Is there a way to put 5 points on the surface of the sphere so that they are indistinguishable?

Question

Is it possible to put 5 points on the surface of the sphere such that, for every pair of them, say A and B, there is an isometric transformation of the space such that A is mapped onto B, B is mapped onto A, and the convex cover of all 5 points is mapped onto itself, so is entire sphere, and the 5 points do not all lie in one plane ?

Answer 1

We will prove that this is not possible. Toward a contradiction, assume that $A,B,C,D,E$ are points on the sphere satisfying the condition, and let $O$ be the center of the sphere. Any isometry which maps convex cover to itself permutes the set $\{A,B,C,D,E\}$, so we can consider this isometry as an element of $Sym(\{A,B,C,D,E\})$ too. Also, the sphere is mapped onto itself, so the center of the sphere $O$ is a fixed point.

Let $I$ be an isometry swapping two points, say $A$ and $B$, and mapping the convex cover onto itself. As an element of $Sym(\{A,B,C,D,E\})$, $I$ can be the transposition $(AB)$ which has fixed points $C,D,E$. It can be double transposition, say $(AB)(CD)$, swapping $C$ and $D$ and with fixed point $E$. Finally, it can be the product of a transposition and a 3-cycle, $(AB)(CDE)$ or $(AB)(CED)$; if this is the case we can consider $I^3=(AB)$ which swaps $A$ and $B$ and fixes $C,D,E$. So for each isometry from the assumptions we can assume that it has at least one fixed points from $A,B,C,D,E$, or more precisely it is a transposition or the product of two transpositions as an element from $Sym(\{A,B,C,D,E\})$.

Also every isometry swapping two points, say $A$ and $B$, fixes the midpoint $M$ of $AB$.

Case 1. Assume that an isometry $I$ swapping two points, say $A$ and $B$, is a double transposition, say $I= (AB)(CD)$. Since $I$ fixes $O$ and $E$, it has two fixed point on the line $OE$, so $I$ fixes every point on $OE$.

Subcase 1. If the midpoint of $AB$ or the midpoint of $CD$ doesn't belong to $OE$, then $I$ has three non-collinear fixed points. So the whole plane $\alpha$ determined by $OE$ and this midpoint is fixed point-wise. Since $I$ is not identity, we conclude that $I$ is the plane reflection with respect to $\alpha$. So, $A$ and $B$, as well as $C$ and $D$, are symmetric with respect to $\alpha$. So $AB,CD\perp\alpha$, hence $AB\parallel CD$, and $A,B,C,D$ are coplanar. Consider now $J$ swapping $A$ and $E$. It is either the transposition $(AE)$ or a double transposition $(AE)(XY)$ for $X,Y\in\{B,C,D\}$. In both cases $J$ maps $\{A,B,C,D\}$ to $\{E,B,C,D\}$ so $E,B,C,D$ are coplanar. Since planes $ABCD$ and $EBCD$ have three non-collinear points in the intersection, they are equal, and $A,B,C,D,E$ are coplanar. A contradiction.

Subcase 2. Both the midpoint of $AB$ and of $CD$ are on $OE$; in particular, $O,E,A,B$ and $O,E,C,D$ are coplanar. Note that none of $A,B,C,D$ is on $OE$ as $OE$ is fixed point-wise and $A$, $B$, $C$ and $D$ are not. Consider $J$ swapping $A$ and $C$. If $J=(AC)$ is a transposition, it maps $\{O,E,A,B\}$ to $\{O,E,C,B\}$, and $\{O,E,C,D\}$ to $\{O,E,A,D\}$, so $O,E,C,B$ are coplanar and $O,E,A,D$ are coplanar. Planes $OEAB$ and $OECB$, and $OECD$ and $OEAD$ have non-collinear intersections, hence they are equal. We further conclude that $A,B,C,D,E$ are coplanar. A contradiction. Assume now that $J$ is a double transposition; we have three possibilities: $J=(AC)(BD)$, $J=(AC)(BE)$ and $J=(AC)(DE)$. If $J=(AC)(BD)$, we may assume that Subcase 2 holds for $J$ instead of $I$ as Subcase 1 doesn't hold by the same arguments. In particular, $O,E,A,C$ are coplanar and $O,E,B,D$ are coplanar, so by using that $O,E,A,B$ are coplanar and $O,E,C,D$ are coplanar, we again get $A,B,C,D,E$ are coplanar; a contradiction. If $J=(AC)(BE)$, we may assume again Subcase 2 for $J$ instead of $I$, so $O,D,A,C$ and $O,D,B,E$ are coplanar. Again we conclude $A,B,C,D,E$ are coplanar; a contradiction. The possibility $J=(AC)(DE)$ is similar.

Case 2. By Case 1 we may assume that none of the isometries swapping two points are double transpositions, i.e. all of them are transpositions. If some four points, say $A,B,C,D$, are coplanar, then a transposition $(AE)$ swapping $A$ and $E$ maps $\{A,B,C,D\}$ to $\{E,B,C,D\}$, so $E,B,C,D$ are coplanar so we conclude that all of them are coplanar; a contradiction. Consider $I=(AB)$ swapping $A$ and $B$. It has a point-wise fixed plane $OCDE$. (Note that $O,C,D,E$ must be coplanar as a non-identity isometry cannot have four non:--coplanar fixed points.) Transpositions $J=(AC)$ and $K=(AD)$ map the plane $OCDE$ to the planes $OADE$ and $OCAE$. Intersection of $OCDE$ and $OADE$ must be the line $ODE$ as there are no four coplanar points among $A,B,C,D,E$; $D$ and $E$ are diametrically opposite. Similarly, intersection of $OCDE$ and $OCAE$ is the line $OCE$; $C$ and $E$ are diametrically opposite. Therefore $C=D$; a contradiction.

Since, none of the two cases is possible, we are done.