# Approximation of convex bodies by polytopes with respect to minimal width and diameter

###### Abstract

Denote by the family of convex bodies in and by the minimal width of . We ask for the greatest number such that every contains a polytope with at most vertices for which . We give a lower estimate of for based on estimates of the smallest radius of antipodal pairs of spherical caps that cover the unit sphere of . We show that , and for every . We also consider the dual question of estimating the smallest number such that every there exists a polytope with at most facets for which . We give an upper bound of for . In particular, for .

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^{†}Key words and phrases: Approximation, convex body, polytope, minimal width, diameter, diametral chord, reduced body

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^{†}Mathematics Subject Classification: Primary 52A27.

## 1 Introduction and results

As usual, by a convex body of the -dimensional Euclidean space we mean a bounded convex set with non-empty interior. Denote by the family of convex bodies in and by the subfamily of centrally symmetric convex bodies. The convex hull of the union of two different parallel hyperplanes and in is called a strip. If and are perpendicular to a direction , then is said to be a strip of direction . The distance between and is called the width of . By the width of a bounded set (in particular, of a convex body) in direction we understand the width of the smallest strip of direction containing . The minimal width of (in some papers called also the thickness of ) is denoted by . We tacitly assume that the considered polytopes are convex.

For any we set

This number makes also sense in the more general situation when are any bounded sets in . For instance, in [12] it is considered for the family of finite subsets of instead of .

For every non-empty family of convex bodies in we define

Clearly, if , then for every convex body . Of course, for any family of convex bodies of .

The main aim of the paper is to give some estimates of and , which means that we deal with the approximation of convex bodies by contained polytopes with at most vertices with respect to the minimal width (it is easy to see that the same estimates are for inscribed polytopes). So the paper presents mostly a continuation of the research from [8], where a number of estimates of this kind are given. For this task is also stated in [2], see Problem 3 on p. 452 there.

In this Section we present our results, the proofs are given in Section 4. Only Proposition 3 is obtained in this Section as a result of a discussion on inscribing even-gons of large minimal width in a disk.

The proof of our main Theorem 1 exhibits a very simple but effective method of estimating for every integer . It is based on estimates of the smallest possible angular radius of pairs of antipodal spherical caps that cover the unit sphere of . It is denoted by . In other words, is the smallest positive real for which there exist straight lines through the origin such that the angle between any straight line through the origin and at least one of these lines is at most . A few estimates of are recalled in Section 3. Also some more are established in Lemma 4. In the following theorem and its application, we take .

###### Theorem 1.

For any integers and we have

The paper [8] gives the estimate for every . In Corollary 1 below for (and thus for every ) this estimate is improved.

###### Corollary 1.

We have for and for .

###### Corollary 2.

, , , , , .

###### Corollary 3.

For every integer we have .

Clearly, this estimate for every even says nothing other than that and are at least . A question is how to get an estimate for better than for an even . In particular, an estimate for better than for .

In [8] it is conjectured that and that this is attained for the regular pentagon. Moreover, it is shown there that . We improve this estimate in the second part of the following proposition.

###### Proposition 1.

Every planar convex body contains a regular triangle of minimal width and (this number is approximately ).

Observe that the regular triangle from Proposition 1 not always may be enlarged to a regular triangle inscribed in .

In connection with the first statement of Proposition 1 we recall two facts. First, for bodies of constant width, the estimate increases up to approximately , as proved by Eggleston and Taylor [4]. Secondly, every planar convex body of unit minimal width contains a square of minimal width as shown by Eggleston [3].

A natural question appears what size regular triangle (respectively, a square) of any given direction of a side can be placed in every planar convex body of unit minimal width. In other words, we ask how large regular triangle (respectively, square) may be “rotated” in every such a body. By later Claim 2, it is sufficient to consider these questions for reduced bodies (see the definition in Section 3). The author conjectures that the regular triangle of minimal width and the square of minimal width may be “rotated inside” every convex body of minimal width , and that in both cases the regular triangle is the worst body (i.e., that the “rotated” triangle and square cannot be enlarged).

