Balanced partitions of colored geometric sets in the plane^{†}^{†}thanks: This paper was published in Discrete Applied Mathematics, 181:21–32, 2015 [8].
Abstract
Let be a finite set of geometric objects partitioned into classes or colors. A subset is said to be balanced if contains the same amount of elements of from each of the colors. We study several problems on partitioning colored sets of points and lines in the plane into two balanced subsets: (a) We prove that for every 3colored arrangement of lines there exists a segment that intersects exactly one line of each color, and that when there are lines of each color, there is a segment intercepting lines of each color. (b) Given red points, blue points and green points on any closed Jordan curve , we show that for every integer with there is a pair of disjoint intervals on whose union contains exactly points of each color. (c) Given a set of red points, blue points and green points in the integer lattice satisfying certain constraints, there exist two rays with common apex, one vertical and one horizontal, whose union splits the plane into two regions, each one containing a balanced subset of .
1 Introduction
Let be a finite set of geometric objects distributed into classes or colors. A subset is said to be balanced if contains the same amount of elements of from each of the colors. Naturally, if is balanced, its complement is also balanced, hence we talk of a balanced bipartition of .
When the point set is in the plane, and the balanced partition is defined by a geometric object splitting the plane into two regions, we say that is balanced (and nontrivial if both regions contain points of ). A famous example of such a partition is the discrete version of the hamsandwich theorem: given a set of red points and blue points in general position in the plane, there always exists a line such that each halfplane bounded by contains exactly red points and blue points. It is well known that this theorem can be generalized to higher dimensions and can be formulated in terms of splitting continuous measures.
There are also plenty of variations of the hamsandwich theorem. For example, it has been proved that given red points and blue points in the plane in general position, there exists a subdivision of the plane into disjoint convex polygons, each of which contains red points and blue points [10]. Also, it was shown in [5] (among other results) that for any two measures in the plane there are rays with common apex such that each of the sectors they define contains of both measures. For many more extensions and detailed results we refer the interested reader to [1, 2], the survey [12] of Kaneko and Kano and to the book [14] by Matoušek.
Notice that if we have a colored set of points in the plane, it is possible that no line produces any nontrivial balanced partition of . Consider for example an equilateral triangle and replace every vertex by a very small disk (so that no line can intersect the three disks), and place red points, green points, and blue points, inside the disks , and , respectively. It is clear for this configuration that no line determines a halfplane containing exactly points of each color, for any value of with .
However, it is easy to show that for every colored set of points in the plane there is a conic that simultaneously bisects the three colors: take the plane to be in , lift the points vertically to the unit paraboloid , use the 3dimensional hamsandwich theorem for splitting evenly the lifted point set with a plane , and use the projection of as halving conic in . On the other hand, instead of changing the partitioning object, one may impose some additional constraints on the point set. For example, Bereg and Kano have recently proved that if all vertices of the convex hull of have the same color, then there exists a nontrivial balanced line [9]. This result was recently extended to sets of points in a space of higher dimension by Akopyan and Karasev [3], where the constraint imposed on the set was also generalized.
Our contribution. In this work we study several problems on balanced bipartitions of colored sets of points and lines in the plane. In Section 2 we prove that for every 3colored arrangement of lines, possibly unbalanced, there always exists a segment intersecting exactly one line of each color. If the number of lines of each color is exactly , we show that there is always a segment intersecting exactly lines of each color. The existence of balanced segments in 3colored line arrangements is equivalent, by duality, to the existence of balanced double wedges in colored point sets.
In Section 3 we consider balanced partitions on closed Jordan curves. Given red points, blue points and green points on any closed Jordan curve , we show that for every integer with there is a pair of disjoint intervals on whose union contains exactly points of each color.
In Section 4 we focus on point sets in the integer plane lattice ; for simplicity, we will refer to as the lattice. We define an line with corner as the union of two different rays with common apex , each of them being either vertical or horizontal. This line partitions the plane into two regions (Figure 7). If one of the rays is vertical and the other ray is horizontal, the regions are a quadrant with origin at and its complement. Note, however, that we allow an line to consist of two horizontal or two vertical rays with opposite direction, in which case the line is simply a horizontal or vertical line that splits the plane into two halfplanes. An line segment can be analogously defined using line segments instead of rays.
lines in the lattice play somehow a role comparable to the role of ordinary lines in the real plane. An example of this is the result due to Uno et al. [18], which extends the hamsandwich theorem to the following scenario: Given red points and blue points in general position in , there always exists an line that bisects both sets of points. This result was also generalized by Bereg [7]; specifically he proved that for any integer and for any red points and blue points in general position in the plane, there exists a subdivision of the plane into regions using at most horizontal segments and at most vertical segments such that every region contains red points and blue points. Several results on sets of points in , using lines or line segments are described in [13].
