Casey s Theorem and its Applications
|
|
- Rudolph Benson
- 7 years ago
- Views:
Transcription
1 Casey s Theorem and its Applications Luis González Maracaibo. Venezuela July 011 Abstract. We present a proof of the generalized Ptolemy s theorem, also known as Casey s theorem and its applications in the resolution of difficult geometry problems. 1 Casey s Theorem. Theorem 1. Two circles Γ 1 (r 1 ) and Γ (r ) are internally/externally tangent to a circle Γ(R) through A, B, respetively. The length δ 1 of the common external tangent of Γ 1, Γ is given by: δ 1 = AB R (R ± r 1 )(R ± r ) Proof. Without loss of generality assume that r 1 r and we suppose that Γ 1 and Γ are internally tangent to Γ. The remaining case will be treated analogously. A common external tangent between Γ 1 and Γ touches Γ 1, Γ at A 1, B 1 and A is the orthogonal projection of O onto O 1 A 1. (See Figure 1). By Pythagorean theorem for O 1 O A, we obtain δ 1 = (A 1 B 1 ) = (O 1 O ) (r 1 r ) Let O 1 OO = λ. By cosine law for OO 1 O, we get (O 1 O ) = (R r 1 ) + (R r ) (R r 1 )(R r ) cos λ By cosine law for the isosceles triangle OAB, we get AB = R (1 cos λ) 1
2 Figure 1: Theorem 1 Eliminating cos λ and O 1 O from the three previous expressions yields δ 1 = (R r 1 ) + (R r ) (r 1 r ) (R r 1 )(R r ) ( 1 AB R ) Subsequent simplifications give δ 1 = AB R (R r 1 )(R r ) (1) Analogously, if Γ 1, Γ are externally tangent to Γ, then we will get δ 1 = AB R (R + r 1 )(R + r ) () If Γ 1 is externally tangent to Γ and Γ is internally tangent to Γ, then a similar reasoning gives that the length of the common internal tangent between Γ 1 and Γ is given by δ 1 = AB R (R + r 1 )(R r ) (3)
3 Theorem (Casey). Given four circles Γ i, i = 1,, 3, 4, let δ ij denote the length of a common tangent (either internal or external) between Γ i and Γ j. The four circles are tangent to a fith circle Γ (or line) if and only if for appropriate choice of signs, δ 1 δ 34 ± δ 13 δ 4 ± δ 14 δ 3 = 0 The proof of the direct theorem is straightforward using Ptolemy s theorem for the quadrilateral ABCD whose vertices are the tangency points of Γ 1 (r 1 ), Γ (r ), Γ 3 (r 3 ), Γ 4 (r 4 ) with Γ(R). We susbtitute the lengths of its sides and digonals in terms of the lenghts of the tangents δ ij, by using the formulas (1), () and (3). For instance, assuming that all tangencies are external, then using (1), we get δ 1 δ 34 + δ 14 δ 3 = ( AB CD+AD BC R ) (R r 1 )(R r )(R r 3 )(R r 4 ) δ 1 δ 34 + δ 14 δ 3 = ( ) AC BD (R r R 1 )(R r 3 ) (R r )(R r 4 ) δ 1 δ 34 + δ 14 δ 3 = δ 13 δ 4. Casey established that this latter relation is sufficient condition for the existence of a fith circle Γ(R) tangent to Γ 1 (r 1 ), Γ (r ), Γ 3 (r 3 ), Γ 4 (r 4 ). Interestingly, the proof of this converse is a much tougher exercise. For a proof you may see [1]. Some Applications. I) ABC is isosceles with legs AB = AC = L. A circle ω is tangent to BC and the arc BC of the circumcircle of ABC. A tangent line from A to ω touches ω at P. Describe the locus of P as ω varies. Solution. We use Casey s theorem for the circles (A), (B), (C) (with zero radii) and ω, all internally tangent to the circumcircle of ABC. Thus, if ω touches BC at Q, we have: L CQ + L BQ = AP BC = AP = L(BQ + CQ) BC The length AP is constant, i.e. Locus of P is the circle with center A and radius AB = AC = L. = L II) (O) is a circle with diameter AB and P, Q are two points on (O) lying on different sides of AB. T is the orthogonal projection of Q onto AB. Let (O 1 ), (O ) be the circles with diameters T A, T B and P C, P D are the tangent segments from P to (O 1 ), (O ), respectively. Show that P C + P D = P Q. []. 3
