On the Feuerbach Triangle

Transcription

1 Forum Geometricorum Volume FORUM GEOM ISSN On the Feuerbach Triangle Dasari Naga Vijay Krishna bstract. We study the relations among the Feuerbach points of a triangle and the feet of the angle bisectors. From these points we construct6points, pairwise on the three sides of the triangle, which lie on a conic. In addition, we also establish some collinearity and perspectivity results. 1. Perspectivity of Feuerbach and incentral triangles In this note we prove some interesting properties of the Feuerbach points of a triangle. Recall that by the famous Feuerbach theorem, the nine-point circle of a triangle is tangent internally to the incircle and externally to each of the excircles. The points of tangency are the Feuerbach points. If a triangle has side lengths a, b, c, its incenter and the excenters are the points I a : b : c, I a a : b : c, I b a : b : c, I c a : b : c in homogeneous barycentric coordinates with reference to. On the other hand, the nine-point center is the point N a 2 b 2 +c 2 b 2 c 2 2 : b 2 2 +a 2 c 2 a 2 2 : c 2 a 2 +b 2 a 2 b 2 2. From the formulas for the circumradius R and the inradiusr R abc 4 and r 2 a+b+c in terms of a, b, c, and the area of the triangle, we obtain the coordinates of the Feuerbach points. Proposition 1. The nine-point circle is tangent to the incircle at F e b c 2 b+c a : c a 2 c+a b : a b 2 a+b c, and to the-,-,-excircles respectively at F a b c 2 a+b+c : c+a 2 a+b c : a+b 2 c+a b, F b b+c 2 a+b c : c a 2 a+b+c : a+b 2 b+c a, F c b+c 2 c+a b : c+a 2 b+c a : a b 2 a+b+c. We call F a F b F c the Feuerbach triangle. Publication Date: June 19, ommunicating Editor: Paul Yiu. The author thanks Editor Paul Yiu for his help in the preparation of this paper.

2 290 D. N. V. Krishna Z c Y b I b X a I c F c Z F e N I X F a Y F b I a Figure 1 We also consider the intersections of the angle bisectors with the sides. Let the internal and external bisectors of angle intersect the line at X and X a respectively. Similarly define Y, Y b, Z, Z c as the intersections of the internal and external bisectors of angles and with their opposite sides see Figure 1. In homogeneous barycentric coordinates, X 0 : b : c X a 0 : b : c Y a : 0 : c Y b a : 0 : c Z a : b : c Z c a : b : 0 We call XYZ the incentral triangle.

3 On the Feuerbach triangle 291 Line Equation YZ x a + y b + z c 0 ZX x a y b + z c 0 x XY a + y b z c 0 F e F a b 2 bc+c 2 a 2 x+cb cy bb cz 0 F e F b cc ax+c 2 ca+a 2 b 2 y +ac az 0 F e F c ba bx aa by +a 2 ab+b 2 c 2 z 0 F b F c b 2 +bc+c 2 a 2 x+cb+cy +bb+cz 0 F c F a cc+ax c 2 +ca+a 2 b 2 y +ac+az 0 F a F b ba+bx+aa+by a 2 +ab+b 2 c 2 z 0 Table 1. Equations of lines. From the equations of the lines in Table 1, it is clear that The line YZ ZX XY F e F a F e F b F e F c F b F c F c F a F a F b contains X a Y b Z c X Y Z X a Y b Z c Table 2: Incidence of points and lines. Note that X a, Y b, Z c are collinear on x a + y b + z c I a : b : c. 0, the trilinear polar of Proposition 2. The triangles F a F b F c and XYZ are perspective at F e and has perspectrix the trilinear polar of I a : b : c. I b F b I c F c Z I N Y F e X F a Figure 2

