Unary Operations on Homogeneous Coordinates in the Plane of a Triangle

Abstract

Suppose that X is a triangle center with homogeneous coordinates (barycentric or trilinear) $x:y:z$. Eight unary operations discussed in this paper include $u_1\left(X\right)=(y-z)/x:(z-x)/y:(x-y)/z$. For each $u_i$, there exist, formally, two points, P and U, such that $u_i\left(P\right)=u_i\left(U\right)=X$. To such pairs of inverses are applied nine binary operations, each resulting in a triangle center. If L is a line, then formally, $u_i\left(L\right)$ is a cubic curve that passes through the vertices $A,B,C$. If L passes through the point $1:1:1$ (the centroid or incenter, assuming that the coordinates are barycentric or trilinear), then the cubic is degenerate as the union of a parabola and the line at infinity. The methods in this work are largely algebraic and computer-dependent.

1. Introduction

Loosely speaking, the history of triangle geometry stretches across three eras: (1) ancient Greek, as in Euclid’s Elements and as represented by four special points—incenter, centroid, circumcenter, and orthocenter; (2) European (and a bit American), represented by points and lines named after Fermat, Euler, Lemoine, Brocard, Feuerbach, Steiner, Morley, and many others; and (3) modern, for which we heartily recommend Paul Yiu’s Introduction to the Geometry of the Triangle [1]. One of the hallmarks of modern triangle geometry is the use of barycentric coordinates instead of the trilinear coordinates that were favored during the 19th and early 20th centuries. Regarding the fundamental role of these two homogeneous coordinate systems, the reader may find several Wikipedia articles to be useful, including “Conway triangle notation” (introduced by John Horton Conway in about 2000); see https://en.wikipedia.org/wiki/Conway_triangle_notation (accessed on 15 June 2024), and “triangle center” (https://en.wikipedia.org/wiki/Triangle_center) (accessed on 15 June 2024) and “circumcircle” (https://en.wikipedia.org/wiki/Circumcircle) (accessed on 15 June 2024).

Suppose that $ABC$ is a triangle with respective sidelengths of $a,b,c$. These are regarded as variables or indeterminates, so that triangle centers in the plane of $ABC$ are determined using homogeneous coordinates, either barycentric or trilinear. In this paper, all the triangle centers to be considered are polynomial centers, in the sense that each has a representation in barycentric (hence also trilinear) coordinates of the form

$$f(a,b,c):f(b,c,a):f(c,a,b),$$

where $f(a,b,c)$ is a homogeneous polynomial satisfying $f(a,c,b)=f(a,b,c)$. To represent such a point, it is sufficient to write merely $f(a,b,c)::$, as in the row for $X_3$ in

Table 1. Polynomial triangle centers (i.e., points).

Point	Barycentrics	Trilinears
$X_1=$ incenter, I	$a:b:c$	$1:1:1$
$X_2=$ centroid, G	$1:1:1$	$bc:ca:ab$
$X_3=$ circumcenter, O	$a^2(b^2+c^2-a^2$)::	$a(b^2+c^2-a^2)::$
$X_6=$ symmedian point, K	$a^2:b^2:c^2$	$a:b:c$

, which includes indexing $X_i$ as in the Encyclopedia of Triangle Centers [2].

A precedent for unary operations on triangle centers occurs in [2] as

$$X_{1981}=\mathrm{Vega}\mathrm{transform}\mathrm{of}{X}_{647},$$

where the Vega transform is defined in trilinears by

$$x:y:z\to (y-z)/x:(z-x)/y:(x-y)/z.$$

In this paper, this formula is represented as the unary operation $u_1\left(X\right)$. In the following definitions of eight unary operations, $X=x:y:z$ is a point given by homogeneous coordinates (barycentric or trilinear).

$$\begin{array}{cc}\hfill u_1\left(X\right)& =(y-z)/x::\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill u_2\left(X\right)& =x/(y-z)::\hfill \end{array}$$

(2)

$$\begin{array}{cc}\hfill u_3\left(X\right)& =(-2x+y+z)/x::\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill u_4\left(X\right)& =x/(-2x+y+z)::\hfill \end{array}$$

(4)

Continuing with definitions, suppose that k is a real number other than $-1$, which allows us to provide the following definitions.

$$\begin{array}{cc}\hfill u_{1,k}\left(X\right)& =\frac{y-z}{-kx+y+z}::\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill u_{2,k}\left(X\right)& =\frac{-kx+y+z}{y-z}::\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill u_{3,k}\left(X\right)& =\frac{-2x+y+z}{-kx+y+z}::,k\ne 2\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill u_{4,k}\left(X\right)& =\frac{-kx+y+z}{-2x+y+z}::,k\ne 2\hfill \end{array}$$

(8)

If the coordinates $x:y:z$ are barycentric, then $u_2$ is the isotomic conjugate of $u_1$ and $u_4$ is the isotomic conjugate of $u_3$; likewise, for each k, the point $u_{2,k}$ is the isotomic conjugate of $u_{1,k}$, and $u_{4,k}$ is the isotomic conjugate of $u_{3,k}$. On the other hand, if the coordinates $x:y:z$ are trilinear, then these pairs are isogonal conjugates rather than isotomic. In the sequel, unless stated otherwise, coordinates are assumed only to be homogeneous; i.e., the results hold for both barycentrics and trilinears.

For each X, the point $u_1\left(X\right)$ is a limiting case of the family of points $u_{1,k}\left(X\right)$; e.g., write (5) as

$$\frac{y-z}{-x+(1/k)(y+z)}$$

and let $k\to \infty$. Likewise, $u_2,u_3,u_4$ are limiting cases of $u_{2,k},u_{3,k},u_{4,k}$.

For a more detailed introduction to triangle geometry using barycentric coordinates, see the excellent online book by Paul Yiu [1], and several other helpful resources [2,3].

