DOCSLIB.ORG

Circles in Neutral Geometry Text Section 3.8 We Will Use the Usual

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Circles in Neutral Geometry Text Section 3.8 We Will Use the Usual Circles in Neutral Geometry Text section 3.8 We will use the usual definition of circle - the one that leads to \square circles" in the Taxicab plane. Many of the familiar properties of circles in the Euclidean plane are inherited from neutral geometry - only those that somehow depend on uniqueness of parallels or a specific angle sum in triangles will not hold. This section gives an assortment of circle definitions and properties in neutral geometry. Definition. A circle is the set of all points in a plane that lie at a fixed distance r from a fixed point O. The point O is the center and the distance r is the radius of the circle. [Note: the term radius is also applied to any segment with one endpoint at the center and the other on the circle, when there is no danger of confusion]. A point P is an interior point of the circle (interior to the circle or in the interior of the circle) if OP < r; P is an exterior point of the circle (exterior to the circle or in the exterior of the circle) if OP > r. Informally, we may say an interior point of a circle is \inside the circle" and an exterior point is \outside the circle". This must be used carefully | it can lead to confusion because an interior point is not \in the circle" according to the formal definition of \circle" (not in the set of points that is the circle). [Exterior points an interior points are not points of the circle in the usual sense]. For convenience in defining a number of terms associated with circles, we will refer to the union of a line with either of its half-planes as a closed half-plane. Definition. In any circle: A chord is a segment whose endpoints are on the circle. A diameter is a chord which passes through the center of the circle. A secant to the circle is a line which meets the circle at exactly two points . A tangent to the circle is a line which meets the circle at exactly one point (called the point of tangency or the point of contact. A central angle is an angle 6 AOB for which O is the center of the circle and A; B are on the circle. An arc of a circle is that part of a circle lying in a closed halfplane determined by a secant; the endpoints of the arc are the points of the arc lying on that secant. For three distinct points A; B; C on a circle, we use the notation ABCd to represent the arc with endpoints A and C which contains point B. [Note that naming an arc with just the endpoints would be ambiguous { there are two arcs with the same endpoints] An angle is inscribed in an arc if it can be named 6 ABC with A; C the endpoints of the arc and B another point of the arc. (Such an angle may also be called an \inscribed angle" without explicit reference to the arc) An arc or chord is subtended by an angle if the endpoints of the arc or chord are on the two sides of the angle and the other points of the arc or chord are in the interior of the angle. Many of the well-known properties of circles in Euclidean geometry also hold in neutral geometry (because they don't fundamentally involve parallelism): • The center of a circle is the midpoint of any diameter. • The perpendicular bisector of any chord of a circle passes through the center of the circle (follows from the "equidistant" characterization of the perpendicular bisector). • A line passing through the center of the circle and perpendicular to a chord of the circle bisects the chord (same reason). • Two congruent central angles subtend congruent chords { and congruent chords are subtended by congruent central angles (SAS axiom). 1 • Two chords equidistant from the center of a circle have equal lengths, and two chords with the same length are the same distance from the center of the circle (HL congruence property with the "perpendicular is bisector" properties, above - recall \distance from center to chord" is "length of the perpendicular segment" since it is the distance from a point to a line). Kay shows proofs of a couple of these. From the definitions given, we can see that every central angle and every diameter in a circle determines two arcs (the two arcs whose endpoints are on the sides of the angle or are endpoints of the diameter). We can also see that every arc determines a central angle or a diameter (the angle whose sides pass through the endpoints of the arc or the diameter whose ends are endpoints of the arc). This correspondence between fcentral angles and diametersg and fpairs of arcsg allows us to use central angles and diameters to classify arcs. Definition. Classification of arcs. An arc is a semicircle if it is the intersection of its circle with a closed halfplane determined by a diameter of the circle. An arc is a minor arc if it lies on and in the interior of its central angle. [If it is subtended by its central