Trigonometric Polynomial Points in the Plane of a Triangle

Abstract

It is well known that the four ancient Greek triangle centers and others have homogeneous barycentric coordinates that are polynomials in the sidelengths $a,b,$ and c of a triangle $ABC$. For example, the circumcenter is represented by the polynomial $a(b^2+c^2-a^2)$. It is not so well known that triangle centers have barycentric coordinates, such as $\tan A : \tan B : \tan C$, that are also representable by polynomials, in this case, by $p(a, b, c) : p(b, c, a) : p(c, a, b)$, where $p(a, b, c)=a(a^2+b^2-c^2)(a^2+c^2-b^2)$. This paper presents and discusses the polynomial representations of triangle centers that have barycentric coordinates of the form $f(a, b, c) : f(b, c, a) : f(c, a, b)$, where f depends on one or more of the functions in the set $\{\cos , \sin , \tan , \sec , \csc , \cot \}$. The topics discussed include infinite trigonometric orthopoints, the n-Euler line, and symbolic substitution.

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:

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

The lengths of the sides opposite the vertex angles (which, like the vertex points, are denoted by $A, B,$ and C) are given the symbols $a, b,$ and c, respectively, and may be regarded as variables or algebraic indeterminates. For an excellent introduction to the subject of barycentrics, see Yiu [1].

Many triangle centers (as defined in [2]) have barycentrics that are polynomials. Following [3], 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 $A,B,$ and C, we call it a trigonometric polycenter. Analogously, we have polylines and trigonometric polylines. Note that if the first barycentric of X is written as $h(a,b,c)$, then the second and third barycentrics are determined (viz., $h(b,c,a)$ and $h(c,a,b)$) such that the shorter notation $h(a,b,c)::$ is sufficient.

Important examples of trigonometric polycenters include the following:

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

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

• $(b^2-c^2)(b^2+c^2-a^2)x+(c^2-a^2)(c^2+a^2-b^2)y+(a^2-b^2)(a^2+b^2-c^2)z=0$.
• $(\tan B-\tan C)x+(\tan C-\tan A)y+(\tan A-\tan B)z=0$.

Of great importance in triangle geometry are the following objects:

• The isotomic conjugate of X, with barycentrics
  $1/x:1/y:1/z=yz:zx:xy$.
• The 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 $x+y+z=0$.
• The Steiner circumellipse, with equation $1/x+1/y+1/z=0$.
• The 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 ([4,5]) 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)

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

(4)

Theorem 1. 

Let $\Lambda =b^2+c^2-a^2$. Then,

$$\cos nA::=a^n({(\Lambda -2iS)}^n+{(\Lambda +2iS)}^n)::$$

(5)

$$\sin nA::=a^n({(\Lambda -2iS)}^n-{(\Lambda +2iS)}^n)::.$$

(6)

Proof. 

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

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

where

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

such 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 $\cos nA=T_n(cosA)$, we have

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

such that (5) holds. Similarly, Equation (6) follows from (4) and the well-known fact that $\sin nA=U_{n-1}(cosA)\sin A$.    □

Our main goal in this section is to represent $\cos nA::$ and $\sin nA::$ as polycenters. To this end, let

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

(7)

so that the expressions in (3) and (4) 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 $a,b,$ and c given by the following three initial terms and a second-order recurrence:

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

and

$$\begin{array}{cc}\hfill c_n(a,b,c):c_n(b,c,a):c_n(c,a,b)& =\cos nA:\cos nB:\cos nC.\hfill \end{array}$$

(10)

Proof. 

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

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

such 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, $\cos nA::=a^n(u^n+v^n)::$, and since $d_n::=c_n::$, we have $c_n::=\cos nA::$.    □

Theorem 3. 

Let $\left(s_n\right)$ be the sequence given by (9). Then, $s_n$ is a polynomial in $a,b,$ and c given by these two initial terms and a second-order recurrence:

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

and

$$\begin{array}{cc}\hfill s_n(a,b,c):s_n(b,c,a):s_n(c,a,b)& =\sin nA:\sin nB:\sin nC.\hfill \end{array}$$

(11)

Proof. 

A proof similar to that of Theorem 2 springs from (4), leading, by way of the identity $\sin nA=U_{n-1}(cosA)\sin A$, to 

$$\sin nA={\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{{(2bc)}^n}}({(\Lambda -2iS)}^n-{(\Lambda +2iS)}^n)),$$

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

Example 1. 

Polycenter representations for $\cos 3A::$ and $\cos 4A::$ are given by

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

Example 2. 