2. Images of Points

In Part 32 of the Encyclopedia of Triangle Centers [2], the preamble just before $X_{62530}$ shows two tables. The first presents 10 examples of each of the eight unary images given by (1)–(8). In this first table, the underlying homogeneous coordinates are barycentric; in the second table, they are trilinear. An examination of the rows of the barycentric table shows that the second four entries of row 1 reappear as the last four entries in row 8. That is, the incenter, $X_1=a:b:c$, is mapped by $u_1,u_2,u_3,u_4$ to the same points ($X_{693},X_{100},X_{4358},X_{88})$ to which the Nagel point, $X_8$, is mapped by $u_{1,k},u_{2,k},u_{3,k},u_{4,k}$ for $k=0$. Actually, much more is true. For example, not only is

$$u_1(a:b:c)=u_5(b+c-a:c+a-b:a+b-c),$$

but if

$$x:y:z=f(a,b,c):f(b,c,a):f(c,a,b)$$

is an arbitrary triangle center, then

$$u_1(x:y:z)=u_5(y+z-x,z+x-y,x+y-z).$$

This sort of symbolic substitution has been studied previously in other settings [4,5]. The next example involving $u_1$ and $u_5$ takes its place in the following theorem.

Theorem 1.

Suppose the $X=x:y:z$ is a triangle center and that ${\widehat{X}}_i=SS\left(X_i:(a,b,c)\to (x,y,z)\right)$; i.e., ${\widehat{X}}_i$ is the result of substituting $x,y,z$, respectively, for $a,b,c$ in the coordinates for $X_i$. Then,

Proof. 

Beginning with row 1 in the table,

$$\begin{array}{cc}\hfill u_5\left({\widehat{X}}_8\right)& =u_5(y+z-x:z+x-y:x+y-z)\hfill \\ \hfill & =\frac{z+x-y-(x+y-z)}{z+x-y+x+y-z}::\hfill \\ \hfill & =\frac{2z-2y}{2x}::\hfill \\ \hfill & =(y-z)/x:(z-x)/y:(x-y)/z\hfill \\ \hfill & =u_1\left({\widehat{X}}_1\right)\hfill \end{array}$$

Taking reciprocals gives $u_2\left({\widehat{X}}_1\right)=u_6\left({\widehat{X}}_8\right)$. For the third entry in row 1, let

$$x^{\prime }=y+z-x,y^{\prime }=z+x-y,z^{\prime }=x+y-z,$$

so that

$$\begin{array}{cc}\hfill u_7\left({\widehat{X}}_8\right)& =(-2x^{\prime }+y^{\prime }+z^{\prime })/(y^{\prime }+z^{\prime })::\hfill \\ \hfill & =\frac{-2(y+z-x)+z+x-y+x+y-z}{z+x-y+x+y-z}::\hfill \\ \hfill & =(2y-2z+4x)/\left(2x\right)::\hfill \\ \hfill & =u_3\left({\widehat{X}}_1\right),\hfill \end{array}$$

and reciprocation gives $u_4\left({\widehat{X}}_1\right)=u_8\left({\widehat{X}}_8\right)$.

The equations in row 1 of

Table 2. Families of identical pairs of images.

$u_1\left({\widehat{X}}_1\right)=u_5\left({\widehat{X}}_8\right)$	$u_2\left({\widehat{X}}_1\right)=u_6\left({\widehat{X}}_8\right)$	$u_3\left({\widehat{X}}_1\right)=u_7\left({\widehat{X}}_8\right)$	$u_4\left({\widehat{X}}_1\right)=u_8\left({\widehat{X}}_8\right)$
$u_1\left({\widehat{X}}_3\right)=u_5\left({\widehat{X}}_4\right)$	$u_2\left({\widehat{X}}_3\right)=u_6\left({\widehat{X}}_4\right)$	$u_3\left({\widehat{X}}_3\right)=u_7\left({\widehat{X}}_4\right)$	$u_4\left({\widehat{X}}_3\right)=u_8\left({\widehat{X}}_4\right)$
$u_1\left({\widehat{X}}_9\right)=u_5\left({\widehat{X}}_7\right)$	$u_2\left({\widehat{X}}_9\right)=u_6\left({\widehat{X}}_7\right)$	$u_3\left({\widehat{X}}_9\right)=u_7\left({\widehat{X}}_7\right)$	$u_4\left({\widehat{X}}_9\right)=u_8\left({\widehat{X}}_7\right)$
$u_1\left({\widehat{X}}_{10}\right)=u_5\left({\widehat{X}}_1\right)$	$u_2\left({\widehat{X}}_{10}\right)=u_6\left({\widehat{X}}_1\right)$	$u_3\left({\widehat{X}}_{10}\right)=u_7\left({\widehat{X}}_1\right)$	$u_4\left({\widehat{X}}_{10}\right)=u_8\left({\widehat{X}}_1\right)$

are now proved, and the method of proof applies to the remaining nine equations. We include here just one more proof, that $u_3\left({\widehat{X}}_3\right)=u_7\left({\widehat{X}}_4\right)$. Let

$$x^{\prime }=x^2(x^2-y^2-z^2),y^{\prime }=y^2(y^2-z^2-x^2),z^{\prime }=z^2(z^2-x^2-y^2).$$

$$\begin{array}{cc}\hfill u_3\left({\widehat{X}}_3\right)& =(-2x^{\prime }+y^{\prime }+z^{\prime })/x^{\prime }::\hfill \\ \hfill & =\frac{-2x^4+y^4+z^4+x^2{y}^2+x^2{z}^2-2y^2{z}^2}{x^2(x^2-y^2-z^2)}::.\hfill \end{array}$$

Now let

$$x^{″}=1/(x^2-y^2-z^2),y^{″}=1/(y^2-z^2-x^2),z^{″}=1/(z^2-x^2-y^2).$$

$$\begin{array}{cc}\hfill u_7\left({\widehat{X}}_4\right)& =(-2x^{″}+y^{″}+z^{″})/(y^{″}+z^{″})::\hfill \\ \hfill & =\frac{-2/(x^2-y^2-z^2)+1/(y^2-z^2-x^2)+1/(z^2-x^2-y^2)}{1/(y^2-z^2-x^2)+1/(z^2-x^2-y^2)}::\hfill \\ \hfill & =\frac{-2x^4+y^4+z^4+x^2{y}^2+x^2{z}^2-2y^2{z}^2}{x^2(x^2-y^2-z^2)}::\hfill \\ \hfill & =u_3\left({\widehat{X}}_3\right).\hfill \end{array}$$