angle] An arc is a major arc if it is not a semicircle and does not lie on and in the interior of its central angle. Once again (as with segments and angles), it will be convenient to have a measure for arcs { to allow us to compare \size". This measure can be defined in every neutral geometry { no new axioms are required to add it to our system: Definition. For any arc ABCd in a circle with center O, we define the measure mABCd of the arc by 8 m6 AOC if ABCd is a minor arc; <> mABCd = 180; if ABCd is a semicircle; > :360 − m6 AOC if ABCd is a major arc: Notice that arc measure is always positive and less than 360, unlike angle measure which is always less than 180 Theorem 1 (Additivity of arc measure). If APd B and BQCd are two arcs in a circle which meet only at B, and if their union ABCd = APd B [ BQCd is an arc of the circle, then mABCd = mAPd B + mBQCd . Theorem 2. (Tangent theorem) A line is tangent to a circle iff it is perpendicular to a radius at a point of the circle. ! ! Corollary. If two tangents PA and PB to a circle with center O have A and B as points of contact with −! the circle, then PA =∼ PB and PO bisects 6 AP B. We would like to say that the interior of a circle is bounded by the circle { that any line that meets the interior of the circle must meet the circle itself. We would like to believe this is already built into our system and does not require another axiom, and this is the case, but we do need to go back to the ruler axiom to prove it, because we need the notions of continuity and completeness. If we did not have some sort of completeness axiom (the Ruler axiom lets us \borrow" the completeness of the real number system), we would need to take this theorem [Theorem 3] as an axiom. −! ! Lemma 1. Let AB be any ray, O a point not on the line AB and f : AB ! R a coordinate system on ! AB for which f(A) = 0 and f(B) > 0. If we define a function d : R ! R by d(x) = OP for the point P with f(P ) = x, then d is a continuous function. The point of this lemma is that we can use the Intermediate Value Theorem to say that if there are −! −! points S and T on AB then, for any distance q with OS < q < QR, there is a point Q on AB with OQ = q. This is closely related to the idea of segment construction { but does not require a segment to be copied, and it does not involve marking off a distance on an already chosen ray (The point O is not on the ray −! AB) 2 Theorem 3. (Secant theorem - or Line/Circle theorem). If a line l passes through a point A which is interior to a circle, then the line is a secant of the circle. Corollary. Any segment joining a point interior to a circle to a point exterior to circle meets the circle. This says, in essence, that a circle, like a line, divides the rest of the plane into two regions, and we have to cross the circle to get from one to the other. This would not be true in a non-ruler geometry such as the Rational Cartesian Plane (which was eliminated from our theory by the ruler axiom) in which the 2 2 line given by y = x contains the point (0; 0) inside the circle given bypx + yp = 1 but doesp not actuallyp meet the circle (only common solutions of y = x and x2 + y2 = 1 are ( 2=2; 2=2) and (− 2=2; − 2=2) - and these do not correspond to points of the Rational Cartesian Plane). In the circle case (unlike the line), the two regions are fundamentally different - one (the interior) is convex and the other (the exterior) is not. The interior has a additional very important property of being \bounded" (there is a finite upper bound on the distance between points in the set { this property is more general than the result in Theorem 3) which also distinguishes it from the exterior.
Load more

Recommended publications
  • Centroids by Composite Areas.Pptx
    Centroids Method of Composite Areas A small boy swallowed some coins and was taken to a hospital. When his grandmother telephoned to ask how he was a nurse said 'No change yet'. Centroids ¢ Previously, we developed a general formulation for finding the centroid for a series of n areas n xA ∑ ii i=1 x = n A ∑ i i=1 2 Centroids by Composite Areas Monday, November 12, 2012 1 Centroids ¢ xi was the distance from the y-axis to the local centroid of the area Ai n xA ∑ ii i=1 x = n A ∑ i i=1 3 Centroids by Composite Areas Monday, November 12, 2012 Centroids ¢ If we can break up a shape into a series of smaller shapes that have predefined local centroid locations, we can use this formula to locate the centroid of the composite shape n xA ∑ ii i=1 x = n A ∑ i 4 Centroids by Composite Areas i=1 Monday, November 12, 2012 2 Centroid by Composite Bodies ¢ There is a table in the back cover of your book that gives you the location of local centroids for a select group of shapes ¢ The point labeled C is the location of the centroid of that shape. 5 Centroids by Composite Areas Monday, November 12, 2012 Centroid by Composite Bodies ¢ Please note that these are local centroids, they are given in reference to the x and y axes as shown in the table. 6 Centroids by Composite Areas Monday, November 12, 2012 3 Centroid by Composite Bodies ¢ For example, the centroid location of the semicircular area has the y-axis through the center of the area and the x-axis at the bottom of the area ¢ The x-centroid would be located at 0 and the y-centroid would be located
