A Generalization of the Convex Kakeya Problem^{†}^{†}thanks: H.K.A. was supported by NRF grant 20110030044 (SRCGAIA) funded by the government of Korea. J.G. is the recipient of an Australian Research Council Future Fellowship (project number FT100100755). O.C. was supported in part by NRF grant 20110030044 (SRCGAIA), and in part by NRF grant 20110016434, both funded by the government of Korea.
Abstract
Given a set of line segments in the plane, not necessarily finite, what is a convex region of smallest area that contains a translate of each input segment? This question can be seen as a generalization of Kakeya’s problem of finding a convex region of smallest area such that a needle can be rotated through 360 degrees within this region. We show that there is always an optimal region that is a triangle, and we give an optimal time algorithm to compute such a triangle for a given set of segments. We also show that, if the goal is to minimize the perimeter of the region instead of its area, then placing the segments with their midpoint at the origin and taking their convex hull results in an optimal solution. Finally, we show that for any compact convex figure , the smallest enclosing disk of is a smallestperimeter region containing a translate of every rotated copy of .
1 Introduction
Let be a family of objects in the plane. A translation cover for is a set such that any object in is contained in a translate of [28]. We are interested in determining a convex translation cover for of smallest possible area or perimeter.
Since the convex hull of a set of objects is the smallest convex figure that contains them, this problem can be reformulated as translating the objects in such that the perimeter or the area of their convex hull is minimized. When consists of objects, we can fix one object and translate the remaining objects. Therefore we can use a vector in to represent the translations of objects. Consider the functions that take a vector in and return the perimeter and the area of the convex hull of the fixed object and the translated copies of the other objects. Ahn and Cheong [1] showed that for the perimeter case, this function is convex. They also showed that for the area case, the function is convex if . However, this is no longer true when , as the following example shows. Let be a vertical segment of length one, and let and be copies of rotated by and . Then the area of their convex hull is minimized when they form an equilateral triangle, so there are two isolated local minima, as shown in Figure 1. This explains why minimizing the perimeter appears to be a much easier problem than minimizing the area of a translation cover.
As a special case of translation covers, we can consider the situation where the family consists of copies of a given compact convex figure , rotated by all angles in . In other words, we are asking for a smallest possible convex set such that can be placed in in every possible orientation. We will call such a translation cover a keyhole for (since a key can be turned fully in a keyhole, it can certainly be placed in every possible orientation).
A classical keyhole or translation cover problem is the Kakeya needle problem. It asks for a minimum area region in the plane, a socalled Kakeya set, in which a needle of length can be rotated through continuously, and return to its initial position. (See Figure 2.) This question was first posed, for convex regions, by Soichi Kakeya in 1917 [14]. Pàl [17] showed that the solution of Kakeya’s problem for convex sets is the equilateral triangle of height one, having area . With our terminology, he characterized the smallestarea keyhole for a line segment.
For the general case, when the Kakeya set is not necessarily convex or even simply connected, the answer was thought to be a deltoid with area . However, Besicovitch gave the surprising answer that one could rotate a needle using an arbitrary small area [3, 4].
Besicovitch’s solution builds upon two basic observations [24]. The first observation is that one can translate any needle to any location using arbitrarily small area. The idea is to slide the needle, rotate it, slide it back and then rotate it back, as illustrated in Fig. 3(a).
The area can be made arbitrarily small by sliding the needle over a large distance. The second observation is that one can construct an open subset of the plane of arbitrary small area, which contains a unit line segment in every direction, as illustrated in Fig. 3(b). The original construction by Besicovitch [3, 4] has been simplified by Perron [18], Rademacher [19], Schoenberg [21, 22], Besicovitch [5, 6] and Fisher [12].