   □

3. Images of Lines

Because each binary operation u is defined on a set of individual points, it is also defined on lines. Thus, if L is a line, given by homogeneous coordinates by an equation $lx+my+nz=0$, then we may—and do—speak of $u\left(L\right)$. Such images are cubic curves (possibly degenerate) that pass through the vertices $A,B,C$, as indicated here:

$$\begin{array}{cc}\hfill u_1\left(L\right)& :l^2(y+z)yz+\left(cyclic\right)-2(mn+nl+lm)xyz=0;\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill u_2\left(L\right)& :(m^{2}y+n^{2}z)yz+\left(cyclic\right)-2(mn+nl+lm)xyz=0;\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill u_3\left(L\right)& :l\left((l+2n)y-(l+2m)z\right)yz+\left(cyclic\right)=0;\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill u_4\left(L\right)& :\left(m(m+2n)y-n(n+2m)z\right)yz+\left(cyclic\right)=0;\hfill \\ \hfill u_{1,k}\left(L\right)& :{\left((k-1)l+m+n\right)}^{2}yz(y+z)+\left(cyclic\right)\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill & -2\left((2k-1)(l^2+m^2+n^2)+(k^2+2)(mn+nl+lm)\right)xyz=0;\hfill \\ \hfill u_{2,k}\left(L\right)& :{\left((k-1)l+m+n\right)}^2{x}^2(y+z)+\left(cyclic\right)\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill & -2\left((2k-1)(l^2+m^2+n^2)+(k^2+2)(mn+nl+lm)\right)xyz=0;\hfill \\ \hfill u_{3,k}\left(L\right)& :\left(k-1)l+m+n\right)\left((k+1)l+3m-n+2kn\right)y^{2}z\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill & -\left(k+1)l+3n-m+2km)y{z}^2\right)+\left(cyclic\right)=0;\hfill \\ \hfill u_{4,k}\left(L\right)& :\left((k-1)m+n+l)\right)\left((k+1)m+(2k-1)n+3l\right)y^{2}z\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill & -\left((k+1)n+m+l)\right)\left(k+1)n+(2k-1)m+3l\right)y{z}^2+\left(cyclic\right)=0.\hfill \end{array}$$

(16)

For Examples (1)–(5), suppose that L is a line that passes through the point 1:1:1 and a point $P=p:q:r$ other than 1:1:1, so that L is provided by the equation $(q-r)x+(r-p)y+(p-q)z=0$. (Here, the coordinates may be barycentric or trilinear.)

Example 1.

$u_1\left(L\right)$ is given by

$$(x+y+z)\left({(q-r)}^{2}yz+{(r-p)}^{2}zx+{(p-q)}^{2}xy\right)=0.$$

(17)

That is, the left-hand side in (9) factors, showing the image of L as a degenerate cubic consisting of the union of $L^{\infty }$ and a circumconic. Specifically, in cases in which all the coordinates are barycentric, the conic has a characteristic

$$p_1^2+q_1^2+r_1^2-2q_1{r}_1-2r_1{p}_1-2p_1{q}_1,$$

where

$$p_1={(q-r)}^2/2,q_1={(r-p)}^2/2,r_1={(p-1)}^2/2.$$

As this characteristic is identically 0, the conic is a parabola ([1], p. 127). Moreover, the infinite point (i.e., the direction of opening) of the parabola is the point $q-r:r-p:p-q$, and the perspector is the point ${(q-r)}^2:{(r-q)}^2:{(p-q)}^2$, which lies on the Steiner inellipse.

The likewise degenerate cubic $u_2\left(L\right)$ is given by

$$(yz+zx+xy)\left({(q-r)}^{2}x+{(r-p)}^{2}y+{(p-q)}^{2}z\right)=0.$$

(18)

This, the isotomic conjugate of $u_1\left(L\right)$, is the union of a line and a conic; viz., the conic $yz+zx+xy=0$, which is the circumcircle if the coordinates are barycentric, and the Steiner circumellipse if the coordinates are trilinear.

Example 2.

Here, we extend the discussion of the parabola given by the barycentric equation

$${(q-r)}^{2}yz+{(r-p)}^{2}zx+{(p-q)}^{2}xy=0,$$

(19)

encountered in Example 1. Let

$$a_1=b^2+c^2-a^2,b_1=c^2+a^2-b^2,c_1=a^2+b^2+c^2.$$

Then, the focus of the parabola is the point that has the first barycentric

$$\begin{array}{cc}\hfill (q-r)(b^2{q}^4& +c^2{q}^4-(a_1+2b^2)q^{3}r-(a_1+2c^2)q{r}^3\hfill \\ \hfill & +c_1(3r-q)p{q}^2+b_1(3q-r)p{r}^2\hfill \\ \hfill & +3a_1{q}^2{r}^2+a^{2}p(2pq+2pr-p^2-6qr))\hfill \end{array}$$

The appearance of $(i,j)$ in the following list means that if $P=X_i$, then the focus is $X_j$.

$$\begin{array}{cc}\hfill & (1,38109),(3,38233),(6,12064),(7,43940),(37,38018),(39,38017),(92,43939),\hfill \\ \hfill & (514,43938),(523,38020),(525,38246),(647,43935),(650,43937),(905,43936).\hfill \end{array}$$

The intersections of the lines tangent to the parabola at the vertices $A,B,C$ form a triangle, T, perspective to $ABC$ at ${(q-r)}^2::$ (the perspector of the parabola). The circumcircle of T passes through the focus.

