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Trigonometric Polynomial Points in the Plane of a Triangle

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18 November 2024

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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; c of a triangle ABC. For example, the circumcenter is represented by the polynomial a(b2 + c2 - a2). It is not so well known that triangle centers having barycentric coordinates such as tanA : tanB : tanC 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(a2 + b2 - c2)(a2 + c2 - b2). This paper presents and discusses 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 fcos; sin; tan; sec; csc; cotg. The topics discussed include innite trigonometric orthopoints, the n-Euler line, and symbolic substitution.
Keywords: 
homogeneous coordinates
;  
barycentric
;  
trilinear
;  
triangle
;  
triangle center
;  
polynomial point
;  
trigonometry
;  
isogonal conjugate
;  
isotomic conjugate
;  
Euler line
;  
Nagel line
;  
symbolic substitution
Subject: 
Computer Science and Mathematics  -   Geometry and Topology

MSC:  51N20; 51M05

1. Introduction

One of the most productive systems of representation for points and lines in the plane of a triangle A B C is a system widely known as homogeneous barycentric coordinates (henceforth simply barycentrics). Serving as the “origin” in this system are the three vertices of A B C , 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 , 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:
G = centroid = 1 : 1 : 1 = 1 : : O = circumcenter = a 2 ( b 2 + c 2 − a 2 ) : : = sin 2 A : : H = orthocenter = ( c 2 + a 2 − b 2 ) ( a 2 + b 2 − c 2 ) : : = tan A : : N = nine - point center = a 2 ( b 2 + c 2 ) − ( b 2 − c 2 ) 2 : : = sin A cos ( B − C ) : :
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
  • ( 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:
  • isotomic conjugate of X, with barycentrics
    1 / x : 1 / y : 1 / z = y z : z x : x y
  • isogonal conjugate of X, with barycentrics
    a 2 / x : b 2 / y : c 2 / z = a 2 y z : b 2 z x : c 2 x y
  • the line at infinity, L ∞ , 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 ( n A ) : : 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 :
T n ( x ) = 2 x T n − 1 ( x ) − T n − 2 ( x ) for n ≥ 2 , T 0 ( x ) = 1 , T 1 ( x ) = x ;
U n ( x ) = 2 x U n − 1 ( x ) − U n − 2 ( x ) for n ≥ 2 , U 0 ( x ) = 1 , U 1 ( x ) = 2 x ,
Another well-known type of recurrence relation for these families of polynomials ([12,13]) depends on complex numbers:
T n ( x ) = ( 1 / 2 ) ( x − x 2 − 1 ) n + ( x + x 2 − 1 ) n
U n ( x ) = ( 1 / ( 2 x 2 − 1 ) ( x + x 2 − 1 ) n + 1 − ( x − x 2 − 1 ) n + 1 .
Theorem 1.
Let Λ = b 2 + c 2 − a 2 . Then
cos n A : : = a n ( Λ − 2 i S ) n + ( Λ + 2 i S ) n : :
sin n A : : = a n ( Λ − 2 i S ) n − ( Λ + 2 i S ) n : : .
Proof. 
We have cos A = Λ / ( 2 b c ) , and (3) gives
T n ( cos A ) = ( 1 / 2 ) ( cos A − i sin A ) n + ( cos A + i sin A ) n = ( 1 / 2 ) ( Λ 2 b c − i S b c ) n + ( Λ 2 b c + i S b c ) n ,
where
S = 2 ( area of ▵ A B C ) = ( 1 / 2 ) ( a + b + c ) ( b + c − 1 ) ( c + a − b ) ( a + b − c ) ,
so that
S 2 = ( 1 / 4 ) ( − a 4 − b 4 − c 4 + 2 b 2 c 2 + 2 c 2 a 2 + 2 a 2 b 2 ) .
Now since cos n A = T n ( cos A ) , we have
cos n A = ( 1 / 2 ) 1 ( 2 b c ) n ( Λ − 2 i S ) n + ( Λ + 2 i S ) n ,
so that (5) holds. Similarly, Equation (6) follows from (4) and the well-known fact that sin n A = U n − 1 ( cos A ) sin A .    □
Our main goal in this section is to represent cos n A : : and sin n A : : as polycenters. To that end, let
u = Λ − 2 i S and v = Λ + 2 i S ,
so that expressions in (3) and () can be recast in order to define sequences ( c n ) and ( s n ) as follows:
c n = c n ( a , b , c ) = a n ( u n + v n )
s n = s n ( a , b , c ) = a n ( u n − v n ) .
Next, we have a lemma about u and v.
Lemma 1.
u 2 − 2 ( b 2 + c 2 − a 2 ) u + 4 b 2 c 2 = v 2 − 2 ( b 2 + c 2 − a 2 ) v + 4 b 2 c 2 = 0 .
Proof. 
The imaginary terms cancel, and the real term is
− ( b 2 + c 2 − a 2 ) 2 − 4 S 2 + 4 b 2 c 2 = 0 .
   □
Theorem 2.
Let ( c n ) be the sequence given by (8). Then c n is a polynomial in a , b , c , given by the following three initial terms and 2nd-order recurrence:
