DOCSLIB.ORG

Metrical Relations in Barycentric Coordinates

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Metrical Relations in Barycentric Coordinates Mathematical Communications 8(2003), 55-68 55 Metrical relations in barycentric coordinates Vladimir Volenec∗ Abstract.Let∆ be the area of the fundamental triangle ABC of barycentric coordinates and let α =cotA, β =cotB, γ =cotC.The vectors vi =[xi,yi,zi](i =1, 2) have the scalar product 2∆(αx1x2 + βy1y2 + γz1z2). This fact implies all important formulas about metrical relations of points and lines. The main and probably new results are Theorems 1 and 8. Key words: barycentric coordinates AMS subject classifications: 51N20 Received December 20, 2002 Accepted April 2, 2003 Let ABC be a given triangle with the sidelengths a = |BC|,b= |CA|,c= |AB|, A, B, C P P the measures of the opposite angles and the area ∆. For any point−→ let be PQ Q−P the radiusvector of this point with respect to any origin. Then−→ we have−→ =−→ . There are uniquely determined numbers y,z ∈ R so that AP = y · AB + z · AC, i.e. P − A = y(B − A)+z(C − A). If we put x =1− y − z, i.e. x + y + z =1, (1) then we have P = xA + yB + zC. (2) Numbers x, y, z, such that (1) and (2) are valid, are uniquely determined by point P and triangle ABC, i.e. these numbers do not depend on the choice of the origin. We say that x, y, z,aretheabsolute barycentric coordinates of point P with respect to triangle ABC and write P =(x, y, z). Obviously A =(1, 0, 0), B =(0, 1, 0), C =(0, 0, 1). Actually, point P is the barycenter of the mass point system of points A, B, C with masses x, y, z, respectively. The centroid of triangle ABC is point G =(1/3, 1/3, 1/3). Any three numbers x ,y ,z proportional to the coordinates x, y, z are said to be relative barycentric coordinates of point P with respect to triangle ABC and we write P =(x : y : z ). Here we have x + y + z =0.Point P is uniquely ∗Department of Mathematics, University of Zagreb, Bijeniˇcka 30, HR-10 000 Zagreb, Croatia, e-mail: [email protected] 56 V. Volenec determined by its relative barycentric coordinates x , y , z because its absolute barycentric coordinates are x y z x = ,y= ,z= . x + y + z x + y + z x + y + z If point P divides two different points P1 =(x1,y1,z1)andP2 =(x2,y2,z2)in −−→ −−→ the ratio (P1P2P )=λ, i.e. if P1P = λ · P2P ,thenfromP − P 1 = λ(P − P 2)with P i = xiA + yiB + ziC (i =1, 2) we obtain (1 − λ)P =(x1 − λx2)A +(y1 − λy2)B +(z1 − λz2)C. Because of xi + yi + zi =1(i =1, 2) we have 1 (x1 − λx2 + y1 − λy2 + z1 − λz2)=1 1 − λ and therefore x1 − λx2 y1 − λy2 z1 − λz2 P = , , , (3) 1 − λ 1 − λ 1 − λ P =((x1 − λx2):(y1 − λy2):(z1 − λz2)) . (4) Specially, with λ = −1, point P is the midpoint of the points P1 and P2 with x1 + x2 y1 + y2 z1 + z2 P = , , , 2 2 2 1 1 i.e. P = 2 P1 + 2 P2.SidesBC, CA, AB have the midpoints 1 1 1 1 1 1 0, , =(0:1:1), , 0, =(1:0:1), , , 0 =(1:1:0). 2 2 2 2 2 2 If λ = 1, then equality (3) has no sense, but equality (4) obtains the form P =((x1 − x2):(y1 − y2):(z1 − z2)) (5) and represents the point at infinity of the straight line P1P2. For this point P we have equality (P1P2P ) = 1 and the relative coordinates in (5) have the zero sum. Therefore, this point does not have the absolute coordinates. Because of P1 = P2 point P at infinity cannot be of the form (0 : 0 : 0). Specially, straight lines BC, CA, AB have the points at infinity (0 : 1 : −1), (−1:0:1),(1:−1 : 0), respectively. v y z v −→For any−→ vector numbers and are uniquely determined such that = y · AB + z · AC, i.e. v = y(B − A)+z(C − A). If we put x = −(y + z), then we have v = xA + yB + zC,x+ y + z =0. (6) Numbers x, y, z are uniquely determined and are said to be the barycentric coordi- nates of vector v with respect to triangle ABC.Wewritev =[x, y, z]. For two points Pi =(xi,yi,zi)(i =1, 2) we have P i = xiA + yiB + ziC and therefore −−→ P1P2 = P 2 −P 1 =(x2 −x1)A+(y2 −y1)B +(z2 −z1)C =[x2 −x1,y2 −y1,z2 −z1]. Metrical relations in barycentric coordinates 57 −→ −→ −→ Specially, BC =[0, −1, 1], CA =[1, 0, −1], AB =[−1, 1, 0]. We conclude that the (relative) barycentric coordinates of the point at infinity of a straight line are pro- portional to the barycentric coordinates of any vector parallel to this line. Therefore, parallel lines have the same point at infinity. The formulas for metrical relations can be written in a more compact form if we use numbers α =cotA, β =cotB, γ =cotC (7) We have e.g. b2 + c2 − a2 =2bc cos A =2bc sin A cot A =4∆α and therefore b2 + c2 − a2 =4∆α, c2 + a2 − b2 =4∆β, a2 + b2 − c2 =4∆γ. Adding, we obtain a2 =2∆(β + γ),b2 =2∆(γ + α),c2 =2∆(α + β). (8) Now, let us prove the most important theorem about metrical relations in barycentric coordinates. Theorem 1. The scalar product of two vectors vi =[xi,yi,zi](i =1, 2) is given by v1 · v2 =2∆(αx1x2 + βy1y2 + γz1z2), where ∆ is the area of the fundamental triangle ABC and numbers α, β, γ are given by (7). −→ 2 2 Proof. Squaringthe equality AB = B − A we obtain c2 = A + B − 2A · B, i.e. 2A · B = A2 + B2 − c2 and analogously 2A · C = A2 + C2 − b2 and 2B · C = 2 2 2 B + C − a . Owingto the equalities xi + yi + zi =0(i =1, 2) and (8) we obtain successively 2v1 · v2 =2(x1A + y1B + z1C)(x2A + y2B + z2C) 2 2 2 2 2 2 =2x1x2A +2y1y2B +2z1z2C +(x1y2 + y1x2)(A + B − c ) 2 2 2 2 2 2 +(x1z2 + z1x2)(A + C − b )+(y1z2 + z1y2)(B + C − a ) 2 2 2 =(x1 + y1 + z1)(x2A + y2B + z2C ) 2 2 2 +(x2 + y2 + z2)(x1A + y1B + z1C ) 2 2 2 −a (y1z2 + z1y2) − b (z1x2 + x1z2) − c (x1y2 + y1x2) = −2∆[(β + γ)(y1z2 + z1y2)+(γ + α)(z1x2 + x1z2) +(α + β)(x1y2 + y1x2)] = −2∆{α [x1(y2 + z2)+(y1 + z1)x2]+β [y1(z2 + x2)+(z1 + x1)y2] +γ[z1(x2 + y2)+(x1 + y1)z2]} = −2∆[α(−2x1x2)+β(−2y1y2)+γ(−2z1z2)] =4∆(αx1x2 + βy1y2 + γz1z2). ✷ 58 V. Volenec Corollary 1. The length of the vector v =[x, y, z] is given by |v|2 =2∆(αx2 + βy2 + γz2). Corollary 2. The angle between two vectors vi =[xi,yi,zi](i =1, 2) is given by 1 cos ∠(v1, v2)= (αx1x2 + βy1y2 + γz1z2), Ω1Ω2 where 2 2 2 2 Ωi = αxi + βyi + γzi (i =1, 2). Corollary 3. Two points Pi =(xi,yi,zi)(i =1, 2) have the distance |P1P2| given by 2 2 2 2 |P1P2| =2∆[α(x1 − x2) + β(y1 − y2) + γ(z1 − z2) ]. Specially, with P1 = P =(x, y, z)andP2 = A =(1, 0, 0) or P2 = B =(0, 1, 0) or P2 = C =(0, 0, 1) we obtain further: Corollary 4. For any point P =(x, y, z) we have equalities |AP | =2∆[α(1 − x)2 + βy2 + γz2], |BP| =2∆[αx2 + β(1 − y)2 + γz2], |CP| =2∆[αx2 + βy2 + γ(1 − z)2]. Theorem 2. For the point P =(x, y, z) and any point S we have |SP|2 = x ·|SA|2 + y ·|SB|2 + z ·|SC|2 − a2yz − b2zx − c2xy. Proof. Let S be the origin. Squaring the equality P = xA + yB + zC and usingthe equalities from the proof of Theorem 1 we get |SP|2 = P 2 = x2A2 + y2B2 + z2C2 + yz(B2 + C2 − a2) 2 2 2 2 +zx(C + A − b2)+xy(A + B − c2) =(x + y + z)(xA2 + yB2 + zC2) − a2yz − b2zy − c2xy and because of x +y + z = 1 the statement of Theorem 2 follows. ✷ 1 1 1 With P = G = 3 , 3 , 3 we obtain: Corollary 5 [Leibniz]. For centroid G of triangle ABC and for any point P we have 1 3 ·|SG|2 = |SA|2 + |SB|2 + |SC|2 − (a2 + b2 + c2). 