1 Introduction

Let H be a hyperplane supporting a convex body C in the hyperbolic space \(\mathbb {H}^d\). After [14], we define the width of C determined by H as the distance between H and any farthest ultraparallel hyperplane supporting C. By compactness arguments, there exists at least one such a most distant one, sometimes there are finitely or even infinitely many of them. The symbol \(\textrm{width}_H (C)\) denotes this width of C determined by H. Let us add that there also are different notions of width in \(\mathbb {H}^d\). For instance, these by Santaló [20], Leichtweiss [16], and Horváth [8] who also presents a survey of earlier notions of width in \(\mathbb {H}^d\). Two more such surveys are given by Böröczky and Sagmeister in [3] and by Böröczky, Csépai, and Sagmeister in [4].

By the thickness \(\Delta (C)\) of a convex body \(C \subset \mathbb {H}^d\) we mean the infimum of \(\textrm{width}_H (C)\) over all hyperplanes H supporting C (see [14]). By compactness arguments, this infimum is realized, so \(\Delta (C)\) is the minimum of the numbers \(\textrm{width}_H (C)\).

The above notions of width and thickness are analogous to the classic notions of width and thickness of a convex body in the Euclidean space \(\mathbb {E}^d\) and in the spherical space \(\mathbb {S}^d\) (for the last one, see [11] and the survey article [13]).

We say that e is an extreme point of a convex body \(C \subset \mathbb {H}^d\) if the set \(C \setminus \{ e \}\) is convex.

A convex body \(R \subset \mathbb {H}^d\) is said to be reduced if \(\Delta (Z) < \Delta (R)\) for every convex body Z properly contained in R. This notion is analogous to the notion of a reduced body in \(\mathbb {E}^d\) introduced by Heil [7] (considered later also in the spherical space in a number of papers, e.g., see [11, 13]).

In Section 2, we define a class of reduced polygons analogous to the classes of all reduced polygons in \(\mathbb {E}^2\) and \(\mathbb {S}^2\). It turns out that this analogy is not full. Just this time there are also additional reduced polygons which do not fit to the patterns from \(\mathbb {E}^2\) and \(\mathbb {S}^2\). Consequently, we call the polygons from the mentioned class ordinary reduced polygons. In Section 3, we present a number of properties concerning the width and thickness of ordinary reduced polygons. Moreover, in Section 4, we estimate the diameter of any ordinary reduced polygon in terms of its thickness. Section 5 is devoted to some questions on the diameter, perimeter, circumradius, and inradius of ordinary reduced polygons.

For the convenience of the reader, let us recall Proposition 1 (as Claim 1) and Theorem 1 from [14] (as Claim 2) which will be applied in this paper.

Claim 1

Let \(C \subset \mathbb {H}^2\) be a convex body and H be any supporting hyperplane of C. Then \(\textrm{width}_H (C)\) is equal to the maximum distance between H and a point of C.

Claim 2

For every convex body \(C \subset \mathbb {H}^d\), we have

$$\begin{aligned} \max \{ { \textrm{width}}_H (C); H \mathrm{\ is \ a \ supporting \ hyperplane \ of} \ C \} = \textrm{diam}(C). \end{aligned}$$

We interpret our considerations in the hyperboloid model of \(\mathbb {H}^d\), so in the model on the upper sheet \(x_{d+1}= \sqrt{x_1^2 + \cdots + x_d^2 +1}\) of the two-sheeted hyperboloid. This approach enables a proper similitude of the achieved results with the analogous facts in the Euclidean space and the spherical space. This model is considered by a number of mathematicians, for instance by Reynolds [19]. Our figures present the orthogonal look on the sheet from the above.

For two points p, q, by pq we denote the segment jointing them and by |pq| its length.

The convex hull V of \(k \ge 3\) points in \(\mathbb {H}^2\) such that each of them does not belong to the convex hull of the remaining points is called a convex k-gon in \(\mathbb {H}^d\). These points are called vertices of V. We write \(V= v_1v_2\dots v_k\) provided \(v_1, v_2, \dots , v_k\) are successive vertices of V, when we go around V on the boundary of V according to the positive orientation. When we take \(k \ge 3\) points \(v_1,\ldots ,v_k\), in a hyperbolic circle of \(\mathbb {H}^2\) such that \(|v_1v_2| =\cdots = |v_{k-1}v_k| = |v_kv_1|\), the convex hull of them is called a hyperbolic regular k-gon.