The vertex of the parabola has barycentric coordinates too long to be shown here; nevertheless, a computer finds similar vertices to those that are listed here; the appearance of $(i,j)$ means that if $P=X\left(i\right)$, then the vertex is $X\left(j\right)$.

$$\begin{array}{cc}\hfill & (1,43943),(3,43945),(6,12065),(37,38242),(39,38241),(514,38244),\hfill \\ \hfill & (523,38245),(525,43941),(647,43942),(650,43944),(905,38243).\hfill \end{array}$$

Other examples of the parabola (19) include the following:

• If $L=X_1{X}_2$ (Nagel line), the parabola $u_1\left(L\right)$ passes through $X_i$ for these i: 514, 693, 927, 3676, 4444, 4555, 4583, 4608, 4817, 6548, 7192, 15634, 17925, 17930, 37143, 43943, 52619, 52620, 53150, 57994, 58860, 59941, 62635.
• If $L=X_2{X}_3$ (Euler line), the parabola $u_1\left(L\right)$ passes through $X_i$ for these i: $524,850,2867,3265,16077,17708,17932,34767,43673,43945,52617,53173,62428,62724$. The list [6] includes more than 6000 triangle centers X on the Euler line; each begets a point $u_1\left(X\right)$ on the cubic $u_1\left(X_2{X}_3\right)$.
• If $L=X_2{X}_6$, then $u_1\left(L\right)$ is known as the X-parabola, introduced in [2] at $X_{12065}$.
• If L is the minor axis of the Steiner circumellipse, then $u_1\left(L\right)$ passes through the Steiner point, $X_{99}$; the first Kiepert infinity point, $X_{3413}$; and $X_{30509}$, which lies on the Kiepert parabola (which is not $u_1\left(L\right)$).
• If L is the major axis of the Steiner circumellipse, then $u_1\left(L\right)$ passes through the Steiner point, $X_{99}$; the second Kiepert infinity point, $X_{3414}$; and $X_{30508}$.

Example 3.

$u_3\left(L\right)$ is given by

$$(x+y+z)\left((2p-q-r)(q-r)yz+(2q-r-p)(r-p)zx+(2r-p-q)(p-q)xy\right)=0,$$

and $u_4\left(L\right)$ by

$$(yz+zx+xy)\left((2p-q-r)(q-r)x+(2q-r-p)(r-p)y+(2r-p-q)(p-q)z\right)=0.$$

Example 4.

Equation (17) also represents $u_{1,0}\left(L\right)$, so that (18) holds for $u_{2,0}\left(L\right)$.

Example 5.

Equation (17) also represents $u_{3,0}\left(L\right)$, so that (18) holds for $u_{4,0}\left(L\right)$.

Example 6.

Here, we discuss the cubic $u_1\left(X_1{X}_6\right)$. If L is the line of the incenter and the symmedian point; i.e., the line $X_1{X}_6$, then (9) represents, in barycentric coordinates, the cubic

$$\begin{array}{cc}\hfill & b^2{c}^2{(b-c)}^2(y+z)yz+c^2{a}^2{(c-a)}^2(z+x)zx+a^2{b}^2{(a-b)}^2(x+y)xy\hfill \\ \hfill & -2abc\left(a^3+b^3+c^3-(b+c)a-(c+a)b-(a+b)c+3abc\right)xyz\hfill \\ \hfill & =0\hfill \end{array}$$

The following thirteen triangle centers are known to lie on $u_1\left(X_1{X}_6\right)$: $X_i$ for $i=$ 693, 850, 7192, 62,725, and 63,216 to 63,224. The first barycentric of each of these has the form $(b-c)p(a,b,c)$, where $p(a,b,c)$ is a polynomial. Consequently, the thirteen points $p(a,b,c)::$ lie on a cubic, and the thirteen points $ap(a,b,c)::$ lie on a cubic, etc. Among the cubics presented in [7] the one most closely related to $u_1\left(X_1{X}_6\right)$ is the cubic through the points $p(a,b,c)::$; it is the isotomic conjugate of the cubic K588; for details, see [8]. For details, see the preamble [2] just before $X_{62530}$.

Further examples of cubics in the family (9) include the following:

• $u_1\left(L^{\infty }\right)$ is the Tucker nodal cubic, K015 in B. Gibert’s indexing [7].
• If $L=X_3{X}_6$, the Brocard axis, then $u_1\left(L\right)$ passes through $X_i$ for $i=$ 850, 58,784, 62,428.
• If $L=X_{230}{X}_{231}$, the orthic axis, then $u_1\left(L\right)$ passes through $X_i$ for $i=$ 69, 877, 883, 5468, 62,645.
• If $L=X_{44}{X}_{513}$, the anti-orthic axis, then $u_1\left(L\right)$ passes through $X_i$ for $i=$ 75, 874, 883, 41,314, 62,638.
• If $L=X_{44}{X}_{513}$, the Lemoine axis, then $u_1\left(L\right)$ passes through $X_i$ for $i=76,877,880,5698$.
• If $L=X_{325}{X}_{523}$, the de Longchamps axis, then $u_1\left(L\right)$ passes through $X_i$ for $i=$ 6, 880, 2284, 2395, 5468, 14,966, 23,968, 23,977, 43,929, 52,131, 52,132.
• If $L=X_{241}{X}_{514}$, the Gergonne line, then $u_1\left(L\right)$ passes through $X_i$ for $i=$ 8, 883, 17,780, 23,836, 53,151.
• If $L=X_6{X}_{13}$, the Fermat line, then $u_1\left(L\right)$ passes through $X_i$ for $i=$ 850, 3268, 5468, 9141, 34,765, 34,767.
• If $L=X_6{X}_{17}$, the Napoleon line, then $u_1\left(L\right)$ passes through $X_i$ for $i=$ 850, 41,298, 62,724.
• If $L=X_7{X}_{883}$, the Garcia-Reznick line, then $u_1\left(L\right)$ passes through $X_i$ for $i=$ 7, 883, 2400, 23,973, 43,930, 56,543.
• If $L=X_1{X}_3$, then $u_1\left(L\right)$ passes through $X_i$ for $i=$ 693, 850, 4391, 4397, 4581, 7253.
• If $L=X_1{X}_5$, then $u_1\left(L\right)$ passes through $X_i$ for $i=693,3265,4131,4397,6332,7253$.

For lists of points on the lines mentioned just above, see the index of central lines [6].

