Generalized Apollonian Circles 1 Introduction

Generalized Apollonian circles Paul Yiu Department of Mathematics Florida Atlantic University

Email:[email protected] January, 2000

1 Introduction

Given a triangle ABC with side lengths a, b, c,weconsiderapointP with homogeneous barycentric coordinates u : v : w with respect to ABC.The traces of P on the side lines are the intersections X of the lines AP and BC, Y of BP and CA,andZ of CP and AB. These have homogeneous barycentric coordinates

X =0:v : w, Y = u :0:w, Z = u : v :0.

The dual (trilinear polar) of P is the line x y z . u + v + w =0

We shall assume u, v, w distinct, so that P does not lie on any of the medians of the triangle. Its dual line, therefore, intersects the side lines at the ﬁnite points

X =0:v : −w, Y = −u :0:w, Z = u : −v :0.

We shall call the three circles with diameters XX, YY,andZZ the Apollonian circles of the point P . These Apollonian circles help determine points whose distances from the vertices of triangle ABC are in given proportions. The one with diameter XX, for example, is the locus of points M for which BM : CM = 1 1 w : v = v : w ; similarly for the other two circles. It follows that if M is a point 1 1 1 common to the three Apollonian circles of P ,thenAM : BM : CM = u : v : w . The centers of the three Apollonian circles of P are respectively the points 0:−v2 : w2, u2 :0:−w2 and −u2 : v2 : 0. It is clear that they all lie on the line x y z , u2 + v2 + w2 =0 the dual (trilinear polar) of the point P 2 = u2 : v2 : w2. It follows that the radical axes of pairs of these circles are parallel. Indeed, they coincide: the three circles are coaxal. This follows by writing down their equations:

1 v2w2 b2 c2 a2 a2 a2yz b2zx c2xy − x y z − − x − y z + + v2 − w2 ( + + )( (v2 w2 ) v2 + w2 )=0; w2u2 b2 c2 a2 b2 a2yz b2zx c2xy − x y z x − − y − z + + w2 − u2 ( + + )(u2 (w2 u2 ) w2 )=0; u2v2 c2 c2 a2 b2 a2yz b2zx c2xy − x y z − x y − − z . + + u2 − v2 ( + + )( u2 + v2 (u2 v2 ) )=0

1.1 Proposition The radical axis of the Apollonian circles of P = u : v : w is the line 1 1 1 1 1 1 1 1 1 1 1 1 (b2( − )+c2( − ))x+(c2( − )+a2( − ))y+(a2 ( − )+b2( − ))z =0. u2 v2 w2 u2 v2 w2 u2 v2 w2 u2 v2 w2

1.2 Remark Note that the coeﬃcients of the equations of the Apollonian circles involve only u2, v2,andw2. This means that the Apollonian circles for the four points ±u : ±v : ±w are the same.

1.3 Corollary The radical axis of the Apollonian circles always passes through the circumcenter.

1.3.1 Remark

While it is routine to verify this, we point out that Weaver [5] made use of the absolute coordinates of Morley and Morley [3], by representing points by complex numbers, with the circumcenter O at the origin and the vertices of the reference triangle on the unit circle. This corollary follows from the observation that the equation of the radical axis has zero constant term.

2Examples 2.1 Example: The incenter The case P = I, the incenter, is classical. The three Apollonian circles have radical axis the Hessian line OK b2 − c2 c2 − a2 a2 − b2 x y z , a2 + b2 + c2 =0 and they intersect at the two isodynamic points, X15 and X16 in Kimberling’s list. The line of centers is the Lemoine axis, the dual of the Lemoine (symmedian) point K = a2 : b2 : c2.

2 2.2 Example: The orthocenter It is very interesting to observe that the radical axis of the generalized Apollo- nian circles for the orthocenter is also the Hessian line. The three Apollonian circles, however, do not always have real common points. They do if and only if the absolute values of cos α,cosβ,andcosγ satisfy the triangle inequality.

2.3 Proposition The radical axis of the Apollonian circles of P is the Hessian line OK if and only if the point P 2 = u2 : v2 : w2 lies on the Kiepert parabola, i.e., a2 b2 − c2 b2 c2 − a2 c2 a2 − b2 ( ) ( ) ( ) . u2 + v2 + w2 =0

P = 1 : 1 : 1 2.4 Example: a b c 1 1 1 If P is the isotomic conjugate of the incenter, the point X63 = a : b : c ,the 2 2 2 line of centers, a x+ b y + c z = 0, is the dual of the third Brocard point X76 = 1 1 1 a2 : b2 : c2 . Weaver [5] was unable to identify this in absolute coordinates. It was pointed out by Ramler [4]. Much more interesting is that the radical axis of the Apollonian circles is the Euler line.

2.5 Proposition The radical axis of the Apollonian circles of P is the Euler line if and only if the point P 2 = u2 : v2 : w2 lies on the Kiepert hyperbola, i.e., b2 − c2 c2 − a2 a2 − b2 . u2 + v2 + w2 =0 3

For the Apollonian circles, Weaver considered the intersection of the radical axis and the line of centers. In most cases, he identiﬁed the inversive image of this intersection in the circumcircle as a triangle center of the tangential triangle. 1 For example, for P = X63, this intersection is the inversive image of the Nagel point of the tangential triangle. For P = X92, Weaver left it unidentiﬁed.

