On Emelyanov's Circle Theorem

Journal for Geometry and Graphics Volume 9 (2005), No. 2, 155–167.

On Emelyanov’s Circle Theorem

Paul Yiu

Department of Mathematical Sciences, Florida Atlantic University Boca Raton, Florida, 33431, USA email: [email protected]

Abstract. Given a triangle and a point T , let +(T ) be the triad of circles each tangent to the circumcircle and to a side line at the trace of T . Assuming T an interior point and each circle tangent the circumcircle internally, Lev Emelyanov has shown that the circle tangent to each of these circles is also tangent to the incircle. In this paper, we study this conguration in further details and without the restriction to interior points. We identify the point of the tangency with the incircle, and derive some interesting loci related this conguration. Key words: Emelyanov circle, homogeneous barycentric coordinates, innite point, nine-point circle, Feuerbach point MSC 2000: 51M04

1. Introduction

Lev Emelyanov has established by synthetic method the following remarkable theorem:

Theorem 1 (Emelyanov [1]) Let A1, B1, C1 be the traces of an interior point T on the a b c side lines of triangle ABC. Construct three circles +, + and + outside the triangle which are tangent to the sides at A1, B1, C1, respectively, and also tangent to the circumcircle of ABC. The circle tangent externally to these three circles is also tangent to the incircle of triangle ABC (see Fig. 1).

In the present paper we study this conguration from the viewpoint of geometric constructions facilitated by a computer software which allows the denitions of macros to perform the constructions eciently and elegantly.1 To achieve this we analyze the underlying geometry by using the language and basic results of triangle geometry. Specically, we make use of barycentric coordinates and their homogenization (see, for example, [3, 4]). We make use of standard notations: a, b, c denote the lengths of the sides of triangle ABC, opposite to A, 1 B, C respectively; s stands for the semiperimeter 2 (a + b + c). Apart from the most basic

ISSN 1433-8157/$ 2.50 c 2005 Heldermann Verlag 156 Paul Yiu: On Emelyanov’s circle theorem

b +

c +

Z+ C1 Y+ 0 Z 0 Y O I B T 1 R

A1 B C 0 a X +

X+

Figure 1: Emelyanov circle

Table 1: Triangle centers and their barycentric coordinates

centroid G (1 : 1 : 1) incenter I (a : b : c) circumcenter O (a2(b2 + c2 a2) : b2(c2 + a2 b2) : (c2(a2 + b2 c2)) 1 1 1 orthocenter H : : b2 + c2 a2 c2 + a2 b2 a2 + b2 c2 ³ ´ symmedian point K (a2 : b2 : c2) 1 1 1 Gergonne point G : : e s a s b s c ³ ´ Nagel point N (s a : s b : s c) Mittenpunkt M (a(s a) : b(s b) : c(s c)) Feuerbach point F ((s a)(b c)2 : (s b)(c a)2 : (s c)(a b)2)

triangle centers listed below with their homogeneous barycentric coordinates, we adopt the labeling of triangle centers in [2].

1Dynamic sketches illustrating results in this paper can be found in the author’s website http://www.math.fau.edu/Yiu/Geometry.html. Paul Yiu: On Emelyanov’s circle theorem 157

a b c The triad of circles +, + and + can be easily constructed with the help of the following simple lemma.

A+

C A+ C

X O X O P ` Q X+ A X+ A Q ` P

Figure 2: Construction of a circle tangent to ` at P and to C

Lemma 2 Given a line ` and a circle C, let A+ and A be the endpoints of the diameter of C perpendicular to `. (a) If a circle touches both ` and C, the line joining the points of tangency passes through one of the points A. (b) If two circles are tangent to the line ` at the same point, the line joining their points of tangency with C passes through the pole of ` with respect to C (see Figs. 2A and 2B).

Computations in this paper were performed with the aids of a computer algebra system. Theorem 4 below characterizes the points of the tangency with the circumcircle. Here, we make use of the notion of the barycentric product of two points. Given two points P1 = (u1 : v2 : w1) and P2 = (u2 : v2 : w2) in homogeneous barycentric coordinates, the barycentric product P1 P2 is the point with coordinates (u1u2 : v1v2 : w1w2). Here is a simple construction ([3]): Let X1, X2 be the traces of P1 and P2 on the line BC, distinct from the vertices B, C. For i = 1, 2, complete the parallelograms AKiXiHi with Ki on AB and Hi on AC. The line joining the intersections of BH1 and CK2, and of BH2 and CK1, passes through the vertex A, and intersects BC at a point X with homogeneous coordinates (0 : v1v2 : w1w2). This is the trace of the point P1 P2 on BC. The traces of the same point on the lines CA and AB can be similarly constructed. From any two of these traces, the barycentric product P1 P2 can be determined (see Fig. 3).

