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1 Introduction

This article is concerned with cyclic polygons, i.e., convex polygons inscribed in a circle. We will provide a new proof of the following elementary theorem in Sect. 2.

Theorem 1.1

There exists a Euclidean cyclic polygon with \(n\ge 3\) sides of lengths \(\ell _{1},\ldots ,\ell _n\in \mathbb R_{>0}\) if and only if they satisfy the polygon inequalities

$$\begin{aligned} \ell _{k}<\sum _{\begin{array}{c} i=1\\ i\not =k \end{array}}^{n}\ell _{i}, \end{aligned}$$
(1)

and this cyclic polygon is unique.

Our proof involves a variational principle with the central angles as variables. The variational principle has a geometric interpretation in terms of volume in 3-dimensional hyperbolic space (see Remark 2.6). Another striking feature of our proof is the use of a fundamental inequality of information theory:

Theorem

(Information Inequality) Let \(p=(p_1, \ldots , p_m)\) and \(q=(q_1, \ldots , q_m)\) be discrete probability distributions, then

$$\begin{aligned} \sum _{k=1}^{m} p_k\log \frac{p_k}{q_k} \ge 0, \end{aligned}$$
(2)

and equality holds if and only if \(p=q\).

The left hand side of inequality (2) is called the Kullback–Leibler divergence or information gain of q from p, also the relative entropy of p with respect to q. The inequality follows from the strict concavity of the logarithm function (see, e.g., Cover and Thomas [3]).

In Sects. 3 and 4 we provide proofs for non-Euclidean versions of Theorem 1.1. The spherical version requires an extra inequality:

Theorem 1.2

There exists a spherical cyclic polygon with \(n\ge 3\) sides of lengths \(\ell _{1},\ldots ,\ell _n\in \mathbb R_{>0}\) if and only if they satisfy the polygon inequalities (1) and

$$\begin{aligned} \sum _{i=1}^{n}\ell _{i} < 2\pi , \end{aligned}$$
(3)

and this cyclic spherical polygon is unique.

Inequality (3) is necessary because the perimeter of a circle in the unit sphere cannot be greater than \(2\pi \), and the perimeter of the inscribed polygon is a lower bound. We require strict inequality to exclude polygons that degenerate to great circles (with all interior angles equal to \(\pi \)).

In Sect. 3, we prove Theorem 1.2 by a straightforward reduction to Theorem 1.1: connecting the vertices of a spherical cyclic polygon by straight line segments in the ambient Euclidean \(\mathbb R^{3}\), one obtains a Euclidean cyclic polygon.

In the case of hyperbolic geometry, the notion of “cyclic polygon” requires additional explanation. We call a convex hyperbolic polygon cyclic if its vertices lie on a curve of constant non-zero curvature. Such a curve is either

  • a hyperbolic circle if the curvature is greater than 1,

  • a horocycle if the curvature is equal to 1,

  • a hypercycle, i.e., a curve at constant distance from a geodesic if the curvature is strictly between 0 and 1.

Theorem 1.3

There exists a hyperbolic cyclic polygon with \(n\ge 3\) sides of lengths \(\ell _{1},\ldots ,\ell _n\in \mathbb R\) if and only if they satisfy the polygon inequalities (1), and this cyclic hyperbolic polygon is unique.

We prove this theorem in Sect. 4. The case of hyperbolic polygons inscribed in circles can be reduced to Theorem 1.1 by considering the hyperboloid model of the hyperbolic plane: Connecting the vertices of a hyperbolic polygon inscribed in a circle by straight line segments in the ambient \(\mathbb R^{2,1}\), one obtains a Euclidean cyclic polygon.

The cases of polygons inscribed in horocycles and hypercycles cannot be reduced to the Euclidean case because the intrinsic geometry of the affine plane of the polygon is not Euclidean: In the horocycle case, the scalar product is degenerate with a 1-dimensional kernel. Hence, this case reduces to the case of degenerate polygons inscribed in a straight line. It is easy to deal with. In the hypercycle case, the scalar product is indefinite. This case reduces to polygons inscribed in hyperbolas in flat \(1+1\) spacetime. The variational principle of Sect. 2 can be adapted for this case (see Sect. 5), but the corresponding target function fails to be concave or convex. It may be possible to base a proof of existence and uniqueness on this variational principle, perhaps using a \(\min \)-\(\max \)-argument, but we do not pursue this route in this article. Instead, we deal with polygons inscribed in hypercycles using a straightforward analytic argument.

Some history, from ancient to recent. Theorems 1.1–1.3 belong to the circle of results connected with the classical isoperimetric problem. As the subject is ancient and the body of literature is vast, we can only attempt to provide a rough historical perspective and ask for leniency regarding any essential work that we fail to mention.

The early history of the relevant results about polygons is briefly discussed by Steinitz [13, Sect. 16]. Steinitz goes on to discuss analogous results for polyhedra, a topic into which we will not go. A more recent and comprehensive survey of proofs of the isoperimetric property of the circle was given by Blåsjö [2].

It was known to Pappus that the regular n-gon had the largest area among n-gons with the same perimeter, and that the area grew with the number of sides. This was used to argue for the isoperimetric property of the circle:

Theorem 1.4

(Isoperimetric Theorem) Among all closed planar curves with given length, only the circle encloses the largest area.

It is not clear who first stated the following theorem about polygons:

Theorem 1.5

(Secant Polygon) Among all n-gons with given side lengths, only the one inscribed in a circle has the largest area.

This was proved by Moula [8], by L’Huilier [5] (who cites Moula), and by Steiner [12] (who cites L’Huilier). L’Huilier also proved the following theorem:

Theorem 1.6

(Tangent Polygon) Among all convex n-gons with given angles, only the one circumscribed to a circle has the largest area when the perimeter is fixed and and smallest perimeter when the area is fixed.