A set is said to be orthoconvex if the intersection of with every horizontal or vertical line is connected. The orthogonal convex hull of a set is the intersection of all connected orthogonally convex supersets of S.
Our main result in Section 4 is in correspondence with the result of Bereg and Kano [9] mentioned above that if the convex hull of a colored point set is monochromatic, then it admits some balanced line. Specifically, we prove here that given a set of red points, blue points and green points in general position (i.e., no two points are horizontally or vertically aligned), whose orthogonal convex hull is monochromatic, then there is always an line that separates a region of the plane containing exactly red points, blue points, and green points from , for some integer in the range .
We conclude in Section 5 with some open problems and final remarks.
2 3colored line arrangements
Let be a set of lines in the plane, such that and are pairwise disjoint. We refer to the elements of , , and as red, green, and blue, respectively. Let be the arrangement induced by the set . We assume that is simple, i.e., there are no parallel lines and no more than two lines intersect at one point. In Section 2.1 we first prove that there always exists a face in that contains all three colors. We also extend this result to higher dimensions. We say that a segment is balanced with respect to if it intersects the same number of red, green and blue lines of . In Section 2.2 we prove that (i) there always exists a segment intersecting exactly one line of each color; and (ii) if the size of each set and is , there always exists a balanced segment intersecting lines of each color. As there are standard duality transformations between points and lines in which segments correspond to double wedges, the results in this section can be rephrased in terms of the existence of balanced double wedges for colored point sets.
2.1 Cells in colored arrangements
In this section we prove that there always exists a colored face in , that is, a face that has at least one side of each color. In fact, we can show that a dimensional arrangement of dimensional hyperplanes, where each hyperplane is colored by one of colors (at least one of each color), must contain a colored cell. This result is tight with respect to the number of colors. If we have only colors, then every cell containing the intersection point of hyperplanes with different colors is colored. On the other hand, it is not difficult to construct examples of arrangements of hyperplanes with colors where no colored cell exists.
An example in the plane is shown in Figure 1. Start with a triangle in which each side has a different color (red, green, or blue), and extend the sides to the colored lines , and that support the sides (Figure 1, left). Then shield each of these lines by two black parallel lines, one on each side (Figure 1, right). Finally perturb the black lines in such a way that the arrangement becomes simple, yet the intersection points between former parallel lines are very far away. Now it is easy to see that no cell can contain all four colors. For example, depending on the specific intersections of the black lines with and , the region may contain the colors green and blue, but cannot contain color red.
This example can be generalized to dimensional space. Start with a simplex in which each of the hyperplanes supporting a facet has a different color, . Then shield each facet with two parallel hyperplanes having color , one on each side, and perturb as above to obtain a simple arrangement in which no cell is colored.
For intuition’s sake, before generalizing the result to higher dimensions, we first prove the result for . Consider the dimensional arrangement as a graph. The dual of a face of is a face that contains a vertex for every bounding line of , and contains an edge between two vertices of if and only if the intersection of the corresponding lines is part of the boundary of (see Figure 2(a)).
Let be a simple cycle of vertices where each vertex is colored either red, green, or blue. Let and be the number of red, green, and blue vertices of , respectively. We simply write , , and if is clear from the context. The type of an edge of is the multiset of the colors of its vertices. Let , , , , , and be the number of edges of the corresponding type. Note that, if is bounded, then is a simple cycle, where each vertex is colored either red, green, or blue. We say a bounded face is complete if holds for .
Lemma 1
Consider a simple cycle in which each vertex is colored either red, green, or blue. Then .
Proof: The result follows from double counting. We can obtain an expression for (twice) the number of vertices of a certain color by summing up over all edges that have vertices of that color. For instance, for we get the equation . This directly implies that . By repeating the same process for the other colors we obtain the claimed result.
Theorem 1
Let be a set of 3colored lines in the plane inducing a simple arrangement , such that each color appears at least once. Then there exists a complete face in .
