Phương pháp giải:

Lời giải chi tiết:

Gọi \(O = AC \cap BD\).

Lấy \(B' \in SB\), trong \(\left( {SAC} \right)\) gọi \(I = A'C' \cap SO\). Trong \(\left( {SBD} \right)\) kéo dài \(B'I\) cắt \(SD\) tại \(D'\).

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:

\(\dfrac{{A'S}}{{A'A}}.\dfrac{{EA}}{{EC}}.\dfrac{{C'C}}{{C'S}} = 1\) \( \Rightarrow \dfrac{1}{2}.\dfrac{{EA}}{{EC}}.4 = 1 \Leftrightarrow \dfrac{{EA}}{{EC}} = \dfrac{1}{2}\) \( \Rightarrow \dfrac{{EA}}{{EO}} = \dfrac{2}{3}\).

\(\dfrac{{A'S}}{{A'A}}.\dfrac{{EA}}{{EO}}.\dfrac{{IO}}{{IS}} = 1\) \( \Rightarrow \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{{IO}}{{IS}} = 1 \Leftrightarrow \dfrac{{IO}}{{IS}} = 3\).

\(\dfrac{{B'S}}{{B'B}}.\dfrac{{FB}}{{FO}}.\dfrac{{IO}}{{IS}} = 1\) \( \Rightarrow \dfrac{x}{{1 - x}}.\dfrac{{FB}}{{FO}}.3 = 1 \Leftrightarrow \dfrac{{FB}}{{FO}} = \dfrac{{1 - x}}{{3x}}\)\( \Leftrightarrow \dfrac{{FO + OB}}{{FO}} = \dfrac{{1 - x}}{{3x}}\).

\(\dfrac{{D'S}}{{D'D}}.\dfrac{{FD}}{{FO}}.\dfrac{{IO}}{{IS}} = 1\) \( \Rightarrow \dfrac{y}{{1 - y}}.\dfrac{{FD}}{{FO}}.3 = 1 \Leftrightarrow \dfrac{{FD}}{{FO}} = \dfrac{{1 - y}}{{3y}}\) \( \Leftrightarrow \dfrac{{FO - OD}}{{FO}} = \dfrac{{1 - y}}{{3y}}\).

\(\begin{array}{l} \Rightarrow \dfrac{{FO + OB}}{{FO}} + \dfrac{{FO - OD}}{{FO}} = \dfrac{{1 - x}}{{3x}} + \dfrac{{1 - y}}{{3y}}\\ \Rightarrow 2 = \dfrac{{1 - x}}{{3x}} + \dfrac{{1 - y}}{{3y}} \Leftrightarrow 2 = \dfrac{{y\left( {1 - x} \right) + x\left( {1 - y} \right)}}{{3xy}}\\ \Leftrightarrow 6xy = y - xy + x - xy = x + y - 2xy\\ \Leftrightarrow x + y = 8xy\,\,\,\left( * \right) \Leftrightarrow \dfrac{1}{x} + \dfrac{1}{y} = 8\end{array}\)

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:

\(\dfrac{1}{y} + \dfrac{1}{y} \le \dfrac{1}{x} + \dfrac{1}{y} = 8 \Leftrightarrow \dfrac{2}{y} \le 8 \Leftrightarrow y \ge \dfrac{1}{4}\).

Ta có: \(\dfrac{{{V_{S.A'B'D'}}}}{{{V_{S.ABD}}}} = \dfrac{{SA'}}{{SA}}.\dfrac{{SB'}}{{SB}}.\dfrac{{SD'}}{{SD}} = \dfrac{1}{3}.x.y = \dfrac{{xy}}{3}\) \( \Rightarrow {V_{S.A'B'D'}} = \dfrac{{xy}}{6}{V_{S.ABCD}}\).

          \(\dfrac{{{V_{S.B'C'D'}}}}{{{V_{S.BCD}}}} = \dfrac{{SB'}}{{SB}}.\dfrac{{SC'}}{{SC}}.\dfrac{{SD'}}{{SD}} = \dfrac{{xy}}{5}\) \( \Rightarrow {V_{S.MNP}} = \dfrac{{xy}}{{10}}{V_{S.ABCD}}\).

\( \Rightarrow \dfrac{{{V_{SMNPQ}}}}{{{V_{S.ABCD}}}} = \dfrac{{xy}}{6} + \dfrac{{xy}}{{10}} = \dfrac{{4xy}}{{15}} = k\).

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:

\(f'\left( y \right) = \dfrac{{2y\left( {8y - 1} \right) - 8{y^2}}}{{{{\left( {8y - 1} \right)}^2}}} = \dfrac{{8{y^2} - 2y}}{{{{\left( {6y - 1} \right)}^2}}}\) ; \(f'\left( y \right) = 0 \Leftrightarrow 8{y^2} - 2y = 0 \Leftrightarrow \left[ \begin{array}{l}y = 0\\y = \dfrac{1}{4}\end{array} \right.\)

BBT:

Vậy \({k_{\min }} = \dfrac{4}{{15}}.\dfrac{1}{{16}} = \dfrac{1}{{60}}\).

Chọn C.