GAZETA MATEMATICĂ ARTICOLE ŞI NOTE MATEMATICE. SERIA B PUBLICAŢIE LUNARĂ PENTRU TINERET Fondată în anul Anul CXXI nr.

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1 GAZETA MATEMATICĂ SERIA B PUBLICAŢIE LUNARĂ PENTRU TINERET Fondată în anul 895 Anul CXXI nr 9 septembrie 06 ARTICOLE ŞI NOTE MATEMATICE A RATIONAL REFINEMENT OF YUN S INEQUALITY IN BICENTRIC QUADRILATERALS Marius Drăgan and Ioan V Maftei Abstract In this paper we shall find the best minimal and maximal αr + βr rational bounds of form with α, β R and γ >, for the R + γr sum sin A cyclic cos B where A, B, C, D represent the angles of a bicentric quadrilateral with circumradius R, inradius r and semiperimeter s Keywords: bounds for cyclic sums of trigonometric functions of the angles of a bicentric quadrilateral MSC: 5M6, 6D05 Introduction In [6], Zhang Yun established the following inequality In every bicentric quadrilateral holds the inequality r R sin A cos B + sin B cos C + sin C cos D + sin D cos A Another proof for this inequality, given by Martin Josefsson, can be found in [5] A refinement of Yun s inequality is given by Vasile Jiglău in [4] In [] appears a refinement of Yun s inequality of the type fr, r sin A cos B gr, r cyclic where fr, R, gr, R represent the best minimal and maximal homogenous Profesor, Liceul,,Mircea cel Bătrân, Bucureşti Profesor, Bucureşti

2 386 Articole şi note matematice functions for the sum sin A cyclic cos B ; gr, r is determined also in [4], where it is stated that In every bicentric quadrilateral holds + rr + 4R + r r sin A cos B 4R + r + r R cyclic The proof of this theorem is based on the monotony of the function see [] E : [S, S ] R, ES = sin A cos B = + x 3 4R + x 3 8R S r cyclic We remember the known facts that sin A = cos C bc = ad + bc x 3 = ac + bd = r r + 4R + r, see [] and S = 8r 4R + r r, S = r + 4R + r are the semiperimeters of the bicentric quadrilaterals A B C D, A B C D which make up the minimal and maximal semiperimeter from Blundon-Eddy inequality S S S see [3] Main results In the following we find the best real constants α, β and γ > such that the inequality αr + βr R + γr sin A cos B + sin B cos C + sin C cos D + sin D cos A 3 is true in every bicentric quadrilateral 3 In every bicentric quadrilateral holds R + r R + r sin A cos B + sin B cos C + sin C cos D + sin D cos A Proof From we have that 4 + rr + 4R + r R sin A cos B + sin B cos C + sin C cos D + sin D cos C

3 M Drăgan, I V Maftei, A refinement of Yun s inequality 387 In order to prove 4 it will be sufficient to prove that R + r R + r + rr + 4R + r R or, after denoting x = R r 3 x + + 4x + x, or that x + x x + x + x, or 8x x + + x + 4x +, + 4x + x, that is or 6x x 4x + x + x +, or, after performing some calculation, that 4x x 4x + 6 x + 5 0, which is true Next, we shall prove that the inequality 4 is the best of type 3 We suppose that other constants α 0, β 0 R, γ 0 > exist, such that α 0 R + β 0 r sin A R + γ 0 r cos B + sin B cos C + sin C cos D + sin D cos A is true in every bicentric quadrilateral and this inequality is the best inequality of type 3 So we have that R + R + r α 0R + β 0 r R + γ 0 r sin A cos B + sin B cos C + sin C cos D + sin D cos A 5 is true in every bicentric quadrilateral If we consider the case of bicentric quadrilateral A B D C which makes 4R up the minimal semiperimeter S = 8r + r r, from 5 it results that the inequality R + r R + r α 0R + β 0 r R + γ 0 r + r r + 4R + r R = = sin A cos B + sin B cos C + sin C cos D + sin D cos A sin A cos B + sin B cos C + sin C cos D + sin D cos A 6

4 388 Articole şi note matematice is true in every bicentric quadrilateral Now yields sin A cos B + sin B cos C + sin C cos D + sin D cos A = = + x 3 4R + x 3 8R r S + x 3 4R + x 3 8R r S = sin A cos B + sin B cos C + sin C cos D + sin D cos A = If we consider the case of square with sides a = b = c = d =, R =, r =, if we replace in 6 we obtain α 0 + β0 or α 0 + β0 = + γ γ0 From 6 we have x + x + α 0x + β 0 x + γ x + x 8 for each x If we take in 8 x, we obtain α 0 or α 0 = From 7 we obtain β 0 = + γ 0 Inequality 8 may be written as x + x + x + + γ 0 x + γ x + x 9 for each x The left side of 9 is equivalent with γ 0 x x, or 0 for each γ 0 0 The right side of 9 may be written as γ 0 + x + γ 0 each x or γ 0 x + 4x + x + 4x + + 4x + x for

5 M Drăgan, I V Maftei, A refinement of Yun s inequality 389 for each x We consider the function f : [, + R, fx = x + 4x + x + 4x + From we have γ 0 min x fx We compute lim x fx = lim x 4x x + 3 x 4x + x + + 4x = 4x + = 4x x + 4x lim + x 4x x = 4x We prove that lim x fx = fx for each x which implies that = lim x fx = min x fx So γ 0 x + 4x + It remains to prove that fx = x + 4x +, that is x + 4x + x + 4x + x, which reduces to 3 x x + + 4x 3 +, or x + + 4x + x, inequality which was proved during the proof of theorem 3 From 0 and follows that γ 0 = and β 0 = γ 0 + =, which represents a contradiction So, that the inequality 4 is the best of type 3 Next, we shall find the best real constants α, β and γ >, such that the inequality sin A cos B + sin B cos C + sin C cos D + sin D cos A αr + βr R + γr 3 is true in every bicentric quadrilateral

