Acute sets in Euclidean spaces

Transcription

1 Acute sets in Eucliean spaces Viktor Harangi April, 011 Abstract A finite set H in R is calle an acute set if any angle etermine by three points of H is acute. We examine the maximal carinality α() of a -imensional acute set. The exact value of α() is known only for 3. For each 4 we improve on the best known lower boun for α(). We present ifferent approaches. On one han, we give a probabilistic proof that α() > c 1.. (This improves a ranom construction given by Erős an Fürei.) On the other han, we give an almost exponential constructive example which outoes the ranom construction in low imension ( 50). Both approaches use the small imensional examples that we foun partly by han ( = 4, 5), partly by computer (6 10). We also investigate the following variant of the above problem: what is the maximal size κ() of a -imensional cubic acute set (that is, an acute set containe in the vertex set of a -imensional hypercube). We give an almost exponential constructive lower boun, an we improve on the best known upper boun. 010 Mathematics Subject Classification: 05D40, 51M16. Keywors: acute sets; probabilistic metho; acute angles; strictly antipoal. Department of Analysis, Eötvös Lorán University, Pázmány Péter sétány 1/c, H-1117 Buapest, Hungary aress: harangi@gmail.com Acknowlegement: The author was supporte by Hungarian Scientific Founation grant no Introuction Aroun 1950 Erős conjecture that given more than points in R there must be three of them etermining an obtuse angle. The vertices of the -imensional cube show that points exist with all angles at most π/. In 196 Danzer an Grünbaum prove this conjecture [7] (their proof can also be foun in []). They pose the following question in the same paper: what is the maximal number of points in R such that all angles etermine are acute? (In other wors, this time we want to exclue right angles as well as obtuse angles.) A set of such points will be calle an acute set in the sequel. 1

2 The exclusion of right angles seeme to ecrease the maximal number of points ramatically: they coul only give 1 points, an they conjecture that this is the best possible. However, this was only prove for =, 3. (For the non-trivial case = 3, see Croft [6], Schütte [10], Grünbaum [9].) Then in 1983 Erős an Fürei isprove the conjecture of Danzer an Grünbaum. They use the probabilistic metho to prove the existence of a -imensional acute set of carinality exponential in. Their iea was to choose ranom points from the vertex set of the -imensional unit cube, that is {0, 1}. We enote the maximal size of acute sets in R an in {0, 1} by α() an κ(), respectively; clearly α() κ(). The ranom construction of Erős an Fürei implie the following lower boun for κ() (thus for α() as well) κ() > 1 ( ) 3 > (1) In their paper they claime (without proof) that a bit more complicate ranom process gives κ() > ( 4 o(1) ) > The best (publishe) lower boun both for α() an for κ() (for large values of ) is ue to Ackerman an Ben-Zwi from 009 [1]. They improve (1) with a factor : α() κ() > c ( ) 3. () In Section we moify the ranom construction of Erős an Fürei to get ( ) α() > c > c 1.. (3) 3 A ifferent approach can be foun in Section 3 where we recursively construct acute sets. These constructions outo (3) up to imension 50. In Theorem 3.10 we will show that this constructive lower boun is almost exponential in the following sense: given any positive integer r, for infinitely many values of we have a -imensional acute set of carinality at least exp(/ log log log()). }{{} r See Table in the Appenix for the best known lower bouns of α() ( 84). These bouns are new results except for 3. Both the probabilistic an the constructive approach use small imensional acute sets as builing blocks. So it is crucial for us to construct small imensional acute sets of large carinality. In Section 4 we present an acute set of 8 points in R 4 an an acute set of 1 points in R 5 (isproving the conjecture of Danzer an Grünbaum for 4 alreay). We use computer to fin acute sets in imension 6 10, see Section 4. Table 1 shows our results compare to the construction of Danzer an Grünbaum ( 1) an the examples foun by Bevan using computer. As far as κ() is concerne, in large imension () is still the best known lower boun. Bevan use computer to etermine the exact values of κ() for 9 [4]. He also gave a recursive construction improving upon the ranom constructions in low imension. The