Steiner also proves versions of Theorems 1.5 and 1.6 for spherical polygons. None of these authors deemed it necessary to prove the existence of a maximizer, an issue that became generally recognized only after Weierstrass [14]. For polygons, the existence of a maximizer follows by a standard compactness argument.

Blaschke [1, Sect. 12], notes that the quadrilateral case (\(n=4\)) of Theorem 1.5 can easily be deduced from the Isoperimetric Theorem 1.4 using Steiner’s four-hinge method. Conversely, one can similarly deduce Theorem 1.4 and the general Theorem 1.5 from the quadrilateral case of Theorem 1.5. He remarks that the quadrilateral case of Theorem 1.5 can be proved directly by deriving the following equation for the area A of a quadrilateral with sides \(\ell _{k}\):

$$\begin{aligned} A^{2} = (s-\ell _{1})(s-\ell _{2})(s-\ell _{3})(s-\ell _{4}) - \ell _{1}\ell _{2}\ell _{3}\ell _{4}\cos ^{2}\theta , \end{aligned}$$
(4)

where \(s=(\ell _{1}+\ell _{2}+\ell _{3}+\ell _{4})/2\) is half the perimeter, and \(\theta \) is the arithmetic mean of two opposite angles.

Neither Blaschke, nor Steiner, L’Huilier, or Moula provide an argument for the uniqueness of the maximizer in Theorem 1.5 or 1.6. It seems that even after Weierstrass, the fact that the sides determine a cyclic polygon uniquely was considered too obvious to deserve a proof.

Penner [9, Theorem 6.2] gives a complete proof of Theorem 1.1. He proceeds by showing that there is one and only one circumcircle radius that allows the construction of a Euclidean cyclic polygon with given sides (provided they satisfy the polygon inequalities).

Schlenker [11] proves Theorems 1.2 and 1.3, and also the isoperimetric property of non-Euclidean cyclic polygons, i.e., the spherical and hyperbolic versions of Theorem 1.5. His proofs of the isoperimetric property are based on the remarkable equation

$$\begin{aligned} \sum \dot{\alpha }_{i}v_{i}=0 \end{aligned}$$
(5)

characterizing the change of angles \(\alpha _{i}\) of a spherical or hyperbolic polygon under infinitesimal deformations with fixed side lengths. Here, \(v_{i}\in \mathbb R^{3}\) are the position vectors of the polygon’s vertices in the sphere or in the hyperboloid, respectively. To prove the uniqueness of spherical and hyperbolic cyclic polygons with given sides he uses separate arguments similar to Penner’s.

2 Euclidean Polygons. Proof of Theorem 1.1

To construct an inscribed polygon with given side lengths \(\ell =(\ell _{1},\ldots ,\ell _n)\in \mathbb R_{>0}^n\) (see Fig. 1) is equivalent to finding a point \((\alpha _1,\ldots \alpha _{n})\) in the set

$$\begin{aligned} D_n=\left\{ \alpha \in \mathbb R_{>0}^n \;\big |\;\sum _{k=1}^n \alpha _k = 2\pi \right\} \subset \mathbb R^{n} \end{aligned}$$
(6)

satisfying, for some \(R\in \mathbb R\) and for all \(k\in \{1,\ldots ,n\}\),

$$\begin{aligned} \frac{\ell _k}{2}=R\sin \frac{\alpha _k}{2}. \end{aligned}$$
(7)
Fig. 1
figure 1

Euclidean polygon inscribed in a circle

Full size image

This problem admits the following variational formulation. Define the function \(f_\ell :\mathbb R^n \rightarrow \mathbb R\) by

$$\begin{aligned} f_\ell (\alpha ) = \sum _{k=1}^n\big ( \text {Cl}_2(\alpha _k) + \log (\ell _k)\,\alpha _k\big ) \end{aligned}$$
(8)

where \(\text {Cl}_2\) denotes Clausen’s integral [4]:

$$\begin{aligned} \text {Cl}_2(x)= -\int _0^x \log \Big |2\sin \frac{t}{2}\Big |\,dt. \end{aligned}$$
(9)

Clausen’s integral is closely related to Milnor’s Lobachevsky function [6]:

The function \(\text {Cl}_2:\mathbb R\rightarrow \mathbb R\) is continuous, \(2\pi \)-periodic, and odd. It is differentiable except at integer multiples of \(2\pi \) where the graph has vertical tangents (see Fig. 2).

Fig. 2
figure 2

Graph of Clausen’s integral \(\text {Cl}_2(x)\)

Full size image

Proposition 2.1

(Variational Principle) A point \(\alpha \in D_n\) is a critical point of \(f_\ell \) restricted to \(D_{n}\) if and only if there exists an \(R\in \mathbb R\) satisfying equations (7).

Proof

A point \(\alpha \in D_n\) is a critical point of \(f_\ell \) restricted to \(D_n\) if and only if there exists a Lagrange multiplier \(\log R\) such that \(\nabla f_\ell (\alpha )=(\log R)\nabla g(\alpha )\) for the constraint function \(g(\alpha )=\sum \alpha _k\), i.e.,

$$\begin{aligned} \begin{pmatrix} -\log \left| 2\sin \frac{\alpha _1}{2}\right| + \log \ell _1\\ \vdots \\ -\log \left| 2\sin \frac{\alpha _n}{2}\right| + \log \ell _n \end{pmatrix} = \log R \begin{bmatrix} 1\\ \vdots \\ 1 \end{bmatrix}{.} \end{aligned}$$

Since \(0<\alpha _k<2\pi \) we may omit the absolute value signs, obtaining equations (7). \(\square \)

Thus, to prove Theorem 1.1, we need to show that \(f_\ell \) has a critical point in \(D_n\) if and only if the polygon inequalities (1) are satisfied, and that this critical point is then unique. The following proposition and corollary deal with the uniqueness claim.

