Trigonometric Polynomial Points in the Plane of a Triangle

1. Introduction

One of the most productive systems of representation for points and lines in the plane of a triangle $ABC$ is a system widely known as homogeneous barycentric coordinates (henceforth simply barycentrics). Serving as the “origin” in this system are the three vertices of $ABC$, shown here with their barycentrics:

$$A=1:0:0B=0:1:0C=0:0:1.$$

The lengths of the sides opposite the vertex angles (which, like the vertex points, are denoted by $A,B,C$) are given the symbols $a,b,c$, respectively, and may be regarded as variables or algebraic indeterminates. For an excellent introduction to the subject of barycentrics, see Yiu [16].

Many triangle centers (as defined in [14]) have barycentrics that are polynomials. Following [5], we refer to a triangle center X that has barycentrics

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

where $p(a,b,c)$ is a polynomial, as a polycenter. If X also has barycentrics

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

where $f(a,b,c)$ involves trigonometric functions of the angles $A,B,C$ as a trigonometric polycenter. Analogously, we have polylines and trigonometric polylines. Note that if the first barycentric of X is written as $h(a,b,c)$, than the second and third barycentrics are determined (viz. $h(b,c,a)$ and $h(c,a,b))$, so that the shorter notation $h(a,b,c)::$ is sufficient.

Important examples of trigonometric polycenters include these:

$$\begin{array}{cc}\hfill G& =\mathrm{centroid}=1:1:1=1::\hfill \\ \hfill O& =\mathrm{circumcenter}=a^2(b^2+c^2-a^2)::=sin2A::\hfill \\ \hfill H& =\mathrm{orthocenter}=(c^2+a^2-b^2)(a^2+b^2-c^2)::=tanA::\hfill \\ \hfill N& =\mathrm{nine}-\mathrm{point}\mathrm{center}=a^2(b^2+c^2)-{(b^2-c^2)}^2::=sinAcos(B-C)::\hfill \end{array}$$

As an example of a trigonometric polyline, the Euler line, which passes through the points $G,O,H,N$, is given in terms of a variable point $x:y:z$ by both of the following equations:

• $(b^2-c^2)(b^2+c^2-a^2)x+(c^2-a^2)(c^2+a^2-b^2)y+(a^2-b^2)(a^2+b^2-c^2)z=0$
• $(tanB-tanC)x+(tanC-tanA)y+(tanA-tanB)z=0$

Of great importance in triangle geometry are the following objects:

• isotomic conjugate of X, with barycentrics
  $1/x:1/y:1/z=yz:zx:xy$
• isogonal conjugate of X, with barycentrics
  $a^2/x:b^2/y:c^2/z=a^{2}yz:b^{2}zx:c^{2}xy$
• the line at infinity, $L^{\infty }$, with barycentric equation $x+y+z=0$
• Steiner circumellipse, with equation $1/x+1/y+1/z=0$
• circumcircle, with equation $a^2/x+b^2/y+c^2/z=0$

2. Trigonometric Polycenters

In this section, we shall see that for all integers n, the triangle centers $f\left(nA\right)::$ for $f=cos,sin,tan$, and others, are polycenters. We begin with the usual recurrences of Chebyshev polynomials of the first kind, $T_n$, and the second kind, $U_n$:

$$\begin{array}{cccccc}\hfill T_n\left(x\right)& =2x{T}_{n-1}\left(x\right)-T_{n-2}\left(x\right)\mathrm{for}n\ge 2,\hfill & \hfill T_0\left(x\right)& =1,\hfill & \hfill T_1\left(x\right)& =x;\hfill \end{array}$$

(1)

$$\begin{array}{cccccc}\hfill U_n\left(x\right)& =2x{U}_{n-1}\left(x\right)-U_{n-2}\left(x\right)\mathrm{for}n\ge 2,\hfill & \hfill U_0\left(x\right)& =1,\hfill & \hfill U_1\left(x\right)& =2x,\hfill \end{array}$$

(2)

Another well-known type of recurrence relation for these families of polynomials ([12,13]) depends on complex numbers:

$$\begin{array}{cc}\hfill T_n\left(x\right)& =(1/2)\left({(x-\sqrt{x^2-1})}^n+{(x+\sqrt{x^2-1})}^n\right)\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill U_n\left(x\right)& =(1/\left(2\sqrt{x^2-1}\right)\left({(x+\sqrt{x^2-1})}^{n+1}-{(x-\sqrt{x^2-1})}^{n+1}\right).\hfill \end{array}$$

(4)

Theorem 1.

Let $\Lambda =b^2+c^2-a^2$. Then

$$\begin{array}{cc}\hfill cosnA::& =a^n\left({(\Lambda -2iS)}^n+{(\Lambda +2iS)}^n\right)::\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill sinnA::& =a^n\left({(\Lambda -2iS)}^n-{(\Lambda +2iS)}^n\right)::.\hfill \end{array}$$

(6)

Proof. 

We have $cosA=\Lambda /\left(2bc\right)$, and (3) gives

$$\begin{array}{cc}\hfill T_n(cosA)& =(1/2)\left({(cosA-isinA)}^n+{(cosA+isinA)}^n\right)\hfill \\ \hfill & =(1/2){({\displaystyle \frac{\Lambda }{2bc}}-{\displaystyle \frac{iS}{bc}})}^n+{({\displaystyle \frac{\Lambda }{2bc}}+{\displaystyle \frac{iS}{bc}})}^n,\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill S& =2(\mathrm{area}\mathrm{of}▵ABC)\hfill \\ \hfill & =(1/2)\sqrt{(a+b+c)(b+c-1)(c+a-b)(a+b-c)},\hfill \end{array}$$

so that

$$S^2=(1/4)(-a^4-b^4-c^4+2b^2{c}^2+2c^2{a}^2+2a^2{b}^2).$$

Now since $cosnA=T_n(cosA)$, we have

$$cosnA=(1/2){\displaystyle \frac{1}{{\left(2bc\right)}^n}}\left({(\Lambda -2iS)}^n+{(\Lambda +2iS)}^n\right),$$

so that (5) holds. Similarly, Equation (6) follows from (4) and the well-known fact that $sinnA=U_{n-1}(cosA)sinA$.    □

Our main goal in this section is to represent $cosnA::$ and $sinnA::$ as polycenters. To that end, let

$$u=\Lambda -2iS\mathrm{and}v=\Lambda +2iS,$$

(7)

so that expressions in (3) and () can be recast in order to define sequences $\left(c_n\right)$ and $\left(s_n\right)$ as follows:

$$\begin{array}{cc}\hfill c_n& =c_n(a,b,c)=a^n(u^n+v^n)\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill s_n& =s_n(a,b,c)=a^n(u^n-v^n).\hfill \end{array}$$

(9)

Next, we have a lemma about u and v.

