1 Several Conclusions on another site setting problem
Yuyang Zhu
Department of Mathematics & Physics, Hefei University, Hefei , P. R. China E-mail: [email protected]
Abstract:
Let },,,{ n AAAS = be a finite point set in m-dimensional Euclidean space m E , and ji AA be the distance between i A and j A . Define = nji ji AAS )( , jinji AASD = max)( , )( )(),( SD Snm = , == nSESSD Snm m ,)( )(max),(sup . This paper proves that, for any point P in an n-dimensional simplex + n AAA in Euclidean space, += ni i PA ≤ =+ nt jinji tttt AA sup . By using this inequality and several results in differential geometry this paper also proves that −+= , )2,(sup + nn ≥ ( ) nnn nC ++ −++ . Keywords:
Setting sites; Discrete geometry; Distance; Supremum
MR(2000)
Subject Classification
1 Introduction
Let },,,{ n AAAS = be a finite point set in m-dimensional Euclidean space m E , and ji AA be the distance between i A and j A . Define = nji ji AAS )( , jinji AASd = min)( , )( )(),( Sd Snm = , == nSESSd Snm m ,)( )(min),(inf . It is a site setting problem to evaluate ),(inf nm . It has been proved with convex geometry theory that [1] nnn Cnn ++ +−=+ , Foundation item: Foundation Project of Hefei University and Anhui Province (12KY04ZD, 2015gxk059) It has also been proved that [2] += . That leads to the following problem. By defining jinji
AASD = max)( , )( )(),( SD Snm = , To evaluate == nSESSD Snm m ,)( )(max),(sup is a conjugate problem of the previous one. These problems enrich the discrete geometry theory and also have real world applications [1-12]. First this paper gives a geometry inequality, and then it proves that −+= , )2,(sup + nn ≥ ( ) nnn nC ++ −++ . Lemmas Lemma 2.1 P AAA E , the following inequality holds: = i i PA ≤ = sup t jiji tttt AA . Proof
Let = AAA . (I) If P , without losing generality, assume point P is on edge AA . Let = PAA , = PAA . Since + AA += , }2,1{ i , i A i holds. Without losing generality, assume A . Since a greater angle in a triangle corresponds to a longer edge, PA ≤ AA , and also considering PA + = PA AA , we have = i i PA ≤ AA + AA ≤ = sup t jiji tttt AA . (II) If point int P , without losing generality, assume ii A min A = . Draw a line parallel to AA through P and let it intersect with AA and AA on points E and E respectively. Also draw a line parallel to AA and let it intersect with AA and AA on points F and F respectively, and then we get EPEAFAEA +=+
111 1 PA and AEPAAEPEPAPEA +=+= A EPA = . Thus EAPA in EPA , and similarly we can prove FAPA . Therefore = i i PA FAEA + FAEA ++ = AA + AA = = sup t jiji tttt AA . Lemma 2.1 can be generalized with convex geometry theory and similar approaches of proving lemma 2.1, and the following conclusion can be drawn: Lemma 2.2
Let be a n-dimensional simplex in n-dimensional Euclidean space n E with vertices ,,, + n AAA . For P , it holds that += ni i PA ≤ =+ nt jinji tttt AA sup . Proof
In the following we only prove the case in which P . For the case of int P , we can prove it with similar approach seen in the proof of Lemma 2.1. If = n , according to lemma 2.1 the conclusion holds. Assume it also holds in the case of )1( = kkn . Then in the case of += kn , since P , there must exist a − k -dimensional simplex − k such that P − k . Without losing generality, assume − k = simplex k AAA , using inductive assumption we get = ki i PA ≤ −= sup kt jikji tttt AA . On the other hand, since is a simplex and P , + k PA ≤ kkkk AAAAAA ,,,max +++ skks AA max + = . then +=+= += kki iki i PAPAPA ≤ −= sup kt jikji tttt AA skks AA max + + ≤ =+ kt jikji tttt AA sup . Using complete induction we conclude that lemma 2.2 is proved if P . Lemma 2.3 [5],[13]
If any point on a curved surface has Gauss curvature K , then the extreme point of the surface must be a maximum point.
