Bài 102454

Question

CMR nếu $ABC$ thỏa mãn $1$ trong các điều kiện sau đây thì nó là tam giác cân
$1/$ $(a - b + c)\cot \frac{C}{2} = (a + b + c)tan\frac{B}{2}$
$2/$ $a(\cot \frac{C}{2} - tanA) = b(tanB - \cot \frac{C}{2})$
$3/$ $a + b = tan\frac{C}{2}(atanA + btanB)$
$4/$ $tanA + tanB = 2\cot \frac{C}{2}$
$5/$ $\frac{1}{2}(tanA + tanB) = \frac{{\sin A + \sin B}}{{\cos A + \cos B}}$
$6/$ $\sin A\sin B = cos^2\frac{C}{2}$
$7/$ $\frac{{\sin A + \sin B + \sin C}}{{\sin A + \sin B - \sin C}} = \cot \frac{A}{2}\cot \frac{C}{2}$
$8/$ $\sin \frac{A}{2}cos^3\frac{B}{2} = \sin \frac{B}{2}cos^3\frac{A}{2}$
$9/$ $\sin \frac{A}{2} = \frac{a}{{2\sqrt {bc} }}$
$10/$ $tan\frac{A}{2} = \frac{a}{{\sqrt {(a + b + c)(b + c - a)} }}$
$11/$ $l_a^2 = \frac{{4bc - {a^2}}}{4}$
$12/$ ${a^2} + {b^2} = tan\frac{C}{2}({a^2}\tan A + b^2\tan B)$
$13/$ $h_a = \sqrt {p(p - a)} $

score 0 · Answer 1

$1/$ ta có        $(a - b + c)\cot g\frac{C}{2} = (a + b + c)tg\frac{B}{2}$
      $ \Leftrightarrow (\sin A + \sin C - \sin B)\cot g\frac{C}{2} = (\sin A + \sin B + \sin C)tg\frac{B}{2}           (1)$
Áp dụng công thức qua các bước sau :
Trong tam giác $ABC$,ta có :
      $\left\{ \begin{array}{l}
\sin A + \sin B + \sin C = 4\cos \frac{A}{2}c{\rm{os}}\frac{B}{2}c{\rm{os}}\frac{C}{2}\\
\sin A + \sin C - \sin B = 4\cos \frac{B}{2}\sin \frac{A}{2}\sin \frac{C}{2}
\end{array} \right.$
Vì thế,($1$) tương đương     $c{\rm{os}}\frac{B}{2}\sin \frac{A}{2}\sin \frac{C}{2}\frac{{c{\rm{os}}\frac{C}{2}}}{{\sin \frac{C}{2}}} = c{\rm{os}}\frac{A}{2}c{\rm{os}}\frac{B}{2}c{\rm{os}}\frac{C}{2}\frac{{\sin \frac{B}{2}}}{{c{\rm{os}}\frac{B}{2}}}$
             $\begin{array}{l}
\Leftrightarrow c{\rm{os}}\frac{B}{2}\sin \frac{A}{2}c{\rm{os}}\frac{C}{2} = c{\rm{os}}\frac{A}{2}c{\rm{os}}\frac{C}{2}\sin \frac{B}{2}\\
\Leftrightarrow \sin \frac{A}{2}c{\rm{os}}\frac{B}{2} = \sin \frac{B}{2}c{\rm{os}}\frac{A}{2}(c{\rm{os}}\frac{C}{2} > 0)
\end{array}$
         $\begin{array}{l}
\Leftrightarrow \sin \frac{{A - B}}{2} = 0\\
\Leftrightarrow A = B
\end{array}$
Ta có $DPCM$
$2/$ ta có : $a(\cot g\frac{C}{2} - tgA) = b(tgB - \cot g\frac{C}{2})$
       $\begin{array}{l}
\Leftrightarrow a(tg\frac{{A + B}}{2} - tgA) = b(tgB - tg\frac{{A + B}}{2})\\
\Leftrightarrow \sin A\frac{{\sin (\frac{{A + B}}{2} - A)}}{{c{\rm{os}}\frac{{A + B}}{2}\cos A}} = \sin B\frac{{\sin (B - \frac{{A + B}}{2})}}{{\cos Bc{\rm{os}}\frac{{A + B}}{2}}}\\
\Leftrightarrow \sin A\frac{{\sin \frac{{B - A}}{2}}}{{\cos A}} = \sin B\frac{{\sin \frac{{B - A}}{2}}}{{\cos B}}\\
\Leftrightarrow \sin \frac{{B - A}}{2}(tgA - tgB) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \frac{{B - A}}{2} = 0\\
tgA = tgB
\end{array} \right. \Leftrightarrow A = B
\end{array}$
Ta có $DPCM$
$3/$     $a + b = tg\frac{C}{2}(atgA + btgB)$
     $\begin{array}{l}
\Leftrightarrow (a + b)\cot g\frac{C}{2} = atgA + btgB\\
\Leftrightarrow a(\cot g\frac{C}{2} - tgA) = b(tgB - \cot g\frac{C}{2})
\end{array}$
Vậy $3/$ chính là dạng tương đương của $2/$

