Identities and inequalities in a quadrilateral
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1 OCTOGON MATHEMATICAL MAGAZINE Vol. 17, No., October 009, pp ISSN , ISBN , Identities inequalities in a quadrilateral Ovidiu T. Pop 3 ABSTRACT. In this paper we will prove some identities inequalities in cyclic tangential quadrilaterals. 1. INTRODUCTION Let ABCD be a convex quadrilateral we note AB = a, BC = b, CD = c, DA = d, BD = e, AC = f, s = a + b + c + d, AC BD = {M}, the measure of the angles AMB is ϕ is the area of quadrilateral ABCD. If ABCD is a cyclic tangential quadrilateral, let R, r be the radii of the circumscribed circle, respectively inscribed circle of quadrilateral ABCD. It is well-known that the sides a, b, c, d are the solutions of the equation (see [4]) x 4 sx 3 +(s +r +r 4R + r )x rs( 4R + r +r)x+r s = 0 (1.1) the following inequalities are true (see [3]) r( 4R + r r) s, (1.) with equality if only if ABCD is a square (when R = r ) isosceles trapezoid (when R ), s 4R + r + r, (1.3) with equality if only if ABCD is a orthodiagonal quadrilateral 3 Received: Mathematics Subject Classification. 51M16. Key words phrases. Cyclic tangential quadrilateral.
2 Identities inequalities in a quadrilateral 755 r( 4R + r r) s 4R + r + r, (1.4) when the equalities hold simultaneous if only if ABCD is a square (when R = r ), if R at least one inequality from (1.4) is strict. On the other h, the L. Fejes Tóth Inequality R (1.5) holds, with equality if only if ABCD is a square the following identities If quadrilateral ABCD is cyclic, then ef = r( 4R + r + r) (1.6) = sr (1.7) e = (ac + bd)(ab + cd), (1.8) ad + bc ef = ac + bd, (1.91) e f = ab + cd ad + bc (1.10) called Girard s relation. 16R = (ab + cd)(ac + bd)(ad + bc), (1.11). IDENTITIES AND INEQUALITIES WITH R 1, R, R 3, R 4 Let ABCD be a convex quadrilateral we note with R 1, R, R 3, R 4 the radii of the circumscribed circles to the triangles AMB, BMC, CMD DMA.
3 756 Ovidiu T. Pop Lemma.1. The following identity a = R 1 + R + R 3 + R 4 = sef sr 1 R 1 + R + R 3 + R 4, b = sr R 1 + R + R 3 + R 4, (.1) c = sr 3 R 1 + R + R 3 + R 4, d = Proof. From the sine theorem we deduce that R 1 = R 3 = c sin ϕ, R 4 = d, from where sin ϕ R 1 + R + R 3 + R 4 = a + b + c + d sin ϕ sr 4 R 1 + R + R 3 + R 4 (.) a sin ϕ, R = b sin ϕ, = s ef sin ϕ. But = then (.1) sin ϕ follows. On the other h, we have that R 1 a = R b = R 3 c = R 4 d = 1 sin ϕ, from where b = R a, c = R 3 a d = R 4 a. From the relations above, by R 1 R 1 R 1 summing, we obtain that s = a + b + c + d = a + R a + R 3 a + R 4 a = a (R 1 + R + R 3 + R 4 ) R 1 R 1 R 1 R 1 then relations from (.) also result. Lemma.. If quadrilateral ABCD is a cyclic tangential quadrilateral, then a = sr 1 4R + r + r. (.3) P roof. From (.), by taking (.1), (1.6) (1.7) into account, relation (.3) is obtained. Theorem.1. If ABCD is a cyclic tangential quadrilateral, then R 1, R, R 3, R 4 are the solutions of the equation 16s x 4 16s ( 4R + r +r)x 3 +4(s +r +r 4R + r )( 4R + r +r) x
