Algebra ∩ Geometry via Desargues' Involution Theorem

(David here.) Recently a problem showed up on the problems feed in the SIMO Retirees' Discord server that made me take a second look at Desargues' Involution Theorem, and I realized it was actually the perfect starting point to discuss the interplay between geometry and algebra!

$\newcommand{\cC}{\mathcal C}$

Introduction

The problem was:

(China TSTST 2017) Let $ABCD$ be a quadrilateral and let $\ell$ be a line. Let $\ell$ intersect the lines $AB,CD,BC,DA,AC,BD$ at points $X,X',Y,Y',Z,Z'$ respectively. Given that these six points on $\ell$ are in the order $X,Y,Z,X',Y',Z'$, show that the circles with diameter $XX',YY',ZZ'$ are coaxal.

The solution is a one-liner, but I'll be intentionally verbose here:

• Desargues' Involution Theorem ("DIT") says that any conic passing through $A,B,C,D$ will cut a fixed line $\ell$ at involutive pairs.
• Thus, applying this to the conics $AB\cup CD, AC\cup BD, AD\cup BC$, we must have $(X,X'), (Y,Y'), (Z, Z')$ describe an involution.
• Thus, there exists $P,Q$ such that $PXQX'$ etc. are harmonic, and so the midpoint $M$ of $PQ$ satisfies $MX\cdot MX' = MP^2$ etc., which implies that the line perpendicular to $\ell$ passing through $M$ is in fact the radical axis of all three circles.

But here's the more interesting question: why Desargues' Involution Theorem? A priori, there's no relationship between conics passing through $A,B,C,D$ and projective geometry.

The "standard" proof

I will shamelessly rip this from MarkBcc168's excellent handout.

We'll use the notation in the TSTST problem, with additionally $(W,W')$ being the intersection of some conic $\cC$ (passing through $A,B,C,D$) with $\ell$.

Then, we can chase the following cross-ratios: $$(X, Y'; W, W') \overset{A}{=} (B, D; W, W') \overset{C}{=} (Y, X'; W, W')$$

Thus, if you had a projective map $f$ on $\ell$ swapping $(W,W')$ and $(X,X')$, it must also swap $(Y,Y')$ and $(Z,Z')$. So the pair $(W,W')$ is part of the same projective involution that swaps $(X,X'), (Y,Y'), (Z,Z')$. This proves DIT.

We'll see two more proofs which have a heavier algebraic flavor.

Pencils

A fact you might know is that pencils of conics are algebraically special. By definition, the points on a conic must satisfy $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$ for some coefficients $A,B,C,D,E,F$. For the set of conics passing through four fixed points, there is a special constraint on their coefficients.

Take three of them $\cC_1, \cC_2, \cC_3$, and let's look at their corresponding quadratic polynomials $P_1,P_2,P_3$ (which are unique up to rescaling). The claim is that there must exists reals $\lambda,\mu$ such that $$P_3 = \lambda P_1 + \mu P_2$$ or in words, that each is a linear combination of the other two. One might even say that they are linearly dependent. And since this holds for any three conics, we can make the stronger claim that the entire set of conics are precisely those whose equations look like $\lambda P_1 + \mu P_2$.

There's a good reason for this. If our four points have coordinates $(x_i,y_i)_{i=1,...,4}$, then we can form a system of four equations for which the coefficients of a conic passing through them must satisfy: $$ \begin{cases} Ax_1^2 + Bx_1y_1 + Cy_1^2 + Dx_1 + Ey_1 + F = 0 \\ Ax_2^2 + Bx_2y_2 + Cy_2^2 + Dx_2 + Ey_2 + F = 0 \\ Ax_3^2 + Bx_3y_3 + Cy_3^2 + Dx_3 + Ey_3 + F = 0 \\ Ax_4^2 + Bx_4y_4 + Cy_4^2 + Dx_4 + Ey_4 + F = 0. \end{cases} $$ If you believe that these equations are non-redundant, then it should cut out a 2-dimensional family of solutions (from the original 6-dimensional space of coefficients), so we're done!