4 Figure : Application II Solution. Let δ 1 denote the length of the common external tangent of (O 1 ), (O ). We use Casey s theorem for the circles (O 1 ), (O ), (P ), (Q), all internally tangent to (O). P C QT + P D QT = P Q δ 1 = P C + P D = P Q δ1 QT = P Q T A T B T Q = P Q. III) In ABC, let ω A, ω B, ω C be the circles tangent to BC, CA, AB through their midpoints and the arcs BC, CA, AB of its circumcircle (not containing A, B, C). If δ BC, δ CA, δ AB denote the lengths of the common external tangents between (ω B, ω C ), (ω C, ω A ) and (ω A, ω B ), respectively, then prove that δ BC = δ CA = δ AB = a + b + c 4 Solution. Let δ A, δ B, δ C denote the lengths of the tangents from A, B, C to ω A, ω B, ω C, respectively. By Casey s theorem for the circles (A), (B), (C), ω B, all tangent to the circumcircle of ABC, we get δ B b = a AE + c CE = δ B = 1 (a + c) Similarly, by Casey s theorem for (A), (B), (C), ω C we ll get δ C = 1 (a + b) 4
5 Now, by Casey s theorem for (B), (C), ω B, ω C, we get δ B δ C = δ BC a + BF BE = δ BC = δ B δ C BF BE a = (a + c)(a + b) bc 4a = a + b + c 4 By similar reasoning, we ll have δ CA = δ AB = 1 (a + b + c). 4 IV) A circle K passes through the vertices B, C of ABC and another circle ω touches AB, AC, K at P, Q, T, respectively. If M is the midpoint of the arc BT C of K, show that BC, P Q, MT concur. [3] Solution. Let R, ϱ be the radii of K and ω, respectively. Using formula (1) of Theorem 1 for ω, (B) and ω, (C). Both (B), (C) with zero radii and tangent to K through B, C, we obtain: T C = CQ R (R ϱ)(r 0) = CQ R R ϱ, T B = BP R (R ϱ)(r 0) = BP R R ϱ = T B T C = BP CQ Let P Q cut BC at U. By Menelaus theorem for ABC cut by UP Q we have UB UC = BP AP AQ CQ = BP CQ = T B T C Thus, by angle bisector theorem, U is the foot of the T-external bisector T M of BT C. V) If D, E, F denote the midpoints of the sides BC, CA, AB of ABC. Show that the incircle (I) of ABC is tangent to (DEF ). (Feuerbach theorem). Solution. We consider the circles (D), (E), (F ) with zero radii and (I). The notation δ XY stands for the length of the external tangent between the circles (X), (Y ), then δ DE = c, δ EF = a, δ F D = b, δ b c DI =, δ a c EI =, δ b a F I = For the sake of applying the converse of Casey s theorem, we shall verify if, for some combination of signs + and, we get ±c(b a)±a(b c)±b(a c) = 0, which is trivial. Therefore, there exists a circle tangent to (D), (E), (F ) and (I), i.e. (I) is internally tangent to (DEF ). We use the same reasoning to show that (DEF ) is tangent to the three excircles of ABC. VI) ABC is scalene and D, E, F are the midpoints of BC, CA, AB. The incircle (I) and 9 point circle (DEF ) of ABC are internally tangent through the Feuerbach point F e. Show that one of the segments F e D, F e E, F e F equals the sum of the other two. [4] 5
