On a New Weighted Erdös-Mordell Type Inequality

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1 Int J Open Problems Compt Math, Vol 6, No, June 013 ISSN ; Copyright c ICSRS Publication, 013 wwwi-csrsorg On a New Weighted Erdös-Mordell Type Inequality Wei-Dong Jiang Department of Information Engineering, Weihai Vocational College Weihai 6410, ShanDong province, P R China jackjwd@163com Abstract In this short note, a new weighted Erdös-Mordell inequality Involving Interior Point of a triangle is established By it s application, some interesting geometric inequalities are derived Keywords: Erdös-Mordell inequality, Geometric inequality, Triangle, spread angle theorem 010 Mathematics Subject Classification: 6D15 1 Introduction Throughout the paper we assume ABC be a Triangle, and denote by a, b, c its sides lengths, be the area Let P be an interior point, Extend AP, BP, CP respectively to meet the opposite sides at D, E and F Let P D = r 1, P E = r, P F = r 3, 1,, 3 denote the areas of BP C, CP A, AP B R a, R b, R c the circumradii of the triangles BP C, CP A, AP B, respectively Let R 1, R, R 3 be the distances from P to A, B, C, and also let r 1, r, r 3 be the distances from P to the sides AB, BC, CA Then Erdös-Mordell inequality is true: Theorem 11 R 1 + R + R 3 (r 1 + r + r 3 ) (1) whereat equality holds if and only if the triangle is equailateral and the point P is its center This inequality was conjectured by Erdös in 1935[1], and was first proved by Mordell in 1937[] In the paper[3], D S Mitrnović etc noted some generaliations of Erdös- Mordell inequality in 1989 Among their results are the following theorem for three-variable quadratic Erdös-Mordell type inequality :

2 New Weighted Erdös-Mordell Type Inequality 45 Theorem 1 If x, y, are three real numbers, then for any point P inside the triangle ABC, we have x R 1 + y R + R 3 (yr 1 + xr + xyr 3 ) () with equality holding if and only if x = y = and P is the center of equilateral ABC Recently, Jiang [6] presented a new weighted Erdös-Mordell type inequality In this note, we give another new weighted Erdös-Mordell type inequality, as application, some interesting geometric inequalities are also established Main results In order to prove Theorem below, we need the following lemma Lemma 1 For any point P inside ABC, x, y, R, then we have x sin A + y sin B + sin C 1 4 Proof We make use of Kooi s inequality [4]: For real numbers λ 1, λ, λ 3 with λ 1 + λ + λ 3 0, ( y x + x y + y ) (3) (λ 1 + λ + λ 3 ) R λ λ 3 a + λ 3 λ 1 b + λ 1 λ c ; (4) Where R be circumradius of triangle ABC, equality holds if and only if the point with homogeneous barycentric coordinates (λ 1 : λ : λ 3 ) with reference to triangle ABC is the circumcenter of the triangle Now, Lemma 1 follows from (4) with λ 1 = y, λ x = x, λ y 3 = xy, and the law of sines: a = R sin A, b = R sin B, c = R sin C Now we are in a position to state and prove our main result Theorem For any point P inside triangle ABC, Extend AP, BP, CP respectively to meet the opposite sides at D, E and F Let R a, R b, R c the circumradiuses of triangles BP C, CP A, AP B, and let P D = r 1, P E = r, P F = r 3 x, y, are positive real numbers, we have + yr + r 3 1 ( y x + x y + xy ) (5) with equality holding if and only if x = y = and P is the center of equilateral ABC

3 46 Wei-Dong Jiang Proof Let BP C = α, CP A = β, AP B = γ It is obvious that 0 < α, β, γ < π and α + β + γ = π By using spread angle theorem, we have sin α r 1 sin(π β) sin(π γ) = + R R 3 = sin β + sin γ R R 3 sin β sin γ, R R 3 Thus, r 1 R R 3 csc β csc γ sin α Make use of b = R b sin β, c = R c sin γ, we get Let Because r 1 R R 3 sin α, bc 1 = sin A sin α A = π α, B = π β, C = π γ 1 A + A sin A sin α (sin A + sin α) = sin cos A A sin A + A we have, xr 1 By the same way, one can get yr 1 A + A x sin, (6) B + B y sin, (7) r 3 3 C + C sin, (8)

