Journal of Inequalities in Pure and Applied Mathematics

Transcription

1 Journal of Inequalities in Pure and Applied Mathematics CERTAIN INEQUALITIES CONCERNING BICENTRIC QUADRILATERALS, HEXAGONS AND OCTAGONS MIRKO RADIĆ University of Rijeka Faculty of Philosophy Department of Mathematics Rijeka, Omladinska 14, Croatia. volume 6, issue 1, article 1, Received 15 April, 2004; accepted 24 November, Communicated by: J. Sándor Abstract Home Page c 2000 Victoria University ISSN (electronic):

2 Abstract In this paper we restrict ourselves to the case when conics are circles one completely inside of the other. Certain inequalities concerning bicentric quadrilaterals, hexagons and octagons in connection with Poncelet s closure theorem are established Mathematics Subject Classification: 51E12 Key words: Bicentric Polygon, Inequality. 1 Introduction Bicentric Quadrilaterals Bicentric Hexagons Bicentric Octagons References Page 2 of 43

3 1. Introduction A polygon which is both chordal and tangential is briefly called a bicentric polygon. The following notation will be used. If A 1 A n is considered to be a bicentric n-gon, then its incircle is denoted by C 1, circumcircle by C 2, radius of C 1 by r, radius of C 2 by R, center of C 1 by I, center of C 2 by O, distance between I and O by d. A 1 C 2 C 1 R d O I A 2 A 3 r A 4 C 2 C 1 Figure 1.1 Figure 1.2 The first person who was concerned with bicentric polygons was the German mathematician Nicolaus Fuss ( ). He found that C 1 is the incircle and C 2 the circumcircle of a bicentric quadrilateral A 1 A 2 A 3 A 4 iff (1.1) (R 2 d 2 ) 2 = 2r 2 (R 2 + d 2 ), (see [4]). The problem of finding this relation has been mentioned in [3] as one of 100 great problems of elementary mathematics. t M O I t m Page 3 of 43

4 Fuss also found the corresponding relations (conditions) for bicentric pentagon, hexagon, heptagon and octagon [5]. For bicentric hexagons and octagons these relations are (1.2) 3p 4 q 4 2p 2 q 2 r 2 (p 2 + q 2 ) = r 4 (p 2 q 2 ) 2 and (1.3) [r 2 (p 2 + q 2 ) p 2 q 2 ] 4 = 16p 4 q 4 r 4 (p 2 r 2 )(q 2 r 2 ), where p = R + d, q = R d. The very remarkable theorem concerning bicentric polygons was given by the French mathematician Poncelet ( ). This theorem is known as Poncelet s closure theorem. For the case when conics are circles, one inside the other, this theorem can be stated as follows: If there is one bicentric n-gon whose incircle is C 1 and circumcircle C 2, then there are infinitely many bicentric n-gons whose incircle is C 1 and circumcircle is C 2. For every point P 1 on C 2 there are points P 2,..., P n on C 2 such that P 1 P n are bicentric n-gons whose incircle is C 1 and circumcircle is C 2. Although the famous Poncelet s closure theorem dates from the nineteenth century, many mathematicians have been working on a number of problems in connection with it. Many contributions have been made, and much interesting information can be found concerning it in the references [1] and [6]. An important role in the following will have the least and the largest tangent that can be drawn from C 2 to C 1. As can be seen from Figure 1.2, the following holds (1.4) t m = (R d) 2 r 2, t M = (R + d) 2 r 2. Page 4 of 43

5 2. Bicentric Quadrilaterals Let A 1 A 2 A 3 A 4 be any given bicentric quadrilateral whose incircle is C 1 and circumcircle C 2 and let (2.1) t i + t i+1 = A i A i+1, i = 1, 2, 3, 4. (Indices are calculated modulo 4.) In [8, Theorem 3.1 and Theorem 3.2] it is proven that the following hold (2.2) t 1 t 3 = t 2 t 4 = r 2, and (2.3) t 1 t 2 + t 2 t 3 + t 3 t 4 + t 4 t 1 = 2(R 2 d 2 ). Reversely, if t 1, t 2, t 3, t 4 are such that (2.2) and (2.3) hold, then there is a bicentric quadrilateral such that (2.1) holds. Theorem 2.1. The tangent-lengths t 1, t 2, t 3, t 4 given by (2.1) satisfy the following inequalities (2.4) 2r t 1 + t 3 t m + t M, Page 5 of 43 (2.5) 2r t 2 + t 4 t m + t M,

6 (2.6) 4r t 1 + t 2 + t 3 + t 4 4r R2 + d 2 R 2 d 2, (2.7) 4r 2 t t t t 2 4 4(R 2 + d 2 r 2 ), and (2.8) t 2k 1 + t 2k 2 + t 2k 3 + t 2k 4 4r 2k, k N. The equalities hold only if d = 0. Proof. First let us remark that t m t M = r 2 since there is a bicentric quadrilateral as shown in Figure 2.1. Now, let C denote a circle whose diameter is t m + t M (Figure 2.2). Then for each t i, i = 1, 2, 3, 4, since t m t i t M, there are points Q and R on C such that (2.9) t i = P Q, t i+2 = P R, where P Q + P R = QR. In this connection let us remark that the power of the circle C at P is t m t M. Therefore P Q P R = t m t M. Obviously t i + t i+2 t m + t M since t m + t M is a diameter of C. Also it is clear that t i + t i+2 2r since r 2 = t m t M. This proves (2.4) and (2.5). In the proof that (2.6) holds we shall use the relations Page 6 of 43 (2.10) t m = r R d R + d, t M = r R + d R d.

