FORUM GEOMETRICORUM A Journal on Classical Euclidean Geometry and Related Areas published by Department of Mathematical Sciences Florida Atlantic University b bbb FORUM GEOM Volume 2 2002 http://forumgeom.fau.edu ISSN 1534-1178 Editorial Board Advisors: John H. Conway Princeton, New Jersey, USA Julio Gonzalez Cabillon Montevideo, Uruguay Richard Guy Calgary, Alberta, Canada George Kapetis Thessaloniki, Greece Clark Kimberling Evansville, Indiana, USA Kee Yuen Lam Vancouver, British Columbia, Canada Tsit Yuen Lam Berkeley, California, USA Fred Richman Boca Raton, Florida, USA Editor-in-chief: Paul Yiu Boca Raton, Florida, USA Editors: Clayton Dodge Orono, Maine, USA Roland Eddy St. John’s, Newfoundland, Canada Jean-Pierre Ehrmann Paris, France Lawrence Evans La Grange, Illinois, USA Chris Fisher Regina, Saskatchewan, Canada Rudolf Fritsch Munich, Germany Bernard Gibert St Etiene, France Antreas P. Hatzipolakis Athens, Greece Michael Lambrou Crete, Greece Floor van Lamoen Goes, Netherlands Fred Pui Fai Leung Singapore, Singapore Daniel B. Shapiro Columbus, Ohio, USA Steve Sigur Atlanta, Georgia, USA Man Keung Siu Hong Kong, China Peter Woo La Mirada, California, USA Technical Editors: Yuandan Lin Boca Raton, Florida, USA Aaron Meyerowitz Boca Raton, Florida, USA Xiao-Dong Zhang Boca Raton, Florida, USA Consultants: Frederick Hoffman Boca Raton, Floirda, USA Stephen Locke Boca Raton, Florida, USA Heinrich Niederhausen Boca Raton, Florida, USA Table of Contents Jean-Pierre Ehrmann, A pair of Kiepert hyperbolas,1 Floor van Lamoen, Some concurrencies from Tucker hexagons,5 Jean-Pierre Ehrmann, Congruent inscribed rectangles,15 Clark Kimberling, Collineation, conjugacies, and cubics,21 Floor van Lamoen, Equilateral chordal triangles,33 Gilles Boute, The Napoleon configuration,39 Bernard Gibert, The Lemoine cubic and its generalizations,47 Kurt Hofstetter, A simple construction of the golden section,65 Lawrence Evans, A rapid construction of some triangle centers,67 Peter Yff, A generalization of the Tucker circles,71 Lawrence Evans, A conic through six triangle centers,89 Benedetto Scimemi, Paper-folding and Euler’s theorem revisited,93 Zvonko Cerin,ˇ Loci related to variable flanks, 105 Barukh Ziv, Napoleon-like configurations and sequencs of triangles, 115 Nicolaos Dergiades, An elementary proof of the isoperimetric inequality, 129 Nicolaos Dergiades, The perimeter of a cevian triangle, 131 Fred Lang, Geometry and group structures of some cubics, 135 Charles Thas, On some remarkable concurrences, 147 Jean-Pierre Ehrmann and Floor van Lamoen, The Stammler circles, 151 Jean-Pierre Ehrmann and Floor van Lamoen, Some similarities associated with pedals, 163 K.R.S. Sastry Brahmagupta quadrilaterals, 167 Darij Grinberg and Paul Yiu, The Apollonius circle as a Tucker circle, 175 Wilfred Reyes, An application of Thebault’s´ theorem, 183 Author Index, 187 Forum Geometricorum b Volume 2 (2002) 1–4. bbb FORUM GEOM ISSN 1534-1178 A Pair of Kiepert Hyperbolas Jean-Pierre Ehrmann Abstract. The solution of a locus problem of Hatzipolakis can be expressed in terms of a simple relationship concerning points on a pair of Kiepert hyperbolas associated with a triangle. We study a generalization. Let P be a finite point in the plane of triangle ABC. Denote by a, b, c the lengths of the sides BC, CA, AB respectively, and by AH , BH , CH the feet of the altitudes. We consider rays through P in the directions of the altitudes AAH , BBH, CCH , and, for a nonzero constant k, choose points A, B , C on these rays such that PA = ka, PB = kb, PC = kc. (1) Antreas P. Hatzipolakis [1] has asked, for k =1, for the locus of P for which triangle ABC is perspective with ABC. C B A P C A B Figure 1 We tackle the general case by making use of homogeneous barycentric coor- dinates with respect to ABC. Thus, write P =(u : v : w). In the notations introduced by John H. Conway, 1 2 A =(uS − k(u + v + w)a :vS + k(u + v + w)SC :wS + k(u + v + w)SB), 2 B =(uS + k(u + v + w)SC :vS − k(u + v + w)b :wS + k(u + v + w)SA), 2 C =(uS + k(u + v + w)SB :vS + k(u + v + w)SA:wS − k(u + v + w)c ). Publication Date: January 18, 2002. Communicating Editor: Paul Yiu. The author expresses his sincere thanks to Floor van Lamoen and Paul Yiu for their help and their valuable comments. 