$4/$ $tgA + tgB = 2\cot g\frac{C}{2} \Leftrightarrow \frac{{\sin (A + B)}}{{\cos A\cos B}} = 2\frac{{c{\rm{os}}\frac{C}{2}}}{{\sin \frac{C}{2}}}$
     $ \Leftrightarrow \frac{{2\sin \frac{C}{2}c{\rm{os}}\frac{C}{2}}}{{\cos A\cos B}} = 2\frac{{c{\rm{os}}\frac{C}{2}}}{{\sin \frac{C}{2}}}         (1)$
Do $c{\rm{os}}\frac{C}{2}\# 0$,nên ($1$) tương đương: $2{\sin ^2}\frac{C}{2} = 2\cos A\cos B$
       $\begin{array}{l}
\Leftrightarrow 1 - \cos C = c{\rm{os}}(A + B) + c{\rm{os}}(A - B)\\
\Leftrightarrow 1 - \cos C = - \cos C + c{\rm{os}}(A - B)\\
\Leftrightarrow 1 = c{\rm{os}}(A - B)\\
\Leftrightarrow A = B
\end{array}$
Ta có $DPCM$

$5/$    $\frac{1}{2}(tgA + tgB) = \frac{{\sin A + \sin B}}{{\cos A + \cos B}}$
    $\Leftrightarrow \frac{{\sin (A + B)}}{{2\cos A\cos B}} = \frac{{2\sin \frac{{A + B}}{2}c{\rm{os}}\frac{{A - B}}{2}}}{{2\cos \frac{{A + B}}{2}c{\rm{os}}\frac{{A - B}}{2}}}$
$ \Leftrightarrow \frac{{2\sin \frac{{A + B}}{2}c{\rm{os}}\frac{{A + B}}{2}}}{{2\cos A\cos B}} = \frac{{\sin \frac{{A + B}}{2}}}{{c{\rm{os}}\frac{{A + B}}{2}}}                      (*)$
Do $\sin \frac{{A + B}}{2} = c{\rm{os}}\frac{C}{2}\# 0$,nên $(*)$ tương đương
         $2{\sin ^2}\frac{C}{2} = 2\cos A\cos B$
     $\begin{array}{l}
\Leftrightarrow 1 - \cos C = c{\rm{os}}(A + B) + c{\rm{os}}(A - B)\\
\Leftrightarrow 1 = c{\rm{os}}(A - B)\\
\Leftrightarrow A = B
\end{array}$
Ta có $DPCM$
$6/$ $\sin A\sin B = c{\rm{o}}{{\rm{s}}^2}\frac{C}{2} \Leftrightarrow c{\rm{os}}(A - B) - c{\rm{os}}(A + B) = 1 + \cos C$
       $\begin{array}{l}
\Leftrightarrow c{\rm{os}}(A - B) + \cos C = 1 + \cos C\\
\Leftrightarrow c{\rm{os}}(A - B) = 1\\
\Leftrightarrow A = B
\end{array}$
Ta có $ĐPCM$
$7/$ $\frac{{\sin A + \sin B + \sin C}}{{\sin A + \sin B - \sin C}} = \cot g\frac{A}{2}\cot g\frac{C}{2}$
Áp dụng các công thức đã biết
       $\begin{array}{l}
\sin A + \sin B + \sin C = 4\cos \frac{A}{2}c{\rm{os}}\frac{B}{2}c{\rm{os}}\frac{C}{2}\\
\sin A + \sin B - \sin C = 4\sin \frac{A}{2}\sin \frac{B}{2}c{\rm{os}}\frac{C}{2}
\end{array}$
Hệ thức đã cho tương đương
      $\frac{{4\cos \frac{A}{2}c{\rm{os}}\frac{B}{2}c{\rm{os}}\frac{C}{2}}}{{4\sin \frac{A}{2}\sin \frac{B}{2}c{\rm{os}}\frac{C}{2}}} = \cot g\frac{A}{2}\cot g\frac{C}{2}$
   $\begin{array}{l}
\Leftrightarrow \cot g\frac{A}{2}\cot g\frac{B}{2} = \cot g\frac{A}{2}\cot g\frac{C}{2}\\
