HYPERBOLAS, ORTHOLOGY, AND ANTIPEDAL TRIANGLES

ZVONKO ČERIN

Abstract. We obtain several characterisations of the Kiepert, Jarabek, and Feuerbach hyperbolas of a triangle ABC using the antipedal triangles of a variable point P in the plane and the notion of orthologic triangles. Our arguments are algebraic and use complex numbers.

1. Introduction

Among conics that pass through the vertices A, B, C of a scalene triangle ABC and its orthocenter H – all of them are equilateral hyperbolas – the most interesting are Feuerbach, Kiepert, and Jarabek hyperbolas. These are hyperbolas that go through the incenter I, the centroid G, and the circumcenter O, respectively. They have been extensively studied in the past. The following are some more recent papers that consider them: [1], [2], [6], [7], [5], [12], [21], [20], and [24].

In this paper we shall present new characterisations of the Kiepert, Jarabek, and Feuerbach hyperbolas associated to a triangle ABC. We shall use the same method for all three hyperbolas. Our idea is to associate to every point P its antipedal triangle P^{a}P^{b}P^{c} and to look for triangles XY Z having the property that P lies on a hyperbola if and only if the triangles P^{a}P^{b}P^{c} and XY Z are orthologic.

Recall that triangles ABC and XY Z are orthologic provided the perpendiculars from the vertices of ABC on the sides Y Z, ZX, and XY of XY Z are concurrent. The point of concurrence of these perpendiculars is denoted by [ABC, XY Z]. It is well-known (see [8] or [17]) that the relation of orthology for triangles is reflexive and symmetric. Hence, the perpendiculars from the vertices of XY Z on the sides BC, CA, and AB of ABC are concurrent at the point [XY Z, ABC].

In this definition and throughout this paper all triangles are nondegenerate, that is, their vertices are not collinear. The last assumption implies that in our approach we must exclude some points P so that ours are characterisations of three named hyperbolas without a small number of their points.

For a triangle ABC, let W (ABC) denote the complement in the plane of the union of the side lines BC, CA, AB and the circumcircle γ_{0} of ABC. Recall that for a point P in W (ABC), the antipedal triangle P^{a}P^{b}P^{c} of P has intersections of perpendiculars from A, B, C on AP, BP, CP as vertices. Let A denote the function that associates to a point P its antipedal triangle P^{a}P^{b}P^{c}.

The domain of the function A is W (ABC) because it follows immediately from Thales Theorem and incidence arguments that the points P^{a}, P^{b}, and P^{c} coincide if and only if P lies on the circumcircle γ_{0}.

Let γ be a curve in the plane. Let F be a function from a subset S of the plane that associates to each point P of S a triangle F(P). A triangle XY Z is called (F, γ)- generating provided XY Z is orthologic to F(P) if and only if a point P is in the set γ ∩ S.

Let γ_{F}, γ_{J}, and γ_{K} denote the Feuerbach, Jarabek, and Kiepert hyperbola of the triangle ABC, respectively. With the above definitions and notation we can formulate the results of this paper as contributions to the following problem.

Problem. For γ ∈ {γ_{F} , γ_{J}, γ_{K}}, find (A, γ)-generating triangles.

Observe that when we know that, for γ ∈ {γ_{F} , γ_{J}, γ_{K}}, a triangle XY Z is (A, γ)- generating, then we have the following characterisation of γ:

The hyperbola γ is the closure of all points P in W (ABC) such that the triangles P^{a}P^{b}P^{c} and XY Z are orthologic.

The vertices of the triangles XY Z which we prove in this paper to provide solutions to the above problem are all endpoints of segments of controlled length perpendicular to sides of ABC. A more formal description uses the following notation.

For a triple h = (s_{1}, s_{2}, s_{3}) of real numbers and for triangles ABC and XY Z, let [ABC, XY Z, h] denote the triangle UV W such that UX, V Y, W Z are perpendicular to BC, CA, AB and the directed distances |UX|, |V Y |, |W Z| are equal to s_{1}, s_{2}, s_{3}, respectively. When s_{1} = 0, we put U = X, and we do similar assignments when s_{2} and s_{3} are zero. For s_{1} > 0 the vector X⃗U points towards outside of ABC while for s_{1} < 0 it points towards inside.

For an expression ε in terms of side lengths a, b, and c of the triangle ABC and a real number h, let ε[h] denote the triple (h ε, h φ(ε), h ψ(ε)). More precisely, the coordinates ε[h]_{1}, ε[h]_{2}, ε[h]_{3} of ε[h] are products with h of ε, the first cyclic permutation φ(ε) of ε, and the second cyclic permutation ψ(ε) of ε, respectively. For example, a[h] = (h a, h b, h c) and if w_{a} = \frac{b+c−a}{2}, w_{b} = \frac{c+a−b}{2}, and w_{c} = \frac{a+b−c}{2}, then w_{a}[h] = (h w_{a}, h w_{b}, h w_{c}).

With this notation at hand, we can describe our task in this paper as a search for expressions ε and points X, Y, and Z in the plane of the triangle ABC such that the triangles [ABC, XY Z, ε[h]] are (A, γ)-generating for γ either γ_{F}, γ_{J}, or γ_{K}.

Recall that the triangles [ABC, XY Z, ε[h]] have already been used for characterisations of Kiepert and Feuerbach hyperbolas. Indeed, the original description of the Kiepert hyperbola is that it is the locus of centers of perspective of triangles ABC and X_{h}Y_{h}Z_{h}, where X_{h}Y_{h}Z_{h} = [ABC, A_{m}B_{m}C_{m}, a[h]] and A_{m}, B_{m}, C_{m} are midpoints of sides (see [6]).

Another application of triangles X_{h}Y_{h}Z_{h} on vertices of similar isosceles triangles build on sides of ABC is a result in [13] which shows that triangles ABC and X_{h}Y_{h}Z_{h} are orthologic and the point [ABC, X_{h}Y_{h}Z_{h}] traces again the Kiepert hyperbola as h goes through the reals.

The Feuerbach hyperbola is the locus of centers of perspective of triangles ABC and P_{h}Q_{h}R_{h}, where P_{h}Q_{h}R_{h} = [ABC, A_{p}B_{p}C_{p}, 1[h]] and A_{p}, B_{p}, C_{p} are projections of the incenter onto sides (see [11]).

Another application of triangles P_{h}Q_{h}R_{h}, whose vertices are intersections of circles concentric to the incircle with perpendiculars through incenter to sides, is a result which shows that triangles ABC and P_{h}Q_{h}R_{h} are orthologic and the point [ABC, P_{h}Q_{h}R_{h}] again traces the Feuerbach hyperbola as h goes through the reals.

2. Preliminaries on complex numbers

We shall use complex numbers because they lead to the simplest expressions. Hence, our proofs are entirely algebraic. Every book on the use of complex numbers in geometry from the references below gives excellent and adequate introductions to this technique of proof. In this section we give only the most basic notions and conventions.

