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Point of Concurrency the Three Perpendicular Bisectors of a Triangle Intersect at a Single Point
3.1 Special Segments and Centers of Classifications of Triangles: Triangles By Side: 1. Equilateral: A triangle with three I CAN... congruent sides. Define and recognize perpendicular bisectors, angle bisectors, medians, 2. Isosceles: A triangle with at least and altitudes. two congruent sides. 3. Scalene: A triangle with three sides Define and recognize points of having different lengths. (no sides are concurrency. congruent) Jul 249:36 AM Jul 249:36 AM Classifications of Triangles: Special Segments and Centers in Triangles By angle A Perpendicular Bisector is a segment or line 1. Acute: A triangle with three acute that passes through the midpoint of a side and angles. is perpendicular to that side. 2. Obtuse: A triangle with one obtuse angle. 3. Right: A triangle with one right angle 4. Equiangular: A triangle with three congruent angles Jul 249:36 AM Jul 249:36 AM Point of Concurrency The three perpendicular bisectors of a triangle intersect at a single point. Two lines intersect at a point. The point of When three or more lines intersect at the concurrency of the same point, it is called a "Point of perpendicular Concurrency." bisectors is called the circumcenter. Jul 249:36 AM Jul 249:36 AM 1 Circumcenter Properties An angle bisector is a segment that divides 1. The circumcenter is an angle into two congruent angles. the center of the circumscribed circle. BD is an angle bisector. 2. The circumcenter is equidistant to each of the triangles vertices. m∠ABD= m∠DBC Jul 249:36 AM Jul 249:36 AM The three angle bisectors of a triangle Incenter properties intersect at a single point. -
Median and Altitude of a Triangle Goal: • to Use Properties of the Medians
Median and Altitude of a Triangle Goal: • To use properties of the medians of a triangle. • To use properties of the altitudes of a triangle. Median of a Triangle Median of a Triangle – a segment whose endpoints are the vertex of a triangle and the midpoint of the opposite side. Vertex Median Median of an Obtuse Triangle A D Point of concurrency “P” or centroid E P C B F Medians of a Triangle The medians of a triangle intersect at a point that is two-thirds of the distance from each vertex to the midpoint of the opposite side. A D If P is the centroid of ABC, then AP=2 AF E 3 P C CP=22CE and BP= BD B F 33 Example - Medians of a Triangle P is the centroid of ABC. PF 5 Find AF and AP A D E P C B F 5 Median of an Acute Triangle A Point of concurrency “P” or centroid E D P C B F Median of a Right Triangle A F Point of concurrency “P” or centroid E P C B D The three medians of an obtuse, acute, and a right triangle always meet inside the triangle. Altitude of a Triangle Altitude of a triangle – the perpendicular segment from the vertex to the opposite side or to the line that contains the opposite side A altitude C B Altitude of an Acute Triangle A Point of concurrency “P” or orthocenter P C B The point of concurrency called the orthocenter lies inside the triangle. Altitude of a Right Triangle The two legs are the altitudes A The point of concurrency called the orthocenter lies on the triangle. -
Exploring Some Concurrencies in the Tetrahedron
Exploring some concurrencies in the tetrahedron Consider the corresponding analogues of perpendicular bisectors, angle bisectors, medians and altitudes of a triangle for a tetrahedron, and investigate which of these are also concurrent for a tetrahedron. If true, prove it; if not, provide a counter-example. Let us start with following useful assumption (stated without proof) for 3D: Three (non-parallel) planes meet in a point. (It is easy to see this in a rectangular room where two walls and the ceiling meet perpendicularly in a point.) Definition 1: The perpendicular edge bisector is the A generalisation to 3D of the concept of a perpendicular bisector. For example, the perpendicular edge bisector of AB is the plane through the midpoint of AB, and perpendicular D to it. Or equivalently, it is the set of points equidistant from A and B. B Theorem 1: The six perpendicular edge bisectors of a C tetrahedron all meet at the circumcentre. Proof: Let S be the intersection of the three perpendicular edge bisectors of AB, BC, CD. Therefore SA = SB = SC = SD. Therefore S must lie on all 6 perpendicular edge bisectors, and the sphere, centre S and radius SA, goes through all four vertices. Definition 2: Let a, b, c, d denote the four faces of a A tetrahedron ABCD. The face angle bisector of ab is the plane through CD bisecting the angle between b the faces a, b. It is therefore also the set of points equidistant from a and b. D a Theorem 2: The six face angle bisectors of a B tetrahedron all meet at the incentre. -
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. -
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. -
Midpoints and Medians in Triangles
