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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