LIBRARY OF THE UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN 510.84 1463 c no. 51-60 ENGINEERING AUG 51976 The person charging this material is re- sponsible for its return to the library from which it was withdrawn on or before the Latest Date stamped below. Theft, mutilation, and underlining of books are reasons for disciplinary action and may result in dismissal from the University. UNIVERSITY OF ILLINOIS LIBRARY AT URBANA-CHAMPAIGN ENfill CON OCT 18 OCT 11 MAY m 7B» FEB 27 MAR 5 m kit I Mp m 31990 Klfff L161 — O-1096 Digitized by the Internet Archive in 2012 with funding from University of Illinois Urbana-Champaign http://archive.org/details/someextremalprob52purd iiNEERING LIBRARY UNIVERSITY OF ILLINOIS URBANA, ILLINOIS UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN URBANA. ILLINOIS 61801 CAC Document No. 52 SOME EXTREMAL PROBLEMS IN GEOMETRY AND ANALYSIS By George B. Purdy July 1, 1972 CAC Document No. 52 SOME EXTREMAL PROBLEMS IN GEOMETRY AND ANALYSIS By George B. Purdy July 1, 1972 Center for Advanced Computation University of Illinois at Urb ana -Champaign Urbana, Illinois 61801 Abstract This report treats three optimization or extremal problems in geometry and analysis. In the first chapter we consider the problem of finding positive minima of certain Dirichlet L-functions with positive real arguments. Optimal triple lattice packing of spheres of equal radius is the subject of the second chapter. Packing and covering problems with spheres are important in information theory for the design of communication networks. In the last chapter we treat an extremal problem about triangles of equal area. TABLE OF CONTENTS Page I. THE REAL ZEROS OF THE EPSTEIN ZETA FUNCTION 1 1. Introduction 1 2. The Programs k 3. The Error Analysis 5 k. Verification of (k) ik 5. Summary lk List of References l6 II. THE LATTICE TRIPLE PACKING OF SPHERES IN EUCLIDEAN SPACE IT 1. Introduction 17 2. An Economical Lattice A 18 o 3. The Lattice A Is Locally Optimal 23 o List of References 37 III. SOME EXTREMAL PROBLEMS IN GEOMETRY 38 1. Introduction 38 2. Notations 38 3. The Article of Erdos and Purdy. 39 k. The Example of Linz Generalized 39 5. Statement of the Main Theorems 1+0 6. Some Graph Theory ho 7. The Relation Between the Main Theorems 1+1 8. Some Lemmas 1+2 9. Proof of Theorem 2 1+5 List of References 1*9 1 I. THE REAL ZEROS OF THE EPSTEIN ZETA FUNCTION §1. Introduction If d > h and -d is a fundamental discriminant — that is, of -d = 1,5,8,9,12 or 13 (modulo l6) and -d is not divisible "by the square of any odd prime, then we define L -,(s) = £ _ n ( - d|n) n (Re s > 0), where (-djn) is the Kronecker symbol. M. E. Low proved in Theorem 2 of [l] that a certain hypothesis, which we will call Low's Hypothesis , implied L n (s) > for s > 0. Low then put an IBM 709^ to work to check this hypo- -d thesis for small values of d and concluded that if -d is a fundamental discriminant and d < 593,000, then L -,(s) > for s > with the possible exception of L.ijcn^y G. J. Rieger, who reviewed Low's paper in the Mathematical Reviews, asserted that Low's results were doubtful because he did no error analysis of his computation. Dr. Paul Bateman suggested to me that I do a computer verification of Low's Hypothesis using higher precision arithmetic than Low did, extend the range, and at the same time do an analysis of the error to ensure the validity of the results. We tested Low's Hypothesis for d _< 800,000 on an IBM 360/75 using double precision arithmetic and discovered one counterexample overlooked by Low — namely d = 357819- Low's Hypothesis is true for 1 < d< 800,000 with three exceptions. We do not wish to quote all the details of [l] here. We give the definition of Low's Hypothesis and refer the reader to [l] for the proof that it implies L (s) > for s > 0. Low's Hypothesis (i) £ a(a,d)//a ^ (ii) y.{a(a,d) - H (a,b,c)}//a > ^Q o = (iii) LO> 2n+1 + a 2n+1 (a,d)/(2n + l)!}//a>0 (n >, 1 ) . Here (1) a(a,d) = log a + log (8ire" Y ) - (l/2) log d, b n = (2 n -l)c(n)/n + 2 n 3 n , and 6 is given by n log{(s-l)ds)} = ^ =1 (-l) n 6 n (s-l) n (|s| < 3), and Q is the set of triples (a,b,c) where a,b,c are the coefficients of a reduced quadratic form of discriminant -d, and H (a,b,c) is an upper bound on the error function, H(s; a,b,c) of Bateman's and Grosswald's approximation to the Epstein zeta function — see [3]. More exactly, 2 Q = {(a,b,c): -d = b -Uac, -a < b <^a < c or < b ^ a = c; a,b,c integers} and (2) H (a,b,c) = 2k" l/2 e" 27Tk cos Ub/a) + (0.08)k~ l/2 e~ 27rk |cos Ub/a)|, o where k = /d/(2a). It is easily seen — see [l] — that H (ajbjc)^ 0.005. Hence, the condition, (3) I Q Wa,d) - 0.005}//a > 0, implies both (i) and (ii) of Low's Hypothesis. As written, condition (iii) is an infinite sequence of inequalities for n = 1,2,3 . . . However, the condition is trivially satisfied for large n. To see this, suppose that d 4 d . Then a(a,d) is bounded and b > 6 . 3k for n >. 3 — for this see [l]. Hence b + a (a, d)/ (2n+l ) ! will be positive for n sufficiently large, depending on d . Summing, we get (iii) for n > n (d ). This leaves o = o o only finitely many sums to check. Since (i) and (ii) are finitely computa- ble, we see that Low's Hypothesis is finitely computable for every d. Low shows in [l] that (iii) is a consequence of (i) for d < 593,000. 3, His proof is in three parts. First of all L.tt^+a (a,d)/6}//a > 2j.a(a,d)//a H 3 Q for d < 593000. Secondly, £ {b +a 5 (a,d)/l20}//a > 2.5La(a,d)//a for d < 1,970,000, and finally b 2n+1 +a 2n+1 (a,d)/(2n+l ) ! > for d < 1,320,000, and n ^ 3. It was therefore possible for us to prove (iii) for d < 1,320,000 by testing on the computer the condition (**) L(b +a 3 (a,d)/6}//a > 0, 593000 4 d< 1,320,000. The number |q| of triples (a,b,c) in Q is the class number of -d, denoted by h(-d). We need the following estimate later: (5) |Q| < ^000 for d 4 80,000. This follows from the following result : Lemma 1 h(-d) = (/d/7r) (L (l)) < (/d/ir) log d < 3900 for k < d 4 800,000, Proof The first equation is well known. Let x(n) = (-d|n), the Kronecker symbol, and let s(n) = x (l) + x(2) +. ••+ x(n). Then L - d (1) = Ci x(n)/n = ir_ n (s(n) - s(n-l))/n L n=l = I~ =1 s(n)(l/n - l/(n+l)) = I~ =1 s(n)/(n(n+l)). Let k = max|s(n)|. Then k 4 (d-l)/2, and n L -d (l) i£l l/(n+D+k^ =k l/(n(n+l)) = ^:il/(n+l)+kl~ =k (l/n-l/(n+l)) 1/2 + 1/3 + . . .+l/k + 1 „ rk+l/2^ . - A /2 dx/x = log(2k+l) < log d. Hence h(-d) = (/d/Tr )L (l) < /d(log d)/w < (/800,000 log(800,000 ) )/tt < 3900, for k < d < 800,000. §2. The Programs In testing (3) and (k) it is not necessary to evaluate the whole sum, because many of the terms are known to be positive. Let Q consist of the triples (a,b,c) of Q such that |b|< 128 and let Q consist of the others. Then I {a(a,d) - 0.005}//a = £ Q {a(a,d) - 0.005}//a + £ Q {a(a,d) - 0.005}//a, and the terms {a(a,d) - 0.005}//a in the second sum are positive for d < 3,000,000. To see this we note that since a j> |b| £ 128, we have a(a,d) -0.005 = log a + log(8Tre" Y ) - 1/2 log d - 0.005 > .035 > 0. Two programs were used to verify (3). Program One tested the condition (6) I {a(a,d) - 0.005>//a > 1/10 Q l in double precision arithmetic for all d 5 1,5,8,9,12,13 (modulo l6) between 5 and 800,000 recording the failure (approximately 2%) onto magnetic tape along with the sum T . The 1/10 allows margin for error due to round-off. Q l Program Two reads the tape and adds terms from J_ until the resulting ^2 sum exceeds 1/10; however if the resulting sum never exceeds 1/10, then d is again written on