ON SOME EXTREMUM PROBLEMS IN ELEMENTARY GEOMETRY


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1 ON SOME EXTREMUM PROBLEMS IN ELEMENTARY GEOMETRY P. ERD& and G. SZEKERES* Mathematical Institute, EBtVBs Lotand University, Budapest and University Adelaide (Received May 20, 7960) Dedicated to the memory of Professor L, FEJ~R 1. Let S denote a set of points in the plane, N(S) the number of points in S. More than 25 years ago we have proved [2] the following conjecture of ESTHER KLEIN SZERERES: There exists a positive integer f (n) with the property that if iv (S) > f(n) then S contains a subset P with N(P) = n such that the points of P form a convex ngon. Moreover we have shown that if lo(n) is the smallest such integer then fcl(n) 5 i n g and conjectured that fo(n) = 2n2 for every n 2 3.l We are unable to prove or disprove this conjecture, but in $ 2 we shall construct a set of 2n2 points which contains no convex ngon. Thus 2*f! If0 {n)< 2rz. I ~ n2 1 A second problem which we shall consider is the following: It was proved by SZEKERES [3] that (i) In any configuration of iv = 2n + 1 points in the plane there are three points which form an angle (I z) greater than (1  l/n + I/n NZ) X. (ii) There exist configurations of 2n points in the plane such that each angle formed by these points is less than (I  I/n) z + E with E > 0 arbitrarily small. The first statement shows that for sufficiently small E > 0 there are no configurations of points which would have the property (ii). Hence in a certain sense this is a best possible result; but it does not determine the exact limiting value of the maximum angle for any given N(S). Let C( (m) denote the greatest positive number with the property that hr every configuration of m points in the plane there is an angle,9 with (1) f3 "u(m). * This paper was written while P. ERDGS was visiting at the University of Adelaide. I The conjecture is trivial for n = 3; it was proved by Miss KLEIN? for n = 4 and by E. MAKAI and P. TURAN for n = 5.
2 54 P. ERDOS AND G. SZEKERES From (i) and (ii) above it follows that CI (~1) exists for every uz 2 3 and that for 2n < m I 2 +1, (2) [ $ l/n (2n + I) ] $I I CL (m) I [ 1  l/(n + l)] z. Two questions arise in this connection: 1. What is the exact value of Q (m), 2. Can the inequality (1) be replaced by (3) P > a (ml For the first few values of m one can easily verify that a (3) = + a (4) = in, CI (5) = $76, a (6) = a (7) = ct (8) = + zr, and that the strict inequality (3) is true for m = 7 and 8. For 3 I m I 6 the regular uz gons represent configurations in which the maximum angle is equal to CL (m); but we know of no other cases in which the equality sign would be necessary in (1). In $ 4 we shalt prove THEOREM 1. Every plane configuration of 2 poinfs cn 2 3) contains an angr e greater than (11 in) S. The theorem shows in conjunction with (ii) above that for n r 3, CL (29 = = (1  I/n) z and that the strict inequality (3) holds for these values of m. The problem is thus completely settled for m = 2, n 2 2. It is not impossible that cx (m) = (1  l/n) YC for 2 l< m < 2, n 2 4, and that (3) holds for every m > 6. However, we can only prove that for 0 ( k < 2 l, cy. (an  k) 2 (11 /n) YE  k x/2 (an k) (Theorem 2). Finally we mention the following conjecture of P. ERDBS : Given 2n+ 1 points in nspace, there is an angle determined by these points which is greater than 1  X. For n = 2 the statement is trivial, for n = 3 it was proved by N. H. 2 KUIPER nad A. II. BOERDIJK. For n > 3 the answer is not known; and it seems that the method of 9 4 is not applicable to this problem. ** 2. In this section we construct a set of 2n2 points in the plane which contains no convex ngon. For representation we use the Cartesian (x, y) plane. All sets to be considered are such that no three points of the set are collinear. A sequence of points!x, V ), v = 091, * , k is said to be convex, of length k, if x, < x, <... < Xfi Yv  yv1 ( YYtcYY for v=l,.... kl; x,jx,,_, xtl+lx,, 2 Unpublished. ** (Added in proof: This conjecture was recently proved by DANZER and GR~~NBAUM in a surprisingly simplex way.)