A point P in the Gauss plane is represented by a complex number p. This number is called the affix of P and we write P̃ = p or P (p) to indicate this. The complex conjugate of p is denoted p̄. However, we shall be avoiding this notation by using next letter (now letter q) for the complex conjugate and sometimes write P (p, q) or P̃ = (p, q) in order to describe affix of a point and to describe its complex conjugate. In order to avoid quotients, we shall use z^{∗} for 1/z.

In the sections on the Kiepert and Jarabek hyperbolas, we follow the standard assumption that the vertexes A, B, and C of the reference triangle are represented by numbers u, v, and w on the unit circle so that the circumcentre O of ABC is the origin. Hence, the affix of O is number 0 (zero) and complex conjugates of u, v, and w are 1/u, 1/v, and 1/w (or, in our notation, u^{∗}, v^{∗}, and w^{∗}).

Most interesting points, lines, circles, curves,... associated with the triangle ABC are expressions that involve symmetric functions of u, v, and w that we denote as follows.

σ = u + v + w, τ = v w + u w + u v, μ = u v w,
σ_{a} = −u + v + w, σ_{b} = u − v + w, σ_{c} = u + v − w,
τ_{a} = −v w + w u + u v, τ_{b} = v w − w u + u v, τ_{c} = v w + w u − u v,
μ_{a} = v w, μ_{b} = w u, μ_{c} = u v, δ_{a} = v − w,
δ_{b} = w − u, δ_{c} = u − v, ζ_{a} = v + w, ζ_{b} = w + u, ζ_{c} = u + v.

For each k ≥ 2, σ_{k}, σ_{ka}, σ_{kb}, and σ_{kc} are derived from σ, σ_{a}, σ_{b}, and σ_{c} with the substitution u = u^{k}, v = v^{k}, w = w^{k}. In a similar fashion we can define analogous expressions using letters τ, μ, δ, and ζ. We shall use corresponding small Latin letters to denote analogous symmetric functions in a, b, and c (lengths of sides of ABC). For example, m = a b c, s = a + b + c, t = b c + c a + a b, z_{a} = b + c, and s_{2a} = b^{2} + c^{2} − a^{2}.

The expressions which appear in triangle geometry usually depend on sets that are of the form {a, b, c, …, x, y, z} (that is, union of triples of letters). Let φ and ψ stand for permutations | b, c, a, …, y, z, x | and | c, a, b, …, z, x, y |.

Let f = f(x, y, …) be an expression that depends on a set S = {x, y, … } of variables and let ϱ : S → S be a permutation of S. Then f^{ϱ} is a short notation for f(ϱ(x), ϱ(y), …). For permutations ϱ, …, ξ of S we shall use 𝕊_{ϱ,..., ξ} f and ℙ_{ϱ, ..., ξ} f to shorten f + f^{ϱ} + ⋯ + f^{ξ} and ff^{ϱ} … f^{ξ}. Finally, 𝕊f and ℙf replace 𝕊_{φ, ψ} f and ℙ_{φ, ψ} f.

For real numbers f, g, and h, let ⟨f, g, h⟩ be a notation for −f μ + 𝕊 u^{2} (g v + h w). For example, ⟨6, 1, 1⟩ = u^{2} v + u^{2} w + v^{2} u + v^{2} w + w^{2} u + w^{2} v − 6 u v w. Let S be the area of ABC.

Since points, lines, conics, ... associated to a triangle often appear in triples in which two members are build from a third by appropriate permutation, we shall often give only one of them while the other two (relatives) are obtained from it by cyclic permutations.

Let us close these preliminaries with few words on analytic geometry that we shall use.

In triangle geometry lines play an important role so that we have special notation [f, g, h] for the set of all points P (p, q) that satisfy the equation f p + g q + h = 0. This set is a line iff there is a complex number z ≠ 0 such that z g is the complex conjugate of z f and z h is a real number.

Let X(x, a), Y (y, b), and Z(z, c) be three points and let ℓ be a line [f, g, h] in the plane. Then the line XY is [a − b, y − x, b x − a y], the parallel to ℓ through X is [f, g, −g a − f x] and the perpendicular to ℓ through X is [f, −g, g a − f x]. The conditions for points X, Y, and Z to be collinear and for lines ℓ_{1} = [f, g, h], ℓ_{2} = [k, m, n], and ℓ_{3} = [r, s, t] to be concurrent are Δ = 0 and Γ = 0, where

Δ = Δ(X, Y, Z) = | \begin{matrix}x & a & 1 \\ y & b & 1 \\ z & c & 1 \end{matrix} | , and Γ = Γ(ℓ_{1}, ℓ_{2}, ℓ_{3}) = | \begin{matrix}f & g & h \\ k & m & n \\ r & s & t \end{matrix} | .

Moreover, the (oriented) area of the triangle XY Z is \frac{I Δ}{4}, where I = \sqrt{−1} is the imaginary unit.

There are some notable exceptions to the rule that complex numbers give simpler expressions than trilinear and barycentric coordinates. As a convenience to readers that are more familiar with these coordinates, we shall describe transformation formulas between all three systems and present these shorter forms in relevant cases.

The above formula for the area of a triangle implies that a point P (p, q) has with respect to the base triangle ABC the absolute barycentric coordinates (α, β, γ), where α = \frac{−u n_{a}}{δ_{b} δ_{c}} and β and γ are relatives of α and n_{a} = p + μ_{a} q − ζ_{a}. It follows that a point P with barycentric coordinates (α, β, γ) has affix

p = \frac{u α + v β + w γ}{α + β + γ} .

In order to have connection between complex and trilinear coordinates it suffices to recall that absolute trilinear coordinates (x, y, z) and absolute barycentric coordinates (α, β, γ) of the same point are related by formulas

α = \frac{a x}{2 S} ,  β = \frac{b y}{2 S} ,  γ = \frac{c z}{2 S} ,

where S is the area of ABC and the lengths of the sides are

a = |δ_{a}| = \sqrt{2 − ζ_{2a} μ_{a}^{∗}},  b = |δ_{b}| = \sqrt{2 − ζ_{2b} μ_{b}^{∗}},  c = |δ_{c}| = \sqrt{2 − ζ_{2c} μ_{c}^{∗}}.

3. Statements of results

Let n_{F} = 6 and n_{J} = n_{K} = 5. For X = K, F, J, let X_{i}, for i = 1, …, n_{X}, denote the following expressions.

K_{1} = a, K_{2} = a z_{a}^{∗}, K_{3} = a^{2} z_{a}^{∗}, K_{4} = a s_{a} z_{a}^{∗}, K_{5} = a^{3}.
F_{1} = 1, F_{2} = s_{a}^{∗}, F_{3} = a s_{a}, F_{4} = a s_{a}^{∗}, F_{5} = s_{2a} z_{a} s_{a}^{∗}, F_{6} = z_{a} s_{a}^{∗}.
J_{1} = a^{∗}, J_{2} = s_{a} z_{a}^{∗}, J_{3} = a s_{2a}^{∗}, J_{4} = z_{a}^{∗} s_{2a}^{∗}, J_{5} = a^{∗} s_{2a}^{∗}.

For i = J_{3}, J_{4}, J_{5} we must assume in addition that ABC has no right angle.

