Characterizations of the sphere by means of visual cones: an alternative proof of Matsuura's theorem
CCharacterizations of the sphere by means ofvisual cones: an alternative proof ofMatsuura’s theorem
E. Morales-Amaya , D. J. Verdusco Hern´andez − Facultad de Matem´aticas-Acapulco,Universidad Aut´onoma de Guerrero, M´exico [email protected], [email protected] July 10, 2020
Abstract
In this work we prove that if there exists a smooth convex body M in the Euclidean space R n , n ≥
3, contained in the interior of theunit ball S n − of R n , and point p ∈ R n such that, for each point of S n − , M looks centrally symmetric and p appears as the centre, then M is an sphere. Introduction.
We have obtained our results while we have been exploring a family ofproblems concerning characterization of spheres and ellipsoids in termsof geometric properties of cones which circumscribes convex bodies,we have considered the papers [1], [2], [7] and [8], but, in particular,the Conjecture 2 in [2] (which we reproduce here as the Conjecture 1).Such conjecture was inspired by the following characterization of thesphere due to S. Matsuura [7]:
If a convex body M ⊂ E n , n ≥ , isstrictly contained in a closed convex surface S and looks spherical fromeach points of S , then M is an sphere . In [5] an important progressto Conjecture 2 in [2] was given, namely, there was proved that if a a r X i v : . [ m a t h . M G ] J u l mooth convex body D ⊂ R n , n ≥
3, looks centrally symmetric fromevery points outside D is because such body is an ellipsoid.Our main result is this work is the Theorem 2, comparing withGruber-Odor’s Theorem [5], for one hand, we have reduced, substan-tially, the quantity of those points where the convex body looks cen-trally symmetric, namely, we have assumed information of the convexbody only from the points in a sphere, and, on the other hand, wehave supposed that there is a point p such that p appears as thecentre of the body from every point in such sphere. As immediateconsequence of Theorem 2, we have obtained the Corollary 1 which isan special case Matsuura’s Theorem, we have just considered the case S = S n − (Of course, first we have proved, under Matsuura’s Theo-rem conditions, that all the axis of the spheric cones where a convexbody is circumscribed are concurrent, see Lemma 3). However, ourproof of the general case, i.e., n >
3, of such restricted version of Mat-suura’s Theorem can be given directly, since in [7] is just indicatedthe procedure to carry out the generalization but he did not providethe complete proof.The Theorem 3 is the natural variant of Theorem 1, we replacedthe sphere, the locus of the set of vertices of the cones circumscribingthe convex body, for a hyperplane. We decided to include in this workTheorem 3 because we did not find an explicit references about thiselemental, however, interesting result.
Preliminars.
Let R n be the Euclidean space of dimension n endowed with usualinterior product (cid:104)· , ·(cid:105) : R n × R n → R . We take a orthogonal coordinatesystem ( x , ..., x n ) for R n . Let B r ( n ) = { x ∈ R n : || x || ≤ r } be the n -ball of ratio r , centered at the origin, and let r S n − = { x ∈ R n : || x || = r } its boundary. For u ∈ S n − and s ∈ R , s non-negative,we denote by Π( u, s ) to the hyperplane { x ∈ R n |(cid:104) u, x (cid:105) = s } whoseunit normal vector is u and distance to the origin is equal to s and byΠ ∗ ( u, s ) to the open half-space { x ∈ R n |(cid:104) u, x (cid:105) < s } . For the points x, y ∈ R n we will denote by aff { x, y } the affine hull of x and y and by[ x, y ] the line segment determined by x and y .The set K ⊂ R n is said to be a convex body if it is a compactconvex set with non empty interior. A convex hypersurface is theboundary of a convex body in R n . As usual int K , bd K will denotethe interior and the boundary of the convex body K . Let Π ⊂ R n bea hyperplane. We denote by S Π : R n → R n the reflection with respect o Π. We say the the body K ⊂ R n is symmetric with respect to Π,or that Π is an hyperplane of symmetry of K , if S Π ( K ) = K . Definition 1