4 New Weighted Erdös-Mordell Type Inequality 47 Let Combining expression (6), (7), (8) and By Cauchy s inequality, we have xr 1 Rb 1 A + A x sin R c 1 x sin A + A, = x sin A + A θ = A + A, φ = B + B, ϕ = C + C Obviously, 0 < θ, φ, ϕ < π and θ + φ + ϕ = π, so θ, φ, ϕ can be angles of a triangle A 1 B 1 C 1 Applying Lemma 1 for the triangle A 1 B 1 C 1 we obtain This conclude that x sin θ + y sin φ + sin ϕ 1 4 xr 1 + yr + r 3 1 ( y x + x y + y ) ( y x + x y + xy ) and with equality holding if and only if x = y =, and P is the center of equilateral ABC The proof of Theorem is completed 3 Some application In this section we give some applications of Theorem Noticed r 1 r 1 etc, we have + yr + r 3 1 ( y x + x y + xy ) (9) By using AM-GM inequality, we have R b R b 1 (R b + R b ), then from (5) we have + yr + r 3 1 ( y R b + R c R c + R a R a + R b 4 x + x y + xy ) (10) By the same way of (9), the following inequality holds R b + R c + yr R c + R a + r 3 R a + R b 1 4 ( y x + x y + xy ) (11)

5 48 Wei-Dong Jiang let x = y = = 1 in (11), we have r 1 r r R b + R c R c + R a R a + R b 4 (1) In fact, (1) was conjectured by Liu in [5] and here we obtained a proof Corollary 31 If x, y, > 0, then x R a + y R b + R c (yr 1 + xr + xyr 3) (13) Proof alter x x R b R c, y y R c R a, R a R b (x, y, > 0) in (5), we obtain x r 1 + y r + r 3 1 ( ) y R x a + x R y b + x y R c (14) and then, let y = x, x = y,, x y x y = in (14), then (13) is obtained (13) is similar to (), that was conjectured by Liu in [5] Obviously x R a + y R b + R c (yr 1 + xr + xyr 3 ) (15) Let x = y = = 1 in (13)and (15), then we have and R a + R b + R c (r 1 + r + r 3) (16) R a + R b + R c (r 1 + r + r 3 ) (17) Note that (17) is similar to (1) let x = y = = 1 in (5) and by AM-GM inequality, we have R a R b R c 8r 1r r 3 (18) and R a R b R c 8r 1 r r 3 (19) 4 Open problem At the end, we pose an open problem Open problem: For an interior point P and positive real numbers x, y,, Let AD = w 1, BE = w, CF = w 3, R and r denote the circumradius and inradius of triangle ABC respectively, then xw 1 + yw + w 3 + r R ( y x + x y + xy ) (0)

6 New Weighted Erdös-Mordell Type Inequality 49 ACKNOWLEDGEMENTS This work is supported, in part, by the Project of Shandong Province Higher Educational Science and Technology Program under grant No J11LA57 References [1] P Erdos, Problem 3740 Amer Math Monthly, 4 (1935), pp 396 [] L J Mordell and D F Barrow Solution 3740, Amer Math Monthly 44(1937), 5-54 [3] D S Mitrinovi`c, J E Pečarć and V Voloneć Recent Advances in Geometric Inequalities Dordrecht, Netherlands: Kluwer 1989 [4] O Bottema etc Geometric Inequalities Groningen: Wolters-Noordhoff 1969 [5] Jian Liu A hundred unsolved Triangle inequality problem, geometric inequalities in china nanjing: Jiangsu education Press, 1996, (in Chinese) [6] Wei-Dong Jiang, A new weighted Erdös-Mordell type inequality, Creative Math & Inf 17(008):