AA “right” path to cyclic polygons
Paolo Dulio ∗† and Enrico Laeng ‡ Dipartimento di Matematica “F. Brioschi”, Politecnico di Milano, Piazza Leonardo daVinci 32, I-20133 Milano, Italy
Abstract
It is well known that Heron’s theorem provides an explicit formula for the area of a triangle, as asymmetric function of the lengths of its sides. It has been extended by Brahmagupta to quadrilateralsinscribed in a circle (cyclic quadrilaterals). A natural problem is trying to further generalize the resultto cyclic polygons with a larger number of edges, which, surprisingly, has revealed to be far fromsimple. In this paper we investigate such a problem by following a new and elementary approach.We start from the simple observation that the incircle of a right triangle touches its hypothenuse ina point that splits it into two segments, the product of whose lengths equals the area of the triangle.From this curious fact we derive in a few lines: an unusual proof of the Pythagoras’ theorem, Heron’stheorem for right triangles, Heron’s theorem for general triangles, and Brahmagupta’s theorem forcyclic quadrangles. This suggests that cutting the edges of a cyclic polygon by means of suitablepoints should be the “right” working method. Indeed, following this idea, we obtain an explicitformula for the area of any convex cyclic polygon, as a symmetric function of the segments split onits edges by the incircles of a triangulation. We also show that such a symmetry can be rediscoveredin Heron’s and Brahmagupta’s results, which consequently represent special cases of the generalprovided formula.MSC: 52A10;52A38
Keywords:
Area; cyclic polygon; incircle; inradius.
A natural and largely considered question in convex geometry is the determination of the area A of aconvex polygon as a function of the lengths of its sides. The problem goes back to Heron of Alexandria,that was able to solve the problem in the case of a triangle. Later, in the seventh century, Brahmaguptaextended the result to cyclic quadrilaterals, namely to quadrilaterals inscribed in a circle (see for instance[1]). Several results concerning the geometry of cyclic polygons have been obtained in different areas ofresearch (see [2, 3, 6, 4, 11]), which points out a general interest for such geometric objects. It is thereforenatural trying to further extend to cyclic polygons with a larger number of edges the nice and ancientformulae by Heron and Brahmagupta. Surprisingly, this has revealed to be far from simple. In [8, 9] analgebraic formulation of the problem led D.P. Robbins to find formulae for cyclic pentagons and cyclichexagons. It was observed that Heron and Brahmagupta’s formulae can be restated in a form where16 A represents a monic polynomial whose coefficients are symmetric polynomials in the squares of thesides. This generalizes to cyclic pentagons and hexagons, where the polynomial have degree 7, but theformulae, even if holding also in the non convex case, do not provide an easy explicit form for the area(see also [7] for interesting comments and remarks). Formulae of the same kind have been conjectured[8], and later proved [5], even for heptagons and octagons, also illuminating some mysterious featuresof Robbins’ formulas for the areas of cyclic pentagons and hexagons (see also [10] for further detail onRobbin’s conjectures). The resulting formulae are interesting, but are very complex and do not seem to ∗ [email protected] † Corresponding author ‡ [email protected] a r X i v : . [ m a t h . HO ] O c t rovide a general picture that could be easily generalizable to polygons with an arbitrary large numberof edges.In this article we follow a different approach, which leads to a complete solution of the consideredproblem. The leitmotif of our paper is to point out that the role played by the edges in Heron’s andBrahmagupta’s theorems must be replaced by the segments cut on the edges of a polygon by the incirclesof the triangles of a triangulation of the polygon. First of all, we show that Pytagoras’, Heron’s andBrahmagupta’s theorems can be linked together thanks to a simple result concerning the area of a righttriangle. Then, we generalize Heron’s and Brahmagupta’s results to a symmetric coordinate free formulathat holds true for any cyclic polygon. Let
ABC be a right triangle and let I be its incenter (see Fig.1). Since ˆ B is a right angle and since theincircle is tangent perpendicularly to the three sides 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 . C B r H A I J K s t s r r t r r Figure 1: A right triangle
ABC .Denoting the area of a triangle with vertical bars we have | ABC | = | AIB | + | BIC | + | CIA | . The half-perimeter of
ABC is p = r + s + t while all the three triangles on the r.h.s. of the aboveequality have altitude r with respect to their sides AB , BC , and AC . Therefore we have | ABC | = r ( r + s + t ) . (1) Remark 1.
