Cho hình chóp `S.ABCD` với bốn điểm bất kì `M,N,P,Q` lần lượt thuộc các cạnh `SA,SB,SC,SD` chứng minh : $\dfrac{V_{S.MNPQ}}{V_{S.ABCD}}=\dfrac{\dfrac{SA}{SM}+\dfrac{SB}{SN}+\dfrac{SC}{SP}+\dfrac{SD}{SQ}}{4.\dfrac{SA}{SM}.\dfrac{SB}{SN}.\dfrac{SC}{SP}.\dfrac{SD}{SQ}}$

Đáp án:

$\dfrac{V_{S.MNPQ}}{V_{S.ABCD}}=\dfrac{\dfrac{SA}{SM}+\dfrac{SB}{SN}+\dfrac{SC}{SP}+\dfrac{SD}{SQ}}{4.\dfrac{SA}{SM}.\dfrac{SB}{SN}.\dfrac{SC}{SP}.\dfrac{SD}{SQ}}$

Giải thích các bước giải:

Ta chứng minh: $\dfrac{SA}{SM}+\dfrac{SC}{SP}=\dfrac{SB}{SN}+\dfrac{SD}{SQ}$

$\overrightarrow{SA}+\overrightarrow{SC}=\overrightarrow{SB}+\overrightarrow{SD}$

$⇒\dfrac{SA}{SM}.\overrightarrow{SM}+\dfrac{SC}{SP}.\overrightarrow{SP}=\dfrac{SB}{SN}.\overrightarrow{SN}+\dfrac{SD}{SQ}.\overrightarrow{SQ}$

Vì M, N, P, Q đồng phẳng nên: $\dfrac{SA}{SM}+\dfrac{SC}{SP}=\dfrac{SB}{SN}+\dfrac{SD}{SQ}$

Nhận xét: $V_{S.ABCD}=2.V_{S.ABD}=2.V_{S.CBD}$

Ta có: $\dfrac{V_{S.MNPQ}}{V_{SABCD}}=\dfrac{V_{S.MNQ}}{2V_{S.ABD}}+\dfrac{V_{S.PNQ}}{2V_{S.CBD}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{2}.\bigg(\dfrac{SM}{SA}.\dfrac{SN}{SB}.\dfrac{SQ}{SD}+\dfrac{SP}{SC}.\dfrac{SN}{SB}.\dfrac{SQ}{SD}\bigg)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{2}.\dfrac{SN}{SB}.\dfrac{SQ}{SD}.\bigg(\dfrac{SM}{SA}+\dfrac{SP}{SC}\bigg)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{2}.\dfrac{1}{\dfrac{SB}{SN}}.\dfrac{1}{\dfrac{SD}{SQ}}.\bigg(\dfrac{1}{\frac{SA}{SM}}+\dfrac{1}{\frac{SC}{SP}}\bigg)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{\dfrac{SA}{SM}+\dfrac{SC}{SP}}{2.\dfrac{SA}{SM}.\dfrac{SB}{SN}.\dfrac{SC}{SP}.\dfrac{SD}{SQ}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{\dfrac{SA}{SM}+\dfrac{SB}{SN}+\dfrac{SC}{SP}+\dfrac{SD}{SQ}}{4.\dfrac{SA}{SM}.\dfrac{SB}{SN}.\dfrac{SC}{SP}.\dfrac{SD}{SQ}}$

Vậy $\dfrac{V_{S.MNPQ}}{V_{S.ABCD}}=\dfrac{\dfrac{SA}{SM}+\dfrac{SB}{SN}+\dfrac{SC}{SP}+\dfrac{SD}{SQ}}{4.\dfrac{SA}{SM}.\dfrac{SB}{SN}.\dfrac{SC}{SP}.\dfrac{SD}{SQ}}$ (đpcm).