2. The triads of circles (T )

Consider a triangle ABC with circumcircle C. We label A the endpoints of the diameter perpendicular to the sideline BC, A+ being the one on the same side of BC as A, A the antipodal point. 158 Paul Yiu: On Emelyanov’s circle theorem A

H2 P1 P2 K1

B X1 X X2 C

Figure 3: Construction of trace X of P1 P2 on BC

Lemma 3 For ε = 1, 2 Aε = a : b(b εc) : c(c εb) ; similarly, for B and C. ¡ ¢ Proof: The innite point of lines perpendicular to BC can be taken as

1 1 a2, (a2 + b2 c2), (c2 + a2 b2) . 2 2 ³ ´ 2 1 2 2 2 1 2 2 2 We seek t such that t(0, 1, 1) + a , 2 (a + b c ), 2 (c + a b ) lies on the circumcircle ¡ a2yz + b2zx + c2xy = 0. ¢ (1)

1 2 2 2 This gives t = 2 (b + ε2bc + c a ) for ε = 1. Thus, the perpendicular bisector of BC intersects the circumcircle Aε given above.

a b c Given a point T , the triad +(T ) consists of the three circles +, +, +, each tangent to the circumcircle and the sidelines at the traces of T , the lines joining the points of tangency passing through A+, B+, C+, respectively. This is the triad considered in [1] when T is an interior point of the triangle. We do not impose this restriction here. Similarly, we also 0 0 0 consider the triad (T ), consisting of the circles Ca, Cb, Cc.

Theorem 4 (a) The points of tangency of the circumcircle with the circles in the triad +(T ) are the vertices of the circumcevian triangle of the barycentric product P = I T , where I is the incenter of triangle ABC (see Fig. 4). (b) Let the trilinear polar of P intersect BC, CA, AB at X 0, Y 0, Z0 respectively. The points of tangency of the circumcircle with the circles in the triad (T ) are the second intersections of AX0, BY 0, CZ0 with the circumcircle. (c) The line joining the points of tangency of the circumcircle with the corresponding circles in the two triads passes through the corresponding vertex of the tangential triangle (see Fig. 5). Paul Yiu: On Emelyanov’s circle theorem 159

b +

c +

O B1 P T

A1 B C a +

Figure 4: The triad +(T ) and P = I T

Proof: Let T be a point with homogeneous barycentric coordinates (u : v : w).

(a) For ε = 1, the line joining Aε to (0 : v : w) intersects the circumcircle again at a point with coordinates (x, y, z) = (a2, b(b εc), c(c εb)) + t(0, v, w) bc(v + w) for some nonzero t satisfying (1). This gives t = , and vw

2 Xε = a vw : bv(bw + εcv) : cw(cv + εbw) .

¡ ¢ a2vw Similarly, we obtain analogous expressions for Y and Z . From this, X = : bv : cw ε ε + bw + cv is on the line joining I T = (au : bv : cw) to A; similarly for Y+ and Z³+. It follows that´ X+Y+Z+ is the circumcevian triangle of I T . a2vw (b) On the other hand, the line joining X = : bv : cw to A intersects BC at bw cv x y z (0 : bv : cw); similarly for Y and Z. These three³ points lie on the´ line + + = 0, au bv cw the trilinear polar of I T . (c) follows from Lemma 2(b).

Remark. The three lines in Theorem 4(c) above are concurrent at a point

b2 c2 a2 c2 a2 b2 a2 b2 c2 P 0 = a2 + : b2 + : c2 + . v2 w2 u2 w2 u2 v2 u2 v2 w2 µ µ ¶ µ ¶ µ ¶¶ 160 Paul Yiu: On Emelyanov’s circle theorem B0

A c B0 a b +

c + C0 P 0 C1 O T B1

A1 C B a b +

Figure 5: The triads +(T ) and (T )

Here are two simple examples. T P P 0 G I O I K K

3. Equations of the circles in +(T )

Since the points of tangency of the circles in +(T ) with the circumcircle are known, the a centers of the circles can be easily located: if the circle + touches the circumcircle at X+, then its center is the intersection of the lines OX+ and the perpendicular to BC at A1; b c similarly for the other two circles + and +.

