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http://www.kalva.demon.co.uk/imo/isoln/isoln991.html Alternate solution : Let us consider the convex hull $S'$ of the points $A_1, A_2, \ldots, A_n$ of $S$. Since the convex hull of a convex polygon is a convex polygon, we have that $S'$ satisfies the conditions of the problem. Then, $S'$ is the set of the vertices of regular polygon. Let's show that : $S = S'$. This means that $S'$ does not contain points inside $S$. Indeed, let us consider an inside point $A \in S$, $S'$ would not be invariant with regard to $s_{AA_i}$, where $AA_i = \min_{1 \le j \le n} AA_j$.

I think the problem statement here is different from the real problem. In the real problem, the points were in a plane, and not in space, so there was nothing about a tetrahedron or an octahedron.

epitomy01 wrote: I think the problem statement here is different from the real problem. In the real problem, the points were in a plane, and not in space, so there was nothing about a tetrahedron or an octahedron. You're right, but the problem appeared as IMO 01 was adopted the main problem...i.e. space to plane.

Proof for the planar case, as asked in the competition. Denote by $\Omega$ the mass center of $S$; clearly $\Omega$ lies on each axis of symmetry. Since for $A,B \in S$ the perpendicular bisector of $AB$ is an axis of symmetry for $S$, it follows $\Omega A = \Omega B$, hence $S$ is made of concyclic points, lying on a circle of centre $\Omega$. Consider any three consecutive (on the circle) points $A,B,C \in S$; since the perpendicular bisector of $AC$ is an axis of symmetry for $S$, it follows $B$ lies on it, hence $BA = BC$. Therefore $S$ is a regular polygon, and clearly all regular polygons fulfill the requirements.

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.

I guess for the 3-d case it suffices to show that 1) convex hull of S = S is circumscribable in a sphere, and 2) every circle formed by taking 3 points of S is congruent. [this, combined with the 2-D case should be sufficient...]

Cute problem! First, the global part. Let $O$ be the centroid of $S$. Since reflection across the perpendicular bisector of $AB$ fixes $S$, it must also fix $O$, so $AO=BO$ and $O$ is the circumcenter of $S$. We now split the problem into two cases: Case 1: $S$ is coplanar. Let $\Omega$ be the circumcircle of $S$. Assume for the sake of contradiction that there are points $A,B,C,D\in S$ in that order around $\Omega$ with $A\ne D$, $A$ and $B$ consecutive, $C$ and $D$ consecutive, and $\overarc{\ensuremath{ADB}}<\overarc{\ensuremath{CAD}}$. [asy][asy] draw(unitcircle); pair A=dir(162),B=dir(234),C=dir(354),D=dir(42); pair Z=(5/4)*sqrt(A*D), Cp=A*D/C; draw(-Z--Z,blue+dashed+(1/2)*bp); draw(C--Cp,blue+(1/2)*bp,Arrow,Margins); dot("$A$",A,A); dot("$B$",B,B); dot("$C$",C,C); dot("$D$",D,D); dot("$C'$",Cp,Cp,blue); [/asy][/asy] Then, reflecting $C$ over the perpendicular bisector of $AD$ gives a point between $A$ and $B$ in $S$, a contradiction. Therefore, the arcs between consecutive points in $S$ must all be congruent, so $S$ consists of the vertices of a regular polygon, as desired. $\square$ Case 2: $S$ is not coplanar. Let $\Omega$ be the circumsphere of $S$. Take the convex polyhedron formed by $S$, i.e. the convex hull of $S$. Let $P\in S$ and let the neighbors of $P$ on the convex hull be $A_1,A_2,\ldots,A_n$ in that order for some $n\ge 3$. By Case 1, the faces of the convex hull are regular polygons, so $$PA_1=PA_2=\ldots=PA_n.$$Thus, $A_1,A_2,\ldots,A_n$ lie on the circle formed by the intersection of $\Omega$ and the sphere centered at $P$ containing $A_1,A_2,\ldots,A_n$. By Case 1, they then must form a regular polygon. Therefore, $$\angle A_1PA_2=\angle A_2PA_3=\ldots=\angle A_nPA_1,$$so the faces containing these angles have the same number of sides and are congruent regular polygons. It follows that the convex hull of $S$ is a platonic solid. The cube, dodecahedron, and icosahedron do not work because they are not preserved when reflecting across the perpendicular bisector of a space diagonal. Thus, $S$ consists of the vertices of either a regular tetrahedron or a regular octahedron, as desired. $\square$

