Cho hình chóp S.ABCD có đáy là hình bình hành và có thể tích là V

* Hướng dẫn giải

Giả sử $\overrightarrow{SD}=m\overrightarrow{SM},\overrightarrow{SB}=n\overrightarrow{SN}$

Ta có $\overrightarrow{SA}+\overrightarrow{SC}=\overrightarrow{SB}+\overrightarrow{SD}\left(=2\overrightarrow{SI}\right)$

Vì $A,M,N,P$ đồng phẳng nên tồn tại các số x;y sao cho $\overrightarrow{AP}=x\overrightarrow{AM}+y\overrightarrow{AN}$

$\iff \frac{1}{2}\left(\overrightarrow{AS}+\overrightarrow{AC}\right)=x\left(\overrightarrow{AS}+\overrightarrow{SM}\right)+y\left(\overrightarrow{AS}+\overrightarrow{SN}\right)$

$\iff \frac{1}{2}\left(\overrightarrow{AS}+\overrightarrow{AS}+\overrightarrow{SB}+\overrightarrow{AS}+\overrightarrow{SD}\right)=x\left(\overrightarrow{AS}+\overrightarrow{SM}\right)+y\left(\overrightarrow{AS}+\overrightarrow{SN}\right)$

$\iff \frac{3}{2}\overrightarrow{AS}+\frac{1}{2}\overrightarrow{SB}+\frac{1}{2}\overrightarrow{SD}=\left(x+y\right)\overrightarrow{AS}+\frac{x}{m}\overrightarrow{SM}+\frac{y}{n}\overrightarrow{SN}$

$\iff \left\{\begin{array}{l}x+y=\frac{3}{2}\\ \frac{x}{m}=\frac{1}{2}\\ \frac{y}{n}=\frac{1}{2}\end{array}\right.\Rightarrow m+n=3.$

 Ta có: $\frac{V_{S.ANP}}{V_{S.ABC}}=\frac{SN}{SB}.\frac{SP}{SC}\Rightarrow V_{S.ANP}=\frac{SN}{SB}.\frac{SP}{SC}.V_{S.ABC}=\frac{SN}{SB}.\frac{1}{2}.\frac{V}{2}\left(1\right)$

$\frac{V_{S.AMP}}{V_{S.ADC}}=\frac{SM}{SD}.\frac{SP}{SC}\Rightarrow V_{S.AMP}=\frac{SM}{SD}.\frac{SP}{SC}.V_{S.ADC}=\frac{SM}{SD}.\frac{1}{2}.\frac{V}{2}\left(2\right)$

Từ (1) và (2) $\frac{V_1}{V_2}=\frac{1}{4}\left(\frac{SB}{SB}+\frac{SM}{SD}\right)=\frac{1}{4}\left(\frac{1}{n}+\frac{1}{m}\right)\ge \frac{1}{m+n}=\frac{1}{3}$