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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 . -
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). -
Group 3: Clint Chan Karen Duncan Jake Jarvis Matt Johnson “The
Group 3: Clint Chan Karen Duncan Jake Jarvis Matt Johnson “The power of a point P with respect to a circle is the product of the two distances PA times PB from the point to the circle measured along a random secant” (B&B 261). When the point P is on the circle, its power is zero; when it is inside the circle, its power is negative; and when it is outside the circle, its power is positive. The power of a point when P is outside the circle is also equal to (PT)^2, where T is the point where the line through P is tangent to the circle. Also of interest, if P is outside the circle and e is a circle which is orthogonal to c and centered at P, then pc(A) = t^2, where t is the radius of e (from "The Power of a Point and Radical Axis" class notes). The radical axis of two circles is, by definition, the locus of points A for which the power of A with respect to c[1] and c[2] are equal. Using some facts about the power of a point, we can see that the radical axis is a line. Fact 1: If A is a point outside the circle and S is a secant intersecting c at M and N, then the power of A with respect to c is equal to _AM__AN_. Fact 2: If A is a point outside the circle and AT s a line tangent to c at T, then the power of A with respect to c is equal to _AT_^2. -
Circles in Areals
Circles in areals 23 December 2015 This paper has been accepted for publication in the Mathematical Gazette, and copyright is assigned to the Mathematical Association (UK). This preprint may be read or downloaded, and used for any non-profit making educational or research purpose. Introduction August M¨obiusintroduced the system of barycentric or areal co-ordinates in 1827[1, 2]. The idea is that one may attach weights to points, and that a system of weights determines a centre of mass. Given a triangle ABC, one obtains a co-ordinate system for the plane by placing weights x; y and z at the vertices (with x + y + z = 1) to describe the point which is the centre of mass. The vertices have co-ordinates A = (1; 0; 0), B = (0; 1; 0) and C = (0; 0; 1). By scaling so that triangle ABC has area [ABC] = 1, we can take the co-ordinates of P to be ([P BC]; [PCA]; [P AB]), provided that we take area to be signed. Our convention is that anticlockwise triangles have positive area. Points inside the triangle have strictly positive co-ordinates, and points outside the triangle must have at least one negative co-ordinate (we are allowed negative masses). The equation of a line looks like the equation of a plane in Cartesian co-ordinates, but note that the equation lx + ly + lz = 0 (with l 6= 0) is not satisfied by any point (x; y; z) of the Euclidean plane. We use such equations to describe the line at infinity when doing projective geometry. -
Power of a Point and Radical Axis
Power of a Point and Radical Axis Tovi Wen NYC Math Team §1 Power of a Point This handout will cover the topic power of a point, and one of its more powerful uses in radical axes. Definition (Power of a Point) Given a circle ! with center O and radius r, and a point P , the power of P with respect to !, which we will denote as (P; !) is OP 2 − r2. Note that • If P is outside ! then (P; !) is positive. • If P is on ! then (P; !) = 0. • If P is inside ! then (P; !) is negative. This definition does not feel particularly motivated, and therefore when taught at a more elementary level, it is often skipped and replaced by the following nice property. Theorem Let ! be a circle and let P be a point not on !. If a line passing through P meets ! at distinct points A and B then ( PA · PB if P lies outside !; (P; !) = −PA · PB if P lies inside ! Proof. It is not immediately obvious why the quantity PA · PB should be fixed for any line passing through P . Draw another chord of ! passing through P as shown. B A ω P O C M D Recall that opposite angles in a cyclic quadrilateral are supplementary. This gives \P AC = \P DB so as \AP C ≡ \BP D is shared, we have 4P AC ∼ 4P DB. In particular, PA PD = =) PA · PB = PC · P D: PC PB 1 Power of a Point and Radical Axis Tovi Wen We now show this quantity is equal to OP 2 − r2. -
Power of a Point and Ceva's Theorem
