Inscribe ellipses and Trilinear coordinates

Question

These days I am self-studying about trilinear coordinate system. While googling about this topic, I found that trilinears are used to describe special inscribed ellipses of a triangle like Lemoine inellipse, Brocard inellipse and Steiner inellipse. Also I found that the general form of an inscribed ellipse is $$l^2x^2+m^2y^2+n^2z^2=2lmxy+2mnyz+2nlzx,$$ for some parameters $l,m,n$ and $x:y:z$ is a point on the ellipse.
I wonder how they came up with this equation. And can we find the center, foci and semi-axis in terms of $l,m,n$?
How about circumscribe ellipses? Do you know a good reference about this?

Answer 1

I'm not sure how determining the equation for an inellipse (or, more generally, an inconic) equation is "really" done, but here's a brute force approach using the fact that a conic through five points $D$, $E$, $F$, $G$, $H$ has its Cartesian equation given by $$\left|\begin{array}{cccccc} x^2 & y^2 & x y & x & y & 1 \\ D_x^2 & D_y^2 & D_x D_y & D_x & D_y & 1 \\ E_x^2 & E_y^2 & E_x E_y & E_x & E_y & 1 \\ F_x^2 & F_y^2 & F_x F_y & F_x & F_y & 1 \\ G_x^2 & G_y^2 & G_x G_y & G_x & G_y & 1 \\ H_x^2 & H_y^2 & H_x H_y & H_x & H_y & 1 \\ \end{array}\right| = 0 \tag{$\star$}$$

Note that, if $(x,y)$ are the coordinates of a sixth point, say, $I$, then $(\star)$ gives the necessary condition for $D$, $E$, $F$, $G$, $H$, $I$ to lie on a common conic.

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Let the vertices of $\triangle ABC$ have Cartesian coordinates $$A := (0,0) \qquad B := (c, 0) \qquad C := ( b \cos A, b \sin A ) \tag{1}$$ and let our conic's point of tangency with the sides of the triangle be given by $$D := \frac{B + d C}{1+d} \qquad E := \frac{C+e A}{1+e} \qquad F := \frac{A+f B}{1+f} \tag{2}$$ (Because we're working with ratios and tangencies and conics, all of which are affine-invariant notions, we could simplify by taking $\triangle ABC$ to have a convenient form, say, $b=c$ and $A = 90^\circ$. I'll continue working at our current level of generality, because (1) I'll be doing the symbol-crunching with Mathematica, which doesn't mind the complexity; and (2) it'll be "obvious" how to convert $xy$-coordinates to trilinears.)

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First, we establish an oddly-familiar identity of which I wasn't previously aware, but is attributed to Brianchon.

Consider the six points $D$, $E$, $F$, $D^\prime$, $E^\prime$, $F^\prime$, where $$D^\prime := D + d^\prime(C-B) \qquad E^\prime := E + e^\prime(A-C) \qquad F^\prime := F + f^\prime(B-A) \tag{3}$$ are points on the the sides of $\triangle ABC$. Feeding the coordinates of these six points into $(\star)$, Mathematica gives an equation with a factor $d^\prime e^\prime f^\prime$; dividing-out that factor, then making $d^\prime$, $e^\prime$, $f^\prime$ vanish, yields the equation

$$\frac{b^4 c^4\sin^4 A}{(1+d)^2(1+e)^2(1+f)^2}\left(1-d^2e^2f^2\right) = 0 \tag{4}$$

Thus, $def = \pm 1$. Note, though, that our $d$, $e$, $f$ are precisely the same ratios one encounters in the theorems of Ceva and Menelaus. So, $def=-1$ invokes Menelaus, making $D$, $E$, $F$ collinear, which isn't allowed in a (non-degenerate) conic. We therefore have

For $d$, $e$, $f$ to be determine points-of-tangency $D$, $E$, $F$ of a conic with the side-lines of $\triangle ABC$, $$d \;e \;f = 1 \tag{$\star\star$}$$

By Ceva, then, $\overline{AD}$, $\overline{BE}$, $\overline{CF}$ concur. Where they meet is called the Brianchon point.

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Now, taking $(\star)$ with points $D$, $E$, $F$, $D^\prime$, $E^\prime$, and $(x,y)$, dividing out a factor $d^\prime e^\prime$, and incorporating $(\star\star)$ by substituting $d = 1/(ef)$, Mathematica gives this equation for the tangent conic through $D$, $E$, $F$: $$\begin{align} 0 &= x^2 b^2 (1 + f)^2 \sin^2 A \\ &+ y^2 \left( b^2 (1 + f)^2 \cos^2 A + c^2 f^2 (1 + e)^2 - 2 b c f (1 - e + f + e f) \cos A \right) \\ &- 2 x y b \sin A \left( b (1+f)^2 \cos A - c f ( 1 + f - e + e f ) \right) \\ &- 2 x b^2 c f(1+f) \sin^2 A \\ &+ 2 y b c f \sin A \left( b (1+f) \cos A - c f (1+e)\right) \\ &+ b^2 c^2 f^2 \sin^2 A \end{align} \tag{5}$$

Somewhat chaotic, but it appears that some terms would prefer to congregate a bit, according to factors $(1+e)$ and $(1+f)$, yielding some interesting terms ...