1 ABC a b c 1 S φ S := S · cot φ Let be a triangle of side lengths , , , and area 2 . For each , φ . Thus, S = 1 (b2 + c2 − a2) S = 1 (c2 + a2 − b2) S = 1 (a2 + b2 − c2) A 2 , B 2 , and C 2 . These satisfy 2 SASB + SB SC + SC SA = S and other simple relations. For a brief summary, see [3, §1]. 2 J.-P. Ehrmann The equations of the lines AA, BB, CC are (wS + k(u + v + w)SB )y−(vS + k(u + v + w)SC )z =0, (2) −(wS + k(u + v + w)SA)x +(uS + k(u + v + w)SC )z =0, (3) (vS + k(u + v + w)SA)x− (uS + k(u + v + w)SB )y =0. (4) These three lines are concurrent if and only if 0 wS + k(u + v + w)SB −(vS + k(u + v + w)SC ) −(wS + k(u + v + w)SA)0 uS + k(u + v + w)SC =0. vS + k(u + v + w)SA −(uS + k(u + v + w)SB)0 This condition can be rewritten as kS(u + v + w)(S · K(u, v, w) − k(u + v + w)L(u, v, w)) = 0, where K(u, v, w)=(b2 − c2)vw +(c2 − a2)wu +(a2 − b2)uv, (5) 2 2 2 2 2 2 L(u, v, w)=(b − c )SAu +(c − a )SBv +(a − b )SC w. (6) Note that K(u, v, w)=0and L(u, v, w)=0are respectively the equations of the Kiepert hyperbola and the Euler line of triangle ABC. Since P is a finite point and k is nonzero, we conclude, by writing k =tanφ, that the locus of P for which ABC is perspective with ABC is the rectangular hyperbola SφK(u, v, w) − (u + v + w)L(u, v, w)=0 (7) in the pencil generated by the Kiepert hyperbola and the Euler line. Floor van Lamoen [2] has pointed out that this hyperbola (7) is the Kiepert hy- perbola of a Kiepert triangle of the dilated (anticomplementary) triangle of ABC. Specifically, let K(φ) be the Kiepert triangle whose vertices are the apexes of sim- ilar isosceles triangles of base angles φ constructed on the sides of ABC.Itis shown in [3] that the Kiepert hyperbola of K(φ) has equation 2 2 2 2 2Sφ( (b − c )yz)+(x + y + z)( (b − c )(SA + Sφ)x)=0. cyclic cyclic If we replace x, y, z respectively by v + w, w + u, u + v, this equation becomes (7) above. This means that the hyperbola (7) is the Kiepert hyperbola of the Kiepert triangle K(φ) of the dilated triangle of ABC. 2 The orthocenter H and the centroid G are always on the locus. Trivially, if P = H, the perspector is the same point H.ForP = G, the perspector is the point 3 1 1 1 : : , 3kSA + S 3kSB + S 3kSC + S 2The Kiepert triangle K(φ) of the dilated triangle of ABC is also the dilated triangle of the Kiepert triangle K(φ) of triangle ABC. 3In the notations of [3], this is the Kiepert perspector K(arctan 3k). A pair of Kiepert hyperbolas 3 the second common point of Kiepert hyperbola and the tangent at P to the locus of P , the Kiepert hyperbola of the dilated triangle of K(φ). Now we identify the perspector when P is different from G. Addition of the equations (2,3,4) of the lines AA, BB, CC gives (v − w)x +(w − u)y +(u − v)z =0. This is the equation of the line joining P to the centroid G, showing that the per- spector lies on the line GP . We can be more precise. Reorganize the equations (2,3,4) as k(SB y − SC z)u+(k(SB y − SC z) − Sz)v+(k(SBy − SC z)+Sy)w =0, (8) (k(SC z − SAx)+Sz)u+(k(SC z − SAx)v+(k(SC z − SAx) − Sx)w =0, (9) (k(SAx − SB y) − Sy)u+(k(SAx − SB y)+Sx)v+ k(SAx − SB y)w =0. (10) Note that the combination x·(8) + y·(9) + z·(10) gives k(u + v + w)(x(SBy − SC z)+y(SC z − SAx)+z(SAx − SBy)=0. Since k and u + v + w are nonzero, we have (SC − SB)yz +(SA − SC )zx +(SB − SA)xy =0, or equivalently, (b2 − c2)yz +(c2 − a2)zx +(a2 − b2)xy =0. It follows that the perspector is also on the Kiepert hyperbola. B C G Q H P A C A B Figure 2 We summarize these results in the following theorem. 4 J.-P. Ehrmann Theorem. Let k =tanφ be nonzero, and points A, B, C be given by (1) along the rays through P parallel to the altitudes. The lines AA, BB, CC are concur- rent if and only if P lies on the Kiepert hyperbola of the Kiepert triangle K(φ) of the dilated triangle. The intersection of these lines is the second intersection of the line GP and the Kiepert hyperbola of triangle ABC. If we change, for example, the orientation of PA, the locus of P is the rectangu- lar hyperbola with center at the apex of the isosceles triangle on BC of base angle φ, 4 asymptotes parallel to the A-bisectors, and passing through the orthocenter H G G (and also the A-vertex A =(−1:1:1)of the dilated triangle).