Geometry Conic and Triangle
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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 Isogonal Tripolar Conic
Forum Geometricorum b Volume 1 (2001) 33–42. bbb FORUM GEOM The Isogonal Tripolar Conic Cyril F. Parry Abstract. In trilinear coordinates with respect to a given triangle ABC,we define the isogonal tripolar of a point P (p, q, r) to be the line p: pα+qβ+rγ = 0. We construct a unique conic Φ, called the isogonal tripolar conic, with respect to which p is the polar of P for all P . Although the conic is imaginary, it has a real center and real axes coinciding with the center and axes of the real orthic inconic. Since ABC is self-conjugate with respect to Φ, the imaginary conic is harmonically related to every circumconic and inconic of ABC. In particular, Φ is the reciprocal conic of the circumcircle and Steiner’s inscribed ellipse. We also construct an analogous isotomic tripolar conic Ψ by working with barycentric coordinates. 1. Trilinear coordinates For any point P in the plane ABC, we can locate the right projections of P on the sides of triangle ABC at P1, P2, P3 and measure the distances PP1, PP2 and PP3. If the distances are directed, i.e., measured positively in the direction of −→ −→ each vertex to the opposite side, we can identify the distances α =PP1, β =PP2, −→ γ =PP3 (Figure 1) such that aα + bβ + cγ =2 where a, b, c, are the side lengths and area of triangle ABC. This areal equation for all positions of P means that the ratio of the distances is sufficient to define the trilinear coordinates of P (α, β, γ) where α : β : γ = α : β : γ. -
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. -
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. -
Lemmas in Euclidean Geometry1 Yufei Zhao [email protected]
IMO Training 2007 Lemmas in Euclidean Geometry Yufei Zhao Lemmas in Euclidean Geometry1 Yufei Zhao [email protected] 1. Construction of the symmedian. Let ABC be a triangle and Γ its circumcircle. Let the tangent to Γ at B and C meet at D. Then AD coincides with a symmedian of △ABC. (The symmedian is the reflection of the median across the angle bisector, all through the same vertex.) A A A O B C M B B F C E M' C P D D Q D We give three proofs. The first proof is a straightforward computation using Sine Law. The second proof uses similar triangles. The third proof uses projective geometry. First proof. Let the reflection of AD across the angle bisector of ∠BAC meet BC at M ′. Then ′ ′ sin ∠BAM ′ ′ BM AM sin ∠ABC sin ∠BAM sin ∠ABD sin ∠CAD sin ∠ABD CD AD ′ = ′ sin ∠CAM ′ = ∠ ∠ ′ = ∠ ∠ = = 1 M C AM sin ∠ACB sin ACD sin CAM sin ACD sin BAD AD BD Therefore, AM ′ is the median, and thus AD is the symmedian. Second proof. Let O be the circumcenter of ABC and let ω be the circle centered at D with radius DB. Let lines AB and AC meet ω at P and Q, respectively. Since ∠PBQ = ∠BQC + ∠BAC = 1 ∠ ∠ ◦ 2 ( BDC + DOC) = 90 , we see that PQ is a diameter of ω and hence passes through D. Since ∠ABC = ∠AQP and ∠ACB = ∠AP Q, we see that triangles ABC and AQP are similar. If M is the midpoint of BC, noting that D is the midpoint of QP , the similarity implies that ∠BAM = ∠QAD, from which the result follows. -
Volume 6 (2006) 1–16
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 6 2006 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: 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 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 Khoa Lu Nguyen and Juan Carlos Salazar, On the mixtilinear incircles and excircles,1 Juan Rodr´ıguez, Paula Manuel and Paulo Semi˜ao, A conic associated with the Euler line,17 Charles Thas, A note on the Droz-Farny theorem,25 Paris Pamfilos, The cyclic complex of a cyclic quadrilateral,29 Bernard Gibert, Isocubics with concurrent normals,47 Mowaffaq Hajja and Margarita Spirova, A characterization of the centroid using June Lester’s shape function,53 Christopher J. -
Degree of Triangle Centers and a Generalization of the Euler Line