6 390 Articole şi note matematice 4 In every bicentric quadrilateral is true the inequality sin A cos B + sin B cos C + sin C cos D + sin D cos A R r R r 4 Proof From we have that sin A cos B + sin B cos C + sin C cos D + sin D cos A 4R + r + r R In order to prove 4, it will be sufficient to prove that 4R + r + r R r, R R r or 4x + + x + 6 4, x x or 4x + + x + α x x α, or x α 4x + x + 4α x + α After squaring and performing some calculations we obtain that or 4x [ 6α x + 3α αx + α ] 0 47 x 0 for each x, inequality which is true Next we shall prove that the inequality 4 is the best of type 3 We suppose that other constants α 0, β 0 R, γ 0 > exist such that sin A cos B + sin B cos C + sin C cos D + sin D cos A α 0R + β 0 r R + γ 0 r is a better inequality of type 3 It follows that sin A cos B + sin B cos C + sin C cos D + sin D cos A α 0R + β 0 r R + γ 0 r R r R r 5

7 M Drăgan, I V Maftei, A refinement of Yun s inequality 39 is true in every bicentric quadrilateral If we consider the bicentric quadrilateral A B C D which makes up the maximal semiperimeter S = 4R + r + r, from 5 it follows that sin A cos B + sin B cos C + sin C cos D + sin D cos A cos B + sin B cos C + sin C cos D + sin D cos A = 4R = + r + r α 0R + β 0 r R r 6 R R + γ 0 r R r sin A is true in every bicentric quadrilateral since we have sin A cos B + sin B cos C + sin C cos D + sin D cos A = = + x 3 4R + x 3 8R r S + x 3 4R + x 3 8R r S = sin A cos B + sin B cos C + sin C cos D + sin D cos A = In the case of the square with sides a = b = c = d =, R =, r =, the relation 6 yields α 0/ + β 0 / / + γ 0 / From 6 we have 4x + x, or α 0 + β0 = γ R r x + γ 0 R r α 0x + β 0 If we take x, we obtain α 0 or α 0 = Replacing in 7 we obtain From 6 we obtain that 4x + + x β 0 = γ x + γ 0 + x x + γ 0 x for each x The right side of inequality 9 may be written as [ 5 ] 7 + γ 0 x 0

8 39 Articole şi note matematice for each x or γ for each x It results that γ The left side of inequality 9 may be written as [ 4x x + x 3 + ] γ 0 x 4x + Consider the function f : [, + R, 4x x + x 3 + fx = x 4x + From we obtain that γ 0 fx for each x It follows that γ 0 max x fx We have lim x fx x 4x = lim + x 3 x x x = 4x + 4x 4x + + 4x + 4x 3 = lim x We prove that 4x 4x + = 3 fx 3 = lim x fx for each x which implies that max x fx = 3 and γ Inequality may be written equivalently as 4x x + x 3 + x 3 4x + or x α 4x + x + 4α x + α, inequality which it was proved in the context of theorem 4 From 0 and 3 it follows that γ 0 = 3 From 8 we obtain β 0 = γ 0 + = 3 4 So, we have α 0 =, β 0 = 3 4, γ 0 = 3 which represent a contradiction So, the inequality 4 is the best of the type 3

9 M Drăgan, I V Maftei, A refinement of Yun s inequality [The rational refinement of Yun s inequaliy] In every bicentric quadrilateral are true the inequalities r R R + r R + r sin A cos B + sin B cos C + sin C cos D + sin D cos A R r R r Proof In the following we consider the functions F : ] [, R, G : 3, + F γ = R + + γ r R + γ r We have F γ = rr r R,, Gγ = R + γ + r R + γ r R + γ r, G γ = rr r R + γ r Since R r, it follows that F γ 0 and G γ 0 which imply that F and G are an increasing functions It results that lim x F γ < F γ F and G 3 Gγ G or F γ R + r R + r [ γ ], and for each γ [ 3, + for each R r R + 4 3r Gγ 4 According to 4, theorem 3 and theorem 4 we obtain the following result 6 In every bicentric quadrilateral are true the following inequalities < R + + γ r R + r R + γ r R + r sin A cos B + sin B cos C + sin C cos D + sin D cos A R r R + 4 3r R + γ + r R + γ r

10 394 Articole şi note matematice for every γ [ ], and γ [ 3, + References [] M Bencze and M Drăgan, Some inequalities in bicentric quadrilateral, Acta Univ Sapientiae, Mathematica, I 03, No 5, 0 38 [] M Bencze and M Drăgan, The best Yun s inequality type for bicentric quadrilateral, Octogon Math Magazine 305, No, [3] W J Blundon, and R H Eddy, Problem 488, Nieuw Archief Wiskunde, 6 978, 3, solution in 6 978, 655 [4] V Jiglău, A refinement of Z Yun inequality, Recreaţii Matematice, No, 04, [5] M Josefsson, A New Proof of Yun s inequality for bicentric Quadrilaterals, Forum Geometricum 0, 79 8 [6] Z Yun, Euler s Inequality Revisited, Mathematical Spectrum , 9