$$\begin{align} 0 &= b^2 (1 + f)^2 ( x \sin A - y \cos A )^2 \\ &+ y^2 c^2 f^2 (1 + e)^2 \\ &+ 2 y b c f ( 1 - e + f + e f ) \left( x \sin A - y \cos A \right) \\ &- 2 b^2 c f ( 1 + f ) \sin A \left( x \sin A - y \cos A \right) \\ &- 2 y b c^2 f^2 (1+e) \sin A \\ &+ b^2 c^2 f^2 \sin^2 A \end{align} \tag{5a}$$

We could clean this up, say, writing $q$ for that curiously-prominent term $x \sin A - y \cos A$, and $T$ for the frequently $b c \sin A$ (which most should recognize at twice the area of our triangle).

$$\begin{align} 0 &= q^2 b^2 (1 + f)^2 + y^2 c^2 f^2 (1 + e)^2 \\ &+ 2 q y b c f ( 1 - e + f + e f ) \\ &- 2 q b f ( 1 + f ) T - 2 y c f^2 (1+e) T + f^2 T^2 \end{align} \tag{5b}$$

We could probably do even more clever grouping, but ... This was supposed to be about trilinears, with coordinates (say) $p$, $q$, $r$ representing distances from our $(x,y)$ to sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ of our triangle. We should get back to that.

Okay, so observe: The signed distance from $(x,y)$ to $\overline{AC}$ is precisely the value we have denoted $q$ above (since the line through $A$ and $C$ has equation $x \sin A - y \cos A = 0$). Moreover, the signed distance to $\overline{AB}$ (that is, the $x$-axis), what we might call $r$, is naturally given by $y$. So, we're actually two-thirds of the way to a trilinear equation already! For the third coordinate, $p$, the distance to $\overline{BC}$, we can use the identity $$p a + q b + r c = 2 |\triangle ABC| = b c \sin A = T \tag{$\star\star\star$}$$

Making appropriate substitutions into $(5b)$, we have

$$\begin{align} 0 &= q^2 b^2 (1 + f)^2 + r^2 c^2 f^2 (1 + e)^2 + 2 q r b c f ( 1 - e + f + e f ) \\ &- 2 q b f ( 1 + f ) (p a + q b + r c ) - 2 r c f^2 (1+e) (p a + q b + r c ) + f^2 ( p a + q b + r c )^2 \end{align} \tag{6}$$

This simplifies to ... $$p^2 a^2 f^2 + q^2 b^2 + r^2 c^2 e^2 f^2 - 2 p q a b f - 2 q r b c e f - 2 r p a c e f^2 = 0 \tag{6a}$$

That's a tad asymmetric, so let's divide-through by ... oh, say ... $\sqrt[3]{e^2f^4}$, while exploiting $(\star\star)$ ...

$$p^2 a^2 \left(\frac{f^{3/3}}{e^{1/3}f^{2/3}}\right)^2 + q^2 b^2\left(\frac{d^{1/3}e^{1/3}f^{1/3}}{e^{1/3}f^{2/3}}\right)^2 + r^2 c^2 \left(\frac{d^{-1/3} e^{2/3} f^{2/3}}{e^{1/3}f^{2/3}}\right)^2 - \cdots = 0 \tag{6b}$$ ... so that we can write ...

$$p^2 \lambda^2 + q^2 \mu^2 + r^2 \nu^2 - 2 p q \lambda \mu - 2 q r \mu \nu - 2 r p \nu \lambda = 0 \tag{7a}$$ ... where $p:q:r$ are the trilinear coordinates of points on the conic, and the "inconic parameters" are ... $$\lambda : \mu : \nu = a \left(\frac{f}{e}\right)^{1/3} : b \left(\frac{d}{f}\right)^{1/3} : c \left(\frac{e}{d}\right)^{1/3} \tag{7b}$$

Finding the center, foci, and other elements of the conic are left as an exercise to the reader. I refer to MathWorld's "Inconic" entry for the fact that the center lies at point $$\mu b + \nu c : \nu c + \lambda a : \lambda a + \mu b$$ and that the inconic is a parabola iff $$\frac{\lambda}{a} + \frac{\mu}{b} + \frac{\nu}{c} = 0$$ with focus given by $$\frac{a}{\lambda^2} : \frac{b}{\mu^2} : \frac{c}{\nu^2}$$

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We can derive "Ceva ratios" $d$, $e$, $f$ from inconic parameters $\lambda$, $\mu$, $\nu$ via $$d = \frac{\mu/b}{\nu/c} \qquad e = \frac{\nu/c}{\lambda/a} \qquad f = \frac{\lambda/a}{\mu/b} \tag{8}$$

Regarding specific cases ...

• The Steiner Inellipse passes through the midpoints of the triangle's sides. This corresponds to $d = e = f = 1$.

• The Brocard Inellipse has inconic parameters $\lambda : \mu :\nu = 1/a : 1/b : 1/c$. This corresponds to $d = (c/b)^2$, $e=(a/c)^2$, $f=(b/a)^2$.

• The Lemoine Inellipse has inconic parameters $\lambda = (2(b^2+c^2)-a^2)/(bc)$, etc. The reader can substitute into $(8)$.

• The Yff Parabola has $d =(c-a)/(a-b)$, etc.

• The Keipert Parabola has $d = (c^2-a^2)/(a^2-b^2)$, etc.