3 Table 1: Results for α() ( 10) im() D,G[7] Bevan[4] Our result = 3 3 = constructive approach of Section 3 yiels a lower boun not only for α() but also for κ(), which further improves the bouns of Bevan in low imension. Table 3 in the Appenix shows the best known lower bouns for κ() ( 8). These bouns are new results except for 1 an = 7. The following notion plays an important role in both approaches. Definition 1.1. A triple A, B, C of three points in R will be calle ba if for each integer 1 i the i-th coorinate of B equals the i-th coorinate of A or C. We enote by κ n () the maximal size of a set S {0, 1,..., n 1} that contains no ba triples. It is easy to see that κ () = κ() but our main motivation to investigate κ n () is that we can use sets without ba triples to construct acute sets recursively (see Lemma.). We give an upper boun for κ n () (Theorem 3.1) an two ifferent lower bouns (Theorem.3 an 3.5). In the special case n = the upper boun yiels κ() 3( ) 1 which improves the boun ( 3 ) given by Erős an Fürei in [8]. Note that for α() the best known upper boun is 1. Although we can make no contribution to it, we mention that there is an affine variant of this problem. A finite set H in R is calle strictly antipoal if for any two istinct points P, Q H there exist two parallel hyperplanes, one through P an the other through Q, such that all other points of H lie strictly between them. Let α () enote the maximal carinality of a -imensional strictly antipoal set. An acute set is strictly antipoal, thus α () α(). For α () Talata gave the following constructive lower boun [11]: α () 4 5 /4 > A weaker result (also ue to Talata) can be foun in [5, Lemma 9.11.]. The probabilistic approach As we mentione in the introuction, in 1983 Erős an Fürei prove the existence of acute sets of exponential carinality [8]. Since then their proof has become a well-known example to emonstrate the probabilistic metho. In this section we use similar arguments 3

4 to prove a better lower boun for α(). The following problem plays a key role in our approach. Question.1. What is the maximal carinality κ n () of a set S {0, 1,..., n 1} that contains no ba triples? (Recall Definition 1.1.) In the case n =, given three istinct points A, B, C {0, 1}, ABC = π/ hols if an only if A, B, C is a ba triple, otherwise ABC is acute. So a set S {0, 1} contains no ba triples if an only if S is an acute set, thus κ () = κ(). If n >, then a triple being ba still implies that the angle etermine by the triple is π/ but we can get right angles from goo triples as well, moreover, we can even get obtuse angles. So for n > the above problem is not irectly relate to acute sets. However, the following simple lemma shows how one can use sets without ba triples to construct acute sets recursively. Lemma.. Suppose that H = {h 0, h 1,..., h n 1 } R m is an acute set of carinality n. If S {0, 1,..., n 1} contains no ba triples, then H S ef = {(h i1, h i,..., h i ) : (i 1, i,..., i ) S} H } H {{... H } R m is also an acute set. Consequently, Proof. Take three istinct points of S: α(m) κ α(m) () an κ(m) κ κ(m) (). (4) i = (i 1, i,..., i ); j = (j 1, j,..., j ); k = (k 1, k,..., k ), an the corresponing points in H S : h i = (h i1, h i,..., h i ) ; h j = (h j1, h j,..., h j ) ; h k = (h k1, h k,..., h k ). We show that h i h j h k is acute by proving that the scalar prouct h i h j, h k h j = h ir h jr, h kr h jr r=1 is positive. Since H is an acute set, the summans on the right-han sie are positive with the exception of those where j r equals i r or k r, in which case the r-th summan is 0. This cannot happen for each r though, else i, j, k woul be a ba triple in S. To prove (4) we set H = n = α(m) an S = κ n () = κ α(m) (). Then α(m) H S = S = κ α(m) (). A similar argument works for κ(m). (Note that if H {0, 1} m, then H S {0, 1} m.) In view of the above lemma, it woul be useful to construct large sets without ba triples. One possibility is using the probabilistic metho. The next theorem is a generalization of the original ranom construction of Erős an Fürei. 4