c 0 = 1 c 1 = a ( b 2 + c 2 − a 2 ) c 2 = a 2 ( a 4 + b 4 + c 4 − 2 a 2 b 2 − 2 a 2 c 2 ) c n = a ( b 2 + c 2 − a 2 ) c n − 1 − a 2 b 2 c 2 c n − 2 for n ≥ 3 ,
and
c n ( a , b , c ) : c n ( b , c , a ) : c n ( c , a , b ) = cos n A : cos n B : cos n C .
Proof. 
It is easy to verify that c 0 , c 1 , c 2 are polycenters as claimed. Suppose now that n ≥ 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 + 4 b 2 c 2 = − v n − 2 v 2 − 2 ( b 2 + c 2 − a 2 ) v + 4 b 2 c 2 ,
so that
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 ) ( a / 2 ) n ( u n + v n ) = a ( b 2 + c 2 − a 2 ) ( a / 2 ) n − 1 ( u n − 1 + v n − 1 ) − 4 b 2 c 2 ( u n − 2 + v n − 2 ) .
This shows that if d k = ( a / 2 ) k ( u k + v k ) for k ≥ 3 , then
d n = a ( b 2 + c 2 − a 2 ) d n − 1 − a 2 b 2 c 2 d n − 2 for n ≥ 3 .
By Theorem 1, cos n A : : = a n ( u n + v n ) : : , and since d n : : = c n : : , we have c n : : = cos n A : : .    □
Theorem 3.
Let ( s n ) be the sequence given by (). Then s n is a polynomial in a , b , c , given by these two initial terms and 2nd-order recurrence:
s 1 = a s 2 = a 2 ( b 2 + c 2 − a 2 ) s n = a ( b 2 + c 2 − a 2 ) s n − 1 − a 2 b 2 c 2 s n − 2 for n ≥ 2 ,
and
s n ( a , b , c ) : s n ( b , c , a ) : s n ( c , a , b ) = sin n A : sin n B : sin n C .
Proof. 
A proof similar to that of Theorem 2 springs from (), leading, by way of the identity sin n A = U n − 1 ( cos A ) sin A , to
sin n A = 1 2 1 ( 2 b c ) n ( Λ − 2 i S ) n − ( Λ + 2 i S ) n ) ,
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 cos 3 A : : and cos 4 A : : are given by
c 3 = a 3 ( a 2 − b 2 − c 2 ) a 4 + b 4 + c 4 − 2 a 2 ( b 2 + c 2 ) − b 2 c 2 ; c 4 = a 4 ( a 8 + b 8 + c 8 − 4 a 6 ( b 2 + c 2 ) + 2 a 4 ( 3 b 4 + 3 c 4 + 4 b 2 c 2 ) − 4 a 2 ( b 4 + c 4 ) ( b 2 + c 2 ) ) .
Example 2.
Polycenter representations for sin n A : : , for n = 3 , 4 , 5 , 6 , are given by
s 3 = a 3 ( a 2 − b 2 − c 2 − b c ) ( a 2 − b 2 − c 2 + b c ) ; s 4 = a 4 ( a 2 − b 2 − c 2 ) a 4 + b 4 + c 4 − 2 a 2 ( b 2 + c 2 ) ; s 5 = a 5 f 1 f 2 , where f 1 = a 4 + b 4 + c 4 + a 2 ( b c − 2 b 2 − 2 c 2 ) + b c ( b c − b 2 − c 2 ) f 2 = a 4 + b 4 + c 4 + a 2 ( − b c − 2 b 2 − 2 c 2 ) + b c ( b c + b 2 + c 2 ) ; s 6 = a 5 g 1 g 2 g 3 g 4 , where g 1 g 2 g 3 = ( a 2 − b 2 − c 2 ) ( a 2 − b 2 − c 2 − b c ) ( a 2 − b 2 − c 2 + b c ) g 4 = a 4 + b 4 + c 4 − 2 a 2 ( b 2 + c 2 ) − b 2 c 2 .
Inductively, c n and s n both have degree 3 n for n ≥ 0 and both are polynomial multiples of a n . By Theorems 2 and 3, the sequences ( c n ) and ( s n ) have the same second-order recurrence signature:
a ( b 2 + c 2 − a 2 ) , − a 2 b 2 c 2 .
Next, let t n = s n / c n , so that t n : : = tan n A : : . 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 n A : : for n = 1 , 2 , 3 .
t 1 = 1 / ( a 2 − b 2 − c 2 ) ; t 2 = ( a 2 − b 2 − c 2 ) / a 4 + b 4 + c 4 − 2 a 2 ( b 2 + c 2 ) ; t 3 = 1 / a 4 + b 4 + c 4 − 2 a 2 ) ( b 2 + c 2 ) − b 2 c 2 ;
The sequence ( t n ) , 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 ( n A ) cos ( n B ) cos ( n C ) , is linearly recurrent, since the three sequences ( sin ( n A ) ) , ( cos ( n B ) ) , ( cos ( n C ) ) are linearly recurrent, and, of course,
u n ( a , b , c ) : u n ( b , c , a ) : u n ( c , a , b ) = tan n A : tan n B : tan n C .
A sequence of associated polycenters derived from on u n ( a , b , c ) is considered in Section 7. Likewise the triangle centers sec n A : : , csc n A : : , cot n A : : 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].
sin A : : = X 1 csc A : : = X 75 cos A : : = X 63 sec A : : = X 92 tan A : : = X 4 cot A : : = X 69 sin 2 A : : = X 3 csc 2 A : : = X 264 cos 2 A : : = X 1993 sec 2 A : : = X 5392 tan 2 A : : = X 68 cot 2 A : : = X 317 sin 3 A : : = X 6149 csc 3 A : : = X 63759 cos 3 A : : = X 63760 sec 3 A : : = X 63764 tan 3 A : : = X 562 cot 3 A : : = X 63761 sin 4 A : : = X 1147 csc 4 A : : = X 55553 cos 4 A : : = X 63762 sec 4 A : : = X 63765 tan 4 A : : = X 43973 cot 4 A : : = X 55552
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.)
sin 2 A : : = a 2 : : = X 6 csc 2 A : : = b 2 c 2 : : = X 76 cos 2 A : : = a 2 ( b 2 + c 2 − a 2 ) 2 : : = X 394 sec 2 A : : = b 2 c 2 ( b 2 + c 2 − a 2 ) − 2 : : = X 2052 cos 2 ( B − C ) : : = b 2 c 2 ( b 4 + c 4 − 2 b 2 c 2 − a 2 b 2 − a 2 c 2 ) 2 : : = X 45793 sin 2 ( B − C ) : : = b 2 c 2 ( b 2 − c 2 ) 2 : : = X 338 csc 2 ( B − C ) : : = a 2 ( b 2 − c 2 ) − 2 : : = X 249

