Perspective sequence that maps collinear points A,B,C,D to D,C,B,A

Question

Find the perspective sequence that maps collinear points A,B,C,D to D,C,B,A.

Attempt: If we need to find a sequence of three perspectives that (A,B,C)->(A,C,B), where A, B, C are collinear, then first we mark p is a line through A, B, C, let S be some point that does not lie on p, let q be a line through A and S, let p' be a line through A different from p and q, let B' be an element of the line BS and C' an element of the line CS, and let S' be the intersection of the lines BC' and B'C. Then the first pespectivity maps us from the line p to the line p' and has its center in S: A->A, B->B', C->C', and the second one from p' to p and has its center in S': A ->A, B'->C, C'->B.

Answer 1

Let us begin by this figure

Fig. 1 : $\mathcal{P}_S \ $ followed by $\mathcal{P}_{S'}$.

(where, in a certain affine map of projective space, $S$ is the centroid of triangle $ABS'$ and $D'A'$ a mid-segment).

Using a projective transformation (see appendix) on Fig. 1, one gets a "general-looking" figure :

Fig. 2.

Now, let us use this figure as a source of inspiration for a projective description (as you have done for three points) by lines and points construction.

Appendix : Matlab program that has generated Fig. 2 out of Fig. 1. :

clear all;close all;hold on;axis off;
% initial coordinates of points :
x=[0 4 8 12 3 5 7 9 6 6]
y=[8 8 8 8 4 4 4 4 (16/3) 0];
% a projective transformation (among others...) :
X=(x+y)./(x+y-3);
Y=(-x+y)./(x+y-3);
% visiting order of the 10 points such 
% that every line segment is plotted :
I=[10,1,10,2,10,3,10,4,5,9,3,
6,9,2,7,9,1,8,9,4,1,8,5];
plot(X(I),Y(I),'-o-')

A key point has to be considered.

(see on Fig. 3) There exists a point $Q$ aligned with $ABCD$ such that, whatever the secant with intersection points $D'C'B'A'$, lines $AA',BB',CC',DD'$ share a common point $S$. Moreover, the locus of these points $S$ is a line issued from $S'$ intersecting axis $ABCD$ in a certain point $Q$ which can be considered as a point in duality with $P$ ; more precisely, point $P$ and $Q$ are such that they are harmonic conjugates with respect to points $A,D$ and with repect to points $B,C$ :

$$(A,D;Q,R)=-1 \ \ \text{and} \ \ (B,C;Q,R)=-1$$

which gives, working with abscissas :

$$\left(\frac{q-a}{q-d}\right)\left(\frac{r-d}{r-a}\right)=-1 \ \ \text{and} \ \ \left(\frac{q-b}{q-c}\right)\left(\frac{r-c}{r-b}\right)=-1\tag{1}$$

giving rise to a system of equations from which the abscissas $q$ and $r$ of $Q$ and $R$ resp. can be obtained.

In fact, some calculations (that I don't display here) show that there is an interesting construction of points $Q$ and $R$. $QR$ is the diameter of the unique circle with center $P$ situated on line $ABCD$ orthogonal to both circles with diameters $AD$ and $BC$. One can begin by constructing its center :

$$R=\left(\frac{ad-bc}{(a+d)-(b+c)},0\right)$$

then plot a tangent $PE$ to one of the circles (for example with diameter $AD$), which, by orthogonality, is the radius of this (dotted) circle, giving at once points $Q$ and $R$.

Remark : This is also connected to pencils of 4 lines $S(ADQR)$ and $S(BCQR)$.

Fig. 3. with an animated version here.