Theorem 1. Let h ≠ 0 be a real number. For any triangle P QR homothetic to the triangle ABC, for X = K, J, F, and for i = 1, …, n_{X}, the triangle [ABC, P QR, X_{i}[h]] is (A, γ_{X})-generating.

Remark. Since there can be at most two values of the parameter h for which the vertices of the ”triangle” [ABC, P QR, X_{i}[h]] are collinear, we must exclude these values in addition to the value h = 0. In the above statement this is implicit in the assumption that we consider only nondegenerate triangles.

In the above theorem the triangle P QR can be, for example, the triangle ABC, the complementary triangle A_{m}B_{m}C_{m}, the anticomplementary triangle A_{a}B_{a}C_{a}, the Euler triangle A_{f}B_{f}C_{f}, and the opposite triangle A_{s}B_{s}C_{s}, where A_{m}, B_{m}, C_{m} denote midpoints of sides of the triangle ABC, A_{a}, B_{a}, C_{a} intersections of parallels through vertices to sides, A_{f}, B_{f}, C_{f} midpoints of segments joining vertices with the orthocenter H, and A_{s}, B_{s}, C_{s} reflections of vertices at the circumcenter O.

Theorem 2. For any triangle P QR homothetic to the triangle ABC, for X = K, F, J, and for i, j = 1, …, n_{X}, the triangle

[ABC, [ABC, P QR, X_{i}[h]], X_{j}[k]]

is (A, γ_{X})-generating for all real numbers h and all real numbers k except the value −h a_{ij}, where a_{ij} is the (i, j) entry of the matrix M_{X} with

M_{K} = [ \begin{matrix}1 & \frac{2 z}{y} & \frac{2 z}{v} & \frac{2 z}{u} & 2 x \\ \frac{y}{2 z} & 1 & \frac{y}{v} & \frac{y}{u} & \frac{x y}{z} \\ \frac{v}{2 z} & \frac{v}{y} & 1 & \frac{v}{u} & \frac{v x}{z} \\ \frac{u}{2 z} & \frac{u}{y} & \frac{u}{v} & 1 & \frac{u x}{z} \\ \frac{1}{2 x} & \frac{z}{x y} & \frac{z}{v x} & \frac{z}{u x} & 1 \end{matrix} ] ,

x = s_{2}^{∗}, y = s^{2}, z = z_{a} z_{b} z_{c}, u = 4 s t − 4 m − s^{3}, v = s^{3} − 2 s t + 2 m,

M_{F} = [ \begin{matrix}1 & \frac{w}{v} & \frac{s}{x} & \frac{y}{x} & \frac{w}{z} & \frac{y}{u} \\ \frac{v}{w} & 1 & \frac{v}{x y} & \frac{v}{s x} & \frac{v}{z} & \frac{v}{s u} \\ \frac{x}{s} & \frac{x y}{v} & 1 & \frac{y}{s} & \frac{x y}{z} & \frac{x y}{s u} \\ \frac{x}{y} & \frac{s x}{v} & \frac{s}{y} & 1 & \frac{s x}{z} & \frac{x}{u} \\ \frac{z}{w} & \frac{z}{v} & \frac{z}{x y} & \frac{z}{s x} & 1 & \frac{z}{s u} \\ \frac{u}{y} & \frac{s u}{v} & \frac{s u}{x y} & \frac{u}{x} & \frac{s u}{z} & 1 \end{matrix} ] ,

x = 2 m, y = s_{a} s_{b} s_{c}, z = s_{2a} s_{2b} s_{2c}, u = 4 s t−6 m−s^{3}, v = u+2 m, w = 16 S^{2},

M_{J} = [ \begin{matrix}1 & \frac{z}{m s} & \frac{y}{4 m^{2}} & \frac{y z}{m s u} & \frac{y}{s x} \\ \frac{m s}{z} & 1 & \frac{s y}{4 m z} & \frac{y}{u} & \frac{m y}{x z} \\ \frac{4 m^{2}}{y} & \frac{4 m z}{s y} & 1 & \frac{4 m z}{s u} & \frac{4 m^{2}}{s x} \\ \frac{m s u}{y z} & \frac{u}{y} & \frac{s u}{4 m z} & 1 & \frac{m u}{x z} \\ \frac{s x}{y} & \frac{x z}{m y} & \frac{s x}{4 m^{2}} & \frac{x z}{m u} & 1 \end{matrix} ] ,

x = s_{a} s_{b} s_{c}, y = s_{2a} s_{2b} s_{2c}, z = z_{a} z_{b} z_{c}, u = 4 s t − 4 m − s^{3}.

Remark. Observe that some important triangles related to the triangle ABC are of the form [ABC, A_{m}B_{m}C_{m}, K_{1}[h]] for a suitable constant h. For example, the first Brocard triangle A_{b}B_{b}C_{b} (for h = 2 S/s_{2}), the Torricelli triangles A_{v}B_{v}C_{v} and A_{u}B_{u}C_{u} on vertices of equilateral triangles build on sides either towards outside or towards inside (for h = ±\sqrt{3}/2), and Napoleon triangles A_{vn}B_{vn}C_{vn} and A_{un}B_{un}C_{un} on centers of these equilateral triangles (for h = ±\sqrt{3}/6).

The orthic triangle A_{o}B_{o}C_{o} and the three images triangle A_{r}B_{r}C_{r} whose vertices are reflections of A, B, and C at opposite sides of ABC are of the form [ABC, ABC, J_{1}[h]]. Also, the tangential triangle A_{t}B_{t}C_{t} (formed by tangents to the circumcircle at vertices of ABC) has the form [ABC, A_{m}B_{m}C_{m}, J_{3}[h]].

Let ν denote the expression a (b^{2} + c^{2} − a^{2}).

Theorem 3. Let k ≠ 0 and h be real numbers. For any triangle P QR homothetic to the triangle ABC, for X = K, F, J, and for j = 1, …, n_{X}, the triangle

[ABC, [ABC, P QR, ν[h]], X_{j}[k]]

is (A, γ_{X})-generating.

Theorem 4. Let k and h ≠ 0 be real numbers. For any triangle P QR homothetic to the triangle ABC, for X = K, F, J, and for j = 1, …, n_{X}, the triangle

[ABC, [ABC, P QR, X_{j}[h]], ν[k]]

is (A, γ_{X})-generating.