7 C 2 t t i M C r t r m 1 t t M m C Q O I P r t i+2 R Figure 2.1 Figure 2.2 It is easy to show that each of the above relations is equivalent to the Fuss relation (1.1). So, for the first of them we can write ( ) 2 R d (R d) 2 r 2 = r 2 R + d (R 2 d 2 ) 2 r 2 (R + d) 2 = r 2 (R d) 2 (R 2 d 2 ) 2 = 2r 2 (R 2 + d 2 ). The proof that (2.6) holds can be written as 2r + 2r t 1 + t 3 + t 2 + t 4, t 1 + t 3 + t 2 + t 4 2(t m + t M ) = 2r ( R d + ) R+d R+d R d = 4r R 2 +d 2. R 2 d 2 The proof that (2.7) holds is as follows. Page 7 of 43

8 Since 2r t 1 + t 3, 2r t 2 + t 4, we have or, since 2t 1 t 3 = 2t 2 t 4 = 2r 2, 4r 2 t t t 1 t 3, 4r 2 t t t 2 t 4 2r 2 t t 2 3, 2r 2 t t 2 4. Thus, 4r 2 t t t t 2 4. From t 1 + t 3 t m + t M, t 2 + t 4 t m + t M it follows that t t 2 3 t 2 m + t 2 M, since 2t 1 t 3 = 2t 2 t 4 = 2t m t M. Thus, we obtain t t 2 4 t 2 m + t 2 M t t t t 2 4 2(t 2 m + t 2 M), where t 2 m + t 2 M = (R d)2 r 2 + (R + d) 2 r 2 = 2(R 2 + d 2 r 2 ). In the same way it can be proved that (2.8) holds. So, starting from 2r t 1 + t 3, since 2t k 1t k 3 = 2r 2k, we can write 2r 2 t t 2 3, 4r 4 t t t 2 1t 2 3 or 2r 4 t t 4 3 and so on. Starting from t 1 + t 3 t m + t M it can be written t t 2 3 t 2 m + t 2 M, t t 4 3 t 4 m + t 4 M, Page 8 of 43

9 and so on. Since t m = t M only if d = 0, it is clear that the relations (2.4) (2.8) become equalities only if d = 0. Thus, if d 0, then in the above relations instead of we may put <. Theorem 2.1 is thus proved. Corollary 2.2. The following holds (2.11) 4 r 1 t i 4 r R2 + d 2 R 2 d 2. Proof. From 2r t 1 + t 3 it follows that 2 r 1 t t 3, since = t 1 + t 3 = t 1 + t 3 t 1 t 3 t 1 t 3 r 2 2r r 2 = 2 r. From t 1 + t 3 t m + t M it follows that 1 t t 3 1 t m + 1 t M, since 1 t t 3 = t 1 + t 3 r 2, = t m + t M. t m t M r 2 Corollary 2.3. Let a = t 1 + t 2, b = t 2 + t 3, c = t 3 + t 4, d = t 4 + t 1.Then (2.12) 8r a + b + c + d 8r R2 d 2 R 2 + d 2. Page 9 of 43

10 Corollary 2.4. Let a, b, c, d be as in Corollary 2.3. Then (2.13) 4(R 2 d 2 + 2r 2 ) a 2 + b 2 + c 2 + d 2 4(3R 2 2r 2 ). Proof. Using relation (2.3) we can write a 2 + b 2 + c 2 + d 2 = 2(t t t t 2 4) + 4(R 2 d 2 ). Now, using relations (2.7) we can write relations (2.13). Corollary 2.5. The following holds (2.14) 2r 2 + d 2 R 2 2r 2 + d 2 + 2rd. Proof. Since t 1 + t 3 2r, t 2 + t 4 2r, we can write (t 1 + t 3 )(t 2 + t 4 ) 4r 2, t 1 t 2 + t 2 t 3 + t 3 t 4 + t 4 t 1 4r 2, 2(R 2 d 2 ) 4r 2, R 2 d 2 2r 2. The fact that R 2 2r 2 + d 2 + 2rd is clear from the quadratic function f(d) = d 2 + 2rd + R 2 2r 2. If d = 0, then f(d) = 0, but if d > 0, then f(d) > 0. Page 10 of 43

11 Remark 1. It may be interesting that relations (2.14) can be obtained directly from Fuss relation (1.1). It was done by L. Fejes Toth in [11]. Namely, relation (1.1) implies (2.15) d 2 = r 2 + R 2 r r 2 + 4R 2, so the left side inequality of (2.14) becomes equivalent to 2r 2 R 2 or (2.16) r 2 R. d) The right side of (2.14) is equivalent to (quadratic polynomial inequality in d r + R 2 r 2 or by using (2.15), after some simple computations, to (2.16), again. Concerning the sign in the relations (2.11) (2.14), it is clear that in the case when d 0, that is, when t m t M, then instead of may be put <. In connection with Theorem 2.1, the following theorem is of some interest. Theorem 2.6. Let P = P 1 P 2 P 3 P 4 and Q = Q 1 Q 2 Q 3 Q 4 be axially symmetric bicentric quadrilaterals whose incircle is C 1 and circumcircle C 2 (Figure 2.3). Denote by 2p M and 2p m respectively the perimeters of P and Q. Then for every bicentric quadrilateral A = A 1 A 2 A 3 A 4 whose incircle is C 1 and circumcircle C 2 it holds that (2.17) p m t i p M, Page 11 of 43

