Triangle Centers Associated with the Malfatti Circles

Forum Geometricorum b

Volume 3 (2003) 83–93. bbb FORUM GEOM ISSN 1534-1178

Milorad R. Stevanovi´c

Abstract. Various formulae for the radii of the Malfatti circles of a triangle are presented. We also express the radii of the excircles in terms of the radii of the Malfatti circles, and give the coordinates of some interesting triangle centers associated with the Malfatti circles.

1. The radii of the Malfatti circles The Malfatti circles of a triangle are the three circles inside the triangle, mutually tangent to each other, and each tangent to two sides of the triangle. See Figure 1. Given a triangle ABC, let a, b, c denote the lengths of the sides BC, CA, AB, s the semiperimeter, I the incenter, and r its inradius. The radii of the Malfatti circles of triangle ABC are given by C

X3 r3

Y3 r3 O3

C3 X2 C2 r I 2 C Y1 1 O2 r1 O1 r2 r1

A Z1 Z2 B Figure 1

r r = (s − r − (IB + IC − IA)) , 1 2(s − a) r r = (s − r − (IC + IA − IB)) , 2 2(s − b) (1) r r = (s − r − (IA + IB − IC)) . 3 2(s − c)

Publication Date: March 24, 2003. Communicating Editor: Paul Yiu. The author is grateful to the referee and the editor for their valuable comments and helps. 84 M. R. Stevanovi´c

According to F.G.-M. [1, p.729], these results were given by Malfatti himself, and were published in [7] after his death. See also [6]. Another set of formulae give the same radii in terms of a, b, c and r:

(IB + r − (s − b))(IC + r − (s − c)) r = , 1 2(IA + r − (s − a)) (IC + r − (s − c))(IA + r − (s − a)) r = , 2 2(IB + r − (s − b)) (2) (IA + r − (s − a))(IB + r − (s − b)) r = . 3 2(IC + r − (s − c)) These easily follow from (1) and the following formulae that express the radii r1, r2, r3 in terms of r and trigonometric functions: 1+tanB 1+tanC r r = 4 4 · , 1 1+tanA 2 4 C A 1+tan 4 1+tan 4 r r2 = · , (3) 1+tanB 2 4 1+tanA 1+tanB r r = 4 4 · . 3 C 2 1+tan 4 These can be found in [10]. They can be used to obtain the following formula which is given in [2, pp.103–106]. See also [12]. 2 1 1 1 1 1 1 = √ + √ + √ − + + . (4) r r1r2 r2r3 r1r2 r1r2 r2r3 r3r1

2. Exradii in terms of Malfatti radii

Antreas P. Hatzipolakis [3] asked for the exradii ra, rb, rc of triangle ABC in terms of the Malfatti radii r1, r2, r3 and the inradius r. Proposition 1. 2 − √ 1 r r2r3 ra − r1 = , 2 − √ 1 2 − √ 1 r r3r1 r r1r2 2 − √ 1 r r3r1 rb − r2 = , (5) 2 − √ 1 2 − √ 1 r r1r2 r r2r3 2 − √ 1 r r1r2 rc − r3 = . 2 − √ 1 2 − √ 1 r r2r3 r r3r1 Triangle centers associated with the Malfatti circles 85

Proof. For convenience we write A B C t := tan ,t:= tan ,t:= tan . 1 4 2 4 3 4 A B C Note that from tan 4 + 4 + 4 =1,wehave 1 − t1 − t2 − t3 − t1t2 − t2t3 − t3t1 + t1t2t3 =0. (6) From (3) we obtain 2 1 t 2 − √ = 1 · , r r2r3 1+t1 r 2 1 t2 2 − √ = · , (7) r r3r1 1+t2 r 2 1 t 2 − √ = 3 · . r r1r2 1+t3 r

For the exradius ra,wehave 2 2 s B C (1 − t2)(1 − t3) ra = · r =cot cot · r = · r. s − a 2 2 4t2t3 It follows that r (1 − t2)(1 − t3) 1 ra − r1 =(1 + t2)(1 + t3) · − 2 2t2t3 1+t1 r (1 + t1)(1 − t2)(1 − t3) − 2t2t3 =(1 + t2)(1 + t3) · · 2 2t2t3(1 + t1) r 2t1 =(1 + t2)(1 + t3) · · (from (6)) 2 2t2t3(1 + t1) t 1+t 1+t r = 1 · 2 · 3 · . 1+t1 t2 t3 2 Now the result follows from (7).

