References for a formula generalizing the "cross ratio = -1" characterization of harmonic division.

Question

It is well known that :

$$\ \ \ \ \ \ \binom{(A,C;B,D)}{\text{harmonic division}}\ \iff \ \binom{\text{cross-ratio}}{ [A,C;B,D]=-1}.$$

(definition of cross-ratio here).

A non-classical formula in the case where $A,B,C,D$ are aligned in this order, expresses the general (negative) cross-ratio under the form :

$$[A,C;B,D]=-\frac{1}{\tan^2(\tfrac{\theta}{2})}\tag{1}$$

as a function of the angle $\theta$ between circles with resp. diameters $[AC] $ and $[BD]$.

Fig. 1 : Case $a=0,b=2,c=3,d=7$, one gets $[A,C;B,D]=-1/\tan^2(93.82°/2)$.

I provide two proofs of formula (1) below.

I hadn't seen previously this formula until I found it, under a different form (up to Cayley transform $Z=\frac{z-i}{z+i}$) in a recent question to which I haven't answered directly because there was no evidence of work. Unlike this previous question, I do not make reference to hyperbolic geometry.

My questions are : where can be found references to this formula in the literature on projective geometry ? To which other properties can it be connected ?

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Proofs of formula (1) :

Proof 1 : (found afterwards, but placed first because it is simpler)

This issue is (classicaly) equivalent to this one (see figure 2), where the coordinates of $A,B,D$ are $0,1,\infty$ resp. More precisely, we keep the left circle as it was and we transform the right circle still making an angle $\theta$ with the first one into a vertical line in $C$, ("circle with an infinite radius").

Fig. 2.

In this way $\angle CEI = \theta$ (mutually orthogonal sides) ; therefore angle $\angle CAI = \theta/2$ (inscribed angle theorem).

As a consequence

$$AH=\tfrac12 \cos \tfrac{\theta}{2} \implies AI=\cos \tfrac{\theta}{2}$$

Then $$AC=\left(\cos \tfrac{\theta}{2}\right)^2$$

The cross ratio is

$$\frac{\overline{CA}}{\overline{CB}}\frac{\overline{DA}}{\overline{DB}} = \frac{-\left(\cos \tfrac{\theta}{2}\right)^2}{1-\left(\cos \tfrac{\theta}{2}\right)^2} \times \frac11$$ $$=-\frac{1}{\left(\tan \tfrac{\theta}{2}\right)^2} $$

Proof 2 : Let lower case letters $a,b,c,d$ be used for the abscissas of $A,B,C,D$ resp.

The angle between two intersecting circles is given by the formula :

$$\cos \theta = \frac{d^2-(r_1^2+r_2^2)}{2 r_1 r_2}\tag{2}$$

where $r_1,r_2$ are the circles' radii and $d$ the distance between the two centers by an immediate application of the law of cosines to triangle $C_1IC_2$ where $I$ is the intersection point of the two half-circles.

With the notations of the figure, (2) can be given the form :

$$\cos \theta = \frac{\left(\tfrac12(b+d)-\tfrac12(a+c)\right)^2-\left(\tfrac12(a-c)\right)^2-\left(\tfrac12(b-d)\right)^2}{2\tfrac12(a-c)\tfrac12(b-d)}$$

Different simplifications yield :

$$\cos \theta = \frac{(a-b)(c-d)+(b-c)(c-a)}{(a-c)(b-d)}$$

$$\cos \theta = \underbrace{[A,D;B,C]}_{\frac{r}{r-1}}-\underbrace{[B,A;C,D]}_{\frac{1}{1-r}}$$

(where $r$ denotes cross-ratio $[A,C;B,D]$). Therefore, we have :

$$\cos \theta = \frac{r}{r-1}-\frac{1}{1-r}= \frac{r+1}{r-1}= \frac{1+\tfrac{1}{r}}{1-\tfrac{1}{r}}.$$

From there, one can deduce formula (1) as a consequence of the following classical formula (see here) :

$$\cos \theta = \frac{1-t^2}{1+t^2} \ \text{where} \ t=\tan(\tfrac{\theta}{2}).$$

Answer 1

This answer is mainly to provide a reference, and also to provide a bit of background and context.

The reference is Coxeter, Inversive Distance.

Coxeter defines inversive distance between two circles and relates it to cross ratio. In section 4 he defines the quantity

$$ \gamma = \dfrac{a^2+b^2-c^2}{2 a b}, $$ where his $a,b,c$ correspond to your $r_1,r_2,d$.

