The Apollonius Problem on the Excircles and Related Circles

Forum Geometricorum b Volume 6 (2006) xx–xx. bbb FORUM GEOM ISSN 1534-1178

Nikolaos Dergiades, Juan Carlos Salazar, and Paul Yiu

Abstract. We investigate two interesting special cases of the classical Apollo- nius problem, and then apply these to the tritangent of a triangle to ﬁnd pair of perspective (or homothetic) triangles. Some new triangle centers are constructed.

1. Introduction This paper is a study of the classical Apollonius problem on the excircles and related circles of a triangle. Given a triad of circles, the Apollonius problem asks for the construction of the circles tangent to each circle in the triad. Allowing both internal and external tangency, there are in general eight solutions. After a review of the case of the triad of excircles, we replace one of the excircles by another (possibly degenerate) circle associated to the triangle. In each case we enumerate all possibilities, give simple constructions and calculate the radii of the circles. In this process, a number of new triangle centers with simple coordinates are constructed. We adopt standard notations for a triangle, and work with homogeneous barycentric coordinates.

2. The Apollonius problem on the excircles

2.1. Let Γ=((Ia), (Ib), (Ic)) be the triad of excircles of triangle ABC. The sidelines of the triangle provide 3 of the 8 solutions of the Apollonius problem, each of them being tangent to all three excircles. The points of tangency of the excircles with the sidelines are as follows. See Figure 2.

BC CA AB

(Ia) Aa =(0:s − b : s − c) Ba =(−(s − b):0:s) Ca =(−(s − c):s :0) (Ib) Ab =(0:−(s − a):s) Bb =(s − a :0:s − c) Cb =(s : −(s − c):0) (Ic) Ac =(0:s : −(s − a)) Bc =(s :0:−(s − c)) Cc =(s − a : s − b :0)

Publication Date: Month day, 2006. Communicating Editor: 2 N. Dergiades et al

2.2. The nine-point circle (N), by the famous Feuerbach theorem, is tangent externally to each of the excircles, and provides a fourth solution. The points of tangency are

2 2 2 Fa =(−(b − c) s :(c + a) (s − c):(a + b) (s − b)), 2 2 2 Fb =((b + c) (s − c):−(c − a) s :(a + b) (s − a)), 2 2 2 Fc =((b + c) (s − b):(c + a) (s − a):−(a − b) a). These points of tangency form a triangle perspective with ABC at the outer Feuerbach point 1 (b + c)2 (c + a)2 (a + b)2 X = : : . 12 s − a s − b s − c

Fc Fc Fb N S

B C Fa

Fa Figure 1

2.3. A simple way to obtain solutions to the Apollonius problem from known solutions is to make use of the radical circle of the triad. Given a non-coaxial triad Γ, the radical circle is the unique circle orthogonal to all three circles. Its center, the radical center of the triad, is the intersection of the three pairwise radical axes, which has equal powers with respect to the circles in the triad. Inversion in the radical circle leaves the circles in the triad invariant, but preserves tangency.

1The labeling of triangle centers follows Kimberling’s Encyclopedia of Triangle Centers [3]. The Apollonius problem on the excircles and related circles 3

Therefore the inversive image of one solution to the Apollonius problem of the triad is another solution.

2.4. The radical circle of the excircles is the Spieker radical circle,√ which has 1 2 2 center the Spieker center S =(b + c : c + a : a + b) and radius 2 r + s . See [2, Theorem 4]. 2 a2yz+b2zx+c2xy+(x+y+z)((s−b)(s−c)x+(s−c)(s−a)y+(s−a)(s−b)z)=0.

