# Prove six cross ratio values

I have to prove that the cross ratio of all permutations of 4 collinear points on a line can at most be one of the following values:

$$\left\{ \tau, \frac{1}{\tau}, 1-\tau, \frac{1}{1-\tau}, \frac{\tau-1}{\tau}, \frac{\tau}{\tau-1}\right\}$$

Given the points $$p_1 < p_2 < p_3 < p_4$$ and the cross ratio $$cr(p_1,p_2,p_3,p_4)=\frac{\Delta_{1,2}\cdot \Delta_{3,4}}{\Delta_{1,3}\cdot \Delta_{2,4}}$$ where $$\Delta_{i,j}$$ is the signed distance.

By setting $$cr(p_1,p_2,p_3,p_4)=\tau$$ it is easy to see that $$cr(p_1,p_3,p_2,p_4)=\frac{\Delta_{1,3}\cdot \Delta_{2,4}}{\Delta_{1,2}\cdot \Delta_{3,4}} = \frac{1}{\tau}$$ which are the first two values. But now i stuck at calculating the other 4.

Is there a way to SEE which permutation corresponds to which value? Or is it really try and error: start calculating and see what i get?

In the last case i would now try to calculate $$cr(p_1,p_3,p_4,p_2) = 1 - cr(p_1,p_2,p_3,p_4)$$ and if it does not work i would try the next permutation.

I know that of 24 permutations each 4 result in the same value, so there are just 4 permutations left.

Edit

New knowledge: Because $$\Delta_{i,j}$$ is just the signed distance i can also write $$\Delta_{i,j} = (-1)(\Delta_{j,i})$$

So the other four cross ratios are: $$cr(p_1,p_2,p_4,p_3)=\frac{\Delta_{1,2}\Delta_{4,3}}{\Delta_{1,4}\Delta_{2,3}}=\frac{\Delta_{1,2}\cdot (-1)\Delta_{3,4}}{\Delta_{1,4}\Delta_{2,3}}$$ $$cr(p_1,p_4,p_2,p_3)=\frac{\Delta_{1,4}\Delta_{2,3}}{\Delta_{1,2}\Delta_{4,3}}=\frac{\Delta_{1,4}\Delta_{2,3}}{\Delta_{1,2}\cdot(-1)\Delta_{3,4}}$$

$$cr(p_1,p_3,p_4,p_2)=\frac{\Delta_{1,3}\Delta_{4,2}}{\Delta_{1,4}\Delta_{3,2}}=\frac{\Delta_{1,3}\cdot (-1)\Delta_{2,4}}{\Delta_{1,4}\cdot (-1)\Delta_{2,3}}$$ $$cr(p_1,p_4,p_3,p_2)=\frac{\Delta_{1,4}\Delta_{3,2}}{\Delta_{1,3}\Delta_{4,2}}=\frac{\Delta_{1,4}\cdot (-1)\Delta_{2,3}}{\Delta_{1,3}\cdot (-1)\Delta_{2,4}}$$