3 If S is the circumcenter O of triangle ABC,then|OA| = |OB| = |OC| = R and by(8)itfollows: Metrical relations in barycentric coordinates 59 Corollary 6. For any point P =(x, y, z) and the circumscribed circle (O, R) of triangle ABC the equality |OP|2 = R2 − a2yz − b2zx − c2xy holds, i.e. |OP|2 = R2 − 2∆Π,where 1 Π=(β + γ)yz +(γ + α)zx +(α + β)xy = (a2yz + b2zx + c2xy). (9) 2∆ With S = P Theorem 2 implies: Corollary 7. For the point P =(x, y, z) the equality x ·|AP |2 + y ·|BP|2 + z2|CP|2 =2∆Π holds, where the number Π is given by (9). The equalities from Corollary 4 can be written in another form because of (9), (8) and the equality x + y + z =1.Weobtaine.g. 1 |AP |2 = α(1 − x)2 + βy2 + γz2 2∆ = α − 2αx + αx(1 − y − z)+βy(1 − z − x)+γz(1 − x − y) = α − αx + βy + γz − (β + γ)yz − (γ + α)zx − (α + β)xy α y z βy γz − 1 γ α zx α β xy − = ( + )+ + Π=x[( + ) +( + ) ] Π 1 − β γ yz − 1 − x − β γ yz = x[Π ( + ) ] Π=x[Π(1 ) ( + ) ] 1 a2 = [Π(y + z) − yz].
Load more

Recommended publications
  • 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.
    [Show full text]
  • 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.
    [Show full text]
  • 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.
    [Show full text]
  • 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.
    [Show full text]
  • Tetrahedra and Relative Directions in Space Using 2 and 3-Space Simplexes for 3-Space Localization
    Tetrahedra and Relative Directions in Space Using 2 and 3-Space Simplexes for 3-Space Localization Kevin T. Acres and Jan C. Barca Monash Swarm Robotics Laboratory Faculty of Information Technology Monash University, Victoria, Australia Abstract This research presents a novel method of determining relative bearing and elevation measurements, to a remote signal, that is suitable for implementation on small embedded systems – potentially in a GPS denied environment. This is an important, currently open, problem in a number of areas, particularly in the field of swarm robotics, where rapid updates of positional information are of great importance. We achieve our solution by means of a tetrahedral phased array of receivers at which we measure the phase difference, or time difference of arrival, of the signal. We then perform an elegant and novel, albeit simple, series of direct calculations, on this information, in order to derive the relative bearing and elevation to the signal. This solution opens up a number of applications where rapidly updated and accurate directional awareness in 3-space is of importance and where the available processing power is limited by energy or CPU constraints. 1. Introduction Motivated, in part, by the currently open, and important, problem of GPS free 3D localisation, or pose recognition, in swarm robotics as mentioned in (Cognetti, M. et al., 2012; Spears, W. M. et al. 2007; Navarro-serment, L. E. et al.1999; Pugh, J. et al. 2009), we derive a method that provides relative elevation and bearing information to a remote signal. An efficient solution to this problem opens up a number of significant applications, including such implementations as space based X/Gamma ray source identification, airfield based aircraft location, submerged black box location, formation control in aerial swarm robotics, aircraft based anti-collision aids and spherical sonar systems.
