Generalization of Heron’s and Brahmagupta’s equalities to any cyclic polygon

It is well known that Heron’s equality provides an explicit formula for the area of a triangle, as a symmetric function of the lengths of its edges. It has been extended by Brahmagupta to quadrilaterals inscribed in a circle (cyclic quadrilaterals). A natural problem is trying to further generalize the result to cyclic polygons with a larger number of edges. Surprisingly, this has proved to be far from simple, and no explicit solutions exist for cyclic polygons having n>4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n>4$$\end{document} edges. In this paper we investigate such a problem by following a new and elementary approach, based on the idea that the simple geometry underlying Heron’s and Brahmagupta’s equalities hides the real players of the game. In details, we propose to focus on the dissection of the edges determined by the incircles of a suitable triangulation of the cyclic polygon, showing that this approach leads to an explicit formula for the area as a symmetric function of the lengths of these segments. We also show that such a symmetry can be rediscovered in Heron’s and Brahmagupta’s results, which consequently represent special cases of the provided general equality.


Introduction
A natural and largely considered question in convex geometry is the determination of the area A of a convex polygon as a symmetric function of the lengths of its edges. The problem goes back to Heron of Alexandria, who was able to solve the problem in the case of a triangle. If a, b, c are the lengths of the edges, and p denotes the semiperimeter, then Later, in the seventh century, Brahmagupta extended the result to cyclic quadrilaterals, namely to quadrilaterals inscribed in a circle (see for instance case when the role of the vertices must be emphasized, we also denote the area by writing the vertices between vertical bars. For instance, |ABC|, means the area of a triangle ABC. In case several triangles T 1 , . . . , T m must be considered simultaneously, then we denote by A j the area of T j , j ∈ {1, . . . , m}.
In view of a clearer presentation of the new proposed approach, we recall Pytagoras' theorem, as well as Heron and Brahmagupta's results concerning the area as a symmetric function of the lengths of the edges.

Theorem 1. (Pythagoras)
In a right triangle the area of the square whose edge is the hypotenuse is equal to the sum of the areas of the squares whose edges are the two legs.

Theorem 2. (Heron's equality for triangles) The area of a triangle is equal
, where a, b, c are the lengths of its edges (taken in any order) and p = (a + b + c)/2 is its half perimeter.
where a, b, c, d are the lengths of its edges (taken in any order) and p = (a + b + c + d)/2 is its half perimeter.
Our main result is Theorem 8, where we prove a general symmetric formula for the area of a cyclic polygon having n ≥ 3 edges, which includes Theorem 2 and Theorem 3 as particular cases.

Heron's equality
Let ABC be a right triangle and let I be its incenter (see Fig. 1). SinceB is a right angle and since the incircle is tangent perpendicularly to the three edges of ABC, we have r = IJ = IK = IH = BJ = BK, where r is the inradius. The internal bisectors of ABC are concurrent in I and this implies AJ = AH = s and CH = CK = t.
We have |ABC| = |AIB| + |BIC| + |CIA|. The half-perimeter of ABC is p = r + s + t while the three triangles on the r.h.s. of the above equality have 944 P. Dulio, E. Laeng AEM altitude r with respect to their edges AB, BC, and AC. Therefore, it results Remark 4. In case ABC is not a right triangle, Formula (3) generalizes to where R is the incircle of ABC, meaning that |ABC| is a symmetric function in r, s, t. Proof. Clearly |ABC| = 1 2 (s + r)(t + r), and by (3) we get st + rs + rt + r 2 = 2(r 2 + rs + rt) which we simplify into st = r 2 + rs + rt = r(r + s + t) = |ABC|.
A new proof of Theorem 2 for right triangles. Using the same notation (as in Fig. 1) we need to show that |ABC| 2 = str(r + s + t), but this is immediate, since |ABC| = st by Lemma 5, and also |ABC| = r(r + s + t) by (3).
The above proof shows that, in any right triangles, Heron's equality can be rediscovered by starting from the symmetric formula provided by Lemma 5. We wish now to extend such a result to any triangle.
The above proof shows that Pythagoras' Theorem can be rediscovered as a consequence of Heron's equality in right triangles. Now, thanks to Pythagoras' Theorem, we can easily extend Heron's equality to any triangle, which consequently follows from Heron's equality for right triangles.
For this, let ABC be a generic triangle, and let CH be the altitude on its edge AB (see Fig. 2), where we assume H between A and B (in any triangle there surely exists an altitude with this property).
Let p = r + s + t be the semiperimeter of ABC, and let By Pythagoras' Theorem in AHC and CHB, we have Vol. 95 (2021) Generalization of Heron's and Brahmagupta's equalities 945 Therefore, from 2|ABC| = cCH, Heron's equality for ABC follows.
Remark 6. By (4) and Heron's equality written in the form |ABC| = rst(r + s + t) we can obtain the incircle R of any triangle as a symmetric function of r, s, t as follows (see Fig. 3).
since p = r + s + t is the half perimeter of ABC.