5 Theorem.3. κ n () > 1 ( n n 1 ) 1 ( n ) > = ( ) + 1 n. Proof. For a positive integer m, we take m (inepenent an uniformly istribute) ranom points in {0, 1,..., n 1} : A 1, A,..., A m. What is the probability that the triple A 1, A, A 3 is ba? For a fixe i, the probability that the i-th coorinate of A is equal to the i-th coorinate of A 1 or A 3 is clearly (n 1)/n. These events are inepenent so the probability that this hols for every i (that is to say A 1, A, A 3 is a ba triple) is ( ) n 1 p =. n We get the same probability for all triples, thus the expecte value of the number of ba triples is ( ) m m(m 1)(m ) 1 3 p = p < 4m 3 p m, where we set m = 3. p Consequently, the m ranom points etermine less than m ba triples with positive probability. Now we take out one point from each ba triple. Then the remaining at least m + 1 points obviously contain no ba triples. So we have prove that there exist m + 1 > 1 p = 1 ( n n 1 points in {0, 1} without a ba triple. (Note that the original m ranom points might contain uplicate points. However, a triple of the form A, A, B is always ba, thus the final (at least) m + 1 points contain no uplicate points.) Combining Lemma. an Theorem.3 we reaily get the following. Corollary.4. Suppose that we have an m-imensional acute set of size n. Then for any positive integer t ( ) α(mt) > 1 t n, n 1 which yiels the following lower boun in general imension: ( ) ( α() α m > 1 ) m m n m ( c m n 1 ) n m n 1 Using this corollary with m = 5 an n = 1 (see Example 4. for a 5-imensional acute set with 1 points) we obtain the following. Theorem.5. ( ) α() > c > c 1., 3 that is, there exist at least c 1. points in R such that any angle etermine by three of these points is acute. (If is ivisible by 5, then c can be chosen to be 1/, for general we nee to use a somewhat smaller c.) 5 ).

6 Remark.6. We remark that one can improve the above result with a factor by using the metho suggeste by Ackerman an Ben-Zwi in [1]. Remark.7. We coul have applie Corollary.4 with any specific acute set. The larger the value m n /(n 1) is, the better the lower boun we obtain. For m = 1,, 3 the largest values of n are known. m = 1 n = } m = n = 3 } m = 3 n = 5 } We will construct small imensional acute sets in Section 4 (see Table 1 for the results). For m = 4, 5, 6 these constructions yiel the following values for m n /(n 1). } } } m = n = m = n = m = n = However, we o not know whether these acute sets are optimal or not. If we foun an acute set of 9 points in R 4, 13 points in R 5 or 18 points in R 6, we coul immeiately improve Theorem.5. 3 The constructive approach 3.1 On the maximal carinality of sets without ba triples In this subsection we investigate Question.1 more closely. Recall that κ n () enotes the maximal carinality of a set in {0, 1,..., n 1} containing no ba triples. We have alreay seen a probabilistic lower boun for κ n () (Theorem.3). Here we first give an upper boun. As we will see, this upper boun is essentially sharp if n is large enough (compare to ). Theorem 3.1. For even an for o κ n () n /, κ n () n (+1)/ + n ( 1)/. Proof. Suppose that S {0, 1,..., n 1} contains no ba triples. Let 0 < r < be an integer, an consier the following two projections: Now we take the set π 1 ((x 1,..., x )) = (x 1,..., x r ) ; π ((x 1,..., x )) = (x r+1,..., x ). S 0 ef = {x S : y (S \ {x}) π 1 (x) = π 1 (y)}. By efinition π 1 is injective on S \ S 0, thus S \ S 0 n r. We claim that π is injective on S 0, so S 0 n r. Otherwise there woul exist x, y S 0 such that π (x) = π (y). Since y S 0, there exists z S such that π 1 (y) = π 1 (z). It follows that the triple x, y, z is ba, contraiction. Consequently, S n r + n r. Setting r = we get the esire upper boun. 6