Lemma 1.

$$\begin{array}{cc}\hfill u^2-2(b^2+c^2-a^2)u+4b^2{c}^2& =v^2-2(b^2+c^2-a^2)v+4b^2{c}^2\hfill \\ \hfill & =0.\hfill \end{array}$$

Proof. 

The imaginary terms cancel, and the real term is

$$-{(b^2+c^2-a^2)}^2-4S^2+4b^2{c}^2=0.$$

   □

Theorem 2.

Let $\left(c_n\right)$ be the sequence given by (8). Then $c_n$ is a polynomial in $a,b,c$, given by the following three initial terms and 2^nd-order recurrence:

$$\begin{array}{cc}\hfill c_0& =1\hfill \\ \hfill c_1& =a(b^2+c^2-a^2)\hfill \\ \hfill c_2& =a^2(a^4+b^4+c^4-2a^2{b}^2-2a^2{c}^2)\hfill \\ \hfill c_n& =a(b^2+c^2-a^2)c_{n-1}-a^2{b}^2{c}^2{c}_{n-2}\mathrm{for}n\ge 3,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill c_n(a,b,c):c_n(b,c,a):c_n(c,a,b)& =cosnA:cosnB:cosnC.\hfill \end{array}$$

(10)

Proof. 

It is easy to verify that $c_0,c_1,c_2$ are polycenters as claimed. Suppose now that $n\ge 3.$ Using u and v as in (7) and Lemma 1, we have

$$u^{n-2}\left(u^2-2(b^2+c^2-a^2)u+4b^2{c}^2\right)=-v^{n-2}\left(v^2-2(b^2+c^2-a^2)v+4b^2{c}^2\right),$$

so that

$$\begin{array}{cc}\hfill u^n+v^n& =2(b^2+c^2-a^2)(u^{n-1}+v^{n-1})-a^2{b}^2{c}^2{(a/2)}^{n-2}(u^{n-2}+v^{n-2})\hfill \\ \hfill {(a/2)}^n(u^n+v^n)& =a(b^2+c^2-a^2){(a/2)}^{n-1}(u^{n-1}+v^{n-1})-4b^2{c}^2(u^{n-2}+v^{n-2}).\hfill \end{array}$$

This shows that if $d_k={(a/2)}^k(u^k+v^k)$ for $k\ge 3$, then

$$d_n=a(b^2+c^2-a^2)d_{n-1}-a^2{b}^2{c}^2{d}_{n-2}\mathrm{for}n\ge 3.$$

By Theorem 1, $cosnA::=a^n(u^n+v^n)::$, and since $d_n::=c_n::$, we have $c_n::=cosnA::$.    □

Theorem 3.

Let $\left(s_n\right)$ be the sequence given by (). Then $s_n$ is a polynomial in $a,b,c$, given by these two initial terms and 2^nd-order recurrence:

$$\begin{array}{cc}\hfill s_1& =a\hfill \\ \hfill s_2& =a^2(b^2+c^2-a^2)\hfill \\ \hfill s_n& =a(b^2+c^2-a^2)s_{n-1}-a^2{b}^2{c}^2{s}_{n-2}\mathrm{for}n\ge 2,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill s_n(a,b,c):s_n(b,c,a):s_n(c,a,b)& =sinnA:sinnB:sinnC.\hfill \end{array}$$

(11)

Proof. 

A proof similar to that of Theorem 2 springs from (), leading, by way of the identity $sinnA=U_{n-1}(cosA)sinA$, to

$$sinnA={\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{{\left(2bc\right)}^n}}\left({(\Lambda -2iS)}^n-{(\Lambda +2iS)}^n)\right),$$

The rest of the proof, using Lemma 1, follows in a manner much as in the proof of Theorem 2.    □

Example 1.

Polycenter representations for $cos3A::$ and $cos4A::$ are given by

$$\begin{array}{cc}\hfill c_3& =a^3(a^2-b^2-c^2)\left(a^4+b^4+c^4-2a^2(b^2+c^2)-b^2{c}^2\right);\hfill \\ \hfill c_4& =a^4(a^8+b^8+c^8-4a^6(b^2+c^2)+2a^4(3b^4+3c^4+4b^2{c}^2)\hfill \\ \hfill & -4a^2(b^4+c^4)(b^2+c^2)).\hfill \end{array}$$

Example 2.

Polycenter representations for $sinnA::,$ for $n=3,4,5,6$, are given by

$$\begin{array}{cc}\hfill s_3& =a^3(a^2-b^2-c^2-bc)(a^2-b^2-c^2+bc);\hfill \\ \hfill s_4& =a^4(a^2-b^2-c^2)\left(a^4+b^4+c^4-2a^2(b^2+c^2)\right);\hfill \\ \hfill s_5& =a^5{f}_1{f}_2,\mathrm{where}\hfill \\ \hfill & f_1=a^4+b^4+c^4+a^2(bc-2b^2-2c^2)+bc(bc-b^2-c^2)\hfill \\ \hfill & f_2=a^4+b^4+c^4+a^2(-bc-2b^2-2c^2)+bc(bc+b^2+c^2);\hfill \\ \hfill s_6& =a^5{g}_1{g}_2{g}_3{g}_4,\mathrm{where}\hfill \\ \hfill & g_1{g}_2{g}_3=(a^2-b^2-c^2)(a^2-b^2-c^2-bc)(a^2-b^2-c^2+bc)\hfill \\ \hfill & g_4=a^4+b^4+c^4-2a^2(b^2+c^2)-b^2{c}^2.\hfill \end{array}$$

Inductively, $c_n$ and $s_n$ both have degree $3n$ for $n\ge 0$ and both are polynomial multiples of $a^n$. By Theorems 2 and 3, the sequences $\left(c_n\right)$ and $\left(s_n\right)$ have the same second-order recurrence signature:

$$\left(a(b^2+c^2-a^2),-a^2{b}^2{c}^2\right).$$

Next, let $t_n=s_n/c_n$, so that $t_n::=tannA::$. For the sake of brevity, we shall sometimes write a polycenter of the form $f(a,b,c)g(b,c,a)g(c,a,b)$ as a quotient: $f(a,b,c)/g(a,b,c)$. Shown here are representations for polycenters $tannA::$ for $n=1,2,3$.

