The Triangle As an Orthocentric Quadrangle
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A Study of Dandelin Spheres: a Second Year Investigation
CALIFORNIA STATE SCIENCE FAIR 2009 PROJECT SUMMARY Name(s) Project Number Sundeep Bekal S1602 Project Title A Study of Dandelin Spheres: A Second Year Investigation Abstract Objectives/Goals The purpose of this investigation is to see if there is a relationship that forms between the radius of the smaller dandelin sphere to the tangent of the angle RLM and distances that form inside the ellipse. Methods/Materials I investigated a conic section of an ellipse that contained two Dandelin Spheres to determine if there was such a relationship. I used the same program as last year, Geometer's Sketchpad, to generate measurements of last year's 2D model of the conic section so that I could better investigate the relationship. After slicing the spheres along the central pole, I looked to see if there were any patterns or relationships. To do this I set the radius of the smaller circle (sphere in 3-D) to 1.09 cm. I changed the radius by .05 cm every time while also noting how the other segments changed or did not change. Materials -Geometer's Sketchpad -Calculator -Computer -Paper -Pencil Results After looking through the data tables I noticed that the radius had the same exact values as (a+c)(tan(1/2) angleRLM), where (a+c) is the distance located in the ellipse. I also noticed that the ratio of the (radius)/(a+c) had the same values as tan((1/2 angleRLM)). I figured that this would be true because the tan (theta)=(opposite)/(adjacent) which in this case, would be tan ((1/2 angleRLM)) = (radius)/(a+c). -
Some Concurrencies from Tucker Hexagons
Forum Geometricorum b Volume 2 (2002) 5–13. bbb FORUM GEOM ISSN 1534-1178 Some Concurrencies from Tucker Hexagons Floor van Lamoen Abstract. We present some concurrencies in the figure of Tucker hexagons to- gether with the centers of their Tucker circles. To find the concurrencies we make use of extensions of the sides of the Tucker hexagons, isosceles triangles erected on segments, and special points defined in some triangles. 1. The Tucker hexagon Tφ and the Tucker circle Cφ Consider a scalene (nondegenerate) reference triangle ABC in the Euclidean plane, with sides a = BC, b = CA and c = AB. Let Ba be a point on the sideline CA. Let Ca be the point where the line through Ba antiparallel to BC meets AB. Then let Ac be the point where the line through Ca parallel to CA meets BC. Continue successively the construction of parallels and antiparallels to complete a hexagon BaCaAcBcCbAb of which BaCa, AcBc and CbAb are antiparallel to sides BC, CA and AB respectively, while BcCb, AcCa and AbBa are parallel to these respective sides. B Cb K Ab B T K Ac Ca KC KA C Bc Ba A Figure 1 This is the well known way to construct a Tucker hexagon. Each Tucker hexagon is circumscribed by a circle, the Tucker circle. The three antiparallel sides are congruent; their midpoints KA, KB and KC lie on the symmedians of ABC in such a way that AKA : AK = BKB : BK = CKC : CK, where K denotes the symmedian point. See [1, 2, 3]. 1.1. Identification by central angles. -
USA Mathematical Talent Search Solutions to Problem 2/3/16
USA Mathematical Talent Search Solutions to Problem 2/3/16 www.usamts.org 2/3/16. Find three isosceles triangles, no two of which are congruent, with integer sides, such that each triangle’s area is numerically equal to 6 times its perimeter. Credit This is a slight modification of a problem provided by Suresh T. Thakar of India. The original problem asked for five isosceles triangles with integer sides such that the area is numerically 12 times the perimeter. Comments Many students simply set up an equation using Heron’s formula and then turned to a calculator or a computer for a solution. Below are presented more elegant solutions. Zachary Abel shows how to reduce this problem to finding Pythagorean triples which have 12 among the