Proposition 5 The centers of the circles in the triad +(T ) are the points 2 O1 = a vw : v(2s(s a)v + b(b + c)w) : w(2s(s a)w + c(b + c)v) , 2 O2 = ¡u(2s(s b)u + a(c + a)w) : b wu : w(2s(s b)w + c(c + a)u)¢ , 2 O3 = ¡u(2s(s c)u + a(a + b)v) : v(2s(s c)v + b(a + b)u) : c uv .¢ ¡ ¢ Paul Yiu: On Emelyanov’s circle theorem 161 Proposition 6 These centers are collinear if and only if T lies on the union of the following curves: (i) the circumconic Em : ayz + bzx + cxy = 0 (ii) the curve

2 2 Cm : (2(s a)(s b)(s c) + abc) xyz(x + y + z) + a(s b)(s c)y z = 0. cyclic X

A Cm Em

B C

Figure 6: The curves Em and Cm

2 Remarks. (1) The circumconic Em is centered at the Mittenpunkt

M = (a(b + c a) : b(c + a b) : c(a + b c)) .

It is an ellipse since it does not contain any interior or innite point. 3 (2) The curve Cm is the I-isoconjugate of the circle

(a2 + b2 + c2 2a 2bc 2ca)(a2yz + b2zx + c2xy) +

+ (x + y + z) cyclic bc(c + a b)(a + b c)x = 0,

P 2s with center at the Mittenpunkt M and radius R(R + r) (see Fig. 6). 4R + r

2The Mittenpunkt M is the intersection of the three lines eacph joining the midpoint of a side to the center of the excircle on that side. a b c 3The I-isoconjugate of a point P = (x : y : z) is the point with coordinates : : . It can be x y z constructed as the barycentric product of the incenter and the isotomic conjugate ofµP . ¶ 162 Paul Yiu: On Emelyanov’s circle theorem

Proposition 7 The equations of the circles in the triad +(T ) are

(v + w)2(a2yz + b2zx + c2xy) = (x + y + z) (cv + bw)2x + a2w2y + a2v2z , 2 2 2 2 2 2 2 2 2 (w + u) (a yz + b zx + c xy) = (x + y + z) ¡b w x + (aw + cu) y + b u z¢ , 2 2 2 2 2 2 2 2 2 (u + v) (a yz + b zx + c xy) = (x + y + z)(¡c v x + c u y + (bu + av) z).¢

a 2 Proof: Since + is tangent to the circumcircle at X+ = (a vw : bv(cv +bw) : cw(bw +cv)), and the tangent of the circumcircle at this point is

(cv + bw)2x + a2w2y + a2v2z = 0,

a the equation of + is

k(a2yz + b2zx + c2xy) = (x + y + z) (cv + bw)2x + a2w2y + a2v2z for a choice of k such that with x = 0, the equation¡ has only one root. It is routine¢ to verify that this is k = (v + w)2. This gives the rst of the equations above. The remaining two b c equations of + and + are easily obtained by cyclically permuting (a, u, x), (b, v, y), and (c, w, z).

4. The radical center QT and the Emelyanov circle C(T )

The radical center of the circles in the triad +(T ) can be determined by a formula given, for example, in [4, §7.3.1].

4 Proposition 8 If T = (u : v : w), the radical center of the triad +(T ) is the point

2 2 2 2 2 2 2 2 QT = a (2a(s a)uvw(u + v + w) + bcu (v + w) a(av w + bw u + cu v )) ³ : : . ´

Corollary 9 The radical center QT is

(1) a point on the circumcircle if and only if T is an innite point or lies on the ellipse Em,

(2) an innite point if T lies on the curve Cm.

For T = G, this radical center is the point

2 X1001 = a(a a(b + c) 2bc) : : , ¡ ¢ which divides the centroid G and X55, the internal center of similitude of the circumcircle and incircle, in the ratio GX1001 : X1001X55 = R + r : 3R. The circle C(T ) Emelyanov established in [1] is the one for which each of the circles in the triad +(T ) is oppositely tangent to both C(T ) and the circumcircle. This circle can be easily constructed since the radical center QT is known. If the line joining QT to X+ a b c 0 (respectively Y+, Z+) intersects the circle + (respectively +, +) again at X (respectively Y 0, Z0), then the circle through X 0, Y 0 and Z0 is the one Emelyanov constructed (see Fig. 1).