12 where t i + t i+1 = A i A i+1, i = 1, 2, 3, 4. Also, if d 0, then p m < p M and (2.18) t i = p M iff A = P, t i = p m iff A = Q. Proof. First we see that (2.19) p M = t m + 2r + t M, p m = 2(ˆt 1 + ˆt 3 ), where r = P 2 H and (2.20) t m = P 1 G = (R d) 2 r 2, t M = P 3 H = (R + d) 2 r 2, (2.21) ˆt 1 = EQ 1 = R 2 (r + d) 2, ˆt 3 = F Q 3 = R 2 (r d) 2. According to Theorem 3.3 in [8], the tangent lengths t 2, t 3, t 4 can be expressed by t 1 as follows: (2.22) t 2 = (R2 d 2 )t 1 + D, t r 2 + t 2 3 = r2, 1 t 1 where (2.23) D = (R 2 d 2 ) 2 t r 2 (r 2 + t 2 1) 2. t 4 = r2 t 2, Page 12 of 43

13 Q 2 P 2 A 1 H R G Q 1 t 1 C 2 C 1 P 3 F O I r E P 1 O I A 4 Q 4 A 2 A 3 Q 3 P 4 Figure 2.3 Figure 2.4 In this connection let us remark that for every point A 1 on C 2 there is a tangent t 1 drawn from C 2 to C 1 (Figure 2.4). If t 1 is given, then quadrilateral A 1 A 2 A 3 A 4 is completely determined by t 1, and t 2, t 3, t 4 can be calculated using expressions (2.22). Let the sum 4 t i, where t 2, t 3, t 4 are expressed by t 1, be denoted by s(t 1 ). d It can be easily found that dt 1 s(t 1 ) = 0 can be written as from which it follows that (t 2 1 r 2 )[t 4 1 2(R 2 d 2 r 2 )t r 4 ] = 0, (t 2 1) 1 = r 2, (t 2 1) 2 = ˆt 2 1, (t 2 1) 3 = ˆt 2 3, where ˆt 1 and ˆt 3 are given by (2.21). In this connection let us remark that ± (R 2 d 2 r 2 ) 2 r 4 = ±2dr Page 13 of 43

14 since, using Fuss relation (1.1), we can write (R 2 d 2 r 2 ) 2 r 4 = (R 2 d 2 ) 2 2(R 2 d 2 )r 2 = 2r 2 (R 2 + d 2 ) 2(R 2 d 2 )r 2 = 4d 2 r 2. as The part of the expression d2 dt 2 1 s(t 1 ) important for discussion can be expressed t 4 1 2(R 2 d 2 r 2 )t r 4 + 2(t 2 1 r 2 )[t 2 1 (R 2 d 2 r 2 )]. For brevity, let the above expression be denoted by S(t 1 ). It is easy to find that (2.24) S(r) = R 2 + 2r 2 + d 2 < 0, (2.25) S(ˆt 1 ) = 2dr > 0, (2.26) S(ˆt 3 ) = (R 2 2r 2 d 2 2rd)( 2dr) > 0, where the relations (2.14) are used. In this connection let us remark that by Theorem 3.3 in [8] the following Page 14 of 43

15 holds: if t 1 = r and t 2, t 3, t 4 are given by (2.22), then if t 1 = ˆt 1 and t 2, t 3, t 4 are given by (2.22), then if t 1 = ˆt 3 and t 2, t 3, t 4 are given by (2.22), then Theorem 2.6 is thus proved. t i = p M, t i = p m, t i = p m. Corollary 2.7. Let A be as in Theorem 2.6, that is, A is any given bicentric quadrilateral whose incircle is C 1 and circumcircle C 2. Then Proof. From (2.17) it follows that area of Q area of A area of P. (2.27) rp m r(t 1 + t 2 + t 3 + t 4 ) rp M. Using relations (2.22) and denoting the area of A by J(t 1 ), the inequalities (2.27) can be written as J(ˆt 1 ) J(t 1 ) J(t m ), where ( ) J(t 1 ) = r t 1 + r2 + (R2 d 2 )t 1 + D r 2 (r 2 + t 2 + 1) t 1 r 2 + t 2 1 (R 2 d 2 )t 1 +. D Page 15 of 43

16 Since, according to Theorem 3.3 in [8], we have J(t m ) = J(r) = J(t M ), J(ˆt 1 ) = J(ˆt 3 ), the graph of J(t 1 ) is like that shown in Figure 2.5. J(t ) m J(t ) 1 J(t ) 1 O t m t 1 r t 2 t M t 1 Figure 2.5 Of course, J(t m ) = r(t m + 2r + t M ), J(ˆt 2 ) = 2r(ˆt 1 + ˆt 2 ). Let us remark that ˆt 2 = ˆt 3. (See Figure 2.3.) Corollary 2.8. The following holds p m r 1 p M 2 t i r. 2 Page 16 of 43

