Chọn D
Áp dụng định lý Menelaus vào tam giác: SOD
B,I,P thẳng hàng \(\Leftrightarrow \frac{PD}{PS} .\frac{IS}{IO} .\frac{BO}{BD} =1\Leftrightarrow PD=4PS\Leftrightarrow \frac{SD}{SP} =5.\)
Đặt:\( \frac{SA}{SM} =a;{\rm \; }\frac{SB}{SB} =b=1;{\rm \; }\frac{SC}{SN} =c;{\rm \; }\frac{SD}{SP} =d=5\)
Ta có \(a+c=b+d=2\frac{SO}{SI} =2.3=6\Rightarrow \left\{\begin{array}{l} {a+c=6} \\ {d=5} \end{array}\right.
\[\frac{V_{S.BMPN} }{V_{S.ABCD} } =\frac{a+b+c+d}{4a.b.c.d} =\frac{12}{20a.c} =\frac{3}{5\left(-a^{2} +6a\right)} \] \)
Xét \(f\left(a\right)=-a^{2} +6a{\rm \; \; \; }\left(1\le a\le 5\right),ta có: f'\left(a\right)=-2a+6;{\rm \; }f'\left(a\right)=0\Leftrightarrow a=3{\rm \; }\left(N\right).\)
Vì hàm số \(f\left(a\right)\) liên tục trên đoạn \(\left[1;5\right]\) nên:
\(\[\begin{array}{l} {\mathop{\min }\limits_{\left[1;5\right]}^{} f\left(a\right)=f\left(1\right)=f\left(5\right)=5\Rightarrow m=\frac{3}{25} } \\ {\mathop{\max }\limits_{\left[1;5\right]}^{} f\left(a\right)=f\left(3\right)=9\Rightarrow n=\frac{1}{15} } \\ {\Rightarrow m+n=\frac{14}{75} .} \end{array}\] \)