In case
ABC is not a right triangle, Formula (1) generalizes to | ABC | = R ( r + s + t ) , (2)where R is the incircle of ABC , meaning that | ABC | is a symmetric function in r, s, t . Theorem 2.
The area of a right triangle
ABC is equal to the area of the rectangle of sides s = AH and t = CH , where H is the point where the incircle is tangent to the hypotenuse AC . Proof
Clearly | ABC | = 12 ( s + r )( t + r ), and by (1) we get st + rs + rt + r = 2( r + rs + rt )which we simplify into st = r + rs + rt = r ( r + s + t ) = | ABC | . (cid:88) heorem 3. [Heron’s formula for right triangles] The area of a right triangle is equal to (cid:112) p ( p − a )( p − b )( p − c ),where a , b , c are the lengths of its sides (taken in any order) and p = ( a + b + c ) / Proof
Using the same notation (as in Fig.1) we need to show that | ABC | = str ( r + s + t ) , but this is immediate, being | ABC | = st by Theorem 2, and also | ABC | = r ( r + s + t ) by (1). (cid:88) Theorem 3 shows that, in any right angle triangles, Heron’s formula can be rediscovered by starting fromthe symmetric formula provided by Theorem 2. We wish now to extend such a result to any triangle.
Theorem 4. [Pythagoras]
In a right triangle the area of the square whose side is the hypotenuse isequal to the sum of the areas of the squares whose sides are the two legs.
Proof
We have r ( r + s + t ) = st . Multiplying by 2 and adding s + t on both sides we get s + t +2 r + 2 rs + 2 rt = s + t + 2 st , i.e.( s + r ) + ( r + t ) = ( s + t ) . (cid:88) Thanks to Pythagoras’ Theorem we can easily extend Heron’s Theorem to any triangle. To this, let
ABC be a generic triangle, and let CH be the altitude on its edge AB (see Figure 2), and assume H isbetween A and B (in any triangle surely exists an altitude with this property). A C B H \\ b c a
Figure 2: A generic triangle
ABC .Let p = r + s + t be the semiperimeter of ABC , and let ϕ = (cid:112) rst ( r + s + t ) = (cid:112) p ( p − a )( p − b )( p − c ).By Pythagoras’ Theorem in AHC and
CHB , we have c = ( AH + HB ) = AH + BH + 2( AH )( HB ) == a + b − CH + 2( AH )( HB ) = a + b − CH + 2 (cid:112) ( a − CH )( b − CH )and, solving for CH we get CH = (cid:112) a b − ( c − a − b ) c = 2 (cid:112) p ( p − a )( p − b )( p − c ) c = 2 c ϕ. Therefore, from 2 | ABC | = cCH , Heron’s theorem for ABC follows. (cid:88)
Remark 5.
By (2) and Heron’s Theorem written in the form | abc | = (cid:112) rst ( r + s + t ) we can obtain theincircle R of any triangle as a symmetric function of r, s, t as follows (see Figure 3) R = (cid:114) rstp . (3)being p = r + s + t the half perimeter of ABC . s R t t C B A r s r Figure 3: Incircle of a generic triangle
ABC . We wish now to show how Brahmagupta’s formula can be obtained by exploiting the same idea ofsymmetry considered in the previous section. with respect to considered a symmetry , we prove thefollowing result.
Theorem 6.
Let
ABC, ADC be two triangles inscribed in a same circumference. If s , t and s , t arethe lengths of the two segments split on the common edge AC by the respective incircles, then | ABC || ACD | = s s t t . Proof.
In the cyclic quadrangle
ABCD the halves of A ˆ BC and A ˆ DC are complementary angles. There-fore the shaded right triangles in Figure 4 are similar, and consequently R r = r R , that is R R = r r . R R r r s s t t t t s s r r A B C D Figure 4: A general cyclic quadrangle
ABCD .By (3) we have p R = r s t and p R = r s t , being p , p the half perimeter of ABC and
ADC respectively, so that( p R p R ) = p ( r s t ) p ( r s t ) = p p ( r r )( s t s t ) = p p ( R R )( s t s t ) , and consequently | ABC || ACD | = p R p R = s t s t . (cid:88) Remark 7.