side lengths. Adam Hesterberg gives us a solution using Heron’s Create PDF with GO2PDFformula. for free, if Finally,you wish to remove Kristin this line, click Cordwell here to buy Virtual shows PDF Printer how to take an intelligent trial-and-error approach to construct the solutions. Solutions edited by Richard Rusczyk. Solution 1 by: Zachary Abel (11/TX) In 4ABC with AB = AC, we use the common notations r = inradius, s = semiperimeter, p = perimeter, K = area, a = BC, and b = AC. The diagram shows triangle ABC with its incircle centered at I and tangent to BC and AC at M and N respectively. A x h N 12 I a/2 12 B M a/2 C The area of the triangle is given by rs = K = 6p = 12s, which implies r = 12. -
Relationships Between Six Excircles
Sangaku Journal of Mathematics (SJM) c SJM ISSN 2534-9562 Volume 3 (2019), pp.73-90 Received 20 August 2019. Published on-line 30 September 2019 web: http://www.sangaku-journal.eu/ c The Author(s) This article is published with open access1. Relationships Between Six Excircles Stanley Rabinowitz 545 Elm St Unit 1, Milford, New Hampshire 03055, USA e-mail: [email protected] web: http://www.StanleyRabinowitz.com/ Abstract. If P is a point inside 4ABC, then the cevians through P divide 4ABC into smaller triangles of various sizes. We give theorems about the rela- tionship between the radii of certain excircles of some of these triangles. Keywords. Euclidean geometry, triangle geometry, excircles, exradii, cevians. Mathematics Subject Classification (2010). 51M04. 1. Introduction Let P be any point inside a triangle ABC. The cevians through P divide 4ABC into six smaller triangles. In a previous paper [5], we found relationships between the radii of the circles inscribed in these triangles. For example, if P is at the orthocenter H, as shown in Figure 1, then we found that r1r3r5 = r2r4r6, where the ri are radii of the incircles as shown in the figure. arXiv:1910.00418v1 [math.HO] 28 Sep 2019 Figure 1. r1r3r5 = r2r4r6 In this paper, we will find similar results using excircles instead of incircles. When the cevians through a point P interior to a triangle ABC are drawn, many smaller 1This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited. -
The Stammler Circles
Forum Geometricorum b Volume 2 (2002) 151–161. bbb FORUM GEOM ISSN 1534-1178 The Stammler Circles Jean-Pierre Ehrmann and Floor van Lamoen Abstract. We investigate circles intercepting chords of specified lengths on the sidelines of a triangle, a theme initiated by L. Stammler [6, 7]. We generalize his results, and concentrate specifically on the Stammler circles, for which the intercepts have lengths equal to the sidelengths of the given triangle. 1. Introduction Ludwig Stammler [6, 7] has investigated, for a triangle with sidelengths a, b, c, circles that intercept chords of lengths µa, µb, µc (µ>0) on the sidelines BC, CA and AB respectively. He called these circles proportionally cutting circles,1 and proved that their centers lie on the rectangular hyperbola through the circumcenter, the incenter, and the excenters. He also showed that, depending on µ, there are 2, 3 or 4 circles cutting chords of such lengths. B0 B A0 C A C0 Figure 1. The three Stammler circles with the circumtangential triangle As a special case Stammler investigated, for µ =1, the three proportionally cutting circles apart from the circumcircle. We call these the Stammler circles. Stammler proved that the centers of these circles form an equilateral triangle, cir- cumscribed to the circumcircle and homothetic to Morley’s (equilateral) trisector Publication Date: November 22, 2002. Communicating Editor: Bernard Gibert. 1Proportionalschnittkreise in [6]. 152 J.-P. Ehrmann and F. M. van Lamoen triangle. In fact this triangle is tangent to the circumcircle at the vertices of the circumtangential triangle. 2 See Figure 1. In this paper we investigate the circles that cut chords of specified lengths on the sidelines of ABC, and obtain generalizations of results in [6, 7], together with some further results on the Stammler circles. -
