On Mixtilinear Incircles and Excircles

Forum Geometricorum Volume 6 (2006) 1 16. FORUM GEOM ISSN 1534-1178 On Mixtilinear Incircles and Excircles Khoa Lu Nguyen and Juan arlos Salazar bstract. mixtilinear incircle (respectively excircle) of a triangle is tangent to two sides and to the circumcircle internally (respectively externally). We study the configuration of the three mixtilinear incircles (respectively excircles). In particular, we give easy constructions of the circle (apart from the circumcircle) tangent the three mixtilinear incircles (respectively excircles). We also obtain a number of interesting triangle centers on the line joining the circumcenter and the incenter of a triangle. 1. Preliminaries In this paper we study two triads of circles associated with a triangle, the mixtilinear incircles and the mixtilinear excircles. For an introduction to these circles, see [4] and 2, 3 below. In this section we collect some important basic results used in this paper. Proposition 1 (d lembert s Theorem [1]). Let O 1 (r 1 ), O 2 (r 2 ), O 3 (r 3 ) be three circles with distinct centers. ccording as ε =+1or 1, denote by 1ε, 2ε, 3ε respectively the insimilicenters or exsimilicenters of the pairs of circles ((O 2 ), (O 3 )), ((O 3 ), (O 1 )), and ((O 1 ), (O 2 )). Forε i = ±1, i =1, 2, 3, the points 1ε1, 2ε2 and 3ε3 are collinear if and only if ε 1 ε 2 ε 3 = 1. See Figure 1. The insimilicenter and exsimilicenter of two circles are respectively their internal and external centers of similitude. In terms of one-dimensional barycentric coordinates, these are the points ins(o 1 (r 1 ),O 2 (r 2 )) = r 2 O 1 + r 1 O 2, (1) r 1 + r 2 exs(o 1 (r 1 ),O 2 (r 2 )) = r 2 O 1 + r 1 O 2. (2) r 1 r 2 Proposition 2. Let O 1 (r 1 ), O 2 (r 2 ), O 3 (r 3 ) be three circles with noncollinears centers. For ε = ±1, let O ε (r ε ) be the pollonian circle tangent to the three circles, all externally or internally according as ε =+1or 1. Then the Monge line containing the three exsimilicenters exs(o 2 (r 2 ),O 3 (r 3 )), exs(o 3 (r 3 ),O 1 (r 1 )), and exs(o 1 (r 1 ),O 2 (r 2 )) is the radical axis of the pollonian circles (O + ) and (O ). See Figure 1. Publication Date: January 18, 2006. ommunicating Editor: Paul Yiu. The authors thank Professor Yiu for his contribution to the last section of this paper.

3+ O + 2 K. L. Nguyen and J.. Salazar 3 2 O 1 2+ O 1 O 2 1+ O 3 Figure 1. Lemma 3. Let be a chord of a circle O(r). Let O 1 (r 1 ) be a circle that touches at E and intouches the circle (O) at D. The line DE passes through the midpoint of the arc that does not contain the point D. Furthermore, D E = 2 = 2. Proposition 4. The perspectrix of the circumcevian triangle of P is the polar of P with respect to the circumcircle. Let be a triangle with circumcenter O and incenter I. For the circumcircle and the incircle, ins((o), (I)) = r O + R I = X 55, R + r exs((o), (I)) = r O + R I = X 56. R r in the notations of [3]. We also adopt the following notations. 0 1 2 point of tangency of incircle with intersection of I with the circumcircle antipode of 1 on the circumcircle Similarly define 0, 1, 2, 0, 1 and 2. Note that (i) 0 0 0 is the intouch triangle of, (ii) 1 1 1 is the circumcevian triangle of the incenter I,

