A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED WITH A TRIANGLE

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1 Global Journal of Advanced Research on Classical and Modern Geometries ISSN: , Vol.6, (207), Issue, pp A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED WITH A TRIANGLE PAUL YIU AND XIAO-DONG ZHANG ABSTRACT. For a given triangle we study the triad of parabolas each with one vertex as focus and its opposite side as directrix. Among the 6 angle bisectors and the perpendicular bisectors of the 3 sides, we show that each of these parabolas is tangent to 4 of the angle bisectors and 2 perpendicular bisectors. The 8 points of tangency fall on another set of 6 lines each joining the antipode (on the circumcircle) of a vertex to the two remaining vertices. Some perspectivity of triangles will also be exhibited. The 6 points of tangency on the external bisectors lie on a conic with a simple barycentric equation with respect to the excentral triangle.. INTRODUCTION In this paper we study a triad of parabolas most naturally associated with a triangle. Given a triangle ABC, we consider the parabolas P a with focus A and directrix BC, P b with focus B and directrix CA, and P c with focus C and directrix AB, and prove some interesting results about these parabolas. In??, we redefine these parabolas in terms of their tangents, and locate the points of tangency. Specifically we show that each of P a, P b, P c is tangent to 4 angle bisectors and 2 perpendicular bisectors of ABC, and establish some perspectivity of triangles with vertices among the points of tangency. In the final section, we show that the 6 points of tangency with the external bisectors lie on a conic. Notations. For triangle ABC, we denote respectively by a, b, c the sides BC, CA, AB, a, b, c the perpendicular bisectors of a, b, c, L a, L b, L c the bisectors of angles A, B, C, L a, L b, L c the external bisectors of angles A, B, C; L (P) the perpendicular to the line L at (or from) the point P. Apart from basic elementary results on parabolas we shall also make use of homogeneous barycentric coordinates. A basic reference is [?]. Let a, b, c denote the lengths of the sides BC, CA, AB respectively. We shall make use of Conway s notations S A := b2 + c 2 a 2 2, S B := c2 + a 2 b Mathematics Subject Classification. 5M04, 5N20. Key words and phrases. Parabolas, barycentric coordinates, collinarity. 45, S C := a2 + b 2 c 2. () 2

2 46 PAUL YIU AND XIAO-DONG ZHANG The use of these symbols greatly simplifies expressions for coordinates involving only even powers of a, b, c. These satisfy basic relations like S B + S C = a 2, S C + S A = b 2, S A + S B = c 2 ; (2) S BC + S CA + S AB = S 2, (3) where S AB := S A S B etc and S stands for twice the area of triangle ABC (see [?, 3.4.]. In fact, S A = S cot A etc. For examples, the homogeneous barycentric coordinates of the circumcenter O are (a 2 S A : b 2 S B : c 2 S C ), the orthocenter and the symmedian point are the points H = (S BC : S CA : S AB ) and K = (S B + S C : S C + S A : S A + S B ) respectively. For later use we record the coordinates of the antipodes A, B, C of A, B, C on the circumcircle: A = ( S BC : b 2 S B : c 2 S C ), B = (a 2 S A : S CA : c 2 S C ), (4) C = (a 2 S A : b 2 S B : S AB ). Instead of using homogeneous linear equations to represent lines, we shall use line coordinates. Thus, the sidelines are a = [ : 0 : 0], b = [0 : : 0], c = [0 : 0 : ], and the Brocard axis is OK = [b 2 c 2 (b 2 c 2 ) : c 2 a 2 (c 2 a 2 ) : a 2 b 2 (a 2 b 2 )]. (5) 2. THREE PAIRS OF POINTS ON THE PERPENDICULAR BISECTORS Consider the following points on the perpendicular bisectors of the sides of triangle ABC. P a P b P c a X b := a b (C) X c := a c (B) b Y a := b a (C) Y c := b c (A) c Z a := c a (B) Z b := c b (A) Proposition. The points Y a, Z a are on the parabola P a ; similarly, Z b, X b are on P b, and X c, Y c are on P c. Proof. Since Y a C = Y a A and Y a C a, Y a is a point on the parabola P a ; so is Z a. Proposition 2. The points X b and X c are inverse in the circumcircle; so are Y c and Y a, Z a and Z b. Proof. Note that in Figure?? triangles OCX b and ABC are oppositely similar, since the directed angles (see [?, 6 9]) (OC, OX b ) = (OC, OB) = (AC, AB) = (AB, AC), 2 (X b O, X b C) = (X b O, BC)+(BC, X b C) = π 2 +(BC, X bc) By the law of sines, = (X b C, AC)+(BC, X b C) = (BC, AC) = (CA, CB). OX b OC = AC AB = b c = OX b = b c R

