Introduction to the Geometry of the Triangle
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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. -
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. -
Deko Dekov, Anticevian Corner Triangles PDF, 101
Journal of Computer-Generated Euclidean Geometry Anticevian Corner Triangles Deko Dekov Abstract. By using the computer program "Machine for Questions and Answers", we study perspectors of basic triangles and triangles of triangle centers of anticevian corner triangles. Given a triangle ABC and a triangle center of kind 1, labeled by P. Let A1B1C1 be the anticevian triangle of P. Construct triangle centers A2, B2, C2 of kind 2 (possibly different from the kind 1) of triangles A1BC, B1CA, C1AB, respectively. We call triangle A2B2C2 the Triangle of the Triangle Centers of kind 2 of the Anticevian Corner triangles of the Triangle Center of kind 1. See the Figure: P - Triangle Center of kind 1; A1B1C1 - Anticevian Triangle of P; A2, B2, C2 - Triangle Centers of kind 2 of triangles A1BC, B1CA, C1AB, respectively; A2B2C2 - Triangle of the Triangle Centers of kind 2 of the Anticevian Corner Triangles of Journal of Computer-Generated Euclidean Geometry 2007 No 5 Page 1 of 14 the Triangle Center of kind 1. In this Figure: P - Incenter; A1B1C1 - Anticevian Triangle of the Incenter = Excentral Triangle; A2, B2, C2 - Centroids of triangles A1BC, B1CA, C1AB, respectively; A2B2C2 - Triangle of the Centroids of the Anticevian Corner Triangles of the Incenter. Known result (the reader is invited to submit a note/paper with additional references): Triangle ABC and the Triangle of the Incenters of the Anticevian Corner Triangles of the Incenter are perspective with perspector the Second de Villiers Point. See the Figure: A1B1C1 - Anticevian Triangle of the Incenter = Excentral Triangle; A2B2C2 - Triangle of the Incenters of the Anticevian Corner Triangles of the Incenter; V - Second de Villiers Point = perspector of triangles A1B1C1 and A2B2C2. -
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. -
On a Construction of Hagge
Forum Geometricorum b Volume 7 (2007) 231–247. b b FORUM GEOM ISSN 1534-1178 On a Construction of Hagge Christopher J. Bradley and Geoff C. Smith Abstract. In 1907 Hagge constructed a circle associated with each cevian point P of triangle ABC. If P is on the circumcircle this circle degenerates to a straight line through the orthocenter which is parallel to the Wallace-Simson line of P . We give a new proof of Hagge’s result by a method based on reflections. We introduce an axis associated with the construction, and (via an areal anal- ysis) a conic which generalizes the nine-point circle. The precise locus of the orthocenter in a Brocard porism is identified by using Hagge’s theorem as a tool. Other natural loci associated with Hagge’s construction are discussed. 1. Introduction One hundred years ago, Karl Hagge wrote an article in Zeitschrift fur¨ Mathema- tischen und Naturwissenschaftliche Unterricht entitled (in loose translation) “The Fuhrmann and Brocard circles as special cases of a general circle construction” [5]. In this paper he managed to find an elegant extension of the Wallace-Simson theorem when the generating point is not on the circumcircle. Instead of creating a line, one makes a circle through seven important points. In 2 we give a new proof of the correctness of Hagge’s construction, extend and appl§ y the idea in various ways. As a tribute to Hagge’s beautiful insight, we present this work as a cente- nary celebration. Note that the name Hagge is also associated with other circles [6], but here we refer only to the construction just described. -
The Role of Euclidean Geometry in High School
JOURNAL OF MATHEMATICAL BEHAVIOR 15, 221-237 (1996) The Role of Euclidean Geometry in High School HUNG-HSI Wu University of California When I was a high school student long ago, the need to study Euclidean geome- try was taken for granted. In fact, the novelty of learning to prove something was so overwhelming to me and some of my friends that, years later, we would look back at Euclidean geometry as the high point of our mathematics education. In recent years, the value of Euclid has depreciated considerably. A vigorous debate is now going on about how much, if any, of his work is still relevant. In this article, I would like to give my perspective on this subject, and in so doing, I will be taking a philosophical tour, so to speak, of a small portion of mathematics. This may not be so surprising because after all, in life as in mathematics, one does need a little philosophy for guidance from time to time. The bone of contention in the geometry curriculum is of course how many of the traditional “two-column proofs” should be retained. Let me first briefly indicate the range of available options in this regard: 1. A traditional text in use in a local high school in Berkeley not too long ago begins on page 1 with the undefined terms of the axioms, and formal proofs start on page 22. There is no motivation or explanation of the whys and hows of an axiomatic system. 