17 Proof. Since t 1 t 3 = t 2 t 4 = r 2, we have (2.28) 1 t i = t 1t 2 t 3 + t 2 t 3 t 4 + t 3 t 4 t 1 + t 4 t 1 t 2 t 1 t 2 t 3 t 4 = t 1 + t 2 + t 3 + t 4 r 2. From the proof it is clear that 1 t i = maximum (minimum) iff Corollary 2.9. The following holds p 2 m 4(R 2 d 2 ) t i = maximum (minimum). t 2 i p 2 M 4(R 2 d 2 ). Proof. Since (t 1 + t 2 + t 3 + t 4 ) 2 = 4 t2 i + 4(R 2 d 2 ), we have p 2 m t 2 i + 4(R 2 d 2 ) p 2 M. From the proof it is clear that t 2 i = maximum (minimum) iff t i = maximum (minimum). Page 17 of 43

18 Corollary When the arithmetic mean A(t 1, t 2, t 3, t 4 ) is maximum, then the harmonic mean H(t 1, t 2, t 3, t 4 ) is minimum and vice versa. Proof. From (2.28) it follows that A(t 1, t 2, t 3, t 4 ) H(t 1, t 2, t 3, t 4 ) = r 2. Corollary Let t 1 be given such that t m t 1 t M. Then the equation J(t 1 )J(x) = J(t m )J(ˆt 1 ) has four positive roots x 1, x 2, x 3, x 4 and we have x 1 x 2 + x 2 x 3 + x 3 x 4 + x 4 x 1 = 2(R 2 d 2 ), x 1 x 2 x 3 x 4 = r 4. Proof. There is a bicentric quadrilateral X 1 X 2 X 3 X 4 whose incircle is C 1 and circumcircle C 2 such that area of A 1 A 2 A 3 A 4 area of X 1 X 2 X 3 X 4 = J(t m )J(ˆt 1 ), x i + x i+1 = X i X i+1, i = 1, 2, 3, 4. In connection with the sum t v 1 + t v 2 + t v 3 + t v 4, where v is a real number, the following theorem will be proved. Theorem If there is a bicentric quadrilateral whose tangent lengths are t 1, t 2, t 3, t 4, then there is a bicentric quadrilateral whose tangent lengths are t v 1, t v 2, t v 3, t v 4, where v may be any given real number. Page 18 of 43

19 Proof. Let A = A 1 A 2 A 3 A 4 be a bicentric quadrilateral whose incircle is C 1 and circumcircle C 2 and let A i A i+1 = t i + t i+1, i = 1, 2, 3, 4. Then (2.29) t v 1t v 3 = t v 2t v 4 = (r v ) 2. According to what we said in connection with the relations (2.2) and (2.3) there is a bicentric quadrilateral A (v) = A (v) 1 A (v) 2 A (v) 3 A (v) 4 such that A (v) i A (v) i+1 = tv i + t v i+1, i = 1, 2, 3, 4. Let its incircle and circumcircle be denoted respectively by C (v) 1 and C (v) 2 and let r v = radius of C (v) 1, R v = radius of C (v) 2, d v = distance between the centers of C (v) 1 and C (v) 2. From (2.29) we see that (2.30) r v = r v. In order to obtain R v and d v we shall use relations (2.31) t v 1t v 2 + t v 2t v 3 + t v 3t v 4 + t v 4t v 1 = 2(R 2 v d 2 v), Page 19 of 43 (2.32) (R 2 v d 2 v) 2 = 2r 2 v(r 2 v + d 2 v),

20 where the second is Fuss relation. If, for brevity, the left-hand side of (2.31) is denoted by s, we can write s 2 = R2 v d 2 v, s 2 4 = 2r2 v(r 2 v + d 2 v) from which follows that (2.33) R v = Theorem 2.12 is thus proved. s2 + 4sr 2 v 4r v, d v = s2 4sr 2 v 4r v. Before we state some of its corollaries here are some examples. Example 2.1. If v = 0, then s = 4, r v = 1, R v = 2, d v = 0. Example 2.2. If v = 1, then s = 2(R2 d 2 ), r r 4 v = 1, R r v = R, d r 2 v = d. r 2 Corollary The following holds where 2(t 1,v + t 3,v ) t v i t m,v + t M,v + 2r v, t 2 m,v = (R v d v ) 2 r 2 v, t 2 M,v = (R v + d v ) 2 r 2 v, t 2 1,v = R 2 v (d v + r v ) 2, t 2 3,v = R 2 v (d v r v ) 2. Page 20 of 43 This corollary is analogous to Theorem 2.6. (See (2.17).)

21 Corollary The following holds 2r v t v 1 + t v 3 t m,v + t M,v 2r v t v 2 + t v 4 t m,v + t M,v 4r v t v 1 + t v 2 + t v 3 + t v 4 4r v R2 v + d 2 v. Rv 2 d 2 v The proof is analogous to the proof that (2.4) (2.6) hold. We can imagine that in Figure 2.2 instead of t i, t i+2, t m, t M, r there are t v i, t v i+2, t m,v, t M,v, r v. Corollary The following holds (2.34) A(t v 1, t v 2, t v 3, t v 4) H(t v 1, t v 2, t v 3, t v 4) = r 2 v. This corollary is analogous to Corollary Theorem Each of the following six sums is maximum (minimum) iff the sum 4 t i is maximum (minimum). a) t 2 i, b) In other words, (2.35) 2(ˆt v 1 + ˆt v 3) t 2 i, c) t 3 i, d) t 3 i, e) t 4 i, f) t 4 i. t v i t v m + 2r v + t v M, v = 2, 2, 3, 3, 4, 4, where t m, t M, ˆt 1, ˆt 3 are given by (2.20) and (2.21). Page 21 of 43