magnetic tape to be read by Program Three. A third program, Program Three, verifies conditions (i) and (ii) directly for those few hundred d's failing Program Two. Only three numbers failed condi- tion (i), and none failed (ii); the numbers were 1151^7, 357819 and 63618U. §3. The Error Analysis The calculations were performed using double precision floating point arithmetic on the IBM 360/75 at the University of Illinois. The numbers representable exactly in this way are of the form r 1&-1\ where r and p are integers, l6 4 |r| 4 l6 -l , and |p| < 6h. Thus small integers and rational numbers with denominator 2 for small k are representable exactly. Any other number x is approximated by (1 + e)x where | e | 4 l6" 13 . Following a notation introduced by Wilkinson, we shall use f£(E) to denote the computed value for the mathematical expression E. When the order of the calculation is ambiguous, such as in E = a+b+cd, it is assumed that the calculation proceeds from left to right; thus E = ((a+b) + cd). For any reasonable machine, f£(a) = (1 + e 1 ) a f£(a+b) = (1 + e 2 ) f£(a) + (l + e^) f£(b), f£(a-b) = (1 + e^) f£(a) - (l + e ) f£(b), f£(axb) = (1 + eg) fJl(a) f&(b), and f£(a/b) = (l + e ) f£(a)/f£(b ) , where | e . | 4 6 for 1 4 i 4 7. There could well be some discussion about the value of 6 . If the machine 360/75 were designed by a numerical analyst, we could assume that 64 16 <2.23 x 10 , but machines today usually are not. We shall see however that 6 does not have to be anywhere near that small for our calculations -12 to be valid - e.g. 6 = 10 would be adequate. The mathematical functions cos x, e , /x , and log x were calculated using the double precision routines in the Fortran Library; we need the following: f£(cos x) = cos x + (0 < x < tt), f£(e X ) = (1 + 2 ) e X + (-100 4 x 4 0), fH(/x) = (1 + 0^) Jx (14x4 1000), aXid f£(log k) = (1 + ) log k (k = 1, 2, ..., 10 6 ), where Jo. I < M, and M is not too large. We do not know whether anyone has done an analytical error analysis of these routines. Perhaps Dr. Kuki of the University of Chicago (now deceased), who designed the routines did one. Kenneth E. Hillstrom of Argonne National Laboratory — see [h] — did a test on these routines, using random arguments over several intervals and comparing the results with higher precision calcula- tions on another machine. His results suggest |0. | 4 1.7 x 10 for 1 4 i 4 k , hut he naturally did not test log x over the range [l, 10 ]. A casual glance at the algorithm used, however, is enough to convince us that the error has little to do with the size of x. (The first step expresses x = 2 m, where < m < 1 ). In what follows we shall take M to he 10 to be safe. Lemma 2 If n (l + e .) = l + nn,n<6 9 and | e . | 46, then | n j 4 6 /(l-n6 ) i=l X X Further, if n 4 100, then | n | 4 6/(l - 1006). Proof jj (1 + £ , ) -l| <_ (1 + 6) n -1 i=l n6 + (§) 8 2 + ... + ( n ) 6 n 4 n6 + (n6) 2 + (n6 ) 3 + .. . = n6/(l - n6 ) 4 n6/(l - 1006). Remark If (l + e ) _1 = 1 + n , where |e| 46 , then | n | < 6/(l - 6). In what follows we shall assume that 6 = 10 , M = 10 , M' = M/(l - 100M), and we shall adopt the convention that all |e. | 4 6, and all |0. | <_ M. In particular, |e. | 46 < M < M' , and |o. | ^M < m' . ¥e now begin the error analysis for Programs One and Two which evaluate the sum L*{a(a,d) - .005)/^, where Q c Q* C Q. We first find the error in calculating each term T = {a(a,d) - 0.005)/Va. We recall (l) that a(a,d) = log a + log (87Te ) - (— ) log d. It is a simple matter to determine from tables such as [5] that 2.6U695576 < log (8^e" Y ) = 3 log 2 + logTT - a < 2.6U695577. We defined C to be the lower value, and fed it into the machine in decimal; using C in place of log (8ire ) will never lead us to erroneously conclude that (3) holds. Let L = C - 0.005 - (|0 log d. Then t!L(l) = (1 + e ± ) tl(C - 0.005) - (l + Eg) f*((|-) log d) = (1 + e ± ) {(1 + e 3 ) f£(C) - (1 + e k ) f£(0.005)} - (1 + e 2 ) (1 + e 5 ) (|) f£(log d) = (1 + e ± ) {(1 + e 3 ) (1 + e 6 )C - (l + e^) (l + e ) 0.005) - (1 + e 2 ) (1 + e ) (|-) (1 + 1 ) log d = (1 + 3n 1 )C - (l + 3n 2 ) 0.005 - (l + 3n 3 ) (|-) log d where |n. | < M(l - 3M)~ (l l/ 10 - 1.2 x 10 _1+ > and therefore satisfies (i) and (ii). We now discuss the rounding error in Program Three, where the sum £{a(a,d) - H (a,h,c)}//a is evaluated for those fundamental discriminants -d H O that failed Programs One and Two. This sum differs from the earlier sum only in that .005 is replaced hy H (a,h,c). Therefore we must estimate the errors in calculating H (a,h,c) and k, which are given by (2). r- 2 2 2 2 We have k ^ /3/2, since d = Uac-b ^ ka -a = 3a , since hy reduction b 2 < a 2 < c 2 , and k 2 = d/(l+a 2 ) >, (3/M- In the follbwing analysis, we still assume that |e.| 46 and |©.| 4 M, and of course the e. and 0. are unrelated to the previous ones. We begin with k, fJl(k) = f£(v^/(2a)) = Sa(l + 1 )(1 + e 1 )/(2a) = /d(l + n 1 )/(2a) = k(l + n ), 10 where |n, | < 2M 1 . We next consider f£(cos(irb/a) )=cos(fJ?,(TTb/a) ) + = cos Cub (l + n 2 )/a) + 2 = cos(iTb/a) cos (n Trb/a) - sin(iTb/a) sinCn^b/a) + p where |n p | <. 2M' . Now |cos(n 2 TTb/a) - l| = 2sin 2 (n 2 *b/(2a))< (l/2) ( n 2 irb/a) 2 < (ir 2 /2)ng <5(2M') 2 < M, Also |sin(n p TTb/a)| 4 tt | r\ | 4 2ttM . Hence, continuing, we have f£(cosUb/a)) = (l + n 3 ) cosUb/a) + r^» where | n | 4 M, and |nj 4 2ttM + |© | 4 8M. Hence f£(2cosUb/a)) = 2 {cos ( Trb/a) (l + nJ + n-> (l + e, ) = 2 (1 + n 5 ) cos UVa) + Hg > where | n | 4 2M* and |iv| < 16m' . We next consider the exponential factor. We have f£(exp(-27Tk)) = exp(f£(-2TTk))(l + 0J + 0^ = (1 + ) exp {-2rrf£(k) (l + n )} + 0^ = (1 + ) exp {-2irk (1 + n 3 ))+ Q k = (1 + ) exp (-2TTk) exp (-2*^) + 0^, I n I 4 2M' and |n Q | < 5M' . Now k = /d/(2a) 4 (l/2) ^800,000 500, and hence |exp(-2Trkrig) - l|. |2uk n 8 l = 1 " |2^k nJ where, 2tt x 300 x 5M 1 - 2tt x 500 x 5M' 11 < m^- < " 9 oo m. That is, exp (-2irkng) = 1 + % 1 where |E 1 | < 15900 M. Therefore f£(exp(-2TTk)) = (1 + E 1 )(l + ) exp (-2irk) + 0^ = (1 + n 9 ) exp (-2Trk) + 0^, where \r\ \ < 15901 m'. We now put n 2 cos (Trb/a) exp (-27Tk) p = Then, using different e., we have Now Hence (l + e.)(l + e )f£(2 cos(TTb/a)) ffc(exp (-2irk)). f£(3) = ■ (1 + ) /fil(k) /fZ(k) = /(l + n )k = (1 + E ) v^~ , where |E | < 2M*. (1 + e 1 )(l + e 2 ){(l + n 5 )2cos(^b/a)+n 6 }{(l+n 9 )exp(-27Tk) + 0^} n(s) (1 + Q^Jd + E 2 ) /k (1 + n c )(l + n n )(l + e n )(l + e ) (l + n, rt )B + n nn , where 1 + n and 10 U 10 (1 + 9 5 )(1 + E 2 ) (1 + e 1 )(l + e 2 ){2(l + n s )0i | cos(^b/a)+ng(l + n g )exp(-27rk)+n 6 1| } ni1 (1 + e 5 )(i + e 2 )v^T Hence |n lQ | 4 16000M, and, since /k > (3/U) > 0.9, we have 2M' + i6m' (0.005) + i6mm' 11 = (1 - M)(l - 2M') (.85) <2^1 < 3 M. 12 We next calculate H (a,b,c) =H=$+O.OU |@| using new e. . fJl(H) = f£(3 + .OU|g|) = f£(6)(l+ ei ) + .0U|fA(B)|(l+e 3 )(l+e 2 ) = B(l+n 10 )(l+ e;L ) + n^il+cj) + .0U|e(i+n 10 ) + n n | (l+e 3 )(l+e 2 ). Therefore, f£(H) - H = 6(n 10 + e 1 (i + n 10 )) + n 11 d + e 1 ) + .0H|3|[(i+n 10 )(i + £ 3 )(i + e 2 ) " 1] ± .OU|n 11 |(l+e 3 )(l+e 2 ). Hence |f£(H) - H| < |s|-|n 12 | + |n 13 | + 0-5 x 16002M' |@| + .05 n^ , where |n 12 | = |n 10 + ^(l+n^) I < |n 10 | + 2| £l | < 16002M', and|n 13 | < 3M» . —1/2 We require abound on |g|. Now |a|= 1 2k exp(-27rk) cos( 2h Tr /a) | , and since k > /3/2, we have | g | < |2( /3/2)" 1/2 exp(-2 7T ( /3/2) ) | < 10~ 2 . Therefore \fl(E) -H| ^ 10"2 x 16002M' + 3M' + .05 x 10 -2 x i6002M' + .05 (3M* ) < 172M' , We now estimate the error for T = {C - H - (|-) log d + log a}//a~ . The analysis is similar to that done previously except that H replaces .005. Making the necessary changes, but using the same numbering on the n's, we have fA(T) = (l + 6n u )c - (l + 6n 5 )f*(H) - (l + 6n 6 )(|0 log d +.