Journal of Inequalities in Pure and Applied Mathematics http://jipam.vu.edu.au/ Volume 6, Issue 1, Article 1, 2005 CERTAIN INEQUALITIES CONCERNING BICENTRIC QUADRILATERALS, HEXAGONS AND OCTAGONS MIRKO RADIC´ UNIVERSITY OF RIJEKA FACULTY OF PHILOSOPHY DEPARTMENT OF MATHEMATICS 51000 RIJEKA,OMLADINSKA 14, CROATIA [email protected] Received 15 April, 2004; accepted 24 November, 2004 Communicated by J. Sándor ABSTRACT. In this paper we restrict ourselves to the case when conics are circles one com- pletely inside of the other. Certain inequalities concerning bicentric quadrilaterals, hexagons and octagons in connection with Poncelet’s closure theorem are established. Key words and phrases: Bicentric Polygon, Inequality. 2000 Mathematics Subject Classification. 51E12. 1. INTRODUCTION A polygon which is both chordal and tangential is briefly called a bicentric polygon. The following notation will be used. If A1 ··· An is considered to be a bicentric n-gon, then its incircle is denoted by C1, circum- circle by C2, radius of C1 by r, radius of C2 by R, center of C1 by I, center of C2 by O, distance between I and O by d. The first person who was concerned with bicentric polygons was the German mathematician Nicolaus Fuss (1755-1826). He found that C1 is the incircle and C2 the circumcircle of a bicentric quadrilateral A1A2A3A4 iff (1.1) (R2 − d2)2 = 2r2(R2 + d2), (see [4]). The problem of finding this relation has been mentioned in [3] as one of 100 great problems of elementary mathematics. 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) 3p4q4 − 2p2q2r2(p2 + q2) = r4(p2 − q2)2 ISSN (electronic): 1443-5756 c 2005 Victoria University. All rights reserved. 118-04 2 MIRKO RADIC´ A1 C2 C2 C1 C1 A t r 4 M t d m O I O I R A3 A2 Figure 1.1 Figure 1.2 and (1.3) [r2(p2 + q2) − p2q2]4 = 16p4q4r4(p2 − r2)(q2 − r2), where p = R + d, q = R − d. The very remarkable theorem concerning bicentric polygons was given by the French math- ematician Poncelet (1788-1867). 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 C1 and circumcircle C2, then there are infin- itely many bicentric n-gons whose incircle is C1 and circumcircle is C2. For every point P1 on C2 there are points P2,...,Pn on C2 such that P1 ··· Pn are bicentric n-gons whose incircle is C1 and circumcircle is C2. 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 C2 to C1. As can be seen from Figure 1.2, the following holds p 2 2 p 2 2 (1.4) tm = (R − d) − r , tM = (R + d) − r . 2. CERTAIN INEQUALITIES CONCERNING BICENTRIC QUADRILATERALS Let A1A2A3A4 be any given bicentric quadrilateral whose incircle is C1 and circumcircle C2 and let (2.1) ti + ti+1 = |AiAi+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 (2.2) t1t3 = t2t4 = r , and 2 2 (2.3) t1t2 + t2t3 + t3t4 + t4t1 = 2(R − d ). J. Inequal. Pure and Appl. Math., 6(1) Art. 1, 2005 http://jipam.vu.edu.au/ BICENTRIC QUADRILATERALS,HEXAGONS AND OCTAGONS 3 Reversely, if t1, t2, t3, t4 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 t1, t2, t3, t4 given by (2.1) satisfy the following inequalities (2.4) 2r ≤ t1 + t3 ≤ tm + tM , (2.5) 2r ≤ t2 + t4 ≤ tm + tM , R2 + d2 (2.6) 4r ≤ t + t + t + t ≤ 4r · , 1 2 3 4 R2 − d2 2 2 2 2 2 2 2 2 (2.7) 4r ≤ t1 + t2 + t3 + t4 ≤ 4(R + d − r ), and 2k 2k 2k 2k 2k (2.8) t1 + t2 + t3 + t4 ≥ 4r , k ∈ N. The equalities hold only if d = 0. 