22 Proof. a) It holds (t 1 + t 2 + t 3 + t 4 ) 2 = t 2 i + 4(R 2 d 2 ). b) Since t 1 t 3 = t 2 t 4 = r 2, we can write c) From t 2 i = r4 (t t t t 2 4) r 8 = t2 1 + t t t 2 4 r 4. (t 1 + t 2 + t 3 + t 4 ) 3 = (t 1 + t 2 + t 3 + t 4 ) 2 (t 1 + t 2 + t 3 + t 4 ) or follows ( ) 3 t i = [t t t t (R 2 d 2 + r 2 )](t 1 + t 2 + t 3 + t 4 ) (t 1 + t 2 + t 3 + t 4 )[(t 1 + t 2 + t 3 + t 4 ) 2 6(R 2 d 2 ) 3r 2 ] = d) It holds t 3 i = t3 1 + t t t 3 4 r 6. t 3 i. Page 22 of 43

23 since (2.36) e) From we get ( ) 2 t 2 i = ( ) 2 t i t i+1 = ( t 4 i = f) It holds t 2 i ( ) t 4 i + 2 t 2 i t 2 i+1 + 2r 4, ( t 2 i t 2 i+1 + 2r 2 ) 2 + 4r 2 ( Theorem 2.16 is proved. t 2 i ) t 4 i = t4 1 + t t t 4 4 r 8. t 2 i ) + 4r 4, 8(R 2 d 2 ) 2 + 4r 4. In connection with b), d), f) in this theorem let us remark that 1 ( ) k ti =. t k i It is easy to see that this is equivalent to r 2 A(t k 1, t k 2, t k 3, t k 4) H(t k 1, t k 2, t k 3, t k 4) = r 2k. Page 23 of 43

24 Corollary Let f i (t 1 ), i = 1, 2, 3, 4, be the functions given by f 1 (t 1 ) = t 1, f 2 (t 1 ) = t 2, f 3 (t 1 ) = r2, f 4 (t 1 ) = r2 t 1 t 2 where t 2 is expressed in (2.22). Then each of the following two equations d d (2.37) f i (t 1 ) = 0, fi k (t 1 ) = 0, k = 2, 3, 4 dt 1 dt 1 has in the interval [t m, t M ] the same solutions t m, ˆt 1, r, ˆt 3, t M given by (2.20) and (2.21). Thus, the graph of the function F (t 1 ) = 4 f k i (t 1 ) is like the graph of the function J(t 1 ) shown in Figure 2.5. If f(t 1 ) and g(t 1 ) are polynomials which respectively correspond to the equations given by (2.37), then f(t 1 ) g(t 1 ). Remark 2. We conjecture that Corollary 2.17 is valid for every real number k 0. Corollary t2 i t 2 i+1 is minimum when 4 t i is maximum and vice versa. In other words, the following holds 4r 2 (R 2 r 2 + d 2 ) t 2 i t 2 i+1 4(R 2 r 2 d 2 ) 2, where t 2 mr 2 + r 2 t 2 M + t 2 Mr 2 + r 2 t 2 m = 4r 2 (R 2 r 2 + d 2 ), ˆt 2 1ˆt ˆt 2 2ˆt ˆt 2 3ˆt ˆt 2 4ˆt 2 1 = 4(R 2 r 2 d 2 ) 2. Page 24 of 43

25 Proof. From (2.36), since 4 t it i+1 = 2(R 2 d 2 ), it follows that ( ) 4(R 2 d 2 ) 2 4r 4 = t 2 i t 2 i+1 + 2r 2 t 2 i. In this connection let us remark that from 4r 2 (R 2 r 2 + d 2 ) 4(R 2 r 2 d 2 ) 2, using Fuss relation (1.1), we get the following inequality (2.38) R 2 2r 2 + 3d 2. (Cf. with (2.14). The equality holds only if d = 0.) Remark 3. W. J. Blundon and R. H. Eddy in [2] have proved that for semiperimeter s of bicentric polygons, the following inequalities hold s r + ( ) r 2 + 4R 2, s 2 8r r2 + 4R 2 r, and two other inequalities in s. (Both inequalities are based upon (2.16) stated in Remark 1.) Inequalities (2.38), using (2.15) stated in Remark 1, can also be proved. Page 25 of 43

26 3. Bicentric Hexagons Let now, in this section, C 1 and C 2 be given circles such that there is a bicentric hexagon whose incircle is C 1 and circumcircle C 2 and let r = radius of C 1, R = radius of C 2, d = distance between centers of C 1 and C 2, (3.1) t m = (R d) 2 r 2, t M = (R + d) 2 r 2. We shall use the following results given in [9, Theorem 1-2]. Let A = A 1 A 6 be any given bicentric hexagon whose incircle is C 1 and circumcircle C 2 and let (3.2) t i + t i+1 = A i A i+1, i = 1,..., 6. Then (3.3) t 1 t 3 + t 3 t 5 + t 5 t 1 = r 2, (3.4) t 2 t 4 + t 4 t 6 + t 6 t 2 = r 2 and (3.5) t 1 t 4 = t 2 t 5 = t 3 t 6 = h, Page 26 of 43