    [Show full text]
  • Archimedes and the Arbelos1 Bobby Hanson October 17, 2007
    Archimedes and the Arbelos1 Bobby Hanson October 17, 2007 The mathematician’s patterns, like the painter’s or the poet’s must be beautiful; the ideas like the colours or the words, must fit together in a harmonious way. Beauty is the first test: there is no permanent place in the world for ugly mathematics. — G.H. Hardy, A Mathematician’s Apology ACBr 1 − r Figure 1. The Arbelos Problem 1. We will warm up on an easy problem: Show that traveling from A to B along the big semicircle is the same distance as traveling from A to B by way of C along the two smaller semicircles. Proof. The arc from A to C has length πr/2. The arc from C to B has length π(1 − r)/2. The arc from A to B has length π/2. ˜ 1My notes are shamelessly stolen from notes by Tom Rike, of the Berkeley Math Circle available at http://mathcircle.berkeley.edu/BMC6/ps0506/ArbelosBMC.pdf . 1 2 If we draw the line tangent to the two smaller semicircles, it must be perpendicular to AB. (Why?) We will let D be the point where this line intersects the largest of the semicircles; X and Y will indicate the points of intersection with the line segments AD and BD with the two smaller semicircles respectively (see Figure 2). Finally, let P be the point where XY and CD intersect. D X P Y ACBr 1 − r Figure 2 Problem 2. Now show that XY and CD are the same length, and that they bisect each other.
    [Show full text]
  • Lesson 20: Composite Area Problems
    NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 20 7•3 Lesson 20: Composite Area Problems Student Outcomes . Students find the area of regions in the coordinate plane with polygonal boundaries by decomposing the plane into triangles and quadrilaterals, including regions with polygonal holes. Students find composite areas of regions in the coordinate plane by decomposing the plane into familiar figures (triangles, quadrilaterals, circles, semicircles, and quarter circles). Lesson Notes In Lessons 17 through 20, students learned to find the areas of various regions, including quadrilaterals, triangles, circles, semicircles, and those plotted on coordinate planes. In this lesson, students use prior knowledge to use the sum and/or difference of the areas to find unknown composite areas. Classwork Example 1 (5 minutes) Scaffolding: For struggling students, display Example 1 posters around the room displaying the visuals and the Find the composite area of the shaded region. Use ퟑ. ퟏퟒ for 흅. formulas of the area of a circle, a triangle, and a quadrilateral for reference. Allow students to look at the problem and find the area independently before solving as a class. What information can we take from the image? Two circles are on the coordinate plane. The diameter of the larger circle is 6 units, and the diameter of MP.1 the smaller circle is 4 units. How do we know what the diameters of the circles are? We can count the units along the diameter of the circles, or we can subtract the coordinate points to find the length of the diameter. Lesson 20: Composite Area Problems 283 This work is derived from Eureka Math ™ and licensed by Great Minds.
    [Show full text]
  • Chapter 6 the Arbelos
    Chapter 6 The arbelos 6.1 Archimedes’ theorems on the arbelos Theorem 6.1 (Archimedes 1). The two circles touching CP on different sides and AC CB each touching two of the semicircles have equal diameters AB· . P W1 W2 A O1 O C O2 B A O1 O C O2 B Theorem 6.2 (Archimedes 2). The diameter of the circle tangent to all three semi- circles is AC CB AB · · . AC2 + AC CB + CB2 · We shall consider Theorem ?? in ?? later, and for now examine Archimedes’ wonderful proofs of the more remarkable§ Theorems 6.1 and 6.2. By synthetic reasoning, he computed the radii of these circles. 1Book of Lemmas, Proposition 5. 2Book of Lemmas, Proposition 6. 602 The arbelos 6.1.1 Archimedes’ proof of the twin circles theorem In the beginning of the Book of Lemmas, Archimedes has established a basic proposition 3 on parallel diameters of two tangent circles. If two circles are tangent to each other (internally or externally) at a point P , and if AB, XY are two parallel diameters of the circles, then the lines AX and BY intersect at P . D F I W1 E H W2 G A O C B Figure 6.1: Consider the circle tangent to CP at E, and to the semicircle on AC at G, to that on AB at F . If EH is the diameter through E, then AH and BE intersect at F . Also, AE and CH intersect at G. Let I be the intersection of AE with the outer semicircle. Extend AF and BI to intersect at D.
