Periodica Mathematica Hungarica Vol. 60 (1 ), 2010, pp. 41–47 DOI: 10.1007/s10998-010-1041-7
¨ ON A CONJECTURE OF GRUNBAUM CONCERNING PARTITIONS OF CONVEX SETS ´ nimo-Castro2 , L. Montejano3 J. Arocha1 , J. Jero ´n-Pensado4 and E. Rolda 1
Instituto de Matem´ aticas, UNAM, Ciudad de M´exico E-mail:
[email protected]
2
Instituto de Matem´ aticas, UNAM Ciudad de M´exico, and Facultad de Matem´ aticas, Universidad Aut´ onoma de Guerrero E-mail:
[email protected] 3
Instituto de Matem´ aticas, UNAM, Ciudad de M´exico E-mail:
[email protected] 4
Instituto de Matem´ aticas, UNAM, Cuernavaca E-mail:
[email protected]
(Received August 17, 2009; Accepted December 15, 2009)
[Communicated by Imre B´ ar´ any]
Abstract 2 In this paper the following is proved: let √ K ∈ R be a convex body and t ∈ [0, 1/4]. If the diameter of K is at least 37 times the minimum width, then there is a pair of orthogonal lines that partition K into four pieces of areas t, t, (1/2 − t), (1/2 − t) in clockwise order. √ Furthermore, if K is centrally symmetric, then we can replace the factor 37 by 3.
1. Introduction Let K be a convex body in the plane. The following result is well known: there exists a pair of orthogonal lines that partition K into four pieces of equal area. Some decades ago, B. Gr¨ unbaum [1] raised the following problem which has never appeared in print:
Mathematics subject classification number : 52A10. Key words and phrases: Gr¨ unbaum’s conjecture, bisector line, reflections, quadrupartitions of convex sets. Research (partially) supported by CONACYT 41340, and SNI 38848 0031-5303/2010/$20.00
c Akad´emiai Kiad´o, Budapest
Akad´ emiai Kiad´ o, Budapest Springer, Dordrecht
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´ ´ J. AROCHA, J. JERONIMO-CASTRO, L. MONTEJANO and E. ROLDAN-PENSADO
Conjecture. Let K be a convex body in the plane with area 1. Then, for each t ∈ [0, 1/4] there exists a pair of orthogonal lines that partition K into four pieces of areas t, t, (1/2 − t), (1/2 − t) in clockwise order. If this is the case for a specific K, we then say that K satisfies Gr¨ unbaum’s conjecture. Using some continuity arguments, it is not difficult to prove the cases t = 0 and t = 1/4, however, we cannot say the same for any other different value of t. It seems that topology or algebraic topology are not enough to prove the conjecture, however, we are able to give a proof for a special kind of convex bodies. That is, we prove that the conjecture is true for convex bodies which are thin enough.
2. Results In order to give the statement of the main results of this paper we introduce some definitions and notation. Given a convex body K we denote by w, w(v), and diam(K), the minimum width of K, the width of K in the direction of the unit vector v, and the diameter of K, respectively. We say that a line is a bisector line of K if it divides K into two pieces of equal areas and we denote by K the reflection of K about . We say that the pair (K, ) satisfies the reflection property if is a bisector line and the boundaries of K and K intersect each other at exactly four points. In what follows, let K be a convex body in the plane with area 1. First we prove the case t = 0 for an arbitrary convex body, and moreover, we prove that the intersection point of the lines partitioning K can be taken on K. Lemma 1. There exists a pair of orthogonal lines and such that their intersection point is contained in K and they divide K into four parts of areas 0, 0, 1/2, 1/2 in clockwise order. Second, we prove Gr¨ unbaum’s conjecture for a certain class of thin enough convex bodies. Lemma 2. Let v ∈ S1 be a unit vector such that w(v) = w and let v ∈ S1 be unbaum’s conjecture. the vector such that v ⊥ v. If w(v ) ≥ 6 · w, then K satisfies Gr¨ Using Lemma 2 we derive the following: √ unbaum’s conjecture. Theorem 1. If diam (K) ≥ 37 · w, then K satisfies Gr¨ In the case when K is centrally symmetric we can greatly improve these results. This can be done because the set of bisector lines of K are all concurrent at the centre of K. This allows us to use a better argument.