If they are redundant, then somehow all conics passing through three of the points must pass through the last, but this cannot happen except for special cases where all four points are collinear.

We call this set a conic pencil. We won't use this immediately, but it helps us not be so afraid of the conic we're dealing with.

Maps

When we say that a map $f$ is involutive (or an involution), we mean that it satisfies $f(f(x)) = x$ for all $x$. In our geometric context, what we really mean is that it is a projective involution - a projective map that is involutive. (For instance, the familiar harmonic conjugation on a line is a projective involution.)

One can ask generally what map is produced in the setup of DIT: so given a point $X$, we would: (1) construct the conic $(ABCDX)$, (2) find the other intersection $X'$ with the line $\ell$, and the map would be $X \mapsto X'$.

With a suitable rotation of coordinates, we can make $\ell$ the $x$-axis, so the coordinates of $X$ are $(x,0)$. Then, we want to solve the system of equations $$ \begin{cases} Ax_1^2 + Bx_1y_1 + Cy_1^2 + Dx_1 + Ey_1 + F = 0 \\ Ax_2^2 + Bx_2y_2 + Cy_2^2 + Dx_2 + Ey_2 + F = 0 \\ Ax_3^2 + Bx_3y_3 + Cy_3^2 + Dx_3 + Ey_3 + F = 0 \\ Ax_4^2 + Bx_4y_4 + Cy_4^2 + Dx_4 + Ey_4 + F = 0. \\ Ax^2 + Dx + F = 0. \end{cases} $$

It should be clear that the result is that each coefficient $A,B,C,D,E,F$ should be some rational function of $x$ (i.e. $A(x) = \frac{A_1(x)}{A_2(x)}$ for some polynomials $A_1,A_2$).

Now, for the second intersection with the $x$-axis, we simply consider the quadratic equation $$A(x) t^2 + D(x) t + F(x) = 0.$$ Because we know $t=x$ to be a solution, the other solution is given by Vieta's formulas as $x' = -\frac{D(x)}{A(x)} - x$, and is thus also a rational function of $x$.

Here comes the punchline:

(Fact) If you have a rational map on a (projective) line and it is invertible, then it is a projective map.

To make sense of this, you need to know that projective maps look like a ratio of linear maps: $$z \mapsto \frac{az+b}{cz+d}.$$ (For this reason they are sometimes called fractional linear transformations.)

So why is this true? Let's say $x \mapsto f(x) = {P(x)}{Q(x)}$. Then, if $f$ is invertible, then there is exactly one solution to $f(x) = w$, so $P(x) - wQ(x) = 0$ has exactly one solution for each $w$. This implies that both $P$ and $Q$ are linear.

Of course, having the map be involutive forces this condition immediately, so we're done!

Reference. I learnt this idea from notzeb's cross-ratio notes. There's a lot of good stuff in here if you like algebra+geometry ideas.

Commentary. If the first proof is the poster child for projective cross-ratio chasing, then this proof is the poster child for the method of moving points. The starting point there is that most configuration-agnostic maps are in fact rational maps, and one can keep track of the degree of such maps. This reduces the problem to checking some finite number of cases (up to the said degree of the map).

RP5

(iykyk)

We return to the conic pencil idea to dig for more connections.

The first is to notice a natural interpretation of conic coefficients in a certain kind of space: $\mathbb R^6$, but "modded out" by rescaling. We will write this as $\mathbb {RP}^5$ ("real projective 5-space"), and we can talk of lines in this space by considering a line in $\mathbb R^6$ and including everything which is scaling-invariant from it (which turns it into a plane, if the original line doesn't pass through the origin).

Now, our original observation looks like:

A conic pencil consists of conics "in a line" (viewed in $\mathbb {RP}^5$).

This is good for us, because much of the problem can be reinterpreted in terms of conic pencils. For example, coaxal circles form a conic pencil (with the two common points, possibly imaginary, and the two circular points at infinity). Then, the fact that pairs $(X,X'), (Y,Y'), (Z,Z')$ form pairs in a projective involution can be rephrased as having $(XX'P), (YY'P), (ZZ'P)$ concur again at a common point $Q$.

