Triangle Centers and Kiepert's Hyperbola Except Where Reference Is Made to the Work of Others, the Work Described in This Thes
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												Angle Chasing
Angle Chasing Ray Li June 12, 2017 1 Facts you should know 1. Let ABC be a triangle and extend BC past C to D: Show that \ACD = \BAC + \ABC: 2. Let ABC be a triangle with \C = 90: Show that the circumcenter is the midpoint of AB: 3. Let ABC be a triangle with orthocenter H and feet of the altitudes D; E; F . Prove that H is the incenter of 4DEF . 4. Let ABC be a triangle with orthocenter H and feet of the altitudes D; E; F . Prove (i) that A; E; F; H lie on a circle diameter AH and (ii) that B; E; F; C lie on a circle with diameter BC. 5. Let ABC be a triangle with circumcenter O and orthocenter H: Show that \BAH = \CAO: 6. Let ABC be a triangle with circumcenter O and orthocenter H and let AH and AO meet the circumcircle at D and E, respectively. Show (i) that H and D are symmetric with respect to BC; and (ii) that H and E are symmetric with respect to the midpoint BC: 7. Let ABC be a triangle with altitudes AD; BE; and CF: Let M be the midpoint of side BC. Show that ME and MF are tangent to the circumcircle of AEF: 8. Let ABC be a triangle with incenter I, A-excenter Ia, and D the midpoint of arc BC not containing A on the circumcircle. Show that DI = DIa = DB = DC: 9. Let ABC be a triangle with incenter I and D the midpoint of arc BC not containing A on the circumcircle. - 
												
												Investigating Centers of Triangles: the Fermat Point
Investigating Centers of Triangles: The Fermat Point A thesis submitted to the Miami University Honors Program in partial fulfillment of the requirements for University Honors with Distinction by Katherine Elizabeth Strauss May 2011 Oxford, Ohio ABSTRACT INVESTIGATING CENTERS OF TRIANGLES: THE FERMAT POINT By Katherine Elizabeth Strauss Somewhere along their journey through their math classes, many students develop a fear of mathematics. They begin to view their math courses as the study of tricks and often seemingly unsolvable puzzles. There is a demand for teachers to make mathematics more useful and believable by providing their students with problems applicable to life outside of the classroom with the intention of building upon the mathematics content taught in the classroom. This paper discusses how to integrate one specific problem, involving the Fermat Point, into a high school geometry curriculum. It also calls educators to integrate interesting and challenging problems into the mathematics classes they teach. In doing so, a teacher may show their students how to apply the mathematics skills taught in the classroom to solve problems that, at first, may not seem directly applicable to mathematics. The purpose of this paper is to inspire other educators to pursue similar problems and investigations in the classroom in order to help students view mathematics through a more useful lens. After a discussion of the Fermat Point, this paper takes the reader on a brief tour of other useful centers of a triangle to provide future researchers and educators a starting point in order to create relevant problems for their students. iii iv Acknowledgements First of all, thank you to my advisor, Dr. - 
												