4. Inverses of Unary Operations

Suppose that $P=p:q:r$. For each unary operation u given by (1)–(8) and each point P, there are formally two points X such that $u\left(X\right)=P$. That is, each P has two inverses. Equations (24) to (32) give the first coordinates of the two inverses using the symbol $\delta$, with the 1st inverse defined by $\delta =1$ and the 2nd by $\delta =-1$. Let

$$\begin{array}{cc}\hfill D_1& =-pqr(p+q+r);\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill D_2& =-qr-rp-pq;\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill D_3& =(1/2)\left(p^2{q}^2{r}^2\right)\left({(1/q-1/r)}^2+{(1/r-1/p)}^2+{(1/p-1/q)}^2\right)\hfill \\ \hfill & =q^2{r}^2+r^2{p}^2+p^2{q}^2-pqr(p+q+r);\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill D_4& =(1/2)\left({(q-r)}^2+{(r-p)}^2+{(p-q)}^2)\right)\hfill \\ \hfill & =p^2+q^2+r^2-qr-rp-pq.\hfill \end{array}$$

(23)

Then

$$\begin{array}{cc}\hfill u_1^{-1}& =qr(2p-q-r)+(q-r)\delta \sqrt{D_1}::\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill u_2^{-1}& =p\left(2qr-rp-pq+(q-r)\delta \sqrt{D_2}\right)::\hfill \end{array}$$

(25)

$$\begin{array}{cc}\hfill u_3^{-1}& =(q-r)(2qr-rp-pq+\delta \sqrt{D_3})::\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill u_4^{-1}& =p(q-r)(2p-q-r+\delta \sqrt{D_4})::\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill u_{1,k}^{-1}& =p^2(q+r)+q^2(p-r)+r^2(p-q)-2pqr\hfill \\ \hfill & +kp(q^2+r^2-4qr+pq+pr)\hfill \\ \hfill & +k^2{(r(q-r)}^2+\delta (k-2)(q-r)\sqrt{D_1}::\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill u_{2,k}^{-1}& =p^2(q+r)-q^2(r+p)-r^2(p+q)+2pqr\hfill \\ \hfill & +kp(2qr-rp-pq)\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill & +\delta (k-2)p(q-r)\sqrt{D_2}::\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill u_{3,k}^{-1}& =(q-r)(3p^2-(qr+rp+pq)+k(2qr-rp-pq)\hfill \\ \hfill & +\delta (k-2)\sqrt{D_3})::\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill u_{4,k}^{-1}& =(q-r)(3qr-p(p+q+r)+kp(2p-q-r)\hfill \\ \hfill & +\delta (k-2)p\sqrt{D_4})::\hfill \end{array}$$

(32)

The foregoing formulas for inverses were found using the following Mathematica code, in which the coordinates for each $u_i$ or $u_{i,k}$ can be used as demonstrated here with $u\left[1\right]$ in the first and third lines.

u[1][{x_, y_, z_}] := {(y - z)/x, (z - x)/y, (x - y)/z}

{x, y, z} /. Factor[Solve[Join[{x + y + z == 1},

Thread[#/Plus @@ # &[u[1][{x, y, z}]]==

#/Plus @@ # &[{p, q, r}]]], {x, y, z}]]

Equations (20) and (24) indicate that there is only one inverse instead of two if $p+q+r=0$; i.e., if the coordinates are barycentric, then P lies on $L^{\infty }$; if they are trilinear, then P lies on the Steiner circumellipse. Possibly $D_1<0$, as when P is the incenter ($a:b:c$, in barycentrics); in that case, the two inverses are a complex conjugate pair. Similarly, Equations (21) and (25) show that there is only one inverse if $1/p+1/q+1/r=0.$ Equations (22) and (26) indicate a single inverse if $P=1:1:1$, and likewise for (23) and (27).

For $i=1,2,3,4$, if $P=u_i\left(F\right)$ for some point $F=f:g:h$, then $\sqrt{D_i}$ is radical-free, and we obtain the following results, which are easily obtained by computer:

$\sqrt{D_1}=-(g-h)(h-f)(f-g)/\left(fgh\right)$; (1st inverse of P)$=F$, and

(2nd inverse of P) $=f(f-g-h)::$, which is the G-Ceva conjugate of F.

$\sqrt{D_2}=1$; (1st inverse of P) $= F$, and (2nd inverse of P) $= f(f-g-h)::$, which is the G-Ceva conjugate of F.

$$\sqrt{D_3}=(f+g+h)\left(f^2+g^2+h^2-(gh+hf+fg)\right)/\left(fgh\right);$$

(1st inverse of P)$=F$, and (2nd inverse of P) $=f(f-g-h)/(2f-g-h)::$.

$$\sqrt{D_4}=\frac{-(f+g+h)\left(f^2+g^2+h^2-(gh+hf+fg)\right)}{(2f-g-h)(2g-h-f)(2h-f-g))};$$

(1st inverse of P) $=f(f-g-h)/(2f-g-h)::$, and (2nd inverse of P) $=F$.

Referring to (20)–(23), note first that $D_1=0$ if and only if P is on the line at infinity, so that

$$u_1^{-1}\left(P\right)=qr(2p-q-r)::.$$

Similarly, $D_2=0$ if and only if P is on the Steiner circumellipse, and then

$$u_2^{-1}\left(P\right)=p(2ar-rp-pq)::.$$

Further, $D_3=0$ if and only if $P=1:1:1$, the centroid of $ABC$, and likewise for $D_4=0$.

Example 7.