Let K ⊂ R n be a convex body, n ≥ , and let x ∈ R n \ K .We call the set (cid:91) y ∈ K aff { x, y } the solid cone generated by K and x . The boundary of the the solidcone generated by K and x will be called the cone that circumscribes K with vertex at x and it will be denoted by C x . Definition 2
Let x be a point in R n such that x = (0 , .., , x n ) and x n (cid:54) = 0 and let M ⊂ R n − be a centrally symmetric convex body in R n − with centre at (0 , ..., . We call the set (cid:91) y ∈ bd M aff { x, y } a symmetric cone with apex x . Such cone will be denoted by T ( M, x ) .In particular, If M is r S n − ⊂ R n − , the symmetric cone T ( M, x ) will be called right circular cone with apex x . The cone C y with apex y is said to be a symmetric cone (right circular cone) if it is a congru-ent copy of the symmetric cone T ( M, x ) for some x ∈ R n and somecentrally symmetric convex body (some sphere r S n − ) M in R n − forsome rigid movement which sent x in y . The image of the x n axisunder such congruence will be called the axis of C y and will be denoteby L y . The results. In virtue of the following fact
Remark 1
Let E ⊂ R n be an ellipsoid, n ≥ . Then, for every x ∈ R n \ E , the cone C x is a symmetric cone, the next interesting problems arise Conjecture 1
Let K ⊂ B ( n ) \ S n − be a convex body, n ≥ . If forevery x ∈ S n − , C x is a symmetric cone, then K is an ellipsoid. Our main result is the Theorem 2, we are able to prove an special caseof Conjecture 1, namely, we assume that all the axis of the cones arepassing for one point. We have decided to present separately the case = 2 of the Theorem 2, the Theorem 1, since it could be consideredas characteristic property of the circle in the direction of the maintheorem of [6].Let M ⊂ R be a convex figure and let x ∈ R \ M be a point. Wedenote by l x the bisector of the angle determined by the two supportinglines of M passing through x . Theorem 1
Let M ⊂ B (2) \ S be a convex figure and let p ∈ R bea point. If for every x ∈ S , l x is passing through p , then M is a circlewith centre at p . Theorem 2
Let K ⊂ B ( n ) \ S n − be a convex body, n ≥ , and let p ∈ R n be a point. If for every x ∈ S n − , C x is a symmetric cone and L x is passing through p , then K is an sphere. We must observe that in the extreme case of Theorems 1, or 2, whenthe radio of the circle, for the case n = 2, or the sphere, for n ≥ Corollary 1 (Matsuura)
Let K ⊂ B ( n ) \ S n − be a convex body, n ≥ . If for every x ∈ S n − , C x is a right circular cone, then K isan sphere. Theorem 3
Let K ⊂ R n be a convex body, n ≥ , and let Π be ahyperplane. If for every x ∈ Π \ K the cone C x is a right circular cone,then K is an sphere. Proof of Theorem 1. Let C and D be two convex figures in R , D ⊂ int C . For every x ∈ bd C , we define the Poncelet’s polygon , P D ( x ), with respect to C and D in the following manner. We take a supporting line L of D passingthrough x = x , and we denote by x the second intersection of L = L with bd C . Given the directed edge −−→ x x , we denote by E , and F , the two half-planes determined by aff { x , x } , been E , the halfsupporting plane of D , and by u interior unit normal vector of E , .Now we require that the set {−−→ x x , u } is a right frame of R . We denotenow by L the supporting line of C , L (cid:54) = L , passing through x andso on. The set of vertices of P D ( x ) is the set of points { x , x , x , ... } and we will assign indices to the vertices in such a way that will givethem a cyclic order if travel the edges of P D ( x ) in a counterclockwise.Therefore, the edges of P D ( x ) are {−−→ x x , −−→ x x , ..., −−−−→ x i x i +1 , ... } . We say hat P D ( x ) is inscribed in C and that circumscribes D . Given thedirected edge −−−−→ x i x i +1 , we denote by E i ( i +1) and F i ( i +1) the two half-planes determined by aff { x i , x i +1 } , been E i ( i +1) the half supportingplane of D and by (cid:92) x i x i +1 ⊂ S the arc S ∩ F i ( i +1) . We denote by Q D the intersection of all the half supporting planes E i ( i +1) of P D ( x ).From now on we will assume that C = S y D = M . Let x ∈ S .We consider a supporting line of M passing through x , say L . Invirtue of the hypothesis, the second supporting line of M passingthrough x , say L , is obtained reflecting L in the line aff { p, x } . Thusthe circle with centre at p and tangent to L is tangent to L too. Let y ∈ S , y (cid:54) = x , the other point given by the intersection of L with S and let L be the second supporting line of M passing through y . L is obtained reflecting L in the line aff { p, y } and the circle with centreat p and tangent to L is also tangent to L and so on. We denote byΣ x the circle with centre at p and tangent to L . Then P M ( x ) = P Σ x ( x ) . (1)We divide the proof in two cases. Firstly, we will assume that, forevery x ∈ bd S , the polygon P M ( x ) has a finite number of vertices. Lemma 1
If for some x ∈ S the polygon P M ( x ) has k vertices, then,for every y ∈ S , y (cid:54) = x , P M ( y ) has k vertices.Proof. Let x ∈ S be a point such that the polygon P M ( x ) has k vertices. The vertices x , x , ..., x k divide S in a collection of k arcs A , ..., A k and we will asigne indices to them in such a way that A i +1 ⊂ E i ( i +1) and x i +1 ∈ bd A i +1 . (2)We claim that there exist an integer r , 0 ≤ r < k , such that foreach i ∈ { , , ..., k } there exists r vertices x i , x i , ..., x i r with theproperty x i , x i , ..., x i r ∈ F i ( i +1) and x i j (cid:54) = x i , x i +1 . (3)Let y ∈ S , y (cid:54) = x . We denote the vertices of P M ( y ) by y = y , y , ..., y s , in the same way as we did with those of P M ( x ), and by G i ( i +1) and H i ( i +1) the two half-planes determined by aff { y i , y i +1 } ,been G i ( i +1) the half supporting plane of M , i.e., the half-plane withinterior unit normal u such that {−−−→ y i y i +1 , u } is a right frame of R .Suppose that for two indices i, m ∈ { , , ..., k } we have that y m ∈ A i . We are going to show that y m +1 ∈ A i +1 . e claim that y m +1 can not belong to A i j for j = 1 , , ..., r . Other-wise, observing that in virtue of (3) we have that A i j ⊂ F i ( i +1) for j = 1 , , ..., r and by (2) A i ⊂ F i ( i +1) , the line segment −−−−−→ y m y m +1 wouldbe contained in F i ( i +1) but since −−−−−→ y m y m +1 is an edge of P M ( y ) wouldexist z ∈ bd M such that z ∈ −−−−−→ y m y m +1 , and from here z ∈ F i ( i +1) but this would contradict that aff { x i , x i +1 } is a supporting line of M .Analogously we can see that for every j = 1 , , ..., rx ( i j +1) ∈ E i ( i +1) and x ( i j +1) (cid:54) = x i , x i + 1 . (4)Now we are going to show that y m +1 can not belong to A i j +1 for j = 1 , , ..., r . If y m +1 ∈ A i j +1 for some j = 1 , , ..., r , then −−−−−→ y m y m +1 ⊂ E i j ( i j +1) and, therefore, −−−−−→ x i j x i j +1 ⊂ H m,m +1 and since aff { x i j , x i j +1 } is a supporting line of M there exist w ∈ −−−−−→ x i j x i j +1 . Thus w ∈ H m,m +1 but this would contradict that aff { y m , y m +1 } is a supporting line of M . Analogously we can show that y m +1 can not belong to A t ∈ E i ( i +1) , t (cid:54) = i j + 1 , i + 1. Hence the unique place where can be y m +1 isin A i +1 . From here we have that s = lk for some integer l > t letΛ t = { y ∈ A i : P M ( y ) has tk edges } . According to this definition we can establish the statement of the firstpart of proof of Lemma 1 in the following way:int A i = ∞ (cid:91) t =1 Λ t (5)We are going to prove that, for every integer t > t (cid:54) = ∅ , the set Λ t is an open set. We will show this by the absurd.Suppose that for some t > t is not empty and is not open.Consequently, there exists y ∈ Λ t and a sequences { y s } ⊂ A i and { t s } ⊂ Z , t s > y s → y when s → ∞ and P M ( y s ) has t s k edges, t s (cid:54) = t . We have the following two possibilities: 1) t s = wk , forsome positive integer w , w (cid:54) = t , and this equality holds for an infinitenumber of indices s , s , ... ; 2) t s → ∞ . In the first case, in virtue ofthe smoothness of M , it follows that P M ( y s ) → P M ( y ), according ofthe Hausdroff’s metric, thus P M ( y ) has wk edges but this contradictsthat y ∈ Λ t . In the second case, in virtue of the smoothness of M again, P M ( y ) has an infinite number of edges but this also contradictsthat y ∈ Λ t . n virtue of (5) and that the sets A t are disjoints and that int A i isconnected, we conclude there exists only one positive integer t suchthat Λ t is non empty and Λ t = int A i . Finally, we claim that t = 1.Since x ∈ bd A i , there exist a sequence { x i } ⊂ Λ t such that x i → x and P M ( x i ) → P M ( x ) which implies that P M ( x ) has t k edges andfrom here t = 1 since we have assumed from the beginning that P M ( x ) has k edges. This completes the prove of Lemma 1. Lemma 2
Let C ⊂ R be a circle and let x ∈ bd C , a ∈ int C be twopoints and, let k be an integer, k ≥ . Then there exist a unique circle W x with centre at a , such that the Poncelet’s poligon P W x ( x ) has k vertices.Proof. The proof follows easily by an argument of continuity. Wedenote by r the ratio of C and by t the distance between the centreof C and a . Let Ω be the circle with centre at a and ratio r − t . Invirtue that, for every x ∈ C , the Poncelet’s polygon P Ω ( x ) has infinitenumber of edges, reducing the ratio of Ω, always with centre at a , wecan make that with the first k edges of the Poncelet’s polygon P Ω ( x ) werich a point y ∈ C , x (cid:54) = y , such that x is not contained in ∪ ki =2 F i ( i +1) .If we continue increasing the ratio of Ω we will get, eventually, thatwith the first k edges of the Poncelet’s polygon P Ω ( x ) we rich a point z ∈ C , x (cid:54) = z , such that x is not contained in ∪ ki =2 F i ( i +1) . In virtueof the continuity of this process, there exists a unique circle as werequired.Let x ∈ S . Let P M ( x ) be the Poncelet’s polygon inscribed in S and circumscribing M . By the case that we are considering, we canassume that P M ( x ) has a finite number of edges, say k edges, i.e., P M ( x ) is a closed polygon. By (1), P M ( x ) = P Σ x ( x ), then P Σ x ( x ) hasalso k edges. We denote by Σ the circle Σ x . We are going to provethat, for every y ∈ S , y (cid:54) = x ,Σ y = Σ . (6)In virtue of the Poncelet’s Porism (see, for example, [3],[4]), since P Σ ( x ) is a closed polygon of k edges, P Σ ( y ) is also a closed polygonof k edges. On the other hand, by the Lemma 1, since P M ( x ) has k edges, P M ( y ) = P Σ y ( y ) has k edges too. By the Lemma 2, Σ y = Σ.From (6) it follows that, for every y ∈ S , Q Σ y ( y ) = Q Σ ( y ) andfrom this we concludeΣ = (cid:92) y ∈ S Q Σ ( y ) = (cid:92) y ∈ S Q Σ y ( y ) . (7) n the other hand, since M ⊂ E i ( i +1) , 1 = 1 , , ..., k ; it follows that M ⊂ Q M ( y ) and by (1) we obtain, for every y ∈ S , M ⊂ Q Σ y ( y ).From this and by (7) M ⊂ (cid:92) y ∈ S Q Σ y ( y ) = Σ . (8)Since every supporting line of M is supporting line of Σ, otherwise,would exist x ∈ S such that Σ x (cid:54) = Σ but this would contradict (6),we conclude Σ ⊂ M . This and (8), implies M = Σ.Suppose now that there exists x ∈ S such that P M ( x ) has aninfinite number of vertices x , x , ... There are two possibilities: 1) theset { x , x , ... } is a dense set in S ; 2) There exists θ ∈ S such that x n → θ when n → ∞ .Let us assume that we are in the case 1). Since, for all i , M, Σ x ⊂ E i ( i +1) , it follows that M, Σ x ⊂ ∞ (cid:92) i =1 E i ( i +1) . (9)On the other hand, let w ∈ (cid:84) ∞ i =1 E i ( i +1) . Let suppose that w / ∈ Σ x .Let L be a supporting line of Σ x which separates w and Σ x . Let m, n the intersection points of L and S . In virtue of the density of theset { x , x , ... } in S , there are two subsequences of vertices x i , x i , ... and x i +1 , x i +1 , ... such that x i t → m and x i t +1 → n when t → ∞ and −−−−−→ x i t x i t +1 is an edge of P M ( x ). Taking t big enough, −−−−−→ x i t x i t +1 isclose enough of L and, consequently, aff { x i t x i t +1 } separates Σ x from x . Hence w / ∈ E i t ( i t +1) . Thus x / ∈ (cid:84) ∞ i =1 E i ( i +1) but this contradictsthe choice of w . Therefore ∞ (cid:92) i =1 E i ( i +1) ⊂ Σ x . (10)Since, for all i , E i ( i +1) is supporting half-plane of Σ x and is also asupporting half-plane of M , in the same way as we did before, weshow that ∞ (cid:92) i =1 E i ( i +1) ⊂ M. (11)Due (9), (10) and (11) Σ x = ∞ (cid:92) i =1 E i ( i +1) = M. hat is, M is a circle.Finally, we assume the case 2), i.e., there exists θ ∈ S such that x n → θ when n → ∞ . Since aff { x i , x i +1 } is a supporting line of M there exists a point α i ∈ −−−−→ x i x i +1 ∩ bd M . From x n → θ it followsthat α i → θ when n → ∞ . In virtue that α i ∈ bd M , we concludethat θ ∈ bd M , because M is closed, however, this contradicts ourassumption that bd M ∩ S = ∅ . Therefore this case is impossible.
Proof of Theorem 2.Proof of Theorem 2 case n = 3. We are going to prove that for eachplane Γ through p the section Γ ∩ K is a circle with centre at p andfrom here we are going to conclude that K is an sphere. Let Γ be aplane through p . In virtue of the hypothesis, for each x ∈ S n − ∩ Γ, C x is a symmetric cone and L x is passing through p , consequently, L x ⊂ Γ, the intersection Γ ∩ C x is a pair of supporting lines of Γ ∩ K passing through x and L x is the bisector of the angle determined bythis two lines. Thus the convex curve Γ ∩ K satisfies the condition ofTheorem 1. Hence Γ ∩ K is a circle with centre at p . Proof of Theorem 2 case n >
3. It follows straightforward fromcase n = 3 by mathematical induction under the dimension.Proof of Corollary 1. Lemma 3
Under the conditions of Corollary 1, there exists a point p ∈ int K such that for every x ∈ S n − the axis L x of C x is passingthrough p .Proof. Let x ∈ S n − . We are going to prove that, for every y ∈ S n − , x (cid:54) = y , we have L x ∩ L y (cid:54) = ∅ . Let us assume first that aff { x, y }∩ K = ∅ .Let Π , Π be two supporting hyperplanes of K containing aff { x, y } .It is clear that Π , Π are supporting hyperplanes planes of C x . LetΠ , be the bisector hyperplane of the solid dihedral angle determinedby Π , Π and containing aff { x, y } . We denote by Σ the hyperplaneaff { L x , Π ∩ Π } . Since for every hyperplane Γ , L x ⊂ Γ, the equality S Γ ( C x ) = C x holds, it follows that S Σ (Π ) is a supporting plane of C x containing Π ∩ Π and different than Π . Thus S Σ (Π ) = Π . HenceΠ , = Σ. Consequently, L x ⊂ Π , . In conclusion, we have that therelation aff { L x , x, y } = (cid:92) Π , holds, where the intersection is taken over all couples Π , Π of su-pporting hyperplanes of K such that aff { x, y } ⊂ Π , Π . Interchan-ging x by y , due the symmetry of this argument, we conclude that elation aff { L y , x, y } = (cid:92) Π , . holds, where again the intersection is taken over all couples Π , Π ofsupporting hyperplanes of K such that aff { x, y } ⊂ Π , Π . Thereforeaff { L x , x, y } = aff { L y , x, y } . Since L x and L y can not be parallel, weget L x ∩ L y (cid:54) = ∅ . Let x, y ∈ S n − such that aff { x, y } ∩ K (cid:54) = ∅ . We can find points z , ..., z k ∈ aff { L x , y } such that, making x = z , y = z k ,aff { z i , z i +1 } ∩ K = ∅ , i = 1 , ..., k − L z i ∩ L z i +1 (cid:54) = ∅ , i = 1 , ..., k − L z i ⊂ aff { L x , y } , i = 2 , ..., k . Hence L x ∩ L y (cid:54) = ∅ .Therefore the family of lines Φ = { L x : x ∈ S n − } has the propertythe every two members intersects. Then either there exists a plane Πsuch that the lines of Φ are contained in Π or there exist a point p ∈ R n so that all the lines of Φ are passing through p . The firstcase is impossible, otherwise, we would have that S n − ⊂ Π which isabsurd. This completes the prove of the Lemma.In virtue of Lemma 3, K satisfies the conditions of Theorem 2.Thus K is an sphere. Proof of Theorem 3.
