Ta có \(\frac{{d\left( {O,\left( {BCNM} \right)} \right)}}{{d\left( {A,\left( {BCNM} \right)} \right)}} = \frac{{CO}}{{CA}} = \frac{1}{2}\) (do O là trung điểm AC).
\( \Rightarrow d\left( {O,\left( {BCNM} \right)} \right) = \frac{1}{2}d\left( {A,\left( {BCNM} \right)} \right)\).
Lại có \({S_{SMN}} = \frac{1}{2}SM.SN.\sin \widehat {MSN} = \frac{1}{8}SB.SC.\sin \widehat {MSN} = \frac{1}{4}{S_{SBC}}\).
Suy ra \({S_{BCNM}} = {S_{SBC}} - {S_{SMN}} = {S_{SBC}} - \frac{1}{4}{S_{SBC}} = \frac{3}{4}{S_{SBC}}\).
Ta có \({S_{ABC}} = \frac{1}{2}d\left( {A,CD} \right).CD\) và SABCD = d(A, CD).CD.
Suy ra \({S_{ABC}} = \frac{1}{2}{S_{ABCD}}\).
Vì vậy \({V_{O.BCNM}} = \frac{1}{3}d\left( {O,\left( {BCNM} \right)} \right).{S_{BCNM}} = \frac{1}{3}.\frac{1}{2}d\left( {A,\left( {BCNM} \right)} \right).\frac{3}{4}{S_{SBC}}\).
\( = \frac{3}{8}{V_{SABC}} = \frac{3}{8}.\frac{1}{3}d\left( {S,\left( {ABC} \right)} \right).{S_{ABC}} = \frac{3}{8}.\frac{1}{3}d\left( {S,\left( {ABCD} \right)} \right).\frac{1}{2}{S_{ABCD}} = \frac{3}{{16}}{V_{S.ABCD}}\).
Suy ra \(\frac{{{V_{O.BCNM}}}}{{{V_{S.ABCD}}}} = \frac{3}{{16}}\).
Vậy tỉ số thể tích giữa hai khối chóp O.BCNM và S.ABCD là \(\frac{3}{{16}}\).