Proposition 2.2

The function \(f_\ell \) is strictly concave on \(D_n\).

Corollary 2.3

If \(f_\ell \) has a critical point in \(D_n\), it is the unique maximizer of \(f_{\ell }\) in the closure \( \bar{D}_n=\{\alpha \in \mathbb R_{\ge 0}^n \;|\;\sum \alpha _k = 2\pi \}. \)

This proves the uniqueness claim of Theorem 1.1.

Proof

(of Proposition 2.2) We will show that

$$\begin{aligned} V_n(\alpha )=\sum _{k=1}^n \text {Cl}_2(\alpha _k) \end{aligned}$$
(10)

is strictly concave on \(D_{n}\). Since \(V_{n}\) differs from \(f_\ell \) by a linear function, this is equivalent to the claim.

Rivin [10, Theorem 2.1] showed that \(V_{3}\) is strictly concave on \(D_{3}\). For \(n>3\) we proceed by induction on n by “cutting off a triangle”: first, note the obvious identity

$$\begin{aligned} V_n(\alpha _1,\dots , \alpha _n)= V_{n-1}(\alpha _1,\dots , \alpha _{n-1}+\alpha _n) -\text {Cl}_2(\alpha _{n-1}&+\alpha _n)\\&+\text {Cl}_2(\alpha _{n-1})+\text {Cl}_2(\alpha _n). \end{aligned}$$

Since Clausen’s integral is \(2\pi \)-periodic and odd,

$$\begin{aligned} -\text {Cl}_2(\alpha _{n-1}+\alpha _n)=\text {Cl}_2(2\pi -\alpha _{n-1}-\alpha _n)=\text {Cl}_2\left( \sum _{k=1}^{n-2}\alpha _k\right) , \end{aligned}$$

so

$$\begin{aligned} V_n(\alpha _1,\dots , \alpha _n)=V_{n-1}(\alpha _1,\dots , \alpha _{n-1}+\alpha _n)+V_3\left( \sum _{k=1}^{n-2}\alpha _k,\alpha _{n-1},\alpha _n \right) . \end{aligned}$$

Hence, if \(V_{n-1}\) and \(V_3\) are strictly concave on \(D_{n-1}\) and \(D_3\), respectively, the claim for \(V_n\) follows. \(\square \)

Since \(f_{\ell }\) attains its maximum on the compact set \(\bar{D}_{n}\), it remains to show that the maximum is attained in \(D_{n}\) if and only if the polygon inequalities (1) are satisfied. This is achieved by the following Propositions 2.4 and 2.5.

Note that \(\bar{D}_{n}\) is an \((n-1)\)-dimensional simplex in \(\mathbb R^{n}\). Its vertices are the points \(2\pi e_{1},\ldots ,2\pi e_{n}\), where \(e_{k}\) are the canonical basis vectors of \(\mathbb R^{n}\). The relative boundary of the simplex \(\bar{D}_n\) is

$$\begin{aligned} \partial \bar{D}_{n}=\{\alpha \in \bar{D}_{n}\;|\;\alpha _{k}=0 \text { for at least one } k\}. \end{aligned}$$
(11)

Proposition 2.4

If the function \(f_\ell \) attains its maximum on the simplex \(\bar{D}_{n}\) at a boundary point \(\alpha \in \partial \bar{D}_{n}\), then \(\alpha \) is a vertex.

Proof

Suppose \(\alpha \in \partial \bar{D}_{n}\) is not a vertex. We need to show that \(f_{\ell }\) does not attain its maximum at \(\alpha \). This follows from the fact that the derivative of \(f_{\ell }\) in a direction pointing towards \(D_{n}\) is \(+\infty \).

Indeed, suppose \(v\in \mathbb R_{\ge 0}^{n}\), \(\sum _{k}v_{k}=0\) and \(v_{k}>0\) if \(\alpha _{k}=0\). Then \(\alpha +tv\in D_{n}\) for small enough \(t>0\), and because \(\lim _{x\rightarrow 0}\text {Cl}_2'(x)=+\infty \),

$$\begin{aligned} \lim _{t\rightarrow 0}\;\frac{d}{dt}\,f_{\ell }(\alpha +tv)=+\infty . \end{aligned}$$
(12)

Hence \(f_{\ell }(\alpha +tv)>f_{\ell }(\alpha )\) for small enough \(t>0\). \(\square \)

Proposition 2.5

The function \(f_\ell \) attains its maximum on \(\bar{D}_{n}\) at a vertex \(2\pi e_{k}\) if and only if

$$\begin{aligned} \ell _{k}\ge \sum _{\begin{array}{c} i=1\\ i\not =k \end{array}}^{n}\ell _{i}. \end{aligned}$$
(13)

Proof

By symmetry, it is enough to consider the case \(k=n\), i.e., to show that the function \(f_\ell \) attains its maximum on \(\bar{D}_{n}\) at the vertex \((0,\ldots ,0,2\pi )\) if and only if \(\ell _{n}\ge \sum _{k=1}^{n-1}\ell _{k}\). To this end, we will calculate the directional derivative of \(f_{\ell }\) in directions \(v\in \mathbb R^{n}\) pointing inside \(D_{n}\), i.e., satisfying

$$\begin{aligned} v_k \ge 0\quad \text {for}\quad k\in \{1,\dots ,n-1\}, \quad v_{n}=-\sum _{k=1}^{n-1}v_{k}<0. \end{aligned}$$

Since we are only interested in the sign, we may assume v to be scaled so that

$$\begin{aligned} \sum _{k=1}^{n-1}v_{k}=1, \qquad v_{n}=-1. \end{aligned}$$

Clausen’s integral has the asymptotic behavior

$$\begin{aligned} \text {Cl}_2(x)= -x \log |x|+x+o(x)\quad \text {as}\quad x\rightarrow 0. \end{aligned}$$
(14)