4The second and third components are obtained from the rst by cyclically permuting (a, b, c) and (u, v, w). Paul Yiu: On Emelyanov’s circle theorem 163 Proposition 10 The equation of the Emelyanov circle C(T ) is

f g h (u + v + w)(a2yz + b2zx + c2xy) (x + y + z) x + y + z = 0, u v w µ ¶ where f = f(u, v, w) = kvw + (s a)(avw + bwu + cuv), (2) g = g(u, v, w) = kwu + (s b)(avw + bwu + cuv), (3) h = h(u, v, w) = kuv + (s c)(avw + bwu + cuv), (4) where k = (s a)(s b) + (s b)(s c) + (s c)(s a). (5)

Examples (1) For the Gergonne point Ge, the Emelyanov circle C(Ge) clearly coincides with the incircle. (2) For the centroid G, the Emelyanov circle C(G) has equation

12(a2yz + b2zx + c2xy) (x + y + z) (a2 + 2a(b + c) (3b2 + 2bc + 3c2))x = 0. cyclic X This has center 5

4 3 2 2 2 2 2 2 2 CG = a a (b + c) 2a (b bc + c ) + a(b + c)(b c) + (b c ) : : . This point¡divides the segment between the incenter and the nine-point center in the¢ ratio 2 : 1. (3) From the equation of the circle, it is clear that C(T ) degenerates into a line if and only if T is an innite point.

5. Point of tangency of the Emelyanov circle and the incircle

The main result of [1] states that the circle C(T ) touches the incircle of the reference triangle. Here we identify the point of tangency explicitly. Note that every point on the incircle has coordinates x2 y2 z2 : : s a s b s c µ ¶ for an innite point (x : y : z). Theorem 11 The circle C(T ) is tangent to the incircle at the point

2 1 1 1 F = : : . (6) T s a (s b)v (s c)w Ã µ ¶ !

Note that FT corresponds to the innite point of the line x y z + + = 0. (s a)u (s b)v (s c)w

Here is an alternative construction of the point FT . 5The center of a circle with given equation can be determined by a formula given in, for example, [4, §10.7.2]. 164 Paul Yiu: On Emelyanov’s circle theorem Proposition 12 Let DEF be the intouch triangle of triangle ABC, i.e., the cevian triangle of the Gergonne point. Given a point T = (u : v : w) with cevian triangle A1B1C1, let

X = B1C1 ∩ EF, Y = C1A1 ∩ F D, Z = A1B1 ∩ DE.

The triangles XY Z and DEF are perspective at FT with coordinates given in (6) (see Fig. 7).

We note an immediate corollary of Theorem 11.

Corollary 13 The Emelyanov circle C(G) touches the incircle at the Feuerbach point

2 2 2 F = X11 = ((s a)(b c) : (s b)(c a) : (s c)(a b) ), the point of tangency of the incircle with the nine-point circle.

Examples: (4) If T is an innite point, the Emelyanov circle degenerates into a line tangent to the incircle. Now, a typical innite point has coordinates

(b c)(a + t) : (c a)(b + t) : (a b)(c + t) . ³ ´ The corresponding Emelyanov circle degenerates into a line tangent to the incircle at the point

(sa)(bc)2(a+t)2(a(a(b+c)(b2 +c2))+t(2a2 a(b+c)(bc)2))2 : : . ³ ´ Z

E C1 X F I B T 1

B D A1 C

Figure 7: Construction of FT Paul Yiu: On Emelyanov’s circle theorem 165

(5) If T is a point on the curve Cm, we have seen in Proposition 6 and Corollary 9 that the centers of the circles in +(T ) are collinear and the radical center QT is an innite point. If ` is the line containing the centers, and if the circles in +(T ) touch the circumcircle at X, Y , Z respectively, then the Emelyanov circle contains the reections of X, Y , Z in `. It is therefore the reection of the circumcircle in `.

6. The locus of T for which C(T ) is tangent to the incircle at a specic point

Theorem 14 Let T be a point other than the Gergonne point. The locus of T 0 for which C(T 0) touches the incircle at the same point as C(T ) does is the circum-hyperbola through T and the Gergonne point. Proof: By the remark following Theorem 11, for T = (u : v : w) and T 0 = (x : y : z), the circles C(T ) and C(T 0) touch the incircle at the same point if and only if

1 1 1 1 1 1 (s b)v (s c)w (s c)w (s a)u (s a)u (s b)v + + = 0. (s a)x (s b)y (s c)z This denes a circumconic, which clearly contains T and the Gergonne point. Since the circumconic contains an interior point (the Gergonne point), it is a hyperbola.