Bezdek and Connelly [7] surveyed results on minimumperimeter and minimumarea translation covers. For the family of closed curves of length at most one, they proved that smallestperimeter translation covers are exactly the convex sets of constant width . The corresponding problem for minimizing the area, known as Wetzel’s problem, is still open, with upper and lower bounds known [7, 28]. For the family of sets of diameter at most one, Bezdek and Connelly [8] proved that the unique minimumperimeter translation cover is the circle of radius . More precisely, they proved that this circle is the unique smallestperimeter keyhole for the equilateral triangle of side length one. By Jung’s theorem [13], this circle contains any set of diameter one, and so the translation cover result follows.
Recently, Kakeyatype problems have received considerable attention due to their many applications. There are strong connections between Kakeyatype problems and problems in number theory [9], geometric combinatorics [29], arithmetic combinatorics [15], oscillatory integrals, and the analysis of dispersive and wave equations [24].
In this paper, we first generalize Pál’s result [17] in the following way: For any family of line segments in the plane, there is a triangle that is a minimumarea translation cover for .
Theorem 1.
Let be a set of line segments in the plane, and let be a convex translation cover for . Then there is a translation cover for which is a triangle, and such that the area of is less than or equal to the area of .
With this characterization in hand, we can efficiently compute a smallest area translation cover for a given family of line segments. Our algorithm runs in time , which we prove to be optimal in the algebraic computation tree model. It is based on the problem of finding a smallestarea affineregular hexagon containing a given centrally symmetric polygon, a problem that is interesting in its own right. As far as we know, except for some trivial cases such as disks or axisaligned squares, previously known algorithms for finding smallestarea translation covers have a running time exponential in , the number of input objects [1, 27].
As observed above, minimizing the perimeter of a translation cover is much easier. Let be a family of centrally symmetric convex figures. We prove that if we translate each figure such that its center of symmetry is the origin, then the convex hull of their union is a smallestperimeter translation cover for .
This immediately implies that a circle with diameter is a smallestperimeter keyhole for the unitlength segment. For figures that are not centrally symmetric, this argument no longer works. We generalize the result by Bezdek and Connelly [8] mentioned above and prove the following theorem (Bezdek and Connelly’s result is the special case where is an equilateral triangle):
Theorem 2.
Let be a compact convex set in the plane, and let be the family of all the rotated copies of by angles in . Then the smallest enclosing disk of is a smallestperimeter translation cover for .
2 Preliminaries
An oval is a compact convex figure in the plane. For an oval , let denote the width function of . The value is the length of the projection of on a line with slope (that is, a line that makes angle with the axis). Let denote the area of .
For two ovals and , we write or to mean pointwise domination, that is for every we have . We also write if and only if both and hold.
The Minkowski symmetrization of an oval is the oval . It is well known and easy to show that is centrally symmetric around the origin, and that .
An oval is a trigonal disk if there is a centrally symmetric hexagon such that contains the triangle and is contained in the hexagon , as illustrated in Figure 4(a). Trigonal disks were called “relative Reuleaux triangles” by Ohmann [16] and Chakerian [10], the term “trigonal disk” being due to Fejes Tóth [25] who used it in the context of packings by convex disks. A trigonal disk has three “main” vertices and three arcs connecting these main vertices. For example, the trigonal disk in Figure 4(a) consists of three vertices , and , and three arcs connecting them.
Ohmann [16] and Chakerian [10] studied sets with a given fixed width function, and obtained the following result (see for instance Theorem in [10] for a proof):
Fact 3.
Given an oval , there is a trigonal disk with such that .
3 Minimum area for a family of segments
In this section we will prove Theorem 1. The proof contains two parts. First we prove that for every oval there exists a triangle with and (Theorem 4). The second part is to prove that for an oval and a closed segment , if then contains a translated copy of (Lemma 5).
Theorem 4.
Given an oval , there exists a triangle with and .
Proof.
Let be the set of trigonal disks such that we have and . The set is nonempty by Fact 3. Consider three arcs connecting the main vertices of a trigonal disk in . Each arc can be straight, or not. We choose a trigonal disk with a maximum number of straight arcs. We show that is a triangle.
