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 1 of 5: Balanced sets of every odd size

Previous step Next step

In plain words

An equilateral triangle is itself balanced, so gluing several of them at a shared vertex should stay balanced.

V=\{O,A_1,B_1,\dots,A_k,B_k\},\quad \triangle OA_iB_i \text{ equilateral}

Detailed analysis

Fix a circle with center $O$ and radius $r$ . For $i=1,\dots,k$ choose points $A_i,B_i$ on the circle so that $\triangle OA_iB_i$ is equilateral, and let $V=\{O,A_1,B_1,\dots,A_k,B_k\}$ , a set of $2k+1$ points. Every point other than $O$ lies at distance $r$ from $O$ , so for two such points $A_i,A_j$ we may take $C=O$ . For the pair $(O,A_i)$ , the third vertex $B_i$ satisfies $B_iO=B_iA_i$ because $\triangle OA_iB_i$ is equilateral. Hence $V$ is balanced, and $|V|=2k+1$ ranges over every odd number $\ge 3$ as $k=1,2,\dots$ .