A Lower Bound for Heilbronn's Problem

We disprove Heilbronn's conjecture—that N points lying in the unit disc necessarily contain a triangle of area less than c/N. Introduction We think that the best account of the problem can be achieved by simply copying the corresponding paragraphs from Roth's paper [6]. "Let Pv, P2,..., Pn (where n ^ 3) be a distribution of n points in a (closed) disc of unit area, such that the minimum of the areas of the triangles PiPjPk (taken over 1 < i < j < k ^ n) assumes its maximum possible value A = A(n). "Heilbronn conjectured that A(n) c2(n/t)(\ogt) . Remark 1. For any 3-graph one has the Turan-type estimate a > n/(3t), since a random (spanned) subgraph of size n/(2t) expectedly contains n/(24t) edges; delete all vertices in edges (see Spencer [8]). This is sharp up to constant multiple, for the Turan 3-graph (n/t disjoint cliques of size t) has t ~ t/2 and a = 2n/t. Thus the condition that G is uncrowded improves the bound on a by a factor (log t). This is again sharp, as is shown by random 3-graphs (choose nt triples at random and delete the few short cycles). Remark 2. Lemma 1 is an analogue of Lemma 1 in [1] or Theorem 2 in [2], which state that for a 2-graph with average valency t = 2e/n the Turan bound a ^ n(t+\) can be improved to a > (i/\00)(n/t)\ogt if only the graph is trianglefree. The proof will also be analogous to the (complicated) one in [1], which uses random methods, rather than to the simple inductive one in [2], which was thoroughly rewritten and simplified by Joel Spencer. The reader is challenged to give a simple, non-probabilistic proof for Lemma 1. 2. The proof of the theorem If we drop N points to the unit disc at random, then (as will be seen shortly) we can select half of them with smallest triangle c/N (an alternative proof for Erdos' lower bound). We shall improve on this method by dropping N points and then selecting an appropriate subset of N points. Define the numbers t and n by the implicit equations t = n / 1 0 0 , N = c2(n/t)(\ogt) 112 . Set A = (1/200) t/n; then n = -Nt(\ogty' 2 and A = c3(log0/iV 2 = Cl(\ogN)/N 2 . Let us drop n points to the unit disc at random, independently of each other, each with a uniform distribution. We define a 3-graph G on these n points (as vertices) by {a, b, c} € G if the points a, b, c form a triangle of area less than A. Now the probability that three random points form a triangle of area less than A is less than 2 2 — d{rn) = — Inrdr = 32TIA < t/n r J r o 16 JANOS KOML6S, JANOS PINTZ AND ENDRE SZEMERED1 (fix two points at a distance r, and then average over r). Hence the expected number of triangles of area less than A is less than nt/6. Thus the expected value of f is less than t/2. Hence, by Markov's inequality (see §4), with probability greater than 1/2, weget a 3-graph with t < t. Now we show that, with large probability, only o(n) short cycles occur in this 3-graph. All calculations will be based on the simple remark (already used above) that once two vertices have been chosen at a distance r, in order to get a triangle of area less than A, the third point has to belong to a strip of area less than 8A/r. The number of pairs of points at a distance less than d = n~° 6 is, with large probability, less than We discard these points. The number of 2-cycles is, with large probability, less than 2 f A c 4 n 4 —Inrdr < c5t \ogn < n . The number of simple 3-cycles is, with large probability, less than