THE CIRCUMCIRCLE AND THE LINE AT INFINITY

Clark Kimberling and Peter Yff

1. Introduction. For a pleasant surprise, pick up the nearest pencil and carry out a little construction: begin with a circle, and mark any three points on it. Label them A, B, C, and draw their triangle. Starting at any fourth point, X, on your circle, draw line AX and reflect it about the bisector of angle A, obtaining a line L_{1}. Then reflect BX about the B-bisector and CX about the C-bisector to obtain lines L_{2} and L_{3}. The surprise is that the three lines are parallel, meeting in a point, X^{−1}, infinitely far from △ABC. The set of all such points is called the line at infinity, denoted L^{∞}, and the plane of △ABC together with L^{∞} is the extended plane of △ABC.

If the initial point X is not on your circle (i.e., the circumcircle, Γ, of △ABC) and not on a sideline of △ABC, then the three reflected lines are not parallel. Try out a couple of X with your pencil, one inside Γ and the other outside. You will discover another pleasing fact: the three lines L_{1}, L_{2}, L_{3} concur in a single point, again written as X^{−1} and called the isogonal conjugate of X. Of course, (X^{−1})^{−1} = X.

We define six more isogonal conjugates by the equations A^{−1} = BC ∩ L^{∞}, B^{−1} = CA∩L^{∞}, C^{−1} = AB ∩L^{∞}, (A^{−1})^{−1} = A, (B^{−1})^{−1} = B, (C^{−1})^{−1} = C, so that now we may describe L^{∞} as follows: (i) each of its points is essentially a family of parallel lines, and (ii) Γ^{−1} = L^{∞}. Here, of course, we are referring to isogonal conjugation as a mapping of one set onto another.

2. Antipedal Lines. Figure 1 shows antipedal lines for a triangle ABC, and Figure 2 represents these antipedal lines by short segments. In this second figure, if you start at any X on Γ and let θ_{0} be the angle (≤ π/2) between the antipedal line XX^{−1} and the tangent to Γ at X, you will encounter two other points on Γ for which the corresponding angle θ equals the original θ_{0}. In fact, as is clear from the constant rate of change of θ as X moves around Γ, the three points form an equilateral triangle. Later we shall discuss two of these triangles: the circumtangential triangle T_{1}T_{2}T_{3}, for which θ = 0, an the circumnormal triangle N_{1}N_{2}N_{3}, for which θ = π/2. These two triangles together form a regular six– pointed star, shown in Figure 3.

Figure 1. Antipedal lines of triangle ABC Figure 2. Segments representing antipedal lines

Figure 3. Circumtangential triangle T_{1}T_{2}T_{3} and circumnormal triangle N_{1}N_{2}N_{3}

The name antipedal line is chosen because of the relationship between such a line and a pedal line, this latter being a standard term synonymous with Simson line and Wallace line–the feet of the perpendiculars from X onto sidelines BC, CA, AB are collinear points on the pedal line of X. Now suppose X on Γ has antipedal line XX^{−1} and pedal line L. In Figure 4, let ^{}X̸AB = φ. Let AY be the reflection of AX about the bisector of ^{}B̸AC. By the definition of isogonal conjugate, X^{−1} is on the line AY. Since X^{−1} is at infinity, XX^{−1} is the line through X parallel to AY. It intersects BC at an angle C − φ, as shown. Now let Q_{1}, Q_{2}, Q_{2} be the feet of the perpendiculars from X to the respective sides of △ABC, so that they are on L. Thus, ^{}X̸Q_{3}A and ^{}X̸Q_{2}A are right angles, making XA a diameter of a circle through Q_{2} and Q_{3}. This means that AXQ_{3}Q_{2} is a cyclic quadrilateral, so that ^{}X̸Q_{2}Q_{3} = ^{}X̸AQ_{3} = φ, and ^{}Q̸_{3}Q_{2}C = π/2 − φ. Also, ^{}C̸Q_{1}Q_{2} = π − C − (π/2 − φ), which makes this angle complementary to the angle between XX^{−1} and BC. Therefore XX^{−1} is perpendicular to L. Next we define two mappings, Λ and Ψ, in terms of pairs of points P and X. The images, Λ(P, X) and Ψ(P, X), always lie on Γ.