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10.4 - Circles, Arcs, Inscribed Angles, Power of a Point We work in Euclidean Geometry for the remainder of these notes. Definition: A minor arc is the intersection of a circle with a central angle and its interior. A semicircle is the intersection of a circle with a closed half-plane whose boundary line passes through its center. A major arc is the intersection of a circle with a central angle and its exterior (that is, the complement of a minor arc, plus its endpoints). Notation: If the endpoints of an arc are A and B, and C is any other point of the arc (which must be used in order to uniquely identify the arc) then we denote such an arc by qACB . Measures of Arcs: We define the measure of an arc as follows: Minor Arc Semicircle Major Arc mqACB = µ(pAOB) mqACB = 180 m=qACB 360 - µ(pAOB) A A A C O O C O C B B B q q Theorem (Additivity of Arc Measure): Suppose arcs A1 = APB and A2 =BQC are any two arcs of a circle with center O having just one point B in common, and such that their union A1 c q A2 = ABC is also an arc. Then m(A1 c A2) = mA1 + mA2. The proof is straightforward and tedious, falling into cases depending on whether qABC is a minor arc or a semicircle, or a major arc. It is mainly about algebra and offers no new insights. We’ll leave it to others who have boundless energy and time, and who can wring no more entertainment out of “24.” Lemma: If pABC is an inscribed angle of a circle O and the center of the circle lies on one of its 1 sides, then μ()∠=ABC mACp . -
Areas of Polygons and Circles
Chapter 8 Areas of Polygons and Circles Copyright © Cengage Learning. All rights reserved. Perimeter and Area of 8.2 Polygons Copyright © Cengage Learning. All rights reserved. Perimeter and Area of Polygons Definition The perimeter of a polygon is the sum of the lengths of all sides of the polygon. Table 8.1 summarizes perimeter formulas for types of triangles. 3 Perimeter and Area of Polygons Table 8.2 summarizes formulas for the perimeters of selected types of quadrilaterals. However, it is more important to understand the concept of perimeter than to memorize formulas. 4 Example 1 Find the perimeter of ABC shown in Figure 8.17 if: a) AB = 5 in., AC = 6 in., and BC = 7 in. b) AD = 8 cm, BC = 6 cm, and Solution: a) PABC = AB + AC + BC Figure 8.17 = 5 + 6 + 7 = 18 in. 5 Example 1 – Solution cont’d b) With , ABC is isosceles. Then is the bisector of If BC = 6, it follows that DC = 3. Using the Pythagorean Theorem, we have 2 2 2 (AD) + (DC) = (AC) 2 2 2 8 + 3 = (AC) 6 Example 1 – Solution cont’d 64 + 9 = (AC)2 AC = Now Note: Because x + x = 2x, we have 7 HERON’S FORMULA 8 Heron’s Formula If the lengths of the sides of a triangle are known, the formula generally used to calculate the area is Heron’s Formula. One of the numbers found in this formula is the semiperimeter of a triangle, which is defined as one-half the perimeter. For the triangle that has sides of lengths a, b, and c, the semiperimeter is s = (a + b + c). -
Power of a Point Yufei Zhao
Trinity Training 2011 Power of a Point Yufei Zhao Power of a Point Yufei Zhao Trinity College, Cambridge [email protected] April 2011 Power of a point is a frequently used tool in Olympiad geometry. Theorem 1 (Power of a point). Let Γ be a circle, and P a point. Let a line through P meet Γ at points A and B, and let another line through P meet Γ at points C and D. Then PA · PB = PC · P D: A B C A P B D P D C Figure 1: Power of a point. Proof. There are two configurations to consider, depending on whether P lies inside the circle or outside the circle. In the case when P lies inside the circle, as the left diagram in Figure 1, we have \P AD = \PCB and \AP D = \CPB, so that triangles P AD and PCB are similar PA PC (note the order of the vertices). Hence PD = PB . Rearranging we get PA · PB = PC · PD. When P lies outside the circle, as in the right diagram in Figure 1, we have \P AD = \PCB and \AP D = \CPB, so again triangles P AD and PCB are similar. We get the same result in this case. As a special case, when P lies outside the circle, and PC is a tangent, as in Figure 2, we have PA · PB = PC2: B A P C Figure 2: PA · PB = PC2 The theorem has a useful converse for proving that four points are concyclic. 1 Trinity Training 2011 Power of a Point Yufei Zhao Theorem 2 (Converse to power of a point). -
