Canadian Mathematical Olympiad 1987, Problem 4 On a large flat field, $n$ people $(n>1)$ are positioned so that for each person the distances to all the other people are different. Each person holds a water pistol and at a given signal fires and hits the person who is closest. When $n$ is odd, show that there is at least one person left dry.
This question is a variant of the question I am asking but I am not using induction in my approach.
My Approach:
Our primary goal is to ensure that no person remains dry.
When a total of $k$ people are present ($k$ is odd), it is evident that if no one remains dry, then a closed chain must have been formed when considering the order of firing. (Since the pairing doesn't change the parity, atleast one dry person will remain in the end)
WLOG, let $P_1$ attack $P_2$, $P_2$ attack $P_3$, $P_3$ attack $P_4$ and so on till $P_{k-1}$ attack $P_k$ and $P_k$ attack $P_1$
Let us denote the distance between $P_i$ and $P_j$ as $i_j$ or $j_i$
Now $2_3<2_1$ since $P_2$ attacks $P_3$, thus $2_3<1_2$. Similarly $3_4<3_2$ since $P_3$ attack $P_4$, thus $3_4<2_3<1_2$
$\therefore $ In the end, we get $k_1<(k-1)_k<(k-2)_{k-1}<\ldots<3_4<2_3<1_2$
From this we can see $k_1=1_k<1_2$ which implies that $P_1$ must have attacked $P_k$ instead of $P_2$ which is a contradiction.
This means that $P_1$ and $P_k$ attacks each other while $P_2$ attacks $P_3$, $P_3$ attacks $P_4$ and so on till $P_{k-1}$ attacks $P_k$ hence leaving an open chain where $P_2$ remains dry.
It can be observed that any pairing will result in an open chain consisting the pair if any of the remaining person attacks a person from the pair. If none of the remaining persons attack any person from pair, then the pair can be isolated and similar argument can be used for remaining $(k-2)$ people.
$\therefore $ We will always get an open chain if the number of persons are odd which means that atleast one person will remain dry.
Please check my solution for any mistakes. Also please suggest any improvements in the solution.
THANKS
 A: As stated in the comments:
The argument as written is not correct.  The initial assumption, that no pair fires on each other, is not possible.  The two people $A,B$ at minimal distance from each other must fire at each other. (of course the case where there is only one person is trivial).
Two ways to solve the problem:
Method I: consider that minimal pair $A,B$.  We distinguish two cases (according to whether anyone else shoots at either $A$ or $B$).
Since the case $n=1$ is trivial it makes sense to proceed by induction.  Let's assume we have a counterexample with minimal $n$ (we will derive a contradiction).
If nobody else shoots at $A,B$ then we can ignore that pair and focus on the $n-2$ remaining people.  By the induction hypothesis, at least one of those stays dry and we are done.
If somebody else, $C$ say, shoots at one of them, say $A$, then at least two people shoot at $A$.  It follows that the map $F: \{1,\cdots, n\}\to \{1,\cdots,n\}$ which maps the $i^{th}$ person to their target is not injective.  Thus it can't be surjective and again we are done.
Method II (sketch).  Suppose we had a collection with odd $n$ in which nobody stayed dry. Then consider the shooting pattern.  Since it must be the case that everyone shoots at (and is shot at by) a unique person, the collection must break up into distinct closed loops.  These can't all have  length $2$ since the collection is odd.  There must, in fact, be an odd loop of length $>2$. But consider the members of that loop.  There must be a minimal distance between any two members of that loop and, as before, we quickly see that those two people can't shoot at anyone else in that loop.  Thus the loop is not possible, and we are done.