7 Setting n = an using that κ () = κ() the next corollary reaily follows. Corollary 3.. For even κ() (+)/ = ( ), an for o κ() (+1)/ + ( 1)/ = 3 ( ). This corollary improves the upper boun ( 3) given by Erős an Fürei in [8]. (We note though that they prove not only that a subset of {0, 1} of size larger than ( 3) must contain three points etermining a right angle but they also showe that such a set cannot be strictly antipoal which is a stronger assertion.) If n is a prime power greater than, then the following constructive metho gives better lower boun than the ranom construction of the previous section. We will nee matrices over finite fiels with the property that every square submatrix of theirs is invertible. In coing theory the so-calle Cauchy matrices are use for that purpose. Definition 3.3. Let F q enote the finite fiel of orer q. A k l matrix A over F q is calle a Cauchy matrix if it can be written in the form A i,j ef = (x i y j ) 1 (i = 1,..., k; j = 1,..., l), (5) where x 1,..., x k, y 1,..., y l F q an x i y j for any pair of inices i, j. In the case k = l = r, the eterminant of a Cauchy matrix A is given by i<j et(a) = (x i x j ) i<j (y i y j ) 1 i,j r (x. i y j ) This well-known fact can be easily prove by inuction. It follows that A is invertible provie that the elements x 1,..., x r, y 1,..., y r are pairwise istinct. Lemma 3.4. Let q be a prime power an k, l be positive integers. Suppose that q k + l. Then there exists a k l matrix over F q any square submatrix of which is invertible. Proof. Let x 1,..., x k, y 1,..., y l be pairwise istinct elements of F q, an take the k l Cauchy matrix A as in (5). Clearly, every submatrix of A is also a Cauchy matrix thus the eterminant of every square submatrix of A is invertible. Now let k + l = an n be a prime power greater than or equal to. Due to the lemma, there exists a k l matrix A over the fiel F n such that each square submatrix of A is invertible. Let us think of {0, 1,..., n 1} as the -imensional vector space F n. We efine an F n -linear subspace of F n: take all points (x, Ax) F n as x runs through F l n (thus Ax F k n). This is an l-imensional subspace consisting n l points. We claim that each of its points has at least k + 1 nonzero coorinates. We prove this by contraiction. Assume that there is a point (x, Ax) which has at most k nonzero coorinates. Let the number of nonzero coorinates of x be r. It follows that the number of nonzero coorinates of Ax is 7

8 at most k r, in other wors, Ax has at least r zero coorinates. Consequently, A has an r r submatrix which takes a vector with nonzero elements to the null vector. This contraicts the assumption that every square submatrix is invertible. Setting k = an l = we get a subspace of imension, every point of which has at least +1 > nonzero coorinates. We claim that this subspace oes not contain ba triples. Inee, taking istinct points x 1, x, x 3 R l, the points (x 1 x, A(x 1 x )) an (x 3 x, A(x 3 x )) are elements of the subspace, thus both have more than nonzero coorinates which means that there is a coorinate where both of them take nonzero value. We have prove the following theorem. Theorem 3.5. If is an integer an n is a prime power, then κ n () n. If n is not a prime power, then there exists no finite fiel of orer n. We can still consier matrices over the ring Z n = Z/nZ. If we coul fin a matrix with all of its square submatrices invertible, it woul imply the existence of a set without ba triples an of carinality n. For example, in the case = 3 the matrix (1 1) over Zn is clearly goo for any n so the next theorem follows. Theorem 3.6. For arbitrary positive integer n it hols that κ n (3) n. Proof. We can prove this irectly by taking all points in the form (i, j, i + j) where i, j run through Z n (aition is meant moulo n). Clearly, there are no ba triples among these n points. Finally we show that the upper boun given in Theorem 3.1 is sharp apart from a constant factor provie that n is sufficiently large compare to. Theorem 3.7. We have κ n () > (1 ε(, n)) n, where for any fixe the error term ε(, n) converges to 0 as n. In fact, for any δ > 0 there exists C δ such that ε(, n) < δ provie that n C δ.11. Proof. It was prove in [3] that for any sufficiently large n there is a prime number q in the interval [n n 0.55, n]. If q, then by Theorem 3.5 we can fin a set S {0, 1,..., q 1} {0, 1,..., n 1} such that S contains no ba triples an S q (n n 0.55 ) = ( 1 n 0.475) n. For fixe the coefficient of n clearly converges to 1 as n. To obtain the stronger claim we use the well-known fact that (1 1/x) x 1 > 1/e for any x > 1. Consequently, if the exponent is at most (n )δ, then we have S > (1/e) δ n > (1 δ)n, whence ε(, n) < δ. A simple calculation completes the proof. 8