27 where (3.6) h = t m t M. If t 1 is given, then t 2,..., t 6 are given by (3.7) t 3 = a ( a 2 2 2) + b, t5 = b, t 3 (3.8) t 2 = h t 5, t 4 = h t 1, t 6 = h t 3 where (3.9) a = (r4 h 2 )t 1 r 2 t h 2, b = h2 (r 2 + t 2 1) r 2 t h 2. Thus, for every t 1 such that t m t 1 t M there is a bicentric hexagon whose tangent lengths are t 1, t 2,..., t 6, where t 2,..., t 6 are given by (3.7) and (3.8). Theorem 3.1. The following results hold (3.10) 2 h t i + t i+3 t m + t M, i = 1, 2, 3 (3.11) 6 h 6 t i 3(t m + t M ) Page 27 of 43

28 and (3.12) 6h 6 t 2 i 6(R 2 + d 2 r 2 + h). Proof. Analogous to the proof of Theorem 2.1. (Here vertices A i and A i+3 are opposite and instead of r 2 we have h.) Theorem 3.2. The following result holds (3.13) r 2 3h, where r 2 = 3h only if d = 0. Proof. Since from (3.5) we have (3.14) t 2 = h, t 4 = h, t 6 = h, t 5 t 1 t 3 the relation (3.4) can be written as ( r ) 2 (3.15) t 1 + t 3 + t 5 = t1 t 3 t 5. h Using this relation and relation (2.4) we can write ( r ) 2 t 3 + t 5 = t1 t 3 t 5 t 1, h t 1 (t 3 + t 5 ) + t 3 t 5 = r 2, Page 28 of 43

29 from which follows that (3.16) t 3 + t 5 = a, t 3 t 5 = b, where a and b are given by (3.9). Thus, we have the equation t 3 + b = a t 3 or t 2 3 at 3 + b = 0. Let the discriminant of the above square equation in t 3 be denoted by D. Then we can write (3.17) D = 4h 2 r 2 t [(r 4 h 2 ) 2 4h 4 4h 2 r 4 ]t 2 1 4h 4 r 2 0. Now, the discriminant of the corresponding quadratic equation in t 2 1 is given by D 1 = [(r 4 h 2 ) 2 4h 4 4h 2 r 4 ] 2 64h 6 r 4. Since D 1 0 must hold, we have the following inequality which can be written as (r 4 h 2 ) 2 4h 4 4h 2 r 4 8h 3 r 2 0, (r 2 3h)(r 2 + h) 0. Thus, r 2 3h 0. If d = 0, then r = R cos 30 = R 3 2 m = t M = R sin 30 = R and 2 r 2 = t m t M. Theorem 3.2 is proved. Page 29 of 43

30 In proving this theorem we also have proved that for every t 1 such that t m t 1 t M the inequality (3.17) holds. It may be of some interest to note that Theorem 3.2 can be readily proved using a connection between arithmetic and harmonic means. Namely, starting from A(t 1, t 3, t 5 ) H(t 1, t 3, t 5 ), we can write or, since t 1 t 3 + t 3 t 5 + t 5 t 1 = r 2, (3.18) r 2 9 (t 1 + t 3 + t 5 ) t1t 3 + t 3 t 5 + t 5 t 1 t 1 t 3 t 5 9 t 1 t 3 t 5 t 1 + t 3 + t 5. Also we have that the relation (3.15) can be written as (3.19) r 2 = h 2 t1 + t 3 + t 5 t 1 t 3 t 5. Now, using relations (3.18) and (3.19) we can write ( ) r 4 9r 2 t 1 t 3 t 5 = 9 h 2 t1 + t 3 + t 5 t 1 t 3 t 5 = 9h 2 t 1 + t 3 + t 5 t 1 t 3 t 5 t 1 + t 3 + t 5 or r 2 3h. Corollary 3.3. The following result holds t 1 t 3 + t 3 t 5 + t 5 t 1 3h, t 2 t 4 + t 4 t 6 + t 6 t 2 3h. Page 30 of 43

31 Proof. Follows from (3.3), (3.4) and (3.13). Corollary 3.4. It holds (3.20) h 3 t 1 t 3 t 5 t 1 + t 3 + t 5. Proof. Since ( r h )2 3, the relation follows from (3.19). h Corollary 3.5. The following holds t 2 t 4 t 6 (3.21) h 3. t 2 + t 4 + t 6 Proof. From (3.3), using (3.5), we get r 2 = h 2 t2 + t 4 + t 6 t 2 t 4 t 6. Corollary 3.6. The following result holds 6 (3.22) t i t i+1 6h. Proof. Starting from r 2 3h, we can write r 4 9h 2, r 4 3h 2 6h 2, r 4 3h 2 h 6h. In [9, Theorem 3] it is proved that 6 t it 1+1 = r4 3h 2 h. Page 31 of 43

32 The following theorem is analogous to Theorem 2.6 but with much more involved calculation. Before its statement we have the following preliminary work. In Figure 3.1a we have drawn an axially symmetric bicentric hexagon P 1 P 6. The marked tangent lengths t 2 and t 3 are given by (3.23) t 2 = t M + R + d and (3.24) t 3 = t m + R d. The proof is as follows. Since t 2 = t 6 and t 4 = t M, the relation (3.4) can be written as (t 2 ) 2 + 2t M t 2 r 2 = 0, from which follows (3.23). Since t 3 = t 5, t 1 = t m, the relation (3.3) can be written as (t 3 ) 2 + 2t m t 3 r 2 = 0, from which follows (3.24). In Figure 3.1b we have drawn an axially symmetric bicentric hexagon Q 1 Q 6. The marked tangent lengths ˆt 1, ˆt 2, ˆt 3 are given by Page 32 of 43 (3.25) ˆt 1 = R 2 (r + d) 2,