Solved with nukelauncher and Th3Numb3rThr33. First, we verify the planar case: Let \(G\) be the centroid of the points in \(S\); then \(G\) lies on each perpendicular bisector \(AB\), so \(GA=GB\) for all \(A\), \(B\) in \(S\); i.e.\ the points in \(S\) lie on a circle centered at \(G\). Let the points of \(S\) lie on the circle in the order \(A_1\), \(A_2\), \ldots, \(A_n\). We can verify for each \(i\) by looking at the perpendicular bisector of \(\overline{A_{i-1}A_{i+1}}\) that \(A_iA_{i-1}=A_iA_{i+1}\), so the polygon is regular, as needed. Then assume the points are not all planar. Let \(G\) be the centroid again, so we can verify analogously that the points in \(S\) lie on a sphere centered at \(G\). Consider the polyhedron formed by the points in \(S\). Observe that any plane containing at least three points must form a regular polygon. We will verify the following claim: Claim: If we have a plane containing at least four points, and there is a point \(P\) not on the plane, then there is a point \(P\) on the other side of the plane as \(P\). Proof. Let the plane contain regular polygon \(A_1A_2\cdots A_k\) with side length \(s\). If there is a point \(P\) ``above'' the plane, then we can select a point \(Q\) on the plane containing \(\triangle PA_1A_2\) yet still ``above'' the plane with \(QA_1=s\). By \(k\ge4\), we have \(QA_1=A_1A_2<A_1A_3\), so the perpendicular bisector plane of \(\overline{QA_3}\) intersects segment \(A_1A_3\); that is, the reflection of \(A_1\) over the perpendicular bisector plane lies ``below'' the plane, and it must also be in \(S\). \(\blacksquare\) It follows that every face of the polyhedron is an equilateral triangle, so the polyhedron is either a regular tetrahedron, regular octahedron, or regular icosahedron (by platonic solids). To rule out the icosahedron, take the space diagonal; the two pentagons do not pair up.

whoa gj raymond and espen :O

Solution from Twitch Solves ISL: Let $G$ be the centroid of $S$. Claim: All points of $S$ lie on a sphere $\Gamma$ centered at $G$. Proof. Each perpendicular bisector plane passes through $G$. So if $A,B \in S$ it follows $GA = GB$. $\blacksquare$ Claim: Consider any plane passing through three or more points of $S$. The points of $S$ in the plane form a regular polygon. Proof. The cross section is a circle because we are intersecting a plane with sphere $\Gamma$. Now if $A$, $B$, $C$ are three adjacent points on this circle, by taking the perpendicular bisector we have $AB=BC$. $\blacksquare$ If the points of $S$ all lie in a plane, we are done. Otherwise, the points of $S$ determine a polyhedron $\Pi$ inscribed in $\Gamma$. All of the faces of $\Pi$ are evidently regular polygons, of the same side length $s$. Claim: Every face of $\Pi$ is an equilateral triangle. Proof. Suppose on the contrary some face $A_1 A_2 \dots A_n$ has $n > 3$. Let $B$ be any vertex adjacent to $A_1$ in $\Pi$ other than $A_2$ or $A_n$. Consider the plane determined by $\triangle A_1 A_3 B$. This is supposed to be a regular polygon, but arc $A_1 A_3$ is longer than arc $A_1 B$, and by construction there are no points inside these arcs. This is a contradiction. $\blacksquare$ Hence, $\Pi$ has faces all congruent equilateral triangles. This implies it is a regular polyhedron --- either a regular tetrahedron, regular octahedron, or regular icosahedron. We can check the regular icosahedron fails by taking two antipodal points as our counterexample. This finishes the problem.

Let $G$ be the centroid. Note that since the entire figure is symmetric across each perpendicular bisector, the centroid must be on every perpendicular bisector. Thus, $S$ is conspheric. $~$ Let $T$ be a subset of $S$ that lies on a circle. Let $A_1$, $A_2$, $A_3$, $A_4$, $A_5$ be adjacent points appearing in that order, taking cyclics if needed. By the perpendicular bisector of $A_3A_4$, $A_2A_3=A_4A_5$. By the perpendicular bisector of $A_2A_4$, $A_1A_2=A_4A_5$ which means $A_1A_2=A_2A_3$, and therefore the points on $T$ form a regular polygon. $~$ Clearly any regular polygon works, so suppose there was some other point $P$. Then its neighbors form a regular polygon and therefore all the angles including $P$ of the convex hull are equal. Furthermore all the faces of this solid are regular polygons so we have ourselves a tetrahedron, cube, octahedron, dodecahedron and icosahedron and it is easy to check only the tetra and octa hedrons work.

First solve the 2D case. If there are at least three points in a plane then consider the convex hull. By choosing a side of the convex hull and taking the "perpendicular bisector plane" we find that every angle of the convex hull is the same. By choosing a segment between points two vertices apart we find that every side of the convex hull has the same length. Hence the convex hull is a regular polygon. Take a point $A$ on the convex hull and interior point $B$. Reflection preserves the condition that a point is on the convex hull, which is a contradiction as $A$ will be sent to $B$. Now we solve the 3D case. We have already found that the points in any nontrivial plane form a convex polygon. Consider the convex hull again. Consider the gravity center (average of vectors) $O$. Since any plane of symmetry passes through the gravity center it follows $O$ is the center of the sphere through all the points. (Oops, could have made this observation earlier.) We can actually repeat this logic. Take a point $A$ and its neighbors $A_1,\dots,A_n$. Reflection about the perpendicular bisector plane of $A_iA_j$ sends neighbors of $A_i$ to neighbors of $A_j$, thus $A$ is preserved. Hence $AA_i$ is constant, meaning $A_1,\dots,A_n$ lie on a circle. Thus they form a regular polygon, implying that the faces of the convex hull containing $A$ are actually congruent. This reduces the search space to platonic solids. Tetrahedrons work. Cubes fail on the space diagonal. Octahedrons work. To disprove dodecahedrons and icosahedrons take two points which are a distance of two edges apart. We are done. $\blacksquare$