Proof: The result clearly holds if . For the general case, we start with one line of each color, and then incrementally add the remaining lines, maintaining a complete face at all times. Without loss of generality, assume that a red line is inserted into . If does not cross , we keep . Otherwise, is split into two faces and (see Figure 2(b)). Similarly, is split into and (with the addition of one red vertex, see Figure 2(c)). Because is red, the number of greenblue edges does not change, that is, . This implies that either or is odd. By Lemma 1 it follows that either or is complete.
We now extend the result to higher dimensions. For convenience we assume that every hyperplane is colored with a “color” in . Consider a triangulation of the surface of the dimensional sphere , where every vertex is colored with a color in . Note that a triangulation of is exactly a simple cycle. As before, we define the type of a simplex (or face) of as the multiset of the colors of its vertices. Furthermore, let be the number of simplices (faces) with type . We say a type is good if does not contain duplicates and . The following analogue of Lemma 1 is similar to Sperner’s lemma [4].
Lemma 2
Consider a triangulation of , where each vertex is colored with a color in . Then either for all good types , or for all good types .
Proof: We again use double counting, following the ideas in the proof of Lemma 1. Consider a subset with (thus includes all but two colors from ). Recall that good types consist of different colors, so there are exactly two good types and that contain : one with each of the two colors of that are not already present in .
Since every dimensional face of is present in exactly two dimensional simplices, we can obtain an expression for (twice) by summing over all face types that contain :
where the summation on the right is done over the types that contain but are not good (the symbol denotes the disjoint union, to allow duplicated colors in the face type).
From the above equation we obtain that . By repeating this procedure for every set , we obtain the claimed result.
Consider a cell of a dimensional arrangement of dimensional hyperplanes. The dual of contains a vertex for every bounding hyperplane of , and contains a simplex on a set of vertices of if and only if the intersection of the corresponding hyperplanes is part of the boundary of . Note that, if is bounded, then is a triangulation of . We say a bounded cell is complete if for all good types . Note that a complete cell is colored.
Theorem 2
Let be a set of colored hyperplanes in inducing a simple arrangement , such that each color appears at least once. Then there exists a complete face in .
Proof: The proof is analogous to the twodimensional case: if there is exactly one hyperplane of each color, then the arrangement has exactly one bounded face , which must be complete (this can be easily shown by induction on ). We maintain a complete face during successive insertions of hyperplanes. Assume we add a hyperplane with color . If does not cross , we can simply keep . Otherwise, is split into two faces and , and is split into and (with the addition of one vertex of color ). Let . Because has color , the number of simplices with type does not change, that is, . This implies that either or is odd. By Lemma 2 it follows that either or is complete.
An immediate consequence of Theorem 2 is the following result:
Corollary 1
Let be a colored set of hyperplanes in inducing a simple arrangement , such that each color appears at least once. Then there exists a segment intersecting exactly one hyperplane of each color.
Proof: Consider a colored cell. By Theorem 2 such a cell must exist and it must also contain an intersection of hyperplanes with different colors. Now we can take the segment from this intersection to a face of the remaining color (in the same cell). By perturbing and slightly extending this segment, we obtain a segment properly intersecting exactly one hyperplane of each color.
2.2 colored point sets and balanced double wedges
We now return to the plane and consider colored point sets. By using the pointplane duality, Corollary 1 implies the following result.
Theorem 3
Let be a 3colored set of points in in general position, such that each color appears at least once. Then there exists a double wedge that contains exactly one point of each color from .
Proof: We apply the standard duality transformation between points and nonvertical lines where a point is mapped to a line with equation , and vice versa [11]. By Corollary 1, there exists a segment that intersects exactly one line of each color. By standard pointline duality properties, the dual of is a double wedge that contains the dual points of the intersected lines.
Since the dual result extends to higher dimensions, so does the primal one. The equivalent statement says that given a set of points colored with colors in , there exists a pencil (i.e., a collection of hyperplanes sharing an affine subspace of dimension ) containing exactly one point of each color.
Next we turn our attention to balanced colored point sets, and prove a hamsandwichlike theorem for double wedges.
Theorem 4
Let be a 3colored balanced set of points in in general position. Then there exists a double wedge that contains exactly points of each color from .
Proof: We call a double wedge satisfying the theorem bisecting. Without loss of generality we assume that the points of have distinct coordinates and distinct coordinates. For two distinct points and in the plane, let denote the line passing through them. Consider the arrangement of all the lines passing through two points from , i.e.