3 ON SOME EXTREMUM PROBLEMS 57 Suppose now that P,, (i = 1,..., T) is a nonempty subset of Ski, 1 I k, < <...<k,<n 1 and such that P = 6 Pi forms a convex polygon. Since i=l the slope of lines within each SI,, is positive, Pi for 1 < i < r consists of a single point and P, must form a concave sequence, P, a convex sequence. But then the total number of points in P is at most k, + (k,  k,  1) + (nkk,) = n The proof of Theorem 1 requires a refinement of the method used in [3], and in this section we set up the necessary graphtheoretical apparatus. We denote by C.(N) the complete graph of order N, i. e. a graph with N vertices in which any two vertices are joined by an edge. Jf G is a graph, then S (G) shall denote the set of vertices of G. If A is a subset of S (G), then G]A denotes the restriction of G to A. An even (odd) circuit of G is a closed circuit containing an even (odd) number of edges. A parfifion of G is a decomposition G = G, G, into subgraphs Gi with the following property: Each Gi consists of all vertices and some edges of G such that each edge of G appears in one and only one Gi (Gi may not contain any edge at all). We call a partition G = G G, even if it has the property that no Gi contains an odd circuit. LEMMA 1. If a graph G contains no odd circuit fken its vertices can be divided in two classes, A and B, suck that every edge of G has one endpoint in A and one in B. This is wellknown and very simple to prove, e. g. [I], p 170. LEMMA 2. If C(N) = G G, is an even parfition of the complefe graph ON) into n parts then N _< 2. This Lemma was proved in 131; for the sake of completeness we repeat the argument. Since G1 contains no odd circuit we can divide S (C(N)) in classes A and B, containing N, and N, vertices respectively, such that each edge of G1 connects a point of A with a point of B. But then G, G, induces an even partition Gi +... I GA of G = ON) j A and since G is a complete graph of ordtr N,, we conclude by induction that N, I 2+l. Similarly N, I 2 l hence N = N, $ N2 K 2. To prove Theorem 1 we shall need more precise information about the structure of even partitions of CY ), particularly in the limiting case of N = 2. The following Lemmas have some interest of their own; they are formulated in greater generality than actually needed for our present purpose. LEMMA 3. Let N = 2 and UN) = G G, an even parfition o/ UN) into n parts. Then the total number of edges emanating from a fixed vertex p E S (UN)) in Gi, Gjj, where 1 < jl < j2 <... < ji I n, is at least 21 and at most Clearly the order in which the components Gi are written is immaterial, therefore we can assume in the proof that jv = Y, v = 1,.., i. We also note that the first statement follows from the second one by applying the latter to the complementary partition G,+l G, and by noting that the number of edges from p in G, +... A G,, is 21. We shall prove the second state
4 58 P. ERDiiS AND G. SZEKERES ment in the form that there are at least 2  vertices in S (Cc )) (including y itself) which are not joined with p in G, Gi. The statement is trivial for n = 1; we may therefore assume the Lemma for n  1. By assumption, G, contains no odd circuit. Therefore by Lemma 1 we can divide its vertices into two classes, A and B, such that no two points of A (or of B) are joined in G,. Both classes A and B contain 2  vertices; for otherwise one of them, say A, would contain more than 2 l vertices and CYN)J A would have an even partition Gh +.., J GA into n  1 parts, contrary to Lemma 2. Let A be the class containing p. Hence there are at least 2 l vertices with which p is not joined in G,. This proves the Lemma for i = 1. Suppose i > 1 and consider the partition GA +.,. + GL of UN)1 A, induced by G, $ G, G,. Since the order of UN)1 A is 2 l, we find by the induction hypothesis that there are at least 2n1([i) = 2  vertices in A to which p is not joined in Gi G:. But p is