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Câu hỏi: Cho hình chóp tứ giác $S.ABCD$ có đáy ABCDlà hình bình hành. Các điểm $A',C'$ thỏa mãn $\overrightarrow{SA'}=\dfrac{1}{3}\overrightarrow{SA},\overrightarrow{SC}'=\dfrac{1}{5}\overrightarrow{SC}$ Mặt phẳng $\left( P \right)$ chứa đường thẳng $A'C'$ cắt các cạnh SB, SDlần lượt tại $B',D'$ và đặt $k=\dfrac{{{V}_{S.A'B'C'D'}}}{{{V}_{S.ABCD}}}$ . Giá trị nhỏ nhất của klà:
A. $\dfrac{\sqrt{15}}{16}$
B. $\dfrac{4}{15}$
C. $\dfrac{1}{60}$
D. $\dfrac{1}{30}$
A. $\dfrac{\sqrt{15}}{16}$
B. $\dfrac{4}{15}$
C. $\dfrac{1}{60}$
D. $\dfrac{1}{30}$
Cách giải:
Gọi O= AC⋂ BD.
Lấy B'∈ SB, trong ( SAC) gọi $I=A'C'\cap SO$. Trong ( SBD) kéo dài $B'I$ cắt SDtại $D'$.
Đặt $\dfrac{SB'}{SB}=x,\dfrac{SD'}{SD}=y$ (Giả sử $0<x<y\le 1$ ).
Gọi $A'C'\cap AC=E,B'D'\cap BD=F$
Áp dụng định lí Menelaus ta có:
$\dfrac{A'S}{A'A}.\dfrac{EA}{EC}.\dfrac{C'C}{C'S}=1\Rightarrow \dfrac{1}{2}.\dfrac{EA}{EC}.4=1\Leftrightarrow \dfrac{EA}{EC}=\dfrac{1}{2}\Rightarrow \dfrac{EA}{EO}=\dfrac{2}{3}$
$\dfrac{A'S}{A'A}.\dfrac{EA}{EO}.\dfrac{IO}{IS}=1\Rightarrow \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{IO}{IS}=1\Leftrightarrow \dfrac{IO}{IS}=3$
$\dfrac{B'S}{B'B}.\dfrac{FB}{FO}.\dfrac{IO}{IS}=1\Rightarrow \dfrac{x}{1-x}.\dfrac{FB}{FO}.3=1\Leftrightarrow \dfrac{FB}{FO}=\dfrac{1-x}{3x}\Leftrightarrow \dfrac{FO+OB}{FO}=\dfrac{1-x}{3x}$
$\dfrac{D'S}{D'D}.\dfrac{FD}{FO}.\dfrac{IO}{IS}=1\Rightarrow \dfrac{y}{1-y}.\dfrac{FD}{FO}.3=1\Leftrightarrow \dfrac{FD}{FO}=\dfrac{1-y}{3y}\Leftrightarrow \dfrac{FO-OD}{FO}=\dfrac{1-y}{3y}$
$\Rightarrow \dfrac{FO+OB}{FO}+\dfrac{FO-OD}{FO}=\dfrac{1-x}{3x}+\dfrac{1-y}{3y}$
$\Rightarrow 2=\dfrac{1-x}{3x}+\dfrac{1-y}{3y}\Leftrightarrow 2=\dfrac{y\left( 1-x \right)+x\left( 1-y \right)}{3xy}$
$\begin{aligned}
& \Leftrightarrow 6xy=y-xy+x-xy=x+y-2xy \\
& \Leftrightarrow x+y=8xy\left( * \right)\Leftrightarrow \dfrac{1}{x}+\dfrac{1}{y}=8 \\
\end{aligned}$
Không mất tính tổng quát, ta giả sử $0<x\le y\le 1\Leftrightarrow \dfrac{1}{x}\ge \dfrac{1}{y}$ , khi đó ta có:
$\dfrac{1}{y}+\dfrac{1}{y}y\le \dfrac{1}{x}+\dfrac{1}{y}=8\Leftrightarrow \dfrac{2}{y}\le 8\Leftrightarrow y\ge \dfrac{1}{4}$
Ta có: $\dfrac{{{V}_{S.A'B'D'}}}{{{V}_{S.ABD}}}=\dfrac{SA'}{SA}.\dfrac{SB'}{SB}.\dfrac{SD'}{SD}=\dfrac{1}{3}.x.y=\dfrac{xy}{3}\Rightarrow {{V}_{S.A'B'D'}}=\dfrac{xy}{6}{{V}_{ABCD}}$
$\dfrac{{{V}_{S.A'B'D'}}}{{{V}_{S.ABD}}}=\dfrac{SB'}{SB}.\dfrac{SC'}{SC}.\dfrac{SD'}{SD}=\dfrac{xy}{5}\Rightarrow {{V}_{S.MNP}}=\dfrac{xy}{10}{{V}_{S.ABCD}}$
$\Rightarrow \dfrac{{{V}_{SMNPQ}}}{{{V}_{S.ABCD}}}=\dfrac{xy}{6}+\dfrac{xy}{10}=\dfrac{4xy}{15}=k$
Từ (*) ta có: $x\left( 8y-1 \right)=y\Leftrightarrow x=\dfrac{y}{8y-1}\left( y\ge \dfrac{1}{4} \right)\Rightarrow k=\dfrac{4}{15}\dfrac{{{y}^{2}}}{8y-1}$
Xét hàm số $f\left( x \right)~=\dfrac{{{y}^{2}}}{8y-1}$, với $y\ge \dfrac{1}{4}$ ta có:
$f'\left( y \right)=\dfrac{2y\left( 8y-1 \right)-8{{y}^{2}}}{{{\left( 8y-1 \right)}^{2}}}=\dfrac{8{{y}^{2}}-2y}{{{\left( 6y-1 \right)}^{2}}};f'\left( y \right)=0\Leftrightarrow 8{{y}^{2}}-2y=0\Leftrightarrow \left[ \begin{aligned}
& y=0 \\
& y=\dfrac{1}{4} \\
\end{aligned} \right.$
BBT:
Vậy ${{k}_{\min }}=\dfrac{4}{15}.\dfrac{1}{16}=\dfrac{1}{60}$
Gọi O= AC⋂ BD.