    [Show full text]
  • The Pythagorean Theorem Crown Jewel of Mathematics
    The Pythagorean Theorem Crown Jewel of Mathematics 5 3 4 John C. Sparks The Pythagorean Theorem Crown Jewel of Mathematics By John C. Sparks The Pythagorean Theorem Crown Jewel of Mathematics Copyright © 2008 John C. Sparks All rights reserved. No part of this book may be reproduced in any form—except for the inclusion of brief quotations in a review—without permission in writing from the author or publisher. Front cover, Pythagorean Dreams, a composite mosaic of historical Pythagorean proofs. Back cover photo by Curtis Sparks ISBN: XXXXXXXXX First Published by Author House XXXXX Library of Congress Control Number XXXXXXXX Published by AuthorHouse 1663 Liberty Drive, Suite 200 Bloomington, Indiana 47403 (800)839-8640 www.authorhouse.com Produced by Sparrow-Hawke †reasures Xenia, Ohio 45385 Printed in the United States of America 2 Dedication I would like to dedicate The Pythagorean Theorem to: Carolyn Sparks, my wife, best friend, and life partner for 40 years; our two grown sons, Robert and Curtis; My father, Roscoe C. Sparks (1910-1994). From Earth with Love Do you remember, as do I, When Neil walked, as so did we, On a calm and sun-lit sea One July, Tranquillity, Filled with dreams and futures? For in that month of long ago, Lofty visions raptured all Moonstruck with that starry call From life beyond this earthen ball... Not wedded to its surface. But marriage is of dust to dust Where seasoned limbs reclaim the ground Though passing thoughts still fly around Supernal realms never found On the planet of our birth. And I, a man, love you true, Love as God had made it so, Not angel rust when then aglow, But coupled here, now rib to soul, Dear Carolyn of mine.
    [Show full text]
  • Extensionalism: the Revolution in Logic
    Nimrod Bar-Am Extensionalism: The Revolution in Logic Bar-Am_Fm.indd iii 1/22/2008 10:19:09 PM N. Bar-Am Head, Rhetoric and Philosophy of Communication Unit Communication Department Sapir College of the Negev M.P. Hof Ashkelon 79165 Israel ISBN: 978-1-4020-8167-5 e-ISBN: 978-1-4020-8168-2 DOI: 10.1007/ 978-1-4020-8168-2 Library of Congress Control Number: 2007941591 All Rights Reserved © 2008 Springer Science + Business Media, B.V. No part of this work may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, microfilming, recording or otherwise, without written permission from the Publisher, with the exception of any material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Printed on acid-free paper. 9 8 7 6 5 4 3 2 1 springer.com Bar-Am_Fm.indd iv 1/22/2008 10:19:09 PM Motto: “… logic … since Aristotle … has been unable to advance a step, and, thus, to all appearances has reached its completion” Immanuel Kant, The Critique of Pure Reason Preface to the 2nd ed., 1787 “Pure mathematics was discovered by Boole … the fact being that Boole was too modest to suppose his book the first ever written on mathematics.… His book was in fact concerned with formal logic, and this is the same thing as mathematics” Bertrand Russell Recent Work in the Philosophy of Mathematics, 1901 Bar-Am_Fm.indd vii 1/22/2008 10:19:09 PM Abstract For a very long time, Aristotelian logic was accepted as a tool (Organon) for the generation of scientific theory.