Figure 4. Antipedal line of X, perpendicular to pedal line, Q_{1}Q_{2}Q_{3}, of X

\underline{Definition 1}. (The Lambda Mapping.) Λ(P, X) is the point (P X ∩ L^{∞})^{−1}, for all P and X for which the line P X meets L^{∞} in one point. Note the connection with antipedal lines; if X ∈ Γ then X^{−1} is the point where the antipedal line XX^{−1} meets L^{∞}, so that Λ(X, X^{−1}) = X.

\underline{Definition 2}. (The Psi Mapping.) Let Y = Λ(P, X), and write Y and Z for the points in which the line Y P^{−1} meets Γ, where Z = Y if and only if XY is tangent to Γ. Define Ψ(P, X) = Z. See Figure 5. Here, too, the connection with antipedal lines is easy to see, for if X ∈ Γ then Y = Λ(X, X^{−1}) = X, so that Ψ(X, X^{−1}) is the point other than X where the antipedal line XX^{−1} meets Γ. We state here some facts about Λ and Ψ, to be proved in greater generality in Section 5.

Figure 5. The mappings Λ and Ψ \underline{Theorem 1}. Suppose P and X are points in the extended plane of △ABC, P ≠

X, and the line of X and Ψ(P, X) meets Γ in two points. Then the two points are Λ(P^{−1}, X^{−1}) and Ψ(P, X). (See Figure 5.)

\underline{Theorem 2}. Suppose X is a point on Γ. Let Y be the antipode of X, and let Q be the point of intersection of the antipedal lines of X and Y. Then Q is the antipode of Λ(X, Y). (See Figure 6.)

Figure 6. Antipodal pairs X, Y and Q, Λ(X, Y) By Theorem 1, if P ∈ Γ and X ≠ P, then the line P X meets Γ again in

the point Ψ(P, X). In particular, if O = circumcenter, then the antipode of P is Ψ(P, O).

3. Trilinear Coordinates. Suppose P is a point in the plane of △ABC having sidelengths a, b, c, and α, β, γ are numbers proportional to the directed distances from P to the sides BC, CA, AB, respectively. Then the actual directed distances from P to the sides are kα, kβ, kγ, where k = 2σ/(aα + bβ + cγ) and σ = area of △ABC. However, mere ratios α : β : γ determine P, and any such triple is called trilinears for P. For example, the incenter, being equidistant from the sidelines of △ABC, has trilinears 1 : 1 : 1. Standard formulas involving trilinears are given in [1]. In particular, the isogonal conjugate of a point P = α : β : γ, where αβγ ≠ 0, is the point P^{−1} = α^{−1} : β^{−1} : γ^{−1}. Also, A^{−1} = 0 : c : −b, B^{−1} = −c : 0 : a, C^{−1} = b : −a : 0. The line at infinity, L^{∞}, has equation aα+bβ +cγ = 0, so that Γ, as the isogonal conjugate of L^{∞}, has equation aβγ + bγα + cαβ = 0. A point X on Γ has trilinears given as in [4] by

(1)  X = csc θ : csc(C − θ) : − csc(B + θ).

Here, 2θ is an angle AOX, oriented in the same sense as △ABC, as indicated in Figure 7. We shall use this parameterization of Γ in a proof of Theorem 3.

Figure 7. The parameter θ \underline{Theorem 3}. The circumtangential triangle, T_{1}T_{2}T_{3}, is an equilateral triangle

with vertices

(2)

T_{1} = csc((C − B)/3) : csc((B + 2C)/3) : − csc((2B + C)/3),
T_{2} = − csc((2C + B)/3 : csc((A − C)/3) : csc((C + 2A)/3),
= csc((A + 2B)/3) : − csc((2A + B)/3) : csc((B − A)/3).

