proving a result in (elementary) planar projective geometry about cross ratios

Question

I'm stuck on the following problem in the projective space $P(\mathbb{R}^3) = \mathbb{R}P^2$.

Say we have three distinct and non-concurrent lines $\mathcal{A}, \mathcal{B}$ and $\mathcal{C}$. Let $P$ be a point outside of these three lines. Show that there exists a unique line $\mathcal{P}$ through the point $P$, such that the cross ratio $(CPAB) = -1$, where $A, B$ and $C$ are respectively the intersections $\mathcal{A} \cap \mathcal{P}$, $\mathcal{B} \cap \mathcal{P}$ and $\mathcal{C} \cap \mathcal{P}$.

My attempt so far: intuitively, I believe this can be proved analytically. "Choosing" $A$ on $\mathcal{A}$ randomly (only one "degree of freedom"), we've also fixed the line $\mathcal{P} = A + P,$ since $A$ and $P$ are necessarily distinct. Then $B$ and $C$ naturally follow. There's only one degree of freedom left, and so filling this in to the definition of cross ratio and setting it equal to $-1$ would uniquely determine the coordinates of $A,$ very loosely speaking. (I am in no way asserting that this is a proof, it's as far as I've got.) Is this how you would do it? Or is there a way to prove it synthetically? I've also tried to look at the dual version of the statement, but I wasn't able to get anything useful from it.

Answer 1

Let $X:=\mathcal{A}\cap \mathcal{C}$ and $Y:=\mathcal{B}\cap \mathcal{C}$. Denote $M:=\mathcal{A}\cap \mathcal{B}$ and let the intersection of the lines $PM$ and $\mathcal{C}$ be $S$. Denote the harmonic conjugate of $S$ with respect to $(X,Y)$ be $C$. (I.e. $C$ is the unique point such that $(XYSC)=-1$.)

Then the line $PC$ has the required property. Let $A:=PC\cap\mathcal{A}$, $B:=PC\cap\mathcal{B}$. Then by the Pappos-Steiner theorem $(ABPC)=(XYSC)=-1$, since they are projected from $M$.

No other lines through $P$ are good: if the intersections with some other line through $P$ are $A_1$, $B_1$ and $C_1$; and $C_2$ denots $MC\cap A_1B_1$, then by the Pappos-Steiner theorem $(A_1B_1PC_2)=-1$. But $C_1$ is different from $C_2$, since $CC_2=MC$ and $CC_1=\mathcal{C}$ are different lines, so by the uniqueness of the harmonic conjugate $(A_1B_1PC_1)$ cannot be $-1$.

Answer 2

Here is another way, using a coordinates setting.

Let us take the following notations:

$$A=L_B \cap L_C, \ \ B=L_C \cap L_A, \ \ C=L_A \cap L_B$$

As any quadrangle in general position like $A,B,C,P$ can be mapped onto any another quadrangle by a projective transformation, and that projective transformation preserve cross ratio, we can assume WLOG that we are in the situation depicted in the following figure.

Consider a variable line with equation $y=tx$ passing through $P$.

The cross ratio:

$$cr=(PD/PF)/(ED/EF) = ((x_D-x_P)/(x_F-x_P))/((x_D-x_E)/(x_F-x_E))$$

is easily computed as a function of $t$:

$$cr = 1-\frac{1}{t}$$

which can take any real value (even value $1$ can be attained for a vertical position of the line...).

In particular, value $-1$ is reached for $t=\frac12$ (the case for which the picture has been drawn).