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**novae** · 27-11-2010, 10:26 AM

Vừa check lại kết quả, chính xác phải là
$(AB.CD.\sin (AB;CD))^2+(AC.BD.\sin (AC;BD))^2+(AD.BC.\sin (AD;BC))^2 = 4(S_1^2+S_2^2+S_3^2+S_4^2) $
Còn đẳng thức $Q_1^2+Q_2^2+Q_3^2= S_1^2+S_2^2+S_3^2+S_4^2 $ vẫn đúng, vì ta có $2Q_1=AB.CD.\sin (AB;CD) $
Còn một cách chứng minh bằng cách sử dụng tích có hướng:
$2S_{ABC}=\left | \overrightarrow{AB} \times \overrightarrow{AC} \right |, 2S_{ABD}=\left | \overrightarrow{AB} \times \overrightarrow{AD} \right |, 2S_{ACD}=\left | \overrightarrow{AC} \times \overrightarrow{AD} \right | $
$2S_{BCD}=\left | \overrightarrow{BC} \times \overrightarrow{BD} \right |=\left | \left (\overrightarrow{AC}-\overrightarrow{AB} \right ) \times \left ( \overrightarrow{AD}-\overrightarrow{AB} \right ) \right |=\left | \overrightarrow{AC} \times \overrightarrow{AD}-\overrightarrow{AB} \times \overrightarrow{AD}-\overrightarrow{AC} \times \overrightarrow{AB} \right | $
$\Rightarrow 4\left( S_{ABC}^2+S_{ABD}^2+S_{ACD}^2+ S_{BCD}^2 \right)=2\left( \overrightarrow{AB} \times \overrightarrow{AC} \right)^2+2\left( \overrightarrow{AB} \times \overrightarrow{AD} \right)^2+2\left( \overrightarrow{AC} \times \overrightarrow{AD} \right)^2 -2 \left( \overrightarrow{AB} \times \overrightarrow{AC} \right) \left( \overrightarrow{AB} \times \overrightarrow{AD} \right) -2 \left( \overrightarrow{AC} \times \overrightarrow{AD} \right) \left( \overrightarrow{AC} \times \overrightarrow{AB} \right) -2 \left( \overrightarrow{AD} \times \overrightarrow{AC} \right) \left( \overrightarrow{AD} \times \overrightarrow{AB} \right) $

$(AB.CD.\sin (AB;CD))^2=\left( \overrightarrow{AB} \times \overrightarrow{AD} - \overrightarrow{AB} \times \overrightarrow{AC} \right)^2=\left( \overrightarrow{AB} \times \overrightarrow{AD} \right)^2+\left( \overrightarrow{AB} \times \overrightarrow{AC} \right)^2 - 2 \left( \overrightarrow{AB} \times \overrightarrow{AC} \right) \left( \overrightarrow{AB} \times \overrightarrow{AD} \right) $
$(AC.BD.\sin (AC;BD))^2=\left( \overrightarrow{AC} \times \overrightarrow{AD} - \overrightarrow{AC} \times \overrightarrow{AB} \right)^2=\left( \overrightarrow{AC} \times \overrightarrow{AD} \right)^2+\left( \overrightarrow{AC} \times \overrightarrow{AB} \right)^2 - 2 \left( \overrightarrow{AC} \times \overrightarrow{AD} \right) \left( \overrightarrow{AC} \times \overrightarrow{AB} \right) $
$(AD.BC.\sin (AD;BC))^2=\left( \overrightarrow{AD} \times \overrightarrow{AC} - \overrightarrow{AD} \times \overrightarrow{AB} \right)^2=\left( \overrightarrow{AD} \times \overrightarrow{AC} \right)^2+\left( \overrightarrow{AD} \times \overrightarrow{AB} \right)^2 - 2 \left( \overrightarrow{AD} \times \overrightarrow{AC} \right) \left( \overrightarrow{AD} \times \overrightarrow{AB} \right) $

Từ đó suy ra đpcm.

