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Volume 10 (2010) 1–6 FORUM GEOMETRICORUM A Journal on Classical Euclidean Geometry and Related Areas published by Department of Mathematical Sciences Florida Atlantic University FORUM GEOM Volume 10 2010 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 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: Nikolaos Dergiades Thessaloniki, Greece Clayton Dodge Orono, Maine, USA Roland Eddy St. John’s, Newfoundland, Canada Jean-Pierre Ehrmann Paris, France 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 Man Keung Siu Hong Kong, China Peter Woo La Mirada, California, USA Li Zhou Winter Haven, Florida, 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 Quang Tuan Bui, A triad of similar triangles associated with the perpendicular bisectors of the sides of a triangle,1 J. Marshall Unger, A new proof of a hard but important Sangaku problem,7 Roger C. Alperin, The Poncelet pencil of rectangular hyperbolas,15 Nicolae Anghel, A maximal parallelogram characterization of ovals having circles as orthoptic curves,21 Martin Josefsson, On the inradius of a tangential quadrilateral,27 Boris Odehnal, Some triangle centers associated with the circles tangent to the excircles,35 Nikolaos Dergiades, Conics tangent at the vertices to two sides of a triangle,41 Benedetto Scimemi, Simple relations regarding the Steiner inellipse of a triangle,55 Maria Flavia Mammana, Biagio Micale, and Mario Pennisi, Orthic quadrilaterals of a convex quadrilateral,79 Bojan Hvala, Diophantine Steiner triples and Pythagorean-type triangles,93 Paris Pamfilos, Three maximal cyclic quadrangles,99 Gotthard Weise, Iterates of Brocardian points and lines, 109 Martin Josefsson, Calculations concerning the tangent lengths and tangency chords of a tangential quadrilateral, 119 Giovanni Lucca, Generalized Fibonacci circle chains, 131 Clark Kimberling, Trilinear distance inequalities for the symmedian point, the centroid, and other triangle centers, 135 Nikolaos Dergiades, Connstruction with inscribed ellipses in a triangle, 141 Allan MacLeod, Integer triangles with R/r = N, 149 Cosmin Pohoata, On the Euler reflection point, 157 Martin Josefsson, Characterizations of bicentric quadrilaterals, 165 Paul Yiu, The circles of Lester, Evans, Parry, and their generalizations, 175 Author Index, 211 Cumulative Index, volumes 1–10, 212 Forum Geometricorum b Volume 10 (2010) 1–6. b b FORUM GEOM ISSN 1534-1178 On a Triad of Similar Triangles Associated With the Perpendicular Bisectors of the Sides of a Triangle Quang Tuan Bui Abstract. We prove some perspectivity and orthology results related to a triad of similar triangles associated with the perpendiculars bisectors of the sides of a given triangle. 1. A triad of triangles from the perpendicular bisectors Given a triangle ABC, let the perpendicular bisectors of the sides AB and AC intersect the sideline BC at Ab and Ac respectively. The angles of triangle AAbAc have measures π − 2A, π − 2B, π − 2C if ABC is acute angled, and 2A − π, 2B, 2C if angle A is obtuse (see Figure 1). A A O Ac B C Ab B Ab Ac C O (a) Acute angle case (b) Obtuse angle case Figure 1. Similarly, we consider the intersections Ba and Bc of CA with the perpendicular bisectors of BA and BC, yielding triangle BaBBc, and the intersections Ca and Cb of AB with the perpendicular bisectors of CA, CB yielding triangle CaCbC. It is clear that the triangles AAbAc, BaBBc and CaCbC are all similar (see Figure 2). We record the homogeneous barycentric coordinates of these points. Ab = (0 : −SA + SB : SA + SB), Ac = (0 : SC + SA : SC − SA); Bc = (SB + SC : 0 : −SB + SC ), Ba = (SA − SB : 0 : SA + SB); Ca = (−SC + SA : SC + SA : 0), Cb = (SB + SC : SB − SC : 0). Given a finite point P with homogeneous barycentric coordinates (x : y : z) relative to ABC, we consider the points with the same coordinates in the triangles AAbAc, BaBBc and CaCbC. These are the points Publication Date: January 25, 2010. Communicating Editor: Paul Yiu. 2 Q. T. Bui Cb A Bc O Ba B Ac C Ab Ca Figure 2. AP = (2SBC x : −SC (SA − SB )y + SB(SC + SA)z : SC (SA + SB)y − SB (SC − SA)z), BP = (SA(SB + SC )z − SC (SA − SB )x : 2SCAy : −SA(SB − SC )z + SC (SA + SB)x), CP = (−SB (SC − SA)x + SA(SB + SC )y : SB (SC + SA)x − SA(SB − SC )y : 2SAB z). with coordinates with respect to triangle ABC (see Figure 3). Cb A C P Bc P AP B P Ba B Ac C Ab Ca Figure 3. Proposition 1. Triangle AP BP CP is oppositely similar to ABC. A triad of similar triangles 3 Proof. The square lengths of the sides are 2 2 2 2 2 2 (BP CP ) = k · a , (CP AP ) = k · b , (AP BP ) = k · c , where 2 4 2 2 S a SAAyz − (x + y + z) b c SBBSCC x = cyclic cyclic k (2 ( + + ))2 . P SABC x y zP From this the similarity is clear. Note that the multiplier k is nonnegative. Now, the areas are related by ∆AP BP CP = −k · ∆ABC. It follows that the two triangles are oppositely oriented. 2. Perspectivity Proposition 2. Triangle AP BP CP is perspective with the medial triangle at x x y z − + + : · · · : · · · . (1) S S S S A A B C Proof. Let DEF be the medial triangle. The line joining DAP has equation SA(SC y − SBz)X + SBC xY − SBC xZ = 0, or y z x − X + (Y − Z) = 0. S S S B C A This clearly contains the point − x : y : z , which is a vertex of the an- SA SB SC ticevian triangle of Q = x : y : z . Similarly, the lines EB and FC SA SB SC P P contain the corresponding vertices of the same anticevian triangle. It follows that AP BP CP and DEF are perspective at the cevian quotient G/Q whose coordi- nates are given by (1) above. Proposition 3. Triangle AP BP CP is perspective with ABC if and only if P lies on the Jerabek hyperbola of ABC. In this case, the perspector traverses the Euler line. Proof. The lines AAP , BBP , CCP have equations (SC (SA + SB)y + SB(SC − SA)z)Y + (SC (SA − SB)y − SB(SC + SA)z)Z = 0, (SA(SB − SC )z − SC (SA + SB)x)X + (SA(SB + SC )z + SC(SA − SB)x)Z = 0, (SB (SC + SA)x + SA(SB − SC )y)X + (SB(SC − SA)x − SA(SB + SC)y)Y = 0. They are concurrent if and only if 2SABC (x + y + z) SA(SBB − SCC )yz = 0. cyclic X 4 Q. T. Bui Since P is a finite point, this is the case when P lies on the conic SA(SBB − SCC )yz = 0, cyclicX which is the Jerabek hyperbola, the isogonal conjugate of the Euler line. Now, if P is the isogonal conjugate of the point (SBC + t : SCA + t : SAB + t), then the perspector is the point 2 (a ((2SABC + t(SA + SB + SC )) · SA − t · SBC ) : · · · : · · · ). 2 2 2 Since (a SBC : b SCA : c SAB) is the point X25 (the homothetic center of the tangential and orthic triangles) on the Euler line, the locus of the perspector is the Euler line. 3. Orthology Proposition 4. The triangles ABC and AP BP CP are orthologic. (a) The perpendiculars from A to BP CP , B to CP AP , and C to AP BP intersect at the point a2S b2S c2S Q = A : B : C . − x + y + z x − y + z x + y − z SABC b2SBB c2SCC a2SAA SABC c2SCC a2SAA b2SBB SABC ! (b) The perpendiculars from AP to BC, BP to CA, and CP to AB intersect at the point 1 2 ( 2 − ) ′ = x − S · S SAA x : · · · : · · · Q 2 2 2 2 . a SAA SA a b c SABC SAA cyclic X Remarks.(1) The orthology center Q lies on the circumcircle of triangle ABC. The orthology centers Q(P ) and Q(P ′) are the same if and only if the line P P ′ contains the circumcenter O. Here are some examples. P G I K Na X110 X69 X184 Q X1294 X1295 X1297 X2734 X2693 X98 X74 ′ (2) The coordinates of Q with respect to AP BP CP are the same as those of Q with respect to ABC. ′ (3) Therefore, Q lies on the circumcircle of triangle AP BP CP . 4. An example: Gergonne points of the triad We assume ABC to be acute angled, so that the triangles AAbAc, BaBBc and CaCbC have angles π − 2A, π − 2B and π − 2C and sidelengths in the proportion 2 2 2 of a SA : b SB : c SC . In this case, the Gergonne point Ag of AAbAc has homogeneous barycentric coordinates (SA : SB : SC ) with respect to the triangle. Therefore, with P = (SA : SB : SC ) (the triangle center X69, we have Ag = AP = (2SA : SB +SC : SB +SC ), and likewise Bg = (SC +SA : 2SB : SC +SA) and Cg = (SA + SB : SA + SB : 2SC ). From the table above, the orthology center A triad of similar triangles 5 Q is X = 1 : 1 : 1 , the Tarry point (see Figure 4).
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