Mathematical Problem Solving Power of a Point A rather simple definition of the power of a point with respect to a circle is: Let C be a circle of radius r. The power of a point P with respect to C is given by d2 − r2, where d is the distance of P to the center of the circle. Just to fix ideas, for example, if C is the circle of radius r centered at the origin, and P has coordinates (x, y), then the power of P with respect to this circle is x2 + y2 − r2. If P is a point, C a circle, I’ll write Π(P, C) to denote the power of P with respect to C. Obviously, Π(P, C) > 0 if and only if P is outside the circle, < 0 if and only if P is inside the circle, and 0 if and only if P is on the circle. The significance of this concept is due to the following result. Theorem 1 Let P,X,X0 be collinear points, let C be a circle. If X,X0 lie on C, and either X 6= X0, or if X = X0 6= P and the line containing X and P is tangent to C, then 0 Π(P, C)= PX · PX , where PX,PX0 are “directed lengths;” to be specific: PX · PX0 < 0 if P is strictly between X,X0, > 0 if either X is strictly between P and X0 or X0 is strictly between P and X, 0 if P coincides with X or X0. -
Prove It Classroom
How to Prove it ClassRoom SHAILESH SHIRALI Euler’s formula for the area of a pedal triangle Given a triangle ABC and a point P in the plane of ABC In this episode of “How (note that P does not have to lie within the triangle), the To Prove It”, we prove pedal triangle of P with respect to ABC is the triangle △ a beautiful and striking whose vertices are the feet of the perpendiculars drawn from formula first found by P to the sides of ABC. See Figure 1. The pedal triangle relates Leonhard Euler; it gives the area of the pedal triangle in a natural way to the parent triangle, and we may wonder of a point with reference whether there is a convenient formula giving the area of the to another triangle. pedal triangle in terms of the parameters of the parent triangle. The great 18th-century mathematician Euler found just such a formula (given in Box 1). It is a compact and pleasing result, and it expresses the area of the pedal triangle in terms of the radius R of the circumcircle of ABC and the △ distance between P and the centre O of the circumcircle. A O: circumcentre of ABC E △ P: arbitrary point Area ( DEF) 1 OP2 F △ = 1 P Area ( ABC) 4 − R2 O △ ( ) B D C Figure 1. Euler's formula for the area of the pedal triangle of an arbitrary point Keywords: Circle theorem, pedal triangle, power of a point, Euler, sine rule, extended sine rule, Wallace-Simson theorem Azim Premji University At Right Angles, July 2018 95 1 A E F P O B D C Figure 2. -
The Radii by R, Ra, Rb,Rc
The Orthopole, by R. GOORMAGHTIGH,La Louviero, Belgium. Introduction. 1. In this little work are presented the more important resear- ches on one of the newest chapters in the modern Geometry of the triangle, and whose origin is, the following theorem due to. Professor J. Neuberg: Let ƒ¿, ƒÀ, y be the projections of A, B, C on a straight line m; the perpendiculars from a on BC, ƒÀ on CA, ƒÁ on ABare concurrent(1). The point of concurrence. M is called the orthopole of m for the triangle ABC. The following general conventions will now be observed : the sides of the triangle of reference ABC will be denoted by a, b, c; the centre and radius of the circumcircle by 0 and R; the orthocentre by H; the centroid by G; the in- and ex-centresby I, Ia 1b, Ic and the radii by r, ra, rb,rc;the mid-points of BC, CA, AB by Am, Bm, Cm;the feet of the perpendiculars AH, BH, OH from A, B, C on BO, CA, AB by An,Bn, Cn; the contact points of the incircle with BC, CA, AB by D, E, F; the corresponding points for the circles Ia, Ib, Ic, by Da, Ea,Fa, Dc, Eb, F0, Dc, Ec, Fc; the area of ABC by 4; the centre of the nine-point circle by 0g; the Lemoine . point by K; the Brocard points by theGergonne and Nagel points by and N(2); further 2s=a+b+c. Two points whose joins to A, B, C are isogonal conjugate in the angles A, B, C are counterpoints. -
A History of Elementary Mathematics, with Hints on Methods of Teaching
;-NRLF I 1 UNIVERSITY OF CALIFORNIA PEFARTMENT OF CIVIL ENGINEERING BERKELEY, CALIFORNIA Engineering Library A HISTORY OF ELEMENTARY MATHEMATICS THE MACMILLAN COMPANY NEW YORK BOSTON CHICAGO DALLAS ATLANTA SAN FRANCISCO MACMILLAN & CO., LIMITED LONDON BOMBAY CALCUTTA MELBOURNE THE MACMILLAN CO. OF CANADA, LTD. TORONTO A HISTORY OF ELEMENTARY MATHEMATICS WITH HINTS ON METHODS OF TEACHING BY FLORIAN CAJORI, PH.D. PROFESSOR OF MATHEMATICS IN COLORADO COLLEGE REVISED AND ENLARGED EDITION THE MACMILLAN COMPANY LONDON : MACMILLAN & CO., LTD. 1917 All rights reserved Engineering Library COPYRIGHT, 1896 AND 1917, BY THE MACMILLAN COMPANY. Set up and electrotyped September, 1896. Reprinted August, 1897; March, 1905; October, 1907; August, 1910; February, 1914. Revised and enlarged edition, February, 1917. o ^ PREFACE TO THE FIRST EDITION "THE education of the child must accord both in mode and arrangement with the education of mankind as consid- ered in other the of historically ; or, words, genesis knowledge in the individual must follow the same course as the genesis of knowledge in the race. To M. Comte we believe society owes the enunciation of this doctrine a doctrine which we may accept without committing ourselves to his theory of 1 the genesis of knowledge, either in its causes or its order." If this principle, held also by Pestalozzi and Froebel, be correct, then it would seem as if the knowledge of the history of a science must be an effectual aid in teaching that science. Be this doctrine true or false, certainly the experience of many instructors establishes the importance 2 of mathematical history in teaching. With the hope of being of some assistance to my fellow-teachers, I have pre- pared this book and have interlined my narrative with occasional remarks and suggestions on methods of teaching. -