(l.+ k^) log a /I where |n . | < M . Since |f£(H) - H| 4172m', and |f£(H)| 4 |H| + |ffc(H) - H| ^ 1 . Oh j 3 1 + 172M* —ft 4 .010U + 1.8 x 10 < .0105, we have |f£(T) - T| = \6r\ k C - (f£(H) - H} - 6r) 5 f&(H) - 3n 6 log d + Uru log a | 13 6 ? 4 (18 + 172 + 6(0.0105) + 3 log 10 + k log 10 J )M* < 26om' . In this particular program (Program Three), a count of the number of terms was recorded and found to "be not more than 600. This is much less than the upper bound U000 that was used on the previous estimate. Let t. (l <^ i < N) be an enumeration of the terms of the sum L{a(a,d) - H o (a,b,c)}//a, N < 600. The error analysis for the summation is similar to that for Programs One and Two. n(f. =1 t.) =mf 1=1 (t. + u.)}= f. =1 (t. + u.)(l + y.), where |u. | < 260m' and |y.| ^N6(l - N6) _1 . Therefore N N l™( I t ) - I t I i=l 1=1 x N = I T (u. + t . y. + u. y. ) I 1=1 4 260NM' + 5N 2 6(1 - W6)" 1 + 260 M* N 2 6(l - N6)" 1 = 260NM' + (5 + 260M*) N 2 6 (l - N6)" 1 4 260 x 600 x 1.1 m + 5.1 x 3.6 x 10 5 6 If we take M = 10 and 6 = 10~ , the error is less than 2 x 10 . Thus the absolute error in calculating J (a(a,d) - H (a,b,c)}//a was less than 2 x 10 . When Program Three was run, the lowest value for the sum printed out was . 7 x 10 . Hence condition (ii) of Low's Hypothesis holds for the exceptional discriminants that failed Programs One and Two; hence (ii) holds for all d up to 800,000. Ik Now Program Three also verified (i) by calculating the sum Y a(a,d)//a. Q The error in this calculation is clearly less than the upper bound for the error in the calculation of L{a(a,d) - H (a,b,c)}//a and therefore less than H O 2 x 10" . Of the values for the sum [ a( a, d)/^a printed out by the computer, three were negative, (d = 115147, 357819, 63618U; and a careful calculation showed them to be indeed negative), but none of the others were less than 1.9 x 10" . Hence condition (i) holds for all d less than 800,000, except for the three exceptions, where it definitely does not hold. §k. Verification of (h) A second series of programs, analogous to Programs One and Two, combined, were run to verify (k) . The first program tested whether L (8/3 + a (a,d)/6}//a ^ 1/10 (we use the fact that b > 8/3), and ^1 3 the second program added on as many terms as necessary to get L*{8/3 +a 3 (a,d)/6}//a > 1/10 for some Q*, 0^ CQ*GQ. There were no failures to this program, and it was more economical of computer time than ones in the first series , because relatively few terms were needed beyond those from Q.. . We shall suppress the error analysis here which is very similar to the one for Programs One and Two of the first series . § 5 » Summary Summarizing our results we have the following table. If T means true, F means false, and U means unknown, we have the following tabulation for fundamental discriminants -d: 15 d 1 < d < 1151U7 d = 1151^7 1151U7 < d < 357819 d = 357819 357819 < d < 63618U d = 63618U 636184 < d ^ 800,000 800,000< d < 1,200,000 As in [l], we can use our results to estimate L , (— ) and L . (l) at -a d -a i) (ii) (iii) T T T F T T T T T F T T T T T F T T T T T U U T the exceptional values. We get ^iisiUt'I' * -°°°°t°65 h(-115lli7) = 32 L -115H.T (1) 4 °- 29626 ^eW^ i °- 0119k0 h(-636l81() = 22k L -636181, (1) * °" 882280 L -357819 ( 2' * °- 0015555 h(-3578l9) = 112 L -357819 (1) i0 - 588215. The two which Low had agree fairly well. He had L -115lU T ( i' ) = °- 000 °T0675 and L -636Wl> = °- 00179 ^ This leads me to think that possibly Low's omission of 357819 was a bookkeeping error. 16 REFERENCES [l] M. E. Low, Real zeros of the Dedekind zeta function of an imaginary quadratic field , Acta Arithmetica, XIV(l969), 117-1 UO. [2] J. H. Wilkinson, Rounding Errors in Algebraic Processes , Prentice Hall, 1963. [3] P. T. Bateman and E. Grosswald, On Epstein's zeta function . Acta Arithmetica, IX (196U), 365-373. [k] K. E. Hillstrom, Performance Statistics for the Fortran IV (H) and PL/I (Version 5) Libraries in IBM OS/360 Release 18 , Argonne National Laboratory Report ANL-7666, August 1970. [5] Handbook of Mathematical Functions vith Formulas, Graphs, and Mathematical Tables , Edited by Milton Abramowitz and Irene A. Stegun, Dover Publica- tions, Inc., New York. 17 II. THE LATTICE TRIPLE PACKING OF SPHERES IN EUCLIDEAN SPACE §1. Introduction Let A be an n-dimensional lattice in n-dimensional Euclidean space E , such that, if open spheres of radius 1 are centered at the points of A, then no point of space is covered more than k times. That is, for any point X in E there do not exist distinct points L ,L , . . . ,L of A such that |X-L |,...,|X-L | < 1. They ve say that A provides a k-fold packing for spheres of radius 1 . The terms single , double and triple are synonymous with k-fold for k=l,2 and 3. (n) Let d(A) denote the determinant of A, and let A, denote the lower bound of d(A), taken over all lattices A that provide a k-fold packing for spheres of radius 1. (Thus A. is the critical determinant of a sphere of radius 2.) It is well known and easy to see (e.g., divide one generator of the lattice by k) that A^ 4 A^/k. It has been shown by Few [l] that A ? 2 = (l/2)A-, 2 , and Heppes [5] showed (3) that A/ 2 = A: 2 /k if and only if k 4 k. In [k] Few and Kanagasab apathy determined the exact value of A namely 3/3/2 , which is less than A, 3 /2 = 2/2. By constructing particular lattices they also showed that A p < A, /2 for every n ^ 3. Few remarks in [2] that A ? 3 is the only multiple packing constant known exactly in three dimensions or more, and in this note I shall prove that A^ 3 4 8/3c727 < A, 3 /3 = U/2/3 and give evidence suggesting that A^ } = 8/3c727. In fact , I prove : Theorem 1 A certain lattice A of determinant d = 8/3B"/27 provides a o o 18 triple packing for the unit sphere S. Also A has generators P,Q,R with |P| = 2/3. Theorem 2 Any lattice A having generators P',Q',R' with |P'| < 0.95 provid- ing a triple packing for S must have determinant d(A) ^ d with equality- only when A = A . Hence A gives a local minimum of d(A) for triple packing of unit spheres . Remark There is extensive numerical evidence that d(A) does not fall below d for any triple packing with S. 62. An Economical Lattice Ap Theorem 1 The best lattice triple packing for spheres in E 3 has determinant d(A) 4 8/3F/27 = 1^32/729 = 1.82649..., since indeed the lattice A gene- rated by P, Q and R where P = (a, 0,0) =(2/3,0,0), Q = (h,b,0) = (l/3,/3,0) and R = (g,f ,c) , where g = 1/3, f = (llv / 3)/27 and c 2 = 3-f 2 , provides a triple packing for the unit sphere S. Proof Convention: The letters X, y and v will denote integers. S(A,r) will be the open sphere of radius r centered at A; S(A) will denote S(A,l) ; thus S = S(origin). Suppose that the point X = (x,y,z) is covered four times Translating X by a lattice point, we may suppose that X e S, and replacing X by -X if necessary we may suppose z ^ 0. The three other spheres covering X can be written S(XP+yQ+vR) = S+XP+yQ+vR where (X,y,v) / (0,0,0). We must have |XP+yQ+vR| < 2 since they must intersect S. Therefore (*) (Xa+yh+vg) + (yb+vf) 2 + v 2 c 2 < k and |v| < 2/c. Since c > 1, we have v e{ -1,0,1}. Now v cannot be -1, since otherwise |X-(XP+yQ-R) | >, |c+z| > 1. Hence v e{0,l}. From (*) we also get I yb+vf I < 2. Since 4 v 4 1 and U/3 this gives -2 < y < 2, y e {-1,0,1}. We divide the proof into two parts. 