3. More Trigonometric Polycenters

In this section, we first present polycenters for triangle centers of the forms cos ( n B − n C ) : : and sin ( n B − n C ) , and follow with a proof-by-computer-code for a recurrence equation for the points cos ( n B − n C ) : : as polycenters . Let M = a 2 ( b 2 + c 2 ) − ( b 2 − c 2 ) 2 . Then
cos ( B − C ) : : = p 1 ( a , b , c ) : : , where p 1 ( a , b , c ) = b c M cos ( 2 ( B − C ) ) : : = p 2 ( a , b , c ) : : , where p 2 ( a , b , c ) = b 2 c 2 ( ( b 2 − c 2 ) 4 + a 4 ( b 4 + c 4 ) + 2 a 2 ( − b 6 + b 4 c 2 + b 2 c 4 − c 6 ) ) : : cos ( n ( B − C ) ) : : = p n ( a , b , c ) : : , where p n ( a , b , c ) = b c M p n − 1 ( a , b , c ) − a 4 b 4 c 4 p n − 2 ( a , b , c ) for n ≥ 3 . sin ( B − C ) : : = q 1 ( a , b , c ) : : , where q 1 ( a , b , c ) = b c ( b 2 − c 2 ) sin ( 2 ( B − C ) ) : : = q 2 ( a , b , c ) : : , where q 2 ( a , b , c ) = b 2 c 2 ( b 2 − c 2 ) M sin ( n ( B − C ) ) : : = q n ( a , b , c ) : : , where q n = b c M q n − 1 − a 4 b 4 c 4 q n − 2 for n ≥ 2 .
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:
b c ( a 2 b 2 − b 4 + a 2 c 2 + 2 b 2 c 2 − c 4 ) , − a 4 b 4 c 4 .
A proof of the recurrence equation, or more precisely, the signature of the recurrence, for sin ( n ( B − C ) ) : : as a polycenter is found in much the same way.
As an alternative to representing the family cos ( n B − n C ) : : by polynomials, there are relatively simpler representations using quotients of polynomials. We begin with
u = cos ( B − C ) = cos B cos C + sin B sin C
= ( b 2 − c 2 ) 2 − a 2 b 2 − a 2 c 2 2 a 2 b c
v = sin ( B − C ) = sin B cos C + sin C cos B
= S ( b 2 − c 2 ) 2 a 2 b c , where
S = ( 1 / 2 ) ( a + b + c ) ( − a + b + c ) ( a − b + c ) ( a + b − c ) .
Let
f n = f n ( u , v ) = ( 1 / 2 ) ( u − i v ) n + ( u + i v ) n = cos ( n B − n C ) .
Then by the binomial theorem,
f n = ∑ k = 0 ⌊ n / 2 ⌋ ( − 1 ) k n 2 k u n − 2 k v 2 k ,
which satisfies the recurrence
f n = 2 u 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 ( n B − n C ) / sin ( B − C ) , we find that
g n = 2 u 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].
sin ( B − C ) : : = X 1577 csc ( B − C ) : : = X 662 cos ( B − C ) : : = X 14213 sec ( B − C ) : : = X 2167 tan ( B − C ) : : = X 15412 cot ( B − C ) : : = X 14570 sin ( 2 B − 2 C ) : : = X 18314 csc ( 2 B − 2 C ) : : = X 18315 cos ( 2 B − 2 C ) : : = X 63763 sec ( 2 B − 2 C ) : : = X 63766

4. Half-Angle Trigonometric Polycenters

The next list shows half-angle functions that involve polynomials (viz. they are “radical multiples of polycenters”). Let
P = ( b + c ) 2 − a 2 / ( b c ) and Q = a 2 − ( b − c ) 2 / ( b c ) .
cos ( A / 2 ) : : = p 1 ( a , b , c ) : : , where p 1 ( a , b , c ) = P cos ( 3 A / 2 ) : : = p 3 ( a , b , c ) : : , where p 3 ( a , b , c ) = P ( a 2 − b 2 − c 2 + b c ) cos ( n A / 2 ) : : = p n ( a , b , c ) : : , where p n ( a , b , c ) = ( − a 2 + b 2 + c 2 ) / ( b c ) p n − 2 − a 4 b 4 c 4 p n − 4 : : , for odd n ≥ 5 ; sin ( A / 2 ) : : = q 1 ( a , b , c ) : : , where q 1 ( a , b , c ) = Q : : sin ( 3 A / 2 ) : : = q 3 ( a , b , c ) : : , where q 3 ( a , b , c ) = Q ( a 2 − b 2 − c 2 − b c ) : : sin ( n A / 2 ) : : = q n ( a , b , c ) : : , where q n ( a , b , c ) = ( − a 2 + b 2 + c 2 ) / ( b c ) q n − 2 − a 4 b 4 c 4 q n − 4 , : : , for odd n ≥ 5 .
Next we show Mathematica code for obtaining trigonometric rational functions (quotients of polynomials) for cos ( ( n B − n C ) / 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 ( ( n B − n C ) / 2 ) : : is a rational function with signature
( a 2 b 2 − b 4 + a 2 c 2 + 2 b 2 c 2 − c 4 ) / ( a 2 b c ) , − 1 .
(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].
cos ( A / 2 ) : : = X 188 sec ( A / 2 ) : : = X 4146 sin ( A / 2 ) : : = X 174 csc ( A / 2 ) : : = X 556 cos ( 3 A / 2 ) : : = X 63779 sin ( 3 A / 2 ) : : = X 63780