Brahmagupta's equality
We wish now to show how Brahmagupta's equality can be rediscovered by exploiting the same idea of symmetry considered in the previous section. First of all, we prove the following result. Proof. In the cyclic quadrilateral ABCD the halves of ABC and ADC are complementary angles. Therefore the shaded right triangles in Fig. 4 are similar, and consequently R1 r1 = r2 R2 , that is R 1 R 2 = r 1 r 2 . By (5) we have p 1 R 2 1 = r 1 s 1 t 1 and p 2 R 2 2 = r 2 s 2 t 2 , since p 1 , p 2 is the half perimeter of ABC and ADC respectively, so that Remark 8. Since R 1 R 2 = r 1 r 2 we have also |ABC||ACD| = p 1 R 1 p 2 R 2 = p 1 p 2 r 1 r 2 .

The area of a circular polygon having an arbitrary number of edges
In this section we generalize the previous results to a cyclic polygon P n , having n + 2 edges for any n ≥ 1. Let us observe that Heron's equality has been extended to Brahmagupta's equality by considering a cyclic quadrilateral Q as the union of two triangles, Q = T 1 ∪ T 2 , and then focusing on the segments r 1 , s 1 , t 1 and r 2 , s 2 , t 2 determined, respectively, on the edges of T 1 and T 2 by the tangent points of the corresponding incircles. This provides the square of the area of Q as a polynomial function, symmetric under the exchange of r 1 , s 1 , t 1 with r 2 , s 2 , t 2 .
As a consequence we are inspired to investigate the square of the area A(n) of a generic cyclic polygon P n by looking at the partitions of the edges of P n determined by the tangent points of the incircles of some triangulation. For this, let us first observe that P n can always be assumed as the union of n consecutive triangles T 1 , T 2 , . . . , T n , all having a common vertex. For i = 1, . . . , n − 1, denote by L i,i+1 the common edge between the two consecutive triangles T i , T i+1 . Let p j , A j , R j be, respectively, the semiperimeter, the area and the radius of the incircle of T j , j = 1, . . . , n. Also, let r j , s j , t j be the segments cut on the edges of T j by its incircle, where L i,i+1 = s i + t i = s i+1 + r i+1 , i = 1, . . . , n − 1 (see Fig. 5).
For the sake of brevity, and in order to avoid heavy notations, in the following theorems we assume all the meaningless products to be equal to 1. AEM Theorem 9. Let P n be a cyclic polygon consisting of n + 2 edges, n ≥ 1. Then Proof. In order to prove the first equality in (6) we apply Theorem 7 iteratively, so that By multiplying on both sides we get By Heron's equality applied to T h+1 , T h+2 , . . . , T k−1 we have and the first equality in (6) is obtained. For the proof of the second equality, let us observe that, by similitude, we have , for all i = 1, . . . , n − 1.
Therefore we get By multiplying on both sides, we obtain and applying (5) to all R 2 i we have risiti pi , and consequently Vol. 95 (2021) Generalization of Heron's and Brahmagupta's equalities 949 The second equality in (6) follows immediately from (7), since Assuming P n = T 1 ∪ T 2 ∪ · · · ∪ T n as in Fig. 5, and using the same notations as above we can now prove a general formula for the area of a cyclic polygon with any number of edges.
Theorem 10. Let P n be a cyclic polygon with n + 2 edges, n ≥ 1, and let A(n) be its area. Then Proof. Since P n is the union of T 1 , . . . , T n , by Heron's equality, and using both equalities in (6), we have Let's rearrange as follows (where each one of the three terms appearing in each bracket comes from the corresponding sum) We can even extend the formula to n = 0 by assuming A(0) = 0, where P 0 is a polygon degenerated in a segment, which can be obtained by progressively removing an edge from a starting polygon P n having n + 2 edges.

Conclusion and remarks
We have shown that Heron's and Brahmagupta's equalities can be extended to a formula that provides the square of the area of any convex cyclic polygon as a symmetric polynomial of the lengths of the segments determined on the edges by the incircles of a suitable triangulation. We remark that the formula is coordinate-free as one should expect from the intrinsic geometric nature of the problem. Otherwise, using for instance Green's theorem, it would be quite easy to provide a coordinate dependent result.
In our opinion the obtained formula is the natural generalization of what happens for triangles and cyclic quadrilaterals, where the lengths of the edges explicitly appear in the computation of the area. This is just because the number of involved edges is small, so that the segments determined by the edge partitions induced by the incircles can be easily related to the original lengths of the edges of the polygon. We also remark that the incircles can be constructed in an elementary way, so that the provided formula also determines an elementary computation of the square of the area of any convex cyclic polygon.
Some related open problems also arise, which would be worth considering in later possible works. For instance, it would be interesting to investigate the functional dependence of the assumed parameters r q , s q , t q on the original edge lengths. Presumably, such functions are not simple, which could explain why an explicit formula for the area as a symmetric function of the edges has not been obtained in the literature. Also, the extension of the result to non-convex cyclic polygons seems to be an appealing and challenging task. A further analysis could be devoted to equality (8) in case it is obtained by using different triangulations of the same polygon, or to its possible simplification in suitable subclasses of cyclic polygons Funding Open Access funding provided by Politecnico di Milano Open Access. This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If