2 On some reduced odd-gons

In a convex n-gon \(v_1 \dots v_n\), we consider indices modulo n.

The following theorem is an analog of the “if” parts of [9, Theorem 7] and [12, Theorem 3.2].

Theorem 1

Let \(V = v_1 \dots v_n\) be a convex odd-gon. Assume that for every \(i \in \{1, \dots , n\}\), the projection of \(v_i\) on the line \(L_i\) containing the side \(S_i = \ v_{i+(n-1)/2} v_{i+(n+1)/2}\) is in the relative interior of this side and the distance between \(v_i\) and \(L_i\) is \(\Delta (V)\). Then V is reduced.

Proof

Consider an odd-gon \(V = v_1v_2\dots v_n\) such that the projection of every \(v_i\) onto the straight line \(L_i\) containing the side \(S_i = \ v_{i+ (n-1)/2}v_{i+ (n+)/2}\) is in the relative interior of this side and that all the distances \(\textrm{dist}(L_i, v_i)\) are equal. Since the projections of vertices \(v_{i+ (n-1)/2}, v_i, v_{i+ (n+1)/2}\), where \(i \in \{1, \dots ,n \}\), onto the lines containing the “opposite” sides \(S_{i+ (n-1)/2}, S_i, S_{i+ (n+1)/2}\), respectively, are in the relative interiors of these sides, the vertex \(v_i\) is the only vertex in the distance \(\Delta (V)\) from the straight line \(L_i\) containing \(S_i\) (so the remaining are closer, i.e, strictly between \(L_i\) and the equidistance curve to \(L_i\) “supporting” V). This is true for every \(i \in \{1,\dots , n\}\). Consequently, the fact that all \(\textrm{dist}(L_i, v_i)\) are equal implies the following. If we take an arbitrary convex body \(Z \subset V\) different from V, then Z does not contain a vertex of V. Consequently, \(\Delta (Z) < \Delta (V)\). Thus V is reduced. \(\square \)

Let us call the polygons described in this theorem ordinary reduced polygons in \(\mathbb {H}^2\). The reason of this name is that the constructions of them are analogous to all reduced polygons in \(\mathbb {E}^2\) and \(\mathbb {S}^2\) (again see [9, Theorem 7] and [12, Theorem 3.2]), and that, surprisingly, there exist some reduced polygons that are not ordinary reduced polygons. Namely, the author recently learned and checked that some hyperbolic rhombi (i.e., convex hulls of two perpendicular segments intersecting each other at midpoints) are reduced. The proof of this statement is a part of an ongoing project on hyperbolic reducedness by K.Jr. Böröczky, A. Freyer, and Á. Sagmeister. By the way, analogously some crosspolytopes in \(\mathbb {H}^d\) (i.e., convex hulls of d perpendicular segments intersecting each other at midpoints) are reduced polytopes. This fact is interesting since still an open question is if there are reduced polytopes in \(\mathbb {E}^d\) for \(d >3\) (see [9, p. 374] and [15, Problem 5]). They exist for \(d=3\) as it follows from the paper [6] by Gonzalez Merino, Jahn, Polyanskii, and Wachsmuth.

We see some ordinary reduced pentagon and heptagon in Fig. 1 and Fig. 2, respectively.

Fig. 1
figure 1

An ordinary reduced pentagon

Full size image

Corollary 1

Every regular hyperbolic odd-gon is reduced.

Corollary 2

The only reduced hyperbolic triangles are the regular ones.

The first corollary is obvious. In order to show the second one, imagine that a non-regular reduced triangle T exists. Its three altitudes are not equal. Take any shortest. Its length is equal to \(\Delta (T)\). Denote by z the vertex at a longer altitude. By a straight line, we cut off from T a sufficiently small piece containing z. We obtain a convex quadrangle of thickness \(\Delta (T)\) in contradiction to the assumption that T is reduced.

Corollary 3

Assume that a straight line L supports an ordinary reduced polygon V. We have \(\textrm{width}_L (V) = \Delta (V)\) if and only if L contains a side of V.