$$\begin{array}{cc}\hfill t_1& =1/(a^2-b^2-c^2);\hfill \\ \hfill t_2& =(a^2-b^2-c^2)/\left(a^4+b^4+c^4-2a^2(b^2+c^2)\right);\hfill \\ \hfill t_3& =1/\left(a^4+b^4+c^4-2a^2)(b^2+c^2)-b^2{c}^2\right);\hfill \end{array}$$

The sequence $\left(t_n\right)$, as well as its equivalent sequence of polynomials, appears—expectedly—to be not linearly recurrent. However, the sequence given by $u_n(a,b,c)=sin\left(nA\right)cos\left(nB\right)cos\left(nC\right)$, is linearly recurrent, since the three sequences $(sin(nA\left)\right),(cos(nB\left)\right),(cos(nC\left)\right)$ are linearly recurrent, and, of course,

$$u_n(a,b,c):u_n(b,c,a):u_n(c,a,b)=tannA:tannB:tannC.$$

A sequence of associated polycenters derived from on $u_n(a,b,c)$ is considered in Section 7. Likewise the triangle centers $secnA::,cscnA::,cotnA::$ are polycenters for all nonzero integers n. Geometrically, these are isotomic conjugates, given by $1/t_n,1/s_n,c_n/s_n$, respectively. As indicated in Example 3, many geometric and algebraic properties of the specific polycenters mentioned above can be found in the Encyclopedia of Triangle Centers (ETC) [2]:

Example 3.

A few trigonometric polycenters in ETC [2].

$$\begin{array}{cccc}\hfill sinA::& =X_1\hfill & \hfill cscA::& =X_{75}\hfill \\ \hfill cosA::& =X_{63}\hfill & \hfill secA::& =X_{92}\hfill \\ \hfill tanA::& =X_4\hfill & \hfill cotA::& =X_{69}\hfill \\ \hfill sin2A::& =X_3\hfill & \hfill csc2A::& =X_{264}\hfill \\ \hfill cos2A::& =X_{1993}\hfill & \hfill sec2A::& =X_{5392}\hfill \\ \hfill tan2A::& =X_{68}\hfill & \hfill cot2A::& =X_{317}\hfill \\ \hfill sin3A::& =X_{6149}\hfill & \hfill csc3A::& =X_{63759}\hfill \\ \hfill cos3A::& =X_{63760}\hfill & \hfill sec3A::& =X_{63764}\hfill \\ \hfill tan3A::& =X_{562}\hfill & \hfill cot3A::& =X_{63761}\hfill \\ \hfill sin4A::& =X_{1147}\hfill & \hfill csc4A::& =X_{55553}\hfill \\ \hfill cos4A::& =X_{63762}\hfill & \hfill sec4A::& =X_{63765}\hfill \\ \hfill tan4A::& =X_{43973}\hfill & \hfill cot4A::& =X_{55552}\hfill \end{array}$$

Barycentric products and quotients ([16], 99-102), denoted by * and /, of the polycenters listed in Example 3 are also trigonometric polycenters; e.g., $X_1*X_{63}=X_3$ and $X_1/X_{63}=X_4$.

In particular, if f is a trigonometic polycenter, then $f^n$, where n is any positive integer, is also a trigonometric polycenter, as represented by these squares:

Example 4.

Trigonometric square polycenters in ETC [2]. (See also Section 7.)

$$\begin{array}{cc}\hfill {sin}^{2}A::& =a^2::=X_6\hfill \\ \hfill {csc}^{2}A::& =b^2{c}^2::=X_{76}\hfill \\ \hfill {cos}^{2}A::& =a^2{(b^2+c^2-a^2)}^2::=X_{394}\hfill \\ \hfill {sec}^{2}A::& =b^2{c}^2{(b^2+c^2-a^2)}^{-2}::=X_{2052}\hfill \\ \hfill {cos}^2(B-C)::& =b^2{c}^2{(b^4+c^4-2b^2{c}^2-a^2{b}^2-a^2{c}^2)}^2::=X_{45793}\hfill \\ \hfill {sin}^2(B-C)::& =b^2{c}^2{(b^2-c^2)}^2::=X_{338}\hfill \\ \hfill {csc}^2(B-C)::& =a^2{(b^2-c^2)}^{-2}::=X_{249}\hfill \end{array}$$

3. More Trigonometric Polycenters

In this section, we first present polycenters for triangle centers of the forms $cos(nB-nC)::$ and $sin(nB-nC)$, and follow with a proof-by-computer-code for a recurrence equation for the points $cos(nB-nC)::$ as polycenters . Let $M=a^2(b^2+c^2)-{(b^2-c^2)}^2$. Then

$$\begin{array}{cc}\hfill cos(B-C)::& =p_1(a,b,c)::,\mathrm{where}{p}_1(a,b,c)=bcM\hfill \\ \hfill cos\left(2\right(B-C\left)\right)::& =p_2(a,b,c)::,\mathrm{where}{p}_2(a,b,c)=b^2{c}^2({(b^2-c^2)}^4\hfill \\ \hfill & +a^4(b^4+c^4)+2a^2(-b^6+b^4{c}^2+b^2{c}^4-c^6))::\hfill \\ \hfill cos\left(n\right(B-C\left)\right)::& =p_n(a,b,c)::,\mathrm{where}\hfill \\ \hfill & p_n(a,b,c)=bcM{p}_{n-1}(a,b,c)-a^4{b}^4{c}^4{p}_{n-2}(a,b,c)\mathrm{for}n\ge 3.\hfill \\ \hfill sin(B-C)::& =q_1(a,b,c)::,\mathrm{where}{q}_1(a,b,c)=bc(b^2-c^2)\hfill \\ \hfill sin\left(2\right(B-C\left)\right)::& =q_2(a,b,c)::,\mathrm{where}{q}_2(a,b,c)=b^2{c}^2(b^2-c^2)M\hfill \\ \hfill sin\left(n\right(B-C\left)\right)::& =q_n(a,b,c)::,\mathrm{where}\hfill \\ \hfill & q_n=bcM{q}_{n-1}-a^4{b}^4{c}^4{q}_{n-2}\mathrm{for}n\ge 2.\hfill \end{array}$$

Instead of a formal proof of the above recurrence equation for $p_n(a,b,c)$, we quote the Mathematica code, which is essentially a proof with the added advantage of usefulness for further explorations.