Note that with the help of (4), the exradii ra, rb, rc can be explicitly written in terms of the Malfatti radii r1, r2, r3. We present another formula useful in the next sections in the organization of coordinates of triangle centers. Proposition 2. 1 1 a (1 + cos B )(1 + cos C ) − = · 2 2 . r r rs A 1 a 1+cos 2 3. Triangle centers associated with the Malfatti circles Let A be the point of tangency of the Malfatti circles C2 and C3. Similarly define B and C. It is known ([4, p.97]) that triangle ABC is perspective with ABC at the first Ajima-Malfatti point X179. See Figure 3. We work out the details here and construct a few more triangle centers associated with the Malfatti circles. In particular, we find two new triangle centers P+ and P− which divide the incenter I and the first Ajima-Malfatti point harmonically. 86 M. R. Stevanović

3.1. The centers of the Malfatti circles. We begin with the coordinates of the centers of the Malfatti circles. Since O1 divides the segment AIa in the ratio AO1 : O1Ia = r1 : ra − r1, O1 1 1 1 rs = − A + · IA ra = we have r1 r1 ra ra . With s−a we rewrite the absolute barycentric coordinates of O1, along with those of O2 and O3, as follows.

O 1 1 s − a 1 = − A + · I , r r r rs a 1 1 a O 1 1 s − b 2 = − B + · I , r r r rs b (8) 2 2 b O3 1 1 s − c = − C + · Ic. r3 r3 rc rs

From these expressions we have, in homogeneous barycentric coordinates,

1 1 O = 2rs − − a : b : c , 1 r r 1 a 1 1 O = a :2rs − − b : c , 2 r r 2 b 1 1 O3 = a : b :2rs − − c . r3 rc

Y3 O3

X2 B

P− O2

O1 C

A Z1 Z2 B Figure 2 Triangle centers associated with the Malfatti circles 87

3.2. The triangle center P−. It is clear that O1O2O3 is perspective with ABC at the incenter (a : b : c). However, it also follows that if we consider A = BO3 ∩ CO2,B= CO1 ∩ AO3,C= AO2 ∩ BO1, then triangle ABC is perspective with ABC at 1 1 1 1 1 1 P = 2rs − − a :2rs − − b :2rs − − c − r r r r r r 1 a 2 b 3 c 1 1 a 1 1 b 1 1 c = − − : − − : − − r r 2rs r r 2rs r r 2rs 1 a 2 b 3 c (1 + cos B )(1 + cos C ) 1 = a 2 2 − : ···: ··· A 2 (9) 1+cos 2 by Proposition 2. See Figure 2.

Remark. The point P− appears in [5] as the first Malfatti-Rabinowitz point X1142. 3.3. The first Ajima-Malfatti point. For the points of tangency of the Malfatti circles, note that A divides O2O3 in the ratio O2A : A O3 = r2 : r3. We have, in absolute barycentric coordinates, 1 1 O O a 1 1 1 1 + A = 2 + 3 = · A + − B + − C; r2 r3 r2 r3 rs r2 rb r3 rc similarly for B and C. In homogeneous coordinates, a 1 1 1 1 A = : − : − , rs r r r r 2 b 3 c 1 1 b 1 1 B = − : : − , r r rs r r (10) 1 a 3 c 1 1 1 1 c C = − : − : . r1 ra r2 rb rs From these, it is clear that ABC is perspective with ABC at 1 1 1 1 1 1 P = − : − : − r r r r r r 1 a 2 b 3 c a(1 + cos B )(1 + cos C ) b(1 + cos C )(1 + cos A ) = 2 2 : 2 2 1+cosA 1+cosB 2 2 c(1 + cos A )(1 + cos B ) : 2 2 1+cosC 2 a b c = : : A 2 B 2 C 2 (11) 1+cos 2 1+cos 2 1+cos 2 88 M. R. Stevanović by Proposition 2. The point P appears as X179 in [4, p.97], with trilinear coordinates A B C sec4 :sec4 :sec4 4 4 4 computed by Peter Yff, and is named the first Ajima-Malfatti point. See Figure 3.