\underline{Proof}. Equation (2) and the trilinears of X^{−1} yield an equation for the line XX^{−1}:

[csc(B + θ) sin(C − θ) − csc(C − θ) sin(B + θ)]α
+ [csc θ sin(B + θ) − csc(B + θ) sin θ]β
+ [csc θ sin(C − θ) − csc(C − θ) sin θ]γ = 0,

or a sin θ sin(C − B − 2θ)α + b sin(C − θ) sin(B + 2θ)β + c sin(B + θ) sin(C − 2θ)γ = 0. It can be shown that the line tangent to Γ at X is given by the equation

(a sin^{2} θ)α + b sin^{2}(C − θ)β + c sin^{2}(B + θ)γ = 0.

If this line is XX^{−1}, the trilinears of X^{−1} must satisfy

a sin^{3} θ + b sin^{3}(C − θ) + c sin^{3}(B + θ) = 0,

or

a sin^{3} θ + b(sin C cos θ − cos C sin θ)^{3} + c(sin B cos θ + cos C sin θ)^{3} = 0.

This equation reduces to

a sin B sin C cos(B − C) sin^{3} θ + 3a sin B sin C sin(B − C) sin^{2} θ cos θ − 3a sin B sin C cos(B − C) sin θ cos^{2} θ − a sin B sin C sin(B − C) cos^{3} θ = 0.

Elimination of the common factor and division by cos^{3} θ yield

(tan^{3} θ − 3 tan θ) cos(B − C) + (3 tan^{2} θ − 1) sin(B − C) = 0,

and the identity

tan 3θ = (3 tan θ − tan^{3} θ)/(1 − 3 tan^{2} θ)

reduces this to

tan 3θ = − tan(B − C).

This equation has three essentially distinct solutions: 3θ = −(B − C), π − (B − C), and 2π − (B − C), corresponding to θ_{1} = −(B − C)/3, θ_{2} = π/3 − (B − C)/3, θ_{3} = 2π/3 − (B − C)/3.

Thus there are three values of θ for which the antipedal line of X is tangent to Γ. They are spaced on the circumcircle Γ at intervals of π/3 (in terms of θ), or 2π/3 (in terms of 2θ), so that the points of tangency are the vertices of an equilateral triangle.

Let T_{1} be the point corresponding to θ_{1}. Then T_{1} and A are on the same arc of BC, since −B < −(B − C)/3 < C, where C is the value of the parameter θ for the point B, and −B is that for C. Similarly, T_{2} and B are on the same arc of CA, while T_{3} and C are on the same arc of AB, therefore justifying the prescribed ordering. Substitution of −(B − C)/3 for θ in (2) gives trilinears for T_{1} as asserted, and cyclic permutations of variables and subscripts give trilinears for T_{2} and T_{3}.

4. Trilinears and Equilateral Triangles. Probably the best known equilateral triangle associated with an arbitrary triangle is Morley’s triangle, described by Morley’s Theorem: The three intersections of the trisectors of the angles of a triangle, lying near the three sides respectively, form an equilateral triangle. “One of the most astonishing and totally unexpected theorems in mathematics,” to quote Oakley and Baker [3], Morley’s Theorem obviously involves the angles A/3, B/3, C/3, as in Figure 8.

\underline{Theorem 4}. The circumtangential triangle, T_{1}T_{2}T_{3}, is homothetic to Morley’s triangle, M_{1}M_{2}M_{3}.

\underline{Proof}. To find trilinears for M_{1}, note that for any point α : β : γ on BM_{1}, we have x_{1}/x_{3} = sin(B/3)/ sin(2B/3) = 1/(2 cos(B/3)). Similarly, if the point is on CM_{1}, then x_{1}/x_{2} = 1/(2 cos(C/3)). Thus, trilinears for M_{1} are now determined, and by cyclic permutation of angles and coordinates, we have M_{1} = 1 : 2 cos(C/3) : 2 cos(B/3), M_{2} = 2 cos(C/3) : 1 : 2 cos(A/3), M_{3} = 2 cos(B/3) : 2 cos(A/3) : 1. From the coordinates, we obtain an equation for the line M_{1}M_{2}.