3 A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED WITH A TRIANGLE 47 where R is the circumradius of triangle ABC. The distances from O to these six points are summarized below. OY a = a c R OZ a = a b R OX b = b c R OZ b = b a R OX c = c b R OY c = c a R From these, it is clear that X b and X c are inverse in the circumcircle; so are Y c and Y a, Z a and Z b. A Y a P c P a P b Z b C Z a F E Y c B O B D X c C A X b FIGURE. Proposition 3. The homogeneous barycentric coordinates of X b, X c, Y c, Y a, Z a, Z b on a, b, c are as follows. a X b = ( a 2 S C : a 2 b 2 : S 2 + S BC ) X c = ( a 2 S B : S 2 + S BC : c 2 a 2 ) b Y c = (S 2 + S CA : b 2 S A : b 2 c 2 ) Y a = (a 2 b 2 : b 2 S C : S 2 + S CA ) c Z a = (c 2 a 2 : S 2 + S AB : c 2 S B ) Z b = (S 2 + S AB : b 2 c 2 : c 2 S A ) Proof. These follow from two sets of line coordinates. () The perpendiculars from the vertices to the sidelines: a (B) = [S B : 0 : S B + S C ] a (C) = [S C : S B + S C : 0] b (A) = [0 : S A : S C + S A ] b (C) = [S C + S A : S C : 0] c (A) = [0 : S A + S B : S A ] c (B) = [S A + S B : 0 : S B ] (2) The perpendicular bisectors of the sides: a = [S B S C : (S B + S C ) : S B + S C ] = [b 2 c 2 : a 2 : a 2 ], b = [S C + S A : S C S A : (S C + S A )] = [ b 2 : c 2 a 2 : b 2 ], c = [ (S A + S B ) : S A + S B : S A S B ] = [c 2 : c 2 : a 2 b 2 ]. For these, it is enough to verify that [S B S C : (S B + S C ) : S B + S C ] contains the midpoint (0 : : ) of BC and the circumcenter O = (S A (S B + S C ) : S B (S C + S A ) : S C (S A + S B )). (6) (7)

4 48 PAUL YIU AND XIAO-DONG ZHANG The first is clear. For the circumcenter, it follows from (S B S C )S A (S B + S C ) (S B + S C )S B (S C + S A )+(S B + S C )S C (S A + S B ) = (S B + S C )(S A (S B S C ) S B (S C + S A )+S C (S A + S B )) = 0. From the coordinates given in (??) and (??), we compute X b = [S B S C : (S B + S C ) : S B + S C ] [S C + S A : S C : 0] = ( S C (S B + S C ) : (S B + S C )(S C + S A ) : (S B S C )S C +(S B + S C )(S C + S A )) = ( S C (S B + S C ) : (S B + S C )(S C + S A ) : 2S BC + S AB + S CA ) = ( S C (S B + S C ) : (S B + S C )(S C + S A ) : S 2 + S BC ) = ( a 2 S C : a 2 b 2 : S 2 + S BC ). The others can be computed similarly. Proposition 4. The lines BX c and CX b intersect at the antipode A of A on the circumcircle. Proof. The intersection of BX c and CX b is the point (0 : : 0) ( a 2 S B : S 2 + S BC : c 2 a 2 ) (0 : 0 : ) ( a 2 S C : a 2 b 2 : S 2 + S BC ) = [c 2 a 2 : 0 : a 2 S B ] [ a 2 b 2 : a 2 S C : 0] = [c 2 : 0 : S B ] [b 2 : S C : 0] = ( S BC : b 2 S B : c 2 S C ). This is A given in (??)). Similarly, CY a AY c = B and AZ b BZ a = C. We shall show in Proposition?? below that b and c are tangent to P a ; similarly for the other two parabolas. 3. THE PARABOLAS P a ETC EACH DEFINED BY FIVE TANGENTS Lemma 5. (a) The reflection of the focus of a parabola in a tangent is a point on the directrix. (b) The circumcircle of the triangle bounded by three tangents to a parabola passes through the focus of the parabola. Theorem 6 ( [?]). The parabola tangent to the L b, L c, L b and L c has focus at vertex A and directrix the line a. It is the parabola P a. Proof. Let I, I b, I c be the incenter and the B-, C-excenters of triangle ABC. The bisectors L b, L b and L c bound the triangle IBI c. The bisectors L c, L c, and L b bound the triangle ICI b. Apart from I, the circumcircles of these triangles intersect at A. This is the focus of the parabola by Lemma??(b). The reflections of A in the L b and L c are points on a. By Lemma??(a), the line a is the directrix. This shows that the parabola tangent to the four bisectors of angles B and C is the parabola P a. The dual conic of P a is a line conic P a containing L b, L b, L c, L c, which have line coordinates [c : 0 : a], [c : 0 : a], [ b : a : 0], [b : a : 0].

