On the area of a polygon inscribed in a circle

On the area of a polygon inscribed in a circle Autor(en): Matsumoto, Y. / Matsuti, Y. / Oda, M. Objekttyp: Article Zeitschrift: L'Enseignement Mathématique Bd (Jahr): 53 (007) Heft 1- PDF erstellt am: 7.08.018 Persistenter Link: http://doi.org/10.5169/seals-10954 Nutzungsbedingungen Die ETH-Bibliothek ist Anbieterin der digitalisierten Zeitschriften. Sie besitzt keine Urheberrechte den Inhalten der Zeitschriften. Die Rechte liegen in der Regel bei den Herausgebern. Die auf der Plattform e-periodica veröffentlichten Dokumente stehen für nicht-kommerzielle Zwecke in Lehre und Forschung sowie für die private Nutzung frei zur Verfügung. Einzelne Dateien oder Ausdrucke aus diesem Angebot können zusammen mit diesen Nutzungsbedingungen und den korrekten Herkunftsbezeichnungen weitergegeben werden. Das Veröffentlichen von Bildern in Print- und Online-Publikationen ist nur mit vorheriger Genehmigung der Rechteinhaber erlaubt. Die systematische Speicherung von Teilen des elektronischen Angebots auf deren Servern bedarf ebenfalls des schriftlichen Einverständnisses der Rechteinhaber. Haftungsausschluss Alle Angaben erfolgen ohne Gewähr für Vollständigkeit oder Richtigkeit. Es wird keine Haftung übernommen für Schäden durch die Verwendung von Informationen aus diesem Online-Angebot oder durch das Fehlen von Informationen. Dies gilt auch für Inhalte Dritter, die über dieses Angebot zugänglich sind. Ein Dienst der ETH-Bibliothek ETH Zürich, Rämistrasse 101, 809 Zürich, Schweiz, www.library.ethz.ch http://www.e-periodica.ch

L Enseignement Mathématique ) 53 007), 17 153 ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE by Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI d T. SHIBUYA ABSTRACT. We prove that if n 5, the area of the general cyclic n-gon cnot be calculated from its side lengths, using only arithmetic operations d k-th roots. To prove this, we apply Galois theory. 1. INTRODUCTION The area of a trigle is given by Heron s formula terms of its side lengths a1 a a3 : before 75 A.D.) in 1) p s(s a1)( s a)(s a3) where s a1 a a3). Obviously, the area of a quadrilateral is not determined by its side lengths a1 a a3 a4 only, but if it is inscribed in a circle, Brahmagupta s formula 68 A.D.) gives the area : ) p s a1)(s a)(s a3)(s a4) where s a1 a a3 a4). See []. Thus the area of a trigle or of a cyclic quadrilateral c be calculated from its side lengths by combining the four arithmetic operations of addition, subtraction, multiplication, d division, together with the operation of taking square roots. Here d in the sequel, a cyclic polygon is a convex polygon whose vertices all lie on the same circle. The purpose of this paper is to prove THEOREM 1. If n 5, there is no formula which expresses the area of the general cyclic n-gon in terms of its side lengths, using only arithmetic operations d k-th roots.

18 Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI AND T. SHIBUYA As a consequence, if n is greater th four, there exists no formula like 1) or ) for the area of a cyclic n-gon. We prove this theorem by applying Galois theory. Blaschke [1] proved that the area of n-gon with given side lengths a1 a attns a maximum if d only if the polygon is cyclic, d that the maximum value is independent of the order of it is easy to see a1 a. To find explicit formula for the area of a cyclic n-gon in terms of its side lengths would be interesting problem. The authors are grateful to Professor Koichi Yo; without his question about the maximum area of polygons with given side lengths, the present investigation would never have been undertaken. The authors are also grateful to the referees for their careful reading d useful comments d suggestions. Note added on May 10th, 006. We recently learned that V. V. Varfo lomeev [6] has proved that the area of a cyclic n-gon is algebrc over the field Q(a1 ) generated by the side lengths a1 d that in other paper [7], he has studied the Galois group of the same equation as our 3) equation 8) in [6]) over the field Q(a1 a5) of rational functions of the sides of a cyclic pentagon d has proved that it is isomorphic to the symmetric group S7 His result, together with the Geometric Theorem in the same paper, immediately implies our Theorem 1 at least for n 5), though this theorem is not stated explicitly in [7]. The merit of the present paper would be that our approach is much more elementary th his.. PROOF OF THEOREM 1 FOR n 5 In this section, we will prove Theorem 1 for n 5. The proof for n 6 will be given in 5. Let ABCDE be a cyclic pentagon, as in Figure 1. Let a b c d e be the side lengths of the pentagon as shown in Figure 1. Let x be the length of the diagonal AC, d let S be the area of the pentagon. LEMMA 1. The diagonal length x satisfies a polynomial equation of degree 4 whose coefficients are rational functions over the rational field Q) of S a b c d e.

ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE 19 A e E a B x d b c D C FIGURE 1 Cyclic pentagon ABCDE LEMMA. The diagonal length x is a solution of the following polynomial equation of degree 7 : 3) cde x 7 c d d e e c a b) x 6 cdef(c d e) (a b) g x 5 fc d e a b(c d e) (a b)(c d d e e c)g x 4 cdef(a b) 4a b (a b)(c d e)g x 3 f( a b)(c d d e e c) c d e(a b ) a b(c d e) g x cde(c d e)(a b ) x c d e(a b) 0 In the special case a degree 5 : b, x is a solution of the following equation of 4) cde x 5 c d d e e c a4) x 4 cdef(c d e) 4a g x 3 fcde a4(c d e) 4a(cd de ec)g x 4acdefa c d e) g x a f4a ( c d d e e c) 4c d e a ( c d e) g 0 Let us consider for example a cyclic pentagon with side lengths a b 1, c, d 3, e 4. Such a cyclic pentagon exists. See Appendix A, Proposition 4.) Then equation 4) becomes 4x5 43x4 600x3 34x 59x 169 0 Dividing out the common factor 3, we obtn

130 Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI AND T. SHIBUYA 5) 8x5 81x4 00x3 114x 864x 73 0 LEMMA 3. The Galois group of equation 5) over Q is S5 the symmetric group of degree 5. In particular, no root of this equation belongs to radical extensions of Q. Proof of Theorem 1 for n 5. We will prove Theorem 1 for n 5, taking Lemmas 1,, 3 momentarily for grted. Suppose that the area S could be calculated from the side lengths a b c d e using only arithmetic operations d k -th roots. Then by Lemma 1, x could also be calculated likewise from the side lengths, because y polynomial equation of degree 4 c be solved by radicals. This would imply that the diagonal x is in a radical extension of the field Q(a b c d e) In particular, equation 5) could be solved by radicals. However, this contradicts Lemma 3. Therefore, Theorem 1 is proved for n 5. 3. PROOFS OF LEMMAS 1 AND Proof of Lemma 1. The area S of the cyclic pentagon ABCDE of Figure 1 is the sum of the areas of the trigle ABC d the cyclic quadrilateral ACDE. Applying formulas 1) d ), we have Hence, S area( ABC) area( ACDE) 1 4pf( a b) x gfx a b) g 1 4pf( x c) d e) gf( d e) x c) g 4S From this, we have pf( a b) x gfx a b) g f( x c) d e) gf( d e) x c) g 6) (a b c d e)x 8cdex 16S a4 b4 c4 d4 e4 ( ab cd de ec) 8Sp x4 (a b)x a b) The required equation of degree 4 for x of 6). is obtned by squaring both sides

ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE 131 Proof of Lemma. Let y denote the length of diagonal AD of the cyclic pentagon ABCDE in Figure 1. Consider the quadrilateral ABCD, d let be the gle ABC. Then ADC We have x a b ab cos y c yc cos( Eliminating cos we get 7) x a b)cy c y)ab ab cy Similarly, considering the quadrilateral ACDE, we have x c)de d e)cx 8) y cx de Eliminating y from 7) d 8), we obtn equation 3). 4. PROOF OF LEMMA 3 The following proposition is well known. For a proof, we refer the reader to [4] Part II, Chap. 3, 5). PROPOSITION 1. Let P(x) be a polynomial of degree 5 with rational coefficients. Suppose that P(x) is irreducible over Q d that the equation 9) P(x) 0 has three real roots d a pr of imaginary roots. Then the Galois group of equation 9) over Q is isomorphic to the symmetric group S5 Therefore, in order to prove Lemma 3, it suffices to prove the following two lemmas. LEMMA 4. The polynomial on the left hd side of equation 5) is irreducible over Q. roots. LEMMA 5. Equation 5) has three real roots d a pr of imaginary

13 Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI AND T. SHIBUYA Both lemmas c be checked insttly by appealing to technological tools. We used Mathematica. Though our use was modest compared to that in [5], we found them very useful. We will give here, however, quite elementary proofs. Proof of Lemma 4. Let Q(x) denote the polynomial on the left hd side of equation 5). To simplify the polynomial, we define R(x) by setting 10) R(x) Q(x ) Obviously, Q(x) is irreducible over Q if d only if R(x) is. We shall prove the irreducibility of R(x) By calculation, R(x) 8x5 x4 18x3 10x 40x 11 As is well known, a polynomial with integral coefficients is irreducible over Q if d only if it is irreducible over Z. First of all, we prove CLAIM 1. The following factorization mod 8 is impossible: 11) R(x) x m)t(x) mod8 where m is integer, d T(x) is a polynomial with integral coefficients. Here, by f x) g(x) mod 8, we me that corresponding coefficients of the polynomials) f x) d g(x) are congruent modulo 8. Proof. We have 1) R(x) x4 x 3 mod8 If we had a factorization mod 8 of the form 11), then from 1) m would be odd integer d therefore, m 1 mod 8. Also from 11), R( m) 0 mod 8. However, this is impossible, because R( m) m) 4 ( m) 3 1 3 4 mod8 This proves Clm 1. Now we prove that R(x) is irreducible over Z.

ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE 133 CASE 1. If R(x) were divisible in Z[x] by a linear polynomial, there would be integers a b c d e k l such that R(x) ax b)(cx4 dx3 ex kx l By comparing coefficients on both sides: x5 : ac 8 x4 : ad bc 1 x3 : ae bd 18 x : ak be 10 x : al bk 40 x0 : bl 11 We shall show that this system of six equations cnot be solved in integers. We may assume that a 0. Since ad bc 1, we have gcd(a c) 1. Since ac 8, we have either a 8 c 1 or a 1 c 8. However, the latter case is excluded by Clm 1. Thus a 8 d c 1. Then ad bc 1 becomes 8d b 1, whence b 1 mod 8. Since b divides 11 d b 1 mod 8, we have b 1. Then from 8d b 1 we have d 0, d ae bd 18 gives e 16. Now ak be 10 becomes 8k 16 10. This yields k 34 a contradiction. CASE. If R(x) were divisible in Z[x] by a quadratic polynomial, there would be integers a b c d e k l such that 13) R(x) ax bx c)(dx3 ex kx l By comparing coefficients on both sides: x5 : ad 8 x4 : ae bd 1 x3 : ak be cd 18 x : al bk ce 10 x : bl ck 40 x0 : cl 11 We shall show that this system of six equations cnot be solved in integers. We may assume that a 0. Since ae bd 1, we have gcd(a d) 1. Since ad 8, we have either a 8 d 1 or a 1 d 8. The former case is impossible. This is proved as follows: In this case, ae bd 1 would

134 Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI AND T. SHIBUYA become 8e b 1, which implies b 1 mod 8. Substituting a 8 d 1 d b 1 mod 8 i3), we would have R(x) x c)(x3 ex kx l) mod8 which is excluded by Clm 1. Thus a 8 d 1 is impossible as asserted, d we have a 1 d 8. Now the above system implies that e 8b 1 k be 8c 18 l bk ce 10 bl ck 40 cl 11 Since cl 11, there are four possiblities for the pr c l : c l 1 11) 1 11) 11 1) 11 1) In each case, c l 10 or c l 10. CLAIM. If c l 10, then b mod 4. If c l 10, then b 0 mod 4. Proof. Calculating mod 8, we have e 1 k b 0 l bk c bl ck 0 According as c l 10 or c l 10, the third equation yields bk 4 or bk 0. The second equation implies that bk b Thus b 6 mod 8 or b 0 4 mod 8, according as c l 10 or c l 10. This proves Clm. Let x) denote the quadratic factor x bx c i3) with a 1. Substituting x in x) d R(x) we have d ) 4 b c ) 4 b c

ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE 135 R( ) 653 R() 73 Thus ) 4 b c resp. ) 4 b c) must divide 653 resp. 73). Note that 653 is a prime number. Thus 14) 4 b c 1 1 653 or 653 Also note that 653 5 mod8 We consider four cases according to the values of c d l CASE i) c l 1 11). Since c l 10, we have b 0 mod 4 by Clm. Then 4 b 1 5 mod 8, d from 14), we have 4 b 1 653. Therefore, b 648, d ) 643. But 643 does not divide 73. CASE ii) c l 1 11). Since c l 10, we have b mod 4 by Clm. Then 4 b 1 1 mod 8, d from 14), we have 4 b 1 1. Therefore, b 4, d ) 7. But 7 does not divide 73. CASE iii) c l 11 1). Since c l 10, we have b mod 4 by Clm. Then 4 b 11 3 mod 8, d from 14), we have 4 b 11 653. Therefore, b 668, d ) 683. But 683 does not divide 73. CASE iv) c l 11 1). Since c l 10, we have b 0 mod 4 by Clm. Then 4 b 11 1 mod 8, d from 14), we have 4 b 11 1. Therefore, b 8, d ) 15. But 15 does not divide 73. We have proved that the factorizatio3) is impossible. Case is done. Now suppose that R(x) were reducible over Z. Then, since R(x) is of degree 5, it would be divisible in Z[x] by a linear or a quadratic factor. However, both factorizations are impossible by Cases 1 d. This completes the proof of Lemma 4. Proof of Lemma 5. Let R(x) be the polynomial defined by 10). Since R(x) has the same number of real roots as Q(x) it suffices to prove that R(x) has exactly 3 real roots. The derivative R x) 4(10x4 x3 96x 5x 10)