2.5. Inversion of the nine-point circle in the Spieker radical circle gives the Apol- lonius circle which is tangent to the excircles internally. The points of tangency are the second intersections of the lines NFa, NFb, NFc with the corresponding excircles. These are

2 2 2 2 2 2 2 2 Fa =(−a (a(b + c)+(b + c )) :4b (c + a) s(s − c): 4c (a + b) s(s − b)), 2 2 2 2 2 2 2 2 Fb =(4a (b + c) s(s − c): −b (b(c + a)+(c + a )) :4c (a + b) s(s − a)), 2 2 2 2 2 2 2 2 Fc =(4a (b + c) s(s − b): 4b (c + a) s(s − a): −c (c(a + b)+(a + b )) ). These form a triangle perspective with ABC at the Apollonius point a2(b + c)2 b2(c + a)2 c2(a + b)2 X = : : . 181 s − a s − b s − c

r2+s2 The circle through them has center X970 and radius 4r . It can be constructed simply as the intersection of the lines NS and IX181.

2.6. Inversion of the line BC gives a circle tangent internally to (Ia) and externally to (Ib) and (Ic). It is the circle through S and the intersections of BC with the Spieker radical circle. The points of tangency are

2 2 2 (Ia): (−(b + c) (s − b)(s − c): c s(s − b): b s(s − c)), 2 2 2 (Ib): ((b + c) (s − a)(s − c): −c s(s − a): (bs + ca) ), 2 2 2 (Ic): ((b + c) (s − a)(s − c): (cs + ab) : −b s(s − a)). The equation of the circle is

2 2 2 (a yz + b zx + c xy)+(x + y + z)fa(x, y, z)=0, where s(a(b + c)+(b2 + c2)) f (x, y, z):=− x +(s − a)(s − c)y +(s − a)(s − b)z. a 2(b + c) a r2+s2 This circle has radius b+c · 4r .

2 The Spieker center S appears as X10 in [3] Here is an easy way to identify this radical center. The radical axis of (Ib) and (Ic) is the perpendicular from the midpoint of BC to the line IbIc.It has equation (b − c)x +(b + c)(y − z)=0; similarly for the other two radical axes. The coordinates of the radical center can be found by solving these three equations simultaneously. 4 N. Dergiades et al

B C

Figure 2

2.7. Similarly, inversion of the lines CA and AB gives two more circles. The points of tangency of has perspector s − a s − b s − c : : . a2 b2 c2 The centers of the circles also form a triangle with perspector 1 a5 + a4(b + c) − a3(b − c)2 − a2(b + c)(b2 + c2) − 2abc(b2 + bc + c2) − 2b2c2(b + c) : ···: ···) .

3. Two excircles and a vertex 3.1. The Apollonius problem for two circles with disjoint closures and a point outside their closures has in general 4 solutions. Nik’s description. The Apollonius problem on the excircles and related circles 5

3.2. In the special case of the triad Γa := (A, (Ib), (Ic)), two of these circles are the sidelines AC and AB. The remaining two circles can be easily constructed as follows. The radical axis of (Ib) and (Ic) is the A-angle bisector of the medial triangle (which contains the Spieker center S). Let X be the intersection with the A-altude of triangle ABC. Then the circle with diameter AX is tangent to both (Ib) and (Ic). The reﬂection of this circle in the line IbIc is the fourth one. The line joining the centers of these two circles passes through the midpoint of IS, which is the triangle center

X1125 =(2a + b + c : a +2b + c : a + b +2c).

Ib X

I S

B C Ac Ab

Figure 3.

3 The radical axis of (Ib) and (Ic) and the A-altitude intersect at X =(−(b + 2 2 2 c) : SC : SB). The circle with diameter AX has center = (a − 2(b + c) : SC : a A R SB) and radius = 4 tan 2 = 2 (1 − cos A). Its equation

1 a2yz+b2zx+c2xy− (x+y+z)((a2−b2+2bc+3c2)y+(a2+3b2+2bc−c2)z)=0. 4 Points of tangency with the excircles are

2 (Ib): ((b + c) : −s(s − c):(s − a)(s − c)) 2 (Ic): ((b + c) :(s − a)(s − b):−s(s − b))