Let be the hexagon from the definition of the trigonal disk , and assume for a contradiction that is not a triangle, that is, there is at least one nonstraight arc among the three arcs connecting , and . See Figure 4(a). Without loss of generality, we assume that the arc connecting and is not straight.
Let the sides and be vertical, with above the line . Let be the point of below with the largest vertical distance from the line . Let be the point vertically above at distance from . Let be the convex hull of the part of above the line and the point . It is not difficult to see that is also a trigonal disk: Let be the point vertically below at distance from . Then the hexagon is centrally symmetric and contains . Clearly contains the triangle . See Figure 4(b).
We show next that . The area of is bounded by the area of the two triangles and , where and are points on such that and are tangent to . This area is equal to times the horizontal distance between and . But the horizontal distance between and is at most the horizontal distance between and , so the area of is bounded by the area of the triangle , and we have .
We also need to argue that . Consider a minimal strip containing . If this strip does not touch from below between and , then the corresponding strip for is at least as wide. Otherwise, it touches from below in a point between and , and touches from above in , as is the only antipodal point of for . A strip with the same direction will be determined either by and , or by and , and in both cases its width is not less than the width of the original strip.
Since and the trigonal disk must be a member of . However, has at least one straight arc more than , contradicting our choice of . It follows that our assumption that is not a triangle must be false. ∎
This finishes the first part. We need the following lemma, which shows that whether or not an oval contains a translated copy of a given segment can be determined by looking at the width functions of and alone:
Lemma 5.
Let be a segment in the plane, and let be an oval such that . Then contains a translated copy of .
Proof.
Without loss of generality, let be a horizontal segment. Let be a horizontal segment of maximal length contained in . Then has a pair of parallel tangents and through and . By the assumption, the distance between and must be large enough to place in between the two lines. But this implies that the segment is at least as long as , and can be placed on the segment in . ∎
4 From triangles to hexagons
We now turn to the computational problem: Given a family of line segments, find a smallestarea convex set that contains a translated copy of every .
By Theorem 1 we can choose the answer to be a triangle. In this section we show that this problem is equivalent to finding a smallestarea affineregular hexagon enclosing some centrally symmetric convex figure. An affineregular hexagon is the image of a regular hexagon under a nonsingular affine transformation. In this paper, we only consider affineregular hexagons that are centrally symmetric about the origin, so by abuse of terminology, we will write affineregular hexagon for an affineregular hexagon that is centrally symmetric about the origin.
In the next section we will then show how to solve that problem, using the tools of computational geometry.
The basic insight is that for centrally symmetric figures, comparing widthfunctions is equivalent to inclusion:
Lemma 6.
Let and be ovals centrally symmetric about the origin. Then if and only if .
Proof.
One direction is trivial, so consider for a contradiction the case where and . Then there is a point . Since is convex, there is a line that separates from . Since and are centrally symmetric, this means that is contained in the strip bounded by the lines and , while contains the points and lying outside this strip. This implies that for the orientation orthogonal to we have , a contradiction. ∎
Recall that denotes the Minkowski symmetrization of an oval .
Lemma 7.
Let be a nondegenerate triangle. Then is an affineregular hexagon, and . Every affineregular hexagon can be expressed in this form.
Proof.
Since every nondegenerate triangle is the affine image of an equilateral triangle, it suffices to observe this relationship for the equilateral triangle and the regular hexagon. ∎
Lemma 8.
Given an oval , a triangle is a smallestarea triangle with if and only if is a smallestarea affine regular hexagon with .
This leads us to the following algorithm. In Section 6, we will show that the time bound is tight.
Theorem 9.
Let be a set of line segments in the plane. Then we can find a triangle in time which is a minimumarea convex translation cover for .
Proof.
