Bài 102475

Question

CMR trong tam giác

$ABC$ có

$a,b,c$ lập thành cấp số cộng tương đương với các điều kiện sau đây:

$a)$

$1. tan\frac{A}{2}tan\frac{C}{2} = \frac{1}{3}$

$3.\frac{{{h_b}}}{{{l_b}}} = \sqrt {\frac{{2r}}{R}}$

$5.{h_b} = 3r$

$2. \sin A\sin C = 3{\sin ^2}\frac{B}{2}$

$4.{h_b} = {r_b}$

$6.$

$S = \frac{2}{3}l_b^2\sin B$

$b)$ CMR nếu

$a,b,c$ lập thành cấp số cộng,thì công sai cấp số cộng đó là

$d = \frac{3}{2}r(tan\frac{C}{2} - tan\frac{A}{2})$

score 0 · Answer 1

$a)$

$1)$ theo giả thiết ta có

$\begin{array}{l} 2b = a + c\\ \Leftrightarrow 2\sin B = \sin A + \sin C\\ \Leftrightarrow 4\sin \frac{B}{2}c{\rm{os}}\frac{B}{2} = 2\sin \frac{{A + C}}{2}c{\rm{os}}\frac{{A - C}}{2}(1) \end{array}$
Do

$\sin \frac{{A + C}}{2} = c{\rm{os}}\frac{B}{2}\# 0$ nên từ (1) suy ra

$2b = a + c \Leftrightarrow 2\cos \frac{{A + C}}{2} = c{\rm{os}}\frac{{A - C}}{2}$

$\begin{array}{l} \Leftrightarrow 2\cos \frac{A}{2}c{\rm{os}}\frac{C}{2} - 2\sin \frac{A}{2}\sin \frac{C}{2} = \cos \frac{A}{2}\cos \frac{C}{2} + \sin \frac{A}{2}\sin \frac{C}{2}\\ \Leftrightarrow tg\frac{A}{2}tg\frac{C}{2} = \frac{1}{3} \end{array}$
Đó là

$dpcm$

$2)$ ,ta có :

$\begin{array}{l} \sin A\sin B = 3{\sin ^2}\frac{C}{2}\\ \Leftrightarrow \frac{a}{{2R}}\frac{b}{{2R}} = 3\frac{{{{\sin }^2}C}}{{2(1 + \cos C)}}\\ \Leftrightarrow \frac{{ab}}{{4{R^2}}} = 3\frac{{{c^2}}}{{4{R^2}.2(1 + \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}})}}\\ \Leftrightarrow ab = \frac{{3{c^2}ab}}{{{{(a + b)}^2} - {c^2}}}\\ \Leftrightarrow {(a + b)^2} - {c^2} = 3{c^2}\\ \Leftrightarrow a + b = 2c\\ \Rightarrow dpcm \end{array}$

$3)$

Và

${\sin ^2}(C + \frac{B}{2}) = c{\rm{o}}{{\rm{s}}^2}\frac{{C - A}}{2}$ ,suy ra

$\frac{{{h_b}}}{{{l_b}}} = \sqrt {\frac{{2r}}{R}}$

$\begin{array}{l} \Leftrightarrow c{\rm{o}}{{\rm{s}}^2}\frac{{C - A}}{2} = 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\\ \Leftrightarrow c{\rm{o}}{{\rm{s}}^2}\frac{{C - A}}{2} = 4\sin B(c{\rm{os}}\frac{{C - A}}{2} - c{\rm{os}}\frac{{C + A}}{2}) \end{array}$

$\begin{array}{l} \Leftrightarrow c{\rm{o}}{{\rm{s}}^2}\frac{{C - A}}{2} - 4\sin \frac{B}{2}c{\rm{os}}\frac{{C - A}}{2} + 4{\sin ^2}\frac{B}{2} = 0\\ \Leftrightarrow {(c{\rm{os}}\frac{{C - A}}{2} - 2\sin \frac{B}{2})^2} = 0\\ \Leftrightarrow c{\rm{os}}\frac{{C - A}}{2} = 2\sin \frac{B}{2}\\ \Leftrightarrow 2\cos \frac{B}{2}c{\rm{os}}\frac{{C - A}}{2} = 4\sin \frac{B}{2}c{\rm{os}}\frac{B}{2}\\ \Leftrightarrow 2\sin \frac{{A + C}}{2}c{\rm{os}}\frac{{C - A}}{2} = 2\sin B\\ \Leftrightarrow \sin C + \sin A = 2\sin B\\ \Leftrightarrow a + c = 2b\\ \Rightarrow dpcm \end{array}$

$4)$ Ta có :

