Cho hình chóp S.ABCD có đáy là hình thoi cạnh bằng 2,

* Hướng dẫn giải

Gọi\[O = AC \cap BD\]ta có:

\[SA = SC \Rightarrow {\rm{\Delta }}SAC\]cân tại \[S \Rightarrow SO \bot AC\]

Tam giác SBD vuông cân tại\[S \Rightarrow SO \bot BD\]

\[ \Rightarrow SO \bot \left( {ABCD} \right)\]

Trong (SBD), gọi\[I = MN \cap BD\]

Đặt \[\frac{{SM}}{{SB}} = x,\,\,\frac{{SN}}{{SD}} = y\,\,(0 < x,\,\,y < 1)\]

Ta có:\[\frac{{{V_{S.AME}}}}{{{V_{S.ABC}}}} = \frac{{SM}}{{SB}}.\frac{{SE}}{{SC}} = \frac{1}{2}x \Rightarrow \frac{{{V_{S.AME}}}}{{{V_{S.ABCD}}}} = \frac{1}{4}x\]

\[\frac{{{V_{S.ANE}}}}{{{V_{S.ADC}}}} = \frac{{SN}}{{SD}}.\frac{{SE}}{{SC}} = \frac{1}{2}y \Rightarrow \frac{{{V_{S.ANE}}}}{{{V_{S.ABCD}}}} = \frac{1}{4}y\]

\[ \Rightarrow \frac{{{V_{S.AMNE}}}}{{{V_{S.ABCD}}}} = \frac{{{V_{S.AME}}}}{{{V_{S.ABCD}}}} + \frac{{{V_{S.ANE}}}}{{{V_{S.ABCD}}}} = \frac{{x + y}}{4}\,\,\,\left( 1 \right)\]

Ta lại có:\[\frac{{{V_{S.AMN}}}}{{{V_{S.ABD}}}} = \frac{{SM}}{{SA}}.\frac{{SN}}{{SD}} = xy \Rightarrow \frac{{{V_{S.AMN}}}}{{{V_{S.ABCD}}}} = \frac{{xy}}{2}\]

\[\frac{{{V_{S.MNE}}}}{{{V_{S.BDC}}}} = \frac{{SM}}{{SB}}.\frac{{SN}}{{SD}}.\frac{{SE}}{{SC}} = \frac{1}{2}xy \Rightarrow \frac{{{V_{S.MNE}}}}{{{V_{S.ABCC}}}} = \frac{{xy}}{4}\]

\[ \Rightarrow \frac{{{V_{S.AMNE}}}}{{{V_{S.ABCD}}}} = \frac{{{V_{S.AMN}}}}{{{V_{S.ABCD}}}} + \frac{{{V_{S.MNE}}}}{{{V_{S.ABCD}}}} = \frac{{xy}}{2} + \frac{{xy}}{4} = \frac{{3xy}}{4}\,\,\left( 2 \right)\]

Từ (1) và (2) \[ \Rightarrow \frac{{x + y}}{4} = \frac{{3xy}}{4} \Leftrightarrow x + y = 3xy\]

\[ \Leftrightarrow x = \left( {3x - 1} \right)y \Leftrightarrow y = \frac{x}{{3x - 1}}\,\,\left( {x \ne \frac{1}{3}} \right)\]

Do \[x,\,\,y > 0 \Rightarrow 3x - 1 > 0 \Leftrightarrow x > \frac{1}{3}\]

Khi đó ta có\[\frac{{{V_{S.AMNE}}}}{{{V_{S.ABCD}}}} = \frac{1}{4}\left( {x + \frac{x}{{3x - 1}}} \right)\]

Xét hàm số \[f\left( x \right) = x + \frac{x}{{3x - 1}}\,\,\left( {x > \frac{1}{3}} \right)\]ta có:

\[f'\left( x \right) = 1 - \frac{1}{{{{\left( {3x - 1} \right)}^2}}} = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x - 1 = 1}\\{3x - 1 = - 1}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{2}{3}}\\{x = 0\left( {ktm} \right)}\end{array}} \right.\]

BBT:

Dựa vào BBT ta thấy\[\min {V_{S.AMNE}} = \frac{1}{4}.\frac{4}{3}{V_{S.ABCD}} = \frac{1}{3}{V_{S.ABCD}}\]

\[ \Rightarrow \max {V_{ABCDNEM}} = \frac{2}{3}{V_{S.ABCD}} \Rightarrow {V_0} = \frac{2}{3}{V_{S.ABCD}}\]

Ta có: \[{\rm{\Delta }}ABD\]đều cạnh 2 \[\left( {AB = AD,\,\angle BAD = {{60}^0}} \right) \Rightarrow {S_{ABD}} = \frac{{{2^2}\sqrt 3 }}{4} = \sqrt 3 \]

\[ \Rightarrow {S_{ABCD}} = 2\sqrt 3 \]

Tam giác ABD đều cạnh 2 ⇒BD=2, lại có tam giác SBD vuông cân tại S nên

\[SO = \frac{1}{2}BD = 1\]

\[ \Rightarrow {V_{S.ABCD}} = \frac{1}{3}SO.{S_{ABCD}} = \frac{1}{3}.1.2\sqrt 3 = \frac{{2\sqrt 3 }}{3}\]

Vậy\[{V_0} = \frac{2}{3}{V_{S.ABCD}} = \frac{{4\sqrt 3 }}{9}\]Đáp án cần chọn là: D