2 Proof. First let us remark that tmtM = r since there is a bicentric quadrilateral as shown in Figure 2.1. Now, let C denote a circle whose diameter is tm + tM (Figure 2.2). Then for each ti, i = 1, 2, 3, 4, since tm ≤ ti ≤ tM , there are points Q and R on C such that (2.9) ti = |PQ|, ti+2 = |PR|, where |PQ| + |PR| = |QR|. In this connection let us remark that the power of the circle C at P is tmtM . Therefore |PQ||PR| = tmtM . Q C2 C t t i r M r t C m 1 t M tm O I P ti+2 r R Figure 2.1 Figure 2.2 Obviously ti + ti+2 ≤ tm + tM since tm + tM is a diameter of C. Also it is clear that 2 ti + ti+2 ≥ 2r since r = tmtM . This proves (2.4) and (2.5). In the proof that (2.6) holds we shall use the relations R − d R + d (2.10) t = r · , t = r · . m R + d M R − d J. Inequal. Pure and Appl. Math., 6(1) Art. 1, 2005 http://jipam.vu.edu.au/ 4 MIRKO RADIC´ 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 R − d2 (R − d)2 − r2 = r2 R + d (R2 − d2)2 − r2(R + d)2 = r2(R − d)2 (R2 − d2)2 = 2r2(R2 + d2). The proof that (2.6) holds can be written as 2r + 2r ≤ t1 + t3 + t2 + t4, R−d R+d R2+d2 t1 + t3 + t2 + t4 ≤ 2(tm + tM ) = 2r R+d + R−d = 4r · R2−d2 . The proof that (2.7) holds is as follows. Since 2r ≤ t1 + t3, 2r ≤ t2 + t4, we have 2 2 2 2 2 2 4r ≤ t1 + t3 + 2t1t3, 4r ≤ t2 + t4 + 2t2t4 2 or, since 2t1t3 = 2t2t4 = 2r , 2 2 2 2 2 2 2r ≤ t1 + t3, 2r ≤ t2 + t4. 2 2 2 2 2 Thus, 4r ≤ t1 + t2 + t3 + t4. From t1 + t3 ≤ tm + tM , t2 + t4 ≤ tm + tM it follows that 2 2 2 2 2 2 2 2 t1 + t3 ≤ tm + tM , t2 + t4 ≤ tm + tM since 2t1t3 = 2t2t4 = 2tmtM . Thus, we obtain 2 2 2 2 2 2 t1 + t2 + t3 + t4 ≤ 2(tm + tM ), 2 2 2 2 2 2 2 2 2 where tm + tM = (R − d) − r + (R + d) − r = 2(R + d − r ). In the same way it can be proved that (2.8) holds. So, starting from 2r ≤ t1 + t3, since k k 2k 2t1t3 = 2r , we can write 2 2 2 2r ≤ t1 + t3, 4 4 4 2 2 4 4 4 4r ≤ t1 + t3 + 2t1t3 or 2r ≤ t1 + t3 and so on. Starting from t1 + t3 ≤ tm + tM it can be written 2 2 2 2 t1 + t3 ≤ tm + tM , 4 4 4 4 t1 + t3 ≤ tm + tM , and so on. Since tm = tM only if d = 0, it is clear that the relations (2.4) – (2.8) become equalities only if d = 0. Thus, if d 6= 0, then in the above relations instead of ≤ we may put <. Theorem 2.1 is thus proved. Corollary 2.2. The following holds 4 4 X 1 4 R2 + d2 (2.11) ≤ ≤ · . r t r R2 − d2 i=1 i 2 1 1 Proof. From 2r ≤ t1 + t3 it follows that ≤ + , since r t1 t3 1 1 t1 + t3 t1 + t3 2r 2 + = = 2 ≥ 2 = . t1 t3 t1t3 r r r J. Inequal. Pure and Appl. Math., 6(1) Art. 1, 2005 http://jipam.vu.edu.au/ BICENTRIC QUADRILATERALS,HEXAGONS AND OCTAGONS 5 1 1 1 1 From t1 + t3 ≤ tm + tM it follows that + ≤ + , since t1 t3 tm tM 1 1 t1 + t3 1 1 tm + tM + = 2 , + = 2 . t1 t3 r tm tM r Corollary 2.3. Let a = t1 + t2, b = t2 + t3, c = t3 + t4, d = t4 + t1.Then R2 − d2 (2.12) 8r ≤ a + b + c + d ≤ 8r · . R2 + d2 Corollary 2.4. Let a, b, c, d be as in Corollary 2.3. Then (2.13) 4(R2 − d2 + 2r2) ≤ a2 + b2 + c2 + d2 ≤ 4(3R2 − 2r2). Proof. Using relation (2.3) we can write 2 2 2 2 2 2 2 2 2 2 a + b + c + d = 2(t1 + t2 + t3 + t4) + 4(R − d ). Now, using relations (2.7) we can write relations (2.13). Corollary 2.5. The following holds (2.14) 2r2 + d2 ≤ R2 ≤ 2r2 + d2 + 2rd. Proof. Since t1 + t3 ≥ 2r, t2 + t4 ≥ 2r, we can write 2 (t1 + t3)(t2 + t4) ≥ 4r , 2 t1t2 + t2t3 + t3t4 + t4t1 ≥ 4r , 2(R2 − d2) ≥ 4r2, R2 − d2 ≥ 2r2. The fact that R2 ≤ 2r2 + d2 + 2rd is clear from the quadratic function f(d) = d2 + 2rd + R2 − 2r2. If d = 0, then f(d) = 0, but if d > 0, then f(d) > 0. Remark 2.6. It may be interesting that relations (2.14) can be obtained directly from Fuss’ relation (1.1).