(* Step 1: trig functions in terms of a, b, c & S*)

   trigRules={Cos[A]->(-a^2+b^2+c^2)/(2 b c),

   Cos[B]->(a^2-b^2+c^2)/(2 a c),

   Cos[C]->(a^2+b^2-c^2)/(2 a b),

   Sin[A]->S/(b c),Sin[B]->S/(a c),Sin[C]->S/(a b)};

(* Step 2: double area powers in terms of a, b, c & S*)

   SRules={S->S,S^x_?EvenQ->2^-x ((a+b-c) (a-b+c)

   (-a+b+c) (a+b+c))^(x/2),

   S^x_?OddQ->2^(1-x)((a+b-c)(a-b+c)(-a+b+c)(a+b+c))^(1/2 (-1+x)) S};

(* Step 3: cyclic permutations of a,b,c *)

   cyclic[coord_]:=Apply[coord/. {a->#1,b->#2,c->#3,A->#4,B->#5,C->#6}&,

   Flatten/@NestList[RotateLeft/@#1&,{{a,b,c},{A,B,C}},2],{1}];

(* Step 4: removal of symmetric factors *)

   removeSym:=(Factor[#1/PolynomialGCD@@#1]&)[Factor[#]]&;

(* Step 5: application of Steps 1-4 *)

   polys = Map[(TrigExpand[cyclic[Cos[#(B-C)]]]//.trigRules

   //.SRules//removeSym//removeSym)[[1]]&,Range[7]]

(* Step 6: find signature of 2nd order recurrence *)

   Factor[FindLinearRecurrence[polys]]

The output of the code is the following signature for the recurrence:

$$\left(bc(a^2{b}^2-b^4+a^2{c}^2+2b^2{c}^2-c^4),-a^4{b}^4{c}^4\right).$$

A proof of the recurrence equation, or more precisely, the signature of the recurrence, for $sin\left(n\right(B-C\left)\right)::$ as a polycenter is found in much the same way.

As an alternative to representing the family $cos(nB-nC)::$ by polynomials, there are relatively simpler representations using quotients of polynomials. We begin with

$$\begin{array}{cc}\hfill u& =cos(B-C)=cosBcosC+sinBsinC\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{{(b^2-c^2)}^2-a^2{b}^2-a^2{c}^2}{2a^{2}bc}}\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill v& =sin(B-C)=sinBcosC+sinCcosB\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{S(b^2-c^2)}{2a^{2}bc}},\mathrm{where}\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill S& =(1/2)\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}.\hfill \end{array}$$

(16)

Let

$$f_n=f_n(u,v)=(1/2)\left({(u-iv)}^n+{(u+iv)}^n\right)=cos(nB-nC).$$

Then by the binomial theorem,

$$f_n=\sum _{k=0}^{\lfloor n/2\rfloor }{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)u^{n-2k}{v}^{2k},$$

which satisfies the recurrence

$$f_n=2u{f}_{n-1}-(u^2+v^2)f_{n-2},$$

with $f_0=1,f_1=u$. Since $f_n$ depends only on u and $v^2$, it is a rational function; i.e., a radical-free quotient of polynomials. Similary, letting $g_n=g_n(u,v)=sin(nB-nC)/sin(B-C)$, we find that

$$g_n=2u{g}_{n-1}-(u^2+v^2)g_{n-2},$$

with $g_0=0,g_1=1$.

Example 5.

A few more trigonometric polycenters in ETC [2].

$$\begin{array}{cccc}\hfill sin(B-C)::& =X_{1577}\hfill & \hfill csc(B-C)::& =X_{662}\hfill \\ \hfill cos(B-C)::& =X_{14213}\hfill & \hfill sec(B-C)::& =X_{2167}\hfill \\ \hfill tan(B-C)::& =X_{15412}\hfill & \hfill cot(B-C)::& =X_{14570}\hfill \\ \hfill sin(2B-2C)::& =X_{18314}\hfill & \hfill csc(2B-2C)::& =X_{18315}\hfill \\ \hfill cos(2B-2C)::& =X_{63763}\hfill & \hfill sec(2B-2C)::& =X_{63766}\hfill \end{array}$$

4. Half-Angle Trigonometric Polycenters

The next list shows half-angle functions that involve polynomials (viz. they are “radical multiples of polycenters”). Let

$$P=\sqrt{\left({(b+c)}^2-a^2\right)/\left(bc\right)}\mathrm{and}Q=\sqrt{\left(a^2-{(b-c)}^2\right)/\left(bc\right)}.$$

$$\begin{array}{cc}\hfill cos(A/2)::& =p_1(a,b,c)::,\mathrm{where}{p}_1(a,b,c)=P\hfill \\ \hfill cos(3A/2)::& =p_3(a,b,c)::,\mathrm{where}{p}_3(a,b,c)=P(a^2-b^2-c^2+bc)\hfill \\ \hfill cos(nA/2)::& =p_n(a,b,c)::,\mathrm{where}{p}_n(a,b,c)=(-a^2+b^2+c^2)/\left(bc\right)p_{n-2}\hfill \\ \hfill & -a^4{b}^4{c}^4{p}_{n-4}::,\mathrm{for}\mathrm{odd}n\ge 5;\hfill \\ \hfill sin(A/2)::& =q_1(a,b,c)::,\mathrm{where}{q}_1(a,b,c)=Q::\hfill \\ \hfill sin(3A/2)::& =q_3(a,b,c)::,\mathrm{where}{q}_3(a,b,c)=Q(a^2-b^2-c^2-bc)::\hfill \\ \hfill sin(nA/2)::& =q_n(a,b,c)::,\mathrm{where}{q}_n(a,b,c)=(-a^2+b^2+c^2)/\left(bc\right)q_{n-2}\hfill \\ \hfill & -a^4{b}^4{c}^4{q}_{n-4},::,\mathrm{for}\mathrm{odd}n\ge 5.\hfill \end{array}$$

Next we show Mathematica code for obtaining trigonometric rational functions (quotients of polynomials) for $cos\left(\right(nB-nC)/2)::$  .

lr = FindLinearRecurrence[

   Map[TrigExpand[Cos[# (B - C)/2]] &, Range[1, 11, 2]]];

cyclic[coord_] :=

  Apply[coord /. {a -> #1, b -> #2, c -> #3, A -> #4, B -> #5,

      C -> #6} &, Flatten /@

    NestList[RotateLeft /@ #1 &, {{a, b, c}, {A, B, C}}, 2], {1}];

trigRules =

  Flatten[{Map[

     cyclic, {Cos[A] -> (-a^2 + b^2 + c^2)/(2  b  c),

      Sin[A] -> S/(b  c),

      Cos[A/2] -> 1/2  Sqrt[((-a + b + c)  (a + b + c))/(b  c)],

      Sin[A/2] -> 1/2  Sqrt[((a + b - c)  (a - b + c))/(b  c)],

      Cos[B/2]*Cos[C/2]*Sin[B/2]*

        Sin[C/2] -> ((-a + b + c)  (a + b - c)  (a - b + c)

        (a + b + c))/(16  a^2  b  c)}]}];

Factor[lr /. trigRules]

This code confirms that $cos\left(\right(nB-nC)/2)::$ is a rational function with signature

$$\left((a^2{b}^2-b^4+a^2{c}^2+2b^2{c}^2-c^4)/\left(a^{2}bc\right),-1\right).$$

(These rational functions can be transformed into polynomials using a technique developed in Section 6.)