Let I_{K} = {b, K, Kb, Km, u, ub, v, vb, un, vn}, I_{J} = {h, o, r, t, tr, w, Hh, Ho, O, Ot}, and I_{F} = {e, ep, er, k, kr, p, pp, Ie, Ik, Ip, Oi}. For each element i of these three sets we define a triangle A_{i}B_{i}C_{i} by describing the vertex A_{i}. The vertices B_{i} and C_{i} have analogous descriptions. Let A_{e} be the centre of the A-excircle, A_{ep} the projection of A_{e} onto BC, the point A_{er} is the reflection of A_{e} at BC, the vertex A_{k} is the second intersection of the bisector of the angle A with the circumcircle, A_{kr} is the reflection of A_{k} at BC, the point A_{p} is the projection of the incentre I onto BC, the vertex A_{pp} is the projection of A_{p} onto AI, A_{Oi} is a projection onto BC of any point different from O on line IO joining the incentre with the circumcentre, A_{Ip} is a projection onto B_{p}C_{p} of any point different from central point X_{65} [10] on line IO, A_{Ie} and A_{Ik} are projections onto B_{e}C_{e} and B_{k}C_{k} of any point on IO different from the incentre I, the point A_{b} is the projection of the Grebe-Lemoine point K onto the perpendicular bisector of BC, the vertices A_{K}, A_{Kb}, and A_{Km} are the projection of any point different from the circumcentre O on the line KO onto BC, B_{b}C_{b}, and OA_{m}, vertices A_{u} and A_{un} are the vertex and the centre of the equilateral triangle build on BC towards inside, A_{v} and A_{vn} are the vertex and the centre of the equilateral triangle build on BC towards outside, A_{ub} and A_{vb} are projections of the Grebe-Lemoine points of A_{u}B_{u}C_{u} and A_{v}B_{v}C_{v} onto perpendicular bisectors of B_{u}C_{u} and B_{v}C_{v}, A_{h} is the second intersection of altitude line AH with the circumcircle, A_{o} is the projection of A onto BC, the point A_{r} is the reflection of A at BC, the intersection of tangents to the circumcircle at B and C is A_{t}, the reflection of A_{t} at BC is A_{tr}, A_{w} is the intersection of common tangents of the A-excircle with B-excircle and C-excircle, A_{Hh} and A_{Ho} are the projections onto B_{h}C_{h} and B_{o}C_{o} of any point X on the Euler line of ABC different from the orthocentre H, and A_{O} and A_{Ot} are the projections onto BC and B_{t}C_{t} of any point X on the Euler line of ABC different from the circumcentre O.

Some of the cases in the following theorem are clearly consequences of the previous theorem (for example, the first Brocard triangle A_{b}B_{b}C_{b} has the form [ABC, A_{m}B_{m}C_{m}, K_{1}[k]], for a suitable k ≠ 0). Moreover, in some cases we must make additional assumptions about the triangle ABC. For example, for i = b, the triangle ABC can not be equilateral and for i = t and i = w it can not have right angle.

Theorem 5. For X = K, J, F, for i ∈ I_{X}, and for all real numbers h, the triangle [ABC, A_{i}B_{i}C_{i}, ν[h]] is (A, γ_{X})-generating.

Theorem 6. For X = K, J, F, for any i ∈ I_{X}, and for all j, j′ = 1, …, n_{X} the triangles

[ABC, [ABC, A_{i}B_{i}C_{i}, ν[h]], X_{j}[k]],
[ABC, [ABC, A_{i}B_{i}C_{i}, X_{j}[h]], ν[k]],
[ABC, [ABC, A_{i}B_{i}C_{i}, X_{j′}[h]], X_{j}[k]],

are (A, γ_{X})-generating for all real values of constants h and k except exactly one value of either h or k. The matrices of exceptions are similar to the matrices M_{K}, M_{J}, and M_{F} from the Theorem 2.

An important source of (A, γ_{X})-generating triangles is the following general result.

Theorem 7. Let Q ∈ W (ABC) be a point different from the orthocentre H. The antipedal triangle Q^{a}Q^{b}Q^{c} of Q with respect to ABC is orthologic with the triangle P^{a}P^{b}P^{c} if and only if P lies on a conic through the points A, B, C, H, and Q.

Corollary . For X = K, J, F, the antipedal triangle Q^{a}Q^{b}Q^{c} with respect to ABC of any point Q on the hyperbola γ_{X} outside the circumcircle γ_{0} and different from the orthocentre H is (A, γ_{X})-generating.

The next result can also be useful in search for (A, γ_{X})-generating triangles.

Theorem 8. For X = F, J, K, if DEF is a (A, γ_{X})-generating in W (ABC), then triangles

[ABC, DEF, ν[h]] and [ABC, DEF, X_{i}[h]]

(i = 1, …, n_{X}) are also (A, γ_{X})-generating for all real numbers h with at most one exception.

4. Preliminaries for proofs

Let us first determine the affixes of points P^{a}, P^{b}, and P^{c}. Since the affix of A is u and the affix of P is p, the line AP is [1 − u q, u(p − u), u^{2} q − p], where q is a complex conjugate of p. The lines BP and CP are relatives of AP. It follows that the perpendicular per_{A} at A to AP is [1 − u q, u(u − p), u^{2} q + p − 2 u], while the perpendiculars per_{B} and per_{C} at B and C to BP and CP are its relatives. The intersection of per_{B} and per_{C} is a point P^{a} with affix n_{a}^{∗}[μ_{a} (p q − 2) + p (ζ_{a} − p)], where n_{a} = p + μ_{a} q − ζ_{a}. The affixes of P^{b} and P^{c} are relatives of the affix of P^{a}.

Recall that points P^{a}, P^{b}, P^{c} coincide if and only if the point P lies on the circumcircle γ_{0} with the equation p q − 1 = 0.

Theorem 9. Triangles XY Z and P QR with affixes of vertices x, y, z, p, q, and r are orthologic if and only if (XY Z, P QR) = 0, where

(XY Z, P QR) = 𝕊 [x (q̄ − r̄) + x̄ (q − r)].

Proof. The line QR is [q̄ − r̄, r − q, q r̄ − q̄r] so that the perpendicular per_{QR}^{X} through X onto QR is the line [q̄ − r̄, q − r, x (r̄ − q̄) + x̄ (r − q)]. The perpendiculars per_{RP}^{Y} and per_{P Q}^{Z} through Y and Z onto RP and P Q are relatives of per_{QR}^{X}. These three perpendiculars are concurrent if and only if Θ = 0, where Θ denotes the determinant

| \begin{matrix}q̄ − r̄ & q − r & x (r̄ − q̄) + x̄ (r − q) \\ r̄ − p̄ & r − p & y (p̄ − r̄) + ȳ(p − r) \\ p̄ − q̄ & p − q & z (q̄ − p̄) + z̄(q − p) \end{matrix} | .

But, Θ = (XY Z, P QR) m, where m = p (q̄ − r̄) + q (r̄ − p̄) + r (p̄ − q̄). Since m = 0 if and only if points P, Q, and R are collinear (and our assumptions exclude this possibility), we conclude that the triangles XY Z and P QR are orthologic if and only if (XY Z, P QR) = 0.

Let us observe that the condition for orthology of two triangles in both barycentric and trilinear coordinates is very complicated because it involves eighteen coordinates of vertices of these triangles. This is the main reason why we are making all calculations with complex numbers.

5. Proof of Theorem 1 for X = K and i = 1

Since triangles ABC and P QR are homothetic, there is a point T (x, y) and a real number ξ ≠ −1 such that P̃ = η (u + ξ x), Q̃ = φ(P̃), and R̃ = ψ(P̃), where η = (ξ + 1)^{∗}.