9 3. Constructive lower bouns for α() an κ() Ranom constructions of acute sets (as the original one of Erős an Fürei or the one given in Section ) give exponential lower boun for α(). However, these only prove existence without telling us exactly how to fin such large acute sets. Also, one can give better (constructive) lower boun if the imension is small. The first (non-linear) constructive lower boun is ue to Bevan[4]: α() κ() > exp (c µ ), where µ = log log 3 = (6) For small this is a better boun than the probabilistic ones. Our goal in this section is to obtain even better constructive bouns. The key will be the next theorem which follows reaily from Lemma., Theorem 3.5 an Theorem 3.6 setting = s 1. (In fact, the special case s = was alreay prove by Bevan, see [4, Theorem 4.]. He obtaine (6) by the repeate application of this special case.) Theorem 3.8. Let s be an integer, an suppose that n s 1 is a prime power. (In the case s = the theorem hols for arbitrary positive integer n.) If H R m is an acute set of carinality n, then we can choose n s points of the set that form an acute set. H } {{ H } R (s 1)m s 1 Remark 3.9. If H is cubic (that is, H {0, 1} m ), then the obtaine acute set is also cubic (that is, it is in {0, 1} (s 1)m ). Now we start with an acute set H of prime power carinality an we apply the previous theorem with the largest possible s. Then we o the same for the obtaine larger acute set (the carinality of which is also a prime power). How large acute sets o we get if we keep oing this? For the sake of simplicity, let us start with the 0 = 4 imensional acute set of size n 0 = 8 that we will construct in Section 4. Let us enote the imension an the size of the acute set we obtain in the k-th step by k an n k, respectively. Clearly n k is a power of, thus at step (k + 1) we can apply Theorem 3.8 with s k = n k /. Setting u k = log n k we get the following: It follows that k+1 /u k+1 k /u k so k+1 = k (s k 1) < k n k ; n k+1 = n s k k = n n k/ k ; u k+1 = u k (n k /) = u k uk 1 uk 1 = u k. k 0 u 0 u k k = 4 3 k u k. It yiels that in imension k we get an acute set of size n k = u k (3/4) k k. 9

10 Due to the factor k in the exponent, n k is not exponential in k. However, the inequality u k+1 u k implies that uk grows extremely fast (an so oes n k an k ) which means that n k is almost exponential. For instance, we can easily obtain that for any positive integer r there exists k 0 such that for k k 0 it hols that n k > exp( k / log log log( }{{} k )). r We have given a constructive proof of the following theorem. Theorem For any positive integer r we have infinitely many values of such that α() > exp(/ log log log ). }{{} r We can also get a constructive lower boun for κ(). We o the same iterate process but this time we start with an acute set in {0, 1} 0. (For instance, we can set 0 = 3 an n 0 = 4.) Then the acute set obtaine in step k will be in {0, 1} k. This way we get an almost exponential lower boun for κ() as well. However, Theorem 3.8 gives acute sets only in certain imensions. In the remainer of this section we consier the problems investigate so far in a slightly more general setting to get large acute sets in any imension. (The proofs of these more general results are essentially the same as the original ones. Thus we coul have consiere this general setting in the first place, but for the sake of better unerstaning we opte not to.) Let n 1, n,..., n be positive integers an consier the n 1 n lattice, that is the set {0, 1,..., n 1 1} {0, 1,..., n 1}. Question What is the maximal carinality of a subset S of the n 1 n lattice containing no ba triples? We claim that if n max{n 1,..., n } an the set S 0 {0, 1,..., n 1} contains no ba triples, then we can get a set in the n 1 n lattice without ba triples an of carinality at least n 1 n n n S 0. Inee, starting with the n... n lattice, we replace the n s one-by-one with the n i s; in each step we keep those n i sections that contain the biggest part of S 0. Combining this argument with Theorem 3.5 an 3.6 we get the following for the o case = s 1. Theorem 3.1. Let s, an suppose that n s 1 is a prime power (in the case s = the theorem hols for arbitrary positive integer n). For positive integers n 1,..., n s 1 n in the n 1 n n s 1 lattice at least n 1 n n s 1 /n s 1 points can be chosen without any ba triple. Also, one can get a more general version of Lemma. with the same proof. Lemma Suppose that the set H t = {h t 0, h t 1,..., h t n t 1} R mt is acute for each 1 t. If S {0, 1,..., n 1 1} {0, 1,..., n 1} contains no ba triples, then the set {( h 1 i1, h i,..., h i ) : (i1, i,..., i ) S } H 1 H H R m 1+ +m is an (m m )-imensional acute set. 10

