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在△ABC中,设a、b、c为其三边;p=1/2(a+b+c);r_a、r_b、r_c为其三个旁切圆的半径;r、R分别为其内切圆、外接圆半径;t_a、t_b、t_c为其三条内角平分线;h_a、h_b、h_c分别为其三边上的高;△为其面积。本文证明了关于r_a、r_b、r_c的下述不等式(为节省篇幅,以下命题只写结论)。 1°9r≤3~(1/2)p≤r_a+r_b+r_c≤(9/2)R (1) 证易证r_a+r_b+r_c=4R+r,以及r_ar_b+r_br_c+r_cr_a=p~2,由欧拉不等式R≥2r,得右边不等式。另方面,由代数不等式χ+y+z≥(3(χy+yz+zχ))~(1/2)及不等式p≥3(3~(1/2))r得r_a+r_b+r_c≥3~(1/2)p≥9r ∴ (1)式成立 2°27r~3≤r_ar_br_c≤(3~(1/2))/9p~3≤(27/8)R~3 (2) 证易证r_ar_br_c=rp~2,p≥3 3~(1/2)r,得r_ar_br_c≥27r~3另方面。r_ar_br_c=(r_ar_b·r_br_c·r_cr_a)~(1/2)≤
In △ABC, let a, b, and c be three sides; p=1/2(a+b+c); r_a, r_b, and r_c are the radii of three adjacent circles; r and R are their Inscribed circles, circumcircle radii; t_a, t_b, t_c are the three internal angle bisectors; h_a, h_b, h_c are the heights of the three sides respectively; △ is the area. This paper proves the following inequalities for r_a, r_b, and r_c (to save space, the following propositions only write conclusions). 1°9r≤3~(1/2)p≤r_a+r_b+r_c≤(9/2)R (1) Evidence proves that r_a+r_b+r_c=4R+r, and r_ar_b+r_br_c+r_cr_a=p~ 2, by Euler inequality R ≥ 2r, the right inequality. On the other hand, the algebraic inequality χ+y+z≥(3(χy+yz+zχ))~(1/2) and the inequality p≥3(3~(1/2))r give r_a+r_b+r_c≥ 3~(1/2)p≥9r ∴ (1) holds 2°27r~3≤r_ar_br_c≤(3~(1/2))/9p~3≤(27/8)R~3 (2) Evidence r_ar_br_c = rp ~ 2, p ≥ 3 3 ~ (1/2) r, r_ar_br_c ≥ 27r ~ 3 other aspects. R_ar_br_c=(r_ar_b·r_br_c·r_cr_a)~(1/2)≤