\Leftrightarrow \cot g\frac{B}{2} = \cot g\frac{C}{2}\\
\Leftrightarrow B = C
\end{array}$
Ta có $ĐPCM$
$8/$ do $c{\rm{o}}{{\rm{s}}^3}\frac{A}{2}c{\rm{o}}{{\rm{s}}^3}\frac{B}{2}\# 0$,nên ta có:
       $\sin \frac{A}{2}c{\rm{o}}{{\rm{s}}^3}\frac{B}{2} = \sin \frac{B}{2}c{\rm{o}}{{\rm{s}}^3}\frac{A}{2}$
                $\begin{array}{l}
\Leftrightarrow \frac{{\sin \frac{A}{2}c{\rm{o}}{{\rm{s}}^3}\frac{B}{2}}}{{c{\rm{o}}{{\rm{s}}^3}\frac{A}{2}c{\rm{o}}{{\rm{s}}^3}\frac{B}{2}}} = \frac{{\sin \frac{B}{2}c{\rm{o}}{{\rm{s}}^3}\frac{A}{2}}}{{c{\rm{o}}{{\rm{s}}^3}\frac{A}{2}c{\rm{o}}{{\rm{s}}^3}\frac{B}{2}}}\\
\Leftrightarrow tg\frac{A}{2}(1 + t{g^2}\frac{A}{2}) = tg\frac{B}{2}(1 + t{g^3}\frac{B}{2})\\
\Leftrightarrow (tg\frac{A}{2} - tg\frac{B}{2}) + (t{g^3}\frac{A}{2} - t{g^3}\frac{B}{2}) = 0\\
\Leftrightarrow (tg\frac{A}{2} - tg\frac{B}{2})(1 + t{g^2}\frac{A}{2} + tg\frac{A}{2}tg\frac{B}{2} + t{g^2}\frac{B}{2}) = 0\\
\Leftrightarrow tg\frac{A}{2} - tg\frac{B}{2} = 0({x^2} + xy + {y^2} \ge 0)\\
\Leftrightarrow A = B
\end{array}$
Ta có $ĐPCM$
$9/$ áp dụng hệ thức   ${\sin ^2}\frac{A}{2} = \frac{{1 - \cos A}}{2} = \frac{{1 - \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}}}{2} = \frac{{{a^2} - {{(b - c)}^2}}}{{4bc}}$
Hệ thức đã cho tương đương:    ${\sin ^2}\frac{A}{2} = \frac{{{a^2}}}{{4bc}}$
        $\begin{array}{l}
\Leftrightarrow \frac{{{a^2} - {{(b - c)}^2}}}{{4bc}} = \frac{{{a^2}}}{{4bc}}\\
\Leftrightarrow {(b - c)^2} = 0\\
\Leftrightarrow b = c
\end{array}$
Ta có $ĐPCM$
$10/$ Áp dụng hệ thức $c{\rm{o}}{{\rm{s}}^2}\frac{A}{2} = \frac{{1 + \cos A}}{2} = \frac{{1 + \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}}}{2} = \frac{{{{(b + c)}^2} - {a^2}}}{{4bc}}$
       $ \Rightarrow t{g^2}\frac{A}{2} = \frac{{{a^2} - {{(b - c)}^2}}}{{{{(b + c)}^2} - {a^2}}}$
Vì thế hệ thức đã cho đưa về dạng tương đương sau:
            $t{g^2}\frac{A}{2} = \frac{{{a^2}}}{{(a + b + c)(b + c - a)}}$
        $\begin{array}{l}
\Leftrightarrow \frac{{{a^2} - {{(b - c)}^2}}}{{{{(b + c)}^2} - {a^2}}} = \frac{{{a^2}}}{{(a + b + c)(b + c - a)}}\\
\Leftrightarrow \frac{{{a^2} - {{(b - c)}^2}}}{{(a + b + c)(b + c - a)}} = \frac{{{a^2}}}{{(a + b + c)(b + c - a)}}\\
\Leftrightarrow b = c
\end{array}$
Ta có $ĐPCM$