Polycenter representations for $\sin nA::$ for $n=3,4,5,6$ are given by

$$\begin{array}{cc}\hfill s_3& =a^3(a^2-b^2-c^2-bc)(a^2-b^2-c^2+bc);\hfill \\ \hfill s_4& =a^4(a^2-b^2-c^2)(a^4+b^4+c^4-2a^2(b^2+c^2));\hfill \\ \hfill s_5& =a^5{f}_1{f}_2,\mathit{where}\hfill \\ \hfill & f_1=a^4+b^4+c^4+a^2(bc-2b^2-2c^2)+bc(bc-b^2-c^2)\hfill \\ \hfill & f_2=a^4+b^4+c^4+a^2(-bc-2b^2-2c^2)+bc(bc+b^2+c^2);\hfill \end{array}$$

$$\begin{array}{cc}\hfill s_6& =a^5{g}_1{g}_2{g}_3{g}_4,\mathit{where}\hfill \\ \hfill & g_1{g}_2{g}_3=(a^2-b^2-c^2)(a^2-b^2-c^2-bc)(a^2-b^2-c^2+bc)\hfill \\ \hfill & g_4=a^4+b^4+c^4-2a^2(b^2+c^2)-b^2{c}^2.\hfill \end{array}$$

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 $t_n::=\tan nA::$. 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 $\tan nA::$ 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 $\left(\sin \right(nA\left)\right),\left(\cos \right(nB\left)\right),$ and $\left(\cos \right(nC\left)\right)$ are linearly recurrent, and, of course,

$$u_n(a,b,c):u_n(b,c,a):u_n(c,a,b)=\tan nA:\tan nB:\tan nC.$$

A sequence of associated polycenters derived from $u_n(a,b,c)$ is considered in Section 7. Likewise the triangle centers $\sec nA::,\csc nA::,$ and $\cot nA::$ are polycenters for all nonzero integers n. Geometrically, these are isotomic conjugates given by $1/t_n,1/s_n,$ and $c_n/s_n$, 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) [6].

Example 3. 

A few trigonometric polycenters in the ETC [6]:

$$\begin{array}{cccc}\hfill \sin A::& =X_1\hfill & \hfill \csc A::& =X_{75}\hfill \\ \hfill \cos A::& =X_{63}\hfill & \hfill \sec A::& =X_{92}\hfill \\ \hfill \tan A::& =X_4\hfill & \hfill \cot A::& =X_{69}\hfill \\ \hfill \sin 2A::& =X_3\hfill & \hfill \csc 2A::& =X_{264}\hfill \\ \hfill \cos 2A::& =X_{1993}\hfill & \hfill \sec 2A::& =X_{5392}\hfill \\ \hfill \tan 2A::& =X_{68}\hfill & \hfill \cot 2A::& =X_{317}\hfill \\ \hfill \sin 3A::& =X_{6149}\hfill & \hfill \csc 3A::& =X_{63759}\hfill \\ \hfill \cos 3A::& =X_{63760}\hfill & \hfill \sec 3A::& =X_{63764}\hfill \\ \hfill \tan 3A::& =X_{562}\hfill & \hfill \cot 3A::& =X_{63761}\hfill \\ \hfill \sin 4A::& =X_{1147}\hfill & \hfill \csc 4A::& =X_{55553}\hfill \\ \hfill \cos 4A::& =X_{63762}\hfill & \hfill \sec 4A::& =X_{63765}\hfill \\ \hfill \tan 4A::& =X_{43973}\hfill & \hfill \cot 4A::& =X_{55552}\hfill \end{array}$$

Barycentric products and quotients ([1], 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 the ETC [6] (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 the quotients of polynomials. We begin with

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

(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)

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

(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

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

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 $f_0=1$ and $f_1=u$. 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 $g_0=0$ and $g_1=1$.

Example 5. 

A few more trigonometric polycenters in the ETC [6]:

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

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 the 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 the 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 the ETC [6]:

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

5. Sums Involving mB + nC and nB + mC

Proofs of the next two theorems can be obtained by adapting the codes in moretrp,halfanglepoly.

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

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

(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 the 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)

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

(24)

with the 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 recurrences with the 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 the recurrence order $p+1$.

Other families to which the procedure applies are represented by

$$\tan nA::,\sec nA::,\csc nA::,\cot nA::.$$

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

$$\tan nA::=\sin nA\cos nB\cos nC::.$$

The resulting sixth-degree recurrence signature for $\sin nA\cos nB\cos nC$ is more efficiently expressed with Conway notation [7] than with $a,b,$ and 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

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

with these being the points at which the lines $X_1{X}_{75},X_1{X}_2,$ and $X_2{X}_3$ 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 $L^{\infty }$ at X. The line through P orthogonal to $PX$ meets $L^{\infty }$ at 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

$$X^{\perp }=y\cos B-z\cos C::.$$

(27)

Thus, if X is a polycenter represented by a polynomial $x(a,b,c)$ as the 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 the ETC [6], occupy the lists in the next two examples.