5 A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED WITH A TRIANGLE 49 Since P a is a parabola, it is tangent to the line of infinity. Therefore, P a also contains [ : : ]. From these five line coordinates we find the barycentric equation of P a : a 2 x 2 b 2 y 2 +(b 2 + c 2 a 2 )yz c 2 z 2 = 0, (8) Proposition 7. The perpendicular bisectors b and c are tangent to the parabola P a. Proof. Rewriting (??) as a 2 (x 2 yz) (y z)(b 2 y c 2 z) = 0, we have, upon substitution of the line coordinates of b = [ b 2 : c 2 a 2 : b 2 ] given in (??), a 2 (b 4 (c 2 a 2 )b 2 ) (c 2 a 2 b 2 )(b 2 (c 2 a 2 ) c 2 b 2 ) = a 2 b 2 (b 2 c 2 + a 2 )+a 2 b 2 (c 2 a 2 b 2 ) = 0. Therefore the dual conic Pa contains b. A similar calculation shows that it also contains c. These two perpendicular bisectors are tangent to the parabola P a. Corollary 8. The parabola P a is tangent to b and c at Y a and Z a respectively. Figure?? shows the 6 points of tangency of the parabola P a with b, c, L b, L b, L c, L c. The counterparts of Theorem??, Proposition??, and Corollary?? hold for the parabolas P b and P c. I ca I ba A Ya Ica Za I ba B O C FIGURE 2. Altogether, the triad of parabolas P a, P b, P c, together with 6 angle bisectors and the 3 perpendicular bisectors, accounts for 8 points of tangency shown in Table below. Table. Points of tangency of the triad of parabolas L a L b L c L a L b L c a b c P a I ba I ca I ba I ca Y a Z a P b I ab I cb I ab I cb X b Z b P c I ac I bc I ac I bc X c Y c

6 50 PAUL YIU AND XIAO-DONG ZHANG 4. 8 POINTS OF TANGENCY ON 6 LINES In this section we find another interesting arrangement of these 8 points of tangency in Table. From the equation of Pa, we find that of P a. The matrix of Pa is a Ma := 0 b 2 S A. 0 S A c 2 This has adjoint matrix S M a = 0 c 2 a 2 a 2 S A 0 a 2 S A a 2 b 2 representing the parabola P a. The polar of B with respect to P a is the line ( 0 0 ) Ma = a 2 [0 c 2 S A ]. (9) It contains the points of tangency of P a withl b andl b, namely, the points I ba and I ba. We can simply find I ba as[c : 0 : a] [0 : c 2 : S A ] and I ba as[c : 0 : a] [0 : c 2 : S A ]; similarly for the remaining points of tangency of the three parabolas with the angle bisectors. We summarize the coordinates in the proposition below. Proposition 9. The coordinates of the points of tangency of P a, P b, P c with the angle bisectors are as follows. P a P b P c L a I ab = ( S B : bc : c 2 ) I ac = ( S C : b 2 : bc) L b I ba = (ca : S A : c 2 ) I bc = (a 2 : S C : ca) L c I ca = (ab : b 2 : S A ) I cb = (a 2 : ab : S B ) L a I ab = (S B : bc : c 2 ) I ac = (S C : b 2 : bc) L b L c I ba = (ca : S A : c 2 ) I ca = (ab : b 2 : S A ) I cb = ( a2 : ab : S B ) I bc = ( a2 : S C : ca) I ac A I ab O B I ab X b C Xc A Iac FIGURE 3. Note that the polar of B given in (??) is the same as c (A) given (??). As such, it contains the antipode B of B on the circumcircle. It also contains the vertex A and the point of tangency Y c of P c with b. In other words, the line AB contains the three points of tangency I ba, I ba, and Y c.

