A Generalization of the Convex Kakeya ProblemH.-K.A. was supported by NRF grant 2011-0030044 (SRC-GAIA) 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 2011-0030044 (SRC-GAIA), and in part by NRF grant 2011-0016434, both funded by the government of Korea.

A Generalization of the Convex Kakeya Problemthanks: H.-K.A. was supported by NRF grant 2011-0030044 (SRC-GAIA) 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 2011-0030044 (SRC-GAIA), and in part by NRF grant 2011-0016434, both funded by the government of Korea.

Hee-Kap Ahn POSTECH, South Korea. Email: heekap@postech.ac.kr.    Sang Won Bae Kyonggi University, South Korea. Email: swbae@kgu.ac.kr.    Otfried Cheong KAIST, South Korea. Email: otfried@kaist.edu.    Joachim Gudmundsson University of Sydney and NICTA, Australia. Email: joachim.gudmundsson@sydney.edu.au.    Takeshi Tokuyama Tohoku University, Japan. Email: tokuyama@dais.is.tohoku.ac.jp.    Antoine Vigneron KAUST, Saudi Arabia. Email: antoine.vigneron@kaust.edu.sa.

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 smallest-perimeter 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.

Figure 1: The area function of the convex hull of segments is not necessarily convex.

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 so-called 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 smallest-area keyhole for a line segment.

Figure 2: Within a Kakeya set (shaded), a needle can be rotated through .

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).

Figure 3: (a) A needle can be translated to any location using arbitrarily small area. (b) There is an open subset of the plane of arbitrary small area which contain a unit line segment in every direction.

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 minimum-perimeter and minimum-area translation covers. For the family of closed curves of length at most one, they proved that smallest-perimeter 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 minimum-perimeter translation cover is the circle of radius . More precisely, they proved that this circle is the unique smallest-perimeter 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, Kakeya-type problems have received considerable attention due to their many applications. There are strong connections between Kakeya-type 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 minimum-area 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 smallest-area affine-regular 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 axis-aligned squares, previously known algorithms for finding smallest-area 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 smallest-perimeter translation cover for .

This immediately implies that a circle with diameter  is a smallest-perimeter keyhole for the unit-length 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 smallest-perimeter 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.

Figure 4: (a) A trigonal disk that is contained in the centrally symmetric hexagon and contains the triangle . (b) The hexagon is centrally symmetric and contains . Since contains the triangle , it is also a trigonal disk.

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 .


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 non-straight 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 .


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 . ∎

To prove Theorem 1, let  be an oval of minimum area that contains a translated copy of every . By Theorem 4 there is a triangle  such that and . Let . Since there is a translated copy of  contained in , we must have . By Lemma 5 there is then a translated copy of contained in .

4 From triangles to hexagons

We now turn to the computational problem: Given a family of line segments, find a smallest-area 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 smallest-area affine-regular hexagon enclosing some centrally symmetric convex figure. An affine-regular hexagon is the image of a regular hexagon under a non-singular affine transformation. In this paper, we only consider affine-regular hexagons that are centrally symmetric about the origin, so by abuse of terminology, we will write affine-regular hexagon for an affine-regular 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 width-functions is equivalent to inclusion:

Lemma 6.

Let  and  be ovals centrally symmetric about the origin. Then if and only if .


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 non-degenerate triangle. Then is an affine-regular hexagon, and . Every affine-regular hexagon can be expressed in this form.


Since every non-degenerate triangle is the affine image of an equilateral triangle, it suffices to observe this relationship for the equilateral triangle and the regular hexagon. ∎

Since , , and by Lemmas 6 and 7, we immediately have

Lemma 8.

Given an oval , a triangle is a smallest-area triangle with  if and only if is a smallest-area 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 minimum-area convex translation cover for .


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 affine-regular 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 affine-regular hexagon

In this section we discuss the following problem: Given a convex polygon , centrally symmetric about the origin, find a smallest-area affine-regular hexagon  such that .

Let us first sketch a simple quadratic-time algorithm: The affine-regular hexagons centered at the origin are exactly the images of a regular hexagon centered at the origin under a non-singular 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 4-dimensional 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 affine-regular hexagon  is the intersection of three strips , , and , as in Figure 5, where the sides of  are supported by , , , , , and in counter-clockwise order.

Figure 5: The hexagon is defined by three strips.

The intersection is a parallelogram . Since is affine-regular, 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 minimum-area affine-regular 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  counter-clockwise while remaining in contact with , then one side rotates about the point , while the opposite side rotates about , see Figure 6.

Figure 6: Rotating strip  counter-clockwise.

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 counter-clockwise.

It follows that while strip  rotates counter-clockwise, 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 minimum-area affine-regular 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 counter-clockwise, then the corresponding  also rotates counter-clockwise.


Consider a configuration where is circumscribed to , and rotate slightly around  in counter-clockwise direction, keeping  fixed. Then and  move downwards along the line , see Figure 7.

Figure 7: Rotating  counter-clockwise.

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 counter-clockwise 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 counter-clockwise 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 minimum-area affine-regular hexagon containing such that a side of  contains an edge of . In addition, if no minimum-area affine-regular hexagon containing shares a vertex with , then each such minimum-area affine-regular hexagon has a side containing a side of .


By Lemma 10, there exists a minimum-area affine-regular 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.

Figure 8: The hexagon , and the convex polygon (shaded).

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


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 single-point 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 minimum-area affine-regular 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 linear-time algorithm to enumerate these hexagons, over all edges of .

Theorem 13.

Given a centrally-symmetric convex -gon , a smallest-area affine-regular hexagon enclosing  can be found in time .


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 counter-clockwise rotation around  by an infinitely small amount.

We then rotate counter-clockwise, until one of the following events occurs:

  • If no longer supports a tangent to  parallel to , replace by the counter-clockwise next vertex of , and continue rotating .

  • If supports an edge of , then replace by the counter-clockwise 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 counter-clockwise next edge of . As long as does not support a tangent to  parallel to , we replace by the counter-clockwise 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 minimum-area 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 minimum-area affine-regular hexagon enclosing is a regular hexagon such that every edge of this hexagon contains an edge of .

Figure 9: Proof of Lemma 14. (a) An optimal enclosing hexagon and the regular 18-gon . (b) When and share two vertices, the area of is larger than the area of . (c) An affine-regular enclosing hexagon . (d) The hexagon .

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 affine-regular 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 affine-regular 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 affine-regular 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 minimum-area translation cover for a family of line segments in the plane.


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 Ben-Or’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 affine-regular hexagon containing .

If on the other hand, for every integer the arc is nonempty, then by Lemma 12, any minimum-area 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 minimum-area hexagon containing has area , where is a minimum-area hexagon containing . Otherwise, if all these arcs are non-empty, then the minimum area is larger than .

Thus, if we could compute in time a minimum-area convex translation cover for the diagonals of , then by Lemma 8 we would also get in time the area of a smallest enclosing affine-regular 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.


By the Cauchy-Crofton 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 smallest-perimeter translation cover for .


We observe first that, if is a segment, then by Theorem 16, the smallest enclosing disk of  is a smallest-perimeter 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


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


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 smallest-perimeter translation cover for either the segment or the triangle, and therefore for . ∎

The minimum enclosing circle is not always the unique minimum-perimeter 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 smallest-perimeter translation cover for the family of all rotated copies of .


We use the same notations as in the proof of Theorem 2: is a smallest-perimeter 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,


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 axis-parallel rectangles. The area of their convex hull is minimized if their bottom left corners coincide (or equivalently if their centers coincide).


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 Brunn-Minkowski 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 . ∎


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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