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Lemma 3. Let K be a centrally symmetric convex body and a line such that the pair (K, ) has the reflection property. Then K satisfies Gr¨ unbaum’s conjecture. In this case we can also give conditions on the width and diameter of K so that K satisfies Gr¨ unbaum’s conjecture. Concretely, we have the following: Theorem 2. Let K be a centrally symmetric convex body such that diam(K) ≥ 3 · w. Then K satisfies Gr¨ unbaum’s conjecture.
3. Proofs Given a planar convex body L and a curve γ ⊂ bd L, we denote by S(γ, L) the set of unit vectors such that there is a support line of L at a point of γ in that direction. Clearly, S(γ, L) consists of two symmetric arcs on S1 . Let γ1 ⊂ bd L1 and γ2 ⊂ bd L2 be two curves. It is easy to see that if S(γ1 , L1 ) and S(γ2 , L2 ) are disjoint then γ1 and γ2 cannot intersect each other at more than one point. Furthermore, given a unit vector v ∈ S1 , there is a unique line parallel to v such that is a bisector line of K. Now, for every point x ∈ R2 we consider the closed half-plane Hx with inner normal vector v and such that its boundary line passes through x. Define f (v, x): S1 × R2 −→ R as the difference between the area of the part of K ∩ Hx to the right of and the part to the left, with respect to v. Similarly we define g(v, x) as the area of K ∩ Hx . It is clear that f and g are continuous functions in the connected set where they are defined. Finally, given any two vectors v1 and v2 we define (v1 , v2 ) as the arc (v1 / v1 , v2 / v2 ) ⊂ S1 in the counterclockwise order. Proof of Lemma 1. Since every convex body can be approximated by convex bodies with smooth boundary, we need only to consider the case when K has smooth boundary. So, we may assume that K has smooth boundary. For every point x ∈ bd K let vx be the inner unit normal vector of K through x and x the line spanned by vx . Also, let x be the unique support line of K through x. Let [p, q] be a diameter of K. We know that p and q are supporting lines of K at p and q, respectively. It is clear that p and q are parallel to each other and both are orthogonal to [p, q], since it is a diameter. Since f (vp , p) = −f (vq , q), we have two possibilities: either f (vp , p) = 0 or f (vp , p) = A, where A is a real number different of 0. In the first case we are done. In the second case we have that there is a point z in the arc pq such that f (vz , z) = 0, because f is continuous. Obviously, z and z satisfy the desired property.
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´ ´ J. AROCHA, J. JERONIMO-CASTRO, L. MONTEJANO and E. ROLDAN-PENSADO
Proof of Lemma 2. In view of Lemma 1 and the comment in the introduction, we need only to consider t ∈ (0, 1/4). Let abcd be the rectangle circumscribed to K, with |bc| = |da| = w, where |xy| denotes the length of the segment [x, y]. Consider the points a , b ∈ [a, b] and d , c ∈ [d, c] such that aa d d and b bcc are −→ −→ squares. Now, let v, v , and v0 , be the unit vectors with directions ad , a d, and −→ aa , respectively. Also, let 1 , 2 be the lines parallel to v passing trough a and c , respectively. Since we know that w(v0 ) ≥ 6 · w, the area of K to the left of 1 is at most 23 (w)2 ≤ 12 w(v0 ) · w ≤ 12 |K|. Similarly, the area of K to the right of 2 is at most 12 |K|. Hence there is a bisector line of K parallel to v between the lines 1 and 2 .
Figure 1
The boundary of K contained in the rectangle a b c d can be expressed as the union of two disjoint curves γ1 and γ2 . It is clear that S(γ1 , K) and S(γ2 , K) are contained in the set (v , −v) ∪ (−v , v). If we reflect K about the line , then the intersection points of bd K and bd K are contained in γ1 ∪ γ2 . Since S(γ1 , K) ⊂ {(v, v ) ∪ (−v, −v )} and S(γ2 , K) ⊂ {(v, v ) ∪ (−v, −v )}, where γ1 and γ2 are the reflected images of γ1 and γ2 , we have that the pair (K, ) has the reflection property. Now, let p, q, r, and s be the intersection points of bd K with bd K , as shown in Figure 2. Since s and q are reflected images of each other, the line sq is perpendicular to v. Without loss of generality we may assume that the upper half-space bounded by sq intersects K in a region of area at least 21 |K|. So, we can choose x ∈ in this upper half-space such that f (v, x) > 0 and g(v, x) = 2t. Analogously, we can find a line orthogonal to and a point x ∈ such that f (v , x ) < 0 and g(v , x ) = 2t. We conclude, by continuity, that there is a unit vector v0 and a point x0 such that f (v0 , x0 ) = 0 and g(v0 , x0 ) = 2t. Therefore, the line parallel to v0 through x0 and the line perpendicular to v0 through x0 divide K into four parts of areas t, t, (1/2 − t), (1/2 − t) in clockwise order.
