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Forum GeometricorumVolume 1 (2001) 99105.

FORUM GEOM

ISSN 1534-1178

Multiplying and Dividing Curves by Points

Clark Kimberling

Abstract. Pointwise products and quotients, defined in terms of barycentric and

trilinear coordinates, are extended to products P and quotients /P, where Pis a point and is a curve. In trilinears, for example, if0 denotes the circumcir-cle, then P0 is a parabola if and only ifP lies on the Steiner inscribed ellipse.Barycentric division by the triangle center X110 carries 0 onto the Kiepert hy-perbola ; ifP is on 0, then the point P

= P/X110 is the point, other thanthe Tarry point, X98, in which the line PX98 meets

, and if1 and 2 denotethe Brocard points, then |P1|/|P

2| = |P1|/|P2|; that is, P and P lie

on the same Apollonian circle with respect to 1 and 2.

1. Introduction

Paul Yiu [7] gives a magnificent construction for a product P Q of points in theplane of triangle ABC. If

P = 1 : 1 : 1 and Q = 2 : 2 : 2 (1)

are representations in homogeneous barycentric coordinates, then the Yiu product

is given by

P Q = 12 : 12 : 12 (2)

whenever {12, 12, 12} = {0}.Cyril Parry [3] constructs an analogous product using trilinear coordinates. In

view of the applicability of both the Yiu and Parry products, the notation in equa-

tions (1) and (2) will represent general homogeneous coordinates, as in [6, Chapter

1], unless otherwise noted. We also define the quotient

P/Q := 122 : 122 : 122

whenever Q / {A ,B,C }. Specialization of coordinates will be communicated byphrases such as those indicated here:

barycentric

trilinear

multiplicationproductdivisionquotient

.

Publication Date: July 12, 2001. Communicating Editor: Floor van Lamoen.

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100 C. Kimberling

If S is a set of points, then P S := {P Q : Q S}. In particular, if S is acurve , then P and /P are curves, except for degenerate cases, such as whenP {A ,B,C }.

In all that follows, suppose P = p : q : r is a point not on a sideline of triangleABC, so that pqr = 0, and consequently, U/P = up :

vq :

wr

for all U = u : v : w.

Example 1. If is a line + m + n = 0, then P is the line (/p) +(m/q)+ (n/r) = 0 and /P is the line p + qm+ rn = 0. Given the lineQR of points Q and R, it is easy to check that P QR is the line ofP Q and P R.In particular, P ABC = ABC, and ifT is a cevian triangle, then P T is acevian triangle.

2. Conics and Cubics

Each conic in the plane of triangle ABC is given by an equation of the form

u2 + v2 + w2 + 2f + 2g + 2h = 0. (3)

That P is the conic

(u/p2)2 + (v/q2)2 + (w/r2)2 + 2(f/qr)+ 2(g/rp) + 2(h/pq) = 0

(4)is clear, since : : satifies (3) if and only ifp : q : r satisfies (4). In thecase of a circumconic given in general form by

f

+

g

+

h

= 0, (5)

the product P is the circumconic

pf

+

qg

+

rh

= 0.

Thus, ifX is the point such that X is a given circumconic u +v +

w = 0, then

X = uf :vg :

wh .

Example 2. In trilinears, the circumconic in (5) is the isogonal transform of theline L given by f + g+ h = 0. The isogonal transform ofP L is /P.

Example 3. Let U = u : v : w. The conic W(U) given in [1, p. 238] by

u22 + v22 + w22 2vw 2wu 2uv = 0

is inscribed in triangle ABC. The conic P W(U) given by

(u/p)22+(v/q)22+(w/r)222(vw/qr)2(wu/rp)2(uv/pq) = 0

is the inscribed conic W(U/P). In trilinears, we start with = incircle, given by

u = u(a,b,c) = a(b + c a), v = u(b,c,a), w = u(c,a,b),

and find 1

1The conics in Example 3 are discussed in [1, p.238] as examples of a type denoted by W(Xi),including incircle = W(X55), Steiner inscribed ellipse = W(X6), Kiepert parabola = W(X512), andYff parabola = W(X647). A list ofXi including trilinears, barycentrics, and remarks is given in [2].

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Multiplying and dividing curves by points 101

Conic Trilinear product Barycentric product

Steiner inscribed ellipse X9 X8 Kiepert parabola X643 X645

Yff parabola X644 X646

Example 4. Here we combine notions from Examples 1-3. The circumcircle, 0,may be regarded as a special circumconic, and every circumconic has the form

P 0. We ask for the locus of a point P for which the circumconic P 0 is aparabola. As such a conic is the isogonal transform of a line tangent to 0, webegin with this statement of the problem: find P = p : q : r (trilinears) for whichthe line L given by p +

q +

r = 0 meets 0, given by

a +

b +

c = 0 in exactly

one point. Eliminating leads to

=

cr ap bq

(ap + bq cr)2 4abpq

2bp.