4 292 D. N. V. Krishna Proof. From Table 2, the lines F a X, F b Y, F c Z are concurrent at the Feuerbach point F e. This means that the triangles F a F b F c and XYZ are perspective at F e see Figure 2. y Desargues theorem, the two triangles F a F b F c and XYZ are also line perspective. This means that the three points F b F c YZ, F a F c XZ, F a F b XY are collinear. From Table 2, these are respectively the points X a, Y b, Z c, they are collinear on the trilinear polar ofi. This is the perspectrix of the triangles. Proposition 3. The following pairs of triangles are perspective. Triangle Triangle Perspector Perspectrix i F e F c F b XY b Z c F a YZ ii F c F e F a X a YZ c F b ZX iii F b F a F e X a Y b Z F c XY Proof. We shall i only. From Table 2, it is clear that the linesf e X, F c Y b, F b Z c concur at F a. lso, F c F b Y b Z c X a, F b F e Z c X Y, F e F c XY b Z. This shows that the lineyz is the perspectrix of the trianglesf e F c F b andxy b Z c. 2. Similarity of the Feuerbach and incentral triangles Proposition 4. Triangles F a F b F c andxyz are similar. Proof. We show that F b F c YZ F cf a ZX F af b XY. 1 For the feet Y, Z of the bisectors of angles,, we have, by applying the law of cosines to triangleyz, YZ 2 Y 2 +Z 2 2 Y Zcos bc 2 cb c+a b+a bc c+a cb b+a b2 +c 2 a 2 2bc bc bca+b 2 c+a 2 a+b 2 +c+a 2 c+aa+bb 2 +c 2 a 2 bc c+a 2 a+b 2 bc2aa+b+c+b 2 +c 2 c+aa+bb 2 +c 2 +a 2 c+aa+b bc c+a 2 a+b 2 2abca+b+c aa+b+cb 2 +c 2 +a 3 a+b+c+a 2 bc

5 On the Feuerbach triangle 293 abc a+b+c 2bc b 2 c+a 2 a+b 2 +c 2 +a 2 +abc abc c+a 2 a+b 2 a+b+ca b+ca+b c+abc 4 R c+a 2 a+b 2 8 r a +4 R Therefore, YZ 16 2 c+a 2 a+b 2 RR+2r a 16 2 OI 2 a c+a 2 a+b 2. b+cr 2 OI b OI c. It follows that 4 OIa c+aa+br abc OIa c+aa+br. From [7, Theorem 3], F bf c F b F c YZ b+cr2 c+aa+br b+cc+aa+br3. OI b OI c abc OI a abc OI a OI b OI c Since this ratio is symmetric in a, b, c, it is also equal to FcFa ZX and FaF b XY. This proves 1, and we conclude that trianglesf a F b F c andxyz are similar. I c F c Z K i H Y G XF a F b I b I c F c Z K i H Y G X F a F b I b Figure 3 Figure 3 Remark. In fact, the similarity of trianglesf a F b F c andxyz is direct. This means that there is a center of similarityp such that PF b F c : PF c F a : PF a F b PYZ : PZX : PXY. In this case, the center of similarity is the Kiepert center K i b 2 c 2 2 : c 2 a 2 2 : a 2 b 2 2

6 294 D. N. V. Krishna which is the center of the Kiepert circum-hyperbola through the orthocenter H and the centroid G of triangle is a center of similarity see Figures 3 and 3. For notational convenience, let Σ : a 4 +b 4 +c 4 b 2 c 2 c 2 a 2 a 2 b 2, 2 Fu,v,w : uvw +u+v +ww +u vu+v w. 3 Note that the coordinate sum of K i is 2Σ, and F is symmetric inv andw. Now, From this, K i YZ a ba cb+cfa,b,c, 2c+aa+bΣ K i F b F c abca ba cb+c3 c+aa+b. 2Fb,c,aFc,a,bΣ is symmetric ina,b,c. This means that K i F b F c K i YZ abcb+c2 c+a 2 a+b 2 Fa,b,cFb,c,aFc,a,b K i F b F c : K i F c F a : K i F a F b K i YZ : K i ZX : K i XY, and the triangles F a F b F c and XYZ are directly similarly with K i as a center of similarity. 3. The Feuerbach conic We consider the points at which the sidelines of the Feuerbach triangle intersect the sidelines of triangle: X b F c F a X c F a F b Y a F b F c Y c F a F b Z a F b F c Z b F c F a Proposition 5. The six points X b,x c,y c,y a,z a,z b lie on a conic. Proof. Since the four points F b, Y a, Z a, F c are collinear, and the line F b F c passes through X a, so does the line Y a Z a. Similarly, the lines Z b X b passes through Y b and X c Y c through Z c. Furthermore, the points X a, Y b, Z c are collinear, being on the trilinear polar of the incenter I. It follows from Pascal s theorem that the six pointsx b,x c,y c,y a,z a,z b are on a conic see Figure 4. We call the conic through these six points the Feuerbach conic of triangle. Proposition 5 is true when the Feuerbach triangle is replaced by any triangle perspective with.