7 A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED WITH A TRIANGLE 5 Theorem 0. The 8 points of tangency of the parabolas P a, P b, P c with the 6 angle bisectors and the 3 perpendicular bisectors fall on 6 lines, each joining the antipode (on the circumcircle) of a vertex of triangle ABC to the remaining two vertices. AB I ba I ba Y c AC I ca I ca Z b BC I cb I cb Z a BA I ab I ab X c CA I ac I ac X b CB I bc I bc Y a Figure?? shows the two lines BA and CA containing I ab, I ab, X c, and I ac, I ac, X b respectively. 5. SOME PERSPECTIVITIES 5.. Triangles with vertices among the tangency points on the perpendicular bisectors. Proposition. The triangle X Y Z bounded by the lines Y a Z a, Z b X b, X c Y c is perspective with ABC. The perspectrix is the Lemoine axis, and the perspector is the Kiepert perspector K ( π 2 ω), where ω is the Brocard angle of triangle ABC. P c P a A Y Ya Yc P b Z b Z Za F E X(262) O B D C Xc X X b FIGURE 4. Proof. The lines Y a Z a, Z b X b, X c Y c are the polars of O with respect to the parabolas P a, P b, P c respectively. They have line coordinates [ S A : c 2 : b 2 ], [c 2 : S B : a 2 ], [b 2 : a 2 : S C ]. These lines intersect a, b, c respectively at (0 : b 2 : c 2 ), ( a 2 : 0 : c 2 ), (a 2 : b 2 : 0),

8 52 PAUL YIU AND XIAO-DONG ZHANG [ ] which are collinear on the Lemoine axis a 2 : b 2 : c 2. This shows that the triangle bounded by the lines is perspective with ABC with the Lemoine axis as perspectrix. The vertices of the triangle are X = [S A + S B : S B : S B + S C ] [S C + S A : S B + S C : S C ] = ( (S BB + S BC + S CC ) : S 2 +(S A + S B + S C )S C : S 2 +(S A + S B + S C )S B ), Y = (S 2 +(S A + S B + S C )S C : (S CC + S CA + S AA ) : S 2 +(S A + S B + S C )S A ), Z = (S 2 +(S A + S B + S C )S B : S 2 +(S A + S B + S C )S A : (S AA + S AB + S BB )). From these, the perspector with ABC is the point ( S 2 : +(S A + S B + S C )S A S 2 +(S A + S B + S C )S B : S 2 +(S A + S B + S C )S C ). Since S A + S B + S C = S ω for the Brocard angle ω of triangle ABC, S 2 S 2 S ω = S tan ω, the perspector is the same as ( ) S A + S tan ω : S B + S tan ω :, S C + S tan ω S A +S B +S C = the Kiepert perspector K ( π 2 ω). Proposition 2 ( [?]). The triangle X Y Z bounded by the Y a Z a, Z a X a, X a Y a is perspective to the triangle of the vertices of P a, P b, P c. The perspector is the point ((S A + S B + S C ) 2 (S BB + 4S BC + S CC ) (S BB + S BC + S CC ) 2 : : ), and the perspectrix is the line x S CA + S AB S BC + y S AB + S BC S CA + z S BC + S CA S AB = 0. Remarks. () Proposition?? appears in [?] under the entry X(262). (2) The line Y a Z a = [ S A : c 2 : b 2 ] is also the directrix of the A-Artzt parabola, which is tangent to b and c at C and B. It has focus (2S A : b 2 : c 2 ), the second intersection of AK with the Brocard circle (of diameter OK). Similar results hold for the B- and C-Artzt parabolas. Therefore, the directrices of the A-, B-, C-Artzt parabolas bound a triangle with perspective with ABC at X(262). (3) The perspector in Proposition?? is the triangle center X(9748) in [?]. Proposition 3. The points X 2 := Y c Z a Y a Z b, Y 2 := Z a X b Z b X c, Z 2 := X b Y c X c Y a are collinear on the Brocard axis, and the lines AX 2, BY 2, CZ 2 are concurrent at the Tarry point on the circumcircle. Proof. The triangles X b Y c Z a and X c Y a Z b are clearly perspective at the circumcenter O. They are also axis-perspective, i.e. the three intersections X 2 = Y c Z a Y a Z b, Y 2 = Z a X b Z b X c, Z 2 = X b Y c X c Y a are collinear. Making use of the

