be placed in pairs, but if the number is odd and the pairwise distances are unique, this is not possible.

b) To be shot $k$ times, a gunman must be the closest member to $k$ others. Thus, what is the maximum packing density at which point no more gunmen can be added without drawing fire?

The measure is circularly symmetric, so we need the maximum packing density of circles, which is hexagonal. However, in hexagonal packing the distances between the centres are all equal; if the distances are adjusted to be inequal, the sixth circle cannot pack more closely to the centre circle than to its neighbours (which would then draw their fire), leaving the maximum packing density around a centre at $5$.

c) Intersecting trajectories would require that distances $AC < BD$, and form a quadrilateral $ABCD$. But in a convex planar quadrilateral $ABCD$...

points $B,D$ are always closer to $C$ than $C$ is to $A$. Only in a concave quadrilateral can $AB < CD$, in which $AC$ and $BD$ do not intersect.