On mixtilinear incircles and excircles 3 (iii) 2 2 2 is the medial triangle of the excentral triangle, i.e., 2 is the midpoint between the excenters I b and I c. It is also the midpoint of the arc of the circumcircle. 2. Mixtilinear incircles The -mixtilinear incircle is the circle (O a ) that touches the rays and at a and a and the circumcircle (O) internally at X. See Figure 2. Define the - and -mixtilinear incircles (O b ) and (O c ) analogously, with points of tangency Y and Z with the circumcircle. See [4]. We begin with an alternative proof of the main result of [4]. Proposition 5. The lines X, Y, Z are concurrent at exs((o), (I)). Proof. Since =exs((o a ), (I)) and X =exs((o), (O a )), the line X passes through exs((o), (I)) by d lembert s Theorem. For the same reason, Y and Z also pass through the same point. Y 2 1 0 Z O c I O O b a 1 0 I a a O a a O a 0 X X Figure 2 1 Figure 3 Lemma 6. (1) I is the midpoint of a a. (2) The -mixtilinear incircle has radius r a = r (3) XI bisects angle X. See Figure 3. cos 2 2 onsider the radical axis l a of the mixtilinear incircles (O b ) and (O c ). Proposition 7. The radical axis l a contains (1) the midpoint 1 of the arc of (O) not containing the vertex, (2) the midpoint M a of I 0, where 0 is the point of tangency of the incircle with the side..

4 K. L. Nguyen and J.. Salazar Y 1 b c 0 1 M b O b Z 0 O c I M c M a c 0 b 1 Figure 4. Proof. (1) y Lemma 3, Z, c and 1 are collinear, so are Y, b, 1. lso, 1 c 1 Z = 1 2 = 1 2 = 1 b 1 Y. This shows that 1 is on the radical axis of (O b ) and (O c ). (2) onsider the incircle (I) and the -mixtilinear incircle (O b ) with common ex-tangents and. Since the circle (I) touches and at 0 and 0, and the circle (O b ) touches the same two lines at b and b, the radical axis of these two circles is the line joining the midpoints of b 0 and b 0. Since b, I, b are collinear, the radical axis of (I) and (O b ) passes through the midpoints of I 0 and I 0. Similarly, the radical axis of (I) and (O c ) passes through the midpoints of I 0 and I 0. It follows that the midpoint of I 0 is the common point of these two radical axes, and is therefore a point on the radical axis of (O b ) and (O c ). Theorem 8. The radical center of (O a ), (O b ), (O c ) is the point J which divides OI in the ratio OJ : JI =2R : r. Proof. y Proposition 7, the radical axis of (O b ) and (O c ) is the line 1 M a. Let M b and M c be the midpoints of I 0 and I 0 respectively. Then the radical axes of (O c ) and (O a ) is the line 1 M b, and that of (O a ) and (O b ) is the line 1 M c. Note that the triangles 1 1 1 and M a M b M c are directly homothetic. Since 1 1 1 is inscribed in the circle O(R) and M a M b M c in inscribed in the circle I( r 2 ), the homothetic center of the triangles is the point J which divides the segment OI in

On mixtilinear incircles and excircles 5 Y 1 b c 0 1 M b O b Z 0 O c M c J I O M a c 0 b 1 Figure 5. the ratio See Figure 5. OJ : JI = R : r =2R : r. (3) 2 Remark. Let T be the homothetic center of the excentral triangle I a I b I c and the intouch triangle 0 0 0. This is the triangle center X 57 in [3]. Since the excentral triangle has circumcenter I, the reflection of I in O, OT : TI =2R : r. omparison with (3) shows that J is the reflection of T in O. 3. The mixtilinear excircles The mixtilinear excircles are defined analogously to the mixtilinear incircles, except that the tangencies with the circumcircle are external. The -mixtilinear excircle (O a) can be easily constructed by noting that the polar of passes through the excenter I a ; similarly for the other two mixtilinear excircles. See Figure 6. Theorem 9. If the mixtilinear excircles touch the circumcircle at X, Y, Z respectively, the lines X, Y, Z are concurrent at ins((o), (I)). Theorem 10. The radical center of the mixtilinear excircles is the reflection of J in O, where J is the radical center of the mixtilinear incircles. Proof. The polar of with respect to (O a) passes through the excenter I a. Similarly for the other two polars of with respect to (O b ) and with respect to (O c).