This can be seen by considering

$$\begin{aligned}\textstyle \text {Cl}_2(x)=-\int _0^x \log \left| (2\sin \frac{t}{2})/t\right| \,dt -\int _0^x \log |t|\,dt. \end{aligned}$$

Using (14) and the \(2\pi \)-periodicity of Clausen’s integral, one obtains

$$\begin{aligned} \begin{aligned} f_{\ell }(2\pi e_{n}+tv)-f_{\ell }(2\pi e_{n})&= \sum _{k=1}^{n}\big (-tv_{k}\log |v_{k}|+tv_{k}\log \ell _{k}\big ) + o(t)\\&= -\sum _{k=1}^{n-1}tv_{k}\log \frac{v_{k}}{\ell _{n}}-t\log \ell _{n}+o(t), \end{aligned} \end{aligned}$$

and hence

$$\begin{aligned} \frac{d}{dt}\Big |_{t=0}f(2\pi e_{n}+tv)= -\sum _{k=1}^{n-1}v_{k}\log \frac{v_{k}}{\ell _{k}}-\log \ell _{n}. \end{aligned}$$

Now we invoke the information inequality (2) for the discrete probability distributions \((v_1,\dots ,v_{n-1})\) and \((\ell _1,\dots ,\ell _{n-1})/\sum _{k=1}^{n-1} \ell _k\). Thus,

$$\begin{aligned} \frac{d}{dt}\Big |_{t=0}f(2\pi e_{n}+tv) = \underbrace{-\sum _{k=1}^{n-1} v_k \log \left( \frac{v_k}{\ell _n/\sum _{m=1}^{n-1} \ell _m}\right) }_{\le 0} +\log \left( \frac{\sum _{k=1}^{n-1} \ell _k}{\ell _n}\right) . \end{aligned}$$

If \(\ell _n \ge \sum _{k=1}^{n-1} \ell _k\), then

$$\begin{aligned} \frac{d}{dt}\Big |_{t=0}f(2\pi e_{n}+tv)\le 0. \end{aligned}$$

With the concavity of \(f_{\ell }\) (Proposition 2.2), this implies that \(f_\ell \) attains its maximum on \(\bar{D}_{n}\) at \((0, \ldots , 0, 2\pi )\).

If, on the other hand, \(\ell _n < \sum _{k=1}^{n-1} \ell _k\), then we obtain, for \(v_k=\ell _k/\sum _{m=1}^{n-1} \ell _m\),

$$\begin{aligned} \frac{d}{dt}\Big |_{t=0}f(2\pi e_{n}+tv) > 0. \end{aligned}$$

This implies that \(f_\ell \) does not attain its maximum at \((0, \ldots , 0, 2\pi )\). \(\square \)

This completes the proof of Theorem 1.1.

Remark 2.6

The function \(V_{n}\) has the following interpretation in terms of hyperbolic volume [6]. Consider a Euclidean cyclic n-gon with central angles \(\alpha _{1},\ldots ,\alpha _{n}\). Imagine the Euclidean plane of the polygon to be the ideal boundary of hyperbolic 3-space in the Poincaré upper half-space model. Then the vertical planes through the edges of the polygon and the hemisphere above its circumcircle bound a hyperbolic pyramid with vertices at infinity. Its volume is \(\frac{1}{2}V_{n}(\alpha _{1},\ldots ,\alpha _{n})\). Together with Schläfli’s differential volume equation (rather, Milnor’s generalization that allows for ideal vertices [7]), this provides another way to prove Proposition 2.1.

3 Spherical Polygons. Proof of Theorem 1.2

The polygon inequalities (1) are clearly necessary for the existence of a spherical cyclic polygon because every side is a shortest geodesic. That inequality (3) is also necessary was already noted in the introduction. It remains to show that these inequalities are also sufficient, and that the polygon is unique.

Fig. 3
figure 3

Spherical and Euclidean polygons

Full size image

We reduce the spherical case to the Euclidean one as shown in Fig. 3. Connecting the vertices of a spherical cyclic polygon with line segments in the ambient Euclidean space, one obtains a Euclidean cyclic polygon whose circumradius is smaller than 1. Conversely, every Euclidean polygon inscribed in a circle of radius less than 1 corresponds to a unique spherical cyclic polygon. The spherical side lengths \(\ell \) are related to the Euclidean lengths \(\bar{\ell }\) by

$$\begin{aligned} {\bar{\ell }} = 2 \sin \frac{\ell }{2}. \end{aligned}$$
(15)

It remains to show the following two propositions:

Proposition 3.1

If the spherical lengths \(\ell \in \mathbb R_{>0}^{n}\) satisfy the inequalities (1) and (3), then the Euclidean lengths \(\bar{\ell }\) defined by (15) satisfy the inequalities (1) as well. By Theorem 1.1 there is then a unique Euclidean cyclic polygon \(P_{\bar{\ell }}\) with side lengths \(\bar{\ell }\).

Proposition 3.2

The circumradius \(\bar{R}\) of the polygon \(P_{\bar{\ell }}\) of Proposition 3.1 is strictly less than 1.

