Fair partitioning by straight lines
aa r X i v : . [ m a t h . M G ] S e p Fair partitioning by straight lines
A. Fruchard and A. Magazinov ∗ Abstract
A pizza is a pair of planar convex bodies A ⊆ B , where B represents the dough and A the topping of the pizza. A partition of a pizza by straight lines is a succession ofdouble operations: a cut by a full straight line, followed by a Euclidean move of one ofthe resulting pieces; then the procedure is repeated. The final partition is said to be fairif each resulting slice has the same amount of A and the same amount of B . This noteproves that, given an integer n ≥
2, there exists a fair partition by straight lines of anypizza (
A, B ) into n parts if and only if n is even. The proof uses the following result: Forany planar convex bodies A, B with A ⊆ B , and any α ∈ ]0 , [ , there exists an α -sectionof A which is a β -section of B for some β ≥ α . (An α -section of A is a straight linecutting A into two parts, one of which has area α | A | .) The question remains open if theword “planar” is dropped. Keywords:
Convex body, alpha-section, fair partitioning.
MSC Classification: 52A10, 52A38, 51M25, 51M04
Let K denote the set of planar convex bodies, endowed with the usual Hausdorf-Pompeiu metric. The area of A ∈ K is denoted by | A | and its boundary is denoted by ∂A .Following [6], what we call a pizza is a pair ( A, B ) of two nested planar convex bodies A ⊆ B ⊂ R . We call A the topping and B the dough . Given a pizza ( A, B ) and aninteger n ≥
2, a fair partition of B in n slices is a family of n internally disjoint convexsubsets B , . . . , B n such that | B | = · · · = | B n | and | A ∩ B | = | A ∩ B | = · · · = | A ∩ B n | . For the sake of clarity, we call pieces the intermediate subsets and slices the final ones.There is a wide literature upon the problem of fair partitioning a convex body, seee.g. [8]. The expressions “equipartition” and “balanced partition” are also used. If thereis no other constraint than to obtain convex slices B i , then it has been proven recentlythat the answer is positive for all n , see e.g. [7, 9, 10].In [2] the authors use k -fans, which are half-lines starting from a common point. Sincethis process is very restrictive, the result is negative for k ≥ B into twopieces with a straight cut. Each of the resulting pieces is a convex body, their interiorsare disjoint, and their union is B . If B is divided into k pieces B , . . . , B k , choose one of ∗ Supported by ERC Advanced Research Grant no. 267165 (DISCONV). hese pieces and divide it into two pieces with one straight cut. After n − B isdivided into n convex slices. We will refer to this rule simply as to the cutting rule ,since no other rule is considered further in this note.Our cutting rule is more restrictive than just partitioning B into n convex bodies. Forexample, a non-degenerate 3-fan partition cannot be obtained with our rule. However,for k ≥
4, our rule becomes somewhat less restrictive than a k -fan partition.Before going further, we need to introduce some notation. The symbol S standsfor the standard unit circle, S := R / (2 π Z ), endowed with its usual metric d ( θ, θ ′ ) =min {| τ − τ ′ | ; τ ∈ θ, τ ′ ∈ θ ′ } . Given θ ∈ S , let ~u ( θ ) denote the unit vector of direction θ , ~u ( θ ) = (cos θ, sin θ ) and let ~u ′ ( θ ) = d~udθ ( θ ) = ( − sin θ, cos θ ).Given an oriented straight line ∆ in the plane, ∆ + denotes the closed half-plane onthe left bounded by ∆, and ∆ − is the closed half-plane on the right. We identify orientedstraight lines with points of the cylinder C = S × R , associating each pair ( θ, t ) ∈ C tothe line oriented by ~u ( θ ) and passing at the signed distance t from the origin. In otherwords, the half-plane ∆ + is given by ∆ + = { x ∈ R ; h x, ~u ′ ( θ ) i ≥ t } . We endow C withthe natural distance d (cid:0) ( θ, t ) , ( θ ′ , t ′ ) (cid:1) = (cid:0) d ( θ, θ ′ ) + | t − t ′ | ) (cid:1) / . The reason to introducethe space C is the following: several times throughout the paper we will say that someoriented line moves continuously. The continuity will always refer to the topology of C .Given α ∈ ]0 ,
1[ and A ∈ K , an α -section of A is an oriented line ∆ such that | ∆ − ∩ A | = α | A | . For all α ∈ ]0 ,
1[ and all θ ∈ [0 , π [, there exists a unique α -section of A of direction θ , denoted by ∆( α, θ, A ). The line ∆( α, θ, A ), treated as a function, depends continuouslyon its three arguments.Our first result has been conjectured in [6]. Theorem 1.