19 Part 1 y ^ . Then y e{0,l}; in fact if y = -1, then for X e S ( AP+yQ+vR ) we would have |x-( AP-Q+vR) | 2 > (b+y-vf) 2 > (|| b ) 2 = |||, since v e{0,l}. Also (y,v) / (1,1), since | AP+Q+R | 2 ^ b 2 + c 2 > k. Hence (y,v) e {(0,0),(l,0),(0,l)}. Type 1 Spheres Suppose (y,v) = (0,0). Then AP+yQ+vR = AP, and S(AP) n S = if |a| > 2, since | 3P | = 3a = 2. The S(AP) such that < |a| ^ 2 are called type 1 spheres. Type 2 Spheres Suppose (y,v) = (l,0). Then AP+yQ+vR = AP+Q, and S(AP+Q) n S = if A *H0,-l}, since then |AP+Q| 2 = b 2 + (Aa+h) 2 = 3+|2A/3+l/3| 2 >, k. The S(Q-P) and S(Q) are called type 2 spheres. Type 3 Spheres Suppose (y,v) = (0,l). Then AP+yQ+vR = R+AP, S(R+AP) n S = if A ^{0,-1}. To see this, observe that if S(R+AP) n S 4 then (Aa+g) 2 + f 2 + c 2 < U; since f 2 + c 2 = 3, |2A/3+l/3| < 1 and A e{0,-l}. We call S(R) and S(R-P) type 3 spheres. It follows from the discussion of type 2 and type 3 spheres that S n S(AP+E) = if A ?f{-l,0} where E i {R,Q}. In particular, = S ° S(E+P) = S(-P) ° S(E) = S(-2P) n S(E-P), = S ° S(E+2P) = S(-2P) n S(E), = S ° S(E-2P) = S(P) n S(E-P) = S(2P) n S(E) , = S ° S(E-3P) = S(2P) n S(E-P). From the discussion of type 1 spheres S n S(AP) = for |a| > 2 so that = S n S(3P) = S(-P) n S(2P) = S(-2P) n S(P), = S n S(1*P) = S(-2P) n S(2P). We now draw a graph, G, where edges A and B are joined only if we know that S(A) n S(B) = 0. 20 G: We next observe that = S(Q+XP) n S n S(R+X"P+vQ). For |Q+XP| > |Q| = A) z + h z > /3. Therefore the height (maximal value of the z coordinate of the closure) of S(Q+XP) n S is less than /1-3A = 1/2 < c-1, since c = 1.5... > 3/2. Since R+X'P has z component c, the above intersection is void. Hence we cannot have X simultaneously inside a sphere of type 2 and a sphere of type 3, so X e sU.jP) n SU 2 P) n S(E+X 3 P) with < | \ ± |" |Xg| < 2,-1 < X 3 < or X e S(X P) n S(E+X 2 P) n S(E+X P) with 0< | X ± | 4 2,-1 4 X 2 , X 3 4 0, where E e {R,Q}. Both of these contradict the graph G, and part 1 follows. Part 2 We now suppose that y < 0. Recall that if S ^ S ( XP+yQ+vR ) i 0, then -1 < p < 1 and 4 v 4 1. For those S( XP+yQ+vR) containing X we must have -1 < p 4 0. For suppose that y = 1; since X = (x,y,z), y < 0, we 21 would have | XP+ Q+vR-X | > |b+vf-y| > b > 1. The spheres S(xP+yQ+vR) contain- ing X other than S may therefore be divided into four types, as follows: Type 1 spheres , when (y,v) = (0,0). As before the only spheres S (xP) intersecting S satisfy < | A | 4 2, i.e., the type 1 spheres are S(2P), S(P), S(-P) and S(-2P). Type 2 spheres , when (y,v) = (-1,0). If S(xP+yQ+vR) is to intersect S we must have k > |xP+yQ+vR| 2 = (xa-h) 2 +b 2 = (2X/3-1/3) 2 + 3. Hence 0< \ |q|,so that the height of S(-Q+ P) n S is not greater than the height of S n S(Q), which is less than 1/2 < c-1, and c is the height of R+X'P. Also, for any X, = S A S(Q+XP) n S(r) = S n S(-Q-XP) n S(-R) = S(R) n S(R-Q-XP) n S. In particular, (1) S(R) n S(R-Q) n S = 0. The following enumeration of possibilites shows that X cannot be contained in the necessary spheres, and theorem 1 follows: Clearly types 2 and 3 cannot both occur by the fourth paragraph above , and two spheres of type 1 cannot occur with anything else by the graph G 1 . Again by the graph G' , if two spheres of type 2 (or two spheres of type 3) occur, the remaining sphere cannot have type 1. Hence the only remaining possibilites are (2) X e S^P) n S(X 2 P-Q) n S(R-Q), < | X ± | 4 2, 4 X g < 1 (3) X e S(X 1 P) n S(X 2 P+R) n S(R-Q) , < JX.J < 2, -1 < \r> < 23 (U) X e S(P-Q) n S(-Q) n S(R-Q) or (5) X e S(R) n S(-P+R) n S(R-Q) Now (2) and (3) contradict G' , and (l) excludes (5). To eliminate (k) we observe that, since S(P) ° S n S(R) n S(Q) = , we must have S(P-Q) n S(-Q) n S(R-Q) H S = . §3. The Lattice A Q is Locally Optimal Remark 1 An arbitraty lattice A in E_ has a basis P,Q,R where |P| <. |q| 4 |r| are the successive minima of the unit sphere, P = (a, 0,0), Q = (h,b,0) , R = (g,f ,c) , a,b,c > 0, 4 h 4 a/2, , |q| 2 = b 2 +h 2 , and the other inequalities of reduction, we have d 2 ^ d 2 (A) = a 2 b 2 c 2 ^ a 2 b 2 (b 2 +h 2 -g 2 -f 2 ) ^ a 2 b 2 ( 3b 2 A-a 2 A) Putting t = b 2 , we get 3a 2 t 2 -a 4 t-4d 2 4 0. Hence b 2 must lie between the roots a 2 /6-2/3/aVl6+3d z /a z and a 2 /6+(2/3) ZaVl6+3d z /a z of the quadratic, o o Let p + denote max {0,p}. Theorem 2 If (A,S) is a triple packing and P = (a, 0,0), Q = (h,b,0) and R = (g,f,c) gives a basis for A reduced in the sense of Gauss, and a < 1, then (6) d(A) = abc > ab/{4-(a+h) 2 -((b 2 -2ah-h 2 )/(2b)) 2 } + = f^a.b.h) when <^ g <. h, and (7) d(A) = abc > ab/{U-9a 2 /U-(b 2 +h 2 -3ah) 2 /(2b) 2 } + = f 2 (a,b,h) 2k when -a/2 4 g < and when h 4 g 4 a/2. Furthermore f^a^h) > f 2 (a,b,h) , so that in fact (8) d(A) > f 2 (a,lD,h) in all cases. Also, if d(A) < d Q and 2/3 < a 4 0.9508, we have (9) f 2 (a,b,h) ^min{p(a), d 2 + 1/100}, where p(a) = 2a 6 -lla 1+ +12a 2 , p(2/3) = d 2 , and (10) p(a) > d 2 for 2/3 < a 4 0.9508. Hence d(A) > d for 2/3, | Q-P | by (ll), it follows that k 4 |Q+P| 2 = b 2 +(a+h) 2 ; hence (12) b 2 ^-(a+h) 2 , and a i 2 ' /3 - Case 1 In this case we assume (13) 4 g 2. Another consequence of (13) is that |R+P | is not less than |R-P | . Consider- ing the parallelogram with vertices P,R,R+P and the origin shows that (15) |r+p| > 2. With a view to proving (6) we imagine a,h,b to be fixed and find the point R = (g,f,c) having least non-negative c such that (13), (1*0 and (15) hold and also (16) < f 2. The problem is somewhat simplified by the fact that the centers P+Q, -P of the spheres lies on the plane z = 0, outside the prism. Case 1. E Q/2 \ P+Q A* 26 If the right hand side of (6) is zero, there is nothing to prove. Let vis suppose, therefore, that it is positive. We shall show that the point X* that lies on the intersection of the "boundary of S(-P,2) and S(P+Q,2) and the plane x = h is the lowest point satisfying (IT) and (l8). We start by finding X*. Let X* = (x*,y*,z*). Let tt be the radical plane of S(-P,2) and S(P+Q,2) (the plane obtained by subtracting the equations of the two spheres) Then it passes through (l/2)Q = (h/2,h/2,0), which is halfway between the cen- ter of the two spheres, and has the equation y-b/2 = (-(2a+h)/b ) (x-h/2) . Putting x* = h, we obtain y* = b/2-(2ah+h 2 )/(2b) . We must show that £_ y* so that (IT) is satisfied. By (12) we have h 2 > Ma+h) 2 ,> 2ah+h 2 + 1/2, since h 4 a/2; hence 2by* = b 2 -(2ah+h 2 ) > and (17) follows. We see that z* = A-(a+h) z -( (b z -2ah-h z )/(2b)) z > by a previous assumption. The first step in showing that X* is optimal is to show that the bottom of the prism is covered by S(-P,2) and S(P+Q,2). This means that there is no X satisfying (17) and (l8) with z = 0. Let A* = (x*,y*,0), E = (0,b/2,0), F = (h,b/2,0) and G = (h,0,0). Then 2 = |P+Q-X*| > |P+Q-A*| > | P+Q-F | , and also |P+Q-(l/2)Q| < 2, since the spheres S(-P,2) and S(P+Q,2) intersect and (l/2)Q is halfway between their centers. Hence the triangle with vertices (l/2)Q, A*,F lies in the interior of S(P+Q,2) . Similarly, 2 = |-P-X*| > |-P-A*| > j -P-G | > |-P| and 2 > |-P-(l/2)Q| > | — P— E | so that the convex pentagon with vertices G, A*,(— )Q, E and the origin lies in the interior of S(-P,2). Hence the bottom of the prism is covered. We now let X = (x ,y ,z ) be a lowest point satisfying (17) and (l8) . We know that X exists, because the set of solutions is non-empty and closed. 