Let h be a real number. Let U, V, and W be vertices of the triangle [ABC, P QR, K_{1}[h]]. Then Ũ = P̃ + I h (v − w), where I = \sqrt{−1}. Also, Ṽ = φ(Ũ) and W̃ = ψ(Ũ).

Since

Δ(U, V, W) = | \begin{matrix}η (u + ξ x) + I h δ_{a} & u^{∗} η (1 + ξ u y) + I h δ_{a} μ_{a}^{∗} & 1 \\ η (v + ξ x) + I h δ_{b} & v^{∗} η (1 + ξ v y) + I h δ_{b} μ_{b}^{∗} & 1 \\ η (w + ξ x) + I h δ_{c} & w^{∗} η (1 + ξ w y) + I h δ_{c} μ_{c}^{∗} & 1 \end{matrix} | ,

has up to a constant factor the form

(h − \frac{I η (σ τ − 9 μ)}{3 ℙ δ_{a}})^{2} + \frac{4 η^{2} (τ^{2} − 3 μ σ) (σ_{2} − τ)}{9 ℙ δ_{a}^{2}},

it follows that points U, V, and W will not be collinear if and only if h is different from

\frac{s_{2} ± 2 \sqrt{s_{4} − t_{2}}}{12 S (ξ + 1)} .

Then (P^{a}P^{b}P^{c}, UV W) = 4 I h j_{K} j_{0} ℙ u^{∗} n_{a}^{∗}, where j_{0} = 1 − p q and

j_{K} = (τ^{2}−3μσ)p^{2}−μ^{2}(σ^{2}−3τ)q^{2}+(4μσ^{2}−σ τ^{2}−3μ τ)p−μ(4τ^{2}−σ^{2}τ −3μ σ)q+τ^{3}−μ σ^{3}.

Notice that j_{0} = 0 is the equation of the circumcircle of ABC while j_{K} = 0 is the equation of the Kiepert hyperbola of ABC since the vertices A(u, u^{∗}), B(v, v^{∗}), and C(w, w^{∗}), the orthocenter H(σ, τ μ^{∗}), and the centroid G(3^{∗} σ, 3^{∗} τ μ^{∗}) satisfy it. This shows that UV W is (A, γ_{K})-generating for all h ≠ 0 except for at most two additional values of h found above when points U, V, and W are collinear.

The polynomial j_{K} is rather complicated. When we transfer it to barycentric coordinates it becomes significantly shorter so that the equation of the Kiepert hyperbola is m_{K} = 0, where m_{K} = 𝕊 δ_{a} (u^{2} − μ_{a}) y z or m_{K} = 𝕊 d_{2a} y z. The last form is simpler than the equation (5) in the excellent recent review article [6] on the Kiepert conics.

6. Proof of Theorem 1 for X = J and i = 1

We first determine P̃, Q̃, and R̃ as above. Let h be a real number. Let U, V, and W be vertices of the triangle [ABC, P QR, J_{1}[h]]. Then Ũ = P̃ + I h v w (v − w)^{∗}, Ṽ = φ(Ũ), and W̃ = ψ(Ũ).

Notice that points U, V, and W will not be collinear if and only if h is different from

\frac{m (3 m ± \sqrt{3 m_{2} − 𝕊 a^{4} s_{2a}})}{4 S (ξ + 1) s_{2}}.

This follows from the fact that

Δ(U, V, W) = | \begin{matrix}η (u + ξ x) + I h δ_{a}^{∗} μ_{a} & u^{∗} η (1 + ξ u y) + I h δ_{a}^{∗} & 1 \\ η (v + ξ x) + I h δ_{b}^{∗} μ_{b} & v^{∗} η (1 + ξ v y) + I h δ_{b}^{∗} & 1 \\ η (w + ξ x) + I h δ_{c}^{∗} μ_{c} & w^{∗} η (1 + ξ w y) + I h δ_{c}^{∗} & 1 \end{matrix} | ,

has up to a constant factor the form

(h + \frac{3 I η ⟨0, 1, −1⟩}{⟨6, 1, 1⟩})^{2} + η^{2} σ τ ℙ δ_{a}^{2}.

Then (P^{a}P^{b}P^{c}, UV W) = 2 I h j_{J} j_{0} ℙ n_{a}^{∗}, where

j_{J} = σ p^{2} − μ τ q^{2} + (τ − σ^{2}) p + (τ^{2} − μ σ)q.

The equation of the Jarabek hyperbola of ABC is j_{J} = 0 since the vertices A, B, and C, the orthocenter H, and the circumcenter O(0, 0) satisfy it.

The polynomial j_{J} which represents the Jarabek hyperbola is quite simple so that its forms in other coordinate systems do not lead to significant simplifications. For completeness, let us observe that in barycentric coordinates the equation of the Jarabek hyperbola is m_{J} = 0, where m_{J} = 𝕊 u δ_{a}^{2} ζ_{a} (u^{2} − μ_{a}) y z or m_{J} = 𝕊 a^{2} d_{2a} s_{2a} y z.

7. Proof of Theorem 1 for X = F and i = 1

In contrast with the previous two sections, in order to avoid square roots, here we shall assume that the vertices A, B, and C of the base triangle have affixes u^{2}, v^{2}, and w^{2}, with the same assumption about u, v, and w. Let ϱ denote a transformation which replaces variables u, v, and w with u^{2}, v^{2}, and w^{2}.

This time P̃ = (ξ + 1)^{∗} (u^{2} + ξ x), Q̃ = φ(P̃), and R̃ = ψ(P̃). Let h be a real number. Let U, V, and W be vertices of the triangle [ABC, P QR, F_{1}[h]]. Then Ũ = P̃ + h v w, Ṽ = φ(Ũ), and W̃ = ψ(Ũ).

Since

Δ(U, V, W) = | \begin{matrix}η (u^{2} + ξ x) + h μ_{a} & \frac{η (1+ξ u^{2} y)}{u^{2}} + h μ_{a}^{∗} & 1 \\ η (v^{2} + ξ x) + h μ_{b} & \frac{η (1+ξ v^{2} y)}{v^{2}} + h μ_{b}^{∗} & 1 \\ η (w^{2} + ξ x) + h μ_{c} & \frac{η (1+ξ w^{2} y)}{w^{2}} + h μ_{c}^{∗} & 1 \end{matrix} | ,

has up to a constant factor the form (h + η)^{2} − η^{2} μ^{∗} σ τ, it follows that points U, V, and W will not be collinear if and only if h is different from

\frac{m ± \sqrt{m (s^{3} − 4 m t + 9 m)}}{4 S (ξ + 1)}.

Then (P^{a}P^{b}P^{c}, UV W) = 2 h j_{F} j_{0} ℙ δ_{a} u^{∗} ϱ(n_{a})^{∗}, where

j_{F} = τ p^{2} − μ_{3} σ q^{2} + (μ σ + 2 μ σ^{2} − σ^{2} τ) p + μ (σ τ^{2} − 2 σ^{2} μ − μ τ) q + τ^{3} − σ^{3} μ.

