Suppose we have two conic section curves (the red one and the blue one) that are each tangent to a third conic (the black one) in two places. The picture shows ellipses, but I don't think that matters. Now suppose we draw four lines as shown: two lines connecting opposite intersection points (the green ones), and two lines connecting opposite tangency points (the pink ones).

It appears that the four lines meet at a point. Is that true? Is this a known result, and, if so, does it have a name? Proof or reference, please?

I just found this, which I'm currently struggling to understand. It's on page 287 of "THE PRINCIPLES OF PROJECTIVE GEOMETRY APPLIED TO THE STRAIGHT LINE AND CONIC" by J. L. S. HATTON, Cambridge University Press, 1913.

There's a legal electronic copy here.

So, the result is known (or, at least, it was known in 1913). • Very nice find! +1. – Kaj Hansen Mar 20 '18 at 9:14

Here's a proof by symbol-crunching.

Since tangency and incidence are affine-invariant, we'll simplify things by having the common tangent conic (black in OP's diagram) be a circle, say with center $C$ and radius $r$.

We define conic $A$ with two points of tangency ($A_{\pm}$) with $\bigcirc C$, and a spare ($A_0$) on the perpendicular bisector of these points. Likewise for conic $B$. Specifically, define

\begin{align} A_0 = C + a (\cos\alpha,\sin\alpha) &\qquad A_{\pm} = C + r (\cos(\alpha\pm\theta),\sin(\alpha\pm\theta)) \\ B_0 = C + b (\cos\beta,\sin\beta) &\qquad B_{\pm} = C + r (\cos(\beta\pm\phi),\sin(\beta\pm\phi)) \end{align}

We can force tangent-point chords $\overline{A_{+}A_{-}}$ and $\overline{B_{+}B_{-}}$ to meet conveniently at the origin by taking $$C = \frac{r}{\sin(\alpha-\beta)} (-\sin\alpha\cos\phi + \sin\beta \cos\theta, \cos\alpha \cos\phi-\cos\beta \cos\theta )$$

The goal will be to show that the intersections of conics $A$ and $B$ lie on chords passing through the origin. We'll need the equations, and to get them, we need to break the points of tangency into double-points, so that we have the requisite five needed for the determinant formula $$\begin{array}{|cccccc|} \ x^2 & y^2 & x y & x & y & 1 \ \\ \ P_x^2 & P_y^2 & P_x P_y & P_x & P_y & 1 \ \\ \ Q_x^2 & Q_y^2 & Q_x Q_y & Q_x & Q_y & 1 \ \\ \ R_x^2 & R_y^2 & R_x R_y & R_x & R_y & 1 \ \\ \ S_x^2 & S_y^2 & S_x S_y & S_x & S_y & 1 \ \\ \ T_x^2 & T_y^2 & T_x T_y & T_x & T_y & 1 \ \\ \end{array} = 0$$ So, we'll define $A^\prime_{\pm} = A_{\pm} + \epsilon (\sin(\alpha\pm\theta),-\cos(\alpha\pm\theta))$ as a point moved just a bit along the tangent at $A_{\pm}$. Likewise for $B^\prime_{\pm}$. Then, with the help of a symbolic calculator like Mathematica, we can calculate the determinant, factor-out $\epsilon^2$, and then set the remaining $\epsilon$s to zero. With clever grouping, we get ...

\begin{align} A:\quad&( a - r \cos\theta )^2 \left( (x - C_x)^2 + (y-C_y)^2 - r^2 \right)^2 = ( a^2 - r^2 ) ( x \cos\alpha + y \sin\alpha )^2 \\ B:\quad&( b - r \cos\phi )^2 \left( (x - C_x)^2 + (y-C_y)^2 - r^2 \right)^2 = ( b^2 - r^2 ) ( x \cos\beta + y \sin\beta )^2 \\ \end{align}

Interestingly, the common factor on left-hand sides is (the square of) the power of $(x,y)$ with respect to $\bigcirc C$; the $xy$ factors on the right-hand sides give (the square of) the distance from $(x,y)$ to those tangent-point chords. (This resulting relationship between distances is probably "obvious" in the geometry from an appropriate point of view, but I don't (yet) see it.) Eliminating that power factor gives an equation that must be satisfied by all points common to $A$ and $B$:

$$(x\cos\alpha+y\sin\alpha)^2 (b - r \cos\phi)^2 (a^2-r^2) = (x \cos\beta + y \sin\beta )^2 (a - r \cos\theta)^2 (b^2 - r^2)$$

Assuming $a^2 - r^2$ and $b^2-r^2$ have the same sign ---that is, if $A_0$ and $B_0$ are either both inside $\bigcirc C$, or both outside (the reader can ponder why we ignore "both on")--- we can take square roots to get ...

$$(x\cos\alpha+y\sin\alpha) (b - r \cos\phi) \sqrt{|a^2-r^2|} = \pm (x \cos\beta + y \sin\beta ) (a - r \cos\theta) \sqrt{|b^2 - r^2|}$$

These describe a pair of lines that pass through the origin! Since any line through a conic contains at most two points, we're assured that, if $A$ and $B$ have four points of intersection, then each line above contains a pair of them. Thus, the four (possibly-extended) chords in the configuration are concurrent, as desired. $\square$

• Thanks. Assuming that the black circle is a conic occurred to me, too. But it didn't seem relevant in my "proof", so I discarded the idea. I think your calculations do confirm my conclusion that the four lines are of the form $L_1=0$, $L_2=0$, $L_1+L_2=0$ and $L_1-L_2=0$. – bubba Mar 19 '18 at 0:30

From numerical experiments, I'm pretty sure that the result is true.

Here is a "proof" that includes some deplorable handwaving. Maybe it can be salvaged.

Suppose the red conic has equation $Q_1=0$, the blue one is $Q_2=0$, and the black one is $R=0$.

Because of the tangency of $Q_1$ and $R$, the equation $Q_1 - R=0$ has double roots (whatever that means), so maybe this implies that there is a linear function $L_1$ such that $Q_1 - R = L_1^2$. This means that the nearly-vertical pink line has equation $L_1=0$.

By similar fuzzy reasoning, we can find a linear function $L_2$ such that $Q_2 - R = L_2^2$. So the other pink line has equation $L_2=0$.

But then we have $Q_1 - Q_2 = L_1^2 - L_2^2 = (L_1 - L_2)(L_1 + L_2)$. This means (major handwaving) that the green lines must have equations $L_1-L_2=0$ and $L_1+L_2=0$.

At the point where $L_1$ and $L_2$ intersect, we have $L_1=0$ and $L_2=0$, so certainly $L_1-L_2=0$ and $L_1+L_2=0$, also. This means that the four lines meet at a point.

I'd still like to know if the result has a name, or find a reference.