6 K. L. Nguyen and J.. Salazar O b O c Z I Y X O a Figure 6. Let 3 3 3 be the triangle bounded by these three polars. Let 4, 4, 4 be the midpoints of 0 3, 0 3, 0 3 respectively. See Figure 7. Since I a I b 3 I c is a parallelogram, and 2 is the midpoint of I b I c, it is also the midpoint of 3 I a. Since 3 3 is parallel to I b I c (both being perpendicular to the bisector 1 ), I a is the midpoint of 3 3. Similarly, I b, I c are the midpoints of 3 3 and 3 3, and the excentral triangle is the medial triangle of 3 3 3. Note also that I is the circumcenter of 3 3 3 (since it lies on the perpendicular bisectors of its three sides). This is homothetic to the intouch triangle 0 0 0 at I, with ratio of homothety r 4R. If 4 is the midpoint of 0 3, similarly for 4 and 4, then 4 4 4 is homothetic to 3 3 3 with ratio 4R r 4R. We claim that 4 4 4 is homothetic to 2 2 2 at a point J, which is the radical center of the mixtilinear excircles. The ratio of homothety is clearly 4R r R. onsider the isosceles trapezoid 0 0 3 3. Since 4 and 4 are the midpoints of the diagonals 0 3 and 0 3, and 3 3 contains the points of tangency a, a of the circle (O a) with and, the line 4 4 also contains the midpoints of 0 a and 0 a, which are on the radical axis of (I) and (O a). This means that the line 4 4 is the radical axis of (I) and (O a).

On mixtilinear incircles and excircles 7 3 O b 4 I b 2 O c I c Z 0 I 0 J Y 2 4 2 0 X 4 3 a I a 3 a O a Figure 7. It follows that 4 is on the radical axis of (O b ) and (O c ). learly, 2 also lies on the same radical axis. This means that the radical center of the mixtilinear excircles is the homothetic center of the triangles 2 2 2 and 4 4 4. Since these two triangles have circumcenters O and I, and circumradii R and 4R r 2, the homothetic center is the point J which divides IO in the ratio J I : J O =4R r :2R. (4) Equivalently, OJ : J I = 2R :4R r. The reflection of J in O divides OI in the ratio 2R : r. This is the radical center J of the mixtilinear incircles. 4. pollonian circles onsider the circle O 5 (r 5 ) tangent internally to the mixtilinear incircles at 5, 5, 5 respectively. We call this the inner pollonian circle of the mixtilinear incircles. It can be constructed from J since 5 is the second intersection of the line JX with the -mixtilinear incircle, and similarly for 5 and 5. See Figure 8. Theorem 11 below gives further details of this circle, and an easier construction.

8 K. L. Nguyen and J.. Salazar Y 5 Z O O 5 5 J 5 X Figure 8. Theorem 11. (1) Triangles 5 5 5 and are perspective at ins((o), (I)). (2) The inner pollonian circle of the mixtilinear incircles has center O 5 dividing the segment OI in the ratio 4R : r and radius r 5 = 3Rr 4R+r. Proof. (1) Let P =exs((o 5 ), (I)), and Q a =exs((o 5 ), (O a )). The following triples of points are collinear by d lembert s Theorem: (1), 5, P from the circles (O 5 ), (I), (O a ); (2), Q a, 5 from the circles (O 5 ), (O a ), (O a ); (3), Q a, P from the circles (O 5 ), (I), (O a ); (4), X, ins((o), (I)) from the circles (O), (I), (O a ). See Figure 9. Therefore the lines 5 contains the points P and ins((o), (I)) (along with Q a, X ). For the same reason, the lines 5 and 5 contain the same two points. It follows that P and ins((o), (I)) are the same point, which is common to 5, 5 and 5. (2) Now we compute the radius r 5 of the circle (O 5 ). From Theorem 8, OJ : JI =2R : r. sj =exs((o), (O 5 )),wehaveoj : JO 5 = R : r 5. It follows that OJ : JI : JO 5 =2R : r : 2r 5, and OO 5 OI = 2(R r 5) 2R r. (5) Since P =exs((o 5 ), (I)) = ins((o), (I)), it is also ins((o), (O 5 )). Thus, OP : PO 5 : PI = R : r 5 : r, and OO 5 OI = R + r 5 R + r. (6)