Observe that j_{F} = 0 is the equation of the Feuerbach hyperbola of ABC since the vertices A(u^{2}, 1/u^{2}), B(v^{2}, 1/v^{2}), and C(w^{2}, 1/w^{2}), the orthocenter H(ϱ(σ, τ μ^{∗})), and the incenter I(−τ, −σ μ^{∗}) satisfy it.

Just as with the Kiepert hyperbola, the polynomial j_{F} which represents the Feuerbach hyperbola is rather complicated. In barycentric and trilinear coordinates its equations are far simpler. More precisely, in barycentric coordinates the Feuerbach hyperbola has the equation m_{F} = 0, where m_{F} = 𝕊 u δ_{a}^{2} ζ_{a} (u^{2} − μ_{a}) y z or m_{F} = 𝕊 a d_{a} s_{a} y z.

8. Proof of Theorem 2 for X = K, i = 1, and j = 5

Let UV W = [ABC, P QR, K_{1}[h]] and LMN = [ABC, UV W, K_{5}[k]]. We know Ũ, Ṽ, and W̃ from the proof of Theorem 1, so that it is not difficult to see that L̃ = Ũ + I k δ_{a}^{3} μ_{a}^{∗}, M̃ = φ(L̃), and Ñ = ψ(L̃).

Let us note that there exist at most two values of k when points L, M, and N are collinear. These values have rather complicated form.

The orthology condition for triangles P^{a}P^{b}P^{c} and LMN is

(P^{a}P^{b}P^{c}, LMN) = 2 I (⟨6, 1, 1⟩k + 2 μ h) j_{K} j_{0} μ_{2}^{∗} ℙ n_{a}^{∗},

This shows that LMN is (A, γ_{K})-generating for all k except the value −2 h s_{2}^{∗} and at most two more values for which points L, M, and N are collinear.

9. Proof of Theorem 3 for X = K and i = 1

Let UV W = [ABC, P QR, ν[h]] and LMN = [ABC, UV W, K_{1}[k]]. It is easy to check that Ũ = P̃ + I h ζ_{a} μ^{∗} ℙ δ_{a}, Ṽ = φ(Ũ), and W̃ = ψ(Ũ). It follows that L̃ = Ũ + I k δ_{a}, M̃ = φ(L̃), and Ñ = ψ(L̃). Once again there exist at most two values of k when points L, M, and N are collinear. These values have complicated expressions in terms of side lengths. Finally, (P^{a}P^{b}P^{c}, LMN) = 4 I k j_{K} j_{0} μ^{∗} ℙ n_{a}^{∗}.

10. Proof of Theorem 5 for X = F and i = e

Assume Ã = u^{2}, B̃ = v^{2}, and C̃ = w^{2}. Recall [14] that Ã_{e} = τ_{a}, B̃_{e} = τ_{b}, and C̃_{e} = τ_{c}. Let U, V, W denote vertices of the triangle [ABC, A_{e}B_{e}C_{e}, ν[h]]. It is easy to check that Ũ = τ_{a} + I h ζ_{2a} μ_{2}^{∗} ℙ δ_{2a}, Ṽ = φ(Ũ), and W̃ = ψ(Ũ).

Once again there exist at most two values of h when points U, V, and W are collinear. These are 2^{∗} S^{∗} (−m ± \sqrt{m (s^{3} − 4 s t + 9 m)}) ℙ s_{a}^{∗}. Finally, (P^{a}P^{b}P^{c}, UV W) is equal to 4 j_{F} j_{0} μ^{∗} ℙ δ_{a} ϱ(n_{a})^{∗}.

11. Proof of Theorem 7

Let Q̃ = (x, y). Then Q̃^{a} = (μ_{a} x y − x^{2} + ζ_{a} x − 2 μ_{a})(x + μ_{a} y − ζ_{a})^{∗}, Q̃^{b} = φ(Q̃^{a}), and Q̃^{c} = ψ(Q̃^{a}). It follows that (P^{a}P^{b}P^{c}, Q^{a}Q^{b}Q^{c}) = 4 (x y − 1) j_{Q} j_{0} ℙ δ_{a} n_{a}^{∗} (x + μ_{a} y − ζ_{a})^{∗}, where j_{Q} = a p^{2} + b q^{2} + c p + d q + e,

a = μ y^{2} − x − τ y + σ, b = μ (σ x − x^{2} + μ y − τ), c = x^{2} − μ σ y^{2} + ζ_{a} ζ_{b} ζ_{c} y − σ_{2} − τ,

d = τ x^{2} − ζ_{a} ζ_{b} ζ_{c} x − μ_{2} y^{2} + τ_{2} + μ σ, e = (σ_{2} + τ) x − σ x^{2} + μ τ y^{2} − (τ_{2} + μ σ) y.

It is obvious that j_{Q} = 0 is an equation of a conic. One can easily check that it goes through A, B, C, H, and Q.

12. Outline of proof of Theorem 8

Let D̃ = (x, a), Ẽ = (y, b), and F̃ = (z, c). The expression (P^{a}P^{b}P^{c}, DEF) has two major factors. The first is j_{0} (the equation of the circumcircle). The second is the polynomial of degree two in p and q. We assume that its coefficients are equal to the product of a constant with corresponding coefficients of the polynomial representing γ_{X}. This gives six equations for variables x, a, y, b, z, c. Let the solutions have index 0 and let D̃_{0} = x_{0}, …. Finally, we check that the triangles

[ABC, D_{0}E_{0}F_{0}, ν[h]] and [ABC, D_{0}E_{0}F_{0}, X_{i}[h]]

(i = 1, …, n_{X}) are (A, γ_{X})-generating for all real numbers h except at most one value.

13. Concluding remarks and an introduction to the appendix

A careful reader should have noticed that we gave proofs of a very few of our results and that no case of both Theorems 4 and 6 was proved. One reason for this is that even with complex numbers we quickly get rather complicated expressions with lots of absolute values that are difficult to handle and hard to write down. In order to overcome these difficulties we must position a base triangle ABC so that lengths of sides are rational functions of three parameters. Only in this way we can avoid square-root and absolute value functions that are creating problems.

In an appendix to this paper we shall describe how one can (with the help from a computer) do all this calculations using only elementary analytic geometry in the plane. We shall limit ourselves to proofs of a few cases of Theorems 4 and 6 hoping that the reader will be able to infer proofs of all our claims following the same technique.

14. Appendix – Preliminaries

For an expression f, let [f] denote a triple (f, φ(f), ψ(f)), where φ(f) and ψ(f) are cyclic permutations of f. For example, if f = sin A and g = b + c, then

[f] = (sin A, sin B, sin C) and [g] = (b + c, c + a, a + b).

Let T denote a function that maps each triple [a] of real numbers to a number

T ([a]) = (a + b + c)(b + c − a)(a − b + c)(a + b − c).