Since R R = r r we have also | ABC || ACD | = p R p R = p p r r . heorem 8. [Brahmagupta’s formula for cyclic quadrangles] The area of a convex quadranglethat can be inscribed in a circle (a cyclic quadrangle) is equal to (cid:112) ( p − a )( p − b )( p − c )( p − d ), where a , b , c , d are the lengths of its sides (taken in any order) and p = ( a + b + c + d ) / Proof.
Let
ABCD be split into
ABC and
ACD , as in Figure 4, and assume a = s + r = AB, b = t + r = BC, c = t + r = CD, d = s + r = AD , so that p = r + r + ( s + t ) = r + r + ( s + t ). Startingfrom | ABCD | = ( | ABC | + | ACD | ) = | ABC | + | ACD | + 2 | ABC || ACD | , we use the previous theorem,and the Remark, to write 2 | ABC || ACD | = s t s t + p p r r , where p , p are the half perimetersof ABC and
ACD , respectively. Moreover, by Heron’s formula, | ABC | = ( r + s + t ) r s t , and | ACD | = ( r + s + t ) r s t . Then, also using s + t = s + t , we get | ABCD | = ( r + s + t ) r s t + ( r + s + t ) r s t ++ s t s t + r r ( r + s + t )( r + s + t )= r ( r + s + t )( r ( r + s + t ) + s t )+ s t ( r ( r + s + t ) + s t )= ( r ( r + s + t ) + s t )( r ( r + s + t ) + s t )= ( r ( r + s + t ) + s t )( r ( r + s + t ) + s t )= ( r + t )( r + s )( r + s )( r + t )= ( p − a )( p − b )( p − c )( p − d ) . (cid:88) 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 formula has been extended to Brahmagupta’s formula by consideringa cyclic quadrilateral Q as the union of two triangles, Q = T ∪ T , and then focusing on the segments r , s , t and r , s , t determined, respectively, on the edges of T and T by the tangent points of thecorresponding incircles. This provides the square of the area of Q as a polynomial function, symmetricunder the exchange of r , s , t with r , s , t .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 ofsome triangulatio. To this, let us first observe that P n can always be assumed as a union of n consecutivetriangles T , T , ..., T n , all having a common vertex. For i = 1 , ..., n −
1, denote by L i,i +1 the commonedge 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 theedges of T j by its incircle, where L i,i +1 = s i + t i = s i +1 + r i +1 , i = 1 , ..., n − Theorem 9.
Let P n be a cyclic polygon consisting of n + 2 edges, n ≥
1. Then it results A h A k = s h t h s k r k k − (cid:89) i = h +1 s i p i = p h r h p k t k k − (cid:89) i = h +1 p i s i , for 1 ≤ h < k ≤ n. (4) Proof.
In order to prove the first equality in (4) we apply Theorem 6 iteratively, so that r i r i t i t i s i s i s s s s s n s n r r t t r r t t r n r n t n t n t i+1 t i+1 r i+1 r i+1 s i+1 s i+1 R R R R i R i R i+1 R i+1 T T T i T i+1 Figure 5: Triangulation of a cyclic polygons. A h A h +1 = s h t h s h +1 r h +1 A h +1 A h +2 = s h +1 t h +1 s h +2 r h +2 ...A k − A k = s k − t k − s k r k . By multiplying on both sides we get A h ( A h +1 A h +2 ...A k − ) A k = s h t h (cid:32) k − (cid:89) i = h +1 r i s i t i (cid:33) s k r k . By Heron’s Theorem applied to T h +1 , T h +2 , ..., T k − we have A h A k = s h t h (cid:16)(cid:81) k − i = h +1 r i s i t i (cid:17) s k r k (cid:81) k − i = h +1 r i s i t i p i = s h t h s k r k k − (cid:89) i = h +1 s i p i , and the first equality in (4) is obtained. For the proof of the second equality, let us observe that, bysimilitude, it results R i r i = t i +1 R i +1 , for all i = 1 , ..., n −
1. Therefore we get R h R h +1 = r h t h +1 R h +1 R h +2 = r h +1 t h +2 ...R k − R k = r k − t k . By multiplying on both sides, it results h ( R h +1 R h +2 ...R k − ) R k = R h R k k − (cid:89) i = h +1 R i = r h t k k − (cid:89) i = h +1 r i t i , and applying (3) to all R i we have R h R k = r h t k k − (cid:89) i = h +1 r i t ik − (cid:89) i = h +1 r i s i t i p i , and consequently R h R k = r h t k k − (cid:89) i = h +1 p i s i for 1 ≤ h < k ≤ n. (5)then The second equality in (4) follows immediately from (5), being A h A k = p h R h p k R k . (cid:88) Assuming P = T ∪ T ∪ ... ∪ T n as in Figure 5, and using the same notations as above we can nowprove a general formula for the area of a cyclic polygon with any number of edges. Theorem 10.