not joined with any vertex of A in G,, therefore it is not joined with at least 2nl vertices in G, +... f G,. In the special case of i = 1 we obtain LEMMA 3.1. Let N = 2n and ) = Gi G, an even partition. Then every p is an endpoint of at least one edge in every Gi. If N < 2n, then Lemma 3.1 is no longer true, but the number of vertices for which it fails cannot exceed 2  N. More precisely we shall prove LEMMA 4. Let N = 2 k, 0 I k < 2, and UN) = G, G, an even partition of UN) into n parts. Denote by Y(P), p E S = S (UN)) the number of graphs Gi in which there is no edge from p. Then ip (29(P)  1) I k. PCS PROOF. For n = 1 the statement is trivial, therefore assume the Lemma for Let ql,..., qj be the exceptional vertices in G, from which there are no edges in G, and denote by Q the union of vertices qi, i = 1,...,. (Q may be empty). By Lemma 1, S is the union of disjoint subsets A, B and 6 such that every edge in G, has one endpoint in A and one in B. Denote by A, the union of A and Q, by B, the union of B and Q. Let a and b be the number of vertices in A and B respectively. Then a + b + i = 2n  k and a + i I 2n1, b + i I 2 l by Lemma 2, applied to UN) IA, and CcN)J Bl. Write a = 2 l  i k,, b = 2+l  j  k, so that k, 2 0, k, 2 0 and 2n  i  k,  k2 = = 2 k 7 (4 k = j + k, + kz. By applying the induction hypothesis to W)JA, and to the partition G;$... + G,: induced by G G,, we find 2 (21*(P)  1) + 2 1) I; k, PEA P?Q for some p(p) 2 v (p). Therefore a fortiori F (2(P) I)+ 2 (2v(Pl11)(x.1 p% PCQ
5 ON SOME EXTREMUM PROBLEMS and similarly Hence 7 (2*(p) 1)  j<k,+k,=kj PT.. by (4), which proves the Lemma. 4. Before proving Theorem 1 we introduce some further notations and definitions. To represent points in the Euclidean plane E we shall sometimes use the complex plane which will also be denoted by E. If ql, p, q2 are points in E, not on one line and in counterclockwise orientation, the angle (< z) formed by the lines pql and pql will be denoted by A (ql p q2). A set of points S in E is said to have the property P, if the angle formed by any three points of S is not greater than (1  l/n) X. We shall briefly say that S is P, or not P, according as it has or has not this property. A direction ti in E is a vector from 0 to elz on the unit circle. An npartition of E with respect to the direction u is a decomposition of E  (0) into sectors T,>, k = 1,...? 2n where T,< consists of all points z = yei(e+o), r > 0, (k  1) 2 < g?<k 2. n n With every set of points S = {pl,..., pn} and every rzpartition of E with respect to some direction w we associate a partition UN) = G, G, of UN) in II parts according to the following rule: pp, py are joined in G, if and only if the vector from p,, to py is in one of the sectors Ti, T,+i. The following Lemma was proved in [3]. LEMMA 5. If the set S is P, then the partition UN) = G, G, associated with any given npartition of E is necessarily even. We shall also need LEMMA 6. If p1 pz... p, are consecutive vertices of a regular ngorz P and q is a point distinct from the centre and inside P then there is a pair of vertices (P,, PJ such that A (p,q Pj) > (1  l/n) n. The proof is quite elementary; if pl is a vertex nearest to q and if q is in the triangle pi pj P,+~ then at least one of the angles A (p, q p,), A (p, q pi1) is >( 1l/n)niT3 Lemma 5 and Lemma 2 give immediately the result that a set of points in the plane cannot be P,. Our purpose, however, is to prove Theorem 1 which can be stated as follows: THEOREM le. A set of 2n points in the plane is not P,. PROOF. Let S be a set of iv = 2 (n > 2) points in the plane, pl, pz,..., pi; the vertices of the least convex polygon of S, in cyclic order and counterclockwise orientation. We can assume that no three vertices of S are collinear, otherwise S is obviously not P,. We distinguish several cases. 3 See Probleln 4056, Amer. Mafh. A4onthl~, 54 (1947), p Solution by C. R. PHELPS.