On mixtilinear incircles and excircles 9 Q a 5 P I O 5 O a X X O a Figure 9. omparing (5) and (6), we easily obtain r 5 = 3Rr 4R+r. onsequently, OO 5 OI = 4R 4R+r and O 5 divides OI in the ratio OO 5 : O 5 I =4R : r. The outer pollonian circle of the mixtilinear excircles can also be constructed easily. If the lines J X, J Y, J Z intersect the mixtilinear excircles again at 6, 6, 6 respectively, then the circle 6 6 6 is tangent internally to each of the mixtilinear excircles. Theorem 12 below gives an easier construction without locating the radical center. Theorem 12. (1) Triangles 6 6 6 and are perspective at exs((o), (I)). (2) The outer pollonian circle of the mixtilinear excircles has center O 6 dividing the segment OI in the ratio 4R :4R + r and radius r 6 = R(4R 3r) r.

10 K. L. Nguyen and J.. Salazar 6 O b O c Y O 6 6 Z J P X O a 6 Figure 10. Proof. (1) Since 6 =exs((o a), (O 6 )) and =exs((i), (O a)), by d lembert s Theorem, the line 6 passes through P =exs((o 6 ), (I)). For the same reason 6 and 6 pass through the same point, the triangles 6 6 6 and are perspective at P =exs((i), (O 6 )). See Figure 10. y Proposition 2,, X Y, 6 6, and O ao b concur at S =exs((o a), (O b )) on the radical axis of (O) and (O 6 ).Now:S S = SX SY = S 6 S 6. Let 6, 6, 6 intersect the circumcircle (O) at,, respectively. Since S S = S 6 S 6, 6 6 is cyclic. Since 6 = = 6 6, is parallel to 6 6. Similarly, and are parallel to 6 6 and 6 6 respectively. Therefore the triangles and 6 6 6 are directly homothetic, and the center of homothety is P =exs((o), (O 6 )). Since P =exs((o), (O 6 )) = exs((i), (O 6 )), it is also exs((o), (I)), and PI : PO = r : R. (2) Since 6 =exs((o 6 ), (O a)) and X =ins((o a), (O)), by d lembert s Theorem, the line 6 X passes through K =ins((o), (O 6 )). For the same reason, 6 Y and 6 Z pass through the same point K. We claim that K is the radical center J of the mixtilinear excircles. Since SX SY = S 6 S 6, we conclude that X 6 Y 6 is cyclic, and KX K 6 = KY K 6. lso, Y 6 6 Z is cyclic, and KY K 6 = KZ K 6. It follows