4 Identities inequalities in a quadrilateral 757 4r( 4R + r + r) 4 x + r ( 4R + r + r) 4 = 0. (.4) Proof. By taking (.3) into account, we have a solution for equation (1.1) by replacing one from (.3) in (1.1), after calculus, we obtain that R 1 verifies equation (.4). Theorem.. If quadrilateral ABCD is a cyclic tangential quadrilateral, then the following identities R1 = 4R + r + r, (.5) R1 R = (s + r + r 4R + r )( 4R + r + r) 4s, (.6) R1 R R = r( 4R + r + r) 4 4s, (.7) R 1 R R R 4 = r ( 4R + r + r) 4 16s, (.8) R 1 = ( 4R + r + r) (s r r 4R + r ) s, (.9) 1 R 1 = 4 r, (.10) 1 = 4(s + r + r 4R + r ) R 1 R r (, (.11) 4R + r + r) 1 R 1 R R 3 = 16s r ( 4R + r + r) 3 (.1) Proof. It results from Theorem.1. Theorem.3. If quadrilateral ABCD is a cyclic tangential quadrilateral, then the following inequalities 4r R 1 R, (.13)
5 758 Ovidiu T. Pop 6r R +r +r 4R + r R 1 R (5 4R + r 3r)( 4R + r + r) 3 64R (5 4R + r 3r)R 4 (10R 3r)R, (.14) 4 4r 3 r( 4R +r +r) 4 R 1 R R 3 ( 4R +r +r) 5 18R R 3, (.15) r 4 r ( 4R +r +r) R 1 R R R 4 r( 4R +r +r) R R4 4, (.16) 8r 4 R ( 4R + r + r) 3 3( 4R + r 5r) 3R R1 R, (.17) 4 R 1 R 1, (.18) 1 R 8(5 4R + r 3r) r( 4R + r + r) 1 R 1 R 16(R + r + r 4R + r ) r ( 4R + r + r) 3R r 4 (.19) 8 rr 51R r( 4R + r + r) 1 4 R 1 R R 3 16 r ( 4R + r + r) 4 r 3 (.0) P roof. It results from Theorem. from inequalities (1.), (1.3) (1.5).
6 Identities inequalities in a quadrilateral 759 IDENTITIES AND INEQUALITIES WITH r 1, r, r 3, r 4 In this section, we consider that ABCD is a cyclic tangential quadrilateral we note with r 1, r, r 3, r 4 the radii of the inscribed circles to the triangles AMB, BMC, CMD DMA. Lemma 3.1. The following identities AM = BM = eda ab + cd, (3.1) eab ab + cd, (3.) CM = ebc ab + cd (3.3) DM = ecd ab + cd (3.4) Proof. From the similarity of triangles ABM DCM, respectively ADM BCM, we have that AM DM = BM CM = AB DC = a c AM BM = DM CM = AD BC = d. From the equalities above, it results that b BM = b d AM, CM = c bc BM = a ad AM DM = c AM. If we choose BM a DM in the equality BM + DM = e, then we obtain b d AM + c AM = e, a from where relation (3.1) is obtained after that, by replacing relations, relations (3.)-(3.4) follow. We note with s 1, T 1 the semiperimeter, respectively the area of the triangle sr ( 4R + r + r) AMB α = R(R + r + 4R + r ).
7 760 Ovidiu T. Pop Lemma 3.. We have that r 1 = α c. (3.5) Proof. By taking s = a + c = b + d, (3.1),(3.) (1.6)-(1.11) into account, we calculate s 1 = 1 (AB + MB + MA) = 1 ( a + eab ab + cd + eda ) ab + cd = a ( ) e(b + d) 1 + = a ( ) s (ac + bd)(ab + cd) 1 + = ab + cd ab + cd ad + bc = a ( ) ef 1 + s = a ( 1 + s ef ) (ab + cd)(ac + bd)(ad + bc) 4R = a ( 1 + ef ) = a ( 1 + r( ) 4R + r + r), 4Rr 4Rr from where s 1 = a R + r + 4R + r. (3.6) 4R MA MB sin ϕ ef sin ϕ We have that T 1 = by taking = (3.1), (3.), (1.10), (1.11), (1.6) into account, it results that from where T 1 = 1 e a bd (ab + cd) ef = e a bd f (ab + cd) = a bd (ab + cd)(ad + bc) = a bd ef (ab + cd)(ac + bd)(ad + bc) = a bd ef 16R, T 1 = a bd( 4R + r + r) 8sR. (3.7) Because abcd = = s r, from (3.6) (3.7) it results that r 1 = T 1 = abcd( 4R + r + r) 4R s 1 8sR c(r + r +, than (3.5) 4R + r ) follows.
