Mathematical Communications 7(2002), 21-34 21 Some relations concerning k-chordal and k-tangential polygons Mirko Radic´∗ Abstract. In papers [6] and [7] the k-chordal and the k-tangential polygons are defined and some of their properties are proved. In this paper we shall consider some of their other properties. Theorems 1-4 are proved. Key words: geometrical inequalities, k-chordal polygon, k-tangential polygon AMS subject classifications: 51E12, 52A40 Received October 8, 2001 Accepted April 4, 2002 1. Preliminaries A polygon with vertices A1 ...An (in this order) will be denoted by A1 ...An and the lengths ofthe sides of A1 ...An will be denoted by a1,... ,an,whereai = | AiAi+1 |, i =1, 2,... ,n. For the interior angle at the vertex Ai we write αi or ∠Ai, i.e. ∠Ai = ∠An−1+iAiAi+1,i=1,... ,n. Ofcourse, indices are calculated modulo n. For convenience we list some definitions given in [6] and [7]. Definition 1. Let A = A1 ...An be a chordal polygon and let C be its circum- circle. By SAi and SAi we denote the semicircles of C such that SAi ∪ SAi = C, Ai ∈ SAi ∩ SAi . The polygon A is said to be of the first kind if the following is fulfilled: 1. all vertices A1 ...An do not lie on the same semicircle, 2. for every three consecutive vertices Ai,Ai+1,Ai+2 it holds Ai ∈ SAi+1 ⇒ Ai+2 ∈ SAi+1 3. any two consecutive vertices Ai,Ai+1 do not lie on the same diameter. Definition 2. Let A = A1 ...An be a chordal polygon and let k be a positive integer. The polygon A is said to be a k-chordal polygon if it is of the first kind and if there holds n ∠AiCAi+1 =2kπ, (1) i=1 ∗Department of Mathematics, University of Rijeka, Omladinska 14, HR-51 000 Rijeka, Croatia 22 M. Radic´ where C is the centre of the circumcircle of the polygon A. Using (1) it is easy to see that the angles ofa k-chordal polygon A1 ...An satisfy the relation: n ∠Ai =(n − 2k)π. (2) i=1 Definition 3. Let A = A1 ...An be a tangential polygon and let k be a positive n−1 n−1 n−2 integer so that k ≤ 2 , that is, k ≤ 2 if n is odd, and k ≤ 2 if n is even. The polygon A will be called a k-tangential polygon if any two of its consecutive sides have only one common point, and if there holds π β1 + ···+ βn =(n − 2k) , (3) 2 where 2βi = ∠Ai,i=1, ..., n. Consequently, a tangential polygon A is k-tangential if ϕ1 + ···+ ϕn =2kπ, (4) where ϕi = ∠AiCAi+1 and C is the centre ofthe circle inscribed into the polygon A. n−1 n−2 The integer k in relations (1)-(4) can be at most 2 if n is odd and 2 if n is even. Remark 1. In the following considerations we shall denote the angles β1,... ,βn such that βi = ∠CAiAi+1, if it is a question of a chordal polygon, 1 βi = ∠Ai, if it is a question of a tangential polygon. 2 2. Some inequalities concerning the radius of k-chordal and k-tangential polygons At first we prove some results concerning a k-chordal polygon. Theorem 1. Let a1,... ,an be the lengths of the sides of a k-chordal polygon A = A1 ...An and let a1 = min{a1 ...an}. If there exist angles γ1,... ,γn such that π π γ1 + ···+ γn =(n − 2k) , 0 <γi < , 1=1,... ,n, (5) 2 2 a1 sin γ1 = a2 sin γ2 = ...= an sin γn. (6) k-chordal and k-tangential polygons 23 Then n 2 ai i=1 2r>n n , (7) 1 1 2 2 π ai − a1 sin (n − 2k) i=1 2 i=1 ai 2n where r is the radius of the circumcircle of the polygon A. Proof. Since βi = ∠CAiAi+1,i=1,... ,n, we have the following relations π π β1 + ···+ βn =(n − 2k) , 0 <βi < ,i=1,... ,n (8) 2 2 2r cos βi = ai,i=1,... ,n (9) from which it follows 2 2rai cos βi = ai ,i=1,... ,n n 2 ai i=1 2r = n . (10) ai cos βi i=1 In addition to the angles β1,... ,βn there are infinitely many angles γ1,... ,γn such that π π γ1 + ···+ γn =(n − 2k) , 0 <γi < ,i=1,... ,n. 2 2 n We shall prove that ai cos γi = maximum ifthe angles γ1,... ,γn satisfy i=1 a1 sin γ1 = a2 sin γ2 = ...