Given a family of line segments, place every with its center at the origin. Let be the convex hull of these translated copies. can be computed in time, and is a centrally symmetric convex polygon with at most vertices. We then compute a smallest area affineregular hexagon containing . In the next section we will show that this can be done in time . Finally, we return a triangle with . The correctness of the algorithm follows from and Lemma 8. ∎
5 Algorithm for computing the smallest enclosing affineregular hexagon
In this section we discuss the following problem: Given a convex polygon , centrally symmetric about the origin, find a smallestarea affineregular hexagon such that .
Let us first sketch a simple quadratictime algorithm: The affineregular hexagons centered at the origin are exactly the images of a regular hexagon centered at the origin under a nonsingular linear transformation. Instead of minimizing the hexagon, we can fix a regular hexagon with center at the origin, and find a linear transformation such that and such that the determinant of is maximized. The transformation can be expressed as a matrix with coefficients . The condition can then be written as a set of linear inequalities in the four unknowns . We want to find a feasible solution that maximizes the determinant , a quadratic expression. This can be done by computing the 4dimensional polytope of feasible solutions, and considering every facet of this polytope in turn. We triangulate each facet, and solve the maximization problem on each simplex of the triangulation.
In the following, we show that the problem can in fact be solved in linear time.
For a set , let denote the mirror image with respect to the origin. A strip is the area bounded by a line and its mirror image .
An affineregular hexagon is the intersection of three strips , , and , as in Figure 5, where the sides of are supported by , , , , , and in counterclockwise order.
The intersection is a parallelogram . Since is affineregular, the sides supported by must be parallel to and half the length of , and so is uniquely defined by and : It supports the sides and of , where is the midpoint of and is the midpoint of . Note that .
It is easy to see that if is a minimumarea affineregular hexagon containing , then two of the three strips must be touching . Without loss of generality, we can assume these to be strips and , so there is a vertex of on the side , and a vertex on the side .
For convenience of presentation, let us choose a coordinate system where is horizontal. If we now rotate counterclockwise while remaining in contact with , then one side rotates about the point , while the opposite side rotates about , see Figure 6.
The triangles and are similar, and since lies above or on the axis, we have . This implies that the area of is nonincreasing during this rotation. Since , the area of decreases or remains constant as well.
Furthermore, the point moves horizontally along the axis to the right. The point moves horizontally to the right with at least twice the speed of point . As is the midpoint of and , this implies that moves horizontally to the right with at least the speed of , and so the line is rotating counterclockwise.
It follows that while strip rotates counterclockwise, the part of lying below the axis and to the left of the line is strictly shrinking. It follows that there is a unique orientation of where the side touches , and the area of is minimized.
Let us say that a polygon is circumscribed to another polygon if and only if and every side of contains a point of . Then we have shown
Lemma 10.
There is a minimumarea affineregular hexagon such that is circumscribed to .
In fact, we have shown that for every there is a unique such that is circumscribed to . We have
Lemma 11.
When rotates counterclockwise, then the corresponding also rotates counterclockwise.
Proof.
Consider a configuration where is circumscribed to , and rotate slightly around in counterclockwise direction, keeping fixed. Then and move downwards along the line , see Figure 7.
The point moves downwards along the line , parallel to . It follows that the new edge now lies strictly outside the old hexagon , and so cannot possibly touch or intersect . By the arguments above, this implies that strip now needs to rotate counterclockwise as well to let be circumscribed to .
Furthermore, similar to the arguments above, we observe that moves with speed at least twice the speed of . Since is the midpoint of , it moves with at least half the speed of , so moves with speed at least equal to the speed of . Since and move on parallel lines, it follows that the line is rotating counterclockwise during the rotation of . ∎
We can now show that we can in fact choose such that one of its sides contains an edge of :
Lemma 12.
There exists a minimumarea affineregular hexagon containing such that a side of contains an edge of . In addition, if no minimumarea affineregular hexagon containing shares a vertex with , then each such minimumarea affineregular hexagon has a side containing a side of .
Proof.