Beitr¨agezur Algebra und Geometrie Contributions to Algebra and Geometry Volume 51 (2010), No. 1, 63-89. Degree of Triangle Centers and a Generalization of the Euler Line Yoshio Agaoka Department of Mathematics, Graduate School of Science Hiroshima University, Higashi-Hiroshima 739–8521, Japan e-mail: [email protected] Abstract. We introduce a concept “degree of triangle centers”, and give a formula expressing the degree of triangle centers on generalized Euler lines. This generalizes the well known 2 : 1 point configuration on the Euler line. We also introduce a natural family of triangle centers based on the Ceva conjugate and the isotomic conjugate. This family contains many famous triangle centers, and we conjecture that the de- gree of triangle centers in this family always takes the form (−2)k for some k ∈ Z. MSC 2000: 51M05 (primary), 51A20 (secondary) Keywords: triangle center, degree of triangle center, Euler line, Nagel line, Ceva conjugate, isotomic conjugate Introduction In this paper we present a new method to study triangle centers in a systematic way. Concerning triangle centers, there already exist tremendous amount of stud- ies and data, among others Kimberling’s excellent book and homepage [32], [36], and also various related problems from elementary geometry are discussed in the surveys and books [4], [7], [9], [12], [23], [26], [41], [50], [51], [52]. In this paper we introduce a concept “degree of triangle centers”, and by using it, we clarify the mutual relation of centers on generalized Euler lines (Proposition 1, Theorem 2). Here the term “generalized Euler line” means a line connecting the centroid G and a given triangle center P , and on this line an infinite number of centers lie in a fixed order, which are successively constructed from the initial center P 0138-4821/93 $ 2.50 c 2010 Heldermann Verlag 64 Y. -
Chapter 10 Isotomic and Isogonal Conjugates
Chapter 10 Isotomic and isogonal conjugates 10.1 Isotomic conjugates The Gergonne and Nagel points are examples of isotomic conjugates. Two points P and Q (not on any of the side lines of the reference triangle) are said to be isotomic conjugates if their respective traces are symmetric with respect to the midpoints of the corresponding sides. Thus, BX = X′C, CY = Y ′A, AZ = Z′B. B X′ Z X Z′ P P • C A Y Y ′ We shall denote the isotomic conjugate of P by P •. If P =(x : y : z), then 1 1 1 P • = : : =(yz : zx : xy). x y z 320 Isotomic and isogonal conjugates 10.1.1 The Gergonne and Nagel points 1 1 1 Ge = s a : s b : s c , Na =(s a : s : s c). − − − − − − Ib A Ic Z′ Y Y Z I ′ Na Ge B C X X′ Ia 10.1 Isotomic conjugates 321 10.1.2 The isotomic conjugate of the orthocenter The isotomic conjugate of the orthocenter is the point 2 2 2 2 2 2 2 2 2 H• =(b + c a : c + a b : a + b c ). − − − Its traces are the pedals of the deLongchamps point Lo, the reflection of H in O. A Z′ L Y o O Y ′ H• Z H B C X X′ Exercise 1. Let XYZ be the cevian triangle of H•. Show that the lines joining X, Y , Z to the midpoints of the corresponding altitudes are concurrent. What is the common point? 1 2. Show that H• is the perspector of the triangle of reflections of the centroid G in the sidelines of the medial triangle. -
Sava Grozdev and Deko Dekov, the Lester Circle Is Orthogonal to The
Journal of Computer-Generated Mathematics Problem 6 The Lester circle is orthogonal to the Orthocentroidal Circle of the Triangle of the Orthocenters of the Triangulation Triangles of the Tarry Point. Sava Grozdev and Deko Dekov Submitted on May 1, 2014 Publication Date: July 10, 2014 At the present time, there are seven notable circles known to be orthogonal to the Lester circle. See [1]. The below problem introduces a new notable circle which is orthogonal to the Lester circle. Prove the following problem, produced by the computer program “Discoverer”: Problem 6. Given ABC . The Lester circle of ABC is the circle passing through the circumcenter, nine-point center and the outer Fermat point. The Tarry point P is the intersection point of the circumcircle and the line passing through the centroid and the midpoint of the circumcenter and the Lemoine (Symmedian) point. Let D, E and F are the orthocenters of triangles PBC , PCA and PAB , respectively, G is the centroid of DEF and H is the orthocenter of DEF . Prove that the circle having as diameter the segment GH is orthogonal to the Lester circle. Short form of the problem: Problem 6. Prove that the Lester circle is orthogonal to the Orthocentroidal Circle of the Triangle of the Orthocenters of the Triangulation Triangles of the Tarry Point. The reader may find the definitions in [2-5]. Please submit the solution of the problem for publication in this journal to the editor of this journal: [email protected] See the figure: Journal of Computer-Generated Mathematics 2014 Problem no 4 Page 1 of 3 In the figure: c – Lester circle, P – Tarry Point, D – Orthocenter of Triangle PBC, E – Orthocenter of Triangle PCA, F – Orthocenter of Triangle PAB, c1 – Orthocentroidal Circle of Triangle DEF, Circle c1 is orthogonal to the Lester circle. -
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. -
Let's Talk About Symmedians!