11 Putting these results together we obtain a more general form of Theorem 3.8. Theorem Let s, an suppose that n s 1 is a prime power (in the case s = the theorem hols for arbitrary positive integer n). Assume that for each t = 1,..., s 1 we have an acute set of n t n points in R mt. Then in R m 1+ +m s 1 there exists an acute set of carinality at least n1 n n s 1/n s 1. The obtaine acute set is cubic provie that all acute sets use are cubic. Remark We also note that in the case s = 3 the theorem can be applie for n = 4 as well. Consier the 4-element fiel F 4 = {0, 1, a, b}. Then the 3 matrix ( ) A = 1 a b has no singular square submatrix which implies that Theorem 3.5 hols for = 5; n = 4, thus Theorem 3.1 an Theorem 3.14 hol for s = 3; n = 4. Now we can use the small imensional acute sets of Section 4 as builing blocks to buil higher imensional acute sets by Theorem Table in the Appenix shows the lower bouns we get this way for 84. (We coul keep oing that for larger values of an up to imension 50 we woul get better boun than the probabilistic one given in Section.) These bouns are all new results except for 3. We can o the same for κ(), see Table 3 in the Appenix for 8. This metho outoes the ranom construction up to imension 00. (We nee small imensional cubic acute sets as builing blocks. We use the ones foun by Bevan who use computer to etermine the exact values of κ() for 9. He also use a recursive construction to obtain bouns for larger s. His metho is similar but less effective: our results are better for 13; 7. In imension = 63 we get a cubic acute set of size This is almost ten times bigger than the one Bevan obtaine which contains 6561 points.) Tables 4 an 5 in the Appenix compare the probabilistic an constructive lower bouns for α() an κ(). Finally we prove the simple fact that α() is strictly monotone increasing. We will nee this fact in Table. Lemma α( + 1) > α() hols for any positive integer. Proof. Assume that we have an acute set H = {x 1,..., x n } R. Let P be the convex hull of H an y be any point in P \ H. We claim that yx i x j < π/ for any i j. Let H i,j be the hyperplane that is perpenicular to the segment x i x j an goes through x i. Let S i,j be the open half-space boune by H i,j that contains x j. For a point z R the angle zx i x j is acute if an only if z S i,j. It follows that H \ {x i } S i,j while x i lies on the bounary of S i,j. Thus y P \ {x i } S i,j which implies that yx i x j < π/. Now let us consier the usual embeing of R into R +1 an let v enote the unit vector (0,..., 0, 1). Consier the point y t = y + tv for sufficiently large t. It is easy to see that y t x i x j < π/ still hols, but now even the angles x i y t x j are acute. It follows that H {y t } R +1 is an acute set. Remark For κ() it is only known that κ( + ) > κ() [4, Theorem 4.1]. In Table 3 we will refer to this result as almost strict monotonicity. 11

12 4 Small imensional acute sets In this section we construct acute sets in imension m = 4, 5 an use computer to fin such sets for 6 m 10. These small imensional examples are important because the ranom construction of Section an the recursive construction of Section 3 use them to fin higher imensional acute sets of large carinality. Danzer an Grünbaum presente an acute set of m 1 points in R m [7]. It is also known that for m =, 3 this is the best possible [6, 10, 9]. Bevan use computer to fin small imensional acute sets by generating ranom points on the unit sphere. For m 7 he foun more than m 1 points [4]. Our approach starts similarly as the construction of Danzer an Grünbaum. We consier the following m points in R m : P ±1 i = (0,..., 0, }{{} ±1, 0,..., 0); i = 1,,..., m 1. i-th What angles o these points etermine? Clearly, P 1 i P ±1 j P +1 i = π/ for i j an all other angles are acute. We can get ri of the right angles by slightly perturbing the points in the following manner: P ±1 i = (0,..., 0, }{{} ±1, 0,..., 0, ε i ); i = 1,,..., m 1, (7) i-th where ε 1, ε,..., ε m 1 are pairwise istinct real numbers. ±1 Our goal is to complement the points P i with some aitional points such that they still form an acute set. In fact, we will complement the points P ±1 i such that all new angles are acute. (Then changing the points P ±1 ±1 i to P i we get an acute set provie that the ε i s are small enough.) Uner what conition can a point x = (x 1,..., x m ) be ae in the above sense? Simple calculation shows that the exact conition is x > 1 an x i + x j < 1 for 1 i, j m 1; i j. (8) For example, the point A = (0,..., 0, a) can be ae for a > 1. This way we get an acute set of size m 1. Basically, this was the construction of Danzer an Grünbaum. We know that this is the best possible for m =, 3. However, we can o better if m 4. Suppose that we have two points x = (x 1,..., x m ) an y = (y 1,..., y m ) both satisfying (8) (that is, they can be separately ae). Both points can be ae (at the same time) if an only if x i + y i < 1 + x, y an x i y i < min ( x, y ) x, y for 1 i m 1. (9) We can fin two such points in the following simple form: A 1 = (a 1, a 1,..., a 1, a ) an A = ( a 1, a 1,..., a 1, a ). Then points A 1 an A can be ae if an only if Such a 1 an a clearly exist if m 4. 1 m 1 < a 1 < 1 an a > 1 (m 1)a 1. (10) 1