We shall position the triangle ABC in the following fashion with respect to the rectangular coordinate system in order to simplify our calculations. The vertex A is the origin with coordinates (0, 0), the vertex B is on the x-axis and has coordinates (r e, 0), and the vertex C has coordinates (g q r/w, 2 f g r/w), where e = f + g, w = f g − 1, p = f^{2} + 1, q = f^{2} − 1, s = g^{2} + 1, t = g^{2} − 1, u = f^{4} + 1, and v = g^{4} + 1. The three parameters r, f, and g are the inradius and the cotangents of half of angles at vertices A and B. Without loss of generality, we can assume that both f and g are larger than 1 (i. e., that angles A and B are acute).

Nice features of this placement are that all central points from Table 1 in [10] have rational functions in f, g, and r as coordinates and that we can easily switch from f, g, and r to side lengths a, b, and c and back with substitutions

a = \frac{r f s}{w},  b = \frac{r g p}{w},  c = r e,

f = \frac{(b + c)^{2} − a^{2}}{\sqrt{T ([a])}} , g = \frac{(a + c)^{2} − b^{2}}{\sqrt{T ([a])}} , r = \frac{\sqrt{T ([a])}}{2 (a + b + c)}.

Moreover, since we use the Cartesian coordinate system, computation of distances of points and all other formulas and techniques of analytic geometry are available and wellknown to widest audience. A price to pay for these conveniences is that symmetry has been lost.

The third advantage of the above position of the base triangle is that we can easily find coordinates of a point with given trilinears. More precisely, if a point P with coordinates x and y has projections P_{a}, P_{b}, and P_{c} onto the side lines BC, CA, and AB and λ = P P_{a}/P P_{b} and μ = P P_{b}/P P_{c}, then

x = \frac{e g (p μ + q) r}{f s λ μ + g p μ + e w}, y = \frac{2 e f g r}{f s λ μ + g p μ + e w}.

This formulas will greatly simplify our exposition because there will be no need to give explicitly coordinates of points but only its first trilinear coordinate. For example, we write X_{6}[a] to indicate that the symmedian point X_{6} has trilinears equal to a : b : c. Then we use the above formulas with λ = a/b and μ = b/c to get the coordinates

(\frac{(f q t + 2 g u) e g r}{2 (f^{2} v + f g q s + g^{2} u)}, \frac{f g e^{2} w r}{f^{2} v + f g q s + g^{2} u}).

of X_{6} in our coordinate system.

Let UV W denote a function which to a triple (P, Q, R) of points P (x, x′), Q(y, y′), and R(z, z′) and a triple (h_{a}, h_{b}, h_{c}) of real numbers associates a triple (U, V, W) of points, where

U (x + \frac{2 g h_{a}}{s} , x′ + \frac{t h_{a}}{s} ), V (y − \frac{2 f h_{b}}{p} , y′ + \frac{q h_{b}}{p} ), W (z, z′ − h_{c}),

On the other hand, for triangles P QR and XY Z, let OR(P QR, XY Z) be the sum

𝕊 [p (y − z) + p′ (z′ − y′)],

where P (p, p′), Q(q, q′), R(r, r′), X(x, x′), Y (y, y′), and Z(z, z′). Observe that triangles P QR and XY Z are orthologic if and only if OR(P QR, XY Z) = 0.

Let P (x, y) be a point. Then the vertices of the antipedal triangle P^{a}P^{b}P^{c} are

P^{a} (\frac{2 g w y^{2} − w t x y + 2 r g e w x + r (2 w + s)(f t − 2 g) y − 2 r^{2} g e^{2} w}{w (2 g x + t y − 2 r g e)} ,

\frac{(r e − x)(2 g w y − t w x + r g (w^{2} − e^{2}))}{w (2 g x + t y − 2 r g e)}) ,

P^{b} (\frac{(w q x + 2 f w y − r g p^{2}) y}{w (q y − 2 f x)} , \frac{(w q x + 2 f w y − r g p^{2}) x}{w (q y − 2 f x)}) , P^{c} (r e − x, \frac{x^{2} − r e x}{y}) .

15. Appendix – Proof of Theorem 4 for X = K and j = 1

Since triangles ABC and P QR are homothetic, there is a point T (x̄, ȳ) and a real number ξ different from −1 such that with η = (ξ + 1)^{∗},

P (η ξ x̄, η ξ ȳ), Q(η (ξ x̄ + e r), η ξ ȳ), R(η (ξ x̄ + g w^{∗} q r), η (ξ ȳ + 2 f g w^{∗} r)).

Let h be a real number. Let (U, V, W) = UV W (P, Q, R, K_{1}[h]). Then U, V, and W are vertices of the triangle [ABC, P QR, K_{1}[h]]. Since K_{1}[h] = (f h r s w^{∗}, g h p r w^{∗}, e h r), it follows that U(P_{x} + 2 f g h r w^{∗}, P_{y} + f h r t w^{∗}), W (R_{x}, R_{y} − e h r), and V (Q_{x}− 2 f g h r w^{∗}, Q_{y} + g h q r w^{∗}), where P_{x} and P_{y} denote the first and the second coordinate of P.

Let k be a real number. Let (L, M, N) = UV W (U, V, W, ν[k]). Then L, M, and N are vertices of the triangle [ABC, UV W, ν[k]]. Since ν[k] = 2 e f g k r^{3} w^{∗2} (q s, p t, e^{2} − w^{2}), it follows that

L(U_{x} + 4 e f g^{2} k q r^{3} w^{∗2}, U_{y} + 2 e f g k q r^{3} t w^{∗2}),
M(V_{x} − 4 e f^{2} g k r^{3} t w^{∗2}, V_{y} + 2 e f g k q r^{3} t w^{∗2}),
N(W_{x}, W_{y} + 2 e f g k r^{3} (w^{2} − e^{2}) w^{∗2}).

The orthology condition is OR(P^{a}P^{b}P^{c}, LMN) = h m_{0} m_{K} q_{a}^{∗} q_{b}^{∗} y^{∗} w^{∗2}, where

q_{a} = 2 g (e r − x) − t y, q_{b} = q y − 2 f x, m_{0} = 2 w (x^{2} + y^{2}) − 2 e r w x + r (w^{2} − e^{2}) y,

m_{K} = 2 f g (f − g) (w + 2) (x^{2} − e r x − y^{2}) −

2 (g^{2} u + f g q t + f^{2} t^{2} − 6 f^{2} g^{2}) x y + e g r (2 g u + f q t − 8 f^{2} g) y.

Notice that m_{0} = 0 is the equation of the circumcircle of ABC because it is equivalent to the equation (x − 2^{∗} e r)^{2} + (y + 4^{∗} (w^{2} − e^{2}) r w^{∗})^{2} − 16^{∗} p^{2} r^{2} s^{2} w^{∗2} = 0 of the circle with center at the circumcenter O(2^{∗} e r, −4^{∗} (w^{2} − e^{2}) r w^{∗}) and the radius equal to the circumradius 4^{∗} p r s w^{∗}. Also, m_{K} = 0 is the equation of the Kiepert hyperbola of ABC since the vertices A(0, 0), B(e h, 0), and C(g q r w^{∗}, 2 f g r w^{∗}), the orthocenter H(g q r w^{∗}, 2^{∗} q r t w^{∗}), and the centroid G(3^{∗} r (f t + 2 g q), 3^{∗} 2 f g r w^{∗}) satisfy it. This shows that LMN is (A, γ_{K})-generating for all real numbers h ≠ 0 and k.