B A

P O2

C O1

A B

Figure 3

3.4. The triangle center P+. Note that the circle through A , B , C is orthogonal to the Malfatti circles. It is the radical circle of the Malfatti circles, and is the incircle of O1O2O3. The lines O1A , O2B , O3C are concurrent at the Gergonne point of triangle O1O2O3. See Figure 4. As such, this is the point P+ given by 1 1 1 O O O + + P = 1 + 2 + 3 r r r + r r r 1 2 3 1 2 3 1 1 I 1 1 I 1 1 I = − A + a + − B + b + − C + c r r r r r r r r r 1 a a 2 b b 3 c c 1 1 1 1 1 1 I = − A + − B + − C + r r r r r r r 1 a 2 b 3 c 1 1 1 1 1 1 1 = − A + − B + − C + (aA + bB + cC) r r r r r r 2rs 1 a 2 b 3 c 1 1 a 1 1 b 1 1 c = − + A + − + B + − + C. r1 ra 2rs r2 rb 2rs r3 rc 2rs Triangle centers associated with the Malfatti circles 89

It follows that in homogeneous coordinates, 1 1 a 1 1 b 1 1 c P = − + : − + : − + + r r 2rs r r 2rs r r 2rs 1 a 2 b 3 c (1 + cos B )(1 + cos C ) 1 = a 2 2 + : ···: ··· A 2 (12) 1+cos 2 by Proposition 2.

Proposition 3. The points P+ and P− divide the segment IP harmonically. Proof. This follows from their coordinates given in (12), (9), and (11).

From the coordinates of P , P+ and P−, it is easy to see that P+ and P− divide the segment IP harmonically.

B A

P+ O2

C O1

A B

Figure 4

3.5. The triangle center Q. Let the Malfatti circle C1 touch the sides CA and AB at Y1 and Z1 respectively. Likewise, let C2 touch AB and BC at Z2 and X2, C3 touch BC and CA at X3 and Y3 respectively. Denote by X, Y , Z the midpoints of the segments X2X3, Y3Y1, Z1Z2 respectively. Stanley Rabinowitz [9] asked if the lines AX, BY , CZ are concurrent. We answer this in the afﬁrmative. Proposition 4. The lines AX, BY , CZ are concurrent at a point Q with homogeneous barycentric coordinates A B C tan :tan :tan . 4 4 4 (13) 90 M. R. Stevanovi´c C

X3 r3

Y3 r3 O3 X

Y X2 I

Q r2 O2

Y1 r1 O1 r2 r1

A Z1 Z Z2 B Figure 5

Proof. In Figure 5, we have 1 BX = (a + BX − X C) 2 2 3 1 r r = a + 2 (s − b) − 3 (s − c) 2 r r 1 = (a + IB − IC) (from (1)) 2 1 r r = 2R sin A + − 2 B C sin 2 sin 2 A B C B + C =4R sin cos sin cos 2 4 4 4 by making use of the formula A B C r =4R sin sin sin . 2 2 2 Similarly, 1 A B C B + C XC = (a − BX + X C)=4R sin sin cos cos . 2 2 3 2 4 4 4 It follows that BX cos B sin C tan C = 4 4 = 4 . XC B C B sin 4 cos 4 tan 4 Likewise, CY tan A AZ tan B = 4 and = 4 , YA C ZB A tan 4 tan 4 Triangle centers associated with the Malfatti circles 91 and it follows from Ceva’s theorem that AX, BY , CZ are concurrent since BX CY AZ · · =1. XC YA ZB In fact, we can easily identify the homogeneous barycentric coordinates of the intersection Q as given in (13) above since those of X, Y , Z are B C X = 0:tan :tan , 4 4 A C Y = tan :0:tan , 4 4 A B Z = tan :tan :0 . 4 4

Remark. The coordinates of Q can also be written as sin A sin B sin C 2 : 2 : 2 A B C 1+cos 2 1+cos 2 1+cos 2 or a b c : : . A A B B C C (1 + cos 2 )cos 2 (1 + cos 2 )cos 2 (1 + cos 2 )cos 2 3.6. The radical center of the Malfatti circles. Note that the common tangent of C2 and C3 at A passes through X. This means that A X is perpendicular to O2O3 at A . This line therefore passes through the incenter I of O1O2O3. Now, the homogeneous coordinates of A and X can be rewritten as a b c A = : : , (1 + cos A )(1 + cos B )(1 + cos C ) (1 + cos B )2 (1 + cos C )2 2 2 2 2 2 b c X = 0: : . B B C C (1 + cos 2 )cos 2 (1 + cos 2 )cos 2 It is easy to verify that these two points lie on the line (1 + cos A )(cos B − cos C ) (1 + cos B )cos B (1 + cos C )cos C 2 2 2 x − 2 2 y + 2 2 z =0, a b c which also contains the point a b c : : . A B C 1+cos 2 1+cos 2 1+cos 2 Similar calculations show that the latter point also lies on the lines BY and CZ. It is therefore the incenter I of triangle O1O2O3. See Figure 6. This point appears in [5] as X483 , the radical center of the Malfatti circles. 92 M. R. Stevanovi´c C