He is concerned with non-intersecting circles, and defines their inversive distance $\delta$ as satisfying $\cosh(\delta)=\gamma.$ This is analogous to $\cos(\theta)=\gamma$ for the case of intersecting circles. (In general, $|\gamma|=1$ for tangential circles, $|\gamma|<1$ for intersecting circles, and $|\gamma|>1$ for disjoint circles.)

(According to one paper I read, $\delta$ is the imaginary part of the circles' imaginary angle of intersection. $\cosh()$ is of course the hyperbolic cosine, and Coxeter and Greizer provide a cheat sheet of hyperbolic functions in another write up of inversive distance and cross ratios in Geometry Revisited, pg 126.)

Back to the reference paper, the equalities in the middle of pg. 76 show that the cross ratio for disjoint circles is equal to $\tanh^2(\tfrac{\delta}{2}),$ which looks analogous to our $\tan^{-2}(\tfrac{\theta}{2})$ for intersecting circles.

I haven't done the work, but I conjecture that you can generalize the formula so that it works when the circles don't intersect. And I think that, had Coxeter not focussed solely on the case of disjoint circles, he would have derived the $\tan^{-2}(\tfrac{\theta}{2})$ formula for intersecting circles.

Answer 2

This answer supplements/complements my earlier answer.

Here, for circles $c_1,c_2$ we define inversive distance $\Delta(c_1,c_2)$ as $$ \Delta = \frac{r_1^2+r_2^2-d^2}{2 r_1 r_2}$$

where $r_1,r_2$ are the circles' radii and $d$ the distance between the two centers. When the circles overlap, $\Delta=\cos(\theta),$ where $\theta$ is the angle between the circles.

This differs from the definition in my earlier answer (where inversive distance was $\cosh^{-1}(\Delta)$), but is more consistent with contemporary usage and literature.

If $A,B,C,D$ are collinear points as defined in the OP, we can write the cross ratio $\chi=[A,C;B,D]$ in terms of $r_1,r_2,d$ as

$$ \chi=\frac{AC\cdot DB}{CB\cdot AD} =\frac{\left(r_1-r_2+d\right) \left(r_1-r_2-d\right)}{\left(r_1+r_2-d\right) \left(r_1+r_2+d\right)}=\cfrac{(r_1-r_2)^2-d^2}{(r_1+r_2)^2-d^2}. $$

Inversive distance is an invariant in inversive and Möbius geometry, while cross ratio is an invariant in projective geometry. But the two are closely related. They are mapped to one another by the involution $\Phi:x \mapsto \dfrac{x+1}{x-1}$, the proof of which we leave as an exercise for the reader.

This allows us to make short work of the proof of formula (1) in the OP. Given the half angle formula $\tan^2(\theta/2)=\dfrac{1-\cos(\theta)}{1+\cos(\theta)}$, we see that $$ \chi=[A,C;B,D] =\Phi(\Delta) =\Phi(\cos(\theta)) =\cfrac{\cos(\theta)+1}{\cos(\theta)-1} =\cfrac{-1}{\tan^2(\theta/2)} . $$

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Switching gears, here is an extension of OP equation (1) to non-overlapping circles.

Circles $(O_1)$ and $(O_2)$ define the non-overlapping intervals $AC,BD$. Circle $(O'_1)$ passes through points $A,C$ and is tangent to $(O_1)$. Similarly Circle $(O'_2)$ passes through points $B,D$ and is tangent to $(O_2)$. Let $\theta=\angle O'_1CO_1$. Then $\theta=\angle O_2BO'_2$ as well. Further, $\Delta((O_1),(O_2))=\sec(\theta)$ and $[A,C;B,D] =\cfrac{1}{\tan^2(\theta/2)}$ (note that for non-overlapping intervals $AC,BD$ the cross ratio is positive, and $|\Delta((AC),(BD))| \gt 1$, where $(XY)$ is the circle with diameter $XY$).

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An reference of interest is Bowers and Hurdal, Planar Conformal Mappings of Piecewise Flat Surfaces, specifically Section 1, titled An Inversive Distance Primer. The framework and definitions are different from here, but the figure in this answer is derived from Figures 3a,3b in that paper.

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Finally, a previous answer of mine on this site speaks to another relationship between inversive and projective geometry.

Answer 3

Comment only

In all probability this is already known to you. In Bipolar Coordinates the Apollonian circles have the ratio of ratios (cross-ratio =-1 ) for internal and external vertex angle bisectors on the inter focal line.

BipolarCordinatesTauConst