Now consider finding wide polygons in a centrally symmetric convex body. Recall that by part (2) of Theorem of [8] the value is attained for balls. In particular, finding -gons of the largest minimal width in a disk gives the lower bound of . Of course, we may limit the consideration to the case when is of the unit minimal width.

###### Proposition 2.

If is odd, then the -gon of the largest minimal width inscribed in the disk is only the regular -gon. We have .

Proposition 2 is not true for even . Just immediately for we may be surprised that the square inscribed in is not the best approximating inscribed quadrangle. The reason is that and for the inscribed regular triangle (still it is a degenerated quadrangle) the value is better. What is more, we have (and we conjecture that ). This estimate is realized for the deltoid inscribed in the disk of the minimal width centered at (later, we always take this ), where , , and , see Figure 1.

Fig 1. Wide quadrangles inscribed in a disk

Also when we change the position of in not too much, the new quadrangle still has the minimal width ; namely the general position of the vertex of the inscribed quadrangle (in Figure 1 with two sides marked by broken lines) must be on the circle bounding so that the distances from to the straight lines carrying and remain at least , i.e., that each of the angles and must be at least . The extreme positions and of give two quadrangles (in Figure 1 with two sides marked by pointed lines) and , each of which is an inscribed trapezium with three sides of length .

A widest found hexagon in the disk has vertices for , where , , , , and . Applying the fact that the minimal width of any polygon is realized for the direction perpendicular to a side of it and using the formula for the distance between a point and a straight line, by an easy but tedious evaluation we find that the minimal width of the hexagagon under study is . The author expects that it gives the value of . By the way, the minimal width of this hexagon does not change when we permit to vary between and . For the first extreme position of the side is parallel to , and for the second the side is parallel to .

Finding even-gons of large minimal width with more vertices becomes more complicated. The author tried to find such large octagons inscribed in with a computer approach. This computation shows that is at least , namely by taking the inscribed octagon with vertices for , where are , , , , , , , .

Recapitulating, we obtain the following proposition.

###### Proposition 3.

We have , , .

An open question remains about analogical estimates of better than those for obtained in Theorem 1 and Corollaries 1 and 2. In particular, one can look for for a better estimate for than the one in the first statement of Corollary 1 (at least for ) in analogy to the -dimensional example of the deltoid (see the paragraph just after Proposition 2) which implied .

Finally, we consider the dual problem of estimating from above the diameter of the polytopes containing a convex body of a given diameter. For it is mentioned in [2] on p. 452. For this aim let us introduce the number

Here , and the symbol diam stands for the diameter. For any family we put

###### Theorem 2.

For any integers and we have

###### Corollary 4.

We have , , , , and . Moreover, for and for .

###### Corollary 5.

For every convex body and every integer there exists a polygon with at most sides (so with at most vertices) containing whose diameter is at most .

If we take only even , this corollary reads: for every convex body and every even there exists a polygon with sides (so with at most vertices) containing whose diameter is at most . It is easy to check that the diameter of the circumscribed regular -gon about , for even, is . So for the regular even-gon circumscribing the value from Corollary 5 is attained. We conjecture that this estimate cannot be improved for the disk in the part of . Let us add that for this is true by [7]. The author believes that also for every odd the best (i.e., of the smallest diameter) convex -gon circumscribing the disk is the regular (for which the ratio of the diameters equals ).

## 2 Auxiliary results on diametral chords and reduced bodies

The longest chords of a convex body in a given direction are called diametral chords of in that direction. By compactness arguments there is at least one diametral chord of in arbitrary direction. By Part 33 of [1] or by Theorem 12.18 of [13] for any convex body , the minimum of lengths of diametral chords of equals , which implies the following claim.

###### Claim 1.

Every diametral chord of any convex body has length at least .

###### Lemma 1.

If two diametral chords of a convex body are not parallel, then they intersect.

###### Proof.

Assume that some two non-parallel diametral chords and do not intersect, where are in this order on the boundary of . Since is convex, is a convex quadrangle. If , then cannot be a longest chord of in its direction. If , then cannot be a longest chord of in its direction. So must be a non-degenerated rectangle. Hence and are parallel diametral chords, which contradicts the assumption of our lemma. ∎

###### Lemma 2.

Assume that for a convex body and every direction there is only one diametral chord in direction . Then the position of the diametral chord of changes continuously as varies.