Consider a vertical line that does not contain any point from . We continuously walk on from to . For any point we define an ordering of as follows: consider the lines and sort them by (increasing values) of slope. Let be the obtained ordering (see Figure 3).
By construction, any consecutive interval of an ordering of corresponds to a set of lines whose points can be covered by a double wedge with apex at (even if the indices are taken modulo ). Likewise, for any , any double wedge with apex at will appear as an interval in the ordering .
Given an ordering of , we construct a polygonal curve as follows: for every let and be the number of blue and green points in the set of points, respectively. We define the corresponding lattice point , and the polygonal curve .
Intuitively speaking, the point indicates how balanced the interval is that starts with point and contains points. By construction, if for some and some , then the associated wedge is balanced (and vice versa). In the following we show that this property must hold for some and . We observe several important properties of :

Path is centrally symmetric (w.r.t. the origin). This follows from the definition of .

Path is a closed curve. Moreover, the interior of any edge of cannot contain the origin: consider the segment between two consecutive vertices and of . Observe that the double wedges associated to and share points. Thus, the orientation and length of the segment only depend on the color of the two points that are not shared. In particular, there are only types of such segments in , see Figure 4. Since these segments do not pass through grid points, the origin cannot appear at the interior of a segment.

If the orderings of two points and are equal, then their paths and are equal. If the orderings and are not equal then either there is a line of separating and or there is a point such that the vertical line passing through separates and .
In our proof we will move along a vertical line, so case will never occur. Consider a continuous vertical movement of , and consider the two orderings and before and after a line of is crossed. Observe that the only difference between the two orderings is that two consecutive points (say, and ) of are reversed in . Thus, the only difference between the two associated curves will be in vertices whose associated interval contains one of the two points (and not the other). Since these two points are consecutive, this situation can only occur at vertices and (recall that, for simplicity, indices are taken modulo ). Since the predecessor and successor of these vertices are equal in both curves, the difference between both curves will be two quadrilaterals (and two more in the second half of the curve).
We claim that the interior of any such quadrilateral can never contain the origin. Barring symmetries, there are two possible ways in which the quadrilateral is formed, depending on the color of and (see Figure 5). Regardless of the case, the interior of any such quadrilateral cannot contain any lattice point, and in particular cannot contain the origin.

Path has a vertex such that , or the curve has nonzero winding with respect to the origin. Intuitively speaking, the winding number of a closed curve with respect to a point measures the net number of clockwise revolutions that a point traveling on makes around the given point (see a formal definition in [17]). By Properties 2 and 3, the only way in which passes through the origin is through a vertex . If this does not happen, then no point of the curve passes through the origin. Recall that is a closed continuous curve that contains the origin and is centrally symmetric around that point. In topological terms, this is called an odd function. It is well known that these functions have odd winding (see for example [6], Lemma 25). In particular, we conclude that the winding of cannot be zero.
Thus, imagine a moving point from to : when is located sufficiently low along , the coordinates of the points of can be ignored, and the resulting order will give first the points to the left of (sorted in decreasing value of the coordinates) and then the points to the right of (also in decreasing value of the coordinates). Similarly, the order we obtain when is sufficiently high will be the exact reverse (see Figure 6).
Now consider the translation of from both extremes and the changes that may happen to the ordering along the translation. By Property 2, the curve will change when we cross a line of , but the differences between two consecutive curves will be very small. In particular, the space between the two curves cannot contain the origin. Consider now the instants of time in which point is at and : if either curve contains the origin as a vertex, we are done (since such vertex is associated to a balanced double wedge). Otherwise, we observe that the orderings must be the reverse of each other, which in particular implies that the associated curves describe exactly the same path, but in reverse direction. By Property 4 both curves have nonzero winding, which in particular it implies that they have opposite winding numbers (i.e. their winding number gets multiplied by 1). Since the winding must change sign, we conclude that at some point in the translation the curve passed through the origin (Otherwise, by Hopf’s degree Theorem [15] they would have the same winding). By Properties 2 and 3 this can only happen at a vertex of , implying the existence of a balanced double wedge.
Using the pointline duality again, we obtain the equivalent result for balanced sets of lines.
Corollary 2
Let be a colored balanced set of lines in inducing a simple arrangement. Then, there always exists a segment intersecting exactly lines of each color.