Lemma 4
Let K ⊂ R n , n ≥ be a convex body and let Σ be a hyper-plane such that Σ ∩ K (cid:54) = ∅ . If for every affine ( n − -plane Γ ⊂ Σ \ K the two supporting hyperplanes of K containing Γ makes equal angleswith Σ , then K is symmetric with respect to Σ .Proof. It is enough to consider the case n = 2. Contrary to thestatement of the Lemma 4, we suppose the body K is not symmetricwith respect to Σ. Consequently, there exists points x, y ∈ bd K suchthat the line segment [ x, y ] is perpendicular to Σ and (cid:107) x − z (cid:107) < (cid:107) y − z (cid:107) (12)where z = [ x, y ] ∩ Σ. Let L x be a supporting line of K at x and let L w be the second supporting line of K passing through L x ∩ Σ (seeFig. 1). In virtue of the hypothesis, the angle between Σ and L x isequal to the angle between Σ and L w . By (12), there exists a point u ∈ [ z, y ] such that (cid:107) x − z (cid:107) = (cid:107) u − z (cid:107) . On the other hand, since L w is supporting line of K , we have that (cid:107) v − z (cid:107) < (cid:107) u − z (cid:107) for all v ∈ [ z, y ] ∩ K . Taking v = y , we have (cid:107) y − z (cid:107) < (cid:107) u − z (cid:107) , i.e. (cid:107) y − z (cid:107) < (cid:107) x − z (cid:107) but this contradicts therelation 12. Lemma 5
Under the conditions of Theorem 3, there exists a point p ∈ int K such that, for every x ∈ Π \ K , the axis L x of C x is passingthrough p .Proof. In analogous way as we did in Lemma 3, it can be shown thatthe family of lines Ψ = { L x : x ∈ Π } has the property the every twomembers intersects. Then either there exists a plane Γ such that thelines of Ψ are contained in Γ or there exist a point p ∈ R n so that allthe lines of Φ are passing through p .If we assume that the former case holds, then it follows that Π = Γand n = 3. Since for every x ∈ Π \ K , L x ∩ int K (cid:54) = ∅ necessarily wehave Γ ∩ K (cid:54) = ∅ , i.e., Π ∩ K (cid:54) = ∅ . Consequently if Π ∩ K = ∅ then,for any n ≥
3, there exist a point p ∈ R so that all the lines of Φ arepassing through p . Now if n = 3 and Π ∩ K (cid:54) = ∅ , then Π ∩ K is a circlewith centre at the point p . For one hand, the curve Π ∩ K is centrallysymmetric, in order to see this, for each unit vector u parallel to Π, wedenote by C u the right circular cylinder circumscribing K and withgeneratrices parallel to u and by L u the corresponding axis, in virtue ofthe hypothesis p ∈ L u and the line L u is situated to the same distancefrom the two lines given by the intersection Π ∩ C u . Thus, in virtue ofthe remark exposed in the paragraph next to the Theorem 2, the curve ∩ K is centrally symmetric. On the other hand, Π ∩ K is of constantwidth. There exists an affine diameter [ a, b ] of K perpendicular toΠ, in fact, it belongs to each cylinder C u , u unit vector parallel toΠ. From this it follows that all the cylinder C u , u parallel to Π, arecongruent. Since L u ⊂ Π and p ∈ L u , for all unit vector u parallel toΠ, we get that the chord [ a, b ] has as his midpoint p and the bandsdetermined by the lines Π ∩ C u are congruent. Consequently, Π ∩ K isa curve of constant width. Thus Π ∩ K is a circle. From here, it easyto see that all the lines of Φ are passing through p . This completesthe prove of the Lemma 5. Proof of Theorem 3 in dimension 3 . In virtue of Lemma 5, thereexists