###### Proof.

As explained in [10] (see the paragraph just after Lemma 1 there) the length of the diametral chord is a continuous function of its direction. So if is the limit of directions , then taking also into account the assumption of our lemma, we conclude that the limit of the diametral chords of in these directions is the diametral chord in direction , which confirms the stated continuity. ∎

Recall that a convex body of Euclidean -space is called reduced if for every convex body different from we have . Denote by the family of reduced bodies in . For a survey of results on reduced bodies see [11]. Later we apply reduced bodies in the proof of Proposition 1. This application is based on the following claim from [11].

###### Claim 2.

Every convex body contains a reduced convex body of the same minimal width.

This claim guarantees that , which permits to work with this narrower class of bodies. What is more, every reduced planar body has diameter at most (see [9]), as opposed to the diameter of an arbitrary convex body which may be arbitrarily large. An additional advantage of reduced bodies is the property shown in the following lemma, which is later applied in the proof of Proposition 1.

###### Lemma 3.

Let be a reduced body. In every direction there is exactly one diametral chord of .

###### Proof.

Assume that there are two different diametral chords of in a direction. Then the boundary of contains the two parallel segments connecting the end-points of these chords. But the boundary of does not contain any two parallel segments, as it follows from Theorem 4, part (b) of [9]. A contradiction. ∎

## 3 Some estimates of

In [6], especially see p. 2286 there, a few values of (denoted there by , where ) are estimated or remembered, in particular from [5]. Recall after [6] that for instance , , , , and .

In the following lemma we present more estimates of the numbers .

###### Lemma 4.

We have for every integer , for , and for .

###### Proof.

The first statement is obvious.

Let us show the second one. Take orthogonal straight lines through the origin (say, the axes of a Cartesian coordinate system). Take an arbitrary straight line through the origin and let be one of the two unit vectors parallel to . Without loss of generality, we may assume that for instance the first axis of the coordinate system is the line from amongst our lines for which the angle with is the smallest. So the angle between and (or ) is the smallest. This and imply that is a largest number from amongst . Hence . Moreover, since the angle between and is at most the angle between and , we have and thus . So which gives , i.e., the second statement.

In order to show the third statement replace the line by the following two lines through the origin: containing , and containing . Take an arbitrary straight line through the origin. We will show that the angle between and at least one of the lines is at most . If the angle between and at least one of the lines fulfills this condition, the situation is clear.

Since now assume the opposite, namely that the angle between and every of the lines is over . So for every we have . Hence .

The projection of the point on the two-dimensional plane is on a circle , where . Without loss of generality, we may assume that it is on the shorter piece of the circle “between” and . Then . Consequently, the angle between and is at most , which ends the proof. ∎

## 4 Proofs of theorems, propositions and corollaries

Proof of Theorem 1.

Let be a convex body. By the definition of for there exist straight lines through the center of the unit sphere, each of which passes through the centers of a pair from amongst of the pairs of antipodal spherical caps that cover the sphere. Take a diametral chord of parallel to for . By Claim 1 the length of every of these diametral chords is at least . Construct the convex hull of the union these diametral chords.

Of course, is a polytope with or fewer vertices inscribed in (just observe that possibly some end-points of our diametral chords coincide).

Take an arbitrary straight line through the origin and the narrowest strip orthogonal to which contains . Of course, contains the diametral chords . By the description of the number for , there exists a straight line , where , such that the angle between and is at most . What is more, the strip contains the diametral chord , and the length of is at least . Consequently, the width of the strip is at least . Since this holds true for every straight line through the origin, we conclude that . By the arbitrariness of the convex body we draw the conclusion of our theorem.

For every there is a wide class of the polytopes constructed in this proof. Still we may take miscellaneous families of lines (in particular by rotating a fixed good family of them). Moreover, diametral chords in a direction may be not are unique.