												Orthocorrespondence and Orthopivotal Cubics
Forum Geometricorum b Volume 3 (2003) 1–27. bbb FORUM GEOM ISSN 1534-1178 Orthocorrespondence and Orthopivotal Cubics Bernard Gibert Abstract. We define and study a transformation in the triangle plane called the orthocorrespondence. This transformation leads to the consideration of a fam- ily of circular circumcubics containing the Neuberg cubic and several hitherto unknown ones. 1. The orthocorrespondence Let P be a point in the plane of triangle ABC with barycentric coordinates (u : v : w). The perpendicular lines at P to AP , BP, CP intersect BC, CA, AB respectively at Pa, Pb, Pc, which we call the orthotraces of P . These orthotraces 1 lie on a line LP , which we call the orthotransversal of P . We denote the trilinear ⊥ pole of LP by P , and call it the orthocorrespondent of P . A P P ∗ P ⊥ B C Pa Pc LP H/P Pb Figure 1. The orthotransversal and orthocorrespondent In barycentric coordinates, 2 ⊥ 2 P =(u(−uSA + vSB + wSC )+a vw : ··· : ···), (1) Publication Date: January 21, 2003. Communicating Editor: Paul Yiu. We sincerely thank Edward Brisse, Jean-Pierre Ehrmann, and Paul Yiu for their friendly and valuable helps. 1The homography on the pencil of lines through P which swaps a line and its perpendicular at P is an involution. According to a Desargues theorem, the points are collinear. 2All coordinates in this paper are homogeneous barycentric coordinates. Often for triangle cen- ters, we list only the first coordinate. The remaining two can be easily obtained by cyclically permut- ing a, b, c, and corresponding quantities. Thus, for example, in (1), the second and third coordinates 2 2 are v(−vSB + wSC + uSA)+b wu and w(−wSC + uSA + vSB )+c uv respectively. - 
												
												Cevians, Symmedians, and Excircles Cevian Cevian Triangle & Circle
10/5/2011 Cevians, Symmedians, and Excircles MA 341 – Topics in Geometry Lecture 16 Cevian A cevian is a line segment which joins a vertex of a triangle with a point on the opposite side (or its extension). B cevian C A D 05-Oct-2011 MA 341 001 2 Cevian Triangle & Circle • Pick P in the interior of ∆ABC • Draw cevians from each vertex through P to the opposite side • Gives set of three intersecting cevians AA’, BB’, and CC’ with respect to that point. • The triangle ∆A’B’C’ is known as the cevian triangle of ∆ABC with respect to P • Circumcircle of ∆A’B’C’ is known as the evian circle with respect to P. 05-Oct-2011 MA 341 001 3 1 10/5/2011 Cevian circle Cevian triangle 05-Oct-2011 MA 341 001 4 Cevians In ∆ABC examples of cevians are: medians – cevian point = G perpendicular bisectors – cevian point = O angle bisectors – cevian point = I (incenter) altitudes – cevian point = H Ceva’s Theorem deals with concurrence of any set of cevians. 05-Oct-2011 MA 341 001 5 Gergonne Point In ∆ABC find the incircle and points of tangency of incircle with sides of ∆ABC. Known as contact triangle 05-Oct-2011 MA 341 001 6 2 10/5/2011 Gergonne Point These cevians are concurrent! Why? Recall that AE=AF, BD=BF, and CD=CE Ge 05-Oct-2011 MA 341 001 7 Gergonne Point The point is called the Gergonne point, Ge. Ge 05-Oct-2011 MA 341 001 8 Gergonne Point Draw lines parallel to sides of contact triangle through Ge. - 
												
												The Apollonius Problem on the Excircles and Related Circles
Forum Geometricorum b Volume 6 (2006) xx–xx. bbb FORUM GEOM ISSN 1534-1178 The Apollonius Problem on the Excircles and Related Circles Nikolaos Dergiades, Juan Carlos Salazar, and Paul Yiu Abstract. We investigate two interesting special cases of the classical Apollo- nius problem, and then apply these to the tritangent of a triangle to find pair of perspective (or homothetic) triangles. Some new triangle centers are constructed. 1. Introduction This paper is a study of the classical Apollonius problem on the excircles and related circles of a triangle. Given a triad of circles, the Apollonius problem asks for the construction of the circles tangent to each circle in the triad. Allowing both internal and external tangency, there are in general eight solutions. After a review of the case of the triad of excircles, we replace one of the excircles by another (possibly degenerate) circle associated to the triangle. In each case we enumerate all possibilities, give simple constructions and calculate the radii of the circles. In this process, a number of new triangle centers with simple coordinates are constructed. We adopt standard notations for a triangle, and work with homogeneous barycen- tric coordinates. 2. The Apollonius problem on the excircles 2.1. Let Γ=((Ia), (Ib), (Ic)) be the triad of excircles of triangle ABC. The sidelines of the triangle provide 3 of the 8 solutions of the Apollonius problem, each of them being tangent to all three excircles. The points of tangency of the excircles with the sidelines are as follows. See Figure 2. BC CA AB (Ia) Aa =(0:s − b : s − c) Ba =(−(s − b):0:s) Ca =(−(s − c):s :0) (Ib) Ab =(0:−(s − a):s) Bb =(s − a :0:s − c) Cb =(s : −(s − c):0) (Ic) Ac =(0:s : −(s − a)) Bc =(s :0:−(s − c)) Cc =(s − a : s − b :0) Publication Date: Month day, 2006. - 
												