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 ( m B + n C ) + sin ( n B + m C ) and let p ( m , n ) = p ( m , n , a , b , c ) be the polycenters given by these recurrences:
p ( m , n ) = a ( a 2 − b 2 − c 2 ) p ( m − 1 , n ) − a 2 b 2 c 2 p ( m − 2 , n ) ; p ( m , n ) = a ( a 2 − b 2 − c 2 ) p ( m , n − 1 ) − a 2 b 2 c 2 p ( m , n − 2 ) ,
where p ( 0 , 0 ) = 0 , and p ( 0 , 1 ) = p ( 1 , 0 ) = b + c . Then
p ( m , n , a , b , c ) : : = sin ( m B + n C ) + sin ( n B + m C ) : : .
Example 7.
The appearance of ( m , n , X k ) in the following list signifies that sin ( m B + n C ) + sin ( n B + m C ) : : = X k .
( 0 , 1 , X 10 ) , ( 0 , 2 , X 5 ) , ( 0 , 3 , X 63803 ) , ( 0 , 4 , X 5449 ) ( 1 , 1 , X 1 ) , ( 1 , 2 , X 63802 ) , ( 1 , 3 , X 44707 ) ( 2 , 2 , X 3 ) , ( 2 , 3 , X 63801 ) , ( 2 , 4 , X 1154 ) ( 3 , 3 , X 6149 ) , ( 3 , 5 , X 63801 )
Theorem 5.
Let g ( m , n ) = g ( m , n , a , b , c ) = cos ( m B + n C ) + cos ( n B + m C ) 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
q ( 0 , 1 ) = q ( 1 , 0 ) = ( b + c ) ( c + a − b ) ( a + b − c ) q ( 0 , 2 ) = ( b 2 + c 2 − a 2 ) ( a 2 ( b 2 + c 2 ) − ( b 2 − c 2 ) 2 ) .
Then
q ( m , n , a , b , c ) : : = cos ( m B + n C ) + cos ( n B + m C ) : : .
Example 8.
The appearance of ( m , n , X k ) here means that cos ( m B + n C ) + cos ( n B + m C ) : : = X k .
( 0 , 1 , X 226 ) , ( 0 , 2 , X 343 ) , ( 0 , 4 , X 63806 ) ( 1 , 1 , X 63 ) , ( 1 , 2 , X 16577 ) , ( 1 , 3 , X 63808 ) ( 2 , 2 , X 1993 ) , ( 2 , 3 , X 73802 ) ( 3 , 3 , X 63760 ) , ( 3 , 5 , X 63762 )