33 C 2 P 3 t 3 Q 3 t 3 P 4 t C 2 2 t 3 C 1 t P 2 t 2 M C 1 r t m Q 1 d P t 1 O I 1 O I Q 6 Q 2 P 6 Q 5 P 5 Figure 3.1a Figure 3.1b Q 4 (3.26) ˆt 3 = R 2 (r d) 2, (3.27) ˆt 2 = r2 ˆt 1ˆt 3 ˆt 1 + ˆt 3. For ˆt 1 and ˆt 3 it is obvious from Figure 3.1b. Also, from Figure 3.1b we see that ˆt 2 = t 5, so relation (3.3) can be written as from which follows t 1 t 3 + t 3 t 5 + t 5 t 1 = r 2, t 5 = r2 ˆt 1ˆt 3 ˆt 1 + ˆt 3. Page 33 of 43

34 Now, let A 1 A 6 be a bicentric hexagon whose incircle is C 1 and circumcircle C 2 and let its area be denoted by J(t 1 ), that is (3.28) J(t 1 ) = r(t 1 + t t 6 ), where t i + t i+1 = A i A i+1, i = 1,..., 6, and t 2,..., t 6 are given by (3.7) and (3.8). According to Theorem 2 in [9], we have (3.29) J(t m ) = J(t 2 ) = J(t 3 ) = J(t M ) and (3.30) J(ˆt 1 ) = J(ˆt 2 ) = J(ˆt 3 ). Theorem 3.7. Let the perimeter of the hexagon P 1 P 6 shown in Figure 3.1a be denoted by 2p M, and let the perimeter of the hexagon Q 1... Q 6 shown in Figure 3.1b be denoted by 2p m. Then (3.31) rp m J(t 1 ) rp M, that is J(t 1 ) = maximum if t 1 { t m, t 2, t 3, t M }, J(t 1 ) = minimum if t 1 {ˆt 1, ˆt 2, ˆt 3 }. Proof. The relation (3.28), using relations given by (3.7) and (3.8), can be written as J(t 1 ) = r2 (h 4 + 2h 2 r 2 t r 4 t ht 2 1(r 2 + t 2 1) 2 ). ht 1 (r 2 + t 2 1)(h 2 + r 2 t 2 1) Page 34 of 43

35 From d dt 1 J(t 1 ) = 0, we obtain the equation r 2 (h t 2 1)[h(h 2 + ht t 4 1) + 2hr 2 t 2 1 r 4 t 2 1][3h 2 t (h 2 + t 4 1)r 2 r 4 t 2 1] ht 2 1(r 2 + t 2 1) 2 (h 2 + r 2 t 2 1) 2 = 0, from which follow (3.32) (t 2 1) 1 = h, (3.33) (t 2 1) 2,3 = (r2 h) 2 ± (r 2 h) 4 4h 4, 2h and (3.34) (t 2 1) 4,5 = r4 3h 2 ± (r 4 3h 2 ) 2 4r 4 h 2 2r 2. Since R, r, d satisfy Fuss relation (1.2), it can be shown, using this relation, that (t 1 ) 1 = ˆt 2, (t 1 ) 2 = ˆt 3, (t 1 ) 3 = ˆt 1 (t 1 ) 4 = t 3, (t 1 ) 5 = t 2. So, for example, it can be found that t 3 = (t 1 ) 4 is equivalent to Page 35 of 43 64r 4 (R d+r)( R+d+r)[3(R 2 d 2 ) 4 4r 2 (R 2 +d 2 )(R 2 d 2 ) 2 16R 2 r 4 d 2 ] = 0,

36 where only the last factor is equal to zero since Fuss relation (1.2) holds, which can be written as 3(R 2 d 2 ) 4 4r 2 (R 2 + d 2 )(R 2 d 2 ) 2 16R 2 r 4 d 2 = 0. In the same way as in Theorem 2.6, only with somewhat more involved calculation, it can be shown that the graph of J(t 1 ) is like that shown in Figure 3.2. J(t ) m J(t ) 1 J(t ) 1 t O t 1 m t 1 t 2 t 2 t t3 3 t M Figure 3.2 Theorem 3.7 is proved. Corollary 3.8. Each of the following three sums a) 6 t 2 i, b) 6 1 t i, c) 6 1 t 2 i Page 36 of 43 has maximum if t 1 = t m and minimum if t 1 = ˆt 1.

37 Proof. a) We have (t t 6 ) 2 = t t (t 1 t 2 + t 2 t t 5 t 6 + t 6 t 1 ) b) We have = + 2(t 1 t 3 + t 3 t 5 + t 5 t 1 ) + 2(t 2 t 4 + t 4 t 6 + t 6 t 2 ) 6 t 2 i + 2 r4 3h 2 + 4r 2. (See [9, Theorem 3].) h 6 1 = h2 (t t 6 ) = t t 6. t i h 3 h Corollary 3.9. Let t 1 be given such that t m t 1 t M. Then the equation J(t 1 )J(x) = J(t m )J(ˆt 1 ) has six positive roots x 1,..., x 6 and we obtain 6 x i x i+1 = r4 3h 2, h 6 x i x i+1 x i+2 x i+3 = r 4 3h 2, x 1 x 2 x 3 x 4 x 5 x 6 = h 3. The proof is analogous to the proof of Corollary Page 37 of 43

