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Given 4 parallel lines $d_1$, $d_2$, $d_3$, $d_4$ , no more than 2 of which can be on a same plane. Plane (P) intersects the 4 lines at 4 points A, B, C, D. Plane (Q) ( not identical to plane (P)) intersects the 4 lines at $A_1$, $B_1$, $C_1$, $D_1$. Proof that the volumes of the 2 tetrahedra ABC $D_1$ and $A_1$$B_1$$C_1$$D$ are equal.

Nguyen Xuan Minh
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  • $ABCD$ and $A_1B_1C_1D_1$ are not tetrahedra, they are quadrilaterals. – Shubham Johri Dec 09 '18 at 06:24
  • Made a mistake, It's fixed now, thank you – Nguyen Xuan Minh Dec 09 '18 at 06:32
  • Counter-example, take $d_1, d_2, d_3, d_4$ be the lines joining origin and $A = (1,0,0)$, $B = (0,1,0)$, $C = (0,0,1)$, $D = (\frac13,\frac13,\frac13)$ respectively. Let $P, Q$ be the planes $x+y+z = 1$ and $x+y+z = 2$. It is clear $A, B, C, D$ are the intersections of lines $d_i$ with $P$ and $A_1 = 2A$, $B_1 = 2B$, $C_1 = 2C$, $D_1 = 2D$ are the intersections with $Q$. However, $${\rm Vol}(A_1B_1C_1D) = 4 {\rm Vol}(ABCD_1) \ne {\rm Vol}(ABCD_1)$$ – achille hui Dec 12 '18 at 05:10
  • I think you need conditions such as that $A$ and $A_1$ are on $d_1,$ $B$ and $B_1$ are on $d_2,$ etc. Inserting the word "respectively" in two places would do this. Otherwise it is still easy to come up with counterexamples. – David K Dec 14 '18 at 14:20

2 Answers2

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Let $(P)||(Q).$

Thus, we need to prove that areas of $\Delta ABC$ and $\Delta A_1B_1C_1$ they are equal,

which is obviously wrong: move $(Q)$ such that $(Q)||(P).$

Michael Rozenberg
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It is not explicitly stated that one of the given four parallel lines passes through both $D$ and $D_1,$ but I will assume this is given since otherwise the two tetrahedra might have different volumes.

If you know that a shear transformation of three-dimensional Euclidean space preserves volume, you can construct a plane $P'$ perpendicular to $d_1$ (hence also perpendicular to the other three given lines) and apply a shear transformation parallel to $d_1$ that maps $P$ to $P'.$ This transformation maps $A,$ $B,$ and $C$ to points $A',$ $B',$ and $C'$ in plane $P'$, and the tetrahedron $ABCD_1$ has volume equal to $A'B'C'D_1,$ which has base $\triangle A'B'C'$ and height $DD_1.$

Likewise you can construct a plane $Q'$ perpendicular to $d_1$ and apply a shear transformation parallel to $d_1$ that maps $Q$ to $Q',$ and in particular maps $A_1,$ $B_1,$ and $C_1$ to $A_1',$ $B_1',$ and $C_1'.$ Then the tetrahedron $A_1B_1C_1D$ has volume equal to $A_1'B_1'C_1'D,$ which has base $\triangle A_1'B_1'C_1'$ and height $DD_1.$ But since $\triangle A_1'B_1'C_1'$ is congruent to $\triangle A'B'C',$ the tetrahedron $\triangle A_1'B_1'C_1'D$ has the same volume as $A'B'C'D_1,$ and hence all four tetrahedra have the same volume.

David K
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