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

Here, we find a sequence p n = p n ( a , b , c ) of polycenters satisfying
p n ( a , b , c ) : p n ( b , c , a ) : p n ( c , a , b ) = j + k cos ( n A ) : j + k cos ( n B ) : j + k cos ( n C ) ,
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
u 0 = j + k
u 1 = j + − a 2 + b 2 + c 2 2 b c k
u 2 = j + a 4 + b 4 + c 4 − 2 a 2 b 2 − 2 a 2 c 2 2 b 2 c 2 k
and recurrence signature ( τ , − τ , 1 ) , where
τ = − a 2 + b 2 + c 2 + b c b c .
The transformation is simply to multiply, where appropriate, by 2 a n b n c n , obtaining
p 0 = 2 j + 2 k
p 1 = 2 a b c j + a ( b 2 + c 2 − a 2 ) k
p 2 = 2 a 2 b 2 c 2 j + a 6 − a 4 ( b 2 + c 2 ) + a 2 ( b 4 + c 4 ) k ,
with third-order recurrence signature
( − a Q a , a 2 b c Q a , a 3 b 3 c 3 ) ,
where Q a = Q a ( a , b , c ) = a 2 − b 2 − c 2 − b c .
Finally, we apply the substitutions ( a , b , c ) → ( b , c , a ) and ( a , b , c ) → ( 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 ( n B − n C ) : : and sin 2 ( n B − n C ) : : . Both cos 2 ( n B − n C ) and sin 2 ( n B − n C ) have third-order recurrence with signature ( k ( a , b , c ) , − k ( a , b , c ) , 1 ) , where
k = ( b 2 − c 2 ) 4 + a 4 ( b 4 + c 4 + b 2 c 2 ) + 2 a 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 ≥ 2 , the polycenters cos p ( n B − n C ) : : and sin p ( n B − n C ) : : have recurrence order p + 1 .
Other families to which the procedure applies are represented by
tan n A : : , sec n A : : , csc n A : : , cot n A : : .
Here we consider only the rational-function recurrence for tan n A . The most direct approach appears to be to use the identity
tan n A : : = sin n A cos n B cos n C : : .
The resulting sixth-degree recurrence signature for sin n A cos n B cos n C 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
k 1 = − 2 ( 3 S A S B S C + S 2 S ω ) D k 2 = − 16 S 6 + 15 S A 2 S B 2 S C 2 + 2 S 2 S A S B S C S ω − S 4 S ω 2 D k 3 = − − 4 ( − 8 S 6 + 5 S A 2 S B 2 S C 2 + 2 S 2 S a S B S C S ω + S 4 S ω 2 ) D ,
where D = ( S A S B S C S 2 S ω ) 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 .
Most of the named points on L ∞ are polycenters. Among the simplest are
X 513 = a ( b − c ) : : X 514 = b − c : : X 30 = cos A - 2 cos B cos C : : = 2 a 4 − ( b 2 − c 2 ) 2 − a 2 ( b 2 + c 2 ) : : ,
these being the points in which the lines X 1 X 75 , X 1 X 2 , X 2 X 3 meet L ∞ , 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 ∞ , then X can be regarded as a direction in the plane of ▵ A B C , since for every point P not on L ∞ , every line parallel to the line P X intersects L ∞ in X. The line through P orthogonal to P X meets L ∞ in a point called the orthopoint (or, in [8], the orthogonal conjugate) of X. We denote the orthopoint of X by X ⊥ . Barycentrics for X ⊥ are given by
X ⊥ = y cos B − z cos C : : .