3 Some properties of ordinary reduced polygons

Here is a lemma for \(\mathbb {H}^2\) (true also in \(\mathbb {E}^2\) and \(\mathbb {S}^2\)) which is needed in the proof of part (iii) of the forthcoming theorem.

Lemma

If \(U \subset W\) are different convex polygons, then the perimeter of U is smaller than this of V.

For the proof, observe that we can get V from W by a finite number of successive cuttings of non-degenerate triangles. By the triangle inequality, each time the perimeter decreases. This implies our thesis.

In the following theorem, we use the notation from Theorem 1 . Moreover, by \(p_i\) denote the projection of \(v_i\) onto the line \(L_i\) containing the segment \(v_{i+(n-1)/2} v_{i+(n+1)/2}|\).

Theorem 2

Let V be any ordinary reduced odd-gon \(v_1 \dots v_n\). Then

  1. (i)

    we have \(|v_ip_{i+(n+1)/2}| = |p_iv_{i+(n+1)/2}|\) for \(i=1, \dots , n\),

  2. (ii)

    for every \(i \in \{1, \dots , n\}\), the segment \(v_ip_i\) halves the perimeter of V,

  3. (iii)

    if V is different from a triangle, then we have \(\beta _i < \alpha _i\) for \(i= 1, \dots ,n\), where \(\alpha _i = \angle v_{i+1}v_ip_i\) and \(\beta _i = \angle p_iv_ip_{i+(n+1)/2}\), and for V as a triangle, we have \(\beta _i = \alpha _i\).

Proof

(i) For every \(i \in \{1, \dots , n\}\), take the line \(L_i\) containing the side \(v_{i+(n-1)/2}v_{i+(n+1)/2}\) (see Fig. 2). By the definition of the ordinary reduced polygons, the projection \(p_i\) of \(v_i\) onto \(L_i\) belongs to the relative interior of this side and the distance of \(v_i\) from \(L_i\), so \(|v_ip_i|\), is \(\Delta (V)\).

Fig. 2
figure 2

Illustration to the proof of Theorem 2

Full size image

Take the line \(L_{i+(n+1)/2}\) containing the side \(v_iv_{i+1}\). We know that the projection \(p_{i+(n+1)/2}\) of \(v_{i+(n+1)/2}\) onto \(L_{i+(n+1)/2}\) belongs to the relative interior of this side. We also know that the distance of \(v_{i+(n+1)/2}\) from \(L_{i+(n+1)/2}\), so \(|v_{i+(n+1)/2}p_{i+(n+1)/2}|\), is \(\Delta (V)\).

By \(w_i\) denote the intersection point of \(L_i\) and \(L_{i+(n+1)/2}\) if they intersect. Since \(|v_{i+(n+1)/2} p_{i+(n+1)/2}| = \Delta (V) = |v_ip_i|\), the rectangular triangles \(w_ip_{i+(n+1)/2}v_{i+(n+1)/2}\) and \(w_ip_iv_i\) are congruent (they are reflected copies of each other about the angular bisector of the triangle \(w_iv_iv_{i +(n+1)/2}\) at \(w_i\)). Hence \(|w_iv_{i+(n+1)/2}| = |w_iv_i|\) and \(|w_ip_{i+(n+1)/2}| = |w_ip_i|\). Consequently, \(|v_ip_{i+(n+1)/2}| = |p_iv_{i+(n+1)/2}|\).

If \(L_i\) and \( L_{i+(n+1)/2}\) do not intersect, provide a straight line that intersects both \(L_i\) and \(L_{i+(n+1)/2}\) at points \(a_i \in L_i\) and \(b_i \in L_{i+(n+1)/2}\) with equal angles \(p_ia_ib_i\) and \(a_ib_ip_{i+(n+1)/2}\) to both these lines. Moreover, let \(p_i\) belong to the relative interior of \(a_iv_{i+(n+1)/2}\) and let \(p_{i+(n+1)/2}\) belong to the relative interior of \(b_iv_i\). Observe that the quadrangles \(v_ip_ia_ib_i\) and \(v_{i+(n+1)/2}p_{i+(n+1)/2}b_ia_i\) are congruent (they still have equal corresponding angles, the common side \(a_ib_i\), and \(|v_ip_i| = |v_{i+(n+1)/2}p_{i+(n+1)/2}|\)). Hence, (i) holds true.