21 The point X must be on the boundary of S(-P,2) or S(P+Q,2) since otherwise it could be lowered and still satisfy (l8) . Let us suppose first that X is on the boundary of S(-P,2). We shall deduce that X is on the boundary of S(P+Q,2). Suppose not. Then (x ,y ) must be the point satisfying (IT) that is farthest from -P, namely (h,b/2). But then the point X = (h,b/2, t4-(h+a) z -b z /4) is easily seen to be inside S(P+Q,2), contrary to (l8) . Suppose that X is not on the boundary of S(-P,2). Then |X -P-Q| = 2, and X = (0,0,A-(a+hj z -b z ) lies inside S(-P,2) contrary to (l8). Hence X lies on the arc of the intersection of S(-P,2) and S(P+Q,2) with z ^ 0. The highest point of the arc is the point directly above (l/2)Q. which is on the boundary of the prism, and the lowest point in the prism is X*, where the arc cuts the x = h plane. Hence X = X* and (6) follows. Case 2 Assume (19) -a/2 £g<0. Since the center of the parallelogram with vertices R,Q-P,R+P,Q must not be covered four times, we know that one of its two diagonals |R-Q| , |R-Q+2P| must be at least 2. Assuming Gauss reduction, we always have | R-Q+2P | >, | R— Q | , since the vectors R-Q+2P and R-Q differ only in the first component, and |g-h+2a| i2a-|g|-h^a^ | g-h | . Hence | R-Q+2P | ^2. As in case 1, one of |r+p| , | R— P | must be at least 2, and from (19) we know that |r~p| >, |r+p| so that |R-P| >, 2. In a manner similar to case 1, we are looking for the lowest point X = (x,y,z) inside the rectangular prism given by (20) -a/2 4x40 4 y 4 b/2, z >_ 28 such that (21) |X-P| > 2, and | X-Q+2P | > 2, ■2P Case 2. A* (Q-P)/2 The procedure is the same as in case 1. We may suppose that the right hand side of (7) is positive. Let X* = (x*,y*,z*) be the point on the two spheres and the plane x = ~a/2, with z* ^ . The radical plane tt of the two spheres passes through (l/2)(Q-P). The equation of tt is 2(h-3a)x+2by = b 2 +(h-3a) (h-a) . Putting x = x* = -a/2 yields y* = (b 2 +h 2 -3ah)/(2b). We must show that 4 y _< b/2. Now b 2 ^ U-(a+h) 2 for a triple packing; hence 2by* = b 2 +h 2 -3ah > ^-a 2 -5ah .> 1/2. On the other hand h 2 _<_ ah/2 <. 3ah, 2by* = b 2 +h 2 -3ah _< b 2 , y* < b/2. We see that z* = / U-(3a/2) z -((b z +h z -3ah)/(2b)) 2 > by assumption. We now show that the bottom of the prism is covered by the two spheres S(P,2) and S(Q-2P,2). Let A* = (x*,y*,0), E = (-a/2 ,b/2,0) , F = (0,b/2,0) and G = (-a/2,0,0). Then 2 = | Q-2P-A | > | Q-2P-A* | > | Q-2P-E | , and also |(Q-2P)-(l/2)(Q-P) I < 2 since the spheres S(P,2) and S(Q-P,2) intersect and (l/2) (Q-P) is halfway between their centers. Hence the triangle with vertices 29 (l/2)(Q-P), E, A* lies in the interior of S(Q-2P,2). Similarly 2 > |P-(1/2)(Q-P) | > |P-F| ^ |P| and 2 > |P-A* | > |P-G | , so that the convex pen- tagon with vertices G, A*, (l/2)(Q-P), F, and the origin lies in the inter- ior of S(P,2). Hence the bottom of the prism is covered. We now let X - (x ,y ,z ) be a lowest point satisfying (20) and (21); X exists because the set of points satisfying (20) and (2l) is closed and non- empty. Since z > 0, X must be on the boundary of one of the spheres. We suppose first that |X -P| = 2, and |X 1 ~Q+2P| > 2 . Then (x ,y ) must be as far from (a,0) as possible still satisfying (20). Hence X = (-a/2,b/2, A-b z /4-(3a/2) z ), and calculation shows that |X -Q+2P| < 2, which is a contradiction. Suppose now that |X -P| >2, so that |X -Q+2P | =2. Then (x,y ) must be as far as possible from (h-2a,b) and still satisfy (20). Now -2a <_ h-2a <_ -(3/2)a<-a/2. Hence X = (0 ,0 , /4-(h-2aJ *-b*) . Hence l^-P^ = a 2 +4-(h-2a) 2 -b 2 < a 2 +4-9a 2 A < 4, which is a contradiction. Hence X must be on the boundary of both spheres. In a manner similar to case 1, X lies on a circular arc whose highest point (l/2)(Q-P) is on the boundary of the prism and whose lowest point inside the prism is X*. Hence X = X* and (7) is proved for this case. Case 3 Assume (22) h_ 2. We are seeking the lowest point X of the prism 30 (23) h < x < a/2, £y < b/2, such that (24) |X+P| > 2, | X-Q+P | > 2, Case 3. Q^2 u. r (h,b/2) i (a/2,b/2) A* ■P If the right hand side of (7) is zero, there is nothing to prove. Let us suppose that it is positive. Let X be the point on the two spheres and the plane x = a/2. To find X = (x,y,z) we solve 9a 2 /4+y 2 +z 2 = 4 (3a/2-h) 2 +(y-b) 2 +z 2 =4 and obtain x* = a/2, y* = (b 2 +h 2 -3ah) /(2b) , and z* = /4-9a z /4-(y*) z . This is the same y* that appears in case 2, so < y* j< b/2 and X* satisfies (23) and (2k). Now let X be a lowest point satisfying (23) and (24), and we shall show that X = X* . We must show that the bottom of the prism is covered. Since x = a/2 maximizes the horizontal distance from both -P and Q-P, it is sufficient to show that the line segment {(a/2,t,0) : 4 t ^ b/2} is covered. The spheres 31 S(-P,2) and S(Q-P,2) intersect the plane x = a/2 in circles which intersect at (a/2,y*,z*). Since z* > 0, the segment is covered. For a lowest point X we must have x = a/2, Since the spheres inter- sect the x = a/2 plane in circles whose centers have y components and b respectively, it is clear that X = X* and inequality (7) holds. This finishes case 3. Hence d( a) ^minff ,f„}. We observe immediately that f _> f . This would follow if ijj(h) ^0 where ^(h) = 9a 2 A+(b/2-(3a-h)h/(2b) ) Mh+a) 2 - {(b/2-(h/2)(h+2a)/b) } 2 . Now ^(0) = 9a 2 A_a 2 =(5/U) a 2 > and ^(a/2) = 0. Differentiating, we have ^'(h) = -{( 5a)/(2b 2 ) } (b 2 +3h 2 -ah) < since b 2 > a 2 _h2 > (3A)a 2 . We now prove (9) f| ^minfd 2 + 1/100, 2a 6 -lla t+ +12a 2 } under the hypothesis that d( a) 4 d and 2/3 (a,h) = a 2{U-( a +h) 2 }{U-9a 2 /U-(lA)(U-(a+h) 2 )-(l/2)h(h-3a)}-(a 2 /ii)h2(h-3a)2 and calculation shows that 9 2 4>/3h 2 = -8a 2 -8a lf < 0. Since cj>(a,0) = (a,a/2) = p(a) where p(x) = 2x 6 -llx l+ +12x 2 , we have > = p(a). To complete the proof of (9), we will show that iKa,h) > d + 1/100. Now h (a), where h = A-b z — m mm 32 To see this, recall that for triple packing we must have b 2 ^^-(a+h) 2 , whereas t> 2 < b 2 , since d(A) < d . The juxtaposition b 2 > U-(a+h) 2 yields — m — o m = h > h . Putting h = a/2 we see that h < a/2. It is conceivable that h — m m = m is negative, even though h never is. For what follows, it is useful to know that h > -a/2. To see this, note that m — b 2 = a 2 /6 + (2/3) ZaVl6 + 3d*/a* m o < l/6+(2/3)/l/l6 + 27 < 3. 6U, since 2/3 4 a < 1 and 3 < d 2 < U, so that Wb 2 > 1/k > a 2 A, h + a = A-b z > a/2, and h > -a/2. We now return to showing that ip(a,h) is a concave function of h for h m 4 h 4 a/2. Since ~^- = 6a 2 h-9a 3 < = 3a 3 -9a 3 = -6a 3 < 0, it is enough to show 9h m Since f(a) is a rather complicated function of the single variable a, we shall simply find an upper bound for its derivative as a function of a and use a computer to evaluate it on a fine grid. Let F (x,y,z) = -z+3y 2 -9xy+9x 2 /*+ so that F(a,h m ,b 2 ) = f(a). Then o J o o where the subscript o indicates that the partial derivatives are evaluated at (x,y,z) = (a,h m ,b 2 ) . Then |f|| = |-9h m+ 9a/2| <9a<9, m < |6hJ + 9a «18. and |f | q = 1. o Let u = a 2 , g(u) = b 2 = £ +(2/3)/u z /l6 + 3d^/u. Then mo o g'(u) = 1/6 +(l/3)(u/8-3d 2 /u 2 )//u z /l6+3d£/u, and 33 |g'(u)| < 1/6+ (1/3X1/8+3 x U/(V9) 2 )MV9)yi6+9 = 1/6 (l/3)(l/8+2U3A)//9" < 7. Therefore 2 |^| g(a 2 )| = |2ag'(a 2 )|< iU. That is, |f^| < lk . Finally, h = A-b^-a, —£ = -l-(l/2)^i/ A-b*", m m da da m and dh -1/2 -1/2 I -—SI < i + 7{li-b 2 } < 1 + 7(0.36} < 13, since b 2 1 da ' = m m |f » (a) | < 9 + 12 x 13 + 1 x lk = 179- Using a computer we verified that (allowing for roundoff error) f(a. )<~0.2 at the points 2/3 = a < a <...min{^(a,h m ), ^(a,a/2)}. The functions ijj(a,h ) and i|i(a,a/2) are