												Stanley Rabinowitz, Arrangement of Central Points on the Faces of A
International Journal of Computer Discovered Mathematics (IJCDM) ISSN 2367-7775 ©IJCDM Volume 5, 2020, pp. 13{41 Received 6 August 2020. Published on-line 30 September 2020 web: http://www.journal-1.eu/ ©The Author(s) This article is published with open access1. Arrangement of Central Points on the Faces of a Tetrahedron Stanley Rabinowitz 545 Elm St Unit 1, Milford, New Hampshire 03055, USA e-mail: [email protected] web: http://www.StanleyRabinowitz.com/ Abstract. We systematically investigate properties of various triangle centers (such as orthocenter or incenter) located on the four faces of a tetrahedron. For each of six types of tetrahedra, we examine over 100 centers located on the four faces of the tetrahedron. Using a computer, we determine when any of 16 con- ditions occur (such as the four centers being coplanar). A typical result is: The lines from each vertex of a circumscriptible tetrahedron to the Gergonne points of the opposite face are concurrent. Keywords. triangle centers, tetrahedra, computer-discovered mathematics, Eu- clidean geometry. Mathematics Subject Classification (2020). 51M04, 51-08. 1. Introduction Over the centuries, many notable points have been found that are associated with an arbitrary triangle. Familiar examples include: the centroid, the circumcenter, the incenter, and the orthocenter. Of particular interest are those points that Clark Kimberling classifies as \triangle centers". He notes over 100 such points in his seminal paper [10]. Given an arbitrary tetrahedron and a choice of triangle center (for example, the circumcenter), we may locate this triangle center in each face of the tetrahedron. We wind up with four points, one on each face. - 
												
												Control Point Based Representation of Inellipses of Triangles∗
Annales Mathematicae et Informaticae 40 (2012) pp. 37–46 http://ami.ektf.hu Control point based representation of inellipses of triangles∗ Imre Juhász Department of Descriptive Geometry, University of Miskolc, Hungary [email protected] Submitted April 12, 2012 — Accepted May 20, 2012 Abstract We provide a control point based parametric description of inellipses of triangles, where the control points are the vertices of the triangle themselves. We also show, how to convert remarkable inellipses defined by their Brianchon point to control point based description. Keywords: inellipse, cyclic basis, rational trigonometric curve, Brianchon point MSC: 65D17, 68U07 1. Introduction It is well known from elementary projective geometry that there is a two-parameter family of ellipses that are within a given non-degenerate triangle and touch its three sides. Such ellipses can easily be constructed in the traditional way (by means of ruler and compasses), or their implicit equation can be determined. We provide a method using which one can determine the parametric form of these ellipses in a fairly simple way. Nowadays, in Computer Aided Geometric Design (CAGD) curves are repre- sented mainly in the form n g (u) = j=0 Fj (u) dj δ Fj :[a, b] R, u [a, b] R, dj R , δ 2 P→ ∈ ⊂ ∈ ≥ ∗This research was carried out as a part of the TAMOP-4.2.1.B-10/2/KONV-2010-0001 project with support by the European Union, co-financed by the European Social Fund. 37 38 I. Juhász where dj are called control points and Fj (u) are blending functions. (The most well-known blending functions are Bernstein polynomials and normalized B-spline basis functions, cf. - 
												