(ii) Applying n times (i), we conclude that the sum of the lengths of the boundary segments of V from \(v_i\) to \(p_i\) (moving according to the positive orientation) is equal to the sum of the boundary segments from \(p_i\) to \(v_i\) (again moving according to the positive orientation). Therefore, we obtain the required thesis.

(iii) First assume that \(L_i\) and \(L_{i+(n+1)/2}\) intersect. By (i) and (ii), the lengths of the pieces \(B_i\) of the boundary of V from \(p_{i + (n+1)/2}\) to \(p_i\) and \(B'_i\) from \(v_{i + (n+1)/2}\) to \(v_i\) are equal. Since \(B_i\) is in the triangle \(T_i= p_{i + (n+1)/2}w_ip_i\), then by Lemma for the triangle \(T_i\) as W and the polygon \(p_{i + (n+1)/2}v_{i+1}\dots v_{i + (n-1)/2}p_i\) as U, we get that the length of \(B_i\) is below \(|p_{i + (n+1)/2}w_i| + |w_ip_i|\) (recall that \(w_i\) is defined in (i) and see Fig. 2). Analogously, the length of the boundary of V from \(v_{i + (n+1)/2}\) to \(v_i\) is over \(|v_iv_{i + (n+1)/2}|\). Hence \(|v_{i + (n+1)/2}v_i| < |p_{i + (n+1)/2}w_i| + |w_ip_i|\). From (i), we obtain \(|p_iv_{i + (n+1)/2}| + |v_iv_{i + (n+1)/2|} < |p_iw_i| + |w_iv_i|\). Since the triangles \(v_ip_iv_{i + (n+1)/2}\) and \(v_iw_ip_i\) have the common side \(v_ip_i\) and right angles at \(p_i\), we get \(|p_ip_{i + (n+1)/2}| < |w_ip_i|\), implying \(\beta _i < \alpha _i\).

Clearly, if V is a triangle, then by Corollary 2, it is a regular triangle and thus \(\beta _i = \alpha _i\).

If \(L_i\) and \(L_{i+(n+1)/2}\) do not intersect, we again deal with the quadrangles, as in the last paragraph of the proof of (i). Applying the observation that the length of \(B_i\) is below \(|p_{i+(n+1)/2}b_i| + |b_ia_i| + |a_ip_i|\), we analogously obtain that \(\beta _i < \alpha _i\). \(\square \)

4 On the diameter of ordinary reduced polygons

Proposition

The diameter of any ordinary reduced n-gon is realized only for some pairs of vertices whose indices (modulo n) differ by \((n-1)/2\) or by \((n+1)/2\).

Proof

Take an ordinary reduced n-gon \(V =v_1 \dots v_n\). Clearly its diameter is realized for a pair of vertices. We have to show that the diameter is \(|v_iv_{i+(n-1)/2}|\) or \(|v_iv_{i+(n+1)/2}|\) for an \(i \in \{1, \dots , n\}\).

Claim 1 says that \(\textrm{diam}(V)\) is equal to the maximum \(\textrm{width}_L (V)\) over all straight lines L supporting V and Claim 2 states that \(\textrm{width}_L (V)\) is equal to the maximum distance of a point, so a vertex, of V from L. Hence \(\textrm{diam}(V)\) is equal to the maximum distance from L, over all supporting straight lines L, to a farthest vertex of V.

Let us continuously change the position of the supporting line L of V according to the positive orientation from the position in which it contains a side \(S_i\) up to the position when it contains the next side \(S_{i+1}\), all the time looking for the farthest points of V from L. During this changing, the farthest points of V from L may be only \(v_i\) or \(v_{i+1}\). More precisely, having in mind the definition of an ordinary reduced polygon, we see that at the beginning positions of L, the most distant vertex from each such L is only \(v_i\), then for exactly one position of L, both \(v_i\) and \(v_{i+1}\) are most distant from L, and finally up to the moment when L contains \(S_{i+1}\), only \(v_{i+1}\) is the most distant vertex to L. Consequently, we see that the most distant vertices from \(v_i\) may be only \(v_i\) or \(v_{i+1}\).

