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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. -
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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. -
Cyclic Quadrilaterals — the Big Picture Yufei Zhao [email protected]
Winter Camp 2009 Cyclic Quadrilaterals Yufei Zhao Cyclic Quadrilaterals | The Big Picture Yufei Zhao [email protected] An important skill of an olympiad geometer is being able to recognize known configurations. Indeed, many geometry problems are built on a few common themes. In this lecture, we will explore one such configuration. 1 What Do These Problems Have in Common? 1. (IMO 1985) A circle with center O passes through the vertices A and C of triangle ABC and intersects segments AB and BC again at distinct points K and N, respectively. The circumcircles of triangles ABC and KBN intersects at exactly two distinct points B and M. ◦ Prove that \OMB = 90 . B M N K O A C 2. (Russia 1995; Romanian TST 1996; Iran 1997) Consider a circle with diameter AB and center O, and let C and D be two points on this circle. The line CD meets the line AB at a point M satisfying MB < MA and MD < MC. Let K be the point of intersection (different from ◦ O) of the circumcircles of triangles AOC and DOB. Show that \MKO = 90 . C D K M A O B 3. (USA TST 2007) Triangle ABC is inscribed in circle !. The tangent lines to ! at B and C meet at T . Point S lies on ray BC such that AS ? AT . Points B1 and C1 lies on ray ST (with C1 in between B1 and S) such that B1T = BT = C1T . Prove that triangles ABC and AB1C1 are similar to each other. 1 Winter Camp 2009 Cyclic Quadrilaterals Yufei Zhao A B S C C1 B1 T Although these geometric configurations may seem very different at first sight, they are actually very related. -
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. -
On a Construction of Hagge
Forum Geometricorum b Volume 7 (2007) 231–247. b b FORUM GEOM ISSN 1534-1178 On a Construction of Hagge Christopher J. Bradley and Geoff C. Smith Abstract. In 1907 Hagge constructed a circle associated with each cevian point P of triangle ABC. If P is on the circumcircle this circle degenerates to a straight line through the orthocenter which is parallel to the Wallace-Simson line of P . We give a new proof of Hagge’s result by a method based on reflections. We introduce an axis associated with the construction, and (via an areal anal- ysis) a conic which generalizes the nine-point circle. The precise locus of the orthocenter in a Brocard porism is identified by using Hagge’s theorem as a tool. Other natural loci associated with Hagge’s construction are discussed. 1. Introduction One hundred years ago, Karl Hagge wrote an article in Zeitschrift fur¨ Mathema- tischen und Naturwissenschaftliche Unterricht entitled (in loose translation) “The Fuhrmann and Brocard circles as special cases of a general circle construction” [5]. In this paper he managed to find an elegant extension of the Wallace-Simson theorem when the generating point is not on the circumcircle. Instead of creating a line, one makes a circle through seven important points. In 2 we give a new proof of the correctness of Hagge’s construction, extend and appl§ y the idea in various ways. As a tribute to Hagge’s beautiful insight, we present this work as a cente- nary celebration. Note that the name Hagge is also associated with other circles [6], but here we refer only to the construction just described. -
The Role of Euclidean Geometry in High School
JOURNAL OF MATHEMATICAL BEHAVIOR 15, 221-237 (1996) The Role of Euclidean Geometry in High School HUNG-HSI Wu University of California When I was a high school student long ago, the need to study Euclidean geome- try was taken for granted. In fact, the novelty of learning to prove something was so overwhelming to me and some of my friends that, years later, we would look back at Euclidean geometry as the high point of our mathematics education. In recent years, the value of Euclid has depreciated considerably. A vigorous debate is now going on about how much, if any, of his work is still relevant. In this article, I would like to give my perspective on this subject, and in so doing, I will be taking a philosophical tour, so to speak, of a small portion of mathematics. This may not be so surprising because after all, in life as in mathematics, one does need a little philosophy for guidance from time to time. The bone of contention in the geometry curriculum is of course how many of the traditional “two-column proofs” should be retained. Let me first briefly indicate the range of available options in this regard: 1. A traditional text in use in a local high school in Berkeley not too long ago begins on page 1 with the undefined terms of the axioms, and formal proofs start on page 22. There is no motivation or explanation of the whys and hows of an axiomatic system. 2. A recently published text does experimental geometry (i.e., “hands-on ge- ometry” with no proofs) all the way until the last 130 pages of its 700 pages of exposition. -
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. -
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. -
Chapter 2 (Circles)
CHAPTER 2 Circles Construct a circle of radius zero. Although it is often an intermediate step, angle chasing is usually not enough to solve a problem completely. In this chapter, we develop some other fundamental tools involving circles. 2.1 Orientations of Similar Triangles You probably already know the similarity criterion for triangles. Similar triangles are useful because they let us convert angle information into lengths. This leads to the power of a point theorem, arguably the most common sets of similar triangles. In preparation for the upcoming section, we develop the notion of similar triangles that are similarly oriented and oppositely oriented. Here is how it works. Consider triangles ABC and XYZ. We say they are directly similar, or similar and similarly oriented,if ABC = XYZ, BCA = YZX, and CAB = ZXY. Wesaytheyareoppositely similar, or similar and oppositely oriented,if ABC =−XYZ, BCA =−YZX, and CAB =−ZXY. If they are either directly similar or oppositely similar, then they are similar. We write ABC ∼XYZ in this case. See Figure 2.1A for an illustration. Two of the angle equalities imply the third, so this is essentially directed AA. Remember to pay attention to the order of the points. T T1 2 T3 Figure 2.1A. T1 is directly similar to T2 and oppositely to T3. 23 24 2. Circles The upshot of this is that we may continue to use directed angles when proving triangles are similar; we just need to be a little more careful. In any case, as you probably already know, similar triangles also produce ratios of lengths.