9 A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED WITH A TRIANGLE 53 coordinates given in Proposition??, we have X 2 = (a 2 (S 3 A + a2 (2S AA S BC ) S A (S BB + S BC + S CC )) : b 2 c 2 (S AB S CC ) : b 2 c 2 (S CA S BB )), Y 2 = (c 2 a 2 (S AB S CC ) : b 2 (S 3 B + b2 (2S BB S CA ) S B (S CC + S CA + S AA )) : c 2 a 2 (S BC S AA )), Z 2 = (a 2 b 2 (S CA S BB ) : a 2 b 2 (S BC S AA ) : c 2 (SC 3 + c2 (2S CC S AB ) S C (S AA + S AB + S BB ))). Substituting the coordinates of X 2 into the equation (??) of the Brocard axis, we have, after ignoring a common factor a 2 b 2 c 2, (S B S C )(S 3 A + 2S AA(S B + S C ) S A (S BB + S BC + S CC ) S BC (S B + S C )) +(S C S A )(S A + S B )(S AB S CC )+(S C + S A )(S A S B )(S CA S BB ) = S 3 A (S B S C )+2S AA (S BB S CC ) S A (S 3 B S3 C ) S BC(S BB S CC ) S 3 A S B S AA (S BB S BC S CC )+S A S C (S BB + S BC S CC ) S B S 3 C + S 3 A S C S AA (S BB + S BC S CC )+S A S B (S BB S BC S CC )+S 3 B S C = 0. This shows that X 2 lies on the Brocard axis. Similar calculations show that Y 2 and Z 2 also lies on the same line. From the coordinates of X 2, Y 2, Z 2, the perspectivity of ABC and X 2 Y 2 Z 2 is ( S BC S AA : S CA S BB : S AB S CC ). This is the Tarry point X(98) on the circumcircle (see Figure??). Z2 A X2 Ya Yc Z b Za E X(98) F O K B Y2 D C Xc X b FIGURE 5.

10 54 PAUL YIU AND XIAO-DONG ZHANG 5.2. Triangles with vertices among the tangency points on the angle bisectors. Proposition 4 (R. Hudson). The triangle bounded by the lines I ba I ca, I cb I ab, I ac I bc is perspective with ABC, the perspectrix is the trilinear polar of the incenter, and the perspector the triangle center X(000). Proof. The lines I ba I ca, I cb I ab, I ac I bc are the polars of the incenter I with respect to the parabolas P a, P b, P c. They have line coordinates[ (c+a b)(a+b c) : 2ca : 2ab], [2ab : (a+b c)(b+c a) : 2bc], [2ca : 2bc : (b+c a)(c+ a b)], and intersect a, b, c respectively at (0 : b : c), ( a : 0 : c), (a : b : 0. Therefore, they bound a triangle perspective with ABC, the perspectrix being the line [ a : b : c], the trilinear polar of the incenter. The vertices of the triangle are X 3 = ( a 4 + 2a 3 (b+c) 2a(b c) 2 (b+c)+(b 2 c 2 ) 2 : 2bc(a 2 + b 2 c 2 4ab) : 2bc(c 2 + a 2 b 2 4ca)), Y 3 = (2ca(a 2 + b 2 c 2 4ab) : b 4 + 2b 3 (c+a) 2b(c a) 2 (c+a)+(c 2 a 2 ) 2 : 2ca(b 2 + c 2 a 2 4bc)), Z 3 = (2ab(c 2 + a 2 b 2 4ca) : 2ab(b 2 + c 2 a 2 4bc) : c 4 + 2c 3 (a+b) 2c(a b) 2 (a+b)+(a 2 b 2 ) 2 ). The perspector with triangle ABC is the triangle center ( b 2 + c 2 a 2 4bc : c 2 + a 2 b 2 4ca : a 2 + b 2 c 2 4ab This is the triangle center X(000) in [?]. Proposition 5. The triangle bounded by the lines I ba I ca, I cb I ab, I aci bc is perspective with ABC, the perspectrix being the trilinear polar of the incenter, and the perspector the orthocenter H. Proof. The lines I ba I ca, I cb I ab, I aci bc are the polars of I a, I b, I c with respect to the parabolas P a, P b, P c respectively. They have line coordinates [(c + a b)(a+b c) : 2ca : 2ab], [2bc : (a+b c)(b+c a) : 2ab], [2bc : 2ca : (b+c a)(c+a b)]. These lines intersect a, b, c respectively at (0 : b : c), ( a : 0 : c),(a : b : 0). Therefore, they bound a triangle perspective with ABC, the perspectrix being the trilinear polar of the incenter. The vertices of the triangle are X 4 = (a 4 2a 3 (b+c)+2a(b c) 2 (b+c) (b 2 c 2 ) 2 : 2bc(a 2 + b 2 c 2 ) : 2bc(c 2 + a 2 b 2 )), Y 4 = (2ca(a 2 + b 2 c 2 ) : b 4 2b 3 (c+a)+2b(c a) 2 (c+a) (c 2 a 2 ) 2 : 2ca(b 2 + c 2 a 2 )), Z 4 = (2ab(c 2 + a 2 b 2 ) : 2ab(b 2 + c 2 a 2 ) : c 4 2c 3 (a+b)+2c(a b) 2 (a+b) (a 2 b 2 ) 2 ). ).

