2 replies to this topic

[Thanh Ngan]

Gọi G là trọng tâm tam giác ABC. Đặt $\widehat{BGC}=\alpha$. Chứng minh: $cot\alpha=\frac{5bc.cosA-2(b^2+c^2)}{3bc.sinA}$

[thuthuybiks]

Mod xóa giùm mình nhé

Edited by thuthuybiks, 05-11-2013 - 22:41.

[Phạm Hữu Bảo Chung]

Giải

Ta có:
$\cos{\alpha} = \dfrac{GB^2 + GC^2 - BC^2}{2GB.GC} = \dfrac{\dfrac{4}{9}(m_b^2 + m_c^2) - a^2}{2\dfrac{2}{3}m_b\dfrac{2}{4}m_c}$

 

$= \dfrac{\dfrac{4}{9}\left (a^2 + \dfrac{b^2 + c^2}{4}\right ) - a^2}{\dfrac{8}{9}m_bm_c } = \dfrac{b^2 + c^2 - 5a^2}{8m_bm_c}$

 

Ta có: $5bc\cos{A} - 2(b^2 + c^2) = \dfrac{5(b^2 + c^2 - a^2)}{2} - 2(b^2 + c^2) = \dfrac{b^2 + c^2 - 5a^2}{2}$

Vì vậy: $\cos{\alpha} = \dfrac{5bc\cos{A} - 2(b^2 + c^2)}{4m_bm_c}$

 

Trong tam giác GBC:
$\dfrac{\sin{\alpha}}{a} = \dfrac{\sin{BCG}}{\dfrac{2}{3}m_b}$

Mặt khác:
$\dfrac{\sin{BCG}}{\dfrac{c}{2}} = \dfrac{\sin{B}}{m_c} = \dfrac{b\sin{A}}{am_c}$

$\Rightarrow \sin{BCG} = \dfrac{bc\sin{A}}{2am_c}$

Vậy:
$\sin{\alpha} = \dfrac{a. \dfrac{bc\sin{A}}{2am_c}}{\dfrac{2}{3}m_b } = \dfrac{3bc\sin{A}}{4m_bm_c}$

Do đó: $\cot{\alpha} = \dfrac{5bc\cos{A} - 2(b^2 + c^2)}{3bc\sin{A}}$