A group of ten bandits stand in a flat desert, with no pair the same distance apart. Tensions grow, and at the crack of dawn each bandit fires a single bullet at the bandit closest to him. All have perfect aim and all those that are hit are killed.
Must one of them live to tell the tale? How many of them could possibly survive?
Up to 7 can survive (but it is possible that none do).
For maximum survival, set up a pentagon and a square such that each gunman is closer to the center of their shape than the other corners. Then, have the same person be closest to each center. All the corners shoot the center, and the two centers shoot the same corner.
All seven other corners survive.
Rough diagram:
For minimum survival, just pair them all off. Everybody shoots their partner, and they all die. (These pairs are not actually parallel, since the distances along each side can't be the same, but that's easy enough to make happen)
The theoretical maximum would be 8 - since the two closest gunmen must shoot each other. However, for this to work, we need everybody else to be closest to one of those two gunmen. There's no way to have two people be closer to the center than each other if their angle around the center is less than or equal to 60*, so hexagons are right out. Thus, we would need two pentagons such that one's corner is another's center. Unfortunately, this causes some of the corners to be nearer to the corners of the other pentagon than their own center.
Example of failure:
The maximum value of dead bandits clearly is $10$ (so that all are dead). For instance, put five bandits at the points $(100,0)$, $(200,0)$, $(400,0)$, $(800,0)$, $(1600,0)$ and five bandits at the points $(100,1)$, $(200,2)$, $(400,4)$, $(800,8)$, $(1600,16)$. Then no pair is the same distance apart, and the five pairs with equal $x$-coordinates kill each other.
The minimum value of dead bandits is more difficult to determine. We reformulate the minimization problem as:
Given a set $S$ of $10$ points in the plane, such that the distances between them are all distinct. For each point $p\in S$ we mark the point $q\in S-\{p\}$ that is nearest to $p$. Find the least possible number of marked points.
(1) We observe that each point $x\in S$ is the nearest to at most five other points in $S$. Indeed, for any six points $p_1,\ldots,p_6$ one of the angles $\angle p_ixp_j$ is at most $60^{\circ}$, in which case the distance $p_ip_j$ is smaller than one of the distances $xp_i$ and $xp_j$.
(2) It follows that at least two points are marked.
(3) Now suppose that exactly two points, say $x$ and $y$, are marked. Then $x$ is the closest point to $y$, and $y$ is the closests point to $x$. So by the observation (1) the remaining eight points in $S$ split into two groups $S_x$ and $S_y$ of four points each, so that their closest points are $x$ and $y$ respectively.
Denote $S_x=\{a_1,a_2,a_3,a_4\}$ and $S_y=\{b_1,b_2,b_3,b_4\}$, so that the angles $\angle a_ixa_{i+1}$ are successively adjacent as well as the angles $\angle b_iyb_{i+1}$, and so that $a_1$ and $b_1$ lie on one side of the line through $xy$ while $a_4$ and $b_4$ lie on the other side of this line.
Since all the angles $\angle a_ixa_{i+1}$ and $\angle b_iyb_{i+1}$ are greater than $60^{\circ}$, it follows that $$ \angle a_1xy + \angle yxa_4 + \angle b_1yx + \angle xyb_4 ~<~ 360^{\circ}.$$ Therefore $\angle a_1xy + \angle b_1yx < 180^{\circ}$ or $\angle yxa_4 + \angle xyb_4 < 180^{\circ}$. Without loss of generality, let us assume the first inequality.
(4) Next, note that the quadrilateral $xyb_1a_1$ is convex because $a_1$ and $b_1$ are on different sides of the perpendicular bisector of $xy$. From $a_1b_1 > a_1x$ and $yb_1>xy$ we get the inequalities $\angle a_1xb_1 > \angle a_1b_1x$ and $\angle yxb_1 > \angle xb_1y$. Adding these two inequalities yields $\angle a_1xy > \angle a_1b_1y$. A symmetric argument yields $\angle b_1yx > \angle b_1a_1x$.
These last two inequalities imply $$180^{\circ} > \angle a_1xy +\angle b_1yx > \angle a_1b_1y + \angle b_1a_1x.$$ Hence the sum of the angles of the quadrilateral $xyb_1a_1$ is less than $360^{\circ}$, which is a contradiction. Consequently at least three points are marked.
(5) Finally, the configuration with ten points $(-100,0)$, $(0,0)$, $(101,0)$, $(213,0)$, $(0,110)$, $(0,-111)$, $(-120,110)$, $(-120,-111)$, $(121,110)$, $(121,-111)$ shows that the minimum with three marked points indeed can be attained.
The maximum number of dead bandits is $10$, and the minimum number of dead bandits is $3$.
I believe the max that can survive is 7. Arrange the bandits in approximately two half hexagons. For one hexagon, it's 5 bandits, one at each point and one in the middle. More than a side length away is the other hexagon, parallel, with one exterior point empty. Between the two is the last bandit.
To satisfy the distance clause, just have the outer bandits be closer to the center than each other, and with very small differences the overall situation stands.
6 survivors is the most. 2 sets of 5. 4 corners and 1 in the middle. All different distances apart from middle, but the 4 outsiders shoot the middle guy and the middle guy shoots the closest corner.
A pentagon and triangle work the same way and if you go with a hexagon you run into issues with the outsiders shooting each other.
I can guarantee at least three will survive.
Take one gunman, called Gunman 1. He is separated from Gunman 2 by a distance $X_1$. Gunman 3 is on the other side of Gunman 2, separated by a distance $X_2$, where $X_2<C_1$. Continue doing this until you reach Gunman 7. Each $X_n<X_{n-1}$. Therefore, Gunman $M$ will shoot Gunman $M+1$, with the exception of Gunman 7, who will shoot Gunman $6$.
Now take the remaining three gunmen and place them 90 degrees apart around Gunman 1, forming a sort of cross. Gunman 8 is a distance $A$ from Gunman 1, Gunman 9 is a distance $B$ from Gunman 1, and Gunman 10 is a distance $C$ from Gunman 1. Here, $C\neq A\neq B\neq C$, all of which are greater than $X_1$. They are separated so that the distance between any of these three gunmen is greater than the distance between each gunman and Gunman 1.
Gunman 8, Gunman 9, and Gunman 10 will all shoot Gunman 1 and survive.
I suspect that you could fit more than these three guys around Gunman 1 - and I don't know just how many - but I don't know for sure.
"Nearest" involves two points, so every pair of bandits would shoot each other. None would survive.
EDIT: Since I was marked down, I'll explain further. The important part is "no pair the same distance apart", which means there would be no overlap on shooting. Folks could only survive if that condition wasn't present, and one or more bandits got two bullets.