Example 6.

A few half-angle trigonometric polycenters in ETC [2].

$$\begin{array}{cccc}\hfill cos(A/2)::& =X_{188}\hfill & \hfill sec(A/2)::& =X_{4146}\hfill \\ \hfill sin(A/2)::& =X_{174}\hfill & \hfill csc(A/2)::& =X_{556}\hfill \\ \hfill cos(3A/2)::& =X_{63779}\hfill & \hfill sin(3A/2)::& =X_{63780}\hfill \end{array}$$

5. Sums Involving mB+nC and nB+mC

Proofs of the next two theorems can be obtained by adapting the codes in Section 3 and Section 4.

Theorem 4.

Let $f(m,n)=f(m,n,a,b,c)=sin(mB+nC)+sin(nB+mC)$ and let $p(m,n)=p(m,n,a,b,c)$ be the polycenters given by these recurrences:

$$\begin{array}{cc}\hfill p(m,n)& =a(a^2-b^2-c^2)p(m-1,n)-a^2{b}^2{c}^{2}p(m-2,n);\hfill \\ \hfill p(m,n)& =a(a^2-b^2-c^2)p(m,n-1)-a^2{b}^2{c}^{2}p(m,n-2),\hfill \end{array}$$

where $p(0,0)=0$, and $p(0,1)=p(1,0)=b+c$. Then

$$p(m,n,a,b,c)::=sin(mB+nC)+sin(nB+mC)::.$$

Example 7.

The appearance of $(m,n,X_k)$ in the following list signifies that $sin(mB+nC)+sin(nB+mC)::=X_k$.

$$\begin{array}{cc}\hfill & (0,1,X_{10}),(0,2,X_5),(0,3,X_{63803}),(0,4,X_{5449})\hfill \\ \hfill & (1,1,X_1),(1,2,X_{63802}),(1,3,X_{44707})\hfill \\ \hfill & (2,2,X_3),(2,3,X_{63801}),(2,4,X_{1154})\hfill \\ \hfill & (3,3,X_{6149}),(3,5,X_{63801})\hfill \end{array}$$

Theorem 5.

Let $g(m,n)=g(m,n,a,b,c)=cos(mB+nC)+cos(nB+mC)$ and let $q(m,n)=q(m,n,a,b,c)$ be the polycenters given by the same recurrences as for $p(m,n)$, where

$$\begin{array}{cc}\hfill q(0,1)& =q(1,0)=(b+c)(c+a-b)(a+b-c)\hfill \\ \hfill q(0,2)& =(b^2+c^2-a^2)(a^2(b^2+c^2)-{(b^2-c^2)}^2).\hfill \end{array}$$

Then

$$q(m,n,a,b,c)::=cos(mB+nC)+cos(nB+mC)::.$$

Example 8.

The appearance of $(m,n,X_k)$ here means that $cos(mB+nC)+cos(nB+mC)::=X_k$.

$$\begin{array}{cc}\hfill & (0,1,X_{226}),(0,2,X_{343}),(0,4,X_{63806})\hfill \\ \hfill & (1,1,X_{63}),(1,2,X_{16577}),(1,3,X_{63808})\hfill \\ \hfill & (2,2,X_{1993}),(2,3,X_{73802})\hfill \\ \hfill & (3,3,X_{63760}),(3,5,X_{63762})\hfill \end{array}$$

6. Polycenters j+k cos(nA)::

Here, we find a sequence $p_n=p_n(a,b,c)$ of polycenters satisfying

$$\begin{array}{cc}\hfill & p_n(a,b,c):p_n(b,c,a):p_n(c,a,b)\hfill \\ \hfill & =j+kcos\left(nA\right):j+kcos\left(nB\right):j+kcos\left(nC\right),\hfill \end{array}$$

(17)

where j and k are nonzero real numbers. The strategy is to determine rational functions $u_n=u_n(a,b,c)$ that can be transformed into the polynomials $p_n$. We begin with the following Mathematica code:

f[a_, b_, c_] := f[a, b, c] = ArcCos[(b^2 + c^2 - a^2)/(2 b  c)];

{a1, b1, c1} = {f[a, b, c], f[b, c, a], f[c, a, b]};

a2[n_] := Collect[Factor[TrigExpand[j + k*Cos[n  a1]]], {j, k}];

b2[n_] := Collect[Factor[TrigExpand[j + k*Cos[n  b1]]], {j, k}];

c2[n_] := Collect[Factor[TrigExpand[j + k*Cos[n  c1]]], {j, k}];

t = Table[a2[n], {n, 0, 10}]; Take[t, 4]

FindLinearRecurrence[t]

t = Table[b2[n], {n, 0, 10}]; Take[t, 4]

FindLinearRecurrence[t]

t = Table[c2[n], {n, 0, 10}]; Take[t, 4]

FindLinearRecurrence[t]

The code gives

$$\begin{array}{cc}\hfill u_0& =j+k\hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill u_1& =j+{\displaystyle \frac{-a^2+b^2+c^2}{2bc}}k\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill u_2& =j+{\displaystyle \frac{a^4+b^4+c^4-2a^2{b}^2-2a^2{c}^2}{2b^2{c}^2}}k\hfill \end{array}$$

(20)

and recurrence signature $(\tau ,-\tau ,1)$, where

$$\tau ={\displaystyle \frac{-a^2+b^2+c^2+bc}{bc}}.$$

(21)

The transformation is simply to multiply, where appropriate, by $2a^n{b}^n{c}^n$, obtaining

$$\begin{array}{cc}\hfill p_0& =2j+2k\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill p_1& =2abcj+a(b^2+c^2-a^2)k\hfill \end{array}$$

(23)

$$\begin{array}{cc}\hfill p_2& =2a^2{b}^2{c}^{2}j+\left(a^6-a^4(b^2+c^2)+a^2(b^4+c^4)\right)k,\hfill \end{array}$$

(24)

with third-order recurrence signature

$$(-a{Q}_a,a^{2}bc{Q}_a,a^3{b}^3{c}^3),$$

(25)

where $Q_a=Q_a(a,b,c)=a^2-b^2-c^2-bc$.

