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A Condition That a Tangential Quadrilateral Is Also Achordalone

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A Condition That a Tangential Quadrilateral Is Also Achordalone View metadata, citation and similar papers at core.ac.uk brought to you by CORE Mathematical Communications 12(2007), 33-52 33 A condition that a tangential quadrilateral is also achordalone Mirko Radic´,∗ Zoran Kaliman† and Vladimir Kadum‡ Abstract. In this article we present a condition that a tangential quadrilateral is also a chordal one. The main result is given by Theo- rem 1 and Theorem 2. Key words: tangential quadrilateral, bicentric quadrilateral AMS subject classifications: 51E12 Received September 1, 2005 Accepted March 9, 2007 1. Introduction A polygon which is both tangential and chordal will be called a bicentric polygon. The following notation will be used. If A1A2A3A4 is a considered bicentric quadrilateral, then its incircle is denoted by C1, circumcircle by C2,radiusofC1 by r,radiusofC2 by R, center of C1 by I, center of C2 by O, distance between I and O by d. A2 C2 C1 r A d 1 O I R A3 A4 Figure 1.1 ∗Faculty of Philosophy, University of Rijeka, Omladinska 14, HR-51 000 Rijeka, Croatia, e-mail: [email protected] †Faculty of Philosophy, University of Rijeka, Omladinska 14, HR-51 000 Rijeka, Croatia, e-mail: [email protected] ‡University “Juraj Dobrila” of Pula, Preradovi´ceva 1, HR-52 100 Pula, Croatia, e-mail: [email protected] 34 M. Radic,´ Z. Kaliman and V. Kadum The first one who was concerned with bicentric quadrilaterals was a German mathematicianNicolaus Fuss (1755-1826), see [2]. He foundthat C1 is the incircle and C2 the circumcircle of a bicentric quadrilateral A1A2A3A4 iff (R2 − d2)2 =2r2(R2 + d2). (1.1) The problem of findings relation (1.1) has ranged in [1] as one of 100 great problems of elementary mathematics. A very remarkable theorem concerning bicentric polygons is given by a French mathematician Poncelet (1788-1867). This theorem is known as the Poncelet’s closure theorem. For the case when conics are circles, one inside the other, this theorem canbe stated as follows: If there is a bicentric n-gonwhose incircle is C1 and circumcircle C2,thenthere are infinitely many bicentric n-gons whose incircle is C1 and circumcircle C2.For every point P1 on C2 there are points P2,...,Pn on C2 such that P1 ...Pn are a bicentric n-gonwhose incircle is C1 and circumcircle C2. Inthe following(Section3) bicentric quadrilaterals will also be considered, where instead of an incircle there is an excircle. As will be seen, there is a great analogy betweenthose two kindsof bicentric quadrilaterals. 2. About one condition concerning bicentric quadrilaterals First, let us briefly discuss the notations to be used. If A1A2A3A4 is a giventangential quadrilateral, thenby t1, t2, t3, t4 we denote its tangent lengths such that ti + ti+1 = |AiAi+1|,i=1, 2, 3, 4. (2.1) By β1, β2, β3, β4 we denote angles IAiAi+1,i=1, 2, 3, 4, where I is the center of the incircle of A1A2A3A4.(SeeFigure 2.1) A1 t1 t1 t2 r t4 A2 I A4 t 2 t4 t3 t3 A3 Figure 2.1. The following theorem will be proved. A condition that ... 35 Theorem 1. Let A1A2A3A4 be any given tangential quadrilateral, and let t1, t2, t3, t4 be its tangent lengths such that (2.1) holds. Then this quadrilateral is also a chordal one if and only if √ |A1A3| |A2A4| = = k, (2.2) t1 + t3 t2 + t4 where 1 <k≤ 2. (2.3) Proof. First we suppose that (2.2) holds. From Figure 2.1 we see that the 2 2 equality |A1A3| = k(t1 + t3) canbe writtenas 2 2 2 |A1A2| + |A2A3| − 2|A1A2||A2A3| cos 2β2 = k(t1 + t3) or t2 − r2 t t 2 t t 2 − t t t t 2 k t t 2, ( 1 + 2) +( 2 + 3) 2( 1 + 2)( 2 + 3) 2 2 = ( 1 + 3) (2.4) t2 + r since 2 1 − tan β2 r cos 2β2 = 2 , tan β2 = . 1+tan β2 t2 2 2 The equality |A1A3| = k(t1 + t3) canalso be writtenas t2 − r2 t t 2 t t 2 − t t t t 4 k t t 2, ( 1 + 4) +( 4 + 3) 2( 1 + 4)( 4 + 3) 2 2 = ( 1 + 3) (2.5) t4 + r where 2 2 2 2 2β4 =measureofA1A4A3, cos 2β4 =(t4 − r )/(t4 + r ). 