orl wrote:

A set $S$ of points from the space will be called completely symmetric if it has at least three elements and fulfills the condition that for every two distinct points $A$ and $B$ from $S$ , the perpendicular bisector plane of the segment $AB$ is a plane of symmetry for $S$ . Prove that if a completely symmetric set is finite, then it consists of the vertices of either a regular polygon, or a regular tetrahedron or a regular octahedron.

Solution

The question can be simplified in dealing with a 2D plane and a 3D plane. We will divide into two cases:
1) 2D plane
Consider 3 points A,B,C. Their perpendicular bisectors meet at a point O that is the center of the circumscribed circle. Reflect, say $A$ over the perpendicular bisector of BC to get another point $A'$ lying on the circle because the perpendicular bisector of $AA'$ passes through $O.$
Essentially this means set $S$ is a set of points that lie on a circle. In an infinite set of points, we can continously add another point by reflecting it over a perpendicular bisector of 2 other points. However, set $S$ is finite, meaning that at some time, the "pattern" will repeat so that a point that was reflected over a perpendicular bisector is the same point as an already existing point. Because the pattern will stop at one point, we can take 3 consecutive points A,B,C in that order on the circle. Assume that AB>BC. Say there exists a perpendicular bisector of any 2 other points that goes through B. Then the reflection of C must lie between the arc $AB$ , but there exists no points, contradiction, meaning that AB=BC. Now say there doesn't exist a perpendicular bisector of any 2points that goes through B but there is a bisector that passes between arc AC, of which we can take the perp bisector of AC. Then B can be reflected to a point $B'$ on AC, which is a contradiction, so AB=BC. In either case, AB=BC, and so we can take any other 3 points, A,B,D (D on the opposite side of C) and arrive at the same conclusion. lnductively, all the sides are equal and it is circumscriptible, so it is a regular polygon. $\blacksquare$
2) 3D plane
Take three points A,B,C and take a crosssection to form a circle around those 3 points. If a perp bisector of some point of the 3 reflects one of the 3 points to a point on the circle, then we have a regular polygon and it reverts to case 1. Take a point D such that, wlog, it is the reflection of C over the perp bisector of AB. argh... too wasted to finish this case, i'll finish it up when I get time, but I feel like you proceed the same way as case 1.