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Câu hỏi: Cho hình chóp $S.ABCD$ có đáy $ABCD$ là hình bình hành. Gọi $M,N$ lần lượt thuộc các cạnh $SA,SD$ sao cho $3SM=2SA,3SN=2SD.$ Mặt phẳng $\left( \alpha \right)$ chứa $MN$ cắt cạnh $SB,SC$ lần lượt tại $Q,P$. Đặt $\dfrac{SQ}{SB}=x,{{V}_{1}}$ là thể tích của khối chóp $S.MNPQ$, $V$ là thể tích khối chóp $S.ABCD$. Tìm $x$ để ${{V}_{1}}=\dfrac{1}{2}V.$
A. $x=\dfrac{-2+\sqrt{58}}{6}.$
B. $x=\dfrac{-1+\sqrt{41}}{4}.$
C. $x=\dfrac{-1+\sqrt{33}}{4}.$
D. $x=\dfrac{1}{2}.$
A. $x=\dfrac{-2+\sqrt{58}}{6}.$
B. $x=\dfrac{-1+\sqrt{41}}{4}.$
C. $x=\dfrac{-1+\sqrt{33}}{4}.$
D. $x=\dfrac{1}{2}.$
Cách 1.
Ta có ${{V}_{1}}={{V}_{S.MNPQ}}={{V}_{S.MNQ}}+{{V}_{S.PNQ}}$
Ta có $\left\{ \begin{aligned}
& \left( \alpha \right)\cap \left( SBC \right)=PQ \\
& MN//BC \\
& MN\subset \left( \alpha \right) \\
& BC\subset \left( SBC \right) \\
\end{aligned} \right.\Rightarrow PQ//MN//BC\Rightarrow \dfrac{SP}{SC}=\dfrac{SQ}{SB}=x.$
Có $\dfrac{{{V}_{S.MNQ}}}{{{V}_{S.ADB}}}=\dfrac{SM}{SA}.\dfrac{SN}{SD}.\dfrac{SQ}{SB}=\dfrac{2}{3}.\dfrac{2}{3}x=\dfrac{4}{9x}\Rightarrow {{V}_{S.MNQ}}=\dfrac{4x}{9}{{V}_{S.ADB}}=\dfrac{4x}{9}.\dfrac{V}{2}=\dfrac{2x}{9}V.$
Đồng thời $\dfrac{{{V}_{S.PNQ}}}{{{V}_{S.CDB}}}=\dfrac{SP}{SC}.\dfrac{SN}{SD}.\dfrac{SQ}{SB}=x.\dfrac{2}{3}.x=\dfrac{2{{x}^{2}}}{3}\Rightarrow {{V}_{S.PNQ}}=\dfrac{2{{x}^{2}}}{3}.{{V}_{S.CDB}}=\dfrac{2{{x}^{2}}}{3}.\dfrac{V}{2}=\dfrac{{{x}^{2}}}{3}V.$
Như vậy ${{V}_{1}}=\left( \dfrac{{{x}^{2}}}{3}+\dfrac{2x}{9} \right)V.$ Mà theo giả thiết ta có ${{V}_{1}}=\dfrac{1}{2}V$ nên ta suy ra:
$\dfrac{{{x}^{2}}}{3}+\dfrac{2x}{9}=\dfrac{1}{2}\Leftrightarrow \left[ \begin{aligned}
& x=\dfrac{-2+\sqrt{58}}{6}\left( Nhan \right) \\
& x=\dfrac{-2-\sqrt{58}}{6}\left( Loai \right) \\
\end{aligned} \right.. $ Vậy $ x=\dfrac{-2+\sqrt{58}}{6}.$
Cách 2:
Đặt $a=\dfrac{SM}{SA}=\dfrac{2}{3};b=\dfrac{SN}{SD}=\dfrac{2}{3};c=\dfrac{SP}{SC}.$ Ta có $\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{1}{b}+\dfrac{1}{x}\Rightarrow c=x.$
Lại có $\dfrac{{{V}_{1}}}{V}=\dfrac{abcx}{4}\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{x} \right)=\dfrac{{{x}^{2}}}{9}\left( 3+\dfrac{2}{x} \right).$