136 Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI AND T. SHIBUYA is a polynomial of degree 4, d x lim R x) R 1) 0 R 0) 0 R 1) 0 lim x Hence R x) has only real roots, one in each of the intervals R x) 1) 1 0) 0 1) 1 We now consider R(x) on each of these intervals. Since x lim R(x) R( 1) 0 R(0) 0 R(1) 0 lim x R(x) R(x) has odd number of roots i), i 0) d i It follows from Rolle s theorem d what we know about the roots of R x) that R(x) has exactly one root in each of these intervals. And R(x) has no root in 0 1), because Indeed, by writing we see that R(x) 0 for 0 x 1 R(x) 8x3(x 1) x4 1) 10x 40x(1 3x) 10 R(x) 40x(1 3x) 10 for 0 x 1 And x(1 3x) attns its maximum on the interval [0 1] at x 13 whence R(x) 40 3 1 1 3 10 80 9 10 0 for 0 x 1 This concludes the proof : the polynomial R(x) has exactly 3 real roots, say x1 x x3 which are such that x1 1 x 0 d x3 1. 5. PROOF OF THEOREM 1 FOR n 6 In, we proved that the area of a cyclic pentagon with side lengths a b 1, c, d 3, e 4 does not belong to y radical extension of Q. In this section, we will prove Theorem 1 for n 6 by showing that the following assumption contradicts the above fact. ASSUMPTION For a certn integer n 6, there exists area formula F(a1 a ) which gives the area of arbitrary cyclic n-gon in terms of the side lengths a1 a using only the four arithmetic operations d k-th roots.

ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE 137 In this section, we will assume d n will always denote the particular integer specified in Let S0 denote the area of the cyclic pentagon with side lengths a1 a 1, a3, a4 3, a5 4. If t is a sufficiently small positive real number, then by Proposition 4 of Appendix A, there exists a cyclic n-gon with side lengths a1 a 1 a3 a4 3 a5 4 a6 t t Note that the radius of the circumscribed circle may depend on t PROPOSITION. 15) lim t 0 F(1 1 3 4 t t) S0 Proof. In general, we will denote by S(c1 c cm) the area of a cyclic m-gon whose side lengths are c1 c cm, where m is y integer with m 3. Then we have 16) S(1 1 3 4 t t) S(1 1 3 u) S(u 4 t t) In this equation, we are considering a cyclic n-gon B1B Bn with B1B BB3 1, B3B4, B4B5 3, B5B6 4, B6B1 t if n 6, or B6B7 t BBn t BnB1 t if n 7. See Figure.) Thus in equatio6), the number of t s on each side is n 5. Also u denotes the diagonal length u B1B5 which is a function of t It is geometrically clear that 17) lim t 0 u 4 d that 18) lim t 0 S(u 4 t t) 0 By Proposition 5 in Appendix A, S(c1 c of c1 c cm) Thus by 17), we have 19) lim t 0 S(1 1 3 u) S(1 1 3 4) S0 Assumption implies that F(1 1 3 4 t t) S(1 1 3 4 t t) Thus by 16), 18) d 19) we have 0) lim t 0 F(1 1 3 4 t t) S0 Proposition is now proved. cm) is a continuous function

138 Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI AND T. SHIBUYA B3 B 1 1 B1 t Bn t B6 B4 3 4 B5 FIGURE Cyclic n-gon B1B BBn By Assumption the value F(1 1 3 4 t t) c be calculated by starting from rational numbers d the variable t d applying the four arith metic operations d taking k-th roots. In other words, F(1 1 3 4 t t) is admissible function as defined in Appendix B. There we also define a restricted admissible function to be admissible function which c be constructed from a finite number of polynomials in t with rational coefficients by using only three arithmetic operations of addition, subtraction, d mul tiplication i.e. without using division), together with the operation of taking k-th roots. For notational simplicity, let us denote F(1 1 3 4 t t) by F(t) By Lemma 7 in Appendix B, admissible function F(t) c be expressed as a quotient of two restricted admissible functions: 1) F(t) f t) g(t) where f t) d g(t) are certn brches of restricted admissible functions which are not identically zero. Note that the domn of F(t) contns a small interval 0 t If is sufficiently small, this interval is contned in unramified domns in the sense of Appendix B) of f t) g(t) d pk t. We c choose a connected d simply connected open set D C) which contns the interval 0 t d serves as unramified domn for all these functions simulteously. We assume that on D a brch denoted by t of pk t is selected so that t 1k 1k 0 for 0 t Then by Proposition 7 in Appendix B, the functions f t) d g(t) have Puiseux expsions

ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE 139 1p p ) f t) c0 c1t ct 0 t 3) g(t) d0 d1t 1q dt q 0 t where p d q are positive integers, d all the coefficients ci a radical extension of Q. dj belong to Since F(t) is the quotient of f t) d g(t) see 1)), d its limit when t 0 is a finite non-zero number S0 see 15)), we infer that the first non-zero terms of ) d 3), say cit ip d djt j q have the same exponents : i p j q Then by ccelling t ip t j q from the numerator d the denominator, we have F(t) ci ci 1t dj dj 1t 1p 1q ci t p dj t q 0 t This implies that 4) lim F(t) t 0 ci dj which belongs to a radical extension of Q. Since by 15) this limit is equal to S0 4) contradicts the fact proved in ) that S0 does not belong to y radical extension of Q. This contradiction shows that Assumption is absurd. This proves Theorem 1 for n 6. We would like to remark that Theorem 1 for n 5 does not trivially imply Theorem 1 for n 6. The following proposition seems to indicate the subtlety of the problem. We have shown in that for certn cyclic pentagons ABCDE with AB AE, there is no formula which gives the area in terms of the side lengths using only arithmetic operations d k-th roots. However, if ABCDE is y cyclic pentagon with AB AE, d if F is y point other th A or E) on the arc AE of the circumscribed circle as in Figure 3), then we c prove : PROPOSITION 3. hexagon ABCDEF arithmetic operations d square roots. There exists a formula which gives the area of the cyclic of Figure 3) in terms of its side lengths, using only