a point p ∈ int K such that, for every x ∈ Π \ K , the axis L x of C x is passing through p . We are going to prove, using Lemma 4, thateach plane Σ passing through p is plane of symmetry of K . Let Σ bea plane, p ∈ Σ. We denote by L the intersection Σ ∩ Π. Let Γ ⊂ Σ \ K be a line. First, we assume that Γ is not parallel to L . Denote by x the intersection Γ ∩ L . If p ∈ Π (see Fig. 2), then for each line L ⊂ Π, p ∈ L , and for each x ∈ L , the axis L x is equal to L . In fact, byLemma 3, for each x ∈ L , the axis L x is determined by the points x and p . Since L = L x and L ⊂ Σ, we have that Σ is plane of symmetryof the cone C x . In the case p / ∈ Π, since p, x ∈ Σ, then L x ⊂ Σ,and again Σ is plane of symmetry of the cone C x . Therefore, in bothcases, there exists two supporting planes of C x and, consequently of K , symmetric with respect to Σ and containing Γ. Thus K satisfiesthe condition of Lemma 4. Hence Σ is plane of symmetry of K . Invirtue of the arbitrariness of Σ, p ∈ Σ, it follows that K is an sphere. Proof of Theorem 3 in dimension >
3. Let K ⊂ R n be a convexbody and let Π be a hyperplane such that the conditions of Theorem3 holds. Let assume that Theorem 3 holds for dimension n −
1, we aregoing to show that Theorem 3 holds for dimension n . By Lemma 5there exists a point p ∈ int K such that, for every x ∈ Π \ K , the axis L x of C x is passing through p . Let Γ be an affine ( n − p . Then, for all x ∈ Γ ∩ Π, the axis L x is equal to aff { x, p } and, therefore, L x ⊂ Γ. It follows that C x ∩ Γ is a circular cone (indimension k ) that circumscribes Γ ∩ K . According to the inductionhypothesis Γ ∩ K is an sphere. Hence all the ( n − K passing through p are spheres. Thus K is an sphere.Finally, we propose the problem: to prove the following conjecture,which could be considered as the following natural step in way of thesolution of Conjecture 1. Conjecture 2
Let K ⊂ B ( n ) \ S n − be a convex body, n ≥ , and let L ⊂ R n be a line. If, for every x ∈ S n − , the cone C x is a symmetriccone such that L x ∩ L (cid:54) = ∅ , then K is an n -ellipsoid and, for every 3-plane Π containing L , thesection Π ∩ K is an ellipsoid of revolution with axis L . The authors want to thank to Jesus Jeronimo-Castro for his commentsand suggestions about this work.
References [1] J. Arocha, L. Montejano and E. Morales,
A quick proof ofHobinger-Burton-Larman Theorem . Geom. Dedicata 63(3). pp331-335, January 1996.[2] G. Bianchi, P. M. Gruber.
Characterizations of ellipsoids . Arch.Math. Vol. 49 Issue 4, pp 344-350. 1987.[3] P. Griffiths, J. Harris.
A Poncelet theorem in the space . Comment.Math. Helvetici 5 (1977) 145-160.[4] P. Griffiths, J. Harris.
On Cayley’s explicit solution to Ponceletporism . L’Enseignement Math´ematique V. 24. 1978.
5] P. Gruber, T. ´Odor,
Ellipsoids are the most symmetric bodies .Arch. Math. 73 (1999) 394-400.[6] J. Jer´onimo-Castro, E. Rold´an-Pensado,
A charecteristic propertyof the Eucliden disc , Periodica Mathematica Hungarica. Vol. 59(2), 2009, pp 2013-222.[7] S. Matsura.
A problem in solid geometry.
J. Math.. Osaka CityUniv. Vol. 12, A 12, pp 89-95. 1961.[8] L. Montejano and E. Morales,
Characterization of ellipsoids andpolarity in convex sets . Mathematika 50 (1). pp 63-72. 2003.. Mathematika 50 (1). pp 63-72. 2003.