3See Footnote 2. 6 N. Dergiades et al

The line joining these two points: (s − b)(s − c)x +4(b + c)(cy + bz)=0. Bound a triangle with perspector b + c : ···: ··· . (s − a)(as +2bc)

3.3. The reﬂection of the circle in the line IbIc 2 2 2 2 2 has center (2s (s − a) + a SA : b SB : c SC ). This means that the circle is also tangent to the circumcircle.

s(s − a) a2yz + b2zx + c2xy − (x + y + z)(c2y + b2z)=0. bc

2 2 2 (Ib): ((b + c) s(s − a):−b (s − a)(s − c):c s(s − c)) 2 2 2 (Ic): ((b + c) s(s − a):b s(s − b):−c (s − a)(s − b)) x y z The line joining these is a(s−a) + b(s−b) + c(s−c) =0. a(b−c)(s−a) perspector: a3−a2(b+c)−a(b−c)2+(b−c)3 .

4. The triangle bounded by the polars of the vertices with respect to the excircles 4.1. The triangle UVW and its perspector. Consider the triangle bounded by the polars of the vertices of ABC with respect to the corresponding excircles. The polar of A with respect to the excircle (Ia) is the line BaCa; similarly for the other two polars.

Bc A

Ic Cc Bb H Ab

Ac B Aa C W

Ca V Ia

Figure 4 The Apollonius problem on the excircles and related circles 7

Lemma 1. The polars of the vertices of ABC with respect to the corresponding excircles bound a triangle with vertices

U =(−a(b + c):SC : SB),

V =(SC : −b(c + a):SA),

W =(SB : SA : −c(a + b)).

Proof. The polar of A with respect to the excircle (Ia) is the line BaCa, whose barycentric equation is xyz −(s − b)0s =0, −(s − c) s 0 or sx +(s − c)y +(s − b)z =0.

Similarly, the polars CbAb and AcBc have equations (s − c)x + sy +(s − a)z =0, (s − b)x +(s − a)y + sz =0.

These intersect at the point U =(−a(2s − a):ab − 2s(s − c):ca − 2s(s − b)) =(−2a(b + c):a2 + b2 − c2 : c2 + a2 − b2)

=(−a(b + c):SC : SB).

The coordinates of V and W can be obtained from these by cyclic permutations of a, b, c.

Corollary 2. Triangles UVW and ABC are perspective at the orthocenter H.

4.2. The circumcircle of UVW.

Proposition 3. The triangle UVW has circumcenter H and circumradius 2R + r.

Proof. Since H, B, V are collinear, HV is perpendicular to CA. Similarly, HW is perpendicular to AB. Since VW makes equal angles with CAand AB, it makes equal angles with HV and HW. This means HV = HW. For the same reason, HU = HV , and H is the circumcenter of UVW. Applying the law of sines to triangle AUBc,wehavewehave

sin 180◦−C C AU = AB · 2 =(s − b)cot = r . c C 2 a sin 2

The circumradius of UVW is HU = HA + AU =2R cos A + ra =2R + r, as a routine calculation shows. 8 N. Dergiades et al

4.3. The homothety of UVW and the intouch triangle. Proposition 4. The triangles UVW and DEF are homothetic at the point b + c c + a a + b J = : : . b + c − a c + a − b a + b − c (1) 2R+r The ratio of homothety is − r . Proof. The homothety follows easily from the parallelism of VW and EF, and of WU, FD, and UV, DE. The homothetic center is also the common point J of the lines DU, EV , and FW. These lines have equations

(b + c)(b + c − a)x +(b − c)(c + a − b)y − (b + c)(a + b − c)z =0, −(c + a)(b + c − a)x +(c − a)(c + a − b)y +(c + a)(a + b − c)z =0, (a + b)(b + c − a)x − (a + b)(c + a − b)y +(a − b)(a + b − c)z =0. It follows that (b + c − x)x :(c + a − b)y :(a + b − c)z c − ac+ a c + a −(c + a) −(c + a) c − a = : : −(a + b) a − b a − ba+ b a + b −(a + b) =2a(b + c):2a(c + a):2a(a + b) =b + c : c + a : a + b. The coordinates of the homothetic center J are therefore as in (1) above. Since the triangles UVF and DEF have circumcircles H(2R + r) and I(r), 2R+r the ratio of homothety is − r . The homothetic center J divides IH in the ratio IJ : JH = r :2R + r.