We conclude that the diameter of V is \(|v_iv_{i+(n-1)/2}|\) or \(|v_iv_{i+(n+1)/2}|\) for an \(i \in \{1, \dots , n\}\). \(\square \)

The following theorem is analogous to the second part of [9, Theorem 9] and [12, Theorem 3.1] (for completeness let us also mention the paper [17] by Liu and Chang). The proof is analogous, but we provide it here since we apply some facts established in the present paper and some formulas from the hyperbolic plane.

Theorem 3

For every ordinary reduced polygon \(V \subset \mathbb {H}^2\), we have

$$\begin{aligned}\textrm{diam}(V) < \mathrm{arccosh \,}\Bigg ( \sqrt{1 + {1 \over 3} \mathrm{sinh \,}\Delta (V)} \cdot \mathrm{cosh \,}\Delta (V) \Bigg ).\end{aligned}$$

Proof

By Proposition, there is an \(i \in \{1, \dots , n\}\) such that \(\textrm{diam}(V)\) is equal to \(|v_iv_{i+(n+1)/2}|\) or \(|v_iv_{i+(n-1)/2}|\). Consider the first possibility (in the second one, the consideration is analogous).

Put \(r_i = |p_iv_{i + (n+1)/2}|\), \(s_i = |v_iv_{i + (n+1)/2}|\), and \(\gamma _i = \angle p_iv_{i + (n+1)/2}v_i\).

By \(o_i\) denote the intersection point of \(v_ip_i\) and \(v_{i + (n+1)/2}p_{i + (n+1)/2}\). From the observation in the brackets in the last paragraph of the proof of (i) in Theorem 2, we conclude that \(|v_io_i| = |v_{i + (n+1)/2}o_i|\). Hence the triangle \(v_io_iv_{i + (n+1)/2}\) is isosceles. Thus \(\angle o_iv_{i + (n+1)/2}v_i = \angle o_iv_{i + (n+1)/2}v_i\). So \(\gamma _i = \alpha _i + \beta _i\). This and (iii) in Theorem 2 imply \(\gamma _i \ge 2\beta _i\). From \(|v_ip_i| = \Delta (V)\) and by the hyperbolic law of sines for the right triangle \(v_ip_iv_{i + (n+1)/2}v_i\), we obtain \({\mathrm{sinh \,}r_i \over \sin \beta _i} = {\mathrm{sinh \,}\Delta (V) \over \sin \gamma _i}\). Thus, by \(\gamma _i \ge 2\beta _i\), we get \({\mathrm{sinh \,}r_i \over \sin \beta _i} \le {\mathrm{sinh \,}\Delta (V) \over \sin 2 \beta _i}\). This and \(\sin 2\beta _i = 2\sin \beta _i \cos \beta _i\) imply

$$\begin{aligned}\mathrm{sinh \,}r_i \le {\mathrm{sinh \,}\Delta (V) \over 2\cos \beta _i}.\end{aligned}$$

Let us show that \(\beta _i < {\pi \over 6}\). Imagine the opposite case when \(\beta _i \ge {\pi \over 6}\). Then from \(\beta _i \le \alpha _i\) (shown in (iii) of Theorem 2) and \(\gamma _i = \alpha _i + \beta _i\), it follows \(\gamma _i \ge 2\beta _i > \frac{\pi }{3}\) and hence \(\gamma _i \ge \frac{\pi }{3}\), which leads to the conclusion that the sum of angles in the triangle \(v_ip_iv_{i +(n+1)/2}\) is at least \(\pi \). This in the hyperbolic triangle is impossible. So \(\beta _i < {\pi \over 6}\).

From \(\beta _i < {\pi \over 6}\), we get \(\cos \beta _i > \frac{\sqrt{3}}{2}\). Thus, by the inequality just before the preceding paragraph, we obtain \(\mathrm{sinh \,}r_i \ < \frac{\mathrm{sinh \,}\Delta (V)}{\sqrt{3}}\). Hence \(\mathrm{sinh \,}^{\hspace{-1.70709pt}2} r_i < \frac{\mathrm{sinh \,}^{\hspace{-1.70709pt}2} \Delta (V)}{3}\). After we apply \(\mathrm{cosh \,}^{\hspace{-1.70709pt}2} r_i - \mathrm{sinh \,}^{\hspace{-1.70709pt}2} r_i = 1\), i.e., \(\mathrm{sinh \,}^{\hspace{-1.70709pt}2} r_i = \mathrm{cosh \,}^{\hspace{-1.70709pt}2} r_i -1\), we obtain \(\mathrm{cosh \,}^{\hspace{-1.70709pt}2} r_i < 1 + \frac{1}{3} \mathrm{sinh \,}^{\hspace{-1.70709pt}2} \Delta (V)\). So \(\mathrm{cosh \,}r_i < \sqrt{1+\frac{1}{3} \mathrm{sinh \,}^{\hspace{-1.70709pt}2} \Delta (V)}\).