140 Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI AND T. SHIBUYA A f a F e E a u B d b c D C FIGURE 3 Cyclic pentagon ABCDE d point F Proof. Consider the cyclic quadrilateral ABEF d its diagonal AE. Let u denote the length of the chord BE. By calculating AE as we did for x in the proof of Lemma, we have 5) AE e f)au a u)ef au ef But AE a ; after some simplifications we get 6) u a(a e f ) ef Since the quadrilaterals ABEF d BCDE are cyclic, their areas c be calculated by Brahmagupta s formula) from the side lengths a u e f d b c d u, respectively, using only arithmetic operations d square roots. This together with 6) completes the proof of Proposition 3. 6. APPENDIX A The purpose of this appendix is to prove two propositions on cyclic polygons, which are probably well-known, but are used in our arguments. PROPOSITION 4. Let n be integer greater th. Let i 1 n, be positive real numbers. The following three conditions are equivalent:

ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE 141 n i) max(a1 a ) i 1, in other words, s 0 for each i 1 n, where s a1 Pa ), ii) there exists n-gon whose side lengths are a1 a iii) there exists a cyclic n-gon whose side lengths are a1 a Proof. The implications iii) ii) d ii) i) are obvious. We will prove that i) iii). Assume condition i). We may assume that max(a1 a ) Then i) is equivalent to 7) a1 a 1 0, let C(r) denote a circle of radius r. If A d B are two points on C(r) d a For r chord AB is AB, then the gle at the center of C(r) subtended by the 8) arcsin a r where we choose the brch of arcsin so that arcsin(x) for 1 x 1 To prove the implication i) iii), we consider three cases : A) B) C) P P P i arcsin 1 i arcsin 1 i 1 arcsin CASE A). Note that if C(r) circumscribes n-gon whose side lengths are a1 a then the diameter r must satisfy r max(a1 a ) that is, r Consider the continuous function f r) defined by 9) f r) n arcsin r r By assumption A), we have

14 Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI AND T. SHIBUYA f arcsin arcsin By 9), we have Since lim r f r) 0 n 30) f r) 0 for r 4r rp the function f r) is monotone decreasing. Therefore, there exists a unique value r0 such that f r0) i.e. 31) n arcsin Equation 31) mes that the sum of the gles at the center of C(r0) subtended by the chords of lengths a1 a is See 8).) Thus there exists n-gon with side lengths a1 a inscribed in the circle C(r0) Case A) is done. r0 CASE B). Take n points A1 A An in this order, on the circle C of radius in such a way that AiAi 1, for i 1. Then by assumption B), the sum of the central gles subtended by the chords A1A AA3 AAn is equal to Thus the chord A1An is a diameter of C d its length is equal to This implies that the n-gon A1A An inscribed in C has the required side lengths a1 a This concludes Case B). CASE C). Take n points A1 A An, in this order, on the circle We take them so that AiAi 1, for C(r) of radius r, where r i 1. The length of the chord A1An depends on r while the lengths of A1A AA3 AAn are fixed as above, independently of r We denote the length of A1An as a continuous function of r, by g(r) It is defined for r Assumption C) implies that if r the sum of the gles at the center of C subtended by A1A AA3 An 1An, is less th Thus if r the chord A1An is shorter th the diameter of C that is, 3) g

ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE 143 As we show below, the function g(r) is monotone increasing, d it is geometrically clear that 33) lim r Recall that by 7), g(r) 34) Then by 3), 33), 34), we infer that there exists a unique value r0 such that r0 d g(r0) ; in other words, we have A1An on the circle C(r0) This implies that the n-gon A1A An inscribed in the circle C(r0) has the required side lengths a1 a This concludes Case C) except for the proof that g(r) is monotone increasing. We now prove this fact. Let O denote the center of C(r) By assump tion C), we have 35) A1OAn arcsin r for r d the function g(r) is written explicitly as 36) g(r) r sin A1OAn r sin arcsin r For simplicity, we set Then we have i arcsin r g r) sin i r cos i r p 4r As is clear from Figure 4, we have 37) p 4r Substituting 37), we have t i

144 Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI AND T. SHIBUYA Ai 4r a i B Ai 1 r FIGURE 4 i AiBAi 1 arcsin r d AiB p4r 38) g r) sin sin i i cos cos Note that we used the fact that 0 i computation. See 35). i i t i t P i 1 i i 0 in the above Thus we have proved that g r) 0 for r i.e that g(r) is monotone increasing, as asserted. This completes the proof of Proposition 4. Dn Let Dn be the open set in n-dimensional space Rn defined as follows: n a1 a ) a1 0 0 max(a1 a ) j By Proposition 4, for each a1 a ) Dn there exists a cyclic n-gon whose side lengths are a1 a. Let S(a1 a ) denote the area of such a cyclic n-gon. n o PROPOSITION 5. S(a1 a ) is a continuous function on Dn Though this proposition seems intuitively clear, we will give a proof for completeness. Before proving Proposition 5, we will prove a closely related lemma.

ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE 145 Let r denote the radius of the circumscribed circle of a cyclic n-gon with side lengths a1 a LEMMA 6. r is a continuous function on Dn Proof. We divide Dn into subsets A C1 C Cn defined as follows: A n a1 a ) n arcsin M o where M max(a1 a ) Cj n a1 a ) aj max(a1 a ) d arcsin ixi j) Note that Dn A C1 C Cn From the arguments of Cases B) d C) in the proof of Proposition 4, it is clear that if a1 a ) Cj, then aj o Thus if j k, then aj ak for k f1 ng n f jg Also note that because if M Suppose that A Cj n a1 a ) j arcsin ixi j) Cj aj, then arcsin aj M Ck arcsin(1) aj o a1 a ) int( A) A n Then from the argument of Case A) in the proof of Proposition 4, r is uniquely determined by the condition f r) where f r) is the function defined by 9). Differentiating f r) we have f r) 0 see 30)). Thus by the implicit function theorem, the value of r satisfying f r) depends smoothly on a1 a ) We denote this function by j A Cj r A(a1 a ) a1 a ) int( A) If a point a1 a ) int( A) approaches the boundary A Cj, if aj max(a1 a ) d that is,

146 Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI AND T. SHIBUYA arcsin ix 0 aj i j) then from Case A) in the proof of Proposition 4, we have aj A(a1 a ) On the other hd, from Case B) in the proof of Proposition 4, it is clear that the radius of the circumscribed circle of a polygon corresponding to a point a1 a ) is equal to A aj Therefore, the smooth function Cj A on int( A) is continuously extended to by defining A aj 39) A(a1 a ) a1 a ) A Cj Next suppose that a1 a ) int( Cj) Cj n A Cj) Then aj max(a1 a ) d the radius r of the circumscribed circle is uniquely determined by the condition g(r) aj, where g(r) is a function explicitly given by g(r) r sin arcsin r i i j) X See equation 36) in Case C) of the proof of Proposition 4, where it was assumed that max(a1 a ) Since g r) 0 by 38), the implicit function theorem tells us that the radius r is a smooth function of a1 a ) int( Cj) Let us denote this function by r j(a1 a ) a1 a ) int( Cj) Cj n A Cj) If a point a1 a ) int( Cj) approaches the boundary A Cj, that is, if arcsin ix 0 aj i j) then from Case C) of the proof of Proposition 4, we have aj j(a1 a ) Thus as in the case of A, the smooth function j on int( Cj) is continuously extended to Cj by defining aj 40) j(a1 a ) a1 a ) A Cj By 39) d 40), we have for j 1 n A(a1 a ) j(a1 a ) a1 a ) A Cj Therefore, by gluing togther A, d j j 1 n, we obtn a continuous function r a1 a ) on Dn This proves Lemma 6.

ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE 147 Once Lemma 6 is proved, Proposition 5 is easy to prove. Proof of Proposition 5. The area S(a1 a ) is calculated as follows : n S(a1 a ) P P i i j) i 1 qr qr 4 4 aj qr aj on A 4 on Cj These two expressions coincide on the boundary because we A Cj aj have there. Since r depends on a1 a ) continuously, r S(a1 a ) also depends continuously on a1 a ) This completes the proof of Proposition 5. 7. APPENDIX B The purpose of this appendix is to discuss Puiseux expsions of complex valued algebrc functions of one complex variable, used in the proof of Theorem 1. For explation of Puiseux expsions, see [8]. We will denote by Q[t] the polynomial ring of a variable t with rational coefficients, d by Q(t) the quotient field of Q[t] In other words, Q(t) is the field of rational functions of a variable t with rational coefficients. In this appendix, our discussion will be confined to rather special types of algebrc functions. DEFINITION. An algebrc function F(t) is sd to be admissible function if it belongs to a radical extension of the field Q(t) This mes that a function F(t) is admissible if d only if it is constructed from a finite number of polynomial functions Q[t] by using the four arithmetic operations, together with the operation of taking k-th roots. Note that when taking a k -th root of a function f t) 41) k f t) we meet the ambiguity that the value is only determined up to multiplication by k-th roots of unity. In other words, the expression 41) may be y one p of the following k functions brches) k k 4) f t) f t) k 1 k f t) p p p