By Lemma 10, there exists a minimumarea affineregular hexagon such that every side of contains a point of . If a side of contains an edge of , then we are done. In the following, we thus assume that every side of intersects in a single point. Also, we assume the vertices of are and their antipodal points . This can be done by applying a nonsingular linear transformation, see Figure 8.
First, we consider the case where no vertex of coincides with a vertex of . We claim that in this case, there exists a nonsingular linear transformation such that and hold, implying that the inverse image of also contains and its area is strictly smaller than , a contradiction. We denote by the three contact points as in Figure 8. The point is a linear combination of and , so we have . Since does not lie in the same quadrant as and , nor the opposite quadrant, then and . As the point lies on the line with equation , we have Then the area of the triangle is given by
(1) 
Assume we apply a linear transformation to such that each point in moves along the side of that currently contains it. Thus, changes, and changes in such a way that remains on the same side of . Then the area of is proportional to the area . As we observed that , then the coefficient of in Equation (1) is positive, so the area cannot be at a local maximum, a contradiction.
Consider now the case where at least one of the contact points lies at a vertex of . Since each side of has a singlepoint intersection with , two of are identical. Using a suitable linear transformation, we can assume in Figure 8. Any linear transformation that keeps fixed and moves along the vertical side of keeps areas unchanged, and so and thus . Hence, there exists a linear transformation such that , , and has one more contact point with the sides of . ∎
We can therefore assume that the minimumarea affineregular hexagon is defined by two strips and , where supports an edge of , and is the unique strip such that the resulting hexagon is circumscribed to . We now give a lineartime algorithm to enumerate these hexagons, over all edges of .
Theorem 13.
Given a centrallysymmetric convex gon , a smallestarea affineregular hexagon enclosing can be found in time .
Proof.
We use a rotating calipers [26] type algorithm. It maintains an edge of defining , a second strip and the vertex of where touches , and a vertex of . Let be the hexagon defined by and , as in Figure 5.
The algorithm proceeds by rotating around as in Figure 6, and maintains the invariant that has a supporting line in that is parallel to .
We initialize to an arbitrary edge of . Let be horizontal for ease of presentation, with below , let be the left endpoint of , and let be the leftmost vertex of . In the initial configuration, is obtained from by a counterclockwise rotation around by an infinitely small amount.
We then rotate counterclockwise, until one of the following events occurs:

If no longer supports a tangent to parallel to , replace by the counterclockwise next vertex of , and continue rotating .

If supports an edge of , then replace by the counterclockwise next vertex of , and continue rotating .

If touches , then we have found the unique such that is circumscribed to . We compute its area and update a running minimum. Then replace by the counterclockwise next edge of . As long as does not support a tangent to parallel to , we replace by the counterclockwise next vertex of . Then continue rotating .
The algorithm ends when edges have been considered. Its running time is clearly linear. ∎
6 Lower bound for computing a translation cover
In this section, we prove an lower bound for the problem of computing a minimumarea translation cover for a set of line segments. We first need the following result on regular gons (see Figure 9(a).):
Lemma 14.
Let denote a regular gon centered at the origin, for some integer . Then any minimumarea affineregular hexagon enclosing is a regular hexagon such that every edge of this hexagon contains an edge of .
Proof.
The statement is trivial for , so assume . Let denote a regular hexagon enclosing , and such that each side of contains a side of . Let denote another smallest affineregular hexagon enclosing . We will argue that is also a regular hexagon whose sides contain sides of .
We first rule out the case where shares a vertex with . For sake of contradiction, assume that shares two opposite vertices and of . Without loss of generality, we assume that are on the axis. The edges of that are not adjacent to are parallel to and have half the length of . In addition, the edges of that are adjacent to and make an angle at most with the axis. (See Figure 9(b).) Then a direct calculation shows that , a contradiction.