Mà $\dfrac{{{V}_{1}}}{V}=\dfrac{1}{2}\Rightarrow 6{{x}^{3}}+4{{x}^{2}}-9x=0\Leftrightarrow \left[ \begin{aligned}
& x=0\left( Loai \right) \\
& x=\dfrac{-2+\sqrt{58}}{6}\left( Nhan \right) \\
& x=\dfrac{-2-\sqrt{58}}{6}\left( Loai \right) \\
\end{aligned} \right..$
Vậy $x=\dfrac{-2+\sqrt{58}}{6}.$
Ta có ${{V}_{1}}={{V}_{S.MNPQ}}={{V}_{S.MNQ}}+{{V}_{S.PNQ}}$
Ta có $\left\{ \begin{aligned}
& \left( \alpha \right)\cap \left( SBC \right)=PQ \\
& MN//BC \\
& MN\subset \left( \alpha \right) \\
& BC\subset \left( SBC \right) \\
\end{aligned} \right.\Rightarrow PQ//MN//BC\Rightarrow \dfrac{SP}{SC}=\dfrac{SQ}{SB}=x.$
Có $\dfrac{{{V}_{S.MNQ}}}{{{V}_{S.ADB}}}=\dfrac{SM}{SA}.\dfrac{SN}{SD}.\dfrac{SQ}{SB}=\dfrac{2}{3}.\dfrac{2}{3}x=\dfrac{4}{9x}\Rightarrow {{V}_{S.MNQ}}=\dfrac{4x}{9}{{V}_{S.ADB}}=\dfrac{4x}{9}.\dfrac{V}{2}=\dfrac{2x}{9}V.$
Đồng thời $\dfrac{{{V}_{S.PNQ}}}{{{V}_{S.CDB}}}=\dfrac{SP}{SC}.\dfrac{SN}{SD}.\dfrac{SQ}{SB}=x.\dfrac{2}{3}.x=\dfrac{2{{x}^{2}}}{3}\Rightarrow {{V}_{S.PNQ}}=\dfrac{2{{x}^{2}}}{3}.{{V}_{S.CDB}}=\dfrac{2{{x}^{2}}}{3}.\dfrac{V}{2}=\dfrac{{{x}^{2}}}{3}V.$
Như vậy ${{V}_{1}}=\left( \dfrac{{{x}^{2}}}{3}+\dfrac{2x}{9} \right)V.$ Mà theo giả thiết ta có ${{V}_{1}}=\dfrac{1}{2}V$ nên ta suy ra:
$\dfrac{{{x}^{2}}}{3}+\dfrac{2x}{9}=\dfrac{1}{2}\Leftrightarrow \left[ \begin{aligned}
& x=\dfrac{-2+\sqrt{58}}{6}\left( Nhan \right) \\
& x=\dfrac{-2-\sqrt{58}}{6}\left( Loai \right) \\
\end{aligned} \right.. $ Vậy $ x=\dfrac{-2+\sqrt{58}}{6}.$
Cách 2:
Đặt $a=\dfrac{SM}{SA}=\dfrac{2}{3};b=\dfrac{SN}{SD}=\dfrac{2}{3};c=\dfrac{SP}{SC}.$ Ta có $\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{1}{b}+\dfrac{1}{x}\Rightarrow c=x.$
Lại có $\dfrac{{{V}_{1}}}{V}=\dfrac{abcx}{4}\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{x} \right)=\dfrac{{{x}^{2}}}{9}\left( 3+\dfrac{2}{x} \right).$
Mà $\dfrac{{{V}_{1}}}{V}=\dfrac{1}{2}\Rightarrow 6{{x}^{3}}+4{{x}^{2}}-9x=0\Leftrightarrow \left[ \begin{aligned}
& x=0\left( Loai \right) \\
& x=\dfrac{-2+\sqrt{58}}{6}\left( Nhan \right) \\
& x=\dfrac{-2-\sqrt{58}}{6}\left( Loai \right) \\
\end{aligned} \right..$
Vậy $x=\dfrac{-2+\sqrt{58}}{6}.$
Đáp án A.