For any planar convex bodies
A, B with A ⊂ B , and any α ∈ ]0 , [ , thereexists an α -section of A which is a β -section of B for some β ≥ α . Proof . By contradiction, if every α -section ∆( α, θ, A ) of A is a β ( θ )-section of B with β ( θ ) < α then, by continuity of θ β ( θ ) and by compactness of S , there exists ε > θ ∈ S , β ( θ ) ≤ α − ε .Choose an integer n > ε . Choose x ∈ ∂A arbitrarily and, for each positive integer i ≤ n , define x i recursively by x i ∈ ∂A and the oriented line D i = ( x i − x i ) is an α -sectionof A . x x x x x A B
Figure 1: Construction of consecutive α -sections We call a cap of B the intersection D − i ∩ B and a cap of A the intersection D − i ∩ A .We thus have n + 1 points on ∂A and n caps of A , resp. B . For each x ∈ B , let K ( x )be the number of caps of B which cover x , i.e. K ( x ) = card { i ∈ { , . . . , n } ; x ∈ D − i } . et k = min { K ( x ) ; x ∈ ∂A } ; it is the number of complete tours made by x , . . . , x n .Observe that, for all x ∈ ∂A we have k ≤ K ( x ) ≤ k + 1 and that, for each 0 < m ≤ n ,the arc \ x m − x m = ∂A ∩ D − m contains K ( x m ) + 1 or K ( x m ) + 2 points among x , . . . , x n (including x m − and x m ). This comes from the fact that, if x i is on the arc \ x m − x m , then x m is on the arc \ x i x i +1 .We claim that K ( x ) ≤ k + 1 for all x ∈ A . Indeed, let x ∈ A and consider a cap A m containing x (if x belongs to no cap, there is nothing to prove). If another cap A i contains x , then x i − or x i must belong to the open arc \ x m − x m \ { x m − , x m } . Now for each ofthe points x j on this open arc, at most one of the caps A j or A j +1 can contain x , hence x is in at most k + 1 caps. It follows that the sum of areas of all caps D − i ∩ A is at most( k + 1) | A | , hence nα ≤ k + 1.Also, for all x ∈ B \ A , we have K ( x ) ≥ k . Hence P ni =1 | D − i ∩ ( B \ A ) | ≥ k | B \ A | .Thus there exists i such that | D − i ∩ ( B \ A ) | ≥ kn | B \ A | ≥ ( α − n ) | B \ A | . It follows that | D − i ∩ B | = | D − i ∩ A | + | D − i ∩ ( B \ A ) | ≥ α | A | + ( α − n ) | B \ A | ≥ ( α − n ) | B | , i.e. D i isa β -section of B , with β ≥ ( α − n ) > α − ε , a contradiction. Remark 2.
A question whether Theorem 1 extends to an arbitrary dimension remainsopen. However, one can show that for every d > α ( d ) > d -dimensional analogue of Theorem 1 holds for all α ∈ ]0 , α [. The idea is similarto the 2-dimensional proof, but instead of an n -fold covering of ∂A by caps we use a 1-foldcovering, namely, the economic cap covering , defined, for example, in [3]. However, thismethod is not very efficient, giving only a very small value of α . Hence we leave thedetails of the proof to the reader.Another equivalent formulation of Theorem 1, which will be more convenient, is asfollows. The proof of the equivalence is easy and left to the reader. Corollary 3.