Thus, by Lemma 12, we know that an edge of contains an edge of . Without loss of generality we assume that this edge is parallel to the axis. (See Figure 9(c).) For sake of contradiction, assume that is not symmetric with respect to the axis. Consider the hexagon that is obtained from by a horizontal shear transformation that moves and the opposite edge parallel to the axis, until they are centered at the axis. Then (see Figure 9(d)) is an affineregular hexagon containing that is symmetric with respect to the axis and that only touches along its top and bottom edges. This implies that strictly contains a regular hexagon enclosing , and hence , a contradiction.
Therefore, is symmetric with respect to the axis, and thus is symmetric with respect to the axis. Only one such affineregular hexagon is circumscribed to , so . ∎
We are now able to prove our lower bound.
Theorem 15.
In the algebraic computation tree model, and in the worst case, it takes time to compute a minimumarea translation cover for a family of line segments in the plane.
Proof.
For an interval , we denote by the arc of the unit circle corresponding with polar angles in the interval , that is . As is the intersection of a circle and a cone, a node of an algebraic computation tree can decide whether a point lies in .
We use a reduction from the following problem. The input is a set of points . The goal is to decide whether there exists an integer such that is empty, that is, this arc does not contain any point . It follows from BenOr’s bound [2] that any algebraic computation tree that decides this problem has depth . (The set of negative instances has at least connected component: To each permutation of , we associate a negative instance where each lies in the ’s arc. In order to move continuously from one of these configuration to another, we must have a crossing , which implies that one interval is empty by the pigeonhole principle, and thus the instance is positive.)
Our construction is as follows. Consider the (fixed) regular gon , whose vertices are for . Let denote the convex gon whose vertices are the vertices of and all the rotated copies of the points by angles around the origin.
If there is an integer such that is empty, then by Lemma 12, the regular hexagon containing whose edges contain the edge and its rotated copies by angles is a minimum area affineregular hexagon containing .
If on the other hand, for every integer the arc is nonempty, then by Lemma 12, any minimumarea affine hexagon containing is a regular hexagon whose edges contain edges of , and thus it cannot contain .
So we have proved that, when some arc is empty, then a minimumarea hexagon containing has area , where is a minimumarea hexagon containing . Otherwise, if all these arcs are nonempty, then the minimum area is larger than .
Thus, if we could compute in time a minimumarea convex translation cover for the diagonals of , then by Lemma 8 we would also get in time the area of a smallest enclosing affineregular hexagon containing , and then we would be able to decide in time whether there exists an empty arc , a contradiction. ∎
7 Minimizing the perimeter
If we wish to minimize the perimeter instead of the area, the problem becomes much easier: it suffices to translate all segments so that their midpoints are at the origin, and take the convex hull of the translated segments. This follows from the following more general result.
Theorem 16.
Let be a family of centrally symmetric convex figures. Under translations, the perimeter of the convex hull of their union is minimized when the centers coincide.
Proof.
By the CauchyCrofton formula [11], the perimeter is the integral of the width of the projection over all directions. We argue that the width is minimized when the centers coincide, for all directions simultaneously, implying the claim.
Assume the objects are placed with their center at the origin. Let be a leftmost point of the convex hull. It belongs to one of the objects . By symmetry, the mirror image of is then a rightmost point of the convex hull. But this implies that the horizontal width of the convex hull is equal to the width of , and therefore as small as possible. ∎
When the figures are not symmetric, our proof of Theorem 16 breaks down. However, we are able to solve the problem for a family consisting of all the rotated copies of a given oval. (Remember that an oval is a compact convex set.) The following theorem was already stated in the introduction.
Theorem 2.
Let be an oval, and let be the family of all the rotated copies of by angles in . Then the smallest enclosing disk of is a smallestperimeter translation cover for .
Proof.
We observe first that, if is a segment, then by Theorem 16, the smallest enclosing disk of is a smallestperimeter translation cover for .
Consider next the case where is an acute triangle. Choose a coordinate system with origin at the center of the circumcircle of , and such that the circumcircle has radius one. We wish to prove that any translation cover for must have perimeter at least , implying that the circumcircle is optimal.