Thus if X is a polycenter represented by a polynomial x ( a , b , c ) as first barycentric, then X ⊥ is the polycenter
X ⊥ = x ( b , c , a ) b ( b 2 − a 2 − c 2 ) − x ( c , a , b ) c ( c 2 − b 2 − a 2 ) : : .
Now for any point X = x : y : z , not necessarily a polycenter and not necessarily on L ∞ , the points
y − z : z − x : x − y and 2 x − y − z : 2 y − z − x : 2 z − x − y
are clearly on L ∞ , as are their orthopoints,
b ( z − x ) ( b 2 − a 2 − c 2 ) − c ( x − y ) ( c 2 − b 2 − a 2 ) : :
and
b ( 2 y − z − x ) ( b 2 − a 2 − c 2 ) − c ( 2 z − x − y ) ( c 2 − b 2 − a 2 ) : : ,
respectively. Moreover, if X is a polycenter, then the orthopoints (29) and (30) are polycenters on L ∞ .
Example 9.
Let
X = x : y : z = cos 2 A : : = a 2 ( a 2 − b 2 − c 2 ) 2 : : = X 394 .
Then
y − z : : = cos 2 B − cos 2 C : : = X 523 2 x − y − z : : = 2 cos 2 A − cos 2 B − cos 2 C : : = X 527 ,
and the two corresponding orthopoints (29) and (30) are respectively
X 30 = 2 a 4 − ( b 2 − c 2 ) 2 − a 2 ( b 2 + c 2 ) : : X 28292 = b c / ( h ( a , b , c ) h ( a , c , b ) ) : : , where h ( a , b , c ) = a 3 + 3 b 3 + c 3 + a 2 ( b − c ) − 5 b 2 ( a + c ) + c 2 ( b − a ) + 2 a b c .
Example 9 typifies infinite polycenters of the forms f ( B ) − f ( C ) : : and f ( 2 A ) − f ( B ) − f ( C ) : : . 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.
cos B − cos C : : = X 522 , X 522 ⊥ = X 515 sin B − sin C : : = X 514 , X 514 ⊥ = X 516 tan B − tan C : : = X 525 , X 525 ⊥ = X 1503 sec B − sec C : : = X 521 , X 521 ⊥ = X 6001 csc B − csc C : : = X 513 , X 513 ⊥ = X 517 cot B − cot C : : = X 523 , X 523 ⊥ = X 30 cos 2 B − cos 2 C : : = X 523 , X 523 ⊥ = X 30 sin 2 B − sin 2 C : : = X 523 , X 523 ⊥ = X 30
Example 10 continued:
tan 2 B − tan 2 C : : = X 520 , X 520 ⊥ = X 6000 sec 2 B − sec 2 C : : = X 520 , X 520 ⊥ = X 6000 csc 2 B − csc 2 C : : = X 512 , X 512 ⊥ = X 511 cot 2 B − cot 2 C : : = X 512 , X 512 ⊥ = X 511 cos 2 B − cos 2 C : : = X 523 , X 523 ⊥ = X 30 sin 2 B − sin 2 C : : = X 525 , X 525 ⊥ = X 1503 sec 2 B − sec 2 C : : = X 924 , X 924 ⊥ = X 13754 cot 2 B − cot 2 C : : = X 6368 , X 6368 ⊥ = X 18400
Example 11.
More pairs of trigonometric orthopoints.
2 cos A − cos B − cos C : : = X 527 , X 527 ⊥ = X 28292 2 sin A − sin B − sin C : : = X 519 , X 527 ⊥ = X 3667 2 tan A − tan B − tan C : : = X 30 , X 30 ⊥ = X 523 2 csc A − csc B − csc C : : = X 536 , X 536 ⊥ = X 28475 2 cot A − cot B − cot C : : = X 524 , X 524 ⊥ = X 1499 2 cos 2 A − cos 2 B − cos 2 C : : = X 524 , X 524 ⊥ = X 1499 2 sin 2 A − sin 2 B − sin 2 C : : = X 524 , X 524 ⊥ = X 1499 2 csc 2 A − csc 2 B − csc 2 C : : = X 538 , X 538 ⊥ = X 32472 2 cot 2 A − cot 2 B − cot 2 C : : = X 538 , X 538 ⊥ = X 32472 2 cos 2 A − cos 2 B − cos 2 C : : = X 524 , X 524 ⊥ = X 1499 2 sin 2 A − sin 2 B − sin 2 C : : = X 30 , X 30 ⊥ = X 523 2 tan 2 A − tan 2 B − tan 2 C : : = X 539 , X 539 ⊥ = X 20184