Remark. The triangle center J appears as X226 in Kimberling’s list [3].

5. Two excircles and a sideline 5.1. We begin with a simple construction of a special case of the classical Apol- lonius problem. Given two circles O(r), O(r) and an external tangent L,to construct a circle O1(r1) tangent to the circles and the line, with point of tangency X A A (O) (O) L between and , those of , and√ . See Figure 5. A simple calculation√ √ √ shows that AX =2 r1r and XA =2 r1r , so that AX : XA = r : r . The radius of the circle is 1 AA 2 r1 = √ √ . 4 r + r From this we design the following construction. Construction 1. On the line L, choose two points P and Q be points on opposite sides of A such that PA = r and AQ = r. Construct the circle with diameter PQ OA F O F L to intersect the line at such that and are on opposite sides√ √ of . The intersection of OF with L is the point X satisfying AX : XA = r : r. The Apollonius problem on the excircles and related circles 9

L A X A Figure 5.

Let M be the midpoint of AX. The perpendiculars to OM at M, and to L at X intersect at the center O1 of the required circle. See Figure 6.

L P A M X Q A

Figure 6.

For a construction in the case when L is not necessarily tangent of (O) and (O), see [1, Problem 471].

5.2. We apply the above construction to the excircles of a triangle ABC. Consider the circle O1(X) tangent to the excircles Ib(rb) and Ic(rc), and to the line BC at a point X between Ac and Ab. See Figure 7.

Lemma 5. The point X has coordinates √ √ √ √ (0 : s s − c − (s − a) s − b : s s − b − (s − a) s − c).

(O ) (I ) (I ) BC X A Proof. If 1 is the circle√ tangent√ to √b , c ,√ and to at between c and Ab, then AcX : XAb = rc : rb = s − b : s − c. Note that AcAb = b + c, 10 N. Dergiades et al

Bb O1

B C Ac X Ab Figure 7 so that

BX =AcX − AcB √ s − b =√ √ · (b + c) − (s − a) s − b + s − c √ √ s s − b − (s − a) s − c = √ √ . s − b + s − c Similarly √ √ s s − c − (s − a) s − b XC = √ √ . s − b + s − c It follows that the point X has coordinates given above.

Similarly, there are circles O2(Y ) and O3(Z) tangent to CA at Y and to AB at Z respectively, each also tangent to a pair of excircles. Their coordinates can be written down from those of X by cyclic permutations of a, b, c. 5.3. Perspectivity of XY Z and UVW. Theorem 6. Triangles XY Z and UVW are perspective at a point with coordinates S S a(b + c) S S b(c + a) S S c(a + b) √ B + √ C − √ : √ C + √ A − √ : √ A + √ B − √ . s − c s − b s − a s − a s − c s − b s − b s − a s − c Proof. With the coordinates of X and U from Lemmas 1 and 2, the line XU has equation xy z −a(b + c) √ SC √ √ SB √ =0. 0 s s − c − (s − a) s − bss − b − (s − a) s − c The Apollonius problem on the excircles and related circles 11

Since the coefﬁcient of x is √ √ (s(SB + SC ) − aSB) s − b − (s(SB + SC ) − aSC ) s − c √ √ =a((as − SB) s − b − (as − SC ) s − c) √ √ =a(b + c)((s − c) s − b − (s − b) s − c). From this, we easily simplify the above equation as √ √ ((s − c) s − b − (s − b) s − c)x √ √ √ √ +(s s − b − (s − a) s − c)y +((s − a) s − b − s s − c)z =0. √ √ √ With u = s − a, v = s − b, and w = s − c, we rewrite this as 2 2 2 2 2 2 2 2 −vw(v −w)x+(v(u +v +w )−u w)y +(u v −w(u +v +w ))z =0. (2)