Special Isocubics in the Triangle Plane
Special Isocubics in the Triangle Plane Jean-Pierre Ehrmann and Bernard Gibert June 19, 2015 Special Isocubics in the Triangle Plane This paper is organized into five main parts : a reminder of poles and polars with respect to a cubic. • a study on central, oblique, axial isocubics i.e. invariant under a central, oblique, • axial (orthogonal) symmetry followed by a generalization with harmonic homolo- gies. a study on circular isocubics i.e. cubics passing through the circular points at • infinity. a study on equilateral isocubics i.e. cubics denoted 60 with three real distinct • K asymptotes making 60◦ angles with one another. a study on conico-pivotal isocubics i.e. such that the line through two isoconjugate • points envelopes a conic. A number of practical constructions is provided and many examples of “unusual” cubics appear. Most of these cubics (and many other) can be seen on the web-site : http://bernard.gibert.pagesperso-orange.fr where they are detailed and referenced under a catalogue number of the form Knnn. We sincerely thank Edward Brisse, Fred Lang, Wilson Stothers and Paul Yiu for their friendly support and help. Chapter 1 Preliminaries and definitions 1.1 Notations We will denote by the cubic curve with barycentric equation • K F (x,y,z) = 0 where F is a third degree homogeneous polynomial in x,y,z. Its partial derivatives ∂F ∂2F will be noted F ′ for and F ′′ for when no confusion is possible. x ∂x xy ∂x∂y Any cubic with three real distinct asymptotes making 60◦ angles with one another • will be called an equilateral cubic or a 60. -
Delaunay Triangulations of Points on Circles Arxiv:1803.11436V1 [Cs.CG
Delaunay Triangulations of Points on Circles∗ Vincent Despr´e † Olivier Devillers† Hugo Parlier‡ Jean-Marc Schlenker‡ April 2, 2018 Abstract Delaunay triangulations of a point set in the Euclidean plane are ubiq- uitous in a number of computational sciences, including computational geometry. Delaunay triangulations are not well defined as soon as 4 or more points are concyclic but since it is not a generic situation, this difficulty is usually handled by using a (symbolic or explicit) perturbation. As an alter- native, we propose to define a canonical triangulation for a set of concyclic points by using a max-min angle characterization of Delaunay triangulations. This point of view leads to a well defined and unique triangulation as long as there are no symmetric quadruples of points. This unique triangulation can be computed in quasi-linear time by a very simple algorithm. 1 Introduction Let P be a set of points in the Euclidean plane. If we assume that P is in general position and in particular do not contain 4 concyclic points, then the Delaunay triangulation DT (P ) is the unique triangulation over P such that the (open) circumdisk of each triangle is empty. DT (P ) has a number of interesting properties. The one that we focus on is called the max-min angle property. For a given triangulation τ, let A(τ) be the list of all the angles of τ sorted from smallest to largest. DT (P ) is the triangulation which maximizes A(τ) for the lexicographical order [4,7] . In dimension 2 and for points in general position, this max-min angle arXiv:1803.11436v1 [cs.CG] 30 Mar 2018 property characterizes Delaunay triangulations and highlights one of their most ∗This work was supported by the ANR/FNR project SoS, INTER/ANR/16/11554412/SoS, ANR-17-CE40-0033. -
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.