We will use the following estimate in the proof of Proposition 3.1:

Lemma 3.3

(Sum of Sines Estimate) If \(\beta _1, \ldots , \beta _n\in \mathbb R_{\ge 0}\) satisfy \(\sum _{k=1}^n \beta _k \le \pi \), then

$$\begin{aligned} \sin \Bigg (\sum _{k=1}^n \beta _k\Bigg ) \le \sum _{k=1}^n \sin \beta _k. \end{aligned}$$
(16)

Proof

(of Lemma 3.3) By induction on n, the base case \(n=1\) being trivial. For the inductive step, use the addition theorem,

$$\begin{aligned} \sin \Bigg (\sum _{k=1}^{n+1} \beta _k\Bigg )= \sin \Bigg (\sum _{k=1}^n \beta _k\Bigg )\cos \beta _{n+1} +\cos \Bigg (\sum _{k=1}^n \beta _k\Bigg )\sin \beta _{n+1}, \end{aligned}$$

and note that the cosines are \(\le 1\). \(\square \)

Remark 3.4

The statement of Lemma 3.3 can be strengthened. Equality holds in (16) if and only if at most one \(\beta _{k}\) is greater than zero. This is easy to see, but we do not need this stronger statement in the following proof.

Proof

(of Proposition 3.1) Suppose \(\ell _1, \ldots , \ell _n\in \mathbb R_{>0}\) satisfy the polygon inequalities (1) and (3). We need to show that \(\bar{\ell }_1, \ldots , \bar{\ell }_n\) defined by (15) satisfy

$$\begin{aligned} \bar{\ell }_{k}<\sum _{i\not =k}\bar{\ell }_{i}. \end{aligned}$$
(17)

To this end, we will show that

$$\begin{aligned} \sin \frac{\ell _k}{2} < \sin \Bigg (\sum _{i\not =k} \frac{\ell _i}{2}\Bigg ), \end{aligned}$$
(18)

from which inequality (17) follows by Lemma 3.3. To prove inequality (18), we consider two cases separately.

  • \(\sum _{i\not =k}\ell _i \le \pi \). Inequality (18) simply follows from the polygon inequality \(\ell _{k}<\sum _{i\not =k}\ell _{i}\) and the monotonicity of the sine function on the closed interval \([0,\frac{\pi }{2}]\).

  • \(\sum _{i\not =k}\ell _i \ge \pi \). Note that \(2\pi >\sum _i\ell _i\) implies \(2\pi -\ell _{k}>\sum _{i\not =k}\ell _i\), and hence

    $$\begin{aligned} 2\pi>2\pi -\ell _{k}>\sum _{i\not =k}\ell _{i}\ge \pi . \end{aligned}$$
    (19)

    Inequality (18) follows from \(\sin \frac{\ell _k}{2}=\sin (\pi -\frac{\ell _k}{2})\) and the monotonicity of the sine function on the closed interval \([\frac{\pi }{2},\pi ]\).

This completes the proof of (18) and hence the proof of Proposition 3.1. \(\square \)

Proof

(of Proposition 3.2) Let \(\alpha _{k}\) be the central angles of the Euclidean cyclic polygon \(P_{\bar{\ell }}\). Then

$$\begin{aligned} \sin \frac{\ell _k}{2} = \frac{\bar{\ell }_k}{2} = \bar{R} \sin \frac{\alpha _k}{2}, \end{aligned}$$
(20)

by (7) and (15). Note that \(\alpha _{k}\) are the central angles of both the Euclidean and the spherical polygon (provided it exists). We consider two cases separately.

First, suppose that \(\alpha _k \le \pi \) for all k. Since \( \sum _k \ell _k < 2 \pi = \sum _k \alpha _k\), there is some k such that \(\ell _k < \alpha _k\). Then \(\sin \frac{\ell _k}{2} < \sin \frac{\alpha _k}{2}\), and equation (20) implies that \(\bar{R} < 1\).

Otherwise, since \(\sum _k\alpha _k = 2\pi \), there is exactly one i such that \(\alpha _i > \pi \), and \(\alpha _k < \pi \) for all \(k\ne i\). By symmetry, it is enough to consider the case

$$\begin{aligned} \alpha _1 > \pi ,\quad \alpha _k < \pi \quad \text {for}\quad k\in \{2,\ldots ,n\}. \end{aligned}$$

For future reference, we note that \(\alpha _1 > \pi \) implies that \(\bar{\ell }_{1}\) is the longest side of \(P_{\bar{\ell }}\). (Use (20) and the monotonicity of the sine function.)

We will show \(\bar{R}<1\) by induction on n. First, assume \(n=3\). Then (18) says

$$\begin{aligned} \sin \frac{\ell _1}{2}<\sin \frac{\ell _2+\ell _3}{2}. \end{aligned}$$

By (20) and using \(2\pi -\alpha _{1}=\alpha _{2}+\alpha _{3}\), we have

$$\begin{aligned} \sin \frac{\ell _1}{2} = \bar{R}\sin \frac{\alpha _2}{2}\cos \frac{\alpha _3}{2} + \bar{R} \cos \frac{\alpha _2}{2}\sin \frac{\alpha _3}{2}, \end{aligned}$$
(21)

and

$$\begin{aligned} {\begin{matrix} \sin \frac{\ell _2+\ell _3}{2} &{}=\sin \frac{\ell _2}{2}\cos \frac{\ell _3}{2} + \cos \frac{\ell _2}{2}\sin \frac{\ell _3}{2}\\ &{}= \bar{R}\sin \frac{\alpha _2}{2}\cos \frac{\ell _3}{2} + \bar{R}\cos \frac{\ell _2}{2}\sin \frac{\alpha _3}{2}\;. \end{matrix}} \end{aligned}$$
(22)

For at least one \(k \in \{2,3\}\), \(\cos \frac{\alpha _k}{2} < \cos \frac{\ell _k}{2}\) and hence \(\sin \frac{\alpha _{k}}{2} > \sin \frac{\ell _{k }}{2}\). Equation (20) implies \(\bar{R}<1\).