7 On the Feuerbach triangle 295 Z b Y b I b I c F c Y a Z a N Y c Z b F b X bx c X a F a Figure 4 From the equations of the lines given in Table 1, we determine the coordinates of the points in Proposition 5: X b 0 : ac+a : c 2 +a 2 b 2 +ca; X c 0 : a 2 +b 2 c 2 +ab : aa+b, Y c a 2 +b 2 c 2 +ab : 0 : ba+b; Y a bb+c : 0 : b 2 +c 2 a 2 +bc, Z a cb+c : b 2 +c 2 a 2 +bc : 0, Z b c 2 +a 2 b 2 +ca : cc+a : 0. Proposition 6. The barycentric equation of the Feuerbach conic is cyclic b 2 +bc+c 2 a 2 b+c x 2 + a+b+cb c2 b+c 2a 2 c+aa+b yz 0. ac+aa+b 4

8 296 D. N. V. Krishna Proof. Withx 0, equation 4 becomes 0 c2 +ca+a 2 b 2 y 2 + a2 +ab+b 2 c 2 z 2 c+a a+b + a+b+cb c2 b+c 2a 2 c+aa+b yz ac+aa+b aa+bc 2 +ca+a 2 b 2 y 2 +ac+aa 2 +ab+b 2 c 2 z 2 The numerator factors as +a+b+cb c 2 b+c 2a 2 c+aa+byz ac+aa+b c 2 +ca+a 2 b 2 y ac+azaa+by a 2 +ab+b 2 c 2 z since the coefficient of yz in this product is equal to a 2 c+aa+b c 2 +ca+a 2 b 2 a 2 +ab+b 2 c 2 a 2 c+aa+b c 2 +ca+a 2 b 2 a 2 +ab+b 2 c 2 2a 2 c+aa+b a 2 c+aa+b ac+a b 2 c 2 aa+b+b 2 c 2 2a 2 c+aa+b aa+bb 2 c 2 ac+ab 2 c 2 +b 2 c 2 2 2a 2 c+aa+b aa+b ac+a+b 2 c 2 b 2 c 2 2a 2 c+aa+b ab c+b 2 c 2 b 2 c 2 2a 2 c+aa+b a+b+cb c 2 b+c 2a 2 c+aa+b. This means that the conic defined by 4 intersects the line at the points 0 : ac+a : c 2 +ca+a 2 b 2 and 0 : a 2 +ab+b 2 c 2 : aa+b. These are the points X b and X c. Similarly the conic intersects at Y c, Y a, and at Z a, Z b. It is therefore the Feuerbach conic. Remark. The coordinates of the center of a conic with known barycentric equation can be computed using the formula in [11, ]. For the Feuerbach conic, the center has homogeneous barycentric coordinates bcb+c 2 ga,b,c : cac+a 2 gb,c,a : aba+b 2 gc,a,b for a polynomialgu,v,w of degree10 symmetric inv andw. It has ET-6,9,13 search number Some collinearity and perspectivity results Proposition 7. The points V a : Y c Z b, V b : Z a X c, V c : X b Y a are collinear and the triangles andv a V b V c are perspective.

9 On the Feuerbach triangle 297 I b Z a Y a F b I c F c Z b V c V b V Y c V a X b X c F a Figure 5 Proof. The linesy c andz b have barycentric equations ba+bx +a 2 +ab+b 2 c 2 z 0, cc+ax +c 2 +ca+a 2 b 2 y 0. They intersect at the point V a 1 : 1 bc : Similarly, V b Z a X c cc+a c 2 +ca+a 2 b 2 : c+a bc 2 +ca+a 2 b 2 : b+c ab 2 +bc+c 2 a 2 : 1 ca : V c X b Y a b+c ab 2 +bc+c 2 a 2 : ba+b a 2 +ab+b 2 c 2 a+b ca 2 +ab+b 2 c 2. a+b ca 2 +ab+b 2 c 2, c+a bc 2 +ca+a 2 b 2 : 1. ab From these coordinates, it is clear that triangles V a V b V c is perspective with at b+c V ab 2 +bc+c 2 a 2 : c+a bc 2 +ca+a 2 b 2 : a+b ca 2 +ab+b 2 c 2. 5