(4 cos \frac{C}{3} cos \frac{A}{3} − 2 cos \frac{B}{3} )α + (4 cos \frac{B}{3} cos \frac{C}{3} − 2 cos \frac{A}{3})β + (1 − 4 cos^{2} \frac{C}{3} )γ = 0.

Figure 8. The circumtangential triangle is homothetic to Morley’s triangle In order to prove that M_{1}M_{2} is parallel to T_{1}T_{2}, it suffices to prove that M_{1}M_{2}

is parallel to the antipedal line of T_{3}. But, since T_{3}^{−1} is the point at infinity on T_{3}T_{3}^{−1}, it suffices to prove that T_{3}^{−1} lies on M_{1}M_{2}. To prove this, we substitute the coordinates of T_{3}^{−1} into the equation of M_{1}M_{2}:

− (4 cos \frac{C}{3} cos \frac{A}{3} − 2 cos \frac{B}{3} ) sin \frac{A + 2B}{3}

+ (4 cos \frac{B}{3} cos \frac{C}{3} − 2 cos \frac{A}{3}) sin \frac{2A + B}{3}

+ (1 − 4 cos^{2} \frac{c}{3}) sin \frac{A − B}{3} ,

which equals zero. This completes a proof that M_{1}M_{2} ∥ T_{1}T_{2}. Symmetry shows that M_{2}M_{3} ∥ T_{2}T_{3} and M_{3}M_{1} ∥ T_{3}T_{1}. (See Figure 8.) Thus, the triangles M_{1}M_{2}M_{3} and T_{1}T_{2}T_{3} are homothetic.

Let z = sin(A/3) sin(B/3) sin(C/3). The sidelength of Morley’s triangle is 8Rz, where R is the circumradius of △ABC. It follows that the ratio of homothety of the circumtangential triangle to Morley’s triangle is 8z/\sqrt{3}.

The circumnormal triangle, N_{1}N_{2}N_{3}, can be described as the triangle whose vertices are the points X on Γ for which the antipedal line XX^{−1} is normal to Γ, or, equivalently, those X for which XX^{−1} passes through the circumcenter, O. An equation for the line OX found via (2) and (1) is [csc(C − θ) cos C + csc(B + θ) cos B]α − [csc(B + θ) cos A + csc θ cos C]β + [csc θ cos B − csc(C − θ) cos A]γ = 0. In order for this to be the antipedal line of X, it must contain the point X^{−1}. When trilinears for X^{−1} are substituted into the equation, the procedure used in proving Theorem 1 yields roots θ′_{1} = π/2 − (B − C)/3, θ′_{2} = 5π/6 − (B − C)/3, θ′_{3} = π/6 − (B − C)/3. These values in (2) give trilinears for the vertices N_{1}N_{2}N_{3}, respectively, and they establish the fact that N_{i} is the antipode of T_{i} for i = 1, 2, 3 so that N_{1}N_{2}N_{3} is also an equilateral triangle. It, too, of course, is homothetic to Morley’s triangle.

5. Trilinears and Antipedal Lines. The envelope of the system of antipedal lines is of some significance. To determine this curve, first consider Figure 9, in which BX is a cevian of △ABC, making an angle φ with BC. Let h be the length of BX, so that the distances from X to BC and AB are h sin φ and h sin(B − φ), respectively. Therefore an equation for BX is sin(B − φ)α − (sin φ)γ = 0. This line meets L^{∞} in the point

(3)  sin φ : − sin(C + φ) : sin(B − φ).

Therefore, (3) is the point at infinity on any line making the angle φ with BC, in the sense shown in Figure 9.

Figure 9. The angle θ associated with cevian BX Since X^{−1} is the point at infinity on the antipedal line of X, its trilinears must

be proportional to those of (3). Hence we have sin φ/ sin θ = − sin(C + φ)/ sin(C − θ) = − sin(B − φ)/ sin(B + θ), for which the solution is φ = −θ or φ = π − θ. Therefore the inclination of XX^{−1} with respect to BC is −θ, which means that as X moves around Γ with angular velocity 2dθ/dt, its antipedal line turns with angular velocity −dθ/dt. The result as in [5] is that the envelope of the antipedal lines is a deltoid, or hypocycloid of three cusps, circumscribed about Γ.