8 Identities inequalities in a quadrilateral 761 Remark 3.1. Similarly, we obtain that r = α d, r 3 = α a r 4 = α b. Remark 3.. Because R 1 = a MB MA 4T 1, with the help of relations (3.1), (3.) (3.7), we can calculate R 1, respectively R, R 3 R 4. Theorem 3.1. If ABCD is a cyclic tangential quadrilateral, then r 1, r, r 3, r 4 are the solutions of the equation 16R 4 (R+r+ 4R + r ) 4 x 4 16rR 3 ( 4R + r +r) (R+r+ 4R + r ) 3 x r R (s + r + r 4R + r )( 4R + r + r) (R + r + 4R + r ) x 4s r 4 R( 4R + r + r) 3 (R + r + 4R + r )x+ +s r 6 ( 4R + r + r) 4 = 0. (3.8) Proof. By taking c from (3.5) as a solution of the equation (1.1), by replacing c from (3.5) in (1.1), after calculus, we obtain that r 1 verifies equation (3.8). Theorem 3.. If quadrilateral ABCD is a cyclic tangential quadrilateral, then r1 = r( 4R + r + r) R(R + r + 4R + r ), (3.9) r1 r = r (s + r + r 4R + r )( 4R + r + r) 4R (R + r + 4R + r ), (3.10) r1 r r = r 1 r r r 4 = s r 4 ( 4R + r + r) 3 4R 3 (R + r + 4R + r ) 3, (3.11) s r 6 ( 4R + r + r) 4 16R 4 (R + r + 4R + r ) 4, (3.1) r 1 = r ( 4R + r + r) (8R + r + r 4R + r s ) R (R + r + 4R + r ), (3.13)
9 76 Ovidiu T. Pop 1 = 4R(R + r + 4R + r ) r 1 r (, (3.14) 4R + r + r) 1 = 4R (s + r + r 4R + r )(R + r + 4R + r ) r 1 r s r 4 ( (3.15) 4R + r + r) 1 = 16R3 (R + r + 4R + r ) 3 r 1 r r 3 s r 5 (. (3.16) 4R + r + r) Proof. It results from Theorem 3.1. Theorem 3.3. If quadrilateral ABCD is a cyclic tangential quadrilateral, then the following inequalities 8r 3 ( 1) R r 1 ( 1)R, (3.17) 4r 6 ( 1) R 4 r 1 r 3R4 ( 1) r, (3.18) 3r 9 ( 1) 3 R 6 r 1 r r 3 R6 ( 1) 3 r 3, (3.19) 64r 1 ( 1) 4 R 8 r 1 r r r 4 R4 ( 1) 4 4 (3.0) 4 ( + 1) R 1 R ( + 1) r 1 r 3 (3.1) P roof. It results from Theorem 3. from inequalities (1.), (1.3) (1.5). REFERENCES [1] Minculete, N., Characterization of a tangential quadrilateral, Forum Geometricorum 9 (009), [] Mitrinović, D. S., Pečarić, J. E. Volonec, V., Recent Advances in Geometric Inequalities, Kluver Academic Publishers, Dordrecht, 1989
10 Identities inequalities in a quadrilateral 763 [3] Pop, O., Inegalităţi în patrulater, Gazeta matematică B, 5-6 (1988), (Romanian) [4] Pop, O., Identităţi şi inegalităţi într-un patrulater, Gazeta matematică B, 8 (1989), (Romanian) National College Mihai Eminescu 5 Mihai Eminescu Street Satu Mare , Romania ovidiutiberiu@yahoo.com
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