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Let's suppose we have eight points $A,B,C,X,A',B',C',X'$ on the same projective line such that $(ABCX) = (A'B'C'X')$. If the first seven points are known, how could I find the eighth point $X'$?

David
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1 Answers1

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Name the given line $L'$. Introduce a point $P$ not on $L'$ and a different line $L''$ not containing $P$ nor any of the other given points.

Define $A'',B'',C'',X''$ as points of intersection of the lines $\overline{AP},\overline{BP},\overline{CP},\overline{XP}$ with $L''$. By construction, these new points have the same cross-ratio as $A,B,C,X$.

Then use this thread to construct $X'$ on $L'$ such that $X''=A''$ would result in $X'=A'$ etc and that $A',B',C',X'$ have the same cross-ratio as $A'',B'',C'',X''$.

ccorn
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