3 Balanced partitions on closed Jordan curves
In this section we consider balanced colored point sets on closed Jordan curves. Our aim is to find a bipartition of the set that is balanced and that can be realized by at most two disjoint intervals of the curve. To prove the claim we use the following arithmetic lemma:
Lemma 3
For a fixed integer , any integer can be obtained from by applying functions and at most times.
Proof: Consider a fixed value . In the following we show that for some where each is either or .
For the purpose, we use the concept of starting points: we say that an integer is an starting point (with respect to ) if number can be obtained from by applying functions from at most times. Note that any number is always a starting point with respect to itself, and our claim essentially says that is a starting point with respect to any .
Instead of explicitly computing all starting points, we compute a consecutive interval of starting points. For any , let and be the left and right endpoints of the interval , respectively. This interval is defined as follows: initially, we set . For larger values of , we use an inductive definition: If we apply as the first operation. That is, any number that, after we apply , falls within should be in . Observe that this implies and . If we apply as the first operation. In this case, we have and . By construction, the fact that all elements of are starting points implies that elements of are starting points.
This sequence will finish at some index such that (that is, is a starting point). In particular, we have that is a starting point since . Thus, to conclude the proof it remains to show that . Each time we use operator as the first operation, the length of the interval is doubled. On the other hand, each time we use operator , the length of the interval does not change. Moreover, function is never applied twice in a row. Thus, after at most steps, the size of interval will be at least , and therefore must contain .
Now we can prove the main result of this section. As we explain below, it is enough to prove the result for the case in which the Jordan curve is the unit circle. Let be the unit circle in . Let be a colored balanced set of points on , and let , , and be the partition of into the three color classes.
Given a closed curve with an injective continuous map and an integer , we say that a set is a carc set if where is the union of at most closed arcs of . Intuitively speaking, if has no crossings, denotes the number of components of . However, can be larger than the number of components if has one or more crossings.
Theorem 5
Let be a closed Jordan curve in the plane, and let be a 3colored balanced set of points on . Then for every positive integer there exists a arc set containing exactly points of each color.
Proof: Using we can map the points on to the unit circle, thus a solution on directly maps to a solution on . Hence, it suffices to prove the statement for the case in which .
Let be the set of numbers such that a subset as in the theorem exists. We prove that using Lemma 3. To apply the lemma it suffices to show that fulfills the following properties: (i) , (ii) If then , and (iii) If then . Once we show that these properties hold, Lemma 3 implies that any integer between 1 and must be in , hence .
Property (i) holds because the whole can be taken as a 2arc set, containing points of each color, thus .
Property (ii) follows from the fact that the complement of any 2arc set containing exactly points of each color is a 2arc set containing exactly points of each color. Thus if there is a 2arc set guaranteeing that , its complement guarantees that .
Proving Property (iii) requires a more elaborate argument. Let be a arc set containing exactly points of each color (such a set must exist by hypothesis, since ). We assume is parameterized as , for . Without loss of generality, we assume that (if necessary we can change the parametrization of by moving the location of the point corresponding to to ensure this).
We lift all points of to using the moment curve, as explained next. Abusing slightly the notation, we identify each point , on with its corresponding parameter . Then, for we define . Also, for any subset of , we define . Recall that we assumed that , thus, forms two disjoint arcconnected intervals in .
Next we apply the hamsandwich theorem to the points in (disregarding other lifted points of ): we obtain a plane that cuts the three colored classes in in half. That is, if is even, each one of the open halfspaces defined by contains exactly points of each color. If is odd, then we can force to pass trough one point of each color and leave points in each open halfspace ([14], Cor. 3.1.3). We denote by and the open halfspaces above and below , respectively. Note that each half space contains exactly points of each color class, as desired.
Let and . Note that both and contain exactly points of each color class, as desired. To finish the proof it is enough to show that either or is a arc set. Since has two connected components (and is lifted to the moment curve), we conclude that any hyperplane (in particular ) can intersect in at most points. Thus, the total number of components of is at most 5. This is also true for the preimages . Then, either or must form a arc set containing exactly points of each color. Thus, as desired.
Our approach generalizes to colors: if contains points of each color on , then for each there exists a arc set such that contains exactly points of each color. In our approach we lift the points to because we have three colors, but in the general case we would lift to . We also note that the bound on the number of intervals is tight. Consider a set of points in in which the points of the first colors are contained in disjoint arcs (one for each color), and each two neighboring disjoint arcs are separated by points of the color. Then, if , it is easy to see that we need at least arcs to get exactly points of each color.
