Common Secants for Plane Convex Sets1
نویسنده
چکیده
A well known theorem of Helly [l] asserts that if $ is a finite family of convex subsets of En in which each re + 1 members have a common point, then there is a point common to all members of g. In this connection, Vincensini [5; 6] has raised the following question: For «^2 and Ogrgn-1, is there an integer k = k(n, r) such that if § is an arbitrary finite family of convex subsets of En in which each k members have a common r-secant (i.e., an r-dimensional linear variety intersecting all k of them), then there is an r-secant common to all members of $? Helly's theorem shows that for r = 0 the answer is affirmative, with k(n, 0)=re + l. But for 0<rg« — 1 the answer is always negative, for Santalo [3] has remarked that for each k^2 there are & + 1 line segments in E2 which have no common 1-secant even though each k of them do have a common 1-secant; from this it follows easily that the assertion is true with 2 and 1 replaced by m and m — l respectively, and hence that no k(n, r) having the required properties can exist for OO^re — 1. (Santalo's result, stated in [3 ] without proof, is not difficult to prove.) It is still of interest to determine whether the desired k(n, r) may exist under additional restrictions on the family $, and for re = 2 the following is known (where "secant" henceforth means "1-secant")— Suppose 5 is a finite family of plane convex sets, each k of which have a common secant. Then each of the following conditions assures the existence of a secant common to all members of ft: (S) k = 3, and the members of $ are line segments, all of which are parallel; (V) k = 4, and there is a line in the plane none of whose parallels intersects more than one member of jj. ((S) has been proved by Santalo [4] and by Rademacher and Schoenberg [2]. (V) is given by Vincensini [6]. Still other conditions have been given by Santalo [3; 4].) The purpose of this note is to show that (V) remains sufficient when "& = 4" is replaced by uk = 3." The resulting theorem includes both of those just mentioned.
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