In some sense, the setup is of the exact same flavor: if $(X,X')$ is a pair in a projective involution, then for any points $A,B,C$ we should perhaps have the conic $(XX'ABC)$ pass through a fixed point $D$ as $(X,X')$ varies.

So we wish to prove a statement of this form:

Suppose that for some three points $A,B,C$ we have that the conic $(XX'ABC)$ passes through a fixed point $D$ as $(X,X')$ varies. Then, for any such three points $A',B',C'$, the conic $(XX'A'B'C')$ also passes through some fixed point $D'$ as $(X,X')$ varies.

In some sense, this is now purely a problem in $\mathbb {RP}^5$. To get started, let's define $S_{\{\text{set}\}}$ as the set of conics in $\mathbb {RP}^5$ passing through a given set of points. I'll handwave a bit here, but this should have dimension $5 - |\{\text{set}\}|$. Each of these sets are a linear subspace of $\mathbb{R}^6$ (and so we could call this a "linear subspace" in $\mathbb{RP}^5$).

Now, $S_{XX'}$ is a 3-dimensional space. However, this space must contain the set with "multiples of $\ell$", which is a 2-dimensional space: $$S_{\ell} = \{\ell(x,y) \cdot (ax+by+c) \mid a,b,c \in \mathbb R\} / \text{(rescaling)}.$$

So, suppose that we had three other points $A',B',C'$. Then $S_{A'B'C'}$ is a 2-dimensional space. Projecting along $S_\ell$, we are left with:

• a 3-dimensional space
• $S_\ell$ becomes a point
• each of $S_{XX'}$ becomes a line
• $S_{ABCD}$ is still a line
• $S_{A'B'C'}$ is a two-dimensional space (where the projection is a one-to-one)

Visually, it's clear that this forces each $S_{A'B'C'XX'}$ to be on the same line, and so this implies that these conics form a pencil and so $D'$ exists. (Technically, we need the converse statement but it's pretty straightforward.)

The original RP5 problem

Consider the following variant:

Consider a pencil of conics and a line $\ell$. For each conic $\cC$, consider the conic $\cC'$ which is tangent to $\cC$ at both points of $\cC\cap \ell$ and passing through a fixed point $P$. Then, all $\cC'$ lie in the same pencil.

This follows from essentially the same argument as before, without having to project away $S_\ell$: $\cC, \cC'$ and $\ell^2$ are collinear (where $\ell^2$ is the "double line" whose equation is that of $\ell$ but squared), and so this is just projecting the pencil onto the "hyperplane" $S_P$.

If $P$ were one of the circular points at infinity, we would get a family of coaxal circles. However, to keep things real, this forces an angle condition on $\ell$ with respect to the four common points on the pencil. Thus we get:

(Poncelet's Coaxial Theorem, generalized) For any $A,B,C,D$, suppose $\ell$ is parallel or perpendicular to the angle bisector of $(AB,CD)$. Then if $P,Q$ are intersections of a conic $\cC$ passing through $A,B,C,D$ with $\ell$, the circle passing through $P,Q$ and tangent to $\cC$ must pass through two fixed points (and thus all such circles are coaxial).

This is a generalization of the problem I had in this handout:

Poncelet's Porisms

Given a fixed circle $C$ and another circle $C_1$, we may define a map $\tau_1: C\to C$ sending a point $P$ to $Q$ such that $PQ$ is tangent to $C_1$ going clockwise. Another way to phrase this theorem in the special case above is:

If circles $C, C_1, C_2$ are coaxial, then $\tau_1\circ \tau_2 = \tau_2 \circ \tau_1$.

I don't know if this fact can be used to prove the Poncelet's porism, which says that

(Poncelet's Porism) If $\tau_1^n$ coincides with the identity map at any point, then $\tau_1^n$ is the identity map.

There's also a generalized version:

(Poncelet's Porism, generalized) If $C, C_1, C_2,..., C_n$ are coaxial, then if $\tau_1\circ \cdots \circ \tau_n$ coincides with the identity map at any point, then it is the identity map.

If you know how to prove these from the coaxial lemma, please let me know!