Ic C c Y B Z b P

A B X A C A c a b W

V Ia

Figure 8

Similarly the equations of the lines VY, WZ are

(v2w − u(u2 + v2 + w2))x − wu(w − u)y +(w(u2 + v2 + w2) − uv2)z =0, (3) (u(u2 + v2 + w2) − vw2)x +(w2u − v(u2 + v2 + w2))y − uv(u − v)z =0. (4) 12 N. Dergiades et al

It is clear that the sum of the coefﬁcients of x (respectively y and z) in (2), (3) and (4) is zero. The system of equations therefore has a nontrivial solution. Solving them, we obtain the coordinates of the common point of the lines XU, YV, ZW as

x : y : z =uv(v2(u2 + v2 + w2) − w2u2)+wu(w2(u2 + v2 + w2) − u2v2) − vw(v2 + w2)(2u2 + v2 + w2) :vw(w2(u2 + v2 + w2) − u2v2)+uv(u2((u2 + v2 + w2) − v2w2) − wu(w2 + u2)(u2 +2v2 + w2) :wu(u2(u2 + v2 + w2) − v2w2)+vw(v2((u2 + v2 + w2) − w2u2) − uv(u2 + v2)(u2 + v2 +2w2)

(s − b)s − (s − c)(s − a) (s − c)s − (s − a)(s − b) a(b + c) = + − w v u (s − c)s − (s − a)(s − b) (s − a)s − (s − b)(s − c) b(c + a) : + − u w v (s − a)s − (s − b)(s − c) (s − b)s − (s − c)(s − a) c(a + b) : + − v u w S S a(b + c) S S b(c + a) S S c(a + b) = B + C − : C + A − : A + B − . w v u u w v v u w

The triangle center constructed in Theorem 3 above does not appear in [3].

6. Two excircles and the incircle

6.1. Given three circles Oi(ri), i =1, 2, 3, on one side of a line L, tangent to the line, we construct a circle O(r), tangent to each of these three circles externally. For i =1, 2, 3, let the circle Oi(ri) touch L at Si and O(r) at Ti. If the line S1T1 meets the circle (O) again at T , then the tangent to (O) at T is a line L parallel to L. Hence, T , T2, S2 are collinear; so are T , T3, S3. Since the line T2T3 is antiparallel to L with respect to the lines TT2 and TT3, it is also antiparallel to L with respect to the lines TS2 and TS3, and the points T2, T3, S3, S2 are concyclic. From TT2 · TS2 = TT3 · TS3, we conclude that the point T lies on the radical axis of the circles O2(r2) and O3(r3), which is the perpendicular from the midpoint of S1S2 to the line O1O2. For the same reason, it also lies on the radical axis of the circles O3(r3) and O1(r1), which is the perpendicular from the midpoint of S1S3 to the line O1O3. Hence T is the radical center of the three given circles Oi(ri), i =1, 2, 3. From this, the required circle (O) can be constructed as follows. The Apollonius problem on the excircles and related circles 13

T L

O T3 T 2 O3

O2 T1

O1 L S2 S1 S3 Figure 9

Construction 2. Construct the perpendicular from the midpoint of S1S2 to O1O2, and from the midpoint of S1S3 to O1O3. Let T be the intersection of these two perpendiculars. For i =1, 2, 3, let Ti be the intersection of the line TSi with the circle (Oi). The required circle (O) is the one through T1, T2, T3. See Figure 9. 6.2. Circles tangent to the incircle and two excircles. We apply Construction 2 to obtain the circle tangent to the incircle (I) and the excircles (Ib) and (Ic). Let the incircle (I) touch the sides BC, CA, AB at D, E, F respectively.

Proposition 7. The radical center of (I), (Ib), (Ic) is the point

Ja =(b + c : c − a : b − a). This is also the midpoint of the segment DU.