Take into account the equality \(\mathrm{cosh \,}s_i = \mathrm{cosh \,}r_i \mathrm{cosh \,}\Delta (V)\), which follows from the hyperbolic Pythagorean theorem for the triangle \(v_ip_iv_{i+ (n+1)/2}\). In other words, \(\mathrm{cosh \,}r_i = \frac{\mathrm{cosh \,}s_i}{\mathrm{cosh \,}\Delta (V)}\).

By the preceding two paragraphs, we get the following inequality implying the next one

$$\begin{aligned} \frac{\mathrm{cosh \,}s_i}{\mathrm{cosh \,}\Delta (V)}< & {} \sqrt{1 + \frac{1}{3} \mathrm{sinh \,}^{\hspace{-1.70709pt}2} \Delta (V)},\\ s_i< & {} \mathrm{arccosh \,}\Bigg (\sqrt{1+ \frac{1}{3} \mathrm{sinh \,}^{\hspace{-1.70709pt}2} \Delta (V)} \cdot \mathrm{cosh \,}\Delta (V)\Bigg ).\end{aligned}$$

Consequently, by Proposition, we obtain the thesis. \(\square \)

5 A few questions

The author expects that \(1< \frac{\textrm{diam}(V)}{\Delta (V)} < 2\) for every ordinary reduced polygon \(V \subset \mathbb {H}^2\) with 2 as the limit of this quotient for the regular triangle when \(\Delta (V)\) tends to \(\infty \). On the other hand, 1 is the limit of this quotient for the limit of the sequence of regular odd-gons of a fixed thickness. This follows from \(\mathrm{cosh \,}r_i = \frac{\mathrm{cosh \,}s_i}{\mathrm{cosh \,}\Delta (V)}\) established in the proof of Theorem 3 and from Proposition. Still \(s_i\) is the diameter of the regular n-gon with n odd and \(r_i \rightarrow 0\).

We conjecture that the diameter of any non-regular ordinary reduced n-gon in \(\mathbb {H}^2\) is greater than the diameter of the regular n-gon of the same thickness, analogously as in \(\mathbb {E}^2\) and on \(\mathbb {S}^2\). For \(\mathbb {E}^2\), this is formulated in [9, Corollary 6] and for \(\mathbb {S}^2\) this is proved in Theorem 4.1 from the paper [5] by Chen, Hou, and Jin.

We may also consider the question about minimizing the diameter for general ordinary convex polygons of a fixed thickness in \(\mathbb {H}^2\), equivalent to maximizing the thickness for convex polygons of a fixed diameter. For this, compare the paper [2] by Bezdek and Fodor on maximizing the thickness (called “width” there) for the unit diameter convex polygons in \(\mathbb {E}^2\). Also see the related paper [1] by Audet, Hansen, and Messine.

Is the perimeter of any ordinary non-regular reduced n-gon in \(\mathbb {H}^2\) larger than the perimeter of the regular n-gon? We could also ask the same question for the “area” in place of the “perimeter”.

Recall that every reduced polygon \(R \subset E^2\) is contained in a disk of radius \(\frac{2}{3} \Delta (R)\) as shown in [10, Proposition]. This generates the following problem for \(\mathbb {H}^2\). What is the smallest radius of a disk which contains every ordinary reduced polygon of a given thickness in \(\mathbb {H}^2\)? Let us add that an analogous question for a reduced polygon on \(\mathbb {S}^2\) is solved in [18] by Liu and Chang. By the way, why do not consider the dual problem on the inradius of an ordinary reduced polygon in \(\mathbb {H}^2\). We could also ask the same question for the reduced polygons in \(\mathbb {E}^2\) and \(\mathbb {S}^2\).