Now assume that \(\bar{R}<1\) has already been shown if \(P_{\bar{\ell }}\) has at most n sides. Suppose \(P_{\bar{\ell }}\) has \(n+1\) sides. The idea of the following argument is to cut off a triangle with sides \(\bar{\ell }_{n}\), \(\bar{\ell }_{n+1}\), and \(\bar{\lambda } = 2\bar{R}\sin \frac{\alpha _n + \alpha _{n+1}}{2}\). Since \(\bar{\lambda }\le \bar{\ell }_{1}\) (the longest side), and \(\bar{\ell }_{1}\le 2\) by (15), we may define \(\lambda = 2 \arcsin \frac{ \bar{\lambda }}{2}\). Now assume \(\bar{R} \ge 1\). Then, by the inductive hypothesis, the polygon inequalities (1) or (3) are violated for the cut-off triangle and the remaining n-gon. Inequality (3) cannot be violated because it was assumed to hold for \(\ell _{1},\ldots ,\ell _{n+1}\). Hence,

$$\begin{aligned} \ell _1 \ge \ell _2 + \dots + \ell _{n-1} + \lambda \quad \text {and}\quad \lambda \ge \ell _n + \ell _{n+1}. \end{aligned}$$

This implies \(\ell _1 \ge \ell _2 + \dots + \ell _{n+1}\). Conversely, if (1) and (3) hold, then \(\bar{R} < 1\). This completes the proof of Proposition 3.2. \(\square \)

4 Hyperbolic Polygons. Proof of Theorem 1.3

The polygon inequalities (1) are clearly necessary for the existence of a hyperbolic cyclic polygon, because every side is a shortest geodesic. It remains to show that they are also sufficient, and that the polygon is unique, i.e., Proposition 4.2. First, we review some basic facts from hyperbolic geometry.

As in the spherical case (Sect. 3), we will connect vertices by straight line segments in the ambient vector space. But instead of the sphere, we consider the hyperbolic plane in the hyperboloid model,

$$\begin{aligned} \mathbb H^2=\{x\in \mathbb R^{2,1} \,|\, \langle x,x\rangle =-1,\, x_3>0\}, \end{aligned}$$

where \(\mathbb R^{2,1}\) denotes the vector space \(\mathbb R^{3}\) equipped with the scalar product

$$\begin{aligned} \langle x,y\rangle =x_1y_1+x_2y_2-x_3y_3, \end{aligned}$$

and lengths and angles in \(\mathbb H^{2}\) are measured using the Riemannian metric induced by this scalar product.

Straight lines (i.e., geodesics) in \(\mathbb H^{2}\) are the intersections of \(\mathbb H^{2}\) with 2-dimensional subspaces of \(\mathbb R^3\). The length \(\ell \) of the geodesic segment connecting points \(p,q\in \mathbb H^{2}\) is determined by

$$\begin{aligned} \cosh \ell = -\langle p,q\rangle . \end{aligned}$$

The length of the straight line segment connecting points \(p,q\in \mathbb H^2\) in the ambient \(\mathbb R^{2,1}\) is

$$\begin{aligned} \bar{\ell }=\sqrt{\langle p-q,p-q\rangle }. \end{aligned}$$

This chordal length \(\bar{\ell }\) and the hyperbolic distance \(\ell \) are related by

$$\begin{aligned} \frac{\bar{\ell }}{2}=\sinh \frac{\ell }{2}. \end{aligned}$$
(23)

An affine plane in \(\mathbb R^{2,1}\) is called spacelike, lightlike, or timelike, if the restriction of the scalar product \(\langle \cdot ,\cdot \rangle \) to (the tangent space of) the affine plane is positive definite, positive semidefinite, or indefinite, respectively. In terms of the standard Euclidean metric on \(\mathbb R^{3}\), a plane is spacelike, lightlike, or timelike if its slope is less than, equal to, or greater than \(45^{\circ }\).

A curve of intersection of \(\mathbb H^{2}\) with an affine plane in \(\mathbb R^{2,1}\) that does not contain 0 is a hyperbolic circle, a horocycle, or a hypercycle, depending on whether the plane is spacelike, lightlike, or timelike.

Thus, connecting the vertices of a hyperbolic cyclic polygon by straight line segments in the ambient \(\mathbb R^{2,1}\), one obtains a planar polygon in \(\mathbb R^{2,1}\), but the intrinsic geometry of the plane will only be Euclidean if the hyperbolic polygon is inscribed in a circle (see Fig. 4). If the polygon is inscribed in a horocycle or hypercycle, then the geometry of the plane will be degenerate with signature \((+,0)\) or a \(1+1\)-spacetime with signature \((+,-)\), respectively.

Fig. 4
figure 4

Hyperbolic polygon inscribed in a circle, shown in the hyperboloid model

Full size image

Proposition 4.1

Let \(P_{\ell }\) be a hyperbolic cyclic polygon with side lengths \(\ell _{1},\ldots ,\ell _n\in \mathbb R_{>0}\), and let \(\bar{\ell }_{k}\) be the chordal lengths (23). If \(P_{\ell }\) is inscribed in

  1. (i)

    a circle then

    $$\begin{aligned} \bar{\ell }_{k}<\sum _{\begin{array}{c} i=1\\ i\not =k \end{array}}^{n}\bar{\ell }_{i} \quad \textit{for} \, \textit{all} \, k. \end{aligned}$$
    (24)
  2. (ii)

    a horocycle then

    $$\begin{aligned} \bar{\ell }_{k}=\sum _{\begin{array}{c} i=1\\ i\not =k \end{array}}^{n}\bar{\ell }_{i} \quad \textit{for} \, \textit{one} \, k. \end{aligned}$$
    (25)
  3. (iii)

    a hypercycle then

    $$\begin{aligned} \bar{\ell }_{k}>\sum _{\begin{array}{c} i=1\\ i\not =k \end{array}}^{n}\bar{\ell }_{i} \quad \textit{for}\, \textit{one} \, k. \end{aligned}$$
    (26)

Proof

(i) If \(P_{\ell }\) is inscribed in a circle, then the chordal polygon obtained by connecting the vertices of \(P_{\ell }\) by straight line segments in \(\mathbb R^{2,1}\) is a Euclidean polygon. Hence, its side lengths \(\bar{\ell }\) satisfy (24).