2. A recently published text does experimental geometry (i.e., “hands-on ge- ometry” with no proofs) all the way until the last 130 pages of its 700 pages of exposition. -
A Conic Through Six Triangle Centers
Forum Geometricorum b Volume 2 (2002) 89–92. bbb FORUM GEOM ISSN 1534-1178 A Conic Through Six Triangle Centers Lawrence S. Evans Abstract. We show that there is a conic through the two Fermat points, the two Napoleon points, and the two isodynamic points of a triangle. 1. Introduction It is always interesting when several significant triangle points lie on some sort of familiar curve. One recently found example is June Lester’s circle, which passes through the circumcenter, nine-point center, and inner and outer Fermat (isogonic) points. See [8], also [6]. The purpose of this note is to demonstrate that there is a conic, apparently not previously known, which passes through six classical triangle centers. Clark Kimberling’s book [6] lists 400 centers and innumerable collineations among them as well as many conic sections and cubic curves passing through them. The list of centers has been vastly expanded and is now accessible on the internet [7]. Kimberling’s definition of triangle center involves trilinear coordinates, and a full explanation would take us far afield. It is discussed both in his book and jour- nal publications, which are readily available [4, 5, 6, 7]. Definitions of the Fermat (isogonic) points, isodynamic points, and Napoleon points, while generally known, are also found in the same references. For an easy construction of centers used in this note, we refer the reader to Evans [3]. Here we shall only require knowledge of certain collinearities involving these points. When points X, Y , Z, . are collinear we write L(X,Y,Z,...) to indicate this and to denote their common line. -
Prove It Classroom
How to Prove it ClassRoom SHAILESH SHIRALI Euler’s formula for the area of a pedal triangle Given a triangle ABC and a point P in the plane of ABC In this episode of “How (note that P does not have to lie within the triangle), the To Prove It”, we prove pedal triangle of P with respect to ABC is the triangle △ a beautiful and striking whose vertices are the feet of the perpendiculars drawn from formula first found by P to the sides of ABC. See Figure 1. The pedal triangle relates Leonhard Euler; it gives the area of the pedal triangle in a natural way to the parent triangle, and we may wonder of a point with reference whether there is a convenient formula giving the area of the to another triangle. pedal triangle in terms of the parameters of the parent triangle. The great 18th-century mathematician Euler found just such a formula (given in Box 1). It is a compact and pleasing result, and it expresses the area of the pedal triangle in terms of the radius R of the circumcircle of ABC and the △ distance between P and the centre O of the circumcircle. A O: circumcentre of ABC E △ P: arbitrary point Area ( DEF) 1 OP2 F △ = 1 P Area ( ABC) 4 − R2 O △ ( ) B D C Figure 1. Euler's formula for the area of the pedal triangle of an arbitrary point Keywords: Circle theorem, pedal triangle, power of a point, Euler, sine rule, extended sine rule, Wallace-Simson theorem Azim Premji University At Right Angles, July 2018 95 1 A E F P O B D C Figure 2. -
Volume 3 2003
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 3 2003 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 Bernard Gibert, Orthocorrespondence and orthopivotal cubics,1 Alexei Myakishev, -
The Radii by R, Ra, Rb,Rc