Corollary 2 follows from Theorem 1 and the estimates listed at the beginning of Section 3 for even between and .

Corollary 3 results from Theorem 1 for and , by considering diametral chords of with successive angles , and by applying the first statement of Lemma 4 for .

Proof of Proposition 1.

Clearly, it is sufficient to show the first statement of the proposition.

By Claim 2 there is a reduced body with .

For every direction take the diametral chord of in this direction. It is unique, which results by Lemma 3. By Claim 1 its length is always at least . Also in the perpendicular direction take the unique diametral chord of . By Lemma 1 these diametral chords intersect. Denote by the point of intersection of them. We will show that there exists a direction for which .

Here , and are the specific positions of for .

Take any particular direction in the role of . Denote by the positions of . If , we are done. Assume that . Let for instance . We rotate starting from until we obtain again after rotating by . All the time the positions of the end-points of change continuously (see Lemmas 2 and 3). At the end of the rotation process we get , which is nothing else but . Clearly . From the continuity we conclude that there exists a direction for which .

Consider the quadrangle which is the convex hull of the diametral chord and the perpendicular diametral chord of . This diametral chord is unique by Lemma 3. By Claim 1 the lengths of both these chords are at least . Of course, intersects at . We will show that contains a regular triangle of minimal width . Clearly, later it is sufficient to consider the worst case when is as small as possible, that is . Let for instance (see Figure 2).

For further evaluation put our points in a rectangular coordinate system so that , , , and , where is a number from the interval .

Case 1, when . Observe that contains the regular triangle of side . Namely the triangle with vertices , and . The minimal width of this triangle is larger than .

Case 2, when . Consider the regular triangle with vertex and the opposite side parallel to with end-points in the segments and . Of course, . Let us show that has the smallest size if the midpoint of is and that then its minimal width is .

Consider the line containing , i.e., the line . Take a point in . An evaluation shows that , where , is a regular triangle for . Of course, the length of the sides of this triangle is . This is a decreasing function of . Hence it attains the smallest value in for , i.e., when the midpoint of is at . Its value is . So the minimal width of this triangle is .

Proof of Proposition 2.

Take any non-regular -gon inscribed in . Clearly, one of its sides must be longer than the sides of the regular -gon inscribed in . Observe that the width of perpendicular to this side is smaller than . The reason is that this width is at most the width of the strip containing between the straight line containing this side and the corresponding parallel supporting line of the disk. We omit an easy calculation which confirms the formula from the second statement.

Proof of Theorem 2.

It is sufficient to show that for arbitrary we have . Of course, we may limit the consideration to the case when the diameter of is one, so let since now. It is well known that there exists a body of unit constant width containing (e.g., see [1], section 64).

By the definition of the number , taking we see that there exists a family of straight lines through the origin such that the angular distance between an arbitrary line through the origin and at least one of these lines is at most . For every , take the strip of width orthogonal to which contains . The intersection of these strips is a polytope with at most facets. Clearly, .

Denote by the image of a convex body under the central symmetrization. Since is a body of constant width, is the ball of unit width.

Clearly, is a centrally symmetric polytope circumscribed about the ball . It has pairs of symmetric facets orthogonal to the lines from . It is well known that every body and its centrally symmetrized body have equal diameters (e.g., see [1], section 42). Thus .

Let be a boundary point of . Let be the line through and . Take a line whose angle with is at most (see Figure 3). By the definition of and since it is centrally symmetric we see that there exists a pair of facets of orthogonal to . Let be one of these two facets. Observe that intersects . The reason is that supports the ball at exactly one point and passes through .

Denote by the hyperplane carrying . Since is a supporting hyperplane of , we see that is on the same side of the hyperplane as . In particular, may be on .

Recall that contains at least lines. Since for we have , and since by Lemma 4 we have , we conclude that the angle between and is below . Thus there exists a point such that .

Of course, . From the right triangle and from (this follows from the fact that is a boundary point of the ball ) we see that . From we get . Thus for the symmetric boundary point of . Since is centrally symmetric, from the arbitrariness of its boundary point we conclude that . Hence . Thus . Since and , this implies , which ends the proof.

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