Lấy B'∈ SB, trong ( SAC) gọi $I=A'C'\cap SO$. Trong ( SBD) kéo dài $B'I$ cắt SDtại $D'$.
Đặt $\dfrac{SB'}{SB}=x,\dfrac{SD'}{SD}=y$ (Giả sử $0<x<y\le 1$ ).
Gọi $A'C'\cap AC=E,B'D'\cap BD=F$
Áp dụng định lí Menelaus ta có:
$\dfrac{A'S}{A'A}.\dfrac{EA}{EC}.\dfrac{C'C}{C'S}=1\Rightarrow \dfrac{1}{2}.\dfrac{EA}{EC}.4=1\Leftrightarrow \dfrac{EA}{EC}=\dfrac{1}{2}\Rightarrow \dfrac{EA}{EO}=\dfrac{2}{3}$
$\dfrac{A'S}{A'A}.\dfrac{EA}{EO}.\dfrac{IO}{IS}=1\Rightarrow \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{IO}{IS}=1\Leftrightarrow \dfrac{IO}{IS}=3$
$\dfrac{B'S}{B'B}.\dfrac{FB}{FO}.\dfrac{IO}{IS}=1\Rightarrow \dfrac{x}{1-x}.\dfrac{FB}{FO}.3=1\Leftrightarrow \dfrac{FB}{FO}=\dfrac{1-x}{3x}\Leftrightarrow \dfrac{FO+OB}{FO}=\dfrac{1-x}{3x}$
$\dfrac{D'S}{D'D}.\dfrac{FD}{FO}.\dfrac{IO}{IS}=1\Rightarrow \dfrac{y}{1-y}.\dfrac{FD}{FO}.3=1\Leftrightarrow \dfrac{FD}{FO}=\dfrac{1-y}{3y}\Leftrightarrow \dfrac{FO-OD}{FO}=\dfrac{1-y}{3y}$
$\Rightarrow \dfrac{FO+OB}{FO}+\dfrac{FO-OD}{FO}=\dfrac{1-x}{3x}+\dfrac{1-y}{3y}$
$\Rightarrow 2=\dfrac{1-x}{3x}+\dfrac{1-y}{3y}\Leftrightarrow 2=\dfrac{y\left( 1-x \right)+x\left( 1-y \right)}{3xy}$
$\begin{aligned}
& \Leftrightarrow 6xy=y-xy+x-xy=x+y-2xy \\
& \Leftrightarrow x+y=8xy\left( * \right)\Leftrightarrow \dfrac{1}{x}+\dfrac{1}{y}=8 \\
\end{aligned}$
Không mất tính tổng quát, ta giả sử $0<x\le y\le 1\Leftrightarrow \dfrac{1}{x}\ge \dfrac{1}{y}$ , khi đó ta có:
$\dfrac{1}{y}+\dfrac{1}{y}y\le \dfrac{1}{x}+\dfrac{1}{y}=8\Leftrightarrow \dfrac{2}{y}\le 8\Leftrightarrow y\ge \dfrac{1}{4}$
Ta có: $\dfrac{{{V}_{S.A'B'D'}}}{{{V}_{S.ABD}}}=\dfrac{SA'}{SA}.\dfrac{SB'}{SB}.\dfrac{SD'}{SD}=\dfrac{1}{3}.x.y=\dfrac{xy}{3}\Rightarrow {{V}_{S.A'B'D'}}=\dfrac{xy}{6}{{V}_{ABCD}}$
$\dfrac{{{V}_{S.A'B'D'}}}{{{V}_{S.ABD}}}=\dfrac{SB'}{SB}.\dfrac{SC'}{SC}.\dfrac{SD'}{SD}=\dfrac{xy}{5}\Rightarrow {{V}_{S.MNP}}=\dfrac{xy}{10}{{V}_{S.ABCD}}$
$\Rightarrow \dfrac{{{V}_{SMNPQ}}}{{{V}_{S.ABCD}}}=\dfrac{xy}{6}+\dfrac{xy}{10}=\dfrac{4xy}{15}=k$
Từ (*) ta có: $x\left( 8y-1 \right)=y\Leftrightarrow x=\dfrac{y}{8y-1}\left( y\ge \dfrac{1}{4} \right)\Rightarrow k=\dfrac{4}{15}\dfrac{{{y}^{2}}}{8y-1}$
Xét hàm số $f\left( x \right)~=\dfrac{{{y}^{2}}}{8y-1}$, với $y\ge \dfrac{1}{4}$ ta có:
$f'\left( y \right)=\dfrac{2y\left( 8y-1 \right)-8{{y}^{2}}}{{{\left( 8y-1 \right)}^{2}}}=\dfrac{8{{y}^{2}}-2y}{{{\left( 6y-1 \right)}^{2}}};f'\left( y \right)=0\Leftrightarrow 8{{y}^{2}}-2y=0\Leftrightarrow \left[ \begin{aligned}
& y=0 \\
& y=\dfrac{1}{4} \\
\end{aligned} \right.$
BBT:
Vậy ${{k}_{\min }}=\dfrac{4}{15}.\dfrac{1}{16}=\dfrac{1}{60}$
Đáp án C.