												Searching for the Center
Searching For The Center Brief Overview: This is a three-lesson unit that discovers and applies points of concurrency of a triangle. The lessons are labs used to introduce the topics of incenter, circumcenter, centroid, circumscribed circles, and inscribed circles. The lesson is intended to provide practice and verification that the incenter must be constructed in order to find a point equidistant from the sides of any triangle, a circumcenter must be constructed in order to find a point equidistant from the vertices of a triangle, and a centroid must be constructed in order to distribute mass evenly. The labs provide a way to link this knowledge so that the students will be able to recall this information a month from now, 3 months from now, and so on. An application is included in each of the three labs in order to demonstrate why, in a real life situation, a person would want to create an incenter, a circumcenter and a centroid. NCTM Content Standard/National Science Education Standard: • Analyze characteristics and properties of two- and three-dimensional geometric shapes and develop mathematical arguments about geometric relationships. • Use visualization, spatial reasoning, and geometric modeling to solve problems. Grade/Level: These lessons were created as a linking/remembering device, especially for a co-taught classroom, but can be adapted or used for a regular ed, or even honors level in 9th through 12th Grade. With more modification, these lessons might be appropriate for middle school use as well. Duration/Length: Lesson #1 45 minutes Lesson #2 30 minutes Lesson #3 30 minutes Student Outcomes: Students will: • Define and differentiate between perpendicular bisector, angle bisector, segment, triangle, circle, radius, point, inscribed circle, circumscribed circle, incenter, circumcenter, and centroid. - 
												
												Circles in Areals
Circles in areals 23 December 2015 This paper has been accepted for publication in the Mathematical Gazette, and copyright is assigned to the Mathematical Association (UK). This preprint may be read or downloaded, and used for any non-profit making educational or research purpose. Introduction August M¨obiusintroduced the system of barycentric or areal co-ordinates in 1827[1, 2]. The idea is that one may attach weights to points, and that a system of weights determines a centre of mass. Given a triangle ABC, one obtains a co-ordinate system for the plane by placing weights x; y and z at the vertices (with x + y + z = 1) to describe the point which is the centre of mass. The vertices have co-ordinates A = (1; 0; 0), B = (0; 1; 0) and C = (0; 0; 1). By scaling so that triangle ABC has area [ABC] = 1, we can take the co-ordinates of P to be ([P BC]; [PCA]; [P AB]), provided that we take area to be signed. Our convention is that anticlockwise triangles have positive area. Points inside the triangle have strictly positive co-ordinates, and points outside the triangle must have at least one negative co-ordinate (we are allowed negative masses). The equation of a line looks like the equation of a plane in Cartesian co-ordinates, but note that the equation lx + ly + lz = 0 (with l 6= 0) is not satisfied by any point (x; y; z) of the Euclidean plane. We use such equations to describe the line at infinity when doing projective geometry. - 
												
												Explaining Some Collinearities Amongst Triangle Centres Christopher Bradley
Article: CJB/2011/138 Explaining some Collinearities amongst Triangle Centres Christopher Bradley Abstract: Collinearities are preserved by an Affine Transformation, so that for every collinearity of triangle centres, there is another when an affine transformation has taken place and vice-versa. We illustrate this by considering what happens to three collinearities in the Euclidean plane when an affine transformation takes the symmedian point into the incentre. C' 9 pt circle C'' A B'' Anticompleme ntary triangle F B' W' W E' L M V O X Z E F' K 4 V' K'' Y G T H N' D B U D'U' C Triplicate ratio Circle A'' A' Fig.1 Three collinearities amongst triangle centres in the Euclidean plane 1 1. When the circumconic, the nine-point conic, the triplicate ratio conic and the 7 pt. conic are all circles The condition for this is that the circumconic is a circle and then the tangents at A, B, C form a triangle A'B'C' consisting of the ex-symmedian points A'(– a2, b2, c2) and similarly for B', C' by appropriate change of sign and when this happen AA', BB', CC' are concurrent at the symmedian point K(a2, b2, c2). There are two very well-known collinearities and one less well-known, and we now describe them. First there is the Euler line containing amongst others the circumcentre O with x-co-ordinate a2(b2 + c2 – a2), the centroid G(1, 1, 1), the orthocentre H with x-co-ordinate 1/(b2 + c2 – a2) and T, the nine-point centre which is the midpoint of OH. - 
												