Finally, we apply the substitutions $(a,b,c)\to (b,c,a)$ and $(a,b,c)\to (c,a,b)$ to (18)-(25) and thereby obtain (17).

7. Applications of the Technique in Section 6

The procedure in Section 6 applies to other families of trigonometric polycenters. Among them are the families ${cos}^2(nB-nC)::$ and ${sin}^2(nB-nC)::$. Both ${cos}^2(nB-nC)$ and ${sin}^2(nB-nC)$ have third-order recurrence with signature $\left(k\right(a,b,c),-k(a,b,c),1)$, where

$$k={\displaystyle \frac{{(b^2-c^2)}^4+a^4(b^4+c^4+b^2{c}^2)+2a^2(b^4{c}^2+b^2{c}^4-b^6-c^6)}{a^4{b}^2{c}^2}}.$$

The resulting polycenters are too long for display here. We do observe, however, that for every integer $p\ge 2$, the polycenters ${cos}^p(nB-nC)::$ and ${sin}^p(nB-nC)::$ have recurrence order $p+1$.

Other families to which the procedure applies are represented by

$$tannA::,secnA::,cscnA::,cotnA::.$$

Here we consider only the rational-function recurrence for $tannA$. The most direct approach appears to be to use the identity

$$tannA::=sinnAcosnBcosnC::.$$

The resulting sixth-degree recurrence signature for $sinnAcosnBcosnC$ is more efficiently expressed with Conway notation [15] than with $a,b,c$:

$$(k_1,k_2,k_3,k_2,k_1,1),$$

where

$$\begin{array}{cc}\hfill k_1& ={\displaystyle \frac{-2(3S_A{S}_B{S}_C+S^2{S}_{\omega })}{D}}\hfill \\ \hfill k_2& =-{\displaystyle \frac{16S^6+15S_A^2{S}_B^2{S}_C^2+2S^2{S}_A{S}_B{S}_C{S}_{\omega }-S^4{S}_{\omega }^2}{D}}\hfill \\ \hfill k_3& =-{\displaystyle \frac{-4(-8S^6+5S_A^2{S}_B^2{S}_C^2+2S^2{S}_a{S}_B{S}_C{S}_{\omega }+S^4{S}_{\omega }^2)}{D}},\hfill \end{array}$$

where $D={\left(S_A{S}_B{S}_C{S}^2{S}_{\omega }\right)}^2$.

8. Infinite Trigonometric Orthopoints

The line at infinity consists of all points $X=x:y:z$ satisfying the linear equation

$$x+y+z=0.$$

(26)

Most of the named points on $L^{\infty }$ are polycenters. Among the simplest are

$$\begin{array}{cc}\hfill X_{513}& =a(b-c)::\hfill \\ \hfill X_{514}& =b-c::\hfill \\ \hfill X_{30}& =\cos \mathrm{A}-2\cos \mathrm{B}\cos \mathrm{C}::=2a^4-(b^2-c^2)2-a^2(b^2+c^2)::,\hfill \end{array}$$

these being the points in which the lines $X_1{X}_{75},X_1{X}_2,X_2{X}_3$ meet $L^{\infty }$, respectively. Of special importance is $X_{30}$, as this is the infinite point on the Euler line, $X_2{X}_3$.

If $X=x:y:z$ is on $L^{\infty }$, then X can be regarded as a direction in the plane of $▵ABC$, since for every point P not on $L^{\infty }$, every line parallel to the line $PX$ intersects $L^{\infty }$ in X. The line through P orthogonal to $PX$ meets $L^{\infty }$ in a point called the orthopoint (or, in [8], the orthogonal conjugate) of X. We denote the orthopoint of X by $X^{\perp }$. Barycentrics for $X^{\perp }$ are given by

$$X^{\perp }=ycosB-zcosC::.$$

(27)

Thus if X is a polycenter represented by a polynomial $x(a,b,c)$ as first barycentric, then $X^{\perp }$ is the polycenter

$$X^{\perp }=x(b,c,a)b(b^2-a^2-c^2)-x(c,a,b)c(c^2-b^2-a^2)::.$$

(28)

Now for any point $X=x:y:z$, not necessarily a polycenter and not necessarily on $L^{\infty }$, the points

$$y-z:z-x:x-y\mathrm{and}2x-y-z:2y-z-x:2z-x-y$$

are clearly on $L^{\infty }$, as are their orthopoints,

$$b(z-x)(b^2-a^2-c^2)-c(x-y)(c^2-b^2-a^2)::$$

(29)

and

$$b(2y-z-x)(b^2-a^2-c^2)-c(2z-x-y)(c^2-b^2-a^2)::,$$

(30)

respectively. Moreover, if X is a polycenter, then the orthopoints (29) and (30) are polycenters on $L^{\infty }$.

Example 9.

Let

$$X=x:y:z={cos}^{2}A::=a^2{(a^2-b^2-c^2)}^2::=X_{394}.$$

Then

$$\begin{array}{cc}\hfill y-z::& ={cos}^{2}B-{cos}^{2}C::=X_{523}\hfill \\ \hfill 2x-y-z::& =2{cos}^{2}A-{cos}^{2}B-{cos}^{2}C::=X_{527},\hfill \end{array}$$

and the two corresponding orthopoints (29) and (30) are respectively

$$\begin{array}{cc}\hfill X_{30}& =2a^4-{(b^2-c^2)}^2-a^2(b^2+c^2)::\hfill \\ \hfill X_{28292}& =bc/\left(h\right(a,b,c\left)h\right(a,c,b\left)\right)::,\mathrm{where}\hfill \\ \hfill h(a,b,c)& =a^3+3b^3+c^3+a^2(b-c)-5b^2(a+c)+c^2(b-a)+2abc.\hfill \end{array}$$

Example 9 typifies infinite polycenters of the forms $f\left(B\right)-f\left(C\right)::$ and $f\left(2A\right)-f\left(B\right)-f\left(C\right)::$. Such polycenters, for which many algebraic and geometric properties are presented in ETC [2], occupy the lists in the next two examples.

Example 10.