$11/$ áp dụng công thức ${l_a} = \frac{{2bc\cos \frac{A}{2}}}{{b + c}} \Rightarrow l_a^2 = \frac{{4{b^2}{c^2}}}{{{{(b + c)}^2}}}c{\rm{o}}{{\rm{s}}^2}\frac{A}{2} = \frac{{4{b^2}{c^2}}}{{{{(b + c)}^2}}}(1 + \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}) \\= \frac{{bc}}{{{{(b + c)}^2}}}\left[ {{{(b + c)}^2} - {a^2}} \right]$
Từ đó suy ra      $l_a^2 = \frac{{4bc - {a^2}}}{4}$
   $\begin{array}{l}
\Leftrightarrow \frac{{bc\left[ {{{(b + c)}^2} - {a^2}} \right]}}{{{{(b + c)}^2}}} = \frac{{4bc - {a^2}}}{4}\\
\Leftrightarrow 4bc{(b + c)^2} - 4bc{a^2} = 4bc{(b + c)^2} - {a^2}{(b + c)^2}\\
\Leftrightarrow 4bc = {(b + c)^2}\\
\Leftrightarrow {(b - c)^2} = 0\\
\Leftrightarrow b = c
\end{array}$
Ta có $ĐPCM$
$12/$ ta có      $\begin{array}{l}
{a^2} + {b^2} = tg\frac{C}{2}({a^2}tgA + {b^2}tgB)\\
\end{array}$
$ \Leftrightarrow {a^2}(1 - tg\frac{C}{2}tgA) + {b^2}(1 - tg\frac{C}{2}tgB) = 0$
   $ \Leftrightarrow \frac{{{a^2}}}{{\cos A}}c{\rm{os}}(\frac{C}{2} + A) + \frac{{{b^2}}}{{\cos B}}c{\rm{os}}(\frac{C}{2} + B) = 0            (1)$
Vì $(\frac{C}{2} + A) + (\frac{C}{2} + B) = \pi \Rightarrow c{\rm{os}}(\frac{C}{2} + B) = - c{\rm{os}}(\frac{C}{2} + A)$
Do đó $(1) \Leftrightarrow c{\rm{os}}(\frac{C}{2} + A)({\sin ^2}A\cos B - {\sin ^2}B\cos A) = 0$
     $ \Leftrightarrow c{\rm{os}}(\frac{C}{2} + A)\left[ {(1 - c{\rm{o}}{{\rm{s}}^2}A)\cos B - (1 - c{\rm{o}}{{\rm{s}}^2}B)\cos A} \right] = 0$
       $\Leftrightarrow c{\rm{os}}(\frac{C}{2} + A)(\cos B - \cos A)(1 + \cos A\cos B) = 0        (2)$
Do $1+cosAcosB>$,nên :
        $(2) \Leftrightarrow \left[ \begin{array}{l}
c{\rm{os}}(\frac{C}{2} + A) = 0\\
\cos A = \cos B
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\frac{C}{2} + A = \frac{A}{2} + \frac{B}{2} + \frac{C}{2}\\
A = B
\end{array} \right. \Leftrightarrow A = B$
Ta có $ĐPCM$
$13/$ ta có ${h_a} = \sqrt {p(p - a)} \Leftrightarrow \frac{{2S}}{4} = \sqrt {p(p - a)} $
   $ \Leftrightarrow \frac{2}{a}\sqrt {p(p - a)(p - b)(p - c)} = \sqrt {p(p - a)}$
      $ \Leftrightarrow 2\sqrt {(p - b)(p - c)} = a                           (1)$
Theo bất đẳng thức Cosi,ta có
   $2\sqrt {(p - b)(p - c)} \le (p - b) + (p - c) \Leftrightarrow 2\sqrt {(p - b)(p - c)} \le a$
Dấu “$=$” xảy ra khi $b=c$
Từ ($1$) suy ra trong ($2$) có dấu “$=$”,từ đó suy ra $ĐPCM$

Bài 102454

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Bài 102454

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