												The Vertex-Midpoint-Centroid Triangles
Forum Geometricorum b Volume 4 (2004) 97–109. bbb FORUM GEOM ISSN 1534-1178 The Vertex-Midpoint-Centroid Triangles Zvonko Cerinˇ Abstract. This paper explores six triangles that have a vertex, a midpoint of a side, and the centroid of the base triangle ABC as vertices. They have many in- teresting properties and here we study how they monitor the shape of ABC. Our results show that certain geometric properties of these six triangles are equivalent to ABC being either equilateral or isosceles. Let A, B, C be midpoints of the sides BC, CA, AB of the triangle ABC and − let G be its centroid (i.e., the intersection of medians AA, BB , CC ). Let Ga , + − + − + Ga , Gb , Gb , Gc , Gc be triangles BGA , CGA , CGB , AGB , AGC , BGC (see Figure 1). C − + Gb Ga B A G + − Gb Ga − + Gc Gc ABC Figure 1. Six vertex–midpoint–centroid triangles of ABC. This set of six triangles associated to the triangle ABC is a special case of the cevasix configuration (see [5] and [7]) when the chosen point is the centroid G.It has the following peculiar property (see [1]). Theorem 1. The triangle ABC is equilateral if and only if any three of the trian- = { − + − + − +} gles from the set σG Ga ,Ga ,Gb ,Gb ,Gc ,Gc have the same either perime- ter or inradius. In this paper we wish to show several similar results. The idea is to replace perimeter and inradius with other geometric notions (like k-perimeter and Brocard angle) and to use various central points (like the circumcenter and the orthocenter – see [4]) of these six triangles. - 
												
												Cyclic Quadrilaterals — the Big Picture Yufei Zhao [email protected]
Winter Camp 2009 Cyclic Quadrilaterals Yufei Zhao Cyclic Quadrilaterals | The Big Picture Yufei Zhao [email protected] An important skill of an olympiad geometer is being able to recognize known configurations. Indeed, many geometry problems are built on a few common themes. In this lecture, we will explore one such configuration. 1 What Do These Problems Have in Common? 1. (IMO 1985) A circle with center O passes through the vertices A and C of triangle ABC and intersects segments AB and BC again at distinct points K and N, respectively. The circumcircles of triangles ABC and KBN intersects at exactly two distinct points B and M. ◦ Prove that \OMB = 90 . B M N K O A C 2. (Russia 1995; Romanian TST 1996; Iran 1997) Consider a circle with diameter AB and center O, and let C and D be two points on this circle. The line CD meets the line AB at a point M satisfying MB < MA and MD < MC. Let K be the point of intersection (different from ◦ O) of the circumcircles of triangles AOC and DOB. Show that \MKO = 90 . C D K M A O B 3. (USA TST 2007) Triangle ABC is inscribed in circle !. The tangent lines to ! at B and C meet at T . Point S lies on ray BC such that AS ? AT . Points B1 and C1 lies on ray ST (with C1 in between B1 and S) such that B1T = BT = C1T . Prove that triangles ABC and AB1C1 are similar to each other. 1 Winter Camp 2009 Cyclic Quadrilaterals Yufei Zhao A B S C C1 B1 T Although these geometric configurations may seem very different at first sight, they are actually very related.