$\begin{array}{l} {h_b} = {r_b} \Leftrightarrow 2R\sin A\sin C = p.tg\frac{B}{2}\\ \Leftrightarrow 2R.4.\sin \frac{A}{2}c{\rm{os}}\frac{A}{2}\sin \frac{C}{2}c{\rm{os}}\frac{C}{2} = 4Rc{\rm{os}}\frac{A}{2}c{\rm{os}}\frac{B}{2}c{\rm{os}}\frac{C}{2}.\frac{{\sin \frac{B}{2}}}{{c{\rm{os}}\frac{B}{2}}}\\ \Leftrightarrow 2\sin \frac{A}{2}\sin \frac{C}{2} = \sin \frac{B}{2}\\ \Leftrightarrow 2\sin \frac{A}{2}\sin \frac{C}{2} = c{\rm{os}}\frac{{A + C}}{2}\\ \Leftrightarrow 2\sin \frac{A}{2}\sin \frac{C}{2} = c{\rm{os}}\frac{A}{2}c{\rm{os}}\frac{C}{2} - \sin \frac{A}{2}\sin \frac{C}{2}\\ \Leftrightarrow 3\sin \frac{A}{2}\sin \frac{C}{2} = c{\rm{os}}\frac{A}{2}c{\rm{os}}\frac{C}{2}\\ \Leftrightarrow tg\frac{A}{2}tg\frac{C}{2} = \frac{1}{3}\\ \Leftrightarrow a + c = 2b \end{array}$
(theo phần

$1$ )Đó là

$dpcm$

$5)$

${h_b} = 3r \Leftrightarrow 2R\sin A\sin C = 3.4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

$\Leftrightarrow 8R\sin \frac{A}{2}\sin \frac{C}{2}c{\rm{os}}\frac{A}{2}c{\rm{os}}\frac{C}{2} = 12R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

$\begin{array}{l} \Leftrightarrow 2\cos \frac{A}{2}c{\rm{os}}\frac{C}{2} = 3\sin \frac{B}{2}\\ \Leftrightarrow 2\cos \frac{A}{2}c{\rm{os}}\frac{C}{2} = 3\cos \frac{{A + C}}{2}\\ \Leftrightarrow 2\cos \frac{A}{2}c{\rm{os}}\frac{C}{2} = 3\cos \frac{A}{2}c{\rm{os}}\frac{C}{2} - 3\sin \frac{A}{2}\sin \frac{C}{2}\\ \Leftrightarrow 3\sin \frac{A}{2}\sin \frac{C}{2} = c{\rm{os}}\frac{A}{2}c{\rm{os}}\frac{C}{2} \end{array}$

$\Leftrightarrow tg\frac{A}{2}tg\frac{C}{2} = \frac{1}{3}$
Theo (

$1$ ) suy ra

$dpcm$

$6)$ ta có

${l_b} = \frac{{2ac\cos \frac{B}{2}}}{{a + c}}$
Vì thế

$l_b^2 = \frac{{4{a^2}{c^2}c{\rm{o}}{{\rm{s}}^2}\frac{B}{2}}}{{{{(a + c)}^2}}} = \frac{{ac}}{{{{(a + c)}^2}}}\left[ {{{(a + c)}^2} - {b^2}} \right]$
Từ đó suy ra

$\begin{array}{l} a + c = 2b\\ \Leftrightarrow l_b^2 = \frac{{ac(4{b^2} - {b^2})}}{{4{b^2}}}\\ \Leftrightarrow l_b^2 = \frac{3}{4}ac\\ \Leftrightarrow l_b^2 = \frac{3}{4}\frac{{2S}}{{\sin B}}\\ \Leftrightarrow S = \frac{2}{3}l_b^2\sin {\bf{B}} \end{array}$
Đó là

$dpcm$

$b)$ Giả sử tam giác

$ABC$ có

$a+c=2b$ ,theo phần

$1$ ta có

$tg\frac{A}{2}tg\frac{C}{2} = \frac{1}{3}$
Rõ ràng

$\frac{3}{2}r(tg\frac{C}{2} - tg\frac{A}{2}) = \frac{3}{2}4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\frac{{\sin \frac{{C - A}}{2}}}{{c{\rm{os}}\frac{A}{2}c{\rm{os}}\frac{C}{2}}}$

$\begin{array}{l} = 6R\sin \frac{B}{2}\sin \frac{{C - A}}{2}tg\frac{A}{2}tg\frac{C}{2}\\ = 2R\cos \frac{{A + C}}{2}\sin \frac{{C - A}}{2}\\ = R(\sin C - \sin A)\\ = \frac{{c - a}}{2} = d \end{array}$
Đó là

$dpcm$

Bài 102475

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Bài 102475

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