Taking T to be the centroid, we obtain the following corollary. Corollary 15 The locus of T 0 for which C(T 0) touches the incircle at the Feuerbach point is the circum-hyperbola through the centroid and the Gergonne point. b c c a a b Remark. This circumconic has equation + + = 0. Its center is the point x y z

2 2 2 X1086 = (b c) : (c a) : (a b)

on the Steiner in-ellipse. The tangen¡ t to the ellipse at this p¢oint is also tangent to the nine-point circle (at the Feuerbach point). p2 q2 r2 More generally, for a given point Q = : : on the incircle, corresponding s a s b s c to an innite point (p : q : r), the locus of ³T for which the Emely´ anov circle C(T ) touches the incircle at Q is the circum-hyperbola p q r H(Q): + + = 0. (s a)x (s b)y (s c)z To construct the hyperbola, we need only locate one more point. This can be easily found with the help of Proposition 12. Let the line DQ intersect AC and AB at Y0 and Z0 respectively. Then T0 = BY0 ∩ CZ0 is a point on the locus. The hyperbola H(Q) is a rectangular hyperbola if it contains the orthocenter. This is the Feuerbach hyperbola with center at the Feuerbach point. For T on this hyperbola, the point of tangency of the Emelyanov circle C(T ) with the incircle is the point

2 2 3 X3022 = a (b c) (s a) : : . ³ ´ 166 Paul Yiu: On Emelyanov’s circle theorem 7. The involution

Theorem 16 Let f(u, v, w), g(u, v, w), and h(u, v, w) be the functions given in (1). The mapping 1 1 1 (u : v : w) = : : f(u, v, w) g(u, v, w) h(u, v, w) µ ¶ is an involution such that the Emelyanov circles C(T ) and C((T )) are identical. Proof: Let (T ) = (T ).6 It is enough to show that is an involution. If T = (u : v : w), it is easy to check that

(u : v : w) = ku + (s a)(au + bv + cw) : kv + (s b)(au + bv + cw) (7) ³ : kw + (s c)(au + bv + cw) , where k is given in (5). The mapping is an involution. In fact,´it N denotes the Nagel point, and P the intersection of the lines NT and ax + by + cz = 0, then (T ) is the harmonic conjugate of T with respect to N and P . Since (T ) = (T ), is also an involution. If T is the centroid, (T ) is the point 1 : : . a2 + 2a(b + c) (3b2 + 2bc + 3c2) µ ¶ 7.1. The xed points of The xed points of are the isotomic conjugates of the xed points of . From (7), it is clear that the xed points of are the Nagel point (s a : s b : s c) and the points on the line ax + by + cz = 0. Therefore, the xed points of are the Gergonne point and points on the circum-ellipse Em. A typical point on Em is a b c T = : : . (b c)(a + t) (c a)(b + t) (a b)(c + t) µ ¶ For such a point T , the circles in the triad +(T ) are all tangent to the circumcircle at a2 b2 c2 P = : : . T (b c)(a + t) (c a)(b + t) (a b)(c + t) µ ¶ Every circle tangent to the circumcircle at this point is clearly tangent to the three circles of the triad +(T ). There are two circles tangent to the incircle and also to the circumcircle at PT . One of them is the Emelyanov circle C(T ). This touches the incircle at the point a2(s a)(bc(2a2 a(b + c) + (b + c)2) 2 2 2 2 +¡t(a (b + c) 2a(b bc + c ) + (b + c)(b c) )) : : . In particular, the point of tangency is the Feuerbach if 7 ¢ 1 1 1 T = X = : : , 673 a(b + c) (b2 + c2) b(c + a) (c2 + a2) c(a + b) (a2 + b2) µ ¶ The circles C(T ) and those in the triad +(T ) are tangent to the circumcircle at X105. 6Here, () stands for isotomic conjugation. 7 X673 can be constructed as the intersection of the lines joining the centroid to the Feuerbach point, and the Gergonne point to the symmedian point. Paul Yiu: On Emelyanov’s circle theorem 167 Acknowledgment

The author thanks the referees for valuable comments leading to improvement of this paper.

References

[1] L. Emelyanov: A Feuerbach type theorem on six circles. Forum Geom. 1, 173–175 (2001). [2] C. Kimberling: Encyclopedia of Triangle Centers. available at http://faculty. evansville. edu/ck6/encyclopedia/ETC.html. [3] P. Yiu: The uses of homogeneous barycentric coordinates in plane euclidean geometry. Int. J. Math. Educ. Sci. Technol. 31, 569–578 (2000). [4] P. Yiu: Introduction to the Geometry of the Triangle. Florida Atlantic University lecture notes, 2001, available at http://www.math.fau.edu/yiu/Geometry.html.

Received April 25, 2005; nal form December 15, 2005