Let P n be a cyclic polygon consisting of n + 2 edges, n ≥
1, and let A be its area. Thenit results A ( n ) = (cid:32) p r + n (cid:88) q =2 r q s q q − (cid:89) m =2 s m p m (cid:33) (cid:32) s t + n (cid:88) q =2 p q t q q − (cid:89) m =2 p m s m (cid:33) . Proof.
Since P n is the union of T , ..., T n , by Heron’s Theorem, and using both equalities in (4), we canwritewe have A = ( A + A + ... + A n ) == n (cid:88) j =1 A j + 2 (cid:88) ≤ h
We emphasize that the formula obtained for A ( n ) is symmetric under mutually exchanging r j with t j , and s j with p j , for all j ∈ { , ..., n } . We also note that only the terms concerning the partitionsof the edges of the polygon P n are in fact necessary. Indeed, p = r + s + t can be immediately computedonce we know the terms r , s , t determined on the edges of P n by the incircle of T . For 1 < q < n − s q = s q − + t q − − r q , so that, by recursion, we get s q = s + q − (cid:88) h =1 t h − q (cid:88) k =2 r k , (6) p q = r q + s q + t q = s + t + t if q = 2 s + q (cid:88) h =1 t h − q − (cid:88) k =2 r k if 2 < q < n − . (7)Consequently, all terms appearing in A ( n ) can be computed once s , s n , r j , t j are known. Examples
We can easily rediscover Heron’s and Brahmagupta’s results from the provided symmetric function. • n = 1 A (1) = ( p r + 0) ( s t + 0) = p r s t Heron’s theorem . • n = 2 A (2) = ( p r + r s )( s t + p t ) == (( r + s + t ) r + r s )( s t + ( r + s + t ) t == (( r + s + r ) r + r s )( s t + ( s + t + t ) t == ( r + s )( r + r )( t + s )( t + t ) . Being s + t = s + r , it is p = r + r + s + t = r + r + t + s , so that, if we denote, respectively,by a, b, c, d the edges s + t , r + t , r + s and r + t , then A (2) = ( p − a )( p − b )( p − c )( p − d ) Brahmagupta’s theorem . • n = 3 generalization of Brahmagupta’s theorem to cyclic pentagons A (3) = (cid:18) p r + r s + r s s p (cid:19) (cid:18) s t + p t + p t p s (cid:19) . • n = 4 generalization of Brahmagupta’s theorem to cyclic hexagons A (4) = (cid:18) p r + r s + r s s p + r s s p s p (cid:19) (cid:18) s t + p t + p t p s + p t p s p s (cid:19) . We can even extend the formula to n = 0 by assuming A (0) = 0, where P is a polygon degenerated ina segment, which can be obtained by progressively removing an edge from a starting polygon P n having n + 2 edges We have shown that Heron’s and Brahmagupta’s theorems can be extended to a formula that providesthe square of the area of a convex cyclic polygon as a symmetric polynomial of the segments determinedon the edges by the incircles of a suitable triangulation. We remark that the formula is coordinate-freeas one should expect from the intrinsic geometric nature of the problem. Differently, using for instanceGreen’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 triangle andcyclic quadrilaterals, where the lengths of the edges explicitly appear in the computation of the area.his is just because the number of involved edges is small, so that the segments determined by theedge partitions induced by the incircles can be easily related to the original lengths of the edges of thepolygon. We also remark that the incircles can be elementary constructed, so that the provided formulaalso determines an elementary computation of the square of the area of any convex cyclic polygon
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