Cho tứ diện S.ABC. Trên cạnh SA, SB lần lượt lấy các điểm M, N sao...

Lời giải:

Đặt ${{V}_{S.ABC}}=V.$
Dựng $NP//SC(P\in BC),MQ//SC(Q\in AC)$
Ta có: $CP=2PB,QC=QA\Rightarrow {{S}_{QBP}}=\dfrac{1}{2}d(Q;BC).BP$
$=\dfrac{1}{2}.\left( \dfrac{1}{2}d(A;BC).\dfrac{1}{3}BC \right)=\dfrac{1}{6}{{S}_{ABC}}\Rightarrow \dfrac{{{V}_{S.QBP}}}{V}=\dfrac{1}{6}$
Tương tự ${{S}_{BAQ}}=\dfrac{1}{2}{{S}_{ABC}}\Rightarrow \dfrac{{{V}_{S.ABQ}}}{V}=\dfrac{1}{2}.$
Lại có: $\dfrac{{{V}_{S.QNP}}}{{{V}_{S.QBP}}}=\dfrac{SN}{SB}=\dfrac{2}{3}\Rightarrow {{V}_{S.QNP}}=\dfrac{2}{3}{{V}_{S.QBP}}=\dfrac{1}{9}V.$
$\dfrac{{{V}_{S.MNQ}}}{{{V}_{S.ABQ}}}=\dfrac{SM}{SA}.\dfrac{SN}{SB}=\dfrac{1}{2}.\dfrac{2}{3}=\dfrac{1}{3}\Rightarrow {{V}_{S.MNQ}}=\dfrac{1}{3}{{V}_{S.ABQ}}=\dfrac{1}{6}V.$
Mặt khác $\dfrac{{{V}_{S.CPQ}}}{V}=\dfrac{{{S}_{CPQ}}}{{{S}_{ABC}}}=\dfrac{CQ}{CA}.\dfrac{CP}{CB}=\dfrac{1}{2}.\dfrac{2}{3}=\dfrac{1}{3}\Rightarrow {{V}_{S.CPQ}}=\dfrac{V}{3}.$
Do đó ${{V}_{SCMNPQ}}=\dfrac{V}{9}+\dfrac{V}{6}+\dfrac{V}{3}=\dfrac{11}{18}V\Rightarrow \left\{ \begin{aligned}
& {{V}_{2}}=\dfrac{11}{18}V \\
& {{V}_{1}}=\dfrac{7}{18}V \\
\end{aligned} \right.\Rightarrow \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{7}{11}.$