Pairs of trigonometric orthopoints.

$$\begin{array}{cc}\hfill cosB-cosC::=X_{522},& X_{522}^{\perp }=X_{515}\hfill \\ \hfill sinB-sinC::=X_{514},& X_{514}^{\perp }=X_{516}\hfill \\ \hfill tanB-tanC::=X_{525},& X_{525}^{\perp }=X_{1503}\hfill \\ \hfill secB-secC::=X_{521},& X_{521}^{\perp }=X_{6001}\hfill \\ \hfill cscB-cscC::=X_{513},& X_{513}^{\perp }=X_{517}\hfill \\ \hfill cotB-cotC::=X_{523},& X_{523}^{\perp }=X_{30}\hfill \\ \hfill {cos}^{2}B-{cos}^{2}C::=X_{523},& X_{523}^{\perp }=X_{30}\hfill \\ \hfill {sin}^{2}B-{sin}^{2}C::=X_{523},& X_{523}^{\perp }=X_{30}\hfill \end{array}$$

Example 10 continued:

$$\begin{array}{cc}\hfill {tan}^{2}B-{tan}^{2}C::=X_{520},& X_{520}^{\perp }=X_{6000}\hfill \\ \hfill {sec}^{2}B-{sec}^{2}C::=X_{520},& X_{520}^{\perp }=X_{6000}\hfill \\ \hfill {csc}^{2}B-{csc}^{2}C::=X_{512},& X_{512}^{\perp }=X_{511}\hfill \\ \hfill {cot}^{2}B-{cot}^{2}C::=X_{512},& X_{512}^{\perp }=X_{511}\hfill \\ \hfill cos2B-cos2C::=X_{523},& X_{523}^{\perp }=X_{30}\hfill \\ \hfill sin2B-sin2C::=X_{525},& X_{525}^{\perp }=X_{1503}\hfill \\ \hfill sec2B-sec2C::=X_{924},& X_{924}^{\perp }=X_{13754}\hfill \\ \hfill cot2B-cot2C::=X_{6368},& X_{6368}^{\perp }=X_{18400}\hfill \end{array}$$

Example 11.

More pairs of trigonometric orthopoints.

$$\begin{array}{cc}\hfill 2cosA-cosB-cosC::=X_{527},& X_{527}^{\perp }=X_{28292}\hfill \\ \hfill 2sinA-sinB-sinC::=X_{519},& X_{527}^{\perp }=X_{3667}\hfill \\ \hfill 2tanA-tanB-tanC::=X_{30},& X_{30}^{\perp }=X_{523}\hfill \\ \hfill 2cscA-cscB-cscC::=X_{536},& X_{536}^{\perp }=X_{28475}\hfill \\ \hfill 2cotA-cotB-cotC::=X_{524},& X_{524}^{\perp }=X_{1499}\hfill \\ \hfill 2{cos}^{2}A-{cos}^{2}B-{cos}^{2}C::=X_{524},& X_{524}^{\perp }=X_{1499}\hfill \\ \hfill 2{sin}^{2}A-{sin}^{2}B-{sin}^{2}C::=X_{524},& X_{524}^{\perp }=X_{1499}\hfill \\ \hfill 2{csc}^{2}A-{csc}^{2}B-{csc}^{2}C::=X_{538},& X_{538}^{\perp }=X_{32472}\hfill \\ \hfill 2{cot}^{2}A-{cot}^{2}B-{cot}^{2}C::=X_{538},& X_{538}^{\perp }=X_{32472}\hfill \\ \hfill 2cos2A-cos2B-cos2C::=X_{524},& X_{524}^{\perp }=X_{1499}\hfill \\ \hfill 2sin2A-sin2B-sin2C::=X_{30},& X_{30}^{\perp }=X_{523}\hfill \\ \hfill 2tan2A-tan2B-tan2C::=X_{539},& X_{539}^{\perp }=X_{20184}\hfill \end{array}$$

9. Trigonometric Infinity Bisectors

Let O denote the circumcenter, $\left(O\right)$ the circumcircle, and $L^{\infty }$ the line at infinity. Suppose that $P=p:q:r$ and $U=u:v:w$ are points on $\left(O\right)$ and that $P,O,U$ are noncollinear. Let $L_1$ be the tangent to $\left(O\right)$ at P and $L_2$ the tangent to $\left(O\right)$ at U. Let $D=L_1\cap L_2$ and $M=OD\cap L^{\infty }$ . As the line $OM$ bisects the angle between $L_1$ and $L_2$, the point M is called the $(P,U)$-infinity bisector. We denote this point by $M(P,U)$. Its barycentrics are given by

$$M(P,U)=(a^2-b^2+c^2)(qu-pv)-(a^2+b^2-c^2)(ru-pw)-2a^2(rv-qw)::$$

(31)

If P and U are trigonometric polycenters, then (31) is also a trigonometric polycenter, since

$$b^2+c^2-a^2:c^2+a^2-b^2:a^2+b^2-c^2=cotA:cotB:cotC.$$

Note that the $M(P,U)$ is the orthopoint of the $PU\cap L^{\infty }$. A few examples follow:

$$\begin{array}{cc}\hfill M(X_{74},X_{477})& =X_{30}=X_{523}^{\perp }\hfill \\ \hfill M(X_{110},X_{476})& =X_{30}=X_{523}^{\perp }\hfill \\ \hfill M(X_{74},X_{476})& =X_{523}=X_{30}^{\perp }\hfill \\ \hfill M(X_{110},X_{477})& =X_{523}=X_{30}^{\perp }\hfill \\ \hfill M(X_{74},X_{98})& =X_{542}=X_{690}^{\perp }\hfill \\ \hfill M(X_{99},X_{110})& =X_{542}=X_{690}^{\perp }\hfill \\ \hfill M(X_{74},X_{99})& =X_{690}=X_{542}^{\perp }\hfill \\ \hfill M(X_{98},X_{110})& =X_{690}=X_{542}^{\perp }\hfill \\ \hfill M(X_{74},X_{110})& =X_{526}=X_{5663}^{\perp }\hfill \\ \hfill M(X_{98},X_{99})& =X_{804}=X_{2782}^{\perp }\hfill \end{array}$$

10. Trigonometric Polylines

Among the many central lines [9] of interest in triangle geometry are the trigonometic polylines n-Euler line and n-Nagel line. The Euler line itself is given by the following barycentric equations:

$$\begin{array}{cc}\hfill 0& =(tanB-tanC)x+(tanC-tanA)y+(tanA-tanB)z\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill & =(sin2B-sin2C)x+(sin2C-sin2A)y+(sin2A-sin2B)z\hfill \end{array}$$

(33)

$$\begin{array}{cc}\hfill & =cosAsin(B-C)x+cosBsin(C-A)y+cosCsin(A-B)z\hfill \\ \hfill & =(b^2-c^2)(b^2+c^2-a^2)x+(c^2-a^2)(c^2+a^2-b^2)y\hfill \\ \hfill & +(a^2-b^2)(a^2+b^2-c^2)z\hfill \end{array}$$

(34)

The n-Euler line is defined by substituting $nA,nB,nC$ for $A,B,C$, respectively, in (32), (), or (). The n-Euler line passes through the following n-polycenters, which, for $n=1$ are indexed in ETC [2] as $X_2,X_3,X_4,X_5$, respectively:

$$\begin{array}{cc}\hfill n-\mathrm{centroid}& =c_0=1:1:1\hfill \\ \hfill n-\mathrm{circumcenter}& =cosnA::\hfill \\ \hfill n-\mathrm{orthocenter}& =tannA::\hfill \\ \hfill n-\mathrm{nine}-\mathrm{point}\mathrm{center}& =sinnAcos(nB-nC)::.\hfill \end{array}$$

These points appear in a little-known paper [7] in a discussion of “layers” in triangle geometry—without mention of the fact that the n-points and n-lines have polynomial representations.