(ii) If \(P_{\ell }\) is inscribed in a horocycle, then the chordal length \(\bar{\ell }_{k}\) of a side is equal to the length of the arc of the horocycle between its vertices (see Fig. 5). Since one horocycle arc comprises all others, this implies (25).

Fig. 5
figure 5

Polygon inscribed in a horocycle, shown in the Poincaré half-plane model. Here, \(\bar{\ell }_{n}=\sum _{i\not =n}\bar{\ell }_{i}\) is the largest chordal length. Since all horocycles are congruent, we may without loss of generality assume that the polygon is inscribed in the horocycle \(y=1\)

Full size image

(iii) If \(P_{\ell }\) is inscribed in a hypercycle at distance R from a geodesic g, then the chordal lengths \(\bar{\ell }_k\), the hypercycle “radius” R, and the distances \(a_{k}\) between the foot points of the perpendiculars from the vertices to g (see Fig. 6) are related by

$$\begin{aligned} \frac{\bar{\ell }_k}{2}= {\cosh (R)}\sinh \frac{a_k}{2}. \end{aligned}$$
(27)
Fig. 6
figure 6

Polygon inscribed in a hypercycle, shown in the Poincaré disk model. Here, \(\ell _{n}\) and \(\bar{\ell }_{n}\) are the largest side length and chordal length, respectively

Full size image

Since one of the segments of g comprises all others,

$$\begin{aligned} a_{k}=\sum _{i\not =k}a_{i}\quad \text {for one } k. \end{aligned}$$

With (27) this implies (26):

$$\begin{aligned} \frac{\bar{\ell }_{k}}{2} = \cosh (R)\sinh \Big (\sum _{i\not =k}\frac{a_i}{2}\Big ) > \sum _{i\not =k}\cosh (R)\sinh \frac{a_i}{2}=\sum _{i\not =k} \frac{\bar{\ell }_i}{2}\,, \end{aligned}$$

where we have used the inequality \(\sinh (x+y)>\sinh (x)+\sinh (y)\), which holds for positive x, y. This follows immediately from the addition theorem for the hyperbolic sine function.

This completes the proof of Proposition 4.1. \(\square \)

Proposition 4.2

If \(\ell \in \mathbb R^n_{>0}\) satisfies the polygon inequalities (1), then there exists a unique hyperbolic cyclic polygon with these side lengths.

Proof

Suppose \(\ell \in \mathbb R^n_{>0}\) satisfies the polygon inequalities (1). Let \(\bar{\ell }\) be the corresponding chordal lengths (23). We will treat each case of Proposition 4.1 separately. In each case, we will tacitly use Proposition 4.1 and its proof. Our treatment of case (iii) is analogous to Penner’s proof [9] of Theorem 1.1 (his Theorem 6.2).

(i) If the chordal lengths \(\bar{\ell }\) satisfy condition (24), then the existence and uniqueness of a hyperbolic cyclic polygon with side lengths \(\ell \) follows from the existence and uniqueness of a Euclidean cyclic polygon with side lengths \(\bar{\ell }\), i.e., from Theorem 1.1 (see Fig. 4).

(ii) If the chordal lengths \(\bar{\ell }\) satisfy condition (25), then the corresponding hyperbolic cyclic polygon can be constructed by marking off the lengths \(\bar{\ell }_{i}\) for \(i\not =k\) along a horocycle (see Fig. 5). To see the uniqueness claim, note that all horocycles are congruent.

(iii) It remains to consider the case that the chordal lengths \(\bar{\ell }\) satisfy condition (26). For simplicity, we will assume that \(\ell _{n}\) is the largest side length. Then \(\bar{\ell }_{n}\) is the largest chordal length and condition (26) says

$$\begin{aligned} \bar{\ell }_{n}>\sum _{k=1}^{n-1}\bar{\ell }_{k}. \end{aligned}$$
(28)

Now suppose \(P_{\ell }\) is a hyperbolic polygon with side lengths \(\ell \) that is inscribed in a hypercycle at distance R from its geodesic g, and let

$$\begin{aligned} \bar{R}=\cosh R. \end{aligned}$$
(29)

Then the distances \(a_{k}\) between the foot points (see Fig. 6) satisfy

$$\begin{aligned} a_{n}=\sum _{k=1}^{n-1}a_{k}, \end{aligned}$$

Using (27), one obtains

$$\begin{aligned} \text {arsinh}\Big (\frac{\bar{\ell }_{n}}{2\bar{R}}\Big )= \sum _{k=1}^{n-1}\text {arsinh}\Big (\frac{\bar{\ell }_{k}}{2\bar{R}}\Big ). \end{aligned}$$
(30)

Conversely, if, for given \(\ell \), a number \(\bar{R}>1\) satisfies (30) then R defined by (29) is the correct hypercycle distance. More precisely, one can then construct a hyperbolic cyclic polygon with side lengths \(\ell \) by marking off the distances \(a_{1},\ldots ,a_{n-1}\) determined by (27) along a geodesic and intersect the perpendiculars in the marked points with a hypercycle at distance R (see Fig. 6).