Case 1a: coordinates in (1) and (24) are barycentric. It is easy to establish that if

$$u_1^{-1}\left(X_i\right)=\{X_j,X_k\},$$

then $X_j$ and $X_k$ are a pair of $X_2$-Ceva conjugates.

$$\begin{array}{cc}\hfill u_1^{-1}\left(X_{693}\right)& =\{X_1,X_9\};\hfill \\ \hfill u_1^{-1}\left(X_{850}\right)& =\{X_3,X_6\};\hfill \\ \hfill u_1^{-1}\left(X_{874}\right)& =\{X_{244},X_{661}\};\hfill \\ \hfill u_1^{-1}\left(X_{877}\right)& =\{X_{125},X_{647}\};\hfill \\ \hfill u_1^{-1}\left(X_{892}\right)& =\{X_{1648},X_{1649}\};\hfill \\ \hfill u_1^{-1}\left(X_{7192}\right)& =\{X_{10},X_{37}\}.\hfill \end{array}$$

Case 1b: coordinates in (1) and (24) are trilinear. Here, if

$$u_1^{-1}\left(X_i\right)=\{X_j,X_j\},$$

then $X_j$ and $X_k$ are a pair of $X_1$-Ceva conjugates.

$$\begin{array}{cc}\hfill u_1^{-1}\left(X_{514}\right)& =\{X_6,X_{55}\};\hfill \\ \hfill u_1^{-1}\left(X_{1577}\right)& =\{X_{31},X_{48}\}.\hfill \end{array}$$

Case 2: coordinates in (2) and (25) are barycentric. If

$$u_1^{-1}\left(X_i\right)=\{X_j,X_k\},$$

then $X_j$ and $X_k$ are a pair of $X_2$-Ceva conjugates. Suppose that X is a point and $X^{*}$ is the isotomic conjugate of X. If $u_1^{-1}\left(X\right)=\{P,U\}$, then $u_2^{-1}\left(X^{*}\right)=\{P,U\}$, as illustrated here:

$$\begin{array}{cc}\hfill u_2^{-1}\left(X_{100}\right)& =\{X_1,X_9\};\hfill \\ \hfill u_2^{-1}\left(X_{110}\right)& =\{X_3,X_6\}.\hfill \end{array}$$

Case 3: coordinates in (3) and (26) are barycentric.

$$\begin{array}{cc}\hfill u_3^{-1}\left(X_{69}\right)& =\{X_{24007},X_{24008}\};\hfill \\ \hfill u_3^{-1}\left(X_{76}\right)& =\{X_{5638},X_{5639}\};\hfill \\ \hfill u_3^{-1}\left(X_{182}\right)& =\{X_{850},X_{46807}\};\hfill \\ \hfill u_3^{-1}\left(X_{468}\right)& =\{X_{69},X_{14977}\};\hfill \\ \hfill u_3^{-1}\left(X_{671}\right)& =\{X_{52722},X_{52723}\};\hfill \\ \hfill u_3^{-1}\left(X_{3266}\right)& =\{X_6,X_{9178}\};\hfill \\ \hfill u_3^{-1}\left(X_{4358}\right)& =\{X_1,X_{1022}\};\hfill \\ \hfill u_3^{-1}\left(X_{11064}\right)& =\{X_4,X_{2394}\};\hfill \\ \hfill u_3^{-1}\left(X_{16704}\right)& =\{X_{10},X_{4049}\}.\hfill \end{array}$$

Case 4: coordinates in (4) and (27) are barycentric. Suppose that X is a point and $X^{*}$ is the isotomic conjugate of X. If $u_3^{-1}\left(X\right)=\{P,U\}$, then $u_4^{-1}\left(X^{*}\right)=\{P,U\}$, as illustrated here:

$$\begin{array}{cc}\hfill u_4^{-1}\left(X_4\right)& =\{X_{24007},X_{24008}\};\hfill \\ \hfill u_4^{-1}\left(X_6\right)& =\{X_{5638},X_{5639}\};\hfill \end{array}$$

Case 5: coordinates in (5) and (28) are barycentric. Here, $k=0$, so that

$$u_{1,0}=(y-z)/(y+z)::.$$

If $u_{1,0}^{-1}\left(X\right)=\{P,U\}$, then U is the isotomic conjugate of P.

$$\begin{array}{cc}\hfill u_{1,0}^{-1}\left(X_{99}\right)& =\{X_{30508},X_{30509}\};\hfill \\ \hfill u_{1,0}^{-1}\left(X_{100}\right)& =\{X_{60476},X_{60477}\};\hfill \\ \hfill u_{1,0}^{-1}\left(X_{476}\right)& =\{X_{2394},X_{2407}\};\hfill \\ \hfill u_{1,0}^{-1}\left(X_{850}\right)& =\{X_4,X_{69}\};\hfill \\ \hfill u_{1,0}^{-1}\left(X_{877}\right)& =\{X_{110},X_{850}\};\hfill \\ \hfill u_{1,0}^{-1}\left(X_{883}\right)& =\{X_{100},X_{693}\};\hfill \\ \hfill u_{1,0}^{-1}\left(X_{4608}\right)& =\{X_{10},X_{86}\};\hfill \\ \hfill u_{1,0}^{-1}\left(X_{7192}\right)& =\{X_1,X_{75}\};\hfill \\ \hfill u_{1,0}^{-1}\left(X_{7253}\right)& =\{X_{63},X_{92}\};\hfill \\ \hfill u_{1,0}^{-1}\left(X_{62631}\right)& =\{X_{13},X_{298}\};\hfill \\ \hfill u_{1,0}^{-1}\left(X_{62632}\right)& =\{X_{14},X_{299}\};\hfill \\ \hfill u_{1,0}^{-1}\left(X_{62724}\right)& =\{X_5,X_{95}\};\hfill \\ \hfill u_{1,0}^{-1}\left(X_{62725}\right)& =\{X_9,X_{85}\};\hfill \\ \hfill u_{1,0}^{-1}\left(X_{62726}\right)& =\{X_{11},X_{4998}\};\hfill \end{array}$$

