A Note on the Barycentric Square Roots of Kiepert Perspectors

Transcription

1 Forum Geometricorum Volume 6 (2006) FORUM GEOM ISSN Note on the arycentric Square Roots of Kiepert erspectors Khoa Lu Nguyen bstract. Let be an interior point of a given triangle. We prove that the orthocenter of the cevian triangle of the barycentric square root of lies on the Euler line of if and only if lies on the Kiepert hyperbola. 1. Introduction In a recent Mathlinks message, the present author proposed the following problem. Theorem 1. Given an acute triangle with orthocenter H, the orthocenter H of the cevian triangle of H, the barycentric square root of H, lies on the Euler line of triangle. H O H H Figure 1. aul Yiu has subsequently discovered the following generalization. Theorem 2. The locus of point for which the orthocenter of the cevian triangle of the barycentric square root lies on the Euler line is the part of the Kiepert hyperbola which lies inside triangle. ublication Date: October 30, ommunicating Editor: aul Yiu. The author is grateful to rofessor Yiu for his generalization of the problem and his help in the preparation of this paper.

2 264 K. L. Nguyen The barycentric square root is defined only for interior points. This is the reason why we restrict to acute angled triangles in Theorem 1 and to the interior points on the Kiepert hyperola in Theorem 2. It is enough to prove Theorem Trilinear polars Let be the cevian triangle of, and 1, 1, 1 be respectively the intersections of and, and, and. y Desargues theorem, the three points 1, 1, 1 lie on a line l, the trilinear polar of Figure 2. If has homogeneous barycentric coordinates (u : v : w), then the trilinear polar is the line x l : u + y v + z w =0. For the orthocenter H =(S : S : S ), the trilinear polar l H : S x + S y + S z =0. is also called the orthic axis. roposition 3. The orthic axis is perpendicular to the Euler line. This proposition is very well known. It follows easily, for example, from the fact that the orthic axis l H is the radical axis of the circumcircle and the nine-point circle. See, for example, [2, 5.4,5]. The trilinear polar l and the orthic axis l H intersect at the point (u(s v S w):v(s w S u):w(s u S v)). In particular, l and l H are parallel, i.e., their intersection is a point at infinity if and only if u(s v S w)+v(s w S u)+w(s u S v)=0. Equivalently, (S S )vw +(S S )wu +(S S )uv =0. (1)

3 note on the barycentric square roots of Kiepert perspectors 265 Note that this equation defines the Kiepert hyperbola. oints on the Kiepert hyperbola are called Kiepert perspectors. roposition 4. The trilinear polar l is parallel to the orthic axis if and only if is a Kiepert perspector. 3. The barycentric square root of a point Let be a point inside triangle, with homogeneous barycentric coordinates (u : v : w). We may assume u, v, w > 0, and define the barycentric square root of to be the point with barycentric coordinates ( u : v : w). aul Yiu [2] has given the following construction of. (1) onstruct the circle with as diameter. (2) onstruct the perpendicular to at the trace of to intersect at X. (3) onstruct the bisector of angle X to intersect at X. Then X is the trace of on. Similar constructions on the other two sides give the traces Y and Z of on and respectively. The barycentric square root is the common point of X, Y, Z. roposition 5. If the trilinear polar l intersects at 1, then 1 X 2 = 1 1. roof. Let M is the midpoint of. Since 1, divide, harmonically, we have M 2 = M 2 = M 1 M (Newton s theorem). Thus, MX 2 = M 1 M. It follows that triangles MX 1 and M X are similar, and MX 1 = M X =90. This means that 1 X is tangent at X to the circle with diameter. Hence, 1 X 2 = 1 1. X 1 X M Figure 3.

4 266 K. L. Nguyen To complete the proof it is enough to show that 1 X = 1 X, i.e., triangle 1 XX is isosceles. This follows easly from 1 X X = 1 X + X X = X + XX = X X 1. orollary 6. If X 1 is the intersection of YZ and, then 1 is the midpoint of XX 1. roof. If X 1 is the intersection of YZ and, then X, X 1 divide, harmonically. The circle through X, X 1, and with center on is orthogonal to the circle. y roposition 5, this has center 1, which is therefore the midpoint of XX roof of Theorem 2 Let be an interior point of triangle, and XYZ the cevian triangle of its barycentric square root. roposition 7. If H is the orthocenter of XY Z, then the line OH is perpendicular to the trilinear polar l. roof. onsider the orthic triangle DEF of XY Z. Since DEXY, EFY Z, and FDZX are cyclic, and the common chords DX, EY, FZ intersect at H, H is the radical center of the three circles, and H D H X = H E H Y = H F H Z. (2) onsider the circles ξ, ξ, ξ, with diameters XX 1, YY 1, ZZ 1. These three circles are coaxial; they are the generalized pollonian circles of the point. See [3]. s shown in the previous section, their centers are the points 1, 1, 1 on the trilinear polar l. See Figure 4. Now, since D, E, F lie on the circles ξ, ξ, ξ respectively, it follows from (2) that H has equal powers with respect to the three circles. It is therefore on the radical axis of the three circles. We show that the circumcenter O of triangle also has the same power with respect to these circles. Indeed, the power of O with respect to the circle ξ is 1 O 2 1 X 2 = O 2 1 R 2 1 X 2 + R 2 = X 2 + R 2 = R 2 by roposition 5. The same is true for the circles ξ and ξ. Therefore, O also lies on the radical axis of the three circles. It follows that the line OH is the radical axis of the three circles, and is perpendicular to the line l which contains their centers. The orthocenter H of XY Z lies on the Euler line of triangle if and only if the trilinear polar l is parallel to the Euler line, and hence parallel to the orthic axis by roposition 3. y roposition 4, this is the case precisely when lies on the Kiepert hyperbola. This completes the proof of Theorem 2.

5 note on the barycentric square roots of Kiepert perspectors Y Z D E H F O X 1 1 X 1 Z 1 Figure 4. Theorem 8. The orthocenter of the cevian triangle of lies on the rocard axis if and only if is an interior point on the Jerabek hyperbola. roof. The rocard axis OK is orthogonal to the Lemoine axis. The locus of points whose trilinear polars are parallel to the rocard axis is the Jerabek hyperbola. 5. oordinates In homogeneous barycentric coordinates, the orthocenter of the cevian triangle of (u : v : w) is the point

6 268 K. L. Nguyen ( ( ( 1 S w + 1 ) ( 1 + S u u + 1 )) ( ( 1 S v v + 1 ) 2 ( 1 + S w u 1 ) ( 1u + S 2 w 2 1v ) ) 2 2 : : ). ( pplying this to the square root of the orthocenter, with (u 2 : v 2 : w 2 ) = S : S : S ), we obtain a 2 S S + S a 2 S : :, cyclic which is the point H in Theorem 1. More generally, if is the Kiepert perspector ( K(θ) = : : S + S θ S + S θ the orthocenter of the cevian triangle of is the point ( a 2 S (S + S θ )(S + S θ )(S + S θ ) S + S θ +S a 2 S + S θ + a 2 S θ S S + S θ : :. cyclic cyclic ), References [1] R.. Johnson, dvanced Euclidean Geometry, 1925, Dover reprint. [2]. Yiu, Introduction to the Geometry of the Triangle, Florida tlantic University lecture notes, [3]. Yiu, Generalized pollonian circles, Journal of Geometry and Graphics, 8 (2004) Khoa Lu Nguyen: Massachusetts Institute of Technology, student, 77 Massachusetts venue, ambridge, M, 02139, US address: treegoner@yahoo.com