also rather complicated func- tions of the single variable a, and they are both above d 2 + 1/100. We shall simply find an upper bound for their derivatives as functions of a 2 or a and use a computer to evaluate them on a fine mesh. 3U Let u = a 2 and f(u) = i|>(a,a/2). Then f(u) = ug(u)fi(u,g(u) )-25u 3 /6U where g(u) = b 2 = u/6+( 2/3 ) /u z /l6+3dg/u , and fl(u,v) = U-(9A)u-v/U+( 5/8)u. Then f'(u) = g(u)fi(u,g(u) )+ug' (u) n(u,g(u) )+ug(u) {3 Q /3u+g' (u) 3ft/3v}-75u 2 /61*. To estimate |f'(u)|, we must estimate g,ft, and their derivatives. We have |g(u) | _< 3.6k < h from before and |g(u) | ^ u/3, trivially. Hence |«(u,g(u)| 4 U+13u/8+|g(u)A| 4^+13/8+147. We also have |g'(u)| 4 7 from before. On the other hand |3fi/3u| = |l3/8| < 2 and|3fi/3v| = 1/U. Putting the estimates together, |f'(u)| d 2 + 1/100 for k/9 4 u 4 1. Let us suppose that a computing machine has verified that f(u, ) >. d 2 + 1/50 + e for k/9 = u < u < ••• < u = 1, where |u. - u. | < ( 50 x 92) _1 . It then follows that f(u) > = d 2 + e for k/9 4U < 1. For ue[u ,u. J for some i, and |f(u)| > |f(u._ 1 )| - |/ U f'(t)dt| i d 2 + 1/50 + £ -92|u. 1 - u.| ^ d Q + e . U i-1 A computing machine was programmed to find the minimum value of f(u. ) for 1 _< i 4 8251, where u. = h/9 + (i-l)/lU850, and the answer was 3.51822. Had there been no round-off error, we could say that f(u. ) 4 d 2 + 1/6. It is certainly safe to say that f(u) 4 d 2 + 1/50 + 1/100. Hence ^a, a/2) = f(a 2 ) > d 2 + 1/100 for 2/3 4 a 4 1. We shall use the same method for \b(a.,h ). Let us rename f(a) = ij>(a,h ) m m = a 2 b 2 {M9A)a 2 - b 2 /U - h (h -3a)/2} - a 2 h 2 (h --3a) 2 A. It is unfortunate m m m m mm that ip(a,h ) cannot be written simply as a function of a 2 ; all our functions are now functions of a. Let g n (a) = b 2 = g(a 2 ), and let h(a) = h . Put 1 m m $(x,y,z) = k - (9/k)x 2 -y/k - z(z-3x)/2 and 0(x,z) = -x 2 z 2 ( z-3x) 2 A. Then f(a) = a 2 g 1 (a)$(a,g (a) ,h(a)) _ e(a,h(a)), and f'(a) = 35 2ag 1 (a)$(a,g 1 (a),h(a)) + a 2 gj(a)<&(a,g 1 (a),h(a)) + a 2 ( a ) { £* + (3$/3y)g-[(a) + (3$/3z)h'(a)} - 30/3x - ( 30/3z)h* ( a) . Using some of the esti- mates from "before and making some new ones, we see that |g (a)| = |g(a 2 )| < k, |h(a)| = |A-b^- a| < I", |*|+ x 8 + 1^x13+ U(6+ll+A+2*L3) + 3 +3 x 13 = ^30. To show that f(a) >_ d 2 + e , therefore, it is enough to show that f(a. ) >^ d 2 + 1/50 + e for 2/3 = a <•••< a =1 where la. _ - a. I < 1/25,000. o o n ' l-l 1 ' A computing machine was programmed to find the minimum value of f(a. ) for 1 < i < 100,001 where a. = 2/3 + i/300,000, and the answer was 3.U03 1 * 1 * . • • Had there been no round-off error we could say that f(a. ) ^ d 2 + 1/15. It is certainly safe to say that f(a.) ^ d + 1/50 + 1/100 so that ^(a,h ) = f(') >, d 2 + 1/100 for 2/3 4 a 4 1. Therefore i|i(a 9 h) ^ min{tj;(a,a/2) , t|j(a,h ) } ^ d 2 + 1/100 and f 2 ^ min{i|j(a,h) ,<{,( a,h) } >_ min{d 2 + 1/100, p(a) } as claimed. Thus (9) is proved. ¥e now prove (10). Let f(t) = 2t 3 - lit 2 + 12t - d 2 . Then f'(t) = 6t 2 - 22t + 12 and f"(t) -12t - 22-< for < t < 1. Hence f(t) is a concave function and has at most two zeroes in the range [0,l]. In fact, f(^/9) = 0, and f(a 2 ) = where a 2 = 0. 901+02... and f(t) > for k/9 < t < a 2 Since f(a 2 ) = p(a) - d 2 , we conclude that p(a) > d 2 for 2/3 < a < a = o o 36 0.950802... and (10) is proved. It now follows from (8), (9) and (10) that d(.\) £ d for 2/3 4 a 4 0.9508 with equality only when a = 2/3. 37 REFERENCES [l] L. Few, Ph.D. Thesis , London, (1953). [2] L. Few, Multiple packing of spheres: a survey , Proc. Coll. Convexity, Copenhagen, 196 5 (1967), 88-93. [3] L. Few, The double packing of spheres , Journal London Math. Society, 28 (1953), 297- 30U. [k] L. Few and P. Kanagasab apathy , The double packing of spheres , Journal London Math. Society, (series l), kh (1969), lUl-lU6. [5] A. Heppes, Mehrfache gitterformige kreislagerungen in der ebene ,Acta Math. Acad. Sci. Hung., 10/1-2 (1959), lUl-lU8. [6] L. E. Dickson, Studies in the Theory of Numbers , University of Chicago Press. 38 III. SOME EXTREMAL PROBLEMS IN GEOMETRY §1. Introduction Let there be given n points X , . . . ,X in k-dimensional Euclidean space E. . Denote by d(X.,X.) the distance between X. and X.. Let A(X, , . . . ,X ) k i j i j 1 n be the number of distinct values of d(X. ,X.) , 1 < i < j < n. Put f (n) = i j k min A(X ,...,X ), where the minimum is taken over all possible choices of distinct X ,...,X . Denote by g,(n) the maximum number of s J. n k olutions of d(X. ,X.) =a, 1 < i < j < n, where the maximum is to be taken over all possi- ble choices of a and n distinct points X ,...,X . The estimation of f (n) and g v (n) are difficult problems even for k=2. It is known that [l,2] (1) en 2 ' 3 < f 2 (n) < Cn//log n and (2) n**(l+c/loglog n) < g (n) < Cn 3 ' 2 , where c and C are positive absolute constants and a**b denotes a . If k>J4 the study of g (n) becomes somewhat simpler [3]. A. Oppenheim posed the problem of investigating the number of triangles chosen from n points in the plane which have the same non-zero area. This question and its generalization were the first investigated in [6]. In this note I support some claims made in [6]. §2. Notations Let n>3s X , . . . ,X be n points in k-dimensional space E , and let A>0. We define g 2 (n; X. X ; A) to be the number of triangles of the form X.X.X, having area A. We let l j k g[ 2) (n; X r ...,X n ) = Max g£ z) (n; X^...^; A) A and 4 2)(n) = MaX 4 2)(n; V-'-'V' x l5 ...,x n 39 Let P be a fixed point and define G 2 (n; X , . . . ,X ; A) to be the num- ber of triangles of the form PX.X. having area A. We let G^(n) = Max G^ 2 '(n; X ,...X ; A). X l5 ...X n> A>0 Clearly g£*[(n) < g£ 2 (n) < nG^(n-l)< nG^in). We see that g^ (n) is analogous to g (n). §3. The Article of Erd8s and Purdy It was shown [6] that (3) cn 2 log log n < g^ } (n) < nG^ 2) (n) < h^ 2 , where c is a positive absolute constant, and (h) g^ 2) (n) < g^n) en 2 and g^ (n) >. en . It is therefore worth asking whether g, 2 (n) and g 2 (n) are o(n 3 ). The object of this note is to support the claim made in [6] that in fact gr 2 (n) 4 g 2 (n) 4 en 3 for some e > 0. ;^ 2) (n) . en 2 . Let n >. 2 be given. Let n = 2m+r where 4 r < 2. Choose a co-ordinate system in E, and put X. = (a. ,b . ,0,0), for 1 4 i 4 m and Y. = (0,0, a. ,b . ) , for 1 4 i 4 m+r, where (a.,b.) are m+r distinct real solutions of a 2 + b 2 = 1. Then the m(m+r) triangles 0X.Y. are all congruent to the triangles with sides 1, 1, /2 and i J therefore have the same (positive) area. Hence G, 2 (n) >, m(m+r) >_ n 2 A-l/U >: en 2 . By choosing the a. ,b . so that some of the triangles 0Y.Y. and 0X.X. ko are congruent to the OX.Y. we may improve this to n 2 /U+cn, but no i J further. We now show that g^ 2 (n) >_ c n 3 . Let n > 3 be given. Let n = 3m+r, where <^ r < 3. Choose a co-ordinate system in E/-, put X. = ( a. ,b . ,0 ,0 ,0,0) for 1 4 i 4 m, put Y. = (0,0, a. ,b . ,0,0) for 1 4 i 4 m, and put Z. = (0,0, 0,0, a. ,b. ) for 1 < i <^ m+r, where (a.,b.) are m+r distinct real solu- tions of a 2 + b 2 = 1. Then the m 2 (m+r) triangles X.Y Z are all equilateral 1 J k triangles of side length /2. Hence g^ 2 (n) >_ m 2 (m+r) > c n 3 . \5. Statement of the Main Theorems Theorem 1 There exist n , e > such that g 2 (n) <_ n 3-e for n _> n . Consequently there exists a positive constant c such that g 2 ( n ) ^ cn 3_e: for all n. Let |s| denote the cardinality of the set S. We shall deduce theorem 1 from the following theorem: Theorem 2 Suppose A,B, and C are finite sets