16. Appendix – Proof of Theorem 4 for X = J and j = 2

We first determine P, Q, and R as above. Let h be a real number. Let U, V, and W be vertices of the triangle [ABC, P QR, J_{2}[h]]. Since

J_{2}[h] = 2 h w ((2 w + s)^{∗}, (2 w + p)^{∗}, w^{∗} (w + 2)^{∗}),

it follows that U(P_{x} + 4 g h w s^{∗} (2 w + s)^{∗}, P_{y} + 2 h t w s^{∗} (2 w + s)^{∗}),

V (Q_{x} − 4 f h w p^{∗} (2 w + p)^{∗}, Q_{y} + 2 h q w p^{∗} (2 w + p)^{∗}), W (R_{x}, R_{y} − 2 h (w + 2)^{∗}).

Let k be a real number. Let (L, M, N) = UV W (U, V, W, ν[k]). Then L, M, and N are vertices of the triangle [ABC, UV W, ν[k]]. It follows that

L(U_{x}+4 e f g^{2} k q r^{3} w^{∗2}, U_{y} +2 e f g k q r^{3} t w^{∗2}), N(W_{x}, W_{y} + 2 e f g k (w^{2} − e^{2}) r^{3} w^{∗2}),

and M(V_{x} − 4 e f^{2} g k r^{3} t w^{∗2}, V_{y} + 2 e f g k q r^{3} t w^{∗2}).

The orthology condition is

OR(P^{a}P^{b}P^{c}, LMN) = 2 e f g h m_{0} m_{J} q_{a}^{∗} q_{b}^{∗} y^{∗} w^{∗} (w + 2)^{∗} p^{∗} s^{∗} (2 w + p)^{∗} (2 w + s)^{∗},

where

m_{J} = 2 w (w + 2) (f − g) (w^{2} − e^{2}) (y^{2} + e r x − x^{2}) +

w (2 g^{2} u+2 f^{2} v+20 f^{2} g^{2}−3 u v) x y+r (2 g v (f^{6}−1)+f t^{3} u−2 f^{2} (v+4 g^{2}) (f t+g q)) y.

Observe that m_{J} = 0 is the equation of the Jarabek hyperbola of ABC since the vertices the orthocenter, and the circumcenter O(2^{∗} e r, −4^{∗} (w^{2} − e^{2}) r w^{∗}) satisfy it. This shows that LMN is (A, γ_{J})-generating for all real numbers h ≠ 0 and k.

17. Appendix – Proof of Theorem 4 for X = F and j = 2

We first determine P, Q, and R as above. Let h be a real number. Let UV W be the triangle [ABC, P QR, F_{2}[h]]. Since F_{2}[h] = 2^{∗} r^{∗} h (f^{∗}, g^{∗}, w e^{∗}), it follows that

U(P_{x} + f^{∗} g h r^{∗} s^{∗}, P_{y} + 2^{∗} f^{∗} h r^{∗} s^{∗} t), V (Q_{x} − f g^{∗} h p^{∗} r^{∗}, Q_{y} + 2^{∗} g^{∗} h p^{∗} q r^{∗}),

and W (R_{x}, R_{y} − 2^{∗} e^{∗} h r^{∗} w).

Let k be a real number. Let (L, M, N) = UV W (U, V, W, ν[k]). Then L, M, and N are vertices of the triangle [ABC, UV W, ν[k]]. It follows that

L(U_{x}+4 e f g^{2} k q r^{3} w^{∗2}, U_{y} +2 e f g k q r^{3} t w^{∗2}), N(W_{x}, W_{y} + 2 e f g k (w^{2} − e^{2}) r^{3} w^{∗2}),

and M(V_{x} − 4 e f^{2} g k r^{3} t w^{∗2}, V_{y} + 2 e f g k q r^{3} t w^{∗2}). The orthology condition is

2^{∗} (e^{2} + f^{2} g^{2} − 1) h m_{0} m_{F} p^{∗} q_{a}^{∗} q_{b}^{∗} r^{∗} s^{∗} w^{∗} y^{∗},

where

m_{F} = 2 (f − g) (x^{2} − e r x − y^{2}) + (4 − q − t + w (2 − w)) x y + r (f q s − 2 g p) y.

Observe that m_{F} = 0 is the equation of the Feuerbach hyperbola of ABC since the vertices, the orthocenter, and the incenter I(f r, r) satisfy it. This shows that LMN is (A, γ_{F})-generating for all real numbers h ≠ 0 and k.

18. Appendix – Proof of Theorem 6 for X = F, i = Oi, j′ = 1, and j = 2

An arbitrary point X different from the circumcenter O on the line IO joining the incenter with the circumcenter can be represented as a point with coordinates

X(2^{∗} η r (e ξ + 2 f), −4^{∗} η r ((w^{2} − e^{2}) ξ − 4 w) w^{∗})

for a real number ξ different from −1, where η = (ξ + 1)^{∗}. Projections P, Q, and R of X onto the sidelines BC, CA, and AB are

P (2^{∗} η r s^{∗} w^{∗} (s (f t + 2 g q) ξ + 2 (f s + 2 g) w), η g r s^{∗} w^{∗} (f s ξ + 2 g w)),

Q(2^{∗} η (g p ξ + 2 f w) p^{∗} q r w^{∗}, η f (g p ξ + 2 f w) p^{∗} r w^{∗}), R(2^{∗} η r (e ξ + 2 f), 0).

Let h be a real number. Let (U, V, W) = UV W (P, Q, R, F_{1}[h]). Then U, V, and W are vertices of the triangle [ABC, P QR, F_{1}[h]]. It follows that

U(P_{x} + 2 g h s^{∗}, P_{y} + h s^{∗} t), V (Q_{x} − 2 f h p^{∗}, Q_{y} + h q p^{∗}), W (R_{x}, −h),

where P_{x} and P_{y} denote the first and the second coordinate of P.

Let k be a real number. Let (L, M, N) = UV W (U, V, W, F_{2}[k]). Then L, M, and N are vertices of the triangle [ABC, UV W, F_{2}[k]]. Hence,

L(U_{x} + f^{∗} g k r^{∗} s^{∗}, U_{y} + 2^{∗} f^{∗} k r^{∗} s^{∗} t), M(V_{x} − f g^{∗} k p^{∗} r^{∗}, V_{y} + 2^{∗} g^{∗} k p^{∗} q r^{∗}),

and N(W_{x}, W_{y} − 2^{∗} e^{∗} k r^{∗} w).

The orthology condition is 2^{∗} η m_{0} m_{1} m_{F} p^{∗} q_{a}^{∗} q_{b}^{∗} r^{∗} s^{∗} w^{∗} y^{∗}, where

m_{1} = (ξ + 1) [4 e f g h r + (e^{2} + f^{2} g^{2} − 1) k] + 4 e f g r^{2},

This shows that LMN is (A, γ_{F})-generating for all real numbers h and k which satisfy the relation m_{1} ≠ 0.

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