Cross ratio of 4 points on circle

Question

I'm in a bit of trouble I want to calculate the cross ratio of 4 points $ A B C D $ that are on a circle

Sadly "officially" it has to be calculated with A B C D as complex numbers and geometers sketchpad ( the geomerty program I am used to) don't know about complex numbers

Now I am wondering The cross ratio of 4 points on a circle is a real number
And is there so much difference between

$$ \frac { (A-C)(B-D) }{ (A-D)(B-C) } \text{ (complex method) }$$ And $$ \frac { |AC||BD| }{ |AD||BC| } \text{ (distance method). }$$

Where X-Y is the complex difference between X and Y

and |XY| is the distance between X andY

If allready is given that the four points are on a circle?

I think that the absolute values of the calculations are the same but am I right (and can we prove it)

ADDED LATER :

on Cave 's suggestion I used a circle inversion to move the four points on the circle to four points on a line and then the formulas give the same value ( if I did it correctly)

What I did:

I took a diameter of the circle This diameter intersects the circle in O and U , u is the line through U perpendicular to OU

Project the 4 points onto line u with centre O (draw a ray through O and the point, the new point is where this ray intersects line u)

And calculate the cross ratio of the four new points.

This "projection" method and the earlier "distance" method give the same value but does this prove anything?

Answer 1

For $A,B,C,D$ on the unit circle, $$ \begin{align} \frac{(A-C)(B-D)}{(A-D)(B-C)} &=\frac{\left(e^{ia}-e^{ic}\right)\left(e^{ib}-e^{id}\right)}{\left(e^{ia}-e^{id}\right)\left(e^{ib}-e^{ic}\right)}\\ &=\frac{\left(e^{i(a-c)}-1\right)\left(e^{i(b-d)}-1\right)}{\left(e^{i(a-d)}-1\right)\left(e^{i(b-c)}-1\right)}\\ &=\frac{e^{i\frac{a-c}2}2i\sin\left(\frac{a-c}2\right)e^{i\frac{b-d}2}2i\sin\left(\frac{b-d}2\right)}{e^{i\frac{a-d}2}2i\sin\left(\frac{a-d}2\right)e^{i\frac{b-c}2}2i\sin\left(\frac{b-c}2\right)}\\ &=\frac{\sin\left(\frac{a-c}2\right)\sin\left(\frac{b-d}2\right)}{\sin\left(\frac{a-d}2\right)\sin\left(\frac{b-c}2\right)}\tag{1} \end{align} $$ which is indeed real.

---

For $A,C$ on the unit circle, $$ \begin{align} |A-C| &=\sqrt{\left(e^{ia}-e^{ic}\right)\left(e^{-ia}-e^{-ic}\right)}\\[3pt] &=\sqrt{2-2\cos(a-c)}\\ &=2\sin\left(\frac{a-c}2\right)\tag{2} \end{align} $$ Therefore, $$ \begin{align} \frac{|A-C||B-D|}{|A-D||B-C|} &=\frac{2\sin\left(\frac{a-c}2\right)2\sin\left(\frac{b-d}2\right)}{2\sin\left(\frac{a-d}2\right)2\sin\left(\frac{b-c}2\right)}\\ &=\frac{\sin\left(\frac{a-c}2\right)\sin\left(\frac{b-d}2\right)}{\sin\left(\frac{a-d}2\right)\sin\left(\frac{b-c}2\right)}\tag{3} \end{align} $$ which is obviously real, but not so obviously, the same as $(1)$.

---

When is the Complex Cross-Ratio Real?

$$ \begin{align} \frac{A-C}{A-D} &=\frac{C-A}{D-A}\\ &=\frac{|C-A|}{|D-A|}\,e^{-i\alpha} \end{align} $$ $$ \begin{align} \frac{B-D}{B-C} &=\frac{D-B}{C-B}\\ &=\frac{|D-B|}{|C-B|}\,e^{-i\beta} \end{align} $$ Therefore, $$ \frac{(A-C)(B-D)}{(A-D)(B-C)} =\frac{|C-A||D-B|}{|D-A||C-B|}\,e^{-i(\alpha+\beta)} $$ $\alpha+\beta$ is an integer multiple of $\pi$ ($\pi$) if and only if $A$ and $B$ are on opposing arcs of a circle with $C$ and $D$.

---

$$ \begin{align} \frac{A-C}{A-D} &=\frac{C-A}{D-A}\\ &=\frac{|C-A|}{|D-A|}\,e^{-i\alpha} \end{align} $$ $$ \begin{align} \frac{B-D}{B-C} &=\frac{D-B}{C-B}\\ &=\frac{|D-B|}{|C-B|}\,e^{i\beta} \end{align} $$ Therefore, $$ \frac{(A-C)(B-D)}{(A-D)(B-C)} =\frac{|C-A||D-B|}{|D-A||C-B|}\,e^{-i(\alpha-\beta)} $$ $\alpha-\beta$ is an integer multiple of $\pi$ ($0$) if and only if $A$ and $B$ are on the same arc of a circle with $C$ and $D$.

Answer 2

Cross ratios are invariant under projective transformations, and projective transformations of the projective complex line $\mathbb{CP}^1$ are Möbius transformations. You can use a Möbius transformation to map the unit circle to the real line. (A circle inversion would do the same for the circle itself, although it would exchange the upper and lower half plane. Either one is fine for the question at hand.) After that you can compute the cross ratio on the line.

Actually there are still 3 real degrees of freedom mapping the unit circle to the real line. In the absense of reasons to do otherwise, I'd suggest the following map: map $1$ to $0$, $i$ to $1$ and $-1$ to $\infty$.

$$z\mapsto\frac{z-1}{iz+i}$$

That map has a strong connection to the tangent half-angle substitution. A point $t$ on the real axis corresponds to $\frac{1-t^2}{1+t^2}+i\frac{2t}{1+t^2}$ on the unit circle. Converseley if you have $x+iy$ as a point on the unit circle, then $t=\frac{y}{x+1}$ is the corresponding real parameter.

As you see, the point at infinity is a relevant point on the real line, so using real numbers only is a poor choice of framework here. Instead use the real projective line, with homogeneous coordinates. A point $x+iy$ on the circle corresponds to a homogeneous coordinate vector

$$\begin{pmatrix}y\\x+1\end{pmatrix}\;.$$

Instead of differences of real numbers, you use $2\times2$ determinants of homogeneous coordinates in the formula for the cross ratio. So you compute

$$\frac{\begin{vmatrix}y_A&y_C\\x_A+1&x_C+1\end{vmatrix}\cdot \begin{vmatrix}y_B&y_D\\x_B+1&x_D+1\end{vmatrix}} {\begin{vmatrix}y_A&y_D\\x_A+1&x_D+1\end{vmatrix}\cdot \begin{vmatrix}y_B&y_C\\x_B+1&x_C+1\end{vmatrix}}=\\ \frac{\bigl(y_A(x_C+1)-y_C(x_A+1)\bigr) \bigl(y_B(x_D+1)-y_D(x_B+1)\bigr)} {\bigl(y_A(x_D+1)-y_D(x_A+1)\bigr) \bigl(y_B(x_C+1)-y_C(x_B+1)\bigr)}$$

This is obviously a real number, computed using real arithmetic only. And in conrtrast to the version using distances, you can get away without square roots, too.