148 Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI AND T. SHIBUYA where is a primitive k -th root of unity. In some special cases, we c remove this ambiguity. For example, take a connected d simply connected region D in the complex number ple C, so that D does not contn y zeros of f t) nor of 1 f t) d restrict the variable t within D, then we c remove the ambiguity of 41) in the sense that we c choose as we like one of the brches from 4) over the domn D without y ambiguity. We will later show how to take a useful domn D in our application. However, before that, we will consider admissible function to be a multi valued function. DEFINITION. An admissible function F(t) is sd to be of restricted type or briefly a restricted admissible function, if we c construct it from a finite number of polynomials Q[t]) by using only the three arithmetic operations of addition, subtraction, d multiplication i.e. without using division), together with the operation of taking k-th roots. For example, a polynomial function Q[t] is a restricted admissible function. It is easy to see that the next lemma holds. LEMMA 7. Every admissible function c be expressed as a quotient of two restricted admissible functions. A polynomial in indeterminate X is sd to be monic, if the coefficient of the leading term Xm is 1. LEMMA 8. A restricted admissible function F(t) which is not identically zero satisfies a monic polynomial equation whose coefficients belong to Q[t] More precisely, given a non-zero restricted admissible function F(t), there exists a monic polynomial equation 43) Xm f1xm 1 fm 1X fm 0 with fi Q[t] i 1 m, such that X F is one of its solutions, that is, 44) Fm f1fm 1 fm 1F fm 0 holds identically as a function of t This lemma c be proved by arguments similar to those in Chapter I, of [3].

ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE 149 For a given F(t) choosing a polynomial equation 43) with the lowest degree m, we may assume that fm is not a zero polynomial. This is because if fm 0, then F(t) would satisfy a polynomial equation of lower degree Fm 1 f1fm fm 1 0 The following proposition is in fact a corollary of Lemma 8. PROPOSITION 6. A restricted admissible function F(t) which is not identically zero has a finite number of zeros. Proof. Suppose that F(t) satisfies equation 43), i.e. that equation 44) holds identically as a function of t Suppose that F(t0) 0 for some t0 C. Then from 44), we have fm(t0) 0 Thus the zero set of F is a subset of the zero set of fm Since a polynomial fm has a finite number of zeros, this proves Proposition 6. Let F(t) be a restricted admissible function which is not identically zero. We define inductive sequence for constructing F(t) to be a sequence consisting of a finite number of non-zero restricted admissible functions 45) I( F(t)) ff1 F FNg which satisfies the following conditions a) d b) : a) F1 is a polynomial in t with rational coefficients, d FN F(t) the given restricted admissible function, b) each Fh(h N) is a polynomial in t with rational coefficients, pk or Fh Fi Fj Fh FiFj, or Fh Fi, where Fi d Fj are functions in the sequence F(t)) which appear before Fh Furthermore, for each such I( h, the indices i d j are explicitly specified. Suppose we are given a non-zero restricted admissible function F(t) Then fixing a certn inductive sequence F(t)) for constructing it, we define the I( set of ramification points of F(t) denoted by Ram(F) C), inductively as follows: i) If Fh( F(t)) I( is a polynomial in t with rational coefficients, we set Ram(Fh) ii) if Fh Fi Fj or Fh FiFj, where Fi d Fj are specified non-zero restricted admissible functions in the sequence F(t)) appearing before Fh I( then we set

150 Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI AND T. SHIBUYA Ram(Fh) Ram(Fi) Ram(Fj) pk Fi with a specified restricted admissible function Fi which iii) if Fh appears before Fh in the sequence I( F(t)) d k 1, then we set Ram(Fh) Ram(Fi) Zero(Fi) where Zero(Fi) is the zero set of Fi REMARK. We adopt the convention that if in the inductive sequence I( F(t)) a function Fh is a polynomial in t with rational coefficients, d at the same time, the construction of Fh is explicitly specified as Fh Fi Fj pk Fh FiFj or Fh Fi, then to define Ram(Fh) we apply rule ii) or iii) rather th i). Although the notation is somewhat imprecise, the set Ram(F(t)) depends not only on F(t) but also on F(t)) When we speak of the set of ramification I( points of F(t) we always assume tacitly that a certn inductive sequence 45) for constructing F(t) has been chosen d fixed. By the definition of a restricted admissible function d Proposition 6, Ram(F) is a finite set of points C). DEFINITION. Let F(t) be a restricted admissible function which is not identically zero. An open set D C) is sd to be unramified domn for Ram(F) F(t) if D is connected d simply connected, d satisfies D If D is unramified domn for a restricted admissible function F(t) then F(t) restricted to D is a disjoint union of a finite number of brches, each of which is a univalent function over D. For example, if D is a connected d simply connected open set which does not contn 0, then D is unramified domn for pk t, for y k 1. Moreover, if D contns open interval 0 R) with a small 0, then we c uniquely select a brch of the function pk t over D such that 46) pk t 0 for each t 0 We will denote this brch by t 1k In the following proposition, D denotes the connected component of ft C j jtj D which contns 0 g

ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE 151 PROPOSITION 7. Let F(t) be a restricted admissible function which is not identically zero. Let D be unramified domn for F(t) Suppose that D does not contn 0, but contns open interval 0 with a sufficiently small 0. Then for each brch of F(t) over D, there exists integer p 0 such that the brch c be expded as follows : 47) F(t) c0 c1t 1p ct p for t Furthermore, for a fixed F(t) all the coefficients ci belong to a radical extension of Q. D The expsion 47) is called the Puiseux expsion of F(t) Proof. Let F(t)) be the inductive sequence I( for constructing F(t) which is tacitly assumed. We will prove Proposition 7 by induction based on I( F(t)) starting from a polynomial Q[t] Note that by the definition of unramified domn for a restricted admissible function, the unramified domn D for F(t) also serves as unramified domn for all the functions which appear in the inductive sequence I( F(t)) A polynomial f t) Q[t] has a natural Puiseux expsion : f t) a0 a1t amtm in which all the coefficients a0 a1 am belong to Q. Suppose that brches of two restricted admissible functions f t) g(t) have Puiseux expsions: 48) 49) f t) a0 a1t g(t) b0 b1t 1p 1q at p bt q in which a0 a1 a belong to a radical extension K1 of Q, d b0 b1 b belong to other radical extension K of Q. If f t) g(t) then the sum f t) g(t) is not identically zero, d has a Puiseux expsion f t) g(t) c0 c1t 1r ct r where r l c m p q) d ci 0 aj bj, or aj bk as the case may be. Thus the coefficients c0 c1 c belong to a radical extension K3 of Q generated by K1 K The argument for the difference f t) g(t) is the same. The product of f t)g(t) has a Puiseux expsion where r pq, d f t)g(t) c0 c1t 1r ct r

15 Y. MATSUMOTO, Y. MATSUTANI, M. ODA, T. SAKAI AND T. SHIBUYA ci X ajbk with the indices j k running over all prs j k) that satisfy j p If for some i no such pr exists, then ci c0 c1 c 0.) Obviously, the coefficients belong to a radical extension K3 of Q generated by K1 K Finally, let us consider a brch of k -th roots of f t) We assume that f t) has Puiseux expsion 48) whose coefficients a0 a1 a belong to a radical extension K1 of Q. Let n be the smallest index such that 0. Then we have k np 50) p p f t) k qt 1t pk t n k pk r1 1 t 1p t Note that the value of pk is determined without ambiguity involving k-th roots of unity by the choice of the brch of pk f t) over D d pk belongs to a radical extension K1 of K1. Here K1 is the radical extension of Q which is generated by K1 d pk Recall the binomial expsion: p kq ir 51) pk 1 z 1ki zi Xi 0 where 5) 1ki 1 1) i! 1k 1k 1k i 1) i 1 1 i 0 In particular, the binomial coefficients 5) belong to Q. The series 51) converges for jzj 1. Thus if is sufficiently small, we have for 0 t : 53) Equation k r1 1 t 1p t p Xi 0 1ki 1 t 1p t p 53) shows that all the coefficients of the Puiseux expsion of i k r1 1 t 1p belong to K1 belong to a radical extension K1 of Q. Therefore, Proposition 7 is proved by induction. t p Thus, by 50), pk f t) has a Puiseux expsion whose coefficients

ON THE AREA OF A POLYGON INSCRIBED IN A CIRCLE 153 REFERENCES [1] BLASCHKE, W. Kreis und Kugel. Walter de Gruyter, Berlin, 1956. Reprint: Chelsea Publishing Compy, New York, 1949. [] BOYER, C. B. A History of Mathematics. nd ed. revised by U. C. Merzbach). John Wiley & Sons, Inc., 1991. [3] LANG, S. Algebrc Number Theory. Graduate Texts in Math., Springer-Verlag, 1986. [4] POSTNIKOV, M. M. Foundations of Galois theory. Trslated by A. Swinfen, trslation ed. P. J. Hilton. Pergamon Press, 196. [5] SWALLOW, J. Exploratory Galois Theory. Cambridge Univ. Press, 004. [6] VARFOLOMEEV, V. V. Inscribed polygons d Heron polynomials. Sb. Mat. 194 003), 311 331. [7] Galois groups of the Heron-Sabitov polynomials for pentagons inscribed in a circle. Sb. Mat. 195 004), 149 16. [8] WALL, C. T. C. Singular Points of Ple Curves. London Math. Soc. Student Texts 63. Cambridge Univ. Press, 004. Reçu le 10 jvier 006; version révisée reç ue le 10 mars 007) Yukio Matsumoto Department of Mathematics Faculty of Science Gakushuin University 1-5-1 Mejiro, Toshima-ku Tokyo 171-8588 Jap e-ml : yukiomat@math.gakushuin.ac.jp Yoshikazu Matsuti 3-1-1 Minamimachi Kokubunji Tokyo 185-001 Jap e-ml : ktbr-amrc@taupe.plala.or.jp Masami Oda Tsuda College, Kodra Tokyo 187-8577 Jap e-ml : oda@tsuda.ac.jp Tsuyoshi Sak Department of Mathematics College of Humities d Sciences Nihon University, Setagaya-ku Tokyo 156-0045 Jap Tsukasa Shibuya -19--301 Nishiwaseda Shinjuku-ku Tokyo 169-0051 Jap