6 Solution. WLOG assume that b a c. Incircle (I, r) touches BC at M. Using formula (1) of Theorem 1 for (I) and (D) (with zero radius) tangent to the 9-point circle (N, R ), we have: F e D = DM ( R ) R ( R r)( R 0) = F (b c) ed = R r By similar reasoning, we have the expressions R (a c) R (b a) F e E =, F e F = R r R r Therefore, the addition of the latter expressions gives R F e E + F e F = R r b c = F e D VII) ABC is a triangle with AC > AB. A circle ω A is internally tangent to its circumcircle ω and AB, AC. S is the midpoint of the arc BC of ω, which does not contain A and ST is the tangent segment from S to ω A. Prove that ST SA = AC AB AC + AB [5] Solution. Let M, N be the tangency points of ω A with AC, AB. By Casey s theorem for ω A, (B), (C), (S), all tangent to the circumcircle ω, we get ST BC + CS BN = CM BS = ST BC = CS(CM BN) If U is the reflection of B across AS, then CM BN = UC = AC AB. Hence ST BC = CS(AC AB) ( ) By Ptolemy s theorem for ABSC, we get SA BC = CS(AB + AC). Together with ( ), we obtain ST SA = AC AB AC + AB 6
7 VIII) Two congruent circles (S 1 ), (S ) meet at two points. A line l cuts (S ) at A, C and (S 1 ) at B, D (A, B, C, D are collinear in this order). Two distinct circles ω 1, ω touch the line l and the circles (S 1 ), (S ) externally and internally respectively. If ω 1, ω are externally tangent, show that AB = CD. [6] Solution. Let P ω 1 ω and M, N be the tangency points of ω 1 and ω with an external tangent. Inversion with center P and power P B P D takes (S 1 ) and the line l into themselves. The circles ω 1 and ω go to two parallel lines k 1 and k tangent to (S 1 ) and the circle (S ) goes to another circle (S ) tangent to k 1, k. Hence, (S ) is congruent to its inverse (S ). Further, (S ), (S ) are symmetrical about P = P C P A = P B P D. By Casey s theorem for ω 1, ω, (D), (B), (S 1 ) and ω 1, ω, (A), (C), (S ) we get: DB = P B P D MN, AC = P A P C MN Since P C P A = P B P D = AC = BD = AB = CD. IX) ABC is equilateral with side length L. Let (O, r) and (O, R) be the incircle and circumcircle of ABC. P is a point on (O, r) and P 1, P, P 3 are the projections of P onto BC, CA, AB. Circles T 1, T and T 3 touch BC, CA, AB through P 1, P, P and (O, R) (internally), their centers lie on different sides of BC, CA, AB with respect to A, B, C. Prove that the sum of the lengths of the common external tangents of T 1, T and T 3 is a constant value. Solution. Let δ 1 denote the tangent segment from A to T 1. By Casey s theorem for (A), (B), (C), T 1, all tangent to (O, R), we have L BP 1 + L CP 1 = δ 1 L = δ 1 = L. Similarly, we have δ = δ 3 = L. By Euler s theorem for the pedal triangle P 1 P P 3 of P, we get: [P 1 P P 3 ] = p(p, (O)) [ABC] = R r [ABC] = 3 4R 4R 16 [ABC] Therefore, we obtain AP AP 3 + BP 3 BP 1 + CP 1 CP = sin 60 ([ABC] [P 1P P 3 ]) = L. ( ) By Casey s theorem for (B), (C), T, T 3, all tangent to (O, R), we get δ δ 3 = L = BC δ 3 + CP BP 3 = L δ 3 + (L AP 1 )(L AP ) By cyclic exchange, we have the expressions: L = L δ 31 + (L BP 3 )(L BP 1 ), L = L δ 1 + (L CP 1 )(L CP ) 7
8 Figure 3: Application VII Adding the three latter equations yields 3L = L(δ 3 + δ 31 + δ 1 ) + 