6 P. ERDOS AND G. SZEKERES Suppose first that there is an angle A (pip1 pi pii1) < (1  l/n) r. Let 01 be the direction pi pi+1 and u (r~) the npartition of E with respect to c(. Then there are no points of S in the sectors T, and Tn.+1 corresponding to a(~), hence there is no edge from the vertex pi in the component G, of the associated partition CtN) = G, I G,. We conclude from Lemma 3.1 that G A G, is not an even partition, hence by Lemma 5, S is not P,. Henceforth we assume that all angles A (pie1 pi JI~+~) are equal to (1l in) x so that the least convex polygon has 2n vertices. For convenience let these vertices be (in cyclic order and counterclockwise orientation) pi q1 p2 qz... pn qn, and denote by ai the side length pi qi and by bj the side qi P~.+.~. Suppose that all angles A (piipipi+l) and A (Qi1qiQi+l) are equal to (12/n) z. Then an elementary argument shows that the triangles pidl Qi1 pi, qi1 piqi, prqipitl etc are similar, hence a,1 b,+, _ uj =  =... ai bi (Ii+1 which implies a, = a, =... = a,, b, = b, =... = b,. We conclude that Pl PL.. * pn is a regular ngon. Now let q be any point of S inside the least convex polygon. If q is inside the triangle pi qi pi+l then clearly A(p, q P!+~) > (1  l/n) R. Therefore we may assume that each q inside the least convex polygon is already inside p1,.. pn. Since n 2 3, there are at least two such points, hence we may assume that q is not the centre. But then by Lemma 6, A (pi q pj) > (1  l/n) n for suitable i and j. The last remaining case to be considered is when not all angles A (pi1pi P~+~) are equal; then for some i, A (pi1 pi P~+~) < (12/n) it. Since A (Pi1 qi1 Pi> = A (Pi qi Pi+d = (11/ 12 ) E, we may assume that there are no points of S inside the triangles pil qia1 pi and pi qi P+~. But then if cx is the direction of pi pi+l and r (CZ) the corresponding npartition of E, UN) = G1+... jg,, the partition of UN) associated with CT (a), then in GnF1 + G, there are only two edges running from pi, namely pi qil and pi qi. By Lemma 3 (with i = 2, jl = n  1, jz = n), UN) = G G, is not an even partition and by Lemma 5, S is not P,. Finally we prove THEOREM 2. In a plane configurafion of N = 2n  there is an angle 2 (11 /n k/2 N) TL k points (0 < k < 2 l) PROOF. Suppose that all angles formed by the points of S are I (11 in) x  f a 6 where 6 = kr/n and 6 >0. Let pe S and y=q(p)=a(q,pq,) I (ll/n)a+y,q,es, qzes the maximum angle at p. If there are several such angles, make an arbitrary but fixed choice.
7 ON SOME EXTREMUM PROBLEMS 61 Let u = CL (p) be the direction of p q1 so that there are no lines pi, q E S in the sectors If S has N = 2  k points, then there are N pairs of directions 4 ~1 (p). Hence there is a such that at least k + 1 directions E,CX (p.), v = 0, 1,.. *, k, F, = 1 1 are in the interval By a slight displacement of p we can achieve that all directions in S should be different from the direction p Consider now the npartition u /? of E with respect to,8 + p 6 I 4i and construct the associated partition ON) = G, G, with the following modification: Every p q with direction between p + $SandB+ S++6 shall be joined in G, instead of G, and every p q with direction between j3 +f n shall be joined in Gz instead of G,. With this modification 4 it is still true tha: the partition UN) = G1 +. a, + G, is even if S has no angle greater or equal to (11 /n) zf 68. But it is easy to see that the vertices p,. are not joined with any vertex of UN) in G,. For if p + <p then the edges emanating fromp, in the sectors T,, Tnfl are count ed to G,; and if p < E, CL (pp) < p + $ 6, the only possible edges from p,, in the sectors T,, T,+l are in directions between p + $ S+Eandj3+ $3+z n n and these are counted to G?. Thus we have a contradiction with Lemma 4 and the Theorem is proved. NOTE ADDED IN PROOF. In the case of k = 1, Theorem 2 can be improved as follows. THEOREM 3. Every plane configuration of 21 points (n 2 2) contains an angle not less than (1  l/n) X. The theorem shows in particular that CI (2N 1) = (11 /n) X, but we cannot decide whether the strict inequality (3) is valid for m = 21.
8 62 P. ERDCiS AND G. SZEKERES: ON SOME EXTREMUM PROBLEMS PROOF. Suppose N (S) = 2n  1 and all angles in S are less than (I 1 In) z. Let q be an interior point of S, that is one which is not on the least convex polygon. Let A (41 q 41) = ( 11 in) x  6, a>0 be the largest angle at q. If p is the direction of qql, a =,8 + i 6, CT (a) the corresponding npartition of E, UN) = G, G, the partition of Oh ) associated with LT (a), then clearly there are no edges from q in G1. We conclude from Lemma 4 and Lemma 5 that from all other vertices of S there is at least one edge in every Gi (i = 1,..., n). We show that this leads to a contradiction. LetP=(p,,.... pm) denote the least convex polygon of S. Since each angle in P is less than (1  I /II) z, we have m I 2 n  1. Therefore if T,,..., T,, are the sectors of CT (a) there is a p, such that if piipi is in T,, then pi plil is not in T,<. But then there is no edge from pi in Gk, as easily seen by elementary geometry. References [I ] K~NIG, D., Theorie der endlichen und unendlichen Graphen, (Leipzig, 1536). (21 ERDBS, I?. and SZEKERES, G., A combinatorial problem in geometry, Composifio Mafh., 2 (1935), [3] SZEKERES, G., On an extremmn problem in the plane, Amer. Journal of Math., 63 (1Q41), 2CB210.
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