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Given a line $l$ through a tetrahedron $ABCD$ (not intersecting any of its edges), take the four points $P_1, P_2, P_3, P_4$ of intersection of the line with the faces of the tetrahedron. Also take the four planes $π_1, π_2, π_3, π_4$ determined by the line and the vertices of the tetrahedron respectively.

Show that the cross ratio of the four points is the same as the cross ratio of the four planes (when choosing a particular ordering), so $$ (P_1, P_2, P_3, P_4) = (π_1, π_2, π_3, π_4) $$

I think i already have an intuitive understanding of why this is true. Each vertice of the tetrahedron has exactly one opposite face. So i would assume as for the particular ordering the plane $π_1$ is that one that goes through the vertice that is opposite of the face which includes $P_1$ and so on. Now i think there could be a projective transformation somehow mapping the planes to the points, thus preserving the cross ratio. Unfortunately i have no idea how to show this actually exist. Was wondering if one could just take clever coordinate representatives for the points and show this using linear algebra, but am also stuck with the particular choice. Appreciate any help!

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Was wondering if one could just take clever coordinate representatives for the points and show this using linear algebra, but am also stuck with the particular choice.

The whole setup is invariant under projective transformations. So you can indeed pick very simple coordinates for the vertices, without loss of generality. Would suggest $$V_1=[1:0:0:0]\quad V_2=[0:1:0:0]\quad V_3=[0:0:1:0]\quad V_4=[0:0:0:1]$$ So the planes of the tetrahedron would be the coordinate planes and the plane at infinity, and the tetrahedron would in fact be the first octant.

Now in this space your line is e.g. spanned by two points $A,B$, i.e. the points on the line can be described as $\lambda_iA+\mu_iB$ and the cross ratio computed as

$$(P_1,P_2;P_3,P_4)=\frac {\begin{vmatrix}\lambda_1&\lambda_3\\\mu_1&\mu_3\end{vmatrix}\cdot \begin{vmatrix}\lambda_2&\lambda_4\\\mu_2&\mu_4\end{vmatrix}} {\begin{vmatrix}\lambda_1&\lambda_4\\\mu_1&\mu_4\end{vmatrix}\cdot \begin{vmatrix}\lambda_2&\lambda_3\\\mu_2&\mu_3\end{vmatrix}}$$

More precisely, you know that the first coordinate of $P_1$ must be zero (since it lies in the plane spanned by $V_2,V_3,V_4$) so pick $\lambda_1=B_1,\mu_1=-A_1$ with $A=[A_1:A_2:A_3:A_4]$ homogeneous coordinates of $A$ and likewise for $B$. Same for other points. So you get

$$(P_1,P_2;P_3,P_4)=\frac {\begin{vmatrix}B_1&B_3\\-A_1&-A_3\end{vmatrix}\cdot \begin{vmatrix}B_2&B_4\\-A_2&-A_4\end{vmatrix}} {\begin{vmatrix}B_1&B_4\\-A_1&-A_4\end{vmatrix}\cdot \begin{vmatrix}B_2&B_3\\-A_2&-A_3\end{vmatrix}}$$

I don't know how you defined the cross ratio of four planes sharing a line, but you should be able to express this in terms of $A$ and $B$ as well, making it accessible to a proof using linear algebra.

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  • $\begingroup$ thank you very much! I calculated the cross ratio of the four planes by first defining each plane $\pi_i$ by a parametric equation using points $A, B, V_i$. Then calculating the corresponding implicit equation. Next taking the coefficients of the implicit equation as the coordinates of a point in the dual space and finally calculating the cross ratio of the four points in the dual space. Maybe there is an easier/quicker way but this at least worked for me. $\endgroup$ – user526159 Jun 10 '18 at 10:38

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