Proof. The radical axis of (I) and (Ib) is the line joining the midpoints of the segments DAb and FCb. These midpoints have coordinates (0 : a − c : a + c) and (c + a : c − a :0). This line has equation −(c − a)x +(c + a)y +(c − a)z =0.

Similarly, the radical axis of (I) and (Ic) is the line (a − b)x − (a − b)y +(a + b)z =0.

The radical center Ja of the three circles is the intersection of these two radical axes. It coordinates are as given above. Since D =(0:a + b − c : c + a − b) and U =(−2a(b + c):a2 + b2 − c2 : c2 +a2 −b2), the midpoint of DU is the point with absolute barycentric coordinates 1 (0,a+ b − c, c + a − b) (−2a(b + c),a2 + b2 − c2,c2 + a2 − b2) J = + a 2 2a −2a(b + c − a) (b + c, c − a, −a + b) = . 2(b + c − a)

This shows that Ja is the midpoint of DU.

The lines JaD, JaAb and JaAc intersect the circles (I), (Ib) and (Ic) respectively again at 14 N. Dergiades et al

2 2 2 A1 =((b + c) (s − b)(s − c):c (s − a)(s − c):b (s − a)(s − b)), 2 2 2 Ab =((b + c) s(s − c):−(ab − c(s − a)) : b s(s − a)), 2 2 2 Ac =((b + c) s(s − b):c s(s − a):−(ca − b(s − a)) ).

The circle through these points is the one tangent to (I), (Ib), and (Ic). See Figure (s−a)2+r2 a · a 10. It has radius b+c 4ra .

A Ja

Ab A c O Ic a

Cc A1

Bb I

Ac B D C Ab

Figure 10.

In the same way, we have a circle (Ob) tangent to (I), (Ic), (Ia) respectively at B1, Bc, Ba, and passing through the radical center Jb of these three circles, and another circle (Oc) tangent to (I), (Ia), (Ib) respectively at C1, Ca, Cb, passing through the radical center Jc of the circles. Jb and Jc are respectively the midpoints of the segments EV and FW. The coordinates of Jb, Jc, B1, C1 are as follows.

Jb =(c − b : c + a : a − b),

Jc =(b − c : a − c : a + b); 2 2 2 B1 =(a (s − b)(s − c):(c + a) (s − c)(s − a):c (s − a)(s − b)), 2 2 2 C1 =(a (s − b)(s − c):b (s − c)(s − a):(a + b) (s − a)(s − b)).

Proposition 8. The triangle JaJbJc is the image of the intouch triangle under the R homothety h(J, − r ). The Apollonius problem on the excircles and related circles 15

A Ja

O Ic a

Cc A1

Bb J I

B C Ac Aa Ab

Jb Ba

Figure 11.

Proof. Since UVW and DEF are homothetic at J, and Ja, Jb, Jc are the midpoints of DU, EV , FW respectively, it is clear that JaJbJc and DEF are also 1 homothetic at the same J. Note that JbJc = 2 (VW − EF). The circumradius of 1 JaJbJc is 2 ((2R + r) − r)=R. The ratio of homothety of JaJbJc and DEF is − R r.

Corollary 9. J is the radical center of the circles (O1), (O2), (O3).

R R Proof. Note that JJa · JA1 = r · DJ · JA1. This is r times the power of J with respect to the incircle. The same is true for JJb · JB1 and JJc · JC1. This shows that J is the radical center of the circles (O1), (O1), (O1).

Since the incircle (I) is the inner Apollonius circle and the circumcircle (Oi), i =1, 2, 3, it follows that JaJbJc is the outer Apollonius circle to the same three circles. See Figure 12. The center O of the circle JaJbJc is the midpoint between the circumcenters of DEF and UVW, namely, the midpoint of IH. It is the triangle center X946 in [3]. 16 N. Dergiades et al

A Ja

O Ic a

Cc A1

I Bb O Q H C1

B1 Jc

B C Ac Aa Ab

Jb Ba

Figure 12.