Pythagorean Theorem
MAC-CPTM Situations Project Situation 36: Pythagorean Theorem Prepared at Pennsylvania State University Mid-Atlantic Center for Mathematics Teaching and Learning June 28, 2005—Patrick Sullivan 27 September 2009 – Kathy Heid, Maureen Grady, Shiv Karunakaran Prompt In both an Algebra I course and an Advanced Algebra course, students were given transparency cutouts of graph paper squares with side lengths from one unit to twenty- five units. Students were asked to create triangles whose sides had the side-lengths of three of the squares. Students began to notice the squares that would create right triangles and the relationship involving the area of those squares. A student asked, “Does this work for every right triangle?” Commentary The Pythagorean Theorem relates the squares of the lengths of the sides of a right triangle. The Law of Cosines establishes a more general relationship between the same quantities that holds for all triangles. Using algebra and geometry together we can prove the Pythagorean Theorem and the Law of Cosines. SIT36_090927.doc Page 1 of 11 Mathematical Focus 1 Visual inspection alone is insufficient for drawing mathematical conclusions The generalization drawn by the student is based on what the student observed using physical models. Although such observations are important for mathematical discovery they cannot replace mathematical proof. The diagram that follows is an example of a case in which the physical representation is illusory. The diagram shown below makes it appear that two right triangles with the same base and height have different areas. However, close examination reveals that neither of the figures is actually a right triangle. -
The Brocard and Tucker Circles of a Cyclic Quadrilateral
ei The Brocard and Tucker Circles of a Cyclic Quadrilateral. By FREDERICK G. W. BROWN. (Received 4th September 1917. Bead 9th November 1917.) 1. The application of the geometrical properties of the Brocard and Tucker circles of a triangle to a quadrilateral appears never to have been adequately worked out, as far as the author can discover. Hence, the object of this paper. Some of the problems involved have been published, under the author's name, as independent questions for solution, and where, in the author's opinion, solutions other than his own have seemed more satisfactory for the logical treatment of the subject, these solutions have been employed, with due acknowledgments to their authors. 2. Condition for Brocard poinU. We shall, first establish the condition necessary for the existence of Brocard points within a quadrilateral. Let ABCD (Fig. 1) be a quadrilateral in which a point X can be found such that the L. ' XAD, XBA, XCB, XDC are all equal; denote each of these angles by o>; the sides BC, CD, DA, AB by a, b, c, d; the diagonals BD, AC by e,f, and the area by Q; then L AXB = ir -to - (A — <a) = TT - A. Similarly LBXC = TT-B, L.CXD = TT-C, LDXA^TT-D. Now AX: sin w = AB : sin AXB = d: sin (JT - A) = d : sin A and AX: sin (Z? - w) = c : sin (ir - D) = c : sin D, hence, eliminating AX by division sin (D -<o): sin <a — d sin D : c sin A, Downloaded from https://www.cambridge.org/core. -
Power of a Point
Power of a Point Ray Li ([email protected]) June 29, 2017 1 Introduction Here are some basic facts about power of a point. 1. (Definition of Power) Let P be a point in the plane and ! be a circle with center O 2 2 and radius R. Then Pow!(P ) = PO − R is the power of P with respect to !. 2. (Power of a Point Theorem) Let P be a point in the plane and ! be a circle. A line through P meets ! at A and B; and another one meets it at C and D: Then PA · PB = PC · PD = Pow!(P ) (assuming directed lengths). 3. (Special Case) When one of the lines is tangent to the circle, we have C = D and PA · PB = PC2: 4. (Converse of Power of a Point) Let A; B; C; D be points in a plane, and let AB meet CD at P: Suppose that P is either on both segments AB and CD; or is on neither of them. If PA · PB = PC · P D; then A; B; C; D lie on a circle. 5. (Power of a Point on coordinates) Let (x − a)2 + (y − b)2 − c2 = 0 be the equation of a circle in the coordinate plane. For a point P = (x; y); we have Pow!(P ) = (x − a)2 + (y − b)2 − c2: 6. (Fact) When P is outside ! we have Pow!(P ) > 0, when P is on ! we have Pow!(P ) = 0, and when P is inside ! we have Pow!(P ) < 0 7. (Radical axis) Given two circles !1 and !2; the locus of points that have equal power to both circles is a line called the radical axis.