38 4. Bicentric Octagons In this section, let C 1 and C 2 be given circles such that there is a bicentric octagon whose incircle is C 1 and circumcircle C 2 and let r = radius of C 1, R = radius of C 2, d = distance between centers of C 1 and C 2, (4.1) t m = (R d) 2 r 2, t M = (R + d) 2 r 2. We shall use some results given in [9, Theorem 4-5]. Let A = A 1 A 8 be any given bicentric octagon whose incircle is C 1 and circumcircle C 2 and let (4.2) t i + t i+1 = A i A i+1, i = 1,..., 8. Then (4.3) r 4 r 2 (t 1 t 3 + t 3 t 5 + t 5 t 7 + t 7 t 1 + t 1 t 5 + t 3 t 7 ) + t 1 t 3 t 5 t 7 = 0, (4.4) r 4 r 2 (t 2 t 4 + t 4 t 6 + t 6 t 8 + t 8 t 2 + t 2 t 6 + t 4 t 8 ) + t 2 t 4 t 6 t 8 = 0, Page 38 of 43 (4.5) t 1 t 5 = t 2 t 6 = t 3 t 7 = t 4 t 8 = h,

39 (4.6) h = t m t M. A bicentric octagon may be convex or non-convex, but the relations (4.3) (4.6) have the same form. The theorem below will now be proved. Theorem 4.1. Let A 1 A 8 be a bicentric octagon. Then ( (4.7) r h ) 2 4h, r where equality holds only if d = 0. Proof. The relation (4.3), using relations t 1 t 5 = t 3 t 7 = h, can be written as ( (4.8) (h + t 2 1)t 2 3 r h ) 2 t 1 t 3 + h(h + t 2 r 1) = 0. The discriminant of the above quadratic equation in t 3 is ( D = r h ) 4 t 2 1 4h(h + t 2 r 1) 2. Since D 0, we obtain 4ht [ ( r h ) ] 4 8h 2 t 2 1 4h 3 0. r Page 39 of 43

40 We shall use the discriminant [ ( D 1 = r h ) ] 4 2 8h 2 64h 4 r of the corresponding quadratic equation in t 2 1 given by [ ( 4ht r h ) ] 4 8h 2 t 2 1 4h 3 = 0. r From D 1 0 it follows that ( r h r ) 4 8h 2 8h 2, i.e. ( r h r ) 2 4h. It remains to prove that (r h r )2 = 4h only if d = 0. Since we have r = R cos 22.5, h = r 2 sin and cos = 2 + 2, sin = ( r h r ) 2 = r 2 (2 2), 4h = r 2 (2 2). Theorem 4.1 is proved. Page 40 of 43

41 Corollary 4.2. The following inequalities hold (4.9) t 1 t 3 + t 3 t 5 + t 5 t 7 + t 7 t 1 4t m t M and (4.10) t 2 t 4 + t 4 t 6 + t 6 t 8 + t 8 t 2 4t m t M. Proof. The relations (4.3) and (4.4), using relations (4.5), can be written as t 1 t 3 + t 3 t 5 + t 5 t 7 + t 7 t 1 = t 2 t 4 + t 4 t 6 + t 6 t 8 + t 8 t 2 = ( r h r ) 2, ( r h r ) 2. Remark 4. It can be shown that for almost every property considered for bicentric quadrilaterals there are analogous properties for bicentric hexagons and octagons. But, since the number of the pages in the paper is limited, we omit some analogous theorems for bicentric hexagons and octagons. So that all we have stated about bicentric hexagons and octagons can be considered as some steps or an insight into possibilities for further research. Page 41 of 43

42 References [1] H.J.M. BOS, C. KERS, F. OORT AND D.W. RAVEN, Poncelet s closure theorem, Exposition. Math., 5 (1987), 289, 364. [2] W.J. BLUNDON AND R.H. EDDY, Problem 488, Nieuw Arch. Wiskunde, 26 (1978), 2321; solution in 26 (1978), 465. [3] H. DÖRRIE, 100 Great Problems of Elementary Mathematics, Dover, [4] N. FUSS, De quadrilateris quibus circulum tam inscribere quam circumscribere licet, Nova Acta Acad. Sci. Petrop., 10 (1797), [5] N. FUSS, De polygonis symmetrice irregularibus circulo simul inscriptis et circumscriptis, Nova Acta Acad. Sci. Imp. Petrop., 13 (1802), [6] P. GRIFFITHS AND J. HARRIS, On Cayley s explicit solution to Poncelet s porism, Enseign. Math., 24(1-2) (1978), [7] J.V. PONCELET, Traité des propriétés projectives des figures, t.i, Paris 1865, (first ed. in 1822). [8] M. RADIĆ, Some relations concerning triangles and bicentric quadrilaterals in connection with Poncelet s closure theorem, Math. Maced., 1 (2003), [9] M. RADIĆ, Some properties and relations concerning bicentric hexagons and octagons in connection with Poncelet s closure theorem, to appear. Page 42 of 43

43 [10] M. RADIĆ, Extreme Areas of Triangles in Poncelet s Closure Theorem, Forum Geometricorum, 4 (2004), [11] L. FEJES TOTH, Inequalities concerning polygons and polyhedra, Duke Math. J., 15(1) (1948), Page 43 of 43