5. Binary Operations on Inverse Pairs

In this section, all coordinates are trilinears. (Analogous results can be written out in case the underlying homogeneous coordinates are barycentrics.) In the Glossary [9] of ETC [2], several binary operations on pairs $P=p:q:r$ and $U=u:v:w$ of points are defined, and geometric constructions are presented. Nine of these are listed here:

$$\begin{array}{cc}\hfill & B_1,\mathrm{trilinear}\mathrm{product}:pu:qv:rw\hfill \\ \hfill & B_2,\mathrm{sum}:p+u:q+v:r+w\hfill \\ \hfill & B_3,\mathrm{difference}:p-u:q-v:r-w\hfill \\ \hfill & B_4,\mathrm{cevapoint}:(pv+qu)(pw+ru)::\hfill \\ \hfill & B_5,\mathrm{crosspoint}:pu(rv+qw):qv(pw+ru):rw(qu+pv)\hfill \\ \hfill & B_6,\mathrm{trilinear}\mathrm{pole}:1/(qw-rv):1/(ru-pw):1/(pv-qu)\hfill \\ \hfill & B_7,P-\mathrm{isoconjugate}\mathrm{of}U:qrvw:rpwu:pquv\hfill \\ \hfill & B_8,\mathrm{crosssum}:qw+rv:ru+pw:pv+qu\hfill \\ \hfill & B_9,\mathrm{crossdifference}:qw-rv:ru-pw:pv-qu\hfill \end{array}$$

If P and U are the inverse pair of a unary operation $u_i$, for $i=1,2,\dots ,8$, then each of the binary operations listed above yields a triangle center. The next two tables depict the first trilinear coordinate of the triangle center $B_i(P,U)$, where P and U are the inverses of $u_j$ (in

Table 3. Binary operations on unary inverse pairs (trilinear coordinates).

	${\mathit{u}}_1^{-1}$	${\mathit{u}}_2^{-1}$	${\mathit{u}}_3^{-1}$	${\mathit{u}}_4^{-1}$
$B_1$	$(y+z)/x$	$x^2(y+z)$	$yz(y+z)$	$x^2(y-z)$
$B_2$	$x(-2x+y+z)$	$x(-2yz+zx+xy)$	$(y-z)(-2yz+zx+xy)$	$x(y-z)(-2yz+zx+xy)$
$B_3$	$y-z$	$x(y-z)$	$y-z$	$x(y-z)$
$B_4$	1	1	1	1
$B_5$	$(y+z)/x$	$y+z$	$x{(y+z)}^2$	$x{(y-z)}^2$
$B_6$	$1/x$	x	$1/x$	x
$B_7$	$x/(y+z)$	$1/\left(x^2(y+z)\right)$	$x(y-z)$	$1/\left(x^2(y-z)\right)$
$B_8$	1	1	$x^2(y-z)$	$x(y-z)$
$B_9$	x	$1/x$	x	$1/x$

) or $u_{j,0}$ (in

Table 4. Binary operations on unary inverse pairs (trilinear coordinates).

	${\mathit{u}}_{1,0}^{-1}$	${\mathit{u}}_{2,0}^{-1}$	${\mathit{u}}_{3,0}^{-1}$	${\mathit{u}}_{4,0}^{-1}$
$B_1$	1	1	$(y-z)Q$	${(y-z)}^2$
$B_2$	1	1	$(y-z)Q$	$(y-z)Q^{*}$
$B_3$	$y-z$	$x(y-z)$	$y-z$	$y-z$
$B_4$	1	$1/\left((y+z)Q^{**}\right)$	1	1
$B_5$	$(y+z)/\left(yz-x(x+y+z)\right)$	$1/\left((y+z)Q^{**}\right)$	${(y-z)}^2(x+y)(x+z)/Q$	$y-z$
$B_6$	$1/(y+z)$	$1/\left(x(y+z)\right)$	$1/(y+z)$	$1/(4x+y+z)$
$B_7$	$1/(y+z)$	1	$(x-y)(x-z)/Q$	$1/{(y-z)}^2$
$B_8$	$(y+z)/\left(yz-x(x+y+z)\right)$	$(y+z)Q^{**}$	$(y-z)/(y+z)$	$1/(y-z)$
$B_9$	$y+z$	$x(y+z)$	$y+z$	$4x+y+z$

), where $i=1,2,\dots ,9$ and $j=1,2,3,4$. For example,

$$B_1\left(u_1^{-1}\right)=(y+z)/x:(z+x)/y:(x+y)/z.$$

In Table 4, a continuation of Table 3, we use the following abbreviations:

$$\begin{array}{cc}\hfill Q& =3x^2-yz+xy+xz;\hfill \\ \hfill Q^{*}& =3yz-x(x+y+z);\hfill \\ \hfill Q^{**}& =x^2-yz-zx-xy.\hfill \end{array}$$

In addition to the results in Table 3 and Table 4, we have a tenth binary operation on the inverse pairs: their midpoint. A convenient way to represent midpoints of the eight pairs in (24) to (32) is to substitute 0 for $\delta$ in those equations. (This works for barycentric coordinates, too.)

6. Constructions

Starting with a point $X=x:y:z$, the points $u_i\left(X\right)$, for $i=1,2,3,4$, are Euclidean constructible in both barycentric and trilinear coordinates.

In barycentrics, the point

$$1/X=1/x:1/y:1/z=yz:zx:xy$$

is the isotomic conjugate of X, with a well-known construction [1] based on midpoints of the sides of the reference triangle $ABC$. In trilinears, $1/X$ is the isogonal conjugate of X, with a similar construction based on angle bisector of $ABC$.

In trilinears, suppose that $\alpha :\beta :\gamma$ is a variable point. Then, the line $x\alpha +y\beta +z\gamma =0$, abbreviated as $[x:y:z]$, is the trilinear polar of $1/X$, constructible as the perspectrix of the cevian triangle of the isogonal conjugate, $1/X$. Similarly, in barycentrics, the line $[x:y:z]$ is the perspectrix of the cevian triangle of the isogonal conjugate, $1/X$. Thus, if a construction of a point X is known, then we have a construction for the line $[x:y:z]$, and conversely; this fact will be used repeatedly below.

In this continuing development of tools for constructing lines and points, we have, in barycentrics, the product $P\ast U=pu:qv:rw$ of points $P=p:q:r$ and $U=u:v:w$, defined if $pu,qv,rv$ are not all 0. An elegant construction for barycentric product was found by Paul Yiu ([1], p. 99). Using the same symbols, a construction for the trilinear product of two points was found by Cyril Parry and is included in Angel Montesdeoca’s online Glossary ([10], item 174: producto trilineal). In both barycentrics and trilinears, the quotient $P/U$ is defined by $P\ast (1/U)$.