in E , such that |a|>M. p|>J!J, and |c|>Jtf, where M and N are certain absolute constants. Then the triangles XYZ for X in A, Y in B, and Z in C cannot all have the same area. unless that area be zero. §6. Some Graph Theory (r) By an r-graph G we mean an object whose basic components are its elements, called vertices, and certain distinguished r-element sets of these (r) elements, called r-sets. When r = 2, G is an ordinary graph. When we (r) say that G is a G (n; m) , we mean that G is an r-graph having n vertices and m r-sets. If G is a G (n;( J), then G is the unique r-graph which has all possible r-element sets as its r-sets . We call this the complete r-graph on n vertices and denote it by K (n). K (n ,...,n ) will denote 141 the r-graph of n n +«««+n vertices and n • • *n r-sets defined as follows: ° -f 1 r 1 r The vertices are 4 1< J n (r ,£) , then o (*) f(n; K (r) (£,. ..,£)) < n**(r - £**(l-r)). We shall ■use this result with r = 3, and we shall refer to the 3-sets of a 3-graph as triples in what follows . §7- The Relation Between the Main Theorems We now prove that theorem 2 implies theorem 1. Let I be the maximum of M and N of theorem 2, let c=l~ 2 , and let X n ,...,X be distinct points in 1 n ^ E with n > n (r,l), where n (r,£) is the function given in Erdtfs's inequality (*). It is an easy consequence of (*) that theorem 2 implies (5) g^Cn; X^...^) < n 3-e. (*) To see this, let A > 0, and let G 6 denote the 3-graph with n vertices X ,...X , where the triple X.X.X is in G 3 if and only if the triangle X.X.X, has area A. Then theorem 2 implies that G 3 does not contain a i j k: K 3 U,J£,£) subgraph, and (5) then follows from (*). Theorem 1 follows, since A was arbitrary. k2 § 8 . Some Lemmas Before proving theorem 2 we must introduce some definitions and lemmas. We shall use the notation {x} to mean the least integer not less than x. Lemma 1 Let triangles PX.Y., 1 < i < n+1 , 1 < j < H all have the same non-zero area A, where X. , Y. are points in real Euclidean n-dimensional i J space. If the n+1 distances d(P,X.) are all different and non-zero, then there are not more than 2 distinct distances d(P,Y.). Hence at least J {H/2 n_1 } of the Y. are equidistant from P. J Proof Let P be the origin of co-ordinates . Let U. be a unit vector parallel to PX. . The area of a triangle OXY can be written in terms of lengths and the inner product as half the square root of |xp|Y| 2 - (X«Y) 2 . For all i and j we have Ua 2 = Ix.| 2 |Y.| 2 - (X.-Y.) 2 , or Iy.I 2 - (U.-Y.) 2 = r 2 , where r. = 2A/|x.|. Let C. be the set of solutions Y of i i ' l ' l (6) |Y| 2 - (U.-Y) 2 = r 2 . In fact, C. is a cylinder with axis U. and radius r. . Let k be the rank of ' i J l l the set {U ,...,U }. By renaming the U. and choosing a suitable co-ordinate system, we may suppose that U. = ( a , . . . ,a. ) for 1 <_ i <. n+1, a. . t* for 1 .1 i <, k, and a. . = if j > k for all i. Putting r = |y| and Y = (y ,...,y ) in (6) and solving for Y«U. , we obtain (7) j 1 . _ a. .y. = ±/r z - r^ (l < i < k) ^.1=1 l.r.i l = = Ej-i Vi, t ' y - ,r 3 ' k+l We shall show that r 2 is the root of a non-zero polynomial of degree at most 2 , and the lemma will follow. Let the system of equations J ._, a. y = z. (l < i < k) have the solution y. =» £ b. .z. (l < i < k) and suppose 1+3 that z, ., = y . . a, . .v.. Then substituting the expression for the y k+1 L j=l k+l.j'j we get z. . = y. n a, , n .7. n "b . . z . = Y c.z. for some c.. & k+1 ^j=l k+1, j *■ i=l ji i L j=l j j j j=l k+1, ^1=1 ji i ^j=l j j k r~—~ r r~ rk /— — There are 2 functions f.(t) of the form /t-r n z _, - V. n ± c./t-rT l k+1 ^j=l j j corresponding to the choices of sign. Let P(t) = f (t)«««f (t), where m = 2 . If Y is a solution of (7), then clearly P(r 2 ) = 0. It is there- fore sufficient to show that P is a non-zero polynomial of degree at most 2 " . Let W = /t-r z . and let W. = c./t-r z for 1 < i < k. By induction o k+1 ill = = j on k we have n (W ± W ±»»«±W. ) = F. (W 2 , W 2 ,...,W 2 ), where F. is a homo- o k koik k k— 1 k genous polynomial of degree 2 and the product is taken over all 2 possi- ble combinations of signs. Hence P(t) = F (t-r 2 ,C 2 (t-r 2 ) ,. . . ,c 2 (t-r 2 ) ) k-1 is a polynomial in t of degree at most 2 To show that P is not the zero polynomial, we proceed as follows: 'i<*> ■ ^£T" &i* c j^T' 2f i (t) " x// ^^ + ^i ±c j //ez? I- Let c n _ = 1, and let R = r be the maximum r. for which c. ^ 0. Then k+1 p J J f.'(t) is of constant sign for R 2 < t < R 2 + e . , some positive e., since the term c //t-r 2 goes to infinity as t 4- R 2 , and the other terms remain bounded. (if the r. were not distinct, considerable difficulty would arise at this point.) Hence f. has at most one zero in that interval, and P has at most m zeros in the interval R 2 < t < R 2 + min e. . The lemma is proved. Definitions If all the points of a set B are equidistant from a point X, then we say that B is equidistant from X. If B is equidistant from every point X of A, then we say that B is equidistant from A. This relation is clearly not symmetric. If all the points of a set B are different dis- tances from a point X, then we say that B is separated by X. If B is Mi separated by every point X of A, then we say that B is separated by A. This relation is also not symmetric. Lemma 2 Let S and T be arbitrary sets of cardinalities M and N respectively and suppose that the elements of S x T are divided into two classes C and C p . (Suppose that the pairs are colored two colors C and C . ) Then there is a subset T' of T of cardinality {N/(2 )} such that for every X in S, the elements (X,Y) for Y in T' are either all in C or all in C ? . (For every X, the color of the pair (X,Y) for Y in T' depends only on X. ) Proof Use induction on M and the pigeon-hole principle. Lemma 3 Given pairwise disjoint finite subsets A, B, C of E , there are subsets A' of A, B' of B, and C of C such that B' is separated by or equidistant from A' and C is separated by or equidistant from A'. Further if |A| = H, we have |A*| = {H/U}, |b ■ | = |b|**2~ H , and |c'| = |c |**2~ {H/2 } . P roof Let B = B, and i > 1. If the elements of A are X.,, X n ,...,X TJ , O — -L d n we define sets B., , B^,. ..B„ as follows. For each X. we color X. and take a 1' 2' H i i subset of B. as follows. If B. has a subset of {/JB. |} points separated by X. , let B. be this subset, and we color X. red. Otherwise, by the pigeon- hole principle there is a subset B. of B. ., of cardinality {/IB. ~ }, l l-l ' l-l ' equidistant from X. , and we color X. blue. If we do this for 1 4 i < H, we get a subset B of B of cardinality |b|**2 that is separated by all n the red X. and equidistant from all the blue X.. Let B' = Bg, Then there is a subset A* of cardinality {H/2 } of A such that B' is either separated by or equidistant from A*. Similarly there is a subset C of C of cardinality |c|**2 , K ={H/2}, and a subset A' of A* cardinality {E/k } such that C is either separated by or equidistant from A'. The lemma follows. 