On mixtilinear incircles and excircles 11 that KX K 6 = KY K 6 = KZ K 6, showing that K =ins((o), (O 6 )) is the radical center J of the mixtilinear excircles. Hence, J O : J O 6 = R : r 6. Note also PO : PO 6 = R : r 6. Then, we have the following relations. OJ : IO =2R :2R r, J O 6 : IO =2r 6 :2R r, OO 6 : IO =r 6 R : R r. Since OJ + J O 6 = OO 6,wehave 2R 2R r + 2r 6 2R r = r 6 R R r. This gives: r 6 = R(4R 3r) r. Since K =ins((o), (O 6 )) = r 6 O+R O 6 R+r 6 and J = (4R r)o 2R I 2R r are the same point, we obtain O 6 = (4R+r)O 4R I r. Remark. The radical circle of the mixtilinear excircles has center J and radius R 2R r (4R + r)(4r 3r). orollary 13. IO 5 IO 6 = IO 2. 5. The cyclocevian conjugate Let P be a point in the plane of triangle, with traces X, Y, Z on the sidelines,, respectively. onstruct the circle through X, Y, Z. This circle intersects the sidelines,, again at points X, Y, Z. simple application of eva s Theorem shows that the X, Y, Z are concurrent. The intersection point of these three lines is called the cyclocevian conjugate of P. See, for example, [2, p.226]. We denote this point by P. learly, (P ) = P. For example, the centroid and the orthocenters are cyclocevian conjugates, and Gergonne point is the cyclocevian conjugate of itself. We prove two interesting locus theorems. Theorem 14. The locus of Q whose circumcevian triangle with respect to XY Z is perspective to X Y Z is the line PP.ForQ on PP, the perspector is also on the same line. Proof. Let Q be a point on the line PP. y Pascal s Theorem for the six points, X,, X, Y,, Y, the intersections of the lines X and Y lies on the line connecting Q to the intersection of XY and X Y, which according to Pappus theorem (for Y, Z, Z and, Y, Y ), lies on PP. Since Q lies on PP,it follows that X, Y, and PP are concurrent. Similarly, and X Y Z are perspective at a point on PP. The same reasoning shows that if and X Y Z are perspective at a point S, then both Q and S lie on the line connecting the intersections XY X Y and YZ Y Z, which is the line PP.

12 K. L. Nguyen and J.. Salazar Y Z Z O S P P Q Y X X Figure 11. For example, if P = G, then P = H. The line PP is the Euler line. If Q = O, the circumcenter, then the circumcevian triangle of O (with respect to the medial triangle) is perspective with the orthic triangle at the nine-point center N. Theorem 15. The locus of Q whose circumcevian triangle with respect to XY Z is perspective to is the line PP. Proof. First note that sin Z X sin X Y = sin Z X sin YY = sin Z Z sin Z sin Y sin YY = Z Y sin X sin YX = sin X sin XZ It follows that sin Z X sin X Y sin X Y sin Y Z sin Y Z sin Z X ( sin YX = sin XZ sin ZY sin YX sin XZ sin ZY )( sin sin sin Z sin Y sin sin. ) sin sin sin sin sin sin sin = sin sin sin. Therefore, is perspective with if and only if it is perspective with X Y Z. y Theorem 14, the locus of Q is the line PP. 6. Some further results We establish some further results on the mixtilinear incircles and excircles relating to points on the line OI.

On mixtilinear incircles and excircles 13 Y Z Z O P P Q S Y X X Figure 12. Theorem 16. The line OI is the locus of P whose circumcevian triangle with respect to 1 1 1 is perspective with XY Z. Proof. We first show that D = 1 Y 1 Z lies on the line 1. pplying Pascal s Theorem to the six points Z, 1, 2, Y, 1, 2 on the circumcircle, the points D = 1 Y 1 Z, I = 2 Y 2 Z, and 1 2 2 1 are collinear. Since 1 2 and 2 1 are parallel to the bisector 1, it follows that D lies on 1. See Figure 13. Now, if E = 1 Z 1 X and F = 1 X 1 Y, the triangle DEF is perspective with 1 1 1 at I. Equivalently, 1 1 1 is the circumcevian triangle of I with respect to triangle DEF. Triangle XY Z is formed by the second intersections of the circumcircle of 1 1 1 with the side lines of DEF. y Theorem 14, the locus of P whose circumcevian triangle with respect to 1 1 1 is perspective with XY Z is a line through I. This is indeed the line IO, since O is one such point. (The circumcevian triangle of O with respect to 1 1 1 is perspective with XY Z at I). Remark. If P divides OI in the ratio OP : PI = t :1 t, then the perspector Q divides the same segment in the ratio OQ : QI =(1+t)R : 2tr. In particular, if P =ins((o), (I)), this perspector is T, the homothetic center of the excentral and intouch triangles. orollary 17. The line OI is the locus of Q whose circumcevian triangle with respect to 1 1 1 (or XY Z) is perspective with DEF. Proposition 18. The triangle 2 2 2 is perspective (1) with XY Z at the incenter I, (2) with X Y Z at the centroid of the excentral triangle.