    [Show full text]
  • Centroids Introduction • the Earth Exerts a Gravitational Force on Each of the Particles Forming a Body
    Centroids Introduction • The earth exerts a gravitational force on each of the particles forming a body. These forces can be replace by a single equivalent force equal to the weight of the body and applied at the center of gravity for the body. • The centroid of an area is analogous to the center of gravity of a body. The concept of the first moment of an area is used to locate the centroid. 5 - 2 Centroids • Centroid of mass – (a.k.a. Center of mass) – (a.k.a. Center of weight) – (a.k.a. Center of gravity) • For a solid, the point where the distributed mass is centered • Centroid of volume, Centroid of area xM xm x dm yM ym y dm Center of Gravity of a 2D Body • Center of gravity of a plate • Center of gravity of a wire M y xW xW x dW M y yW yW y dW 5 - 7 Centroids &First Moments of Areas & Lines • Centroid of an area • Centroid of a line xW x dW xgM g x dM xW x dW M V tA) dM dV tdA x La x adL xAt x tdA xL x dL xA x dA Q y yL y dL first moment with respect to y yA y dA Q x first moment with respect to x 5 - 8 First Moments of Areas and Lines • An area is symmetric with respect to an axis BB’ if for every point P there exists a point P’ such that PP’ is perpendicular to BB’ and is divided into two equal parts by BB’.
    [Show full text]
  • Theta Circles 2019 Maθ National Convention
    Theta Circles 2019 MAθ National Convention “For all questions, answer choice “E. NOTA" means none of the above answers is correct.” 1. The area of a circle in square yards is the same as the circumference of the circle in feet. What is the radius in feet? A. 2 B. 6 C. 12 D. 18 E. NOTA Answer D. Let the radius in yards be 푟. Then 3∙2휋푟 = 휋푟, so 푟 =6, or 18 feet. For #2-3 us the following two circles: 푥 +(푦 − 4) = 16 푥 +(푦 − 4) = 32 2. Which of the following gives an accurate relationship between the two circles: A. The radius of the second is 16 units larger than that of the first B. The radius of the second is 4 units larger than that of the first C. The center of the second is 16 units higher than that of the first D. The radius of the second is twice as large as that of the first E. NOTA Answer E. The center is (0,4) for both circles, but the radius of the second is 4√2 whereas the radius of the first is 4. 3. What is the area between the two circles? A. 4√2 −4 휋 B. 4휋 C. 16휋 D. 256휋 E. NOTA Answer C. The area of the second is 32휋 and the area of the first is 16휋. For #4-7 use the following function: 푓(푥) = √−푥 + 18푥 + 19 4. If I formed an isosceles triangle with one side on the x-axis and all its vertices on the function, what would be the sum of the base angles? A.
    [Show full text]
  • The Thirteen Books of Euclid's Elements Sketch of Contents Book
    The Thirteen Books of Euclid’s Elements Sketch of contents book by book Book I Triangles: Proposition 32, the sum of the angles in any triangle is 180 degrees. Proposition 47, the final proposition in this book, is the Theorem of Pythagoras. Proposition 48, the converse of the Theorem of Pythagoras Book II Areas: Proposition 11 constructs the division of a line segment, A C into two 2 segments, A B and B C , so that (A B ) = (B C )⋅ (A C ). This is called division into mean and extreme ratio. † Proposition 14† shows† how to construct a square with area equal to any figure formed by† a number of straight lines which close up. Book III Circles, chords, angles in circles: Proposition 31 proves that an angle inscribed in a semicircle is a right angle. Book IV Circles and Polygons: Inscribe regular triangles, squares, pentagons (Proposition 11), hexagons and 15-gons (Proposition 16) in a given circle, and also circumscribe circles about given polygons. Book V Proportions and magnitudes. Book VI Similar figures. Book VII Number Theory: Prime numbers, the Euclidean algorithm to find the GCD Proposition 2, How to find the greatest Common Divisor of two integers. Book VIII Large sets of numbers, squares and cubes of integers. Book IX More on numbers, even and odd numbers. Proposition 14. The Fundamental theorem of arithmetic. Any integer is represented uniquely as a product of powers of primes Proposition 20, There is an infinity of prime numbers. Proposition 36, The formula for all even perfect numbers, 2n - 1(2n - 1) if 2n - 1 is a prime number.