We write the discriminant as

(p, q, r) = a2p2 + b2q2 + c2r2 2bcqr 2carp 2abpq.

In view of Example 3 and [5, p.81], we conclude that ifW(X6) denotes the Steinerinscribed ellipse, with trilinear equation ( , , ) = 0, then

P 0 is a

hyperbolaparabola

ellipse

according as P lies

inside W(X6)on W(X6)

outside W(X6)

.

(6)

Returning to the case that L is tangent to 0, it is easy to check that the point oftangency is (X1/P) cX6. (See Example 7 for Ceva conjugacy, denoted by c.)

If the method used to obtain statement (6) is applied to barycentric multiplica-

tion, then a similar conclusion is reached, in which the role of W(X6) is replacedby the inscribed conic whose barycentric equation is

2 + 2 + 2 2 2 2 = 0,

that is, the ellipse W(X2).

Example 5. Suppose points P and Q are given in trilinears: P = p : q : r, andU = u : v : w. We shall find the locus of a point X = : : such that P Xlies on the line U X. This on-lying is equivalent to the determinant equation

u v w

p q r

= 0,expressible as a circumconic:

u(q r)

+

v(r p)

+

w(p q)

= 0. (7)

One may start with the line X1P, form its isogonal transform , and then recognize(7) as U . For example, in trilinears, equation (7) represents the hyperbolas of

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102 C. Kimberling

Kiepert, Jerabek, and Feuerbach according as (P, U) = (X31, X75), (X6, X48),and (X1, X3); or, in barycentrics, according as (P, U) = (X6, X76), (X1, X3),and (X2, X63).

Example 6. Again in trilinears, let be the self-isogonal cubic Z(U) given in [1,p. 240] by

u(2 2) + v(2 2) + w(2 2) = 0.

This is the locus of points X such that X, X1/X, and U are collinear; the point Uis called the pivot ofZ(U). The quotient /P is the cubic

up(q22 r22) + vq(r22 p22) + wr(p22 q22) = 0.

Although /P is not generally self-isogonal, it is self-conjugate under the P2-isoconjugacy defined (e.g., [4]) by X X1/(X P

2).

Example 7. Let X cP denote the X-Ceva conjugate ofP, defined in [1, p.57] forX = x : y : z and P = p : q : r by

X cP = p(p

x+

q

y+

r

z) : q(

q

y+

r

z+

p

x) : r(

r

z+

p

x+

q

y).

Assume thatX = P

. It is easy to check that the locus of a pointX

for which

X cP lies on the line XP is given by

p(

2

q2

2

r2) +

q(

2

r2

2

p2) +

r(

2

p2

2

q2) = 0. (8)

In trilinears, equation (8) represents the product P where is the cubic Z(X1).The locus ofX for which P cX lies on XP is also the cubic (8).

3. Brocard Points and Apollonian Circles

Here we discuss some special properties of the triangle centers X98 (the Tarrypoint) and X110 (the focus of the Kiepert parabola). X98 is the point, other than A,B, C, that lies on both the circumcircle and the Kiepert hyperbola.

Let be the Brocard angle, given bycot = cot A + cot B + cot C.

In trilinears,

X98 = sec(A + ) : sec(B + ) : sec(C+ ),

X110 =a

b2 c2:

b

c2 a2:

c

a2 b2.

Theorem. Barycentric division by X110 carries the circumcircle 0 onto the Kieperthyperbola . For every point P on 0, the line joining P to the Tarry point X98(viz., the tangent atX98 ifP = X98) intersects

again atP = P/X110. Further-more, P/X110 lies on the Apollonian circle of P with respect to the two Brocard

points 1 and 2; that is |P1|

|P2|=

|P1|

|P2|. (9)

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Multiplying and dividing curves by points 103

C

B

A

O

K

1

2

X98

X110

P

P

Figure 1

Proof. In barycentrics, 0 and are given by

a2

+

b2

+

c2

= 0 and

b2 c2

+

c2 a2

+

a2 b2

= 0,

and, also in barycentrics,

X110 =a

b2 c2:

b

c2 a2:

c

a2 b2

so that = 0/X110.For the remainder of the proof, we use trilinears. A parametric representation

for 0 is given by

P = P(t) = a(1 t) : bt : ct(t 1), (10)

for < t < , and the barycentric product P/X110 is given in trilinears by

(1 t)b2 c2

a : tc2 a2

b : t(t 1)a2 b2

c .