3.1 Proposition Let P = u : v : w. The inversive image, in the circumcircle, of the intersection of the radical axis and line of centers of the Apollonian circles of P has, with respect to the tangential triangle, homogeneous barycentric coordinates, a b c , u2 : v2 : w2 1Did Weaver consider the case of incenter?

3 where

a : b : c = a2(b2 + c2 − a2):b2(c2 + a2 − b2):c2(a2 + b2 − c2) are the proportions of the sides of the tangential triangle.

3.1.1 Remark

In identifying this point with reference to the tangential triangle, the following equivalence of (??) is useful:

a2 : b2 : c2 = a(b + c − a):b(c + a − b):c(a + b − c).

3.1.2 Proof of Proposition

It is routine to work out the intersection of the radical axis and the line of centers. This is the point

Q =[u2((b2 + c2 − 2a2)u2v2w2 − u2(c2v4 + b2w4)+a2v2w2(v2 + w2))].

The inversive image of Q in the circumcircle O(R)isthepointQ which divides the segment OQ in the ratio OQ : QQ = R2 : |OQ|2 − R2. It turns out that the homogeneous barycentric coordinates of Q are surprisingly simple: b2 c2 a2 c2 a2 b2 a2 b2 c2 Q a2 − b2 − c2 − . = (v2 + w2 u2 ): (w2 + u2 v2 ): (u2 + v2 w2 )

This is the P 2−Ceva conjugate of the symmedian point K = a2 : b2 : c2.As such, this is the point of concurrency of the lines joining the traces of P 2 to the corresponding vertices of the anticevian triangle of K, which is the tangential triangle. We shall label the vertices of the tangential triangle by A, B, C respectively. Their opposite sides a, b, c are in the proportions of

sin 2α :sin2β :sin2γ = a2(b2 + c2 − a2):b2(c2 + a2 − b2):c2(a2 + b2 − c2).

With respect to ABC, the vertices A, B, C have homogeneous coordinates

A =0:c2 + a2 − b2 : a2 + b2 − c2, B = b2 + c2 − a2 :0:a2 + b2 − c2, C = b2 + c2 − a2 : c2 + a2 − b2 :0.

Now, the trace of P 2 = u2 : v2 : w2 on the side BC being the point X =0: v2 : w2, its homogeneous barycentric coordinates with respect to the tangential triangle ABC are b2 c2 b2 c2 a2 − b2 c2 a2 b2 − c2 b2 c2 − a2 ( + ) ( + ). v2 + w2 ( + ): v2 : w2

4 Similarly, we write down the coordinates of the traces Y , Z of P 2 on the lines CA, AB. With respect to the tangential triangle, these have coordinates a2 b2 c2 − a2 c2 a2 c2 a2 b2 − c2 ( + ) c2 a2 − b2 ( + ) 2 : 2 + 2 ( + ): 2 ; u w u w a2 b2 c2 − a2 b2 c2 a2 − b2 a2 b2 ( + ) ( + ) a2 b2 − c2 . u2 : v2 : u2 + v2 ( + )

From these, it is clear that the lines AX, BY ,andCZ are perspective at the point a2 b2 c2 − a2 b2 c2 a2 − b2 c2 a2 b2 − c2 a b c Q ( + ) ( + ) ( + ) . = u2 : v2 : w2 = u2 : v2 : w2 This completes the proof of the proposition.

3.2 Corollary Consider a pair of points isotomically conjugate with respect to triangle ABC. The inversive images (in the circumcircle) of their intersections of the radical axis and line of centers of their triples of Apollonian circles are isogonally conjugate with respect to the tangential triangle.

3.2.1 Proof

2 2 2 a b c The points a u : b v : c w and u2 : v2 : w2 are isogonal conjugates with respect to the tangential triangle.

3.3 Examples 3.3.1 P = incenter With respect to the tangential triangle, the inversive image Q has coordinates a b c 1 1 1 a2 : b2 : c2 = b+c−a : c+a−b : a+b−c , the Gergonne point.

3.3.2 P = X92 The point unidentiﬁed in Weaver [5], being the isogonal conjugate of the Nagel point, is the external center of similitude of the circumcircle and the incircle (of the tangential triangle).

4 Intersections of the Apollonian circles 4.1 Proposition The Apollonian circles of P = u : v : w intersect at real points if and only if the point b2 c2 a2 c2 a2 b2 a2 b2 c2 Q a2 − b2 − c2 − = (v2 + w2 u2 ): (w2 + u2 v2 ): (u2 + v2 w2 )

5 lies inside or on the circumcircle.