    [Show full text]
  • 1 Proposition XII
    Connemara Doran, HS 206r, April 2009 Propositions XII-XV of the Banu Musa (9th century Arabic text attributed to the three brothers, translated into Latin by Gerard of Cremona in the 12th century): comparison with the equivalent (related) propositions of Archimedes in Sphere and Cylinder Book I. Proposition XII: The statement of the Banu Musa Proposition XII amounts to a very precise verbal explanation of what is depicted in the following diagram (Fig. 46, although without reference to any letters in the diagram which followed as an “example”), and the assertion that that the sum of the verbally-described (green) lines in the left picture is equal to the (green) line DE in the duplicated picture on the right. As the left picture suggests, the Banu Musa Proposition XII covers the same geometric consideration of parallel chords of a circle (and has the same ultimate end – to establish the surface area and volume of a sphere) as did Archimedes’ Props. 21 and 22 in SCI, although the proofs are significantly different and the Banu Musa advanced additional ends as well in the proof. Proposition XII in fact serves as a lemma to what follows in the proof – namely, an equivalent to Archimedes’ Propositions 24 and 29: the Banu Musa establish the plane figure (rectangle) whose area equals the surface of a solid inscribed in (circumscribed about) a hemisphere in terms of the diagram of Prop. XII. A key difference is that the Banu Musa treat a hemisphere rather than a full sphere, so their summation is one-half of Archimedes’ in Prop.21, and since they also use one-half the side (i.e., ) of the inscribed regular polygon rather than whole side (Archimedes’ ), they obtain a rectangle whose area is equal to one-fourth the surface area of the full sphere (surface area of a quadrant) which is then doubled to get the area of the full hemisphere.
    [Show full text]
  • Power of a Point Solutions Yufei Zhao Trinity College, Cambridge [email protected] April 2011
    Trinity Training 2011 Power of a Point Yufei Zhao Power of a Point Solutions Yufei Zhao Trinity College, Cambridge [email protected] April 2011 Practice problems: 1. Let Γ1 and Γ2 be two intersecting circles. Let a common tangent to Γ1 and Γ2 touch Γ1 at A and Γ2 at B. Show that the common chord of Γ1 and Γ2, when extended, bisects segment AB. B A Solution. Let the common chord extended meet AB at M. Since M lies on the radical 2 2 axis of Γ1 and Γ2, it has equal powers with respect to the two circles, so MA = MB . Hence MA = MB. 2. Let C be a point on a semicircle of diameter AB and let D be the midpoint of arc AC. Let E be the projection of D onto the line BC and F the intersection of line AE with the semicircle. Prove that BF bisects the line segment DE. Solution. E D F C A B Let Γ denote the circle with diameter AB, and Γ1 denote the circle with diameter BE. ◦ ◦ Since \AF B = 90 ,Γ1 passes through F . Also since \DEB = 90 ,Γ1 is tangent to DE. From Problem 1, we deduce that the common chord BF of Γ and Γ1 bisects their common tangent DE. 3. Let A; B; C be three points on a circle Γ with AB = BC. Let the tangents at A and B meet at D. Let DC meet Γ again at E. Prove that the line AE bisects segment BD. Solution. 1 Trinity Training 2011 Power of a Point Yufei Zhao A C E D B Let Γ1 denote the circumcircle of ADE.
    [Show full text]
  • Circle Theorems
    Proving circle theorems Angle in a semicircle We want to prove that the angle subtended at the circumference by a semicircle is a right angle. Step 1: Create the problem Draw a circle, mark its centre and draw a diameter through the centre. Use the diameter to form one side of a triangle. The other two sides should meet at a vertex somewhere on the circumference. Step 2: Split the triangle Divide the triangle in two by drawing a radius from the centre to the vertex on the circumference. Step 3: Two isosceles triangles Recognise that each small triangle has two sides that are r radii. All radii are the same in a particular circle. r r This means that each small triangle has two sides the same length. They must therefore both be isosceles triangles. Step 4: Angles in isosceles triangles a b Because each small triangle is an isosceles triangle, they a b must each have two equal angles. Step 5: Angles in the big triangle add up to 180° The sum of internal angles in any triangle is 180°. By comparison with the diagram in step 4, we notice that the a + b three angles in the big triangle are a, b and a + b. a b We can set up an equation: 2a + 2b = 180! ! a + b = 90! a + b is therefore a right angle – proven as required. Angles at the centre and circumference We want to prove that the angle subtended by an arc at the centre of a circle is twice the angle subtended at any point on the circumference.
    [Show full text]