This hypocycloid resembles Steiner’s deltoid, which is the envelope of the pedal lines and is circumscribed about the nine–point circle. In fact, it is easy to prove that is the nine–point circle of the excentral triangle, whose vertices are the excenters (I_{1} = −1 : 1 : 1, I_{2} = 1 : −1 : 1, I_{3} = 1 : 1 : −1) of △ABC. Finally, it can be shown that the antipedal lines of △ABC are the pedal lines of I_{1}I_{2}I_{3}.

6. Trilinears and the Two Mappings. We begin by formulating the mappings Λ and Ψ introduced in Section 1. Let P = u : v : w and X = x : y : z. Form D = vz − wy, E = wx − uz, F = uy − vx, and define

(4) Λ(P, X) = (aE − bD)(aF − cD) : (bF − cE)(bD − aE) : (cD − aF)(cE − bF)
(5)  Ψ(P, X) = avwEF : bwuF D : cuvDE.

Formula (4) comes directly from the equation Dα + Eβ + F γ = 0 for line P X and that the fact that this line meets L^{∞} in the point whose first coordinate is bF −cE. Formula (5) is a bit more troublesome. The point on the right–hand side clearly lies on the circumcircle a/α + b/β + c/γ = 0 and is also collinear with P^{−1} and Λ(P, X), as attested by (1), since

(6)  | \begin{matrix}1/u & 1/v & 1/w \\ a/uD & b/vE & c/wF \\ 1/(bF − cE) & 1/(cD − aF) & 1/(aE − bD) \end{matrix} | = 0.

It appears that the most practical way to evaluate the determinant is by a computer algebra system, treating a, b, c, u, v, x, y, z as indeterminates. We prefer (6) to a purely geometric argument because (6) includes the geometric result as a special case of a larger truth; that is, a, b, c need not be sidelengths of a triangle — indeed, they and the other six numbers can be any complex numbers, with certain obvious exceptions.

We now give a few details about each mapping. To form Λ(P, X), we must have X ≠ P, and P and X cannot both lie on L^{∞} unless they are identical. The set of all Y such that Λ(P, Y) = Λ(P, X) is simply the line P X with the point P deleted. Thus, if C is any simple closed curve such that P lies in the interior of C, then as X moves around C once, the point Λ(P, X) moves around Γ once. To form Ψ(P, X), we must again have X ≠ P, and P and X both must not lie on either L^{∞} or Γ. The set of all Y satisfying Ψ(P, Y) = Ψ(P, X) is again the line P X with P deleted.

\underline{Theorem 1}. (Generalized.) Suppose a, b, c; u, v, w; x, y, z are indeterminates over the complex number field, and D = vz − wy, E = wx − uz, F = uy − vx. Then

| \begin{matrix}x & y & z \\ a/uD & b/vE & c/wF \\ u/(bwzF − cvyE) & v/(cuxD − awzF) & w/(avyE − buxD) \end{matrix} | = 0.

\underline{Proof}. That the determinant vanishes can be easily established using a computer algebra system. We shall now see why this vanishing proves Theorem 1 as stated in Section 1. Suppose a, b, c are sidelengths of a △ABC, P = u : v : w, and X = x : y : z. Then row 1 consists of trilinears for X, row 2, those of Ψ(P, X), and row 3, those of Λ(P^{−1}, X^{−1}). By (1), the points given by these rows are collinear. Since Ψ(P, X) and Λ(P^{−1}, X^{−1}) lie on the circumcircle, Γ, of △ABC, they must be the two points in which the line of X and Ψ(P, X) meets Γ.

\underline{Theorem 2}. (Generalized.) Suppose a, b, c, x′, y′ are indeterminates over the complex number field. Let

(7)  q = ay′ + bx′, x = qx′, y = qy′, z = −cx′y′

(8)  a′ = a′(a, b, c) = (b^{2} + c^{2} − a^{2})/(2bc), b′ = a′(b, c, a), c′ = a′(c, a, b)