It remains to show that there is exactly one \(\bar{R}>1\) satisfying (30). To this end, consider the function

$$\begin{aligned} \varPhi (x)=\text {arsinh}\Big (\frac{\bar{\ell }_{n}}{2x}\Big ) -\sum _{k=1}^{n-1}\text {arsinh}\Big (\frac{\bar{\ell }_{k}}{2x}\Big ). \end{aligned}$$
(31)

We need to show that \(\varPhi \) has exactly one zero in the interval \((1,\infty )\). Using (23), we see

$$\begin{aligned} \varPhi (1) = \frac{1}{2}\Big (\ell _{n}-\sum _{k=1}^{n-1}\ell _{k}\Big )<0. \end{aligned}$$
(32)

For \(x\rightarrow \infty \),

$$\begin{aligned} \varPhi (x)=\frac{1}{2x}\Big (\bar{\ell }_{n}-\sum _{k=1}^{n-1}\bar{\ell }_{k}\Big ) + o\Big (\frac{1}{x}\Big ), \end{aligned}$$
(33)

so

$$\begin{aligned} \varPhi (x)>0\quad \text {for large }x. \end{aligned}$$

By continuity, \(\varPhi \) has at least one zero in the interval \((1,\infty )\).

Finally, we will show that the derivative

$$\begin{aligned} \varPhi '(x)= \frac{1}{x}\Bigg ( -\frac{\frac{\bar{\ell }_{n}}{2x}}{\sqrt{1+\big (\frac{\bar{\ell }_{n}}{2x}\big )^{2}}} + \sum _{k=1}^{n-1} \frac{\frac{\bar{\ell }_{k}}{2x}}{\sqrt{1+\big (\frac{\bar{\ell }_{k}}{2x}\big )^{2}}} \Bigg ), \end{aligned}$$
(34)

is positive at the positive zeroes of \(\varPhi \). This implies that \(\varPhi \) has at most one zero in \(\mathbb R_{>0}\). Let us define

$$\begin{aligned} a_{k}(x)=\text {arsinh}\Big (\frac{\bar{\ell }_{k}}{2x}\Big ), \end{aligned}$$

so

$$\begin{aligned} \varPhi (x)=a_{n}(x)-\sum _{k=1}^{n-1}a_{k}(x), \end{aligned}$$

and

$$\begin{aligned} \varPhi '(x)=\frac{1}{x}\Big ( -\tanh a_{n}(x)+\sum _{k=1}^{n-1}\tanh a_{k}(x) \Big ). \end{aligned}$$

The claim follows from the inequality

$$\begin{aligned} \tanh \sum _{j=1}^{m}a_{j}<\sum _{j=1}^{m}\tanh a_{j}, \end{aligned}$$

which holds for \(m\ge 2\) and positive numbers \(a_{j}\). (Use induction on m and the addition theorem for \(\tanh \) for the base case \(m=2\).)

This concludes the proof of Proposition 4.2. \(\square \)

5 Concluding Remarks on 1 + 1 Spacetime

The scalar product of \(\mathbb R^{1,1}\) is

$$\begin{aligned} \langle x,y\rangle _{1,1}=x_{1}y_{1}-x_{2}y_{2}, \end{aligned}$$

and the length of a spacelike vector x is \(\ell =\sqrt{\langle x,x\rangle _{1,1}}\). The proof of Theorem 1.3 for polygons inscribed in hypercycles (Sect. 4) also proves the following theorem about “cyclic” polygons in \(1+1\) spacetime.

Theorem 5.1

There exists a polygon in \(\mathbb R^{1,1}\) with \(n\ge 3\) spacelike sides with lengths \(\ell _{1},\ldots ,\ell _n>0\) that is inscribed in one branch of a hyperbola \(\langle x,x\rangle _{1,1}=-R^{2}\) if and only if

$$\begin{aligned} \ell _{k}>\sum _{\begin{array}{c} i=1\\ i\not =k \end{array}}^{n}\ell _{i}\quad \textit{for} \, \textit{one} \, k, \end{aligned}$$
(35)

and this polygon is unique.

Without loss of generality, we will assume that the nth side is the longest, i.e., \(k=n\) in (35). Like in the Euclidean case (Sect. 2), the construction of such an inscribed polygon in \(\mathbb R^{1,1}\) is equivalent to the following analytic problem: Find a point \(a\in \mathbb R_{>0}^{n}\) satisfying

$$\begin{aligned} a_{n}=\sum _{i=1}^{n-1}a_{i} \end{aligned}$$
(36)

and

$$\begin{aligned} \frac{\ell _{k}}{2}=R\sinh \frac{a_{k}}{2} \end{aligned}$$
(37)

for some \(R\in \mathbb R\) and all \(k\in \{1,\ldots , n\}\).

This problem admits the following variational formulation. Define the function \(\varphi _{\ell }:\mathbb R^{n}\rightarrow \mathbb R\) by

$$\begin{aligned} \varphi _{\ell }(a)=\sum _{k=1}^{n-1}\big ( \text {Clh}_2(a_k) + \log (\ell _k)\,a_k\big ) -\big (\text {Clh}_2(a_n) + \log (\ell _n)\,a_n\big ), \end{aligned}$$

where \(\text {Clh}_2\) denotes the “hyperbolic version” of Clausen’s integral,

$$\begin{aligned} \text {Clh}_2(x)= -\int _0^x \log \Big |2\sinh \frac{t}{2}\Big |\,dt. \end{aligned}$$

(This notation is not standard.) The function \(\text {Clh}_2(x)\) can be expressed in terms of the real part of the dilogarithm function:

$$\begin{aligned} \text {Clh}_2(x)={\text {Re}}{\text {Li}}_{2}(e^{x})+\frac{x^{2}}{4}-\frac{\pi ^{2}}{6}. \end{aligned}$$

Like in the Euclidean case, one sees that \(a\in \mathbb R_{>0}^{n}\) is a critical point of \(\varphi _{\ell }\) under the constraint (36) if and only if there is an R satisfying equations (37). However, the function \(\varphi _{\ell }\) is neither concave nor convex on the subspace (36), so any proof of Theorem 5.1 (or the hypercycle case of Theorem 1.3) based on this variational principle would have to be more involved.