Three Properties of the Symmedian Point / Darij Grinberg
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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. -
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. -
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. -
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. -
Advanced Euclidean Geometry What Is the Center of a Triangle?
Advanced Euclidean Geometry What is the center of a triangle? But what if the triangle is not equilateral? ? Circumcenter Equally far from the vertices? P P I II Points are on the perpendicular A B bisector of a line ∆ I ≅ ∆ II (SAS) A B segment iff they PA = PB are equally far from the endpoints. P P ∆ I ≅ ∆ II (Hyp-Leg) I II AQ = QB A B A Q B Circumcenter Thm 4.1 : The perpendicular bisectors of the sides of a triangle are concurrent at a point called the circumcenter (O). A Draw two perpendicular bisectors of the sides. Label the point where they meet O (why must they meet?) Now, OA = OB, and OB = OC (why?) O so OA = OC and O is on the B perpendicular bisector of side AC. The circle with center O, radius OA passes through all the vertices and is C called the circumscribed circle of the triangle. Circumcenter (O) Examples: Orthocenter A The triangle formed by joining the midpoints of the sides of ∆ABC is called the medial triangle of ∆ABC. B The sides of the medial triangle are parallel to the original sides of the triangle. C A line drawn from a vertex to the opposite side of a triangle and perpendicular to it is an altitude. Note that in the medial triangle the perp. bisectors are altitudes. Thm 4.2: The altitudes of a triangle are concurrent at a point called the orthocenter (H). Orthocenter (H) Thm 4.2: The altitudes of a triangle are concurrent at a point called the orthocenter (H). -
Where Are the Conjugates?
Forum Geometricorum b Volume 5 (2005) 1–15. bbb FORUM GEOM ISSN 1534-1178 Where are the Conjugates? Steve Sigur Abstract. The positions and properties of a point in relation to its isogonal and isotomic conjugates are discussed. Several families of self-conjugate conics are given. Finally, the topological implications of conjugacy are stated along with their implications for pivotal cubics. 1. Introduction The edges of a triangle divide the Euclidean plane into seven regions. For the projective plane, these seven regions reduce to four, which we call the central re- gion, the a region, the b region, and the c region (Figure 1). All four of these regions, each distinguished by a different color in the figure, meet at each vertex. Equivalent structures occur in each, making the projective plane a natural back- ground for fundamental triangle symmetries. In the sense that the projective plane can be considered a sphere with opposite points identified, the projective plane di- vided into four regions by the edges of a triangle can be thought of as an octahedron projected onto this sphere, a remark that will be helpful later. 4 the b region 5 B the a region 3 the c region 1 the central region C A 6 2 7 the a region the b region Figure 1. The plane of the triangle, Euclidean and projective views A point P in any of the four regions has an harmonic associate in each of the others. Cevian lines through P and/or its harmonic associates traverse two of the these regions, there being two such possibilities at each vertex, giving 6 Cevian (including exCevian) lines. -
The Symmedian Point
Page 1 of 36 Computer-Generated Mathematics: The Symmedian Point Deko Dekov Abstract. We illustrate the use of the computer program "Machine for Questions and Answers" (The Machine) for discovering of new theorems in Euclidean Geometry. The paper contains more than 100 new theorems about the Symmedian Point, discovered by the Machine. Keywords: computer-generated mathematics, Euclidean geometry "Within ten years a digital computer will discover and prove an important mathematical theorem." (Simon and Newell, 1958). This is the famous prediction by Simon and Newell [1]. Now is 2008, 50 years later. The first computer program able easily to discover new deep mathematical theorems - The Machine for Questions and Answers (The Machine) [2,3] has been created by the author of this article, in 2006, that is, 48 years after the prediction. The Machine has discovered a few thousands new mathematical theorems, that is, more than 90% of the new mathematical computer-generated theorems since the prediction by Simon and Newell. In 2006, the Machine has produced the first computer-generated encyclopedia [2]. Given an object (point, triangle, circle, line, etc.), the Machine produces theorems related to the properties of the object. The theorems produced by the Machine are either known theorems, or possible new theorems. A possible new theorem means that the theorem is either known theorem, but the source is not available for the author of the Machine, or the theorem is a new theorem. I expect that approximately 75 to 90% of the possible new theorems are new theorems. Although the Machine works completely independent from the human thinking, the theorems produced are surprisingly similar to the theorems produced by the people. -
