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Câu hỏi: Cho hình lăng trụ $ABC.{A}'{B}'{C}'$ và M, N là hai điểm lần lượt bên cạnh CA, CB sao cho MN song song với AB và $\dfrac{CM}{CA}=k$. Mặt phẳng $\left( MN{B}'{A}' \right)$ chia khối lăng trụ $ABC.{A}'{B}'{C}'$ thành hai phần có thể tích ${{V}_{1}}$ (phần chứa điểm C) và ${{V}_{2}}$ sao cho $\dfrac{{{V}_{1}}}{{{V}_{2}}}=2$. Khi đó giá trị của k là:
A. $k=\dfrac{-1+\sqrt{5}}{2}$
B. $k=\dfrac{1}{2}$
C. $k=\dfrac{1+\sqrt{5}}{2}$
D. $k=\dfrac{\sqrt{3}}{3}$
A. $k=\dfrac{-1+\sqrt{5}}{2}$
B. $k=\dfrac{1}{2}$
C. $k=\dfrac{1+\sqrt{5}}{2}$
D. $k=\dfrac{\sqrt{3}}{3}$
Ta có: $\left\{ \begin{aligned}
& \left( MN{B}'{A}' \right)\cap \left( AC{C}'{A}' \right)={A}'M \\
& \left( MN{B}'{A}' \right)\cap \left( BC{C}'{B}' \right)={B}'N \\
& \left( AC{C}'{A}' \right)\cap \left( BC{C}'{B}' \right)=C{C}' \\
\end{aligned} \right.\Rightarrow {A}'M,{B}'N,C{C}'$ đồng quy tại S
Áp dụng định lí Ta-lét ta có: $\dfrac{SM}{S{A}'}=\dfrac{MN}{{A}'{B}'}=\dfrac{MN}{AB}=\dfrac{CM}{CA}=k=\dfrac{SN}{S{B}'}=\dfrac{SC}{S{C}'}$
$\Rightarrow \dfrac{{{V}_{S.MNC}}}{{{V}_{S.{A}'{B}'{C}'}}}=\dfrac{SM}{S{A}'}.\dfrac{SN}{S{B}'}.\dfrac{SC}{S{C}'}={{k}^{3}}\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{S.{A}'{B}'{C}'}}}=1-{{k}^{3}}\Rightarrow {{V}_{1}}=\left( 1-{{k}^{3}} \right){{V}_{S.{A}'{B}'{C}'}}$
Ta có: $\dfrac{SC}{S{C}'}=k\Rightarrow \dfrac{S{C}'-C{C}'}{S{C}'}=k\Leftrightarrow \dfrac{C{C}'}{S{C}'}=1-k$
$\dfrac{{{V}_{S.{A}'{B}'{C}'}}}{{{V}_{ABC.{A}'{B}'{C}'}}}=\dfrac{\dfrac{1}{3}S{C}'}{C{C}'}=\dfrac{1}{3\left( 1-k \right)}\Rightarrow {{V}_{S.{A}'{B}'{C}'}}=\dfrac{{{V}_{ABC.{A}'{B}'{C}'}}}{3\left( 1-k \right)}$
$\Rightarrow {{V}_{1}}=\left( 1-{{k}^{3}} \right){{V}_{S.{A}'{B}'{C}'}}=\left( 1-{{k}^{3}} \right)\dfrac{{{V}_{ABC.{A}'{B}'{C}'}}}{3\left( 1-k \right)}=\dfrac{1+k+{{k}^{2}}}{3}{{V}_{ABC.{A}'{B}'{C}'}}$
Ta có: $\dfrac{{{V}_{1}}}{{{V}_{2}}}=2\Leftrightarrow {{V}_{2}}=\dfrac{2}{3}{{V}_{ABC.{A}'{B}'{C}'}}\Rightarrow \dfrac{1+k+{{k}^{2}}}{3}=\dfrac{2}{3}\Leftrightarrow 1+k+{{k}^{2}}=2\Leftrightarrow k=\dfrac{\sqrt{5}-1}{2}$.
& \left( MN{B}'{A}' \right)\cap \left( AC{C}'{A}' \right)={A}'M \\
& \left( MN{B}'{A}' \right)\cap \left( BC{C}'{B}' \right)={B}'N \\
& \left( AC{C}'{A}' \right)\cap \left( BC{C}'{B}' \right)=C{C}' \\
\end{aligned} \right.\Rightarrow {A}'M,{B}'N,C{C}'$ đồng quy tại S
Áp dụng định lí Ta-lét ta có: $\dfrac{SM}{S{A}'}=\dfrac{MN}{{A}'{B}'}=\dfrac{MN}{AB}=\dfrac{CM}{CA}=k=\dfrac{SN}{S{B}'}=\dfrac{SC}{S{C}'}$
$\Rightarrow \dfrac{{{V}_{S.MNC}}}{{{V}_{S.{A}'{B}'{C}'}}}=\dfrac{SM}{S{A}'}.\dfrac{SN}{S{B}'}.\dfrac{SC}{S{C}'}={{k}^{3}}\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{S.{A}'{B}'{C}'}}}=1-{{k}^{3}}\Rightarrow {{V}_{1}}=\left( 1-{{k}^{3}} \right){{V}_{S.{A}'{B}'{C}'}}$
Ta có: $\dfrac{SC}{S{C}'}=k\Rightarrow \dfrac{S{C}'-C{C}'}{S{C}'}=k\Leftrightarrow \dfrac{C{C}'}{S{C}'}=1-k$
$\dfrac{{{V}_{S.{A}'{B}'{C}'}}}{{{V}_{ABC.{A}'{B}'{C}'}}}=\dfrac{\dfrac{1}{3}S{C}'}{C{C}'}=\dfrac{1}{3\left( 1-k \right)}\Rightarrow {{V}_{S.{A}'{B}'{C}'}}=\dfrac{{{V}_{ABC.{A}'{B}'{C}'}}}{3\left( 1-k \right)}$
$\Rightarrow {{V}_{1}}=\left( 1-{{k}^{3}} \right){{V}_{S.{A}'{B}'{C}'}}=\left( 1-{{k}^{3}} \right)\dfrac{{{V}_{ABC.{A}'{B}'{C}'}}}{3\left( 1-k \right)}=\dfrac{1+k+{{k}^{2}}}{3}{{V}_{ABC.{A}'{B}'{C}'}}$
Ta có: $\dfrac{{{V}_{1}}}{{{V}_{2}}}=2\Leftrightarrow {{V}_{2}}=\dfrac{2}{3}{{V}_{ABC.{A}'{B}'{C}'}}\Rightarrow \dfrac{1+k+{{k}^{2}}}{3}=\dfrac{2}{3}\Leftrightarrow 1+k+{{k}^{2}}=2\Leftrightarrow k=\dfrac{\sqrt{5}-1}{2}$.
Đáp án A.