13 Example 4.1. For sufficiently small an pairwise istinct ε i s the 8 points below form an acute set in R 4. ( ε 1 ) ( ε 1 ) ( ε ) ( ε ) ( ε 3 ) ( ε 3 ) ( ) ( ) For m = 5, we can even a four points of the following form: A 1 = (a 1, a 1, a 1, a 1, a ); A = ( a 1, a 1, a 1, a 1, a ); B 1 = (b 1, b 1, b 1, b 1, b ); B = ( b 1, b 1, b 1, b 1, b ). We have seen that 1/4 < a 1, b 1 < 1/ must hol so we set a 1 = 1/4 + δ an b 1 = 1/ δ. Then we set a = 3/ an b = δ so that A i an B i are slightly bigger than 1. Example 4.. Let us fix a positive real number δ < 1/48 an consier the points below. A 1 = ( 1/4 + δ 1/4 + δ 1/4 + δ 1/4 + δ 3/ ) A = ( 1/4 δ 1/4 δ 1/4 δ 1/4 δ 3/ ) B 1 = ( 1/ δ 1/ δ 1/ + δ 1/ + δ δ ) B = ( 1/ + δ 1/ + δ 1/ δ 1/ δ δ ) ±1 Then the set { P i : i = 1,, 3, 4} {A 1, A, B 1, B } is an acute set of 1 points in R 5 assuming that ε i s are sufficiently small an pairwise istinct. (This specific example is important because the ranom metho presente in Section gives the best result starting from this example.) Proof. We nee to prove that A 1, A, B 1, B can be ae to P ±1 i s in such a way that all new angles are acute. First we prove that any pair of these 4 points can be ae. Since each of them satisfies (8), we only have to check that each pair satisfies (9). For the pair A 1, A we are one since they satisfy (10). It goes similarly for the pair B 1, B. For the pairs A i, B j (9) yiels the conition 3/4 < 1 3δ δ < 1/48. Now we have checke all new angles except those that are etermine by three new points. The squares of the istances between the 4 new points are: (A 1, A ) = 1 + 8δ + 16δ ; (B 1, B ) = 4 16δ + 16δ ; (A i, B j ) = + 3δ + δ + 8δ. Now for any triangle in {A 1, A, B 1, B } the square of each sie is less than the sum of the squares of the two other sies which means that the triangle is acute-angle. For m 6 we use computer to a further points to the system (7). We generate ranom points on the sphere with raius 1 + δ an we ae the point whenever it was possible. Table 1 shows the carinality of acute sets we foun this way compare to previous results. The reaer can also fin examples for m = 6, 7, 8 in the Appenix. For m 11, the recursive construction presente in Section 3 gives better result than the computer search (see Table in the Appenix for the best known lower bouns of α() for 84). 13

14 Appenix A Small imensional acute sets We have seen in Section 4 that the following m points form an acute set in R m : P ±1 i = (0,..., 0, }{{} ±1, 0,..., 0, ε i ); i = 1,,..., m 1, i-th where ε 1, ε,..., ε m 1 enote sufficiently small, pairwise istinct real numbers. We use computer to a points to this system in such a way that they still form an acute set. Below the reaer can fin the obtaine aitional points in imension m = 6, 7, 8. In orer ±1 to get integer coorinates, we start with the enlarge system 999 P i ; i = 1,,..., m 1. With the following aitional points we have an acute set of 16 points in R 6. ( ) ( ) ( ) ( ) ( ) ( ) With the following aitional points we have an acute set of 0 points in R 7. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) With the following aitional points we have an acute set of 3 points in R 8. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 14