1*5 Lemma h Let PX.Y. have area A > for 1 < i < 3 and 1 <, j < 3, where X , X , X are distinct and Y , Y , Y are distinct. Then the X. are not collinear. By symmetry, the Y. are not collinear. Proof Suppose the X. are collinear. Let 1 < j 4 3 . The X. lie on the surface of a cylinder with axis PY . . This can happen only if the X. lie on a line parallel to PY . . Consequently, P and the Y. are collinear. J J The distances d(P,Y.) cannot all he equal. Suppose, without loss of J generality, that d(P,Y ) > d(P ,Y ) . Then triangle PX-.Y has a greater area than triangle PX..Y , contrary to the hypothesis. Lemma 5 If in E^ the cylinder C = (X: |x| 2 - (U-X) 2 = c 2 }, where |u| = 1, intersects the hyperplane tt, then there are three possibilities. (i) If OU is parallel to tt , then C intersects it in a cylinder. (ii) If OU is perpendicular to tt , then C intersects it in a sphere. (iii) If neither of the above, then C intersects tt in an ellipsoid of revolution whose axis is the projection of OU onto tt . Proof Choose the origin to be on tt and choose the X, axis normal to tt. Then tt is the set of points (x ,x ,x ,x, ) such that x, =0 . Now choose the X axis lying in tt, in the direction of the projection OU* of OU onto tt . Then U = (a, 0,0, 3) where a 2 + 3 2 = 1. The cylinder C has the equation I x 2 - (ax + gx, ) 2 = c 2 and C intersects tt in a surface with equation 3 2 x 2 + x| + x 2 = c 2 . If 3 = we have (i), if 3 = 1, we have (ii), and if < 3 < 1, then we have (iii). §9. Proof of Theorem 2 Let A,B,C be sets of cardinality M,N,N respectively such that the tri- angles XYZ for X in A, Y in B and Z in C all have a common positive area A. We shall show that this leads to a contradiction if M and N are large 1+6 enough, and the theorem will follow. Let us assume that N >_ M >_ 3; then, by lemma k no three points of A (or B or C) are collinear. By lemma 3 and the symmetry "between B and C, we may suppose without loss of generality that one of the following holds : (i) B is separated by A, but C is equidistant from A. (II) B is separated by A and C is separated by A. (ill) B is equidistant from A, and C is equidistant from A. This application of lemma 3 reduces M and N. From now on, M and N are for the new sets. Let A = {X 1 ,...,X M }, B = {Y 1 ,...,Y N >. First of all, (II ) leads imme- diately to a contradiction. Take one point X of A and six points Y , . . . ,Y^ of B. Then by lemma 1, there are at most l6 points Z ,...,Z >- such that {Z , ...,Z ,-} is separated by X. We get a contradiction if N >_ 17. Secondly, (ill) leads to a contradiction; (ill) implies that the affine hull of A is orthogonal to the affine hull of B and the affine hull of C. We shall find subsets B' of B and C of C whose affine hulls are orthogonal. Let X be a fixed point of A, let B ' be a subset of order three of B, let d be the common distance of B from X, and let e be the common dis- tance of C from X. Then *+A 2 = e 2 d 2 - { (Y-X) • (Z-X) } 2 or | (Y-X) • (Z-X) | = /e z d z -4A z for all (Y,Z) in B' x C. By lemma 2 there is a subset C of C of order {N/8} such that for each Y in B', (Y-X)-(Z-X) has a constant sign as Z ranges over C». Hence for Z , Z* in C ' and Y , Y ' in B ' , (Y-X)-(Z-X) = (Y-X) -(Z'-X) , (Y-X)-(Z-Z') = 0, (Y-Y')-(Z-Z') = , and the affine covers of C and B' are orthogonal. Since no three points of A (or B or C) can be collinear, and three pairwise orthogonal planes cannot exist in R 5 , we ob- tain a contradiction for M :> 3 and N :> 17- We next show that (i) leads to a contradiction. This is the last and hi hardest case. We start by reducing B to be {Y , ...,Y } by throwing away N-M points . Now 4A 2 = |X-Z| 2 |Y-X| 2 -{(Y-X)-(Z-X)} 2 for all (X,Y,Z) in A x B x C. Hence |(Z-X)-(Y-X) | =/|X-Z| z |Y-X| z -i+A z , and the right hand side is independent of Z, since |X-Z| is independent of Z. Let y , • • • ,y where r = M 2 , be an enu- meration of A x B. Let us 2-color G 2 = (A x B) x C as follows: If (Z-X)-(Y-X) ^ then color ((X,Y),Z) red. Otherwise color ((X,Y),Z) blue. By lemma 2, there is a subset C of C of cardinality {N/(2 )} such that (Z-X)«(Y-X) is of constant sign as Z ranges over C with (X,Y) fixed. Hence for (X,Y) in A x B and Z, Z' in C , (Z-Z')-(Y-X) = 0; for Y,Y' in B, Z,Z' in C and X,X* in A we have (Z-Z ' ) • (Y-Y ' ) = ( Z-Z » ) • (X-X 1 ) = 0. Hence the affine hull of C ' is orthogonal to the affine hull of A y B. Let us assume that N > 2**(M 2 +l)+l, so that the order, {N/(2 r )}, of C is at least 3, so that C contains three non-collinear points. Hence the dimension of the affine hull of C is at least two, and this forces A yj B to lie in a three dimensional subspace it. If C' is also contained in tt , then the whole configuration is in R 3 and if M and N are large enough, we have a contradiction by (h) . We may therefore suppose the existence of a point Z of C that is not in tt . Let Z* be the orthogonal projection of Z onto tt . The points of B lie on cylin- ders with axes X.Z (l <^ i < H), which by lemma 5 intersect tt in surfaces £ . which are either cylinders, spheres, or ellipsoids of revolution with axes X. Z* . Call these surfaces £ . . By the same lemma, £ . cannot be a cylinder, since X.Z is not parallel to tt . Also by lemma 5 5 £. is a sphere only if X.=Z*. Since no three points of the set A are collinear, there exist two points, say X and X , of A such that X ,X ,Z* are not collinear. This implies that neither X nor X hH coincide with Z*, so that by lemma 5, 5, and £ are ellipsoids of revolution with axes of revolution X Z* and X Z* . Suppose that B has a nine-point subset B* that is equidistant from Z. Then B* lies on a sphere S* having center Z* and lying in tt , and for i=l,2 each K. intersects the sphere S* in a pair of circles C. and C* whose centers 1 11 lie on the line X^*. For i,j=l or 2 and j^i, C is distinct from C and C, J J since the normals X.Z* and X.Z* are not parallel, due to the fact that X ,X ,Z* are non collinear. Two distinct circles on the surface of a sphere in R intersect in at most two points. Hence (C y C ) (C (j C') = C ° C 2 y C^ ° C 2 u C n C 2 y C^ ° C 2 is a set of order less than nine containing a set B* of order nine, which is absurd. Hence there exists a set B' of cardinality {M/8}, which is separated by Z. Let us suppose that M ^ ^1 and take B' to have cardinality at least 6. By lemma 1 there exists a subset A' of A of cardinality R = {M/l6} that is equidistant from Z; let d be the common distance of the points of A' from Z. Let B" be a subset of three elements of B'. Then 4A 2 = d 2 | Y-Z | 2 - {(Y-Z)-(X-Z)} 2 or |(Y-Z).(X-Z) | = /d z | Y-Z | z -Ua 2 for all (X,Y) e A' x B". The right hand side is independent of X. By lemma 2, there is a subset A" of A of order S = {R/8} such that (Y-Z)-(X-Z) is of constant sign for fixed Y in B" as X ranges over A". Hence (Y-Z)«(X-X") = for all X, X' in A" and Y in B". Hence (Y-Y 1 ) • (X-X 1 ) = for all X, X» in A" and Y, Y" in B", and the affine hull of A" is orthogonal to the affine hull of B". Combining this with our earlier result, we see that the affine hulls of A", B", and C are pairwise orthogonal. To get a contradic- tion, it is sufficient to ensure that each of these sets has at least three elements. If M = 257, then R = IT and S = 3. If N > 2**(M 2 +l)+l we obtain the desired contradiction. h9 REFERENCES [l] P. Erd8s , On sets of distances of n points , Amer. Math. Monthly, 53(19^6) , 2U8-250. [2] L. Moser, On the different distances determined by n points , Amer. Math Monthly, 59(1952), 85-91. [3] P. Erd8s , On some applications of graph theory to geometry , Canadian Journal Math., 19(1967), 968-971, see also P. Erdos, On sets of distances of n points in euclidean space , Publ. Math. Inst. Hung. Acad. Sci. , 5(1960), 165-169. [k] T. Kovari , V. T. Sos and P. Turan, On a problem of K. Zarankiewicz , Coll. Math., 3(19^5), 50-57. [5] P. ErdiBs , On extremal problems of graphs and generalized graphs , Israel J. Math., 2(l96U), 183-190 , Theorem 1. [6] P. Erdtfs and G. Purdy , Some extremal problems in geometry , Journal of Combinatorial Theory, 10(l97l) , 2^6-252.