Example 10. 

Pairs of trigonometric orthopoints:

$$\begin{array}{cc}\hfill \cos B-\cos C::=X_{522},& X_{522}^{\perp }=X_{515}\hfill \\ \hfill \sin B-\sin C::=X_{514},& X_{514}^{\perp }=X_{516}\hfill \\ \hfill \tan B-\tan C::=X_{525},& X_{525}^{\perp }=X_{1503}\hfill \\ \hfill \sec B-\sec C::=X_{521},& X_{521}^{\perp }=X_{6001}\hfill \\ \hfill \csc B-\csc C::=X_{513},& X_{513}^{\perp }=X_{517}\hfill \\ \hfill \cot B-\cot C::=X_{523},& X_{523}^{\perp }=X_{30}\hfill \\ \hfill {\cos }^{2}B-{\cos }^{2}C::=X_{523},& X_{523}^{\perp }=X_{30}\hfill \\ \hfill {\sin }^{2}B-{\sin }^{2}C::=X_{523},& X_{523}^{\perp }=X_{30}\hfill \end{array}$$

Example 10 continued:

$$\begin{array}{cc}\hfill {\tan }^{2}B-{\tan }^{2}C::=X_{520},& X_{520}^{\perp }=X_{6000}\hfill \\ \hfill {\sec }^{2}B-{\sec }^{2}C::=X_{520},& X_{520}^{\perp }=X_{6000}\hfill \\ \hfill {\csc }^{2}B-{\csc }^{2}C::=X_{512},& X_{512}^{\perp }=X_{511}\hfill \\ \hfill {\cot }^{2}B-{\cot }^{2}C::=X_{512},& X_{512}^{\perp }=X_{511}\hfill \\ \hfill \cos 2B-\cos 2C::=X_{523},& X_{523}^{\perp }=X_{30}\hfill \\ \hfill \sin 2B-\sin 2C::=X_{525},& X_{525}^{\perp }=X_{1503}\hfill \\ \hfill \sec 2B-\sec 2C::=X_{924},& X_{924}^{\perp }=X_{13754}\hfill \\ \hfill \cot 2B-\cot 2C::=X_{6368},& X_{6368}^{\perp }=X_{18400}\hfill \end{array}$$

Example 11. 

More pairs of trigonometric orthopoints:

$$\begin{array}{cc}\hfill 2\cos A-\cos B-\cos C::=X_{527},& X_{527}^{\perp }=X_{28292}\hfill \\ \hfill 2\sin A-\sin B-\sin C::=X_{519},& X_{527}^{\perp }=X_{3667}\hfill \\ \hfill 2\tan A-\tan B-\tan C::=X_{30},& X_{30}^{\perp }=X_{523}\hfill \\ \hfill 2\csc A-\csc B-\csc C::=X_{536},& X_{536}^{\perp }=X_{28475}\hfill \\ \hfill 2\cot A-\cot B-\cot C::=X_{524},& X_{524}^{\perp }=X_{1499}\hfill \\ \hfill 2{\cos }^{2}A-{\cos }^{2}B-{\cos }^{2}C::=X_{524},& X_{524}^{\perp }=X_{1499}\hfill \\ \hfill 2{\sin }^{2}A-{\sin }^{2}B-{\sin }^{2}C::=X_{524},& X_{524}^{\perp }=X_{1499}\hfill \\ \hfill 2{\csc }^{2}A-{\csc }^{2}B-{\csc }^{2}C::=X_{538},& X_{538}^{\perp }=X_{32472}\hfill \\ \hfill 2{\cot }^{2}A-{\cot }^{2}B-{\cot }^{2}C::=X_{538},& X_{538}^{\perp }=X_{32472}\hfill \\ \hfill 2\cos 2A-\cos 2B-\cos 2C::=X_{524},& X_{524}^{\perp }=X_{1499}\hfill \\ \hfill 2\sin 2A-\sin 2B-\sin 2C::=X_{30},& X_{30}^{\perp }=X_{523}\hfill \\ \hfill 2\tan 2A-\tan 2B-\tan 2C::=X_{539},& X_{539}^{\perp }=X_{20184}\hfill \end{array}$$

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 $P,O$, and 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

$$b^2+c^2-a^2:c^2+a^2-b^2:a^2+b^2-c^2=\cot A:\cot B:\cot C.$$

Note that $M(P,U)$ 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:

$$\begin{array}{cc}\hfill 0& =(\tan B-\tan C)x+(\tan C-\tan A)y+(\tan A-\tan B)z\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill & =(\sin 2B-\sin 2C)x+(\sin 2C-\sin 2A)y+(\sin 2A-\sin 2B)z\hfill \end{array}$$