15 Appenix B Best known bouns in low imension The following tables show the best known lower bouns for α() an κ(). Besie the imension an the boun itself, we state the value of s, n an the prouct n 1 n s 1 /n s 1 with which Theorem 3.14 is applie. From the n i s the reaer can easily obtain the m i s. Str. mon. an a. str. mon. stan for strict monotonicity (cf. Lemma 3.16) an almost strict monotonicity (cf. Remark 3.17). For example, in imension 39 in Table we see that s = 5 an n = 9. (Note that n is inee a prime power an n s 1 hols.) The expression /9 4 means that we nee to apply Theorem 3.14 with n 1 = n =... = n 6 = 8 an n 7 = n 8 = n 9 = 9. (Note that they are all inee at most n.) Then for each i we take the smallest imension m i in which we have an acute set containing at least n i points. In our case the corresponing imensions are m 1 = m =... = m 6 = 4 an m 7 = m 8 = m 9 = 5. Consequently, the total imension is = 39. We obtain that in R 39 there exists an acute set of carinality at least /9 4 = 918. Recall that in the case s = we can take arbitrary n (it oes not nee to be a prime power). See imension 14 an 15 in Table. Also, accoring to Remark 3.15, in the case s = 3 we can have n = 4 (even though n s 1 oes not hol). See imension 13 an 15 in Table 3. 15

16 im l. b. s n Table : Best known lower boun for α() (1 84) construction 5 1 construction 6 16 computer 7 0 computer 8 3 computer 9 7 computer computer / / str. mon / / str. mon / / / / str. mon. 514 str. mon / / / / / / str. mon str. mon str. mon / / / / / / / / / / /16 3 im l. b. s n / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / str. mon str. mon / / / /

17 im l. b. s n Table 3: Best known lower boun for κ() (1 8) Bevan 5 6 Bevan 6 8 Bevan 7 9 Bevan 8 10 Bevan / / / / / a. str. mon / / / a. str. mon / / / / / / / a. str. mon / / / / / / / / /8 3 im l. b. s n / a. str. mon / / / / / / / / / / / / / / / / / / a. str. mon a. str. mon / / / / / / / / / / / / / /

18 Appenix C Comparing the two approaches To compare the lower bouns given by the probabilistic an the constructive approach we o the following. For a small value of m we take an m-imensional acute set of prime power carinality n, then we apply Theorem 3.8 with the largest possible s to get an acute set of size n s in imension = (s 1)n. Then we compare this to the probabilistic boun α() > (1/)(144/3) /10 (in fact, we obtaine this result only for ivisible by 5; for general it only hols with a somewhat smaller constant factor). For the sake of simplicity we consier the base-10 logarithm of the bouns. (See Table for values of n use here.) Table 4: Comparing constructive an probabilistic lower boun of α() m n s imension constructive l.b. probabilistic l.b. = (s 1)m s lg n lg lg We can o the same for κ(). We apply Theorem 3.8 for small imensional acute sets in {0, 1} with the largest possible s an compare what we get to the boun κ() > (1/)(4/3) / given by Erős an Fürei. (See Table 3 for values of n use here.) Table 5: Comparing constructive an probabilistic lower boun of κ() m n s imension constructive l.b. probabilistic l.b. = (s 1)m s lg n lg 1 + lg

19 References [1] E. Ackerman, O. Ben-Zwi, On sets of points that etermine only acute angles, European J. Combin. 30 (009) no. 4, [] M. Aigner, G.M. Ziegler, Proofs from THE BOOK, 3r e. Springer-Verlag (003), [3] R.C. Baker, G. Harman, J. Pintz, The ifference between consecutive primes, II., Proc. Lonon Math. Soc. 83 (001) no. 3, [4] D. Bevan, Sets of points etermining only acute angles an some relate colouring problems, Electron. J. Combin. 13 (006). [5] K. Böröczky Jr., Finite packing an covering, Cambrige Tracts in Math., vol. 154, Cambrige Univ. Press (004), [6] H.T. Croft, On 6-point configurations in 3-space, J. Lonon Math. Soc. 36 (1961), [7] L. Danzer, B. Grünbaum, Über zwei Probleme bezüglich konvexer Körper von P. Erős un von V.L. Klee, Math. Zeitschrift 79 (196), [8] P. Erős, Z. Fürei, The greatest angle among n points in the -imensional Eucliean space, Ann. Discrete Math. 17 (1983), [9] B. Grünbaum, Strictly antipoal sets, Israel J. Math. 1 (1963), [10] K. Schütte, Minimale Durchmesser enlicher Punktmengen mit vorgeschriebenem Minestabstan, Math. Ann. 150 (1963), [11] I. Talata, Personal communication. 19