For any planar convex bodies
A, B with A ⊂ B , and any α ∈ ]0 , [ , thereexists an α -section of B which is a β -section of A for some β ≤ α . Our next result, Theorem 4, concerns a fair pizza partition problem using the cuttingrule . It has been already mentioned in [6] as a consequence of Theorem 1, but withouta proof of implication. Here we give a proof, and thus confirm the result.
Theorem 4.
Let n be a positive integer. Then If n is even, then for every pair A ⊆ B of nested planar convex bodies there exists afair partition obeying the cutting rule . If n is odd, then for some pairs A ⊆ B such a partition may not exist. Proof . It is easy to check that two concentric disks A and B cannot be divided in a fairway into an odd number of slices: The first cut divides the pizza in two pieces, containing k , resp. l final slices, with k + l = n odd, hence k = l , and the smaller piece will not haveenough topping.To construct a fair partition for all even n , we proceed by induction.For n = 2, this follows from the intermediate value theorem. Given n ∈ N , n even ≥
4, and a pair of nested convex bodies A ⊆ B , assume that a fair partitioning exists forany pair of nested convex bodies and any even integer i < n .If n = 4 k , then the intermediate value theorem yields a fair partitioning of two equalhalves, and, by induction hypothesis, each of these halves admits a fair partitioning in 2 k slices.Let n = 4 k + 2. Then we set α = k k +2 and consider two subcases. Suppose that we can cut B into two convex pieces B and B of areas | B | = α | B | , | B | = (1 − α ) | B | so that | A ∩ B | = α | A | , | A ∩ B | = (1 − α ) | A | . hen, by induction, B and B have both a fair partitioning in 2 k , resp. 2 k + 2, slices,and this gives a fair partitioning of B in n slices. If we are not in subcase then no α -section of B contains an α -portion of A . Thenfrom Corollary 3 it follows that each α -section of B (with this α ) is a β -section of A forsome β < α .Cut B into two fair halves B ′ and B ′′ . We claim that there is a cut of B ′ (and,similarly, of B ′′ ) such that it produces a slice of area n | B | with the topping part of area n | A | (i.e., a fair slice).Consider a piece C ⊂ B ′ between two parallel lines, one of which is the initial cut, andthe other one is chosen so that | C | = n | B | . By the construction, B ′ \ C is an α -sectionof B , so | A ∩ ( B ′ \ C ) | < α | A | and hence | A ∩ C | > (cid:0) − α (cid:1) | A | = n | A | .On the other hand, by Corollary 3, there exists a n -section of B ′ , which is at most n -section of A ∩ B ′ . If C is the piece of B ′ obtained by that section, then | C | = n | B | ,and | A ∩ C | ≤ n | A | .Using the intermediate value theorem for n -sections of B ′ , we obtain that there is aslice C , which is cut from B ′ by a single line, such that | C | = n | B | , and | A ∩ C | = n | A | .By induction hypothesis, the piece B ′ \ C admits a fair partition into 2 k slices. Asa result, there is a fair partition of B ′ into 2 k + 1 slices. The same can be done for B ′′ ,yielding a fair partition of the whole pizza. Acknowledgements.
The authors thank Maud Chavent from Plougonver who askedthe question of partition, and Nicolas Chevallier, Costin Vˆılcu, Imre B´ar´any, and AttilaP´or for fruitful discussions.
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58 (2012),71–76. ddresses of the authors:Augustin FruchardLaboratoire de Math`ematiques, Informatique et ApplicationsFacult`e des Sciences et TechniquesUniversit`e de Haute Alsace2 rue des Fr´eres Lumi´ere68093 Mulhouse cedex, FRANCEE-mail: [email protected]
Alexander MagazinovSteklov Mathematical Institute8 Gubkina Str.Moscow 119991, RussiaE-mail: [email protected]@yandex.ru