We borrow an idea of Bezdek and Connelly [8]. Let , , be the three vertices of . By our assumptions, the origin lies in the interior of their convex hull, and the three vectors have length one. The origin can be expressed as a convex combination with and . Let , for , be the angle formed by and the positive axis.
Let be a translation cover for and let be the support function [20] of . That is, for any unit vector . We denote by the unit vector making angle with the positive axis, so that .
The length of the perimeter of is equal to the integral over the support function [23]
Since is a periodic function with period , we have
It follows that
(2) 
Consider now a fixed orientation . The translation cover must contain a rotated copy of such that, for some translation vector , the vertices of are the points for .
Since lies in , the value of the support function is lower bounded by
(3) 
and thus
Plugging this into Eq. (2) gives .
Consider finally the general case where is an arbitrary compact convex figure, and let be the smallest enclosing disk of . Either touches in two points that form a diameter of , or touches in three points that form an acute triangle. In both cases, our previous results imply that is a smallestperimeter translation cover for either the segment or the triangle, and therefore for . ∎
The minimum enclosing circle is not always the unique minimumperimeter keyhole: For instance, when is a unit line segment, then any set of constant width is a solution. In the theorem below, we show that when is an acute triangle, then its circumcircle is the unique solution. This generalizes directly to any figure that touches its circumcircle at 3 points.
Theorem 17.
If is an acute triangle, then its smallest enclosing disk is the unique smallestperimeter translation cover for the family of all rotated copies of .
Proof.
We use the same notations as in the proof of Theorem 2: is a smallestperimeter translation cover for . For any , it contains a copy of rotated by angle . The vertices of are the points , for .
We will prove that all the triangles have the same circumcircle. Our strategy is to show that the function is differentiable and its derivative is . Without loss of generality, we only prove that , and we assume that .
For sake of contradiction, assume that is not differentiable at , or it is differentiable at and its derivative is nonzero. This means that we do not have . Hence, there exists an such that for any integer , there exists with . This implies , and so is a sequence of unit vectors. Since the set of unit vectors is compact, there is a subsequence such that converges to a unit vector . We denote this subsequence again as .
Since span , there exists such that . So
As for all , this implies that for large enough,
hence
Thus, for large enough , we have . Since for all , this implies . But since , this means , and so Inequality (3) in the proof of Theorem 2 is not tight. Since the support function is continuous [20], this implies that , a contradiction. ∎
8 Conclusions
In practice, it is an important question to find the smallest convex container into which a family of ovals can be translated. For the perimeter, this is answered by the previous lemma for centrally symmetric ovals. For general ovals, it is still not difficult, as the perimeter of the convex hull is a convex function under translations [1]. This means that the problem can be solved in practice by numerical methods.
For minimizing the area, the problem appears much harder, as there can be multiple local minima. The following lemma solves a very special case.
Lemma 18.
Let be a family of axisparallel rectangles. The area of their convex hull is minimized if their bottom left corners coincide (or equivalently if their centers coincide).
Proof.
Let be the convex hull of some placement of the rectangles. For any , let be the length of the intersection of the vertical line at coordinate with . The function is concave (by the BrunnMinkowski theorem in two dimensions). For any , we define to be the length of the interval of all where .
We observe that the area of is equal to , which is again equal to . We will now argue that is minimized for every when the bottom left corners of the rectangles coincide, implying the claim.
To see this, consider the placement with coinciding bottom left corners at the origin, and the line . It intersects the convex hull at and at some convex hull edge defined by two rectangles and . is equal to the length of this intersection. It remains to observe that for any placement of and , the convex hull of these two rectangle already enforces this value of . ∎
Acknowledgments
We thank Helmut Alt, Tetsuo Asano, Jinhee Chun, Dong Hyun Kim, Mira Lee, Yoshio Okamoto, János Pach, Günter Rote, and Micha Sharir for helpful discussions.
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