4.2 We exploit the simplicity of the expressions for the coordinates of Q to determine the coordinates of the common points of the Apollonian circles. Recalling that the radical axis of these circles passes through the circumcircle, we write the coordinates of a common point of the Apollonian circles in the form b2 c2 a2 a2 − ta2 b2 c2 − a2 (v2 + w2 u2 )+ ( + ) c2 a2 b2 b2 − tb2 c2 a2 − b2 : (w2 + u2 v2 )+ ( + ) a2 b2 c2 c2 − tc2 a2 b2 − c2 . : (u2 + v2 w2 )+ ( + )

Such a point lies on any of the Apollonian circles if and only if

(a + b + c)(b + c − a)(c + a − b)(a + b − c)u4v4w4t2 =(avw + bwu + cuv)(bwu + cuv − avw)(avw − bwu + cuv)(avw + bwu − cuv).

Equivalently, a b c a b c a b c a b c ( u + v + w )(− u + v + w )( u − v + w )( u + v − w ) t = ± . (a + b + c)(b + c − a)(c + a − b)(a + b − c)

4.3 Theorem The common points of the Apollonian circles, if real, are inverse to each other with respect to the circumcircle.

4.3.1 Proof

The intersection points divide the segment OQ harmonically. . . .

4.4 Corollary The Apollonian circles of P = u : v : w have real intersections if and only if a b c b c a c a b a b c − − − ≥ . u + v + w v + w u w + u v u + v w 0

This can be rewritten as b2 c2 a4 · · − ≥ . 2 v2 w2 u4 0

6 4.5 Corollary The Apollonian circles of P intersect at real points if and only if the point a2 b2 c2 u2 : v2 : w2 lies inside or on the inscribed Steiner ellipse.

5 Mutually tangent Apollonian circles 5.1 Theorem The following statements are equivalent. (a) The Apollonian circles of P are tangent to each other. (b) The isogonal conjugate of P 2 lies on the inscribed Steiner ellipse.

a2(b2+c2−a2) b2(c2+a2−b2) c2(a2+b2−c2) (c) The point Q = u2 : v2 : w2 lies on the circumcircle. (d) The point P lies on one of the four circumconics a b c ± ± . x y z =0

5.1.1

The equivalence of these statements is clear from the preceding section. Suppose the Apollonian circles of P are tangent to each other. Among the points in Kimberling’s ﬁrst list of triangle centers [1], X88 and X100, X162 and X190 satisfy this condition.

P Point of Tangency Q a a2 X88 =[b+c−2a ] X106 =[b+c−2a ] a a2 X100 =[b−c ] X101 =[b−c ] a a2 X162 =[(b2−c2)(b2+c2−a2) ] X112 =[(b2−c2)(b2+c2−a2) ] 1 a X190 =[b−c ] X100 =[b−c ] These data clearly suggest the following theorem, which can be easily veriﬁed.

5.2 Theorem The Apollonian circles of P are tangent at the point a2 b2 c2 Q = q − r : r − p : p − q on the circumcircle if and only if a2 b2 c2 P 2 = : : . (q − r)2 (r − p)2 (p − q)2

7 Proof. Suppose the Apollonian circles of P = u : v : w are tangent at Q given above. Write a2 b2 c2 k − , u2 + v2 + w2 = q − r a2 b2 c2 k − , u2 v2 + w2 = r − p a2 b2 c2 k − , u2 + v2 w2 = p − q for a constant k. Solving these equations we obtain

a2 −k(q − r) b2 −k(r − p) c2 −k(p − q) = ; = ; = . u2 2(r − p)(p − q) v2 2(p − q)(q − r) w2 2(q − r)(r − p)

From these, we have a2 r − p p − q b2 p − q q − r c2 q − r r − p u2 v2 w2 ( )( ) ( )( ) ( )( ) : : = q − r : r − p : p − q a2 b2 c2 = : : . (q − r)2 (r − p)2 (p − q)2

The veriﬁcation for the converse is routine.

We conclude with a theorem on the construction of points with speciﬁed proportions of distances from the vertices of the reference triangle.

6.1 Theorem

Given lengths da, db, dc,therearetwopointsM satisfying AM : BM : CM = da : db : dc if and only if there is a triangle of side lengths ada, bdb and cdc.If this condition is satisﬁed, the two points M are the intersections of the Apol- P 1 1 1 lonian circles of the point = da : db : dc . The two points coincide (on the circumcircle) if such a triangle degenerates into a line segment.

7 References:

1. Kimberling, C. [1994]: Central points and central lines in the plane of a triangle, Math. Mag. 67, 163 – 187. 2. Kimberling, C. [1998]: Triangle centers and central triangles, Congressus Numerantium, vol. 140, 295 + xxv pages. 3. Morley, F. and Morley F.V. [1931]: Inversive Geometry, ...

8 4. Ramler, O.J. [1938]: The trilinear polar of the third Brocard point, Amer. Math. Monthly, 45, 540. 5. Weaver, J.H. [1938]: A generalization of the circles of Apollonius, Amer. Math. Monthly, 45, 17 – 21. 6. Yiu, P. [1999]: The uses of homogeneous barycentric coordinates in plane euclidean geometry, to appear in Int. J. Math. Edu. Sci. Technol.