9. Trigonometric Infinity Bisectors

Let O denote the circumcenter, ( O ) the circumcircle, and L ∞ the line at infinity. Suppose that P = p : q : r and U = u : v : w are points on ( O ) and that P , O , U are noncollinear. Let L 1 be the tangent to ( O ) at P and L 2 the tangent to ( O ) at U. Let D = L 1 ∩ L 2 and M = O D ∩ L ∞ . As the line O M 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 ) ( q u − p v ) − ( a 2 + b 2 − c 2 ) ( r u − p w ) − 2 a 2 ( r v − q w ) : :
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 the M ( P , U ) is the orthopoint of the P U ∩ L ∞ . A few examples follow:
M ( X 74 , X 477 ) = X 30 = X 523 ⊥ M ( X 110 , X 476 ) = X 30 = X 523 ⊥ M ( X 74 , X 476 ) = X 523 = X 30 ⊥ M ( X 110 , X 477 ) = X 523 = X 30 ⊥ M ( X 74 , X 98 ) = X 542 = X 690 ⊥ M ( X 99 , X 110 ) = X 542 = X 690 ⊥ M ( X 74 , X 99 ) = X 690 = X 542 ⊥ M ( X 98 , X 110 ) = X 690 = X 542 ⊥ M ( X 74 , X 110 ) = X 526 = X 5663 ⊥ M ( X 98 , X 99 ) = X 804 = X 2782 ⊥