27-11-2010, 10:26 AM	#5
novae +Thành Viên Danh Dự+ Tham gia ngày: Jul 2010 Đến từ: Event horizon Bài gởi: 2,453 Thanks: 53 Thanked 3,057 Times in 1,288 Posts	Vừa check lại kết quả, chính xác phải là $(AB.CD.\sin (AB;CD))^2+(AC.BD.\sin (AC;BD))^2+(AD.BC.\sin (AD;BC))^2 = 4(S_1^2+S_2^2+S_3^2+S_4^2) $ Còn đẳng thức $Q_1^2+Q_2^2+Q_3^2= S_1^2+S_2^2+S_3^2+S_4^2 $ vẫn đúng, vì ta có $2Q_1=AB.CD.\sin (AB;CD) $ Còn một cách chứng minh bằng cách sử dụng tích có hướng: $2S_{ABC}=\left \| \overrightarrow{AB} \times \overrightarrow{AC} \right \|, 2S_{ABD}=\left \| \overrightarrow{AB} \times \overrightarrow{AD} \right \|, 2S_{ACD}=\left \| \overrightarrow{AC} \times \overrightarrow{AD} \right \| $ $2S_{BCD}=\left \| \overrightarrow{BC} \times \overrightarrow{BD} \right \|=\left \| \left (\overrightarrow{AC}-\overrightarrow{AB} \right ) \times \left ( \overrightarrow{AD}-\overrightarrow{AB} \right ) \right \|=\left \| \overrightarrow{AC} \times \overrightarrow{AD}-\overrightarrow{AB} \times \overrightarrow{AD}-\overrightarrow{AC} \times \overrightarrow{AB} \right \| $ $\Rightarrow 4\left( S_{ABC}^2+S_{ABD}^2+S_{ACD}^2+ S_{BCD}^2 \right)=2\left( \overrightarrow{AB} \times \overrightarrow{AC} \right)^2+2\left( \overrightarrow{AB} \times \overrightarrow{AD} \right)^2+2\left( \overrightarrow{AC} \times \overrightarrow{AD} \right)^2 -2 \left( \overrightarrow{AB} \times \overrightarrow{AC} \right) \left( \overrightarrow{AB} \times \overrightarrow{AD} \right) -2 \left( \overrightarrow{AC} \times \overrightarrow{AD} \right) \left( \overrightarrow{AC} \times \overrightarrow{AB} \right) -2 \left( \overrightarrow{AD} \times \overrightarrow{AC} \right) \left( \overrightarrow{AD} \times \overrightarrow{AB} \right) $ $(AB.CD.\sin (AB;CD))^2=\left( \overrightarrow{AB} \times \overrightarrow{AD} - \overrightarrow{AB} \times \overrightarrow{AC} \right)^2=\left( \overrightarrow{AB} \times \overrightarrow{AD} \right)^2+\left( \overrightarrow{AB} \times \overrightarrow{AC} \right)^2 - 2 \left( \overrightarrow{AB} \times \overrightarrow{AC} \right) \left( \overrightarrow{AB} \times \overrightarrow{AD} \right) $ $(AC.BD.\sin (AC;BD))^2=\left( \overrightarrow{AC} \times \overrightarrow{AD} - \overrightarrow{AC} \times \overrightarrow{AB} \right)^2=\left( \overrightarrow{AC} \times \overrightarrow{AD} \right)^2+\left( \overrightarrow{AC} \times \overrightarrow{AB} \right)^2 - 2 \left( \overrightarrow{AC} \times \overrightarrow{AD} \right) \left( \overrightarrow{AC} \times \overrightarrow{AB} \right) $ $(AD.BC.\sin (AD;BC))^2=\left( \overrightarrow{AD} \times \overrightarrow{AC} - \overrightarrow{AD} \times \overrightarrow{AB} \right)^2=\left( \overrightarrow{AD} \times \overrightarrow{AC} \right)^2+\left( \overrightarrow{AD} \times \overrightarrow{AB} \right)^2 - 2 \left( \overrightarrow{AD} \times \overrightarrow{AC} \right) \left( \overrightarrow{AD} \times \overrightarrow{AB} \right) $ Từ đó suy ra đpcm. __________________ M.