10 298 D. N. V. Krishna The three pointsv a,v b,v c are collinear. The line containing them has barycentric equation b cb 2 +bc+c 2 a 2 x 0, 6 cyclic This line is the perspectrix of the triangles andv a V b V c. Remarks. 1 The perspector V given in 5 is the triangle center X6757 of [5]. It lies on the perpendicular to the Euler line at the nine-point center: a 2 b 2 +bc+c 2 a 2 b 2 bc+c 2 a 2 x 0. cyclic 2 The perspectrix the linev a V b V c contains, among others, the triangle centers 1 X79 : :, which is the perspector of and b 2 +bc+c 2 a 2 the reflection triangle of the incenter, a X2160 : :, which is the perspector of b 2 +bc+c 2 a 2 and the triangle bounded by the radical axes of the circumcircle with the circles tangent to two sides of the reference triangle and center on the third side see Figure 6. Proof. These circles have barycentric equations 4b+c 2 a 2 yz +b 2 zx+c 2 xy x+y +z a+b+c 2 b+c a 2 x+c 2 +a 2 b 2 2 y +a 2 +b 2 c 2 2 z 0, 4c+a 2 a 2 yz +b 2 zx+c 2 xy x+y +z b 2 +c 2 a 2 2 x+a+b+c 2 c+a b 2 y +a 2 +b 2 c 2 2 z 0, 4a+b 2 a 2 yz +b 2 zx+c 2 xy x+y +z b 2 +c 2 a 2 2 x+c 2 +a 2 b 2 2 y +a+b+c 2 a+b c 2 z 0. Their radical axes with the circumcircle are the lines a+b+c 2 b+c a 2 x+c 2 +a 2 b 2 2 y +a 2 +b 2 c 2 2 z b+c 2 0, b 2 +c 2 a 2 2 x+a+b+c 2 c+a b 2 y +a 2 +b 2 c 2 2 z c+a 2 0, b 2 +c 2 a 2 2 x+c 2 +a 2 b 2 2 y +a+b+c 2 a+b c 2 z a+b 2 0. These lines bound a triangle with vertices b fa,b,c : c 2 +ca+a 2 b 2 : c a 2 +ab+b 2 c 2, a b 2 +bc+c 2 a 2 : fb,c,a : c a 2 +ab+b 2 c 2, a b 2 +bc+c 2 a 2 : b c 2 +ca+a 2 b 2 : fc,a,b,

11 On the Feuerbach triangle 299 where fu,v,w : Fu,v,wu+v +wu3 v +wv w 2 +2u 2 vw v 2 +w 2 u 2 2 w 2 +wu+u 2 v 2 u 2 +uv +v 2 w 2, and F is defined in 3. From the coordinates of,,, it is clear that and are perspective at a b 2 +bc+c 2 a 2 : b c 2 +ca+a 2 b 2 : c a 2 +ab+b 2 a 2. Z Y X2160 I X Figure 6 3 On the other hand, the points Y a Z a, Z b X b, and X c Y c are on the bisectors of angles,, respectively, as is easily verified. References [1] Lev Emelyanov and Tatiana Emelyanova, round of Feet of isectors, Introduction, Some theoretical facts. [2] Lev Emelyanov and Tatiana Emelyanova, round the Feet of isectors, Main problems Theory. [3] Lev Emelyanov and Tatiana Emelyanova, round the Feet of isectors, Main problems, Using solutions by azhov I. and hekalkin S. [4] Lev Emelyanov and Tatiana Emelyanova, Note on the Feuerbach Point, Forum Geom., [5]. Kimberling, Encyclopedia of Triangle enters, available at

12 300 D. N. V. Krishna [6] S. N. Kiss, Distance Property of the Feuerbach Point and Its Extension, Forum Geom., [7] S. N. Kiss, Distances mong the Feuerbach Points, Forum Geom., [8] T. Kodera, New proofs of two theorems concerning the Feuerbach point of the triangle, Tohoku Mathematical Journal, [9] M. Palej, simple proof for the theorems of Pascal and Pappus, Journal for Geometry and Graphics, [10] I. F. Sharygin. Problems in plane Geometry, problem 586 in Russian, mir publications. [11] P. Yiu, Introduction to the Geometry of the Triangle, Florida tlantic University Lecture Notes, 2001; with corrections, 2013, available at Dasari Naga Vijay Krishna: Department of Mathematics, Narayana Educational Institutions, Machilipatnam, engalore, India address: vijay @gmail.com