14 K. L. Nguyen and J.. Salazar D 2 Y 1 1 O b Z O c I 2 2 1 Figure 13. Proof. (1) follows from Lemma 6(b). (2) Referring to Figure 7, the excenter I a is the midpoint a a. Therefore X I a is a median of triangle X a a, and it intersects 2 2 at its midpoint X. Since 2 2 I a 2 is a parallelogram, 2, X, X and I a are collinear. In other words, the line 2 X contains a median, hence the centroid, of the excentral triangle. So do 2 Y and 2 Z. Let 7 be the second intersection of the circumcircle with the line l a, the radical axis of the mixtilinear incircles (O b ) and (O c ). Similarly define 7 and 7. See Figure 14. Theorem 19. The triangles 7 7 7 and XY Z are perspective at a point on the line OI. Remark. This point divides OI in the ratio 4R r : 4r and has homogeneous barycentric coordinates ( ) a(b + c 5a) b(c + a 5b) c(a + b 5c) : :. b + c a c + a b a + b c 7. Summary We summarize the triangle centers on the OI-line associated with mixtilinear incircles and excircles by listing, for various values of t, the points which divide OI in the ratio R : tr. The last column gives the indexing of the triangle centers in [2, 3].

On mixtilinear incircles and excircles 15 Y 7 1 1 O b Z O c O J O a 7 7 X 1 Figure 14. t first barycentric coordinate X n 1 a 2 (s a) ins((o), (I)) X 55 perspector of and X Y Z perspector of and 5 5 5 a 1 2 s a exs((o), (I)) perspector of and XYZ X 56 perspector of and 6 6 6 2R 2R+r a s a homothetic center of excentral X 57 and intouch triangles 1 2 a 2 (b 2 + c 2 a 2 4bc) radical center of mixtilinear X 999 incircles 1 4 a 2 (b 2 + c 2 a 2 +8bc) center of pollonian circle of mixtilinear incircles 4R r 2r a 2 f(a, b, c) radical center of mixtilinear excircles 4R+r 4r a 2 g(a, b, c) center of pollonian circle of mixtilinear excircles 4R r a(3a 2 2a(b + c) (b c) 2 ) centroid of excentral triangle X 165 perspector of 7 7 7 and XY Z 4R 4R r a(b+c 5a) b+c a The functions f and g are given by

16 K. L. Nguyen and J.. Salazar f(a, b, c) =a 4 2a 3 (b + c)+10a 2 bc +2a(b + c)(b 2 4bc + c 2 ) (b c) 2 (b 2 +4bc + c 2 ), g(a, b, c) =a 5 a 4 (b + c) 2a 3 (b 2 bc + c 2 )+2a 2 (b + c)(b 2 5bc + c 2 ) + a(b 4 2b 3 c +18b 2 c 2 2bc 3 + c 4 ) (b c) 2 (b + c)(b 2 8bc + c 2 ). References [1] F. G.-M., Exercices de Géométrie, 6th ed., 1920; Gabay reprint, Paris, 1991. [2]. Kimberling, Triangle centers and central triangles, ongressus Numerantium, 129 (1998) 1 285. [3]. Kimberling, Encyclopedia of Triangle enters, available at http://faculty.evansville.edu/ck6/encyclopedia/et.html. [4] P. Yiu, Mixtilinear incircles, mer. Math. Monthly, 106 (1999) 952 955. [5] P. Yiu, Mixtilinear incircles II, unpublished manuscript, 1998. Khoa Lu Nguyen: 806 andler Dr., Houston, Texas, 77037-4122, US E-mail address: treegoner@yahoo.com Juan arlos Salazar: alle Maturín N o 19, Urb. Mendoza, Puerto Ordaz 8015, Estado olívar, Venezuela E-mail address: caisersal@yahoo.com