    [Show full text]
  • Theta Circles and Polygons Test #112
    Theta Circles and Polygons Test #112 Directions: 1. Fill out the top section of the Round 1 Google Form answer sheet and select Theta- Circles and Polygons as the test. Do not abbreviate your school name. Enter an email address that will accept outside emails (some school email addresses do not). 2. Scoring for this test is 5 times the number correct plus the number omitted. 3. TURN OFF ALL CELL PHONES. 4. No calculators may be used on this test. 5. Any inappropriate behavior or any form of cheating will lead to a ban of the student and/or school from future National Conventions, disqualification of the student and/or school from this Convention, at the discretion of the Mu Alpha Theta Governing Council. 6. If a student believes a test item is defective, select “E) NOTA” and file a dispute explaining why. 7. If an answer choice is incomplete, it is considered incorrect. For example, if an equation has three solutions, an answer choice containing only two of those solutions is incorrect. 8. If a problem has wording like “which of the following could be” or “what is one solution of”, an answer choice providing one of the possibilities is considered to be correct. Do not select “E) NOTA” in that instance. 9. If a problem has multiple equivalent answers, any of those answers will be counted as correct, even if one answer choice is in a simpler format than another. Do not select “E) NOTA” in that instance. 10. Unless a question asks for an approximation or a rounded answer, give the exact answer.
    [Show full text]
  • Application of Nine Point Circle Theorem
    Application of Nine Point Circle Theorem Submitted by S3 PLMGS(S) Students Bianca Diva Katrina Arifin Kezia Aprilia Sanjaya Paya Lebar Methodist Girls’ School (Secondary) A project presented to the Singapore Mathematical Project Festival 2019 Singapore Mathematic Project Festival 2019 Application of the Nine-Point Circle Abstract In mathematics geometry, a nine-point circle is a circle that can be constructed from any given triangle, which passes through nine significant concyclic points defined from the triangle. These nine points come from the midpoint of each side of the triangle, the foot of each altitude, and the midpoint of the line segment from each vertex of the triangle to the orthocentre, the point where the three altitudes intersect. In this project we carried out last year, we tried to construct nine-point circles from triangulated areas of an n-sided polygon (which we call the “Original Polygon) and create another polygon by connecting the centres of the nine-point circle (which we call the “Image Polygon”). From this, we were able to find the area ratio between the areas of the original polygon and the image polygon. Two equations were found after we collected area ratios from various n-sided regular and irregular polygons. Paya Lebar Methodist Girls’ School (Secondary) 1 Singapore Mathematic Project Festival 2019 Application of the Nine-Point Circle Acknowledgement The students involved in this project would like to thank the school for the opportunity to participate in this competition. They would like to express their gratitude to the project supervisor, Ms Kok Lai Fong for her guidance in the course of preparing this project.
    [Show full text]
  • 3. Adam Also Needs to Know the Altitude of the Smaller Triangle Within the Sign
    Lesson 3: Special Right Triangles Standard: MCC9‐12.G.SRT.6 Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles. Essential Question: What are the relationships among the side lengths of a 30° - 60° - 90° triangle? How can I use these relationships to solve real-world problems? Discovering Special Triangles Learning Task (30-60-90 triangle) 1. Adam, a construction manager in a nearby town, needs to check the uniformity of Yield signs around the state and is checking the heights (altitudes) of the Yield signs in your locale. Adam knows that all yield signs have the shape of an equilateral triangle. Why is it sufficient for him to check just the heights (altitudes) of the signs to verify uniformity? 2. A Yield sign created in Adam’s plant is pictured above. It has the shape of an equilateral triangle with a side length of 2 feet. If you draw the altitude of the triangular sign, you split the Yield sign in half vertically. Use what you know about triangles to find the angle measures and the side lengths of the two smaller triangles formed by the altitude. a. What does an altitude of an equilateral triangle do to the triangle? b. What kinds of triangles are formed when you draw the altitude? c. What did the altitude so to the base of the triangle? d. What did the altitude do to the angle from which it was drawn? 3. Adam also needs to know the altitude of the smaller triangle within the sign.