Proposition 10. A1B1C1 is perspective with ABC at the point 1 1 1 Q = : : , a2(s − a) b2(s − b) c2(s − c) which is the isotomic conjugate of the insimilicenter of the circumcircle and the incircle.

This is clear from the coordinates of A1, B1, C1. The perspector Q is the isotomic conjugate of the insimilicenter of the circumcircle and the incircle. It is not in the current listing in [3].

6.3. There is a circle tangent to incircle and the B- and C-excircles. It is the inversion of the nine-point circle in the radical center Ja. This circle has equation. bc − s(s − a) ca − s(s − b) ab − s(s − c) a2yz+b2ax+c2xy+(s−a)(x+y+z) − x + y + z =0. a b c

It touches the circles at The Apollonius problem on the excircles and related circles 17

(I): (a2(a(b + c) − (b2 + c2))2 :4b2(c − a)2(s − a)(s − b):4c2(a − b)2(s − a)(s − c)), 2 2 2 2 2 2 2 2 (Ib): (4a (b + c) (s − a)(s − c):−4b (c − a) s(s − a):c (c(a − b) − (a + b )) ), 2 2 2 2 2 2 2 2 (Ic): (4a (b + c) (s − a)(s − b):b (b(−c − a) − (c + a )) : −4c (a − b) s(s − a)).

The radius of the circle is

(s − a)3 + s(s − b)(s − c) (s − a)2 + r2 = a . 4∆ 4ra The points of tangency with the incircle have perspector a2(b − c)2(s − a):···: ··· .

7. Two excircles and the circumcircle 7.1. The circle in §through the vertex A is tangent to the circumcircle. It has 2 2 2 2 2 center (2s (s − a) + a SA : b SB : c SC ).

s(s − a) a2yz + b2zx + c2xy − (x + y + z)(c2y + b2z)=0. bc

2 (O): (−a SBC :2b(b + c)(s − b)(s − c)SC :2c(b + c)(s − b)(s − c)SB) 2 2 2 (Ib): ((b + c) s(s − a):−b (s − a)(s − c):c s(s − c)) 2 2 2 (Ic): ((b + c) s(s − a):b s(s − b):−c (s − a)(s − b))

a2yz + b2zx + c2xy s(s − a) (b + c)2(s − b)2(s − c)2 − (x + y + z)( x + S y + S z)=0. 4∆2 + a2bc a2 BB CC

Ra3+∆(a2+(b+c)2) A radius = 4a(Ra+∆) tan 2 .

7.2. The radical center of ((O), (Ib), (Ic)). The radical center of the triad is the intersection of the radical axes of (Ib) and (Ic) with the circumcircle. This is the point

(−(b + c)(a2 +(b + c)2):b(a2 + b2 − c2):c(c2 + a2 − b2)).

The point of tangency with the circumcircle is a2 b c − : : . (b + c)(s − b)(s − c) c2 + a2 − b2 a2 + b2 − c2 18 N. Dergiades et al

References [1] F. G.-M., Exercices de G´eom´etrie, 6th ed., 1920; Gabay reprint, Paris, 1991. [2] D. Grinberg and P. Yiu, The Apollonius circle as a Tucker circle, Forum Geom., 2 (2002) 175– 182. [3] C. Kimberling, Encyclopedia of Triangle Centers, available at http://faculty.evansville.edu/ck6/encyclopedia/ETC.html.

Nikolaos Dergiades: I. Zanna 27, Thessaloniki 54643, Greece E-mail address: [email protected]

Juan Carlos Salazar: Calle Matur´ın NoC 19, Urb. Mendoza, Puerto Ordaz 8015, Estado Bol´ıvar, Venezuela E-mail address: [email protected]

Paul Yiu: Department of Mathematical Sciences, Florida Atlantic University, Boca Raton, Florida 33431-0991, USA E-mail address: [email protected]