We note that there exist several results in the literature that are similar to Theorem 5. For example, in [16] they show that given probability measures on , we can find a arc set whose measure is exactly in the measures. The methods used to prove their result are topological, while our approach is combinatorial. Our result can also be seen as a generalization of the wellknown necklace theorem for closed curves [14].
4 lines in the plane lattice
We now consider a balanced partition problem for colored point sets in the integer plane lattice . For simplicity, we will refer to and as the plane and the lattice, respectively. Recall that a set of points in the plane is said to be in general position if no three of them are collinear. When the points lie in the lattice, the expression is used differently: we say instead that a set of points in the lattice is in general position when every vertical line and horizontal line contains at most one point from .
An line with corner is the union of two different rays with common apex , each of them being either vertical or horizontal. An line partitions the plane into two regions (Figure 7 shows a balanced line with apex ). Since we look for balanced lines, we will only consider lines that do not contain any point of . Note that an line can always be slightly translated so that its apex is not in the lattice, thus its rays do not go through any lattice point.
lines in the lattice often play the role of regular lines in the Euclidean plane. For example, the classic hamsandwich theorem (in its discrete version) bisects a 2colored finite point set by a line in the plane. Uno et al. [18] proved that, when points are located in the integer lattice, there exists a bisecting line as well.
Recently, the following result has been proved for bisecting lines in the plane.
Theorem 6 (Bereg and Kano, [9])
Let be a 3colored balanced set of points in general position in the plane. If the convex hull of is monochromatic, then there exists a nontrivial balanced line.
As a means to further show the relationship between lines in the plane, and lines in the lattice, the objective of this section is to extend Theorem 6 to the lattice. Replacing the term line for line in the above result does not suffice (see a counterexample in Figure 8). In addition we must also use the orthogonal convex hull.
Theorem 7
Let be a 3colored balanced set of points in general position in the integer lattice. If the orthogonal convex hull of is monochromatic, then there exists a nontrivial balanced line.
Proof: Recall that a set is said to be orthoconvex if the intersection of with every horizontal or vertical line is connected. The orthogonal convex hull of a set is the intersection of all connected orthogonally convex supersets of S.
Without loss of generality, we assume that the points on the orthogonal convex hull are red. We use a technique similar to that described in the proof of Theorem 4; that is, we will create a sequence of orderings, and associate a polygonal curve to each such ordering. As in the previous case, we show that a curve can pass through the origin only at a vertex, which will correspond to the desired balanced line. Finally, we find two orderings that are reversed, which implies that some intermediate curve must pass through the origin. The difficulty in the adaptation of the proof lies in the construction of the orderings. This is, to the best of our knowledge, the first time that such an ordering is created for the lattice.
Given a point we define the ordering of as follows. Consider the points above (including ) and sort them by decreasing coordinate, i.e., from top to bottom. Let be the sorting obtained (notice that ). The remaining points (i.e., those strictly below ) are sorted by increasing coordinate, i.e., from left to right. Let denote the sorting obtained. Similarly, we define the , and sided orderings , and , respectively. Each ordering can be obtained by computing after having rotated clockwise the point set by , or radians, respectively. As an example, Figure 9 shows . Let
By construction, any prefix of an ordering of corresponds to a set of points that can be separated with an line. Likewise, any line will appear as prefix of some sorting (for example, one of the sortings associated with the apex of the line).
Given a sorting , we associate it with a polygonal curve in the lattice as follows: for every let and be the number of blue and green points in , respectively. Further, define the point . Based on these points we define a polygonal curve . Similarly to the construction of Theorem 4, the fact that for some is equivalent to the fact that the corresponding line is balanced. Therefore the goal of the proof is to show that there is always some ordering in for which some has .
We observe several important properties of :

is centrally symmetric (w.r.t. the origin). This follows from the definition of .

The interior of any segment of cannot contain the origin. The segment connecting two consecutive vertices in only depends on the color of the added element. Thus, there are only possible types of segments, see Figure 10. Since these segments do not pass through grid points, the origin cannot appear at the interior of a segment.

For any , we have , and . Notice that corresponds to an line having exactly one point on its upper/left side, while corresponds to an line leaving exactly one point on its lower/right side. In particular, these points must belong to the orthogonal convex hull, and thus must be red. That is, is a continuous polygonal curve that starts at , travels to . The curve is symmetric and returns to . Thus, as in the proof of Theorem 4 we conclude that either passes through the origin or it has nonzero winding.