Next, in both coordinate systems, the point

$$u_1\left(X\right)=(y-z)/x:(z-x)/y:(x-y)/z$$

is simply the intersection of the lines $[x:y:z]$ and $[x^2:y^2:z^2]$. Consequently, we have constructions of $u_2\left(X\right)$ as the reciprocal of $u_1\left(X\right)$; that is, the isotomic conjugate of $u_1\left(X\right)$ in barycentrics, and isogonal conjugate of $u_1\left(X\right)$ in trilinears.

To construct $u_3\left(X\right)$, we first obtain $1/y-1/z::$ as $X^2\ast u_1\left(X\right)$. The line of X and $X^2\ast u_1\left(X\right)$ is then $\left[\right(-2x+y+z)/x::]$, from which we construct the point

$$(-2x+y+z)/x:(-2y+z+x)/y:(-2z+x+y)/z,$$

which is $u_3\left(X\right)$. Then, $u_4\left(X\right)$ is constructed as the reciprocal of $u_3\left(X\right)$.

We turn now to constructions of points $B_i$ in Table 3 and Table 4.

First, $y-z::=X\ast u_1\left(X\right)$, and then we have a construction for the point $y+z::$ as the intersection of lines

$$[y-z:z-x:x-y]\mathrm{and}[x^2(y-z):y^2(z-x):z^2(x-y)].$$

Another construction of $y+z::$ is to intersect the lines $[x:y:z]$ and $[1/x:1/y:1/z]$ and then divide by $y-z::$.

Constructions for all the points in Table 3 are now established. For Table 4, the intersection of $[2x-y-z::]$ and $[1/(y+z)::]$ is $(y^2-z^2)(4x+y+z)::$, from which we obtain $4x+y+z::$, which accounts for the entry in the lower-right corner of Table 4. Constructions for other entries are left to the reader.

7. Unary Operations on Triangles

Suppose that $P=p:q:r$ is a triangle center. Then, the points

$$A^{\prime }=q+r:-q:-r,B^{\prime }=-p:r+p:-r,C^{\prime }=-p:-q:p+q,$$

(33)

lie on the line at infinity. Applying the unary operations $u_i$ to the vertices in (33) results in the triangles that are considered in this section.

Case 1. Let $A^{″}{B}^{″}{C}^{″}$ be the image of $A^{\prime }{B}^{\prime }{C}^{\prime }$ under $u_1$, so that

$$\begin{array}{cc}\hfill A^{″}& =(q-r)/(q+r):-(q+2r)/q:(2q+r)/r\hfill \\ \hfill B^{″}& =(2r+p)/p:(r-p)/(r+p):-(r+2p)/r\hfill \\ \hfill C^{″}& =-(p+2q)/p:(2p+q)/q:(p-q)/(p+q).\hfill \end{array}$$

If P lies on the Steiner circumellipse (i.e., $D_1=qr+rp+pq=0)$, then the lines $A{A}^{″},B{B}^{″},$

$C{C}^{″}$ concur, which is to say that the triangles $A^{″}{B}^{″}{C}^{″}$ and $ABC$ are perspective. The perspector is the barycentric quotient $U/P$, where U is the reflection of P in the centroid, 1:1:1. This perspector lies on the Tucker nodal cubic, indexed as K015 and discussed in [7].

Case 2. Again taking $A^{\prime }{B}^{\prime }{C}^{\prime }$ as in (33), we apply $u_2$ to obtain

$$\begin{array}{cc}\hfill A^{*}& =(q+r)/q-r):-q/(q+2r):r/(2q+r)=\mathrm{isotomic}\mathrm{conjugate}\mathrm{of}{A}^{″}\hfill \\ \hfill B^{*}& =p/(2r+p):(r+p)/(r-p):-r/(r+2p)=\mathrm{isotomic}\mathrm{conjugate}\mathrm{of}{B}^{″}\hfill \\ \hfill C^{*}& =-p/(p+2q):q/(2p+q):(p+q)/(p-q)=\mathrm{isotomic}\mathrm{conjugate}\mathrm{of}{C}^{″}.\hfill \end{array}$$

Here, if P lies on the Steiner circumellipse (i.e., $D_2=qr+rp+pq=0)$, then the lines $A{A}^{*},B{B}^{*},C{C}^{*}$ concur, which is to say that the triangles $A^{*}{B}^{*}{C}^{*}$ and $ABC$ are perspective. The perspector is the barycentric quotient $P/U$, where U is the reflection of P in the centroid, 1:1:1. This perspector lies on the Tucker nodal cubic, indexed as K015 and discussed in [7].

Cases 3 and 4. Trivially, $u_3^{-1}\left(A^{\prime }{B}^{\prime }{C}^{\prime }\right)=u_4^{-1}\left(A^{\prime }{B}^{\prime }{C}^{\prime }\right)$ if and only if $P=1:1:1$.

8. Other Unary Operations

In the preceding sections, we have sampled four specific unary operations, ((1)–(4)) and four families ((5)–(8)) of unary operations. These were all selected for this paper because of the distinctive duality of their inverses. Other unary operations defined on triangle centers—for which inversion is much more complicated—have been sampled in both barycentric and trilinear coordinates, as in the following list ([2], Part 32, preamble near $X_{62530}$):

$$\begin{array}{cc}\hfill u_9(x:y:z)& =(yz-zx-xy)/(y^2-z^2)::\hfill \\ \hfill u_{10}(x:y:z)& =(y^2-z^2)/(yz-zx-xy)::\hfill \\ \hfill u_{11}(x:y:z)& =(yz-zx-xy)/(y^2+z^2)::\hfill \\ \hfill u_{12}(x:y:z)& =(y^2+z^2)/(yz-zx-xy)::\hfill \end{array}$$

All the unary operations discussed in the paper are homogeneously 0 degrees in $x,y,z$. Much remains to be discovered for other unary operations of various non-negative degrees of homogeneity.