11 A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED WITH A TRIANGLE 55 ( The perspector with triangle ABC is the orthocenter. b 2 + c 2 a 2 : c 2 + a 2 b 2 : a 2 + b 2 c 2 Proposition 6. Let X 5 = I b I ba I ci ca, Y 5 = I c I cb I ai ab, and Z 5 = I a I ac I b I bc. The triangle X 5 Y 5 Z 5 is perspective with ABC at the incenter, and the perspectrix is the trilinear polar of the Nagel point. Proof. In homogeneous coordinates, the points are X 5 = I b I ba I ci ca = ( a(b 2 + 6bc+c 2 a 2 ) : b(c+a b)(a+b c) : c(c+a b)(a+b c)), Y 5 = (a(a+b c)(b+c a) : b(c 2 + 6ca+a 2 b 2 ) : c(a+b c)(b+c a)), Z 5 = (a(b+c a)(c+a b) : b(b+c a)(c+a b) : c(a 2 + 6ab+b 2 c 2 )). From these, it is clear that triangles ABC and ABC are perspective at the incenter. The lines Y 5 Z 5, Z 5 X 5, X 5 Y 5 are [(a+b+c)(a(b+c) (b c) 2 ) : a(a+b c)(b+c a) : a(b+c a)(c+a b)], [b(c+a b)(a+b c) : (a+b+c)(b(c+a) (c a) 2 ) : b(b+c a)(c+a b)], [c(c+a b)(a+b c) : c(a+b c)(b+c a) : (a+b+c)(c(a+b) (a b) 2 )]. They intersect a, b, c respectively at (0 : (c+a b) : a+b c), (b+c a : 0 : (a+b c)), ( (b+c a) : c+a b : 0), x which are collinear on b+c a + y c+a b + z = 0, the trilinear polar a+b c of the Nagel point. This is the perspectrix of the triangles ABC and X 5 Y 5 Z A CONIC THROUGH THE POINTS OF TANGENCY WITH THE EXTERNAL ANGLE BISECTORS Consider the excentral triangle I a I b I c. From the coordinates of I ab given in Proposition?? with coordinate sum 2 (c+a b)(a+b c), and the expression 2a(c 2 + a 2 b 2, 2bc, 2c 2 ) = (a+b c)(b+c a)(a, b, c)+(a+b+c)(c+a b)(a, b, c), we have I ab = (b+c a)i b +(a+b+c)i c. 2a From this we have the homogeneous barycentric coordinates of I ab with respect to the excentral triangle I a I b I c ; similarly for the other five points of tangency on the external angle bisectors. Proposition 7. With respect to the excentral triangle, I ab = (0 : a+b+c : (b+c a)), I ac = (0 : (b+c a) : a+b+c), I bc = ( (c+a b) : 0 : a+b+c), I ba = (a+b+c : 0 : (c+a b)), I ca = (a+b+c : (a+b c) : 0), I cb = ( (a+b c) : a+b+c : 0). ),