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:
0 = ( tan B − tan C ) x + ( tan C − tan A ) y + ( tan A − tan B ) z
= ( sin 2 B − sin 2 C ) x + ( sin 2 C − sin 2 A ) y + ( sin 2 A − sin 2 B ) z
= cos A sin ( B − C ) x + cos B sin ( C − A ) y + cos C sin ( A − B ) z = ( 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
The n-Euler line is defined by substituting n A , n B , n C 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:
n - centroid = c 0 = 1 : 1 : 1 n - circumcenter = cos n A : : n - orthocenter = tan n A : : n - nine - point center = sin n A cos ( n B − n C ) : : .
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
0 = ( sin B − sin C ) x + ( sin C − sin A ) y + ( sin A − sin B ) z . = ( b − c ) x + ( c − a ) y + ( a − b ) z ,
and the n-Nagel line by
0 = ( sin n B − sin n C ) x + ( sin n C − sin n A ) y + ( sin n A − sin n B ) z .
The Nagel line passes through the incenter, X 1 = sin A : sin B : sin C and the centroid, X 2 = 1 : 1 : 1 , so that the n-Nagel line passes through the centroid and the point sin n A : sin n B : sin n C . Thus, for every n, the n-Euler line and n-Nagel line meet in the centroid. Moreover, by () and (36), the 2 n -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:
X 2 = 1 : 1 : 1 X 54 = Kosnita point = sin A sec ( B − C ) : : = isogonal conjugate of the nine - point center , X 5 X 68 = Prasolov point = tan 2 A : : X 1147 = sin 4 A : : X 5449 = 2 nd Hatzipolakis - Moses point = sin 4 B + sin 4 C : : = midpoint of X 68 and X 1147 X 6193 = tan B + tan C − tan A : : X 16032 = csc A sec ( B − C ) : :
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
    n A + r π , n B + r π , n C + r π
    for some nonzero number r, such as the Fermat point,
    X 13 = sin A csc ( A + π / 3 ) : sin B csc ( B + π / 3 ) : sin C csc ( C + π / 3 ) = a 4 − 2 ( b 2 − c 2 ) 2 + a 2 ( b 2 + c 2 + Ψ ) : : ,
    where Ψ = 4 3 × (area of triangle A B C ), and related points X i for i = 14 , … , 18 in ETC [2].
  • Bicentric pairs [1] of points, such as the Brocard points,
    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 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 ,
    leading to Brocard n-points by substituting n A , n B , N C 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 → cos n A , b → cos n B , c → cos n C )
    transforms this “polynomial cubic” into a “trigonometric cubic”, and likewise for the substitution
    ( a → sin n A , b → sin n B , c → sin n C ) ,
    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 cos n A ,
    where A, as usual, is the angle opposite side B C in a triangle A B C having sidelengths | B C | = a , | C A | = b , | A B | = 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
    X 2 = a : b : c = sin A : sin B : sin C X 3 = a ( b 2 + c 2 − a 2 ) : : = cos A : cos B cos C : : ,
    which lead to trigonometric polycenters by substituting n A , n B , n C for A , B , C . The resulting trilinear representations are equivalent to the barycentric representations sin A sin n A : : and sin A cos n A : : , these being trigonometric polycenters not previously mentioned in this article.

References

  1. Bicentric Pairs of Points. Available online: https://faculty.evansville.edu/ck6/encyclopedia/BicentricPairs.html.
  2. Encyclopedia of Triangle Centers. Available online: https://faculty.evansville.edu/ck6/encyclopedia/etc.html.
  3. Bernard Gibert, Cubics in the Triangle Plane. Available online: http://bernard-gibert.fr/.
  4. Clark Kimberling, “Triangle Centers and Central Triangles”, jCongressus Numerantium 129 (1998) i-xxv, 1-295.
  5. Clark Kimberling, “Polynomial Points in the Plane of a Triangle”, in Polynomials—Exploring Fundamental Mathematical Expressions, Intech Open, forthcoming.
  6. Clark Kimberling, “Symbolic substitutions in the transfigured plane of a triangle,” Aequationes Mathematicae 73 (2007) 156-171.
  7. Clark Kimberling and Cyril Parry, “Products, square roots, and layers in triangle geometry”, Mathematics and Informatics Quarterly 10 (2000) 9-22.
  8. Clark Kimberling and Peter J. C. Moses, “Line conjugates in the plane of a triangle”, Aequationes Mathematicae 97 (2023) 161-184.
  9. Peter J. C. Moses, “Central Lines”. Available online: https://faculty.evansville.edu/ck6/encyclopedia/CentralLines.html.
  10. Peter J. C. Moses and Clark Kimberling, “Unary Operations on Homogeneous Coordinates in the Plane of a Triangle”, Geometry 1 (2024) 3-15. Available online: https://www.mdpi.com/3042-402X/1/1/2.
  11. Online Encyclopedia of Integer Sequences. Available online: https://oeis.org.
  12. R. O. Rayes, V. Trevisan, and P. S. Wang, “Factorization properties of Chebyshev Polynomials”, Computers & Mathematics with Applications 50 (2005) 1231-1240. Available online: https://www.sciencedirect.com/science/article/pii/S0898122105003767.
  13. Theodore J. Rivlin, The Chebyshev Polynomials, Wiley, New York, 1974.
  14. Eric W. Weisstein, “Triangle Center”, MathWorld. Available online: https://mathworld.wolfram.com/TriangleCenter.html.
  15. Eric W. Weisstein, “Conway Triangle Notation”, MathWorld. Available online: https://mathworld.wolfram.com/ConwayTriangleNotation.html.
  16. Paul Yiu, Introduction to the Geometry of the Triangle, 2001, 2013. Available online: https://web.archive.org/web/20180422091419id_/http://math.fau.edu/Yiu/YIUIntroductionToTriangleGeometry130411.pdf.
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