Y3 O3 X

Y B A X2 I

Y1 C O1

A Z1 Z Z2 B Figure 6

Remarks. (1) The line joining Q and I has equation (1 + cos A )(cos B − cos C ) (1 + cos B )(cos C − cos A ) 2 2 2 x + 2 2 2 y A B sin 2 sin 2 (1 + cos C )(cos A − cos B ) + 2 2 2 z =0. C sin 2 A B C This line clearly contains the point sin 2 :sin 2 :sin 2 , which is the point X174, the Yff center of congruence in [4, pp.94–95]. (2) According to [4], the triangle ABC in §3.3 is also perspective with the excentral triangle. This is because cevian triangles and anticevian triangles are always perspective. The perspector a (2 + cos A +cosB +cosC )2 +cos A (cos2 B +cos2 C − (2 + cos A )2) 2 2 2 2 2 2 2 : ···: ··· 1+cosA 2 is named the second Ajima-Malfatti point X180. For the same reason, the triangle XY Z in §3.5 is also perspective with the excentral triangle. The perspector is the point A A B B C C a − cos 1+cos +cos 1+cos +cos 1+cos : ···: ··· . 2 2 2 2 2 2

This point and the triangle center P+ apparently do not appear in the current edition of [5]. Editor’s endnote. The triangle center Q in §3.5 appears in [5] as the second Malfatti- Rabinowitz point X1143. Its coordinates given by the present editor [13] were not correct owing to a mistake in a sign in the calculations. In the notations of [13], if Triangle centers associated with the Malfatti circles 93

α, β, γ are such that a b c sin2 α = , sin2 β = , sin2 γ = , s s s 1 and λ = 2 (α + β + γ), then the homogeneous barycentric coordinates of Q are (cot(λ − α):cot(λ − β):cot(λ − γ)). These are equivalent to those given in (13) in simpler form.

References [1] F. G.-M., Exercices de G«eom«etrie, 6th ed., 1920; Gabay reprint, Paris, 1991. [2] H. Fukagawa and D. Pedoe, Japanese Temple Geometry Problems, Charles Babbage Research Centre, Winnipeg, 1989.,p.p.103-106. [3] A. P. Hatzipolakis, Hyacinthos message 2375, January 8, 2001. [4] C. Kimberling, Triangle centers and central triangles, Congressus Numerantium, 129 (1998) 1 – 285. [5] C. Kimberling, Encyclopedia of Triangle Centers, August 22, 2002 edition, available at http://www2.evansville.edu/ck6/encyclopedia/; February 26, 2003 edition available at http://faculty.evansville.edu/ck6/encyclopedia/ETC.html. [6] U. J. Knisely, Solution to problem Problem 186. The Mathematical Visitor, 1 (1881) 189; reprinted in [8, pp.148–149]. [7] Malfatti, Annales math«ematiques de Gergonne, 1 (1811) 347. [8] S. Rabinowitz, Problems and solutions from the Mathematical Visitor, 1877Ð1896, MathPro Press, Westford, MA, 1996. [9] S. Rabinowitz, Hyacinthos message 4610, December 30, 2001. [10] E. B. Seitz, Solution to problem 186. The Mathematical Visitor, 1 (1881) 190; reprinted in [8, p.150]. [11] M. R. Stevanovi´c, Hyacinthos message 4613, December 31, 2001. [12] G. Tsintsifas, Solution to problem 618, Crux Math., 8 (1982) 82–85. [13] P. Yiu, Hyacinthos message 4615, December 30, 2001.

Milorad R. Stevanovi´c: Technical Faculty, Svetog Save 65, 32000 Caˇˇ cak, Serbia, Yugoslavia E-mail address: [email protected]