3 Theorems and proof Theorem 3.1 −+= . Proof
If three of four points ,,,
AAAA in a plane are collinear, it is straightforward to prove that )4,2( −+ . So the only problem left is the case that no three points are collinear. If one point A is in the simplex composed of the other three points AAA , then according to lemma 2.1 or lemma 2.2, =
31 4 i i AA ≤ = sup t jiji tttt AA , thus )4,2( )( )( SD S = jijiji ji AAAA max = jijii ji jii AAAAAA max = += jijiji jit jiji
AAAAAA tttt maxsup = + = jijijiji AAAA −+ . In the following we prove that ,,, AAAA are vertices of planar convexity, and )4,2( −+ . ( I ) If any three points of ,,,
AAAA are vertices of a regular triangle, and the side length of the triangle jiji
AASD max)( = , then in the following we prove )4,2( −+ . Without losing generality, let = SD , vertices of the regular triangle be ,, AAA . Since quadrilateral
AAAA is convex, A is outside the regular triangle and the distances between it and ,, AAA are all ≤ = SD . Let ,, AAA be centers and draw arc AA , AA , AA with radius 1 respectively, then these three arcs form three bow shapes with the edges of the regular triangle, and A must fall into one of them. Regarding symmetry of this problem, assume A falls into the bow shape with AA as the bowstring. Since the edge lengths of regular triangle AAA == SD , let us assume the coordinates of three vertices of the triangle are )0,0( A , )0,1( A , ),( A , and that of A be ),( yx . According to the previous statements, +−+= yxxyyxyx , which is the bow shape where A falls into. Since AA ≤ = SD , the remaining task is to prove the maximum of AA AA + is −+ . Let ),( yxf = AA AA + , and we have ),( yxf = )1()()( yxyx +−+−+− , and )1()()( )( yx yyx yx fr +−+−+− −== , )1( )1()()( )( yx xyx xy ft +− −+−+− −== , +− −+−+− −−−== )1( )1()()( ))(( yx yxyx yxyx fs . Since A is not collinear with AA , or equivalently −−− x yxy , according to Cauchy inequality, − srt . According to differential geometry theory on curved surface, the function ),( yxf has its Gauss curvature K everywhere in . According to lemma2.3, the extreme point of ),( yxf in must be a maximum point. In the following we evaluate the stable point of ),( yxf in . =+−+−+− −= =+− −+−+− −= .0)1()()( ,0)1( 1)()( yx yyx yyf yx xyx xxf (1) Solve (1) and we get yxyx )())(1( −=−− , (2) or yxyx )())(1( −=−− . (3) Since A , the left hand side of (3) while the right hand side , then (3) does not hold. Only (2) holds and according to (2), =−+ xy , which is the equation of line passing through points A and A . Since A is not collinear with AA , function ),( yxf has no stable point in . According to the continuity of ),( yxf in , ),( yxf must reach its maximum on the boundary of , or . According to previous statement, if ),( yxf reaches its extreme value on , then we can evaluate its maximum there. Since ( , ) 3 3 0, 1 x y y x x = + − = ( , ) 1, 1, 0 x y x y x y + = , it is obvious that if point A ),( yx is collinear with AA , then =−+ xy , and if ≤ x ≤ , then ),( yxf = − . If A ),( yx =+ yxyxyx , then according to Lagrange multiplier method the only maximum point is ),( , where the maximum of ),( yxf is reached, or == ),(),(max fyxf yx − . Hence, we proved that in the convex quadrilateral if there are three vertices which are also the vertices of a regular triangle and let )( SD equals to the side length of the triangle, then )4,2( = ji ji AA AA + ),( yxf + ≤ ++ − = −+ . The equality holds if and only if A ),( yx bisects