= an sin γn. First we shall prove the following lemma. Lemma 1. If a1 and a2 are positive numbers and π γ1 + γ2 = s, 0 <s<π, 0 <γi < ,i=1, 2 2 then the function f(γ1,γ2)=a1 cos γ1 + a2 cos γ2 assumes maximum if a1 sin γ1 = a2 sin γ2. Proof. Let g(γ1)=a1 cos γ1 + a2 cos(s − γ1), then g (γ1)=−a1 sin γ1 + a2 sin(s − γ1), g (γ1)=−a1 cos γ1 − a2 cos(s − γ1) < 0, 24 M. Radic´ a1 sin γ1 + a2 sin(s − γ1)=0 ⇒ a1 sin γ1 = a2 sin(s − γ2). ✷ n From the above lemma it is clear that the sum ai cos γi assumes maximum i=1 iffor each sum ai cos γi + aj cos γj,i,j∈{1,... ,n} there holds ai sin γi = aj sin γj, since we can put γi + γj = s. Now, we are going to prove that the inequality (7) is valid if(6) is fulfiled. Based on the assumption that equations (6) exist, we can write ai sin γi = λ, i =1,... ,n from which it follows 2 2 λ 1 λ cos γi = 1 − < 1 − ,i=1,... ,n, ai 2 ai n 2 n n 1 λ ai 1 − > ai cos γi ≥ ai cos βi i=1 2 ai i=1 i=1 so that instead of(10) we can write n 2 ai i=1 2r>n n . (11) 1 1 2 ai − λ i=1 2 i=1 ai λ i Since γ =arcsinai ,i=1,... ,n we have the equation n λ π arcsin =(n − 2k) , (12) i=1 ai 2 or 3 3 λ λ 1 λ λ π + ···+ + + ···+ + ···=(n − 2k) . (13) a1 an 6 a1 an 2 Since by assumption a1 =min{a1,... ,an}, from (13) it follows that 3 3 λ λ 1 λ λ π + ···+ + + ···+ + ···≥(n − 2k) a1 a1 6 a1 a1 2 k-chordal and k-tangential polygons 25 or λ π arcsin ≥ (n − 2k) . a1 2n Hence π λ ≥ a1 sin(n − 2k) . (14) 2n Now using (11) and (14) we readily get (7). So, Theorem 1 is proved. ✷ Before stating some of its corollaries here is an example. If A1 ...A5 is a l-chordal pentagon as shown in Figure 1, then there are angles γ1,... ,γ5 such that π γ1 + ···+ γ5 =(5− 2) ,a1 sin γ1 = ...= a5 sin γ5 2 ifinstead ofthe drawn circles these can be drawn greater such that the above 3π equalities are valid. (For these drawn ones it is γ1 + ···+ γ5 < 2 . Let us remark that in the case when a side is small enough, then there are no angles γ1,... ,γ5 3π such that γ1 + ...+ γ5 = 2 .) A3 a2 A2 γ2 γ3 a1 γ1 a3 C A1 A4 γ4 γ5 a5 A5 Figure 1. Now we state some ofthe corollaries of Theorem 1. Corollary 1. There are angles γ1,... ,γn such that (5) and (6) hold if and only if 3 a1 1 a1 ···≥ − π + 3 3 + (n 2k) (15) H(a1,... ,an) 6 H(a1,... ,an) 2n 26 M. Radic´ i i i i where H(a1,... ,an) is the harmonic mean of a1,... ,an. Proof. It is clear from (13) since λ may be at most a1. ✷ Corollary 2. A sufficient condition for the existence of the angles γ1,... ,γn such that (5) and (6) hold is the inequality π a1 ≥ H(a1,... ,an)sin(n − 2k) (16) 2n Proof. If(16) holds, then obviously (15) holds, too. Namely, if 3 a1 1 a1 π + + ···≥(n − 2k) , H(a1,... ,an) 6 H(a1,... ,an) 2n then certainly (15) is valid because ofthe property ofthe arithmetics mean. ✷ Corollary 3. If there exists a k-chordal polygon whose sides have the lengths 1 1 2k π ≥ − 1 n a1 ,... , an and n sin(n 2k) 2n , then there exist angles γ ,... ,γ such that (5) and (6) hold. Proof. We shall use Corollary 2 in [6]. If a1,... ,an are the lengths ofthe sides ofthe k-chordal polygon A,then n ai > 2kaj,j=1,... ,n. (17) i=1 1 1 If ,... , are also the lengths ofthe sides ofa k-chordal polygon, then a1 an 1 1 2k + ···+ > a1 an a1 or 2k a1 > H(a1,... ,an). (18) n 2k π Accordingly, if ≥ sin(n − 2k) then (16) is valid. ✷ n 2n n − 1 Corollary 4. If n is odd and k is maximal, i.e. k = , then there exist the 2 angles γ1,... ,γn such that (5) and (6) hold. n − 1 Proof. If k = , then equation (5) can be written as 2 π γ1 + ...+ γn = , 2 n λ π and obviously there is λ such that arcsin = . ✷ i=1 ai 2 Corollary 5. If n =3and a, b, c are the lengths of the sides of an acute triangle, then a2 + b2 + c2 2r> (19) 3 a2 a + b + c − 8 H(a, b, c) k-chordal and k-tangential polygons 27 where a =min{a, b, c}.