The Orthopole, by R. GOORMAGHTIGH,La Louviero, Belgium. Introduction. 1. In this little work are presented the more important resear- ches on one of the newest chapters in the modern Geometry of the triangle, and whose origin is, the following theorem due to. Professor J. Neuberg: Let ƒ¿, ƒÀ, y be the projections of A, B, C on a straight line m; the perpendiculars from a on BC, ƒÀ on CA, ƒÁ on ABare concurrent(1). The point of concurrence. M is called the orthopole of m for the triangle ABC. The following general conventions will now be observed : the sides of the triangle of reference ABC will be denoted by a, b, c; the centre and radius of the circumcircle by 0 and R; the orthocentre by H; the centroid by G; the in- and ex-centresby I, Ia 1b, Ic and the radii by r, ra, rb,rc;the mid-points of BC, CA, AB by Am, Bm, Cm;the feet of the perpendiculars AH, BH, OH from A, B, C on BO, CA, AB by An,Bn, Cn; the contact points of the incircle with BC, CA, AB by D, E, F; the corresponding points for the circles Ia, Ib, Ic, by Da, Ea,Fa, Dc, Eb, F0, Dc, Ec, Fc; the area of ABC by 4; the centre of the nine-point circle by 0g; the Lemoine . point by K; the Brocard points by theGergonne and Nagel points by and N(2); further 2s=a+b+c. Two points whose joins to A, B, C are isogonal conjugate in the angles A, B, C are counterpoints. -
Fermat Points and Euler Lines Preliminaries 1 the Fermat
Fermat Points and Euler Lines N. Beluhov, A. Zaslavsky, P. Kozhevnikov, D. Krekov, O. Zaslavsky1 Preliminaries Given a nonequilateral 4ABC, we write M, H, O, I, Ia, Ib, and Ic for its medicenter, orthocenter, circumcenter, incenter, and excenters opposite A, B, and C. The points O, M, and H lie on a line known as the Euler line. The point M divides the segment OH in ratio 1 : 2. The midpoints of the sides of 4ABC, the feet of its altitudes, and the midpoints of the segments AH, BH, and CH lie on a circle known as the Euler circle (or the nine-point circle). Its center is the midpoint E of the segment OH. Let X be an arbitrary point in the plane of 4ABC. Then the reflections of the lines AX, BX, and CX in the bisectors of 6 A, 6 B, and 6 C meet at a point X∗ (or are parallel, in which case X∗ is a point at infinity) known as the isogonal conjugate of X. The point L = M ∗ is known as Lemoine's point. For every triangle there exists an affine transformation mapping it onto an equi- lateral triangle. The inverse transformation maps the circumcircle and the incircle of that equilateral triangle onto two ellipses known as the circumscribed and inscribed Steiner ellipses. 1 The Fermat, Napoleon, and Apollonius points Let 4ABTc, 4BCTa, and 4CATb be equilateral triangles constructed externally on the sides of 4ABC and let Na, Nb, and Nc be their centers. Analogously, let 0 0 0 4ABTc, 4BCTa, and 4CATb be equilateral triangles constructed internally on the 0 0 0 sides of 4ABC and let Na, Nb, and Nc be their centers. -
From Euler to Ffiam: Discovery and Dissection of a Geometric Gem
From Euler to ffiam: Discovery and Dissection of a Geometric Gem Douglas R. Hofstadter Center for Research on Concepts & Cognition Indiana University • 510 North Fess Street Bloomington, Indiana 47408 December, 1992 ChapterO Bewitched ... by Circles, Triangles, and a Most Obscure Analogy Although many, perhaps even most, mathematics students and other lovers of mathematics sooner or later come across the famous Euler line, somehow I never did do so during my student days, either as a math major at Stanford in the early sixties or as a math graduate student at Berkeley for two years in the late sixties (after which I dropped out and switched to physics). Geometry was not by any means a high priority in the math curriculum at either institution. It passed me by almost entirely. Many, many years later, however, and quite on my own, I finally did become infatuated - nay, bewitched- by geometry. Just plane old Euclidean geometry, I mean. It all came from an attempt to prove a simple fact about circles that I vaguely remembered from a course on complex variables that I took some 30 years ago. From there on, the fascination just snowballed. I was caught completely off guard. Never would I have predicted that Doug Hofstadter, lover of number theory and logic, would one day go off on a wild Euclidean-geometry jag! But things are unpredictable, and that's what makes life interesting. I especially came to love triangles, circles, and their unexpectedly profound interrelations. I had never appreciated how intimately connected these two concepts are. Associated with any triangle are a plentitude of natural circles, and conversely, so many beautiful properties of circles cannot be defined except in terms of triangles.