MPM1D Midpoints and Medians in Triangles Midpoint: B A C 1) Use a ruler to create a midpoint for AB. Label it X. 2) Use a ruler to create a midpoint for AC. Label it Y. 3) Join the points X and Y with a line. 4) Measure all the sides. Write them on the diagram. Q: What do you notice about the lenGth of XY and BC? Q: What do you notice about the direction of the lines XY and BC? Q: How could we prove this? The line seGments joining the midpoint of two sides of a triangle is _________________________ the third side and ___________________________ . Recall: The area of a trianGle is: � = Median: A line from one vertex to the midpoint of the opposite side. A B C 1) Draw the midpoint of BC, label it M 2) Draw the median from A to BC 3) Calculate the area of rABM and rAMC (MeasurinG any lenGths you need to find the area) Medians of a trianGle ________ its area. Example 2 The area of rXYZ is 48 ��(. A median is drawn from X, and the midpoint of the opposite side is labelled M. What is the area of rXYM? 1. Find the length of line segment MN in each triangle. a) b) c) d) 2. Find the lengths of line segments AD and DE in each triangle. a) b) c) d) 3. The area of rABC is 10 cm2. Calculate the area of each triangle. a) rABD b) rADC 4. Calculate the area of each triangle given the area of rMNQ is 12 cm2. -
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. -
Arxiv:2101.02592V1 [Math.HO] 6 Jan 2021 in His Seminal Paper [10]
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 arXiv:2101.02592v1 [math.HO] 6 Jan 2021 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. -
The Uses of Homogeneous Barycentric Coordinates in Plane Euclidean Geometry by PAUL
The uses of homogeneous barycentric coordinates in plane euclidean geometry by PAUL YIU Department of Mathematical Sciences, Florida Atlantic University, Boca Raton, FL 33431, USA The notion of homogeneous barycentric coordinates provides a pow- erful tool of analysing problems in plane geometry. In this paper, we explain the advantages over the traditional use of trilinear coordinates, and illustrate its powerfulness in leading to discoveries of new and in- teresting collinearity relations of points associated with a triangle. 1. Introduction In studying geometric properties of the triangle by algebraic methods, it has been customary to make use of trilinear coordinates. See, for examples, [3, 6, 7, 8, 9]. With respect to a ¯xed triangle ABC (of side lengths a, b, c, and opposite angles ®, ¯, °), the trilinear coordinates of a point is a triple of numbers proportional to the signed distances of the point to the sides of the triangle. The late Jesuit mathematician Maurice Wong has given [9] a synthetic construction of the point with trilinear coordinates cot ® : cot ¯ : cot °, and more generally, in [8] points with trilinear coordinates a2nx : b2ny : c2nz from one with trilinear coordinates x : y : z with respect to a triangle with sides a, b, c. On a much grandiose scale, Kimberling [6, 7] has given extensive lists of centres associated with a triangle, in terms of trilinear coordinates, along with some collinearity relations. The present paper advocates the use of homogeneous barycentric coor- dinates instead. The notion of barycentric coordinates goes back to MÄobius. See, for example, [4]. With respect to a ¯xed triangle ABC, we write, for every point P on the plane containing the triangle, 1 P = (( P BC)A + ( P CA)B + ( P AB)C); ABC 4 4 4 4 and refer to this as the barycentric coordinate of P . -
The Euler Line in Non-Euclidean Geometry
California State University, San Bernardino CSUSB ScholarWorks Theses Digitization Project John M. Pfau Library 2003 The Euler Line in non-Euclidean geometry Elena Strzheletska Follow this and additional works at: https://scholarworks.lib.csusb.edu/etd-project Part of the Mathematics Commons Recommended Citation Strzheletska, Elena, "The Euler Line in non-Euclidean geometry" (2003). Theses Digitization Project. 2443. https://scholarworks.lib.csusb.edu/etd-project/2443 This Thesis is brought to you for free and open access by the John M. Pfau Library at CSUSB ScholarWorks. It has been accepted for inclusion in Theses Digitization Project by an authorized administrator of CSUSB ScholarWorks. For more information, please contact [email protected]. THE EULER LINE IN NON-EUCLIDEAN GEOMETRY A Thesis Presented to the Faculty of California State University, San Bernardino In Partial Fulfillment of the Requirements for the Degree Master of Arts in Mathematics by Elena Strzheletska December 2003 THE EULER LINE IN NON-EUCLIDEAN GEOMETRY A Thesis Presented to the Faculty of California State University, San Bernardino by Elena Strzheletska December 2003 Approved by: Robert Stein, Committee Member Susan Addington, Committee Member Peter Williams, Chair Terry Hallett, Department of Mathematics Graduate Coordinator Department of Mathematics ABSTRACT In Euclidean geometry, the circumcenter and the centroid of a nonequilateral triangle determine a line called the Euler line. The orthocenter of the triangle, the point of intersection of the altitudes, also belongs to this line. The main purpose of this thesis is to explore the conditions of the existence and the properties of the Euler line of a triangle in the hyperbolic plane. -
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.