The cross-ratio of four collinear points $A,B,C,D$ in the Euclidean Plane is defined by $$(A,B,C,D) = \frac{AC}{AD}\frac{BD}{BC}$$

And the wikipedia article states it already appeared in the works of Pappus of Alexandria. Since I can't find any geometrical interpretation for this ratio I am really curious to know how Pappus constructed it, and what his intuition was for it; as I understand, in his epoch constructions of that kind were only made using straightedge-and-compass.

  • $\begingroup$ The first picture in the Wikipedia article you refer on shows its geometric meaning. $\endgroup$ – Alexandre Eremenko Nov 21 '14 at 18:41
  • $\begingroup$ Geometrical in the sense of a measure of some constructible segment, of course it has meaning as the invariant under a certain group, but thats not the meaning I asked about. $\endgroup$ – Jonas Gomes Nov 21 '14 at 18:50

It appears in the demonstration of Pappus's hexagon theorem as a tool in the demonstration.

given one set of collinear points A, B, C, and another set of collinear points a, b, c, then the intersection points X, Y, Z of line pairs Ab and aB, Ac and aC, Bc and bC are collinear, lying on the Pappus line

(illustrations from the wikipedia article)

(Note: Pappus' proof is invalid in the case C,c and X are colinear, but the theorem is true nonetheless. )

enter image description here


Proposition 129 of Pappus' Book VII says that the projection from one line to another keeps the "cross ratio". Geometric interpretation is the hexagon theorem, geometric construction can be found possibly in von Staudt.

A nice lecture about Pappus and the cross ratio is on youtube.


To construct a segment equal to the cross-ratio, you slightly modify the first picture in the Wikipedia article. Let us call $P$ the point in the top where the 4 black lines cross.

Suppose you want to construct two segments whose ratio equals to the cross-ratio of 4 given points on a line. Referring to this picture, suppose that your given points are $A',B',C',D'$, and they are on the red line $L'$. Then draw a second red line L, and choose the point $P$ so that $PD'$ is parallel to $L$. (Or choose $P$ arbitrarily, and then choose $L$ parallel to $PD$). Then use this $P$ as the center of projection and project $A',B',C'$ on $L$ from $P$. The ratio $AC/BC$ will be equal to the cross ratio of $A',B',C',D'$.

To obtain the "numerical" value (the notion foreign to the Greek mathematics; after the discovery of non-commensurable segments they did not measure the segments with numbers; they were only speaking of proportions, see Euclid), you have to choose the unit of length. Suppose that $BC$ has length $1$, then $AC$ is the cross ratio.

I have not checked with Pappus, but I suppose he had some similar construction, if he had any.


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