the arc. ( II ) Since there must be two points between which the distance reaches maximum in a convex quadrilateral, without losing generality, let AA = )( SD and the coordinates of A and A be )0,0( A and )0,1( A respectively, while other two points be in the region += yxyx +− yxyx . Since any point P not collinear with AA must be on a circular arc AA . In other words, ),( yxP and y , + RrRb , such that ),( yxP satisfies the following equation: )()( rbyx =−+− . (4) This arc is denoted by ),( rbS . In the following we prove that if ),( yxP ),( rbS , to make PAPA + reaches its maximum on the circle arc, P must bisect ),( rbS and its coordinate satisfies (4), and ( ) PAPA + ≤ ( ) PAPA + ( ) )1(2 yxyx +−++= . Let L = ( ) PAPA + ( ) )1(2 yxyx +−++= , then according to (4) L ( ) )1(2 yxyx +−++= ( )( ) )(2)1(2 −+++−+= xrbxx . If −+−−−−−= )()(4)(4242 xrxbxxL x = , since P , b and = x , in other words, if P bisects ),( rbS , then L reaches its maximum. If = x , then PAPA = , hence PAPA + L = also reaches its maximum. According to this conclusion, only if points , AA are on the midnormal of AA , then ( ) ( ) AAAAAAAA +++ can reach its maximum. On the other hand, since AA ≤ = SD , only if AA = = SD and points , AA are on the midnormal of AA , then ji ji AA can reaches its maximum. Thus we can let the coordinates of , AA be ),( yA and ),( yA respectively, where =− yy and ),(, − yy , and use similar approach as in (I) to prove that )4,2( ≤ −+ . To conclude, sup )4,2( ≤ −+ , and according to (I) the equality does conditionally hold, hence sup )4,2( = −+ . The theorem is proved. From definition we can also conclude that Theorem 3.2 )1,(sup + =+ n Cnn . Using the method of constitution, the following theorem can be proved: Theorem 3.3 )2,(sup + nn ≥ ( ) nnn nC ++ −++ . Proof
Consider + n points ,,,, ++ nn AAAA in n-dimensional Euclidean space n E . If ,,, + n AAA are the vertices of n-dimensional regular simplex with unit side length, then draw a n-dimensional hyper-sphere O whose center is A and with unit radius length. − n -dimensional simplex + n AAA gives a − n -dimensional hyper plane and divide O into two parts of sphere with different sizes. Denote the median point of the smaller one face by + n A , then according to convex geometry theory we have )},1,,2,1{,(1 jinjiAA ji += , = + n AA , ( ) ( ) −−++ +−= nnnnnnni AA ( ) nn + −= )1,,3,2( += ni , and ( ) − + nn , we conclude that += +++ ++=+
12 22111 )2,( ni ninnji ji
AAAAAAnn = ( ) nnn nC ++ −++ , thus )2,(sup + nn ≥ ( ) nnn nC ++ −++ . The theorem is proved.
4 Conclusions and guesses
The proof of sup )4,2( = −+ is more complicated compared with the proof of += ([9]). Therefore the evaluation of ),(sup nm may be more difficult than that of ),(inf nm . According to the conclusion of [10], if )( mnn points are vertices of a convexity in m -dimensional Euclidean space, and their distribution is such that ),(inf nm can be reached, then the distances between neighboring points must be equal. However, this is not case for ),(sup nm , which is shown in the proof of sup )4,2( = −+ . According to theorem 2.2 and [1], )1,(inf)1,(sup + =+=+ n Cnnnn . On the other hand according to the conclusion in [9] += )4,2(sup , can prove )5,2(inf )5,2(sup . Guess 4.1 If ,1 + mn and m ≥ , then ),(inf nm ),(sup nm . Guess 4.2 )2,(sup + nn = ( ) nnn nC ++ −++ . References [1] YuYang ZHU. Several Results on a Problem of Setting Site.
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