Problem 1

We say that a finite set

S

of points in the plane is balanced if, for any two different points

A

and

B

S

, there is a point

C

S

such that

AC=BC

. We say that

S

is center-free if for any three different points

A

B

C

S

, there is no point

P

S

such that

PA=PB=PC

. (a) Show that for all integers

n\ge3

, there exists a balanced set consisting of

n

points. (b) Determine all integers

n\ge3

for which there exists a balanced center-free set consisting of

n

points.

Step 4 of 5: Even

n

is impossible: a pigeonhole argument

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In plain words

Too many pairs must share too few possible "witness" points, so some point ends up equidistant from three others.

\binom n2 = \tfrac{n(n-1)}2 \text{ pairs},\quad \Big\lceil \tfrac{n(n-1)/2}{n}\Big\rceil=\tfrac n2 \text{ for even } n

Detailed analysis

Suppose $V$ is balanced, center-free, and $|V|=n$ is even. For each of the $\binom n2=\frac{n(n-1)}2$ unordered pairs $\{A,B\}\subset V$ , fix one witness $C(A,B)\in V$ with $C(A,B)A=C(A,B)B$ . These $\frac{n(n-1)}2$ pairs are distributed among the $n$ points of $V$ as possible witnesses, so by pigeonhole some point $P$ witnesses at least $\big\lceil\frac{n(n-1)/2}{n}\big\rceil=\big\lceil\frac{n-1}2\big\rceil=\frac n2$ pairs (the last equality uses that $n$ is even, so $n-1$ is odd).