That this point lies on line P X98 is equivalent to the following easily verifiedidentity:

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104 C. Kimberling

(1 t) b

2c2

a tc2a2

b t(t 1)a2b2

c

a(1 t) bt ct(t 1)sec(A + ) sec(B + ) sec(C+ )

= 0.We turn now to a formula [1, p.31] for the distance between two points expressed

in normalized2 trilinears (,,) and (, , ):

1

2

abc[a cos A( )2 + b cos B( )2 + c cos C( )2], (11)

where denotes the area of triangle ABC. Let

D = c2t2 (c2 + a2 b2)t + a2,S = a2b2 + b2c2 + c2a2.

Normalized trilinears for (10) and the two Brocard points follow:

P = ((1 t)ha,thb,t(t 1)hc),

where h = 2D , and

1 =

h1cb

, h1ac

, h1ba

, 2 =

h1b

c, h1c

a, h1a

b

,

where and h1 =2abcS

.

Abbreviate a cos A, b cos B, c cos C, and 1 t as a, b, c, and t respectively,and write

E = a

tha h1c

b

2

+ b

thb h1a

c

2

+ c

tthc h1b

a

2

, (12)

F = a

tha h1b

c

2

+ b

thb h1c

a

2

+ c

tthc h1a

b

2

. (13)

Equation (11) then gives

|P1|2|P2|2

= EF

. (14)

In (12) and (13), replace cos A by (b2 + c2 a2)/2bc, and similarly for cos B and

cos C, obtaining from (14) the following:

|P1|2

|P2|2=

t2a2 t(a2 + b2 c2) + b2

t2b2 t(b2 + c2 a2) + c2.

2Sometimes trilinear coordinates are called normal coordinates. We prefer trilinears, so that

we can say normalized trilinears, not normalized normals. One might say that the latter double

usage of normal can be avoided by saying actual normal distances, but this would be unsuitable

for normalization of points at infinity. Another reason for retaining trilinear and quadriplanarnot replacing both with normal is that these two terms distinguish between lines and planes as

the objects with respect to which normal distances are defined. In discussing points relative to a

tetrahedron, for example, one could have both trilinears and quadriplanars in the same sentence.

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Multiplying and dividing curves by points 105

Note that if the numerator in the last fraction is written as f(t,a,b,c), then thedenominator is t2f(1t ,c ,b ,a). Similarly,

|P1|2

|P2|2=

g(t,a,b,c)

t4g(1t ,c ,b ,a),

where

g(t,a,b,c) = t4

e4 + t3

e3 + t2

e2 + te1 + e0,and

e4 = a4b2(a2 b2)2,

e3 = a2(a2 b2)(b6 + c6 + 2a2b2c2 2a4b2 2a2c4 2b2c4 + a4c2 + a2b4),

e2 = b2c2(b2 c2)3 + a2c2(c2 a2)3 + a2b2(a6 + 2b6 3a2b4)

+ a2b2c2(b4 + c4 2a4 4b2c2 + 2a2c2 + 2a2b2),

e1 = b2(c2 b2)(a6 + c6 3b2c4 + 2b4c2 2a4b2 2a4c2 + 2a2b2c2 + a2b4),

e0 = b4c2(b2 c2)2.

One may now verify directly, using a computer algebra system, or manually with

plenty of paper, that

t2f(t,a,b,c)g(1

t,c ,b ,a) = f(

1

t,c ,b ,a)g(t,a,b,c),

which is equivalent to the required equation (9).

References

[1] C. Kimberling, Triangle Centers and Central Triangles, Congressus Numerantium, 129 (1998) 1

285.

[2] C. Kimberling, Encyclopedia of Triangle Centers, 2000

http://cedar.evansville.edu/ck6/encyclopedia/.

[3] C. Kimberling and C. Parry, Products, square roots, and layers in triangle geometry, Mathematics

and Informatics Quarterly, 10 (2000) 9-22.

[4] C. Kimberling, Conjugacies in the plane of a triangle, Aequationes Mathematicae, 61 (2001)

forthcoming.[5] S.L. Loney, The Elements of Coordinate Geometry, Part II: Trilinear Coordinates, Etc., Macmil-

lan, London, 1957.

[6] E.A. Maxwell, The Methods of Plane Projective Geometry Based on the Use of General Homo-

geneous Coordinates, Cambridge University Press, 1957.

[7] P. Yiu, The uses of homogeneous barycentric coordinates in plane euclidean geometry, Int. J.

Math. Educ. Sci. Technol., 31 (2000) 569 578.

Clark Kimberling: Department of Mathematics, University of Evansville, 1800 Lincoln Avenue,

Evansville, Indiana 47722, USA

E-mail address: ck6@evansville.edu