Let be the points of sorted from left to right (analogously, for the points sorted from bottom to top). Observe that and and their reverses are and , respectively.
Analogous to the proof of Theorem 4, our aim is to transform to its reverse ordering through a series of small transformations of polygonal curves such that the winding number between the first and last curve is of different sign. If we imagine the succession of curves as a continuous transformation, a change in the sign of the winding number can only occur if at some point on the transformation the origin is contained in some curve. We will argue below that the only way in which the transformation passes through the origin is by having it as a vertex of one of the intermediate curves, which immediately leads to a balanced line.
Recall from Property 2 that the origin cannot be at the interior of an edge of any intermediate curve. Thus, the only way the origin can be swept during the transformation is (i) if it is a vertex of one of the curves, or (ii) if it is contained in the space between two consecutive curves (“swept by the curves”). In the remainder of the proof we show that the latter case cannot occur, that is, the origin is never swept by the local changes between two consecutive curves. This implies that the origin must be a vertex of some curve for some intermediate ordering , in turn implying that the associated line would be balanced.
The transformation we use is the following:
(1) 
First we give a geometric interpretation of this sequence. Imagine sweeping the lattice with a horizontal line (from top to bottom). At any point of the sweep, we sort the points below the line from bottom to top, and the remaining points are sorted from right to left. By doing so, we would obtain the orderings , and (once we reach ) (i.e., the reverse of ). Afterwards, we rotate the line clockwise by radians, keeping all points of to the right of the line, and sweep from left to right. During this second sweep, we sort the points to the right of the line from right to left, and those to the left from top to bottom. By doing this we would obtain the orderings , which is the reverse of . . Once all points have been swept, this process will finish with the ordering
Thus, to complete the proof it remains to show that the difference between any two consecutive orderings and in the above sequence cannot contain the origin.
Observe that two consecutive orderings differ in at most the position of one point (the one that has just been swept by the line). Thus, there exist two indices and such that and , and (that is, point moved immediately after ).
Abusing slightly the notation, we denote by the vertices of (respectively, denotes the vertices of ). Since only one point has changed its position in the ordering, we can explicitly obtain the differences between the two orderings. Given an index , we define if is red, if is blue, or if is green. Then, we have the following relationship between the vertices of and .
(2) 
Observe that the ordering of the first points and the last points is equal in both permutations. In particular, the points to and to do not change between consecutive polygonal curves. For the intermediate indices, the transformation only depends on the color of ; it consists of a translation by the vector .
We now show that the origin cannot be contained in the interior of the quadrilateral of vertices , and , for any . Consider the case in which (that is, neither nor satisfy the first line of Equation 2). Observe that the shape of the quadrilateral only depends on the color of and . It is easy to see that when has the same color as , is degenerate and cannot contain the origin. Thus, there are six possible color combinations for and that yield three nondegenerate different quadrilaterals (see Figure 11).
Assume that for some index we have that is red, is blue (as shown in Figure 11, left) and that the quadrilateral contains the origin. Note that in this case, must be either or . From the definition of the coordinate of , we have , and thus we conclude that . Consider now the coordinate of ; recall that this coordinate is equal to , which cannot be or whenever . The proof for the other quadrilaterals is identical; in all cases, we show that either the or coordinate of a vertex of is zero must simultaneously satisfy: it is congruent to zero modulo 3, and it is either or , resulting in a similar contradiction.
Thus, in order to complete the proof it remains to consider the cases in which or . In the former case we have , and in the latter. Whichever the case, quadrilateral collapses to a triangle, and we have three nontrivial possible color combinations (see Figure 12). The same methodology shows that none of them can sweep through the origin.
That is, we have transformed the curve from to its reverse in a way that the origin cannot be contained between two successive curves. However, since these curves have winding number of different sign, at some point in our transformation one of the curves must have passed through the origin. The previous arguments show that this cannot have happened at the interior of an edge or at the interior of a quadrilateral between edges of two consecutive curves. Thus it must have happened at a vertex of , for some . In particular, the corresponding line must be balanced.
5 Concluding remarks
In this paper we have studied several problems about balanced partitions of colored sets of points and lines in the plane. As a final remark we observe that our results on double wedges can be viewed as partial answers to the following interesting open problem: Find all such that, for any colored balanced set of points in general position in the plane, there exists a double wedge containing exactly points of each color. We have given here an affirmative answer for and (Theorems 3 and 4). Theorem 5 gives the affirmative answer for all values of under the constraint that points are in convex position.