A New Way to Think About Triangles
A New Way to Think About Triangles Adam Carr, Julia Fisher, Andrew Roberts, David Xu, and advisor Stephen Kennedy June 6, 2007 1 Background and Motivation Around 500-600 B.C., either Thales of Miletus or Pythagoras of Samos introduced the Western world to the forerunner of Western geometry. After a good deal of work had been devoted to the ¯eld, Euclid compiled and wrote The Elements circa 300 B.C. In this work, he gathered a fairly complete backbone of what we know today as Euclidean geometry. For most of the 2,500 years since, mathematicians have used the most basic tools to do geometry|a compass and straightedge. Point by point, line by line, drawing by drawing, mathematicians have hunted for visual and intuitive evidence in hopes of discovering new theorems; they did it all by hand. Judging from the complexity and depth from the geometric results we see today, it's safe to say that geometers were certainly not su®ering from a lack of technology. In today's technologically advanced world, a strenuous e®ort is not required to transfer the ca- pabilities of a standard compass and straightedge to user-friendly software. One powerful example of this is Geometer's Sketchpad. This program has allowed mathematicians to take an experimental approach to doing geometry. With the ability to create constructions quickly and cleanly, one is able to see a result ¯rst and work towards developing a proof for it afterwards. Although one cannot claim proof by empirical evidence, the task of ¯nding interesting things to prove became a lot easier with the help of this insightful and flexible visual tool. -
The Conics of Ludwig Kiepert: a Comprehensive Lesson in the Geometry of the Triangle R
188 MATHEMATIC5 MAGAZINE The Conics of Ludwig Kiepert: A Comprehensive Lesson in the Geometry of the Triangle R. H. EDDY Memorial University of Newfoundland 5t. John's, Newfoundland, Canada A1C 557 R. FRITSCH Mathematisches Institut Ludwig-Maximilians-Unlversität D-8000 Munich 2, Germany 1. Introduction If a visitor from Mars desired to leam the geometry of the triangle but could stay in the earth' s relatively dense atmosphere only long enough for a single lesson, earthling mathematicians would, no doubt, be hard-pressed to meet this request. In this paper, we believe that we have an optimum solution to the problem. The Kiepert conics, though seemingly unknown today, constitute a significant part of the geometry of the triangle and to study them one has to deal with many fundamental concepts related to this geometry such as the Euler line, Brocard axis, circumcircle, Brocard angle, and the Lemoine line in addition to weIl-known points including the centroid, circumcen tre, orthocentre, and the isogonic centres. In the process, one comes into contact with not so weIl known, but no less important concepts, such as the Steiner point, the isodynamie points and the Spieker circle. In this paper, we show how the Kiepert' s conics are derived using both analytic and projective arguments and discuss their main properties, which we have drawn together from several sourees. We have applied some modem technology, in this case computer graphics, to produce aseries of pictures that should serve to increase the reader' s appreciation for this interesting pair of conics. In addition, we have derived some results that we were unable to locate in the available literature. -
Some Lines Through the Centroid G Christopher Bradley Abstract: Using Areal Co-Ordinates We Catalogue Some Lines Through the Centroid G of a Triangle ABC
Article: CJB/2010/121 Some Lines through the Centroid G Christopher Bradley Abstract: Using areal co-ordinates we catalogue some lines through the centroid G of a triangle ABC. Co-ordinates of points and equations of lines are given. 1. The Symmedian line A C' B' N M Ka KmG K B L C G(1,1,1) Ka K (b^2+c^2 (a^2,b^2,c^2) Km -a^2,..) (b^2+c^2,..) A' Fig. 1 The equation of the line through the three symmedian points is (b2 – c2)x + (c2 – a2)y + (a2 – b2)z = 0. (1) Km is the symmedian of the Medial triangle and Ka is the symmedian of the Anticomplementary triangle. Ka has co-ordinates (b2 + c2 – a2, c2 + a2 – b2, a2 + b2 – c2) and Km has co-ordinates (b2 + c2, c2 + a2, a2 + b2). K, of course, has co-ordinates (a2, b2, c2). 1 2. The Euler line A deL O Sch G N H B C Fig. 2 The Centroid G(1, 1, 1), the Circumcentre O(a2(b2 + c2 – a2), .., ..), the Orthocentre H(1/(b2 + c2 – a2), .., ..), the Nine-Point Centre N(a2(b2 + c2) – (b2 – c2)2, .., ..), deLongchamps point deL(– 3a4 + 2a2(b2 + c2) + (b2 – c2)2, .., ..), Schiffler’s point Sch(a/(cos B + cos C), .., ..). The equation of the Euler line is (b2 + c2 – a2)(b2 – c2)x + (c2 + a2 – b2)(c2 – a2)y + (a2 + b2 – c2)(a2 – b2)z = 0. (2) Schiffler’s point is the intersection of the Euler’s line of ABC, IBC, ICA, IAB, where I is the incentre.