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Figure 2 Proof of Theorem 1. Consider v ∈ S1 as defined in Lemma 2 and also consider the rectangle abcd defined in the √ previous proof. If we have that diam(K) ≥ √ 37 · w, we then also have that |ac| ≥ 37 · w. It follows that |ab|2 = w(v )2 = |ac|2 − |bc|2 ≥ 37 · w2 − w2 = 36 · w2 , so Lemma 2 applies and therefore K satisfies Gr¨ unbaum’s conjecture.
Proof of Lemma 3. Let O be the centre of K. Since K is centrally symmetric, every bisector line of K passes through O. Let be the line such that (K, ) has the reflection property and let be the line perpendicular to through O. It is easy to see that (K, ) also has the reflection property. From here, the proof is very similar to the last paragraphs of the proof of Lemma 2.
Proof of Theorem 2. It is sufficient to prove that if diam(K) ≥ 3 · w, then there is a line such that (K, ) has the reflection property. Let O be the centre of K and a, b, c, d be points in the boundary of K such that |ac| = diam(K) and |bd| = w. We define θ = aOb and suppose that θ ≤ 90◦ . Instead of proving the assertion of the theorem, we prove the following, a little more general, statement: If |ac| · sin2 (θ/2) ≥ |bd|, then K satisfies the reflection property.
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´ ´ J. AROCHA, J. JERONIMO-CASTRO, L. MONTEJANO and E. ROLDAN-PENSADO
Figure 3 Let be the internal angle bisector of aOb, and a and c the reflections of a and c about . The perpendicular lines to bd through b and d (which we know to be support lines of K) and their reflections about determine a parallelogram pqrs as shown in Figure 3. Note that |Oa|·cos(θ/2) is equal to the distance from O to the line aa and that |Op| = |Ob| · cos−1 (θ/2). Since |Oa| · cos2 (θ/2) ≥ |Oa| · sin2 (θ/2) ≥ |Ob| we have that |Oa| · cos(θ/2) ≥ |Ob|, then p is contained in the triangle a Oa. Now, using that |Oq| = |Ob| · sin−1 (θ/2), we may prove that q is contained in the triangle
a Oc. It follows that the parallelogram pqrs is contained in the parallelogram aa cc . The part of bd K contained in pqrs can be expressed as the disjoint union of two curves γ1 and γ2 . We have that S(γ1 , K) and S(γ2 , K) are contained in − → −→ − → − → (Os, Op) ∪ (Oq, Or), and since bd K and bd K intersect each other inside pqrs, we conclude that (K, ) has the reflection property. Now, for the particular case established in the theorem we proceed as follows: As we know that |Oa| ≥ 3 · |Ob| and aOb < π/2, we have that cos(θ) < |Ob|/|Oa|. It follows that θ 1 1 |Ob| 1 = (1 − cos(θ)) > 1− ≥ . sin2 2 2 2 |Oa| 3 Therefore, we conclude that θ θ 1 ≥ |ac| · sin2 ≥ |ac| ≥ |bd|, |ac| · cos2 2 2 3 which implies that K has the reflection property, hence it satisfies Gr¨ unbaum’s conjecture.
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3.1. Acknowledgements We thank the referee, Attila P´ or, for noticing and telling us that the idea of the proof for centrally symmetric bodies, after some simple modifications, is still valid for not necessarily centrally symmetric bodies. References ´ NY and B. GRU ¨ NBAUM, Quadrupartitions, Building Bridges II, G. O. H. [1] I. B´ ARA Katona and M. Dezs˝ o (eds.), Springer, to appear.