Acknowledgments Most of this research took place during the 5th International Workshop on Combinatorial Geometry held at UPCBarcelona from May 30 to June 22, 2012. We thank Inês Matos, Pablo PérezLantero, Vera Sacristán and all other participants for useful discussions.
F.H., M.K., C.S., and R.S. were partially supported by projects MINECO MTM201230951, Gen. Cat. DGR2009SGR1040 and DGR2014SGR46, and ESF EUROCORES programme EuroGIGA, CRP ComPoSe: MICINN Project EUIEURC20114306. F.H. and C.S. were partially supported by project MICINN MTM200907242. M.K. also acknowledges the support of the Secretary for Universities and Research of the Ministry of Economy and Knowledge of the Government of Catalonia and the European Union. R. I. S. was funded by Portuguese funds through CIDMA and FCT, within project PEstOE/MAT/UI4106/2014, and by FCT grant SFRH/BPD/88455/2012.
References
References
 [1] J. Akiyama, G. Nakamura, E. RiveraCampo, and J. Urrutia. Perfect division of a cake, Proceedings of the Tenth Canadian Conference on Computational Geometry (CCCG), 114115, 2008.
 [2] J. Akiyama, A. Kaneko, M. Kano, G. Nakamura, E. RiveraCampo, S. Tokunaga, and J. Urrutia. Radial Perfect Partitions of Convex Sets in the Plane. Revised notes of the Japanese Conference of Discrete and Computational Geometry (JCDCG), 113, 2000.
 [3] A. Akopyan and R. Karasev. Cutting the same fraction of several measures. Discrete and Computational Geometry, 49:402–410, 2013.
 [4] K. T. Atanassov. On Sperners lemma. Studia Sci. Math. Hungar, 32:71–74, 1996.
 [5] I. Bárány, J. Matoušek. Simultaneous Partitions of Measures by kFans. Discrete and Computational Geometry, 25(3):317–334, 2001.
 [6] W. F. Basener. Topology and Its Applications. A Wiley Series of Texts, Monographs and Tracts. WileyInterscience; 1st edition, 2006.
 [7] S. Bereg. Orthogonal equipartitions. Computational Geometry: Theory and Applications, 42(4):305–314, 2009.
 [8] S. Bereg, F. Hurtado, M. Kano, M. Korman, D. Lara, C. Seara, R. I. Silveira, J. Urrutia, and K. Verbeek. Balanced partitions of 3colored geometric sets in the plane. Discrete Applied Mathematics, 181:21–32, 2015.
 [9] S. Bereg and M. Kano. Balanced line for a 3colored point set in the plane. The Electronic Journal of Combinatorics, 19:P33, 2012.
 [10] S. Bespamyatnikh, D. Kirkpatrick, J. Snoeyink. Generalizing Ham Sandwich Cuts to Equitable Subdivisions. Discrete and Computational Geometry, 24(4):605–622, 2000.
 [11] M. de Berg, O. Cheong, M. van Kreveld, M. Overmars. Computational Geometry: Algorithms and Applications. SpringerVerlag; 3rd ed., 2008.
 [12] A. Kaneko and M. Kano. Discrete geometry on red and blue points in the plane  a survey. In Discrete and Computational Geometry, The GoodmanPollack Festschrift; edited by B. Aronov, S. Basu, J. Pach, and M. Sharir, Springer, 551–570, 2003.
 [13] M. Kano and K. Suzuki. Discrete geometry on red and blue points in the plane lattice. In Thirty Essays on Geometric Graph Theory; edited by J. Pach, Springer (in print).
 [14] J. Matoušek. Using the BorsukUlam Theorem: Lectures on Topological Methods in Combinatorics and Geometry, Chapter 3. Springer, 2007.
 [15] T. Needham. Visual Complex Analysis. Oxford University Press, USA, 1999.
 [16] W. Stromquist and D. Woodall. Sets on which several measures agree. Journal of Mathematical Analysis and Applications, 108(1):241–248, 1985.
 [17] T. Needham. Winding Numbers and Topology. In Visual Complex Analysis, chapter 7. Oxford University Press, 338–341, 1999.
 [18] M. Uno, T. Kawano, and M. Kano. Bisections of two sets of points in the plane lattice. IEICE Transactions, 92A(2):502–507, 2009.