2 2 Inthe same way cansee that the equality |A2A4| = k(t2 + t4) canbe written inthe followingtwo ways: t2 − r2 t t 2 t t 2 − t t t t 1 k t t 2, ( 1 + 2) +( 1 + 4) 2( 1 + 2)( 1 + 4) 2 2 = ( 2 + 4) (2.6) t1 + r t2 − r2 t t 2 t t 2 − t t t t 3 k t t 2. ( 3 + 2) +( 3 + 4) 2( 3 + 2)( 3 + 4) 2 2 = ( 2 + 4) (2.7) t3 + r Solving equation (2.4) for t2 we get 2 2 (t2)1 = − 4r t1 − 4r t3 − 2 2 2 2 2 2 2 2 (4r t1 +4r t3) − 4(4r + t1 − kt1 − 2t1t3 − 2kt1t3 + t3 − kt3) 1 2 2 2 2 2 2 2 2 2 2 2 (r t1 − kr t1 +2r t1t3 − 2kr t1t3 + r t3 − kr t3) / 2 2 2 2 2 2(4r + t1 − kt1 − 2t1t3 − 2kt1t3 + t3 − kt3) , (2.8) 36 M. Radic,´ Z. Kaliman and V. Kadum 2 2 (t2)2 = − 4r t1 − 4r t3 + 2 2 2 2 2 2 2 2 (4r t1 +4r t3) − 4(4r + t1 − kt1 − 2t1t3 − 2kt1t3 + t3 − kt3) 1 2 2 2 2 2 2 2 2 2 2 2 (r t1 − kr t1 +2r t1t3 − 2kr t1t3 + r t3 − kr t3) / 2 2 2 2 2 2(4r + t1 − kt1 − 2t1t3 − 2kt1t3 + t3 − kt3) . (2.9) It is easy to see that equation(2.4) in t2 has the same solutions as equation (2.5) in t4,thatis {(t2)1, (t2)2} = {(t4)1, (t4)2}. Since equation (2.4) has t2 as one solution, and equation (2.5) has t4 as one solution, it follows that {(t2)1, (t2)2} = {(t4)1, (t4)2} = {t2,t4}. (2.10) Putting t2 =(t2)1, t4 =(t2)2 in(2.6) we get − k r2 t t 2 ( 1+ ) ( 1 + 3) t t . 2 2 2 = 1 3 (2.11) −4r +(−1+k)t1 +2(1+k)t1t3 +(−1+k)t3 Solving this equation for t3 yields √ √ 2 r −t1 − 2 kt1 − kt1 −t1 +2 kt1 − kt1 t3 ∈ , , t1 −1+k −1+k Thus, the only positive t3 is givenby r2 t3 = . (2.12) t1 Now we find that from (2.8) and (2.9) there follows r2 t t 2 − k t · t ( 1 + 3) (1 ) , ( 2)1 ( 2)2 = 2 2 2 (t1 − t3) +4r − k(t1 + t3) which according to (2.10) and (2.12) can be written as 2 t2t4 = r . (2.13) 2 That also t1t3 = r ,thatis 2 t1t3 = t2t4 = r , (2.14) 2 2 follows from (t1 +t2 +t3 +t4)r = t1t2t3 +t2t3t4 +t3t4t1 +t4t1t2 putting t4 = r /t2. 2 2 Namely, we get r (t2 + t4)=t1t3(t2 + t4), from which follows t1t3 = r . We shall prove that these relations are sufficient for a tangential quadrilateral to be a chordal one. The proof is as follows. Since t2 − r2 t2 − r2 β 2 , β 4 cos 2 2 = 2 2 cos 2 4 = 2 2 t2 + r t4 + r A condition that ... 37 using (2.14) we can write r2/t 2 − r2 t2 − r2 β ( 2) − 2 − β . cos 2 4 = 2 2 2 = 2 2 = cos 2 2 (r /t2) + r t2 + r Inthe same way we findthat cos 2 β3 = − cos 2β1. Thus, from (2.2) it follows that the giventangential quadrilateral A1A2A3A4 is also a chordal one since 2β1 +2β3 =2β2 +2β4 = π. In this connection let us remark that it is not difficult to check that identically holds r(t1 + t2 + t3 + t4)= (t1 + t2)(t2 + t3)(t3 + t4)(t4 + t1) 2 for every positive numbers r, t1, t2, t3, t4 such that t1t3 = t2t4 = r . Now we prove that relations (2.14) are necessarily for a tangential quadrilateral to be a chordal one. The proof is easy; namely, it is easy to see that cos 2β2 = − cos 2β4 or t2 − r2 t2 − r2 2 − 4 2 2 = 2 2 t2 + r t4 + r 2 is valid only if t2t4 = r . 2 Inthe same way it canbe seenthat cos 2 β1 = − cos 2β3 only if t1t3 = r . Here let us remark that the following holds. If A1A2A3A4 and B1B2B3B4 are two bicentric quadrilaterals which have the same incircle and ti + ti+1 = |AiAi+1| ,i=1, 2, 3, 4 ui + ui+1 = |BiBi+1| ,i=1, 2, 3, 4 2 t1t3 = t2t4 = r , 2 u1u3 = u2u4 = r , then these quadrilateral need not have the same circumcirle. It will be only if 2 2 t1t2 + t2t3 + t3t4 + t4t1 = u1u2 + u2u3 + u3u4 + u4u1 =2(R − d ). (See Theorem 3.2 in[3].) In this connection may be interesting how radius R can be obtained and some other relations. In short about this. Let C1 and C2 denote the incircle and the circumcircle of the considered bicentric quadrilateral A1A2A3A4, and let the other notation be as stated in the introduction. The radius of C2 can be obtained using well-known relations which hold for a bicentric quadrilateral: 2 (a1a2 + a3a4)(a1a3 + a2a4)(a1a4 + a2a3) 2 R = ,J= a1a2a3a4 16J 2 where a1 = t1 + t2, a2 = t2 + t3, a3 = t3 + t4, a4 = t4 + t1, J =areaofA1A2A3A4. It canbe foundthat 2 2 2 2 2 a1a2a3 a2a3a4 a3a4a1 a4a1a2 16R = a1 + a2 + a3 + a4 + + + + (2.15) a4 a1 a2 a3 38 M.
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