Let's Talk About Symmedians! Sammy Luo and Cosmin Pohoata Abstract We will introduce symmedians from scratch and prove an entire collection of interconnected results that characterize them. Symmedians represent a very important topic in Olympiad Geometry since they have a lot of interesting properties that can be exploited in problems. But first, what are they? Definition. In a triangle ABC, the reflection of the A-median in the A-internal angle bi- sector is called the A-symmedian of triangle ABC. Similarly, we can define the B-symmedian and the C-symmedian of the triangle. A I BC X M Figure 1: The A-symmedian AX Do we always have symmedians? Well, yes, only that we have some weird cases when for example ABC is isosceles. Then, if, say AB = AC, then the A-median and the A-internal angle bisector coincide; thus, the A-symmedian has to coincide with them. Now, symmedians are concurrent from the trigonometric form of Ceva's theorem, since we can just cancel out the sines. This concurrency point is called the symmedian point or the Lemoine point of triangle ABC, and it is usually denoted by K. //As a matter of fact, we have the more general result. Theorem -1. Let P be a point in the plane of triangle ABC. Then, the reflections of the lines AP; BP; CP in the angle bisectors of triangle ABC are concurrent. This concurrency point is called the isogonal conjugate of the point P with respect to the triangle ABC. We won't dwell much on this more general notion here; we just prove a very simple property that will lead us immediately to our first characterization of symmedians. -
The Brocard and Tucker Circles of a Cyclic Quadrilateral
ei The Brocard and Tucker Circles of a Cyclic Quadrilateral. By FREDERICK G. W. BROWN. (Received 4th September 1917. Bead 9th November 1917.) 1. The application of the geometrical properties of the Brocard and Tucker circles of a triangle to a quadrilateral appears never to have been adequately worked out, as far as the author can discover. Hence, the object of this paper. Some of the problems involved have been published, under the author's name, as independent questions for solution, and where, in the author's opinion, solutions other than his own have seemed more satisfactory for the logical treatment of the subject, these solutions have been employed, with due acknowledgments to their authors. 2. Condition for Brocard poinU. We shall, first establish the condition necessary for the existence of Brocard points within a quadrilateral. Let ABCD (Fig. 1) be a quadrilateral in which a point X can be found such that the L. ' XAD, XBA, XCB, XDC are all equal; denote each of these angles by o>; the sides BC, CD, DA, AB by a, b, c, d; the diagonals BD, AC by e,f, and the area by Q; then L AXB = ir -to - (A — <a) = TT - A. Similarly LBXC = TT-B, L.CXD = TT-C, LDXA^TT-D. Now AX: sin w = AB : sin AXB = d: sin (JT - A) = d : sin A and AX: sin (Z? - w) = c : sin (ir - D) = c : sin D, hence, eliminating AX by division sin (D -<o): sin <a — d sin D : c sin A, Downloaded from https://www.cambridge.org/core.