ABSTRACT. In this paper we consider families of distinct ovals in the plane, with the property that certain subfamilies have stabbing lines (transversals).
A. B E Z D E K ,
K. B E Z D E K A N D T. B I S Z T R I C Z K Y
ON THE (n-2)-TRANSVERSALS
OF n CONVEX
SUBSETS
OF THE PLANE
ABSTRACT.In this paper we consider families of distinct ovals in the plane, with the property that certain subfamilies have stabbing lines (transversals). O u r main result says that if any k member of the family can be stabbed by a line avoiding all the other ovals and k is large enough, then the family consists of at most k + 1 ovals. For any n/> 4 we show a family of n ovals, whose n - 2 element subfamilies have, but the n - 1 element subfamilies do not have, transversals.
be a Euclidean plane with points p, q . . . . and lines L, M . . . . . An oval is a compact convex subset of E 2, and we denote ovals by A, B , . . . . Let ~ be a family of (mutually) disjoint ovals, and ~ ~ ~ ' ~ ~-. We set Let E 2
F'=conv( U A ) \ AE~'
/
and say that A ' E ~ ' is a vertex of F' if A' ¢ conv(UA~,\{A,}A ). If F' has exactly k vertices, we say that F' is a k-gon and that ~-' determines a k-gon. Three pairwise disjoint ovals are collinear if the convex hull of their union is a 2-gon. Finally, a line is a transversal of ~ if it intersects each member of ~-, and it is a k-transversal of ~ if it is a transversal of a k-element subset of ~ . Let k s Z ÷, the set of positive integers. Then ~ m a y have the following properties:
Gk: Any
k-element subset of ~ has a transversal which is not a ( k + 1)transversal of ~ . Hk: For any ~ ~ ~-' ~ ~ , F' is at most a k-gon. Jk: There is a k-transversal of ~ . Tk: Any k-element subset of ~ has a transversal. Let I~1 = n ~> 3. Then J, and T, are equivalent. We recall from [4] that, independent of the size of n, T._ 1 does not imply Jn(T.). It is now natural to ask if T,_ 2 implies J,-1, say, for large n. If yes, does T,_ 2 imply T._I? We remark that if ~ satisfies T,_ 2 but not J , _ 1 then ~ has the property G,-2. Let m=n--2 and let ~-* be any family of disjoint ovals with the property Gin. It was shown in [2] that I~*1 ~< m + 4 6 . In [1], this bound was reduced to m + 2 for m ~> 12. Since I~1 = n = m + 2, the answer to the first question is yes if, in fact, I~*1 ~< m + 1 for sufficiently large m. Geometriae Dedicata 40:. 263-268, 1991. © 1991 K l u w e r Academic Publishers. Printed in the Netherlands.
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A. BEZDEK
ET AL.
We state our results. T H E O R E M 1. Let ~ be a family of n disjoint ovals in E 2. I f ~ has the property G,_ 2, then there is an integer fi, independent of n, such that n < h. C O R O L L A R Y 1. Let ~ be a family of disjoint ovals in E 2 with the property G~. Then Io~l .< m + 1 for m >>.~. C O R O L L A R Y 2. Let o~ be a family of n >1 fi disjoint ovals in E 2 with the property T~_ 2. Then o~ satisfies J,_ 1. T H E O R E M 2. For any n ~ 7/+, n >1 4, there exists a family ~ of n disjoint ovals in E 2 which satisfies T~_ 2 but not T~_ I. Proof of Theorem 1. Let o~ = {A 1. . . . . A,} be a family of n ~> 4 disjoint ovals in E 2 with the property Gn_ 2" We may suppose that n ~> 4. We prove (ii) and (iii), and then obtain h by Ramsey's theorem and the following result from [3]: (i) For any k e 77+, k/> 4, there is a smallest integer g(k) such that if o~- is a family of disjoint ovals in E 2, n o three of which are collinear, and I ~ l > g(k), then some k-element subset of ~" determines a k-gon. For A i # Aj in o~, let B o = c o n v ( A i u A j) and Lii be a transversal of o~\{Ai, A j} which is disjoint from A~ u2 Aj. The property Gn_ 2 implies that if one freely chooses three members of o~ and distinguishes two of them, then there is a line meeting exactly the distinguished two sets. Thus, (1)
F \ A k and
Bij\A k a r e connected sets for any A k ~ .
(ii) o~ has the property H s. We assume that there is a o ~ ' = {At,... ' A6} _c ~- such that F' is a 6-gon, and seek a contradiction. From (1), it follows that int F ' # ~ , ~ = bd F' is a convex curve and (6~C3 A k is connected for A k ~ o~'. We orient (ff and assume that (up to reversal) ~f meets the vertices o f F ' in the cyclic order At, A 2 . . . . . A 6 , A t. Let li --j] ~ 1 (mod 6). Then there exist A k and Am in ~ ' such that ~f c~ A k and ~ n Am are in distinct components of F ' \ B o. Since Lq meets Ak and Am, it follows from (1) that Lo strictly separates A~ and Aj. We now consider Lla and L,,6. Since L46 meets A1 and Aa, and L13 strictly separates the two sets, we have that Lt3 c~ L46 is a point q and
(2)
q~B~3\{At u a3}.
Similarly, we have that (3)
q ~ B,6\{A,, u A6}.
Since L t 3 strictly separates the sets A ~ and A3, and since L46 strictly separates
TRANSVERSALS OF CONVEX SUBSETS
265
the sets A# and A6, it follows from (2) and (3) that cg meets the vertices A 1, A3, A 4 and A 6 of F' in the cyclic order A4, A3, A6, A1, a contradiction. [] (iii) N o eight element subset of ~ has the property H E. We assume that {A1 . . . . . As} _ ~ has the property H2, and seek a contradiction. We note first that HE yields that there are three sets, say, At, A 2 and A 3 such that (4)
A 2 c B13
and
A i ¢ B13
for i = 4 , . . . , 8 .
Then L13 strictly separates A~ and A 3. Since int B13 56 ~ , there exists distinct lines M and N which support each of AI, A 3 and B13. Let M* = M n B13
and
N' = N c3 B13.
We note that L13 intersects int B13 , M* and N'. Let LI*3 and L't3 be the components L~3\(int B13 ) such that L* 3 meets M* and L'~3 meets N', and let L~3 N M* = p* and L~13~ N' = p'. Then L13 = L~'a u p*p' u L'13 where p*p' = conv{p*, p'}. Finally, let
~ * = {AiIA in(M*UL~'3)¢ ~
and
4~