    [Show full text]
  • Saccheri and Lambert Quadrilateral in Hyperbolic Geometry
    MA 408, Computer Lab Three Saccheri and Lambert Quadrilaterals in Hyperbolic Geometry Name: Score: Instructions: For this lab you will be using the applet, NonEuclid, developed by Castel- lanos, Austin, Darnell, & Estrada. You can either download it from Vista (Go to Labs, click on NonEuclid), or from the following web address: http://www.cs.unm.edu/∼joel/NonEuclid/NonEuclid.html. (Click on Download NonEuclid.Jar). Work through this lab independently or in pairs. You will have the opportunity to finish anything you don't get done in lab at home. Please, write the solutions of homework problems on a separate sheets of paper and staple them to the lab. 1 Introduction In the previous lab you observed that it is impossible to construct a rectangle in the Poincar´e disk model of hyperbolic geometry. In this lab you will study two types of quadrilaterals that exist in the hyperbolic geometry and have many properties of a rectangle. A Saccheri quadrilateral has two right angles adjacent to one of the sides, called the base. Two sides that are perpendicular to the base are of equal length. A Lambert quadrilateral is a quadrilateral with three right angles. In the Euclidean geometry a Saccheri or a Lambert quadrilateral has to be a rectangle, but the hyperbolic world is different ... 2 Saccheri Quadrilaterals Definition: A Saccheri quadrilateral is a quadrilateral ABCD such that \ABC and \DAB are right angles and AD ∼= BC. The segment AB is called the base of the Saccheri quadrilateral and the segment CD is called the summit. The two right angles are called the base angles of the Saccheri quadrilateral, and the angles \CDA and \BCD are called the summit angles of the Saccheri quadrilateral.
    [Show full text]
  • 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.
    [Show full text]
  • Circles JV Practice 1/26/20 Matthew Shi
    Western PA ARML January 26, 2020 Circles JV Practice 1/26/20 Matthew Shi 1 Warm-up Questions 1. A square of area 40 is inscribed in a semicircle as shown: Find the radius of the semicircle. 2. The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle? 3. Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The 8 area of the shaded region in the diagram is 13 of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: π radians is 180 degrees.) 4. Let C1 and C2 be circles of radius 1 that are in the same plane and tangent to each other. How many circles of radius 3 are in this plane and tangent to both C1 and C2? 1 Western PA ARML January 26, 2020 2 Problems 1. Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region. 2. A square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the circle? 3.
    [Show full text]
  • Talking Geometry: a Classroom Episode
    Learning and Teaching Mathematics, No. 6 Page 3 Talking Geometry: a Classroom Episode Erna Lampen Wits School of Education The Geometry course forms part of the first year early in the course and they used the programme mathematics offering for education students at to investigate their conjectures in an informal way Wits School of Education. The course is outside class time. The course has a distinctly compulsory for all first years who want to major problem-centred approach. Students did not get in mathematics education at FET level. Most of demonstrations of the constructions or guidelines these students obtained about 60% on standard that show “How to construct a line perpendicular grade for mathematics at Grade 12 level and to another line.” Instead, we discussed what it many confessed not to have done geometry means to ask “spatial” questions, and decided beyond Grade 9 level. A few students who together that spatial questions involve questions specialize in Foundation Phase as well as about shape, structure, size and position. The Intermediate and Senior Phase also attended. This course is dedicated to asking and seeking answers class consisted of 85 students of whom close to to spatial questions. We started by investigating 70 attended regularly. All eleven official languages relative positions of points and lines in a plane. were represented in the class. There was also a We did constructions to answer questions like: deaf student and a Chinese student who “Where are all points that are equidistant from a immigrated in 2005 – imagine the need to have given point A? Where are all points that are shared objects to refer to when we discussed the equidistant from two given points A and B?” and mathematics! There was always a tutor, a fellow so on.
    [Show full text]