The Nagel line is given by the equations

$$\begin{array}{cc}\hfill 0& =(sinB-sinC)x+(sinC-sinA)y+(sinA-sinB)z.\hfill \\ \hfill & =(b-c)x+(c-a)y+(a-b)z,\hfill \end{array}$$

(35)

and the n-Nagel line by

$$0=(sinnB-sinnC)x+(sinnC-sinnA)y+(sinnA-sinnB)z.$$

(36)

The Nagel line passes through the incenter, $X_1=sinA:sinB:sinC$ and the centroid, $X_2=1:1:1$, so that the n-Nagel line passes through the centroid and the point $sinnA:sinnB:sinnC$. Thus, for every n, the n-Euler line and n-Nagel line meet in the centroid. Moreover, by () and (36), the $2n$-Nagel line and n-Euler line are identical for every positive integer n. Among the notable trigonometric polycenters on the 2-Euler line, alias 4-Nagel line, are the following:

$$\begin{array}{cc}\hfill X_2& =1:1:1\hfill \\ \hfill X_{54}& =\mathrm{Kosnita}\mathrm{point}=sinAsec(B-C)::\hfill \\ \hfill & =\mathrm{isogonal}\mathrm{conjugate}\mathrm{of}\mathrm{the}\mathrm{nine}-\mathrm{point}\mathrm{center},X_5\hfill \\ \hfill X_{68}& =\mathrm{Prasolov}\mathrm{point}=tan2A::\hfill \\ \hfill X_{1147}& =sin4A::\hfill \\ \hfill X_{5449}& =2\mathrm{nd}\mathrm{Hatzipolakis}-\mathrm{Moses}\mathrm{point}=sin4B+sin4C::\hfill \\ \hfill & =\mathrm{midpoint}\mathrm{of}{X}_{68}\mathrm{and}{X}_{1147}\hfill \\ \hfill X_{6193}& =tanB+tanC-tanA::\hfill \\ \hfill X_{16032}& =cscAsec(B-C)::\hfill \end{array}$$

Each of these points, and others on the 2-Euler line, has a list of properties in ETC [2], involving many other trigonometric polycenters and their interrelationships.

11. Concluding Remarks

The notion of trigonometric polycenter extends to various subjects, other than those mentioned above. Several examples follow:

• Triangle centers whose barycentrics depend on angles of the form

  $$nA+r\pi ,nB+r\pi ,nC+r\pi$$

  for some nonzero number r, such as the Fermat point,

  $$\begin{array}{cc}\hfill X_{13}& =sinAcsc(A+\pi /3):sinBcsc(B+\pi /3):sinCcsc(C+\pi /3)\hfill \\ \hfill & =a^4-2{(b^2-c^2)}^2+a^2(b^2+c^2+\Psi )::,\hfill \end{array}$$

  where $\Psi =4\sqrt{3}\times$(area of triangle $ABC$), and related points $X_i$ for $i=14,\dots ,18$ in ETC [2].
• Bicentric pairs [1] of points, such as the Brocard points,

  $$\begin{array}{cc}\hfill 1/b^2:1/c^2:1/a^2& ={sin}^{2}C{sin}^{2}A:{sin}^{2}A{sin}^{2}B:{sin}^{2}B{sin}^{2}C\hfill \\ \hfill 1/c^2:1/a^2:1/b^2& ={sin}^{2}B{sin}^{2}A:{sin}^{2}C{sin}^{2}B:{sin}^{2}A{sin}^{2}C,\hfill \end{array}$$

  leading to Brocard n-points by substituting $nA,nB,NC$ for $A,B,C$, respectively.
• Cubic curves such as those indexed and elegantly described by Bernard Gibert [3]. Here we sample just one of more than one thousand: K007, the Lucas cubic, consisting of all points $x:y:z$ that satisfy

  $$(b^2+c^2-a^2)x(y^2-z^2)+(c^2+a^2-b^2)y(z^2-x^2)+(a^2+b^2-c^2)z(x^2-y^2)=0.$$
  For every n, the symbolic substitution

  $$(a\to cosnA,b\to cosnB,c\to cosnC)$$

  transforms this “polynomial cubic” into a “trigonometric cubic”, and likewise for the substitution

  $$(a\to sinnA,b\to sinnB,c\to sinnC),$$

  etc. For details regarding symbolic substitutions see [6].
• Triangle centers that result from unary operations on trigonometric polycenters, such as

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

  where $x:y:z$ is a trigonometric polycenter. See [10].
• For specific numbers $a,b,c$, such as $(a,b,c)=(2,3,4)$, representing the smallest integer-sided isosceles triangle, we have integer sequences such as given by

  $$a_n=2^{2^n+1}cosnA,$$

  where A, as usual, is the angle opposite side $BC$ in a triangle $ABC$ having sidelengths $\left|BC\right|=a,\left|CA\right|=b,\left|AB\right|=c$. Such sequences have interesting divisibility properties, such as the fact that if p is a prime that divides a term, then the indices n such that p divides n comprise an arithmetic sequence. For this sequence and access to related sequences, see A375880.
• A final comment may be loosely summarized by the observation that, throughout this paper, the role of homogeneous coordinates can be taken by trilinear coordinates [4], but with different results. For example, in trilinear coordinates, we have

  $$\begin{array}{cc}\hfill X_2& =a:b:c=sinA:sinB:sinC\hfill \\ \hfill X_3& =a(b^2+c^2-a^2)::=cosA:cosBcosC::,\hfill \end{array}$$

  which lead to trigonometric polycenters by substituting $nA,nB,nC$ for $A,B,C$. The resulting trilinear representations are equivalent to the barycentric representations $sinAsinnA::$ and $sinAcosnA::$, these being trigonometric polycenters not previously mentioned in this article.