Given the tetrahedron $ ABCD $ whose edges $ AB $ and $ CD $ have lengths $ a $ and $ b $ respectively. The distance between the skew lines $ AB $ and $ CD $ is $ d $, and the angle between them is $ \omega $. Tetrahedron $ ABCD $ is divided into two solids by plane $ \epsilon $, parallel to lines $ AB $ and $ CD $. The ratio of the distances of $ \epsilon $ from $ AB $ and $ CD $ is equal to $ k $. Compute the ratio of the volumes of the two solids obtained. Construction. Through the point $ B $, we draw the line $ n $, $ n \parallel CD $, see picture 1. Through the point $ C $, we draw the line $ m $, $ m \parallel AB $. Then, plane $ (AB,n)$ $ \parallel $ plane $ (CD,m) $. There are many lines, perpendicular to plane $ (AB,n)$ and plane $ (CD,m) $, there is only one line, perpendicular to $ AB $ and $ CD $, who is also crossing $ AB $ and $ CD $. Suppose that $ EF $ is that special perpendicular. On $ EF $, we put the point $ M $, with the condition: \[\frac{\left|ME\right|}{\left|MF\right|}=k\] On the picture, that ratio $ k=1 $. Through the point $ M $, we construct the plane $ \epsilon $, parallel to lines $ AB $ and $ CD $, hence $ \epsilon \parallel (AB,n) \parallel (CD,m) $. That plane $ \epsilon $ is crossing $ AD $ in the point $ P $, $ AC $ in the point $ Q $, $ BC $ in the point $ R $ and $ BD $ in the point $ S $. The points $ P,Q,R,S $ divide $ AD \ , AC \ , BC \ , BD $ with the same ratio $ k $. These points $ P,Q,R,S $ form a parallelogram and: \[AB \parallel PS \parallel QR \ \ ,\ \ CD \parallel PQ \parallel SR \] The plane $ \epsilon $ divide our tetrahedron $ ABCD $ into two solids, two truncated prism $ (PS,QR,AB) $ (green on the picture 1 and rotated in picture 2) and $ (PQ,RS,CD) $ (red on the picture 1). Calculation. An old theorem says: the volume of a truncated prism equals one third of the product of two factors: first factor is the sum of the lenght of the lateral edges, second factor is the surface of the perpendicular cross-section of these lateral edges. If \[\frac{\left|ME\right|}{\left|MF\right|}=k\] then \[\frac{\left|EF\right|}{\left|MF\right|}=k+1\] or \[\left|EF\right|=(k+1)\left|MF\right|\] In $\triangle ABD$ and $\triangle ABC$, on the same way: \[\left|AB\right|=(k+1)\left|PS\right| \]and \[\left|AB\right|=(k+1)\left|QR\right|\] The sum of the lenght of the lateral edges in the truncated prism $ (PS,QR,AB) $: \[\left|AB\right|+\left|PS\right|+\left|QR\right|=\left|AB\right| \cdot (1+\frac{2}{k+1})\] In $\triangle ACD$ and $\triangle BCD$: \[\left|PQ\right|=\frac{k}{k+1}\left|CD\right| \]and \[\left|SR\right|=\frac{k}{k+1}\left|CD\right|\] The sum of the lenght of the lateral edges in the truncated prism $ (PQ,RS,CD) $: \[\left|CD\right|+\left|PQ\right|+\left|RS\right|=\left|CD\right| \cdot (1+\frac{2k}{k+1})\] Now the surface of the perpendicular cross-section of the truncated prism $ (PS,QR,AB) $ and $ (PQ,RS,CD) $; the cross-section of the truncated prism $ (PS,QR,AB) $ is perpendicular to $ AB $ and the cross-section of the truncated prism the cross-section of the truncated prism is perpendicular to $ CD $; the angle between them is $ \omega $; the ratio of these surfaces contains $ \cos \omega $. Because $\frac{\left|ME\right|}{\left|MF\right|}=k$, this ratio contains also $ k^{2}$. The surface of the perpendicular cross-section of the truncated prism $ (PS,QR,AB) = k^{2} \cdot \cos \omega \cdot T $ ; $ T $ is the surface of the perpendicular cross-section of the truncated prism $ (PQ,RS,CD) $. Conclusion. The volume of the truncated prism $ (PS,QR,AB)= \frac{1}{3}\left|AB\right|\frac{k+3}{k+1} \cdot k^{2} \cdot \cos \omega \cdot T $. The volume of the truncated prism $ (PQ,RS,CD)= \frac{1}{3}\left|CD\right|\frac{3k+1}{k+1} \cdot T $. The ratio of the volumes is: \[ \frac{a}{b} \cdot k^{2} \frac{k+3}{3k+1}\cdot \cos \omega \]

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