(33)

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

(34)

The n-Euler line is defined by substituting $nA,nB,$ and $nC$ for $A,B,$ and C, respectively, in (32), () or (). The n-Euler line passes through the following n-polycenters, which, for $n=1$, are indexed in the ETC [6] as $X_2,X_3,X_4,$ and $X_5$, respectively:

$$\begin{array}{cc}\hfill n\text{-}\mathrm{centroid}& =c_0=1:1:1\hfill \\ \hfill n\text{-}\mathrm{circumcenter}& =\cos nA::\hfill \\ \hfill n\text{-}\mathrm{orthocenter}& =\tan nA::\hfill \\ \hfill n\text{-}\mathrm{nine}\text{-}\mathrm{point}\mathrm{center}& =\sin nA\cos (nB-nC)::.\hfill \end{array}$$

These points appear in a little-known paper [10] 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

$$\begin{array}{cc}\hfill 0& =(\sin B-\sin C)x+(\sin C-\sin A)y+(\sin A-\sin B)z.\hfill \\ \hfill & =(b-c)x+(c-a)y+(a-b)z,\hfill \end{array}$$

(35)

and the n-Nagel line by

$$0=(\sin nB-\sin nC)x+(\sin nC-\sin nA)y+(\sin nA-\sin nB)z.$$

(36)

The Nagel line passes through the incenter $X_1=\sin A:\sin B:\sin C$ and the centroid $X_2=1:1:1$ such that the n-Nagel line passes through the centroid and the point $\sin nA:\sin nB:\sin nC$. Thus, for every n, the n-Euler line and n-Nagel line meet at the centroid. Moreover, by (33) 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, and thus, the 4-Nagel line, are the following:

$$\begin{array}{cc}\hfill X_2& =1:1:1\hfill \\ \hfill X_{54}& =\mathrm{Kosnita}\mathrm{point}=\sin A\sec (B-C)::\hfill \\ \hfill & =\mathrm{isogonal}\mathrm{conjugate}\mathrm{of}\mathrm{the}\mathrm{nine}\text{-}\mathrm{point}\mathrm{center},X_5\hfill \\ \hfill X_{68}& =\mathrm{Prasolov}\mathrm{point}=\tan 2A::\hfill \\ \hfill X_{1147}& =\sin 4A::\hfill \\ \hfill X_{5449}& =2\mathrm{nd}\mathrm{Hatzipolakis}\text{-}\mathrm{Moses}\mathrm{point}=\sin 4B+\sin 4C::\hfill \\ \hfill & =\mathrm{midpoint}\mathrm{of}{X}_{68}\mathrm{and}{X}_{1147}\hfill \\ \hfill X_{6193}& =\tan B+\tan C-\tan A::\hfill \\ \hfill X_{16032}& =\csc A\sec (B-C)::\hfill \end{array}$$

Each of these points, and others on the 2-Euler line, has a list of properties in the ETC [6] involving many other trigonometric polycenters and their interrelationships.

11. Concluding Remarks

The notion of a 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

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

  where $\Psi =4\sqrt{3}\times$(area of triangle $ABC$), and related points $X_i$ for $i=14,\dots ,18$ in the ETC [6].
• Bicentric pairs [11] 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 $nA,nB,$ and $nC$ for $A,B,$ and C, respectively.
• Cubic curves, such as those indexed and elegantly described by Bernard Gibert [12]. 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

  $$(a\to \cos nA,b\to \cos nB,c\to \cos nC)$$

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

  $$(a\to \sin nA,b\to \sin nB,c\to \sin nC),$$

  etc. For details regarding symbolic substitutions, see [13].
• 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 [14].
• For specific numbers $a,b,$ and c, such as $(a,b,c)=(2,3,4)$, representing the smallest integer-sided isosceles triangle, we have integer sequences, such as that given by

  $$a_n=2^{2^n+1}\cos nA,$$

  where A, as usual, is the angle opposite side $BC$ in a triangle $ABC$ having sidelengths $\left|BC\right|=a,\left|CA\right|=b,$ and $\left|AB\right|=c$. 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 https://oeis.org/A375880 (accessed on 1 December 202) [15].
• 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 [16], but with different results. For example, in trilinear coordinates, we have

  $$\begin{array}{cc}\hfill X_2& =a:b:c=\sin A:\sin B:\sin C\hfill \\ \hfill X_3& =a(b^2+c^2-a^2)::=\cos A:\cos B:\cos C,\hfill \end{array}$$

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