12 56 PAUL YIU AND XIAO-DONG ZHANG Proposition 8. The six points are on a conic C with barycentric equation (a+b+c)(x 2 + y 2 + z 2 )+ cyclic relative to the excentral triangle I a I b I c. 2(a 2 +(b+c) 2 ) yz = 0 (0) b+c a I ca I ab I ba A I ac B C I bc I cb FIGURE 6. Proof. Substituting x = 0 into (??) we obtain 0 = (a+b+c)(y 2 + z 2 )+ 2(a2 +(b+c) 2 ) yz b+c a = b+c a ((a+b+c)(b+c a)(y2 + z 2 )+2(a 2 +(b+c) 2 )yz) = ((b+c a)y+(a+b+c)z)((a+b+c)y+(b+c a)z). b+c a Therefore the conic intersects the line I b I c at (0 : a+b+c : (b+c a)) and (0 : (b+c a) : a+b+c). These are the points I ab and I ac. Similarly, the points I bc, I ba, I ca, I cb are also on the same conic. The center of a conic with given barycentric equation can be computed using the formula given in [?, 0.7.2]. Applying this to the conic C we have Proposition 9. Relative to the excentral triangle, the conic C through the six points I ab, I ac, I bc, I ba, I ca, I cb has center with homogeneous barycentric coordinates (a 2 (c+a b)(a+b c)(a 4 2abc(b+c a) (b 2 c 2 ) 2 ) : b 2 (a+b c)(b+c a)(b 4 2abc(c+a b) (c 2 a 2 ) 2 ) : c 2 (b+c a)(c+a b)(c 4 2abc(a+b c) (a 2 b 2 ) 2 )). Remark. With reference to ABC, the conic C has barycentric equation f a yz+ f b zx+ f c xy+4abc(x+y+z)g(x, y, z) = 0,

13 where A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED WITH A TRIANGLE 57 f a = a 6 a 4 (b c) 2 a 2 (b c) 2 (b 2 + 0bc+c 2 ) + 8abc(b c) 2 (b+c)+(b c) 4 (b+c) 2, g(x, y, z) = bc(a+b c)(a b+c)x, cyclic and f b, f c are obtained from f a by cyclic rotations of a, b, c. The center of the conic has homogeneous barycentric coordinates (a(a 0 a 8 (b 2 6bc+c 2 ) 6a 7 bc(b+c) 2a 6 (b 4 + 3b 3 c 2b 2 c 2 + 3bc 3 + c 4 ) + 6a 5 bc(b c) 2 (b+c)+2a 4 (b c) 2 (b 4 b 3 c 8b 2 c 2 bc 3 + c 4 ) + 2a 3 bc(b c) 2 (b+c)(3b+c)(b+3c)+a 2 (b c) 4 (b+c) 2 (b 2 + 8bc+c 2 ) 2abc(b c) 2 (b+c) 3 (3b 2 2bc+3c 2 ) (b 2 c 2 ) 4 (b 2 + c 2 )) : : ). This has ETC-search number REFERENCES [] R. A. Johnson, Advanced Euclidean Geometry, Dover reprint, [2] C. Kimberling, Encyclopedia of Triangle Centers, available at [3] P. Yiu, Introduction to the Geometry of the Triangle, Florida Atlantic University Lecture Notes, 200; revised with corrections 203, math.fau.edu/yiu/geometry.html. [4] P. Yiu, M. Bataille, and O. Kouba, Problem 3854, Crux Math., 39 (203) 274, 276; solution, 40 (204) DEPARTMENT OF MATHEMATICAL SCIENCES, FLORIDA ATLANTIC UNIVERSITY, 777 GLADES ROAD, BOCA RATON, FLORIDA , USA. address: yiu@fau.edu DEPARTMENT OF MATHEMATICAL SCIENCES, FLORIDA ATLANTIC UNIVERSITY, 777 GLADES ROAD, BOCA RATON, FLORIDA , USA. address: xzhang@fau.edu

The Apollonius Circle and Related Triangle Centers

Forum Geometricorum Qolume 3 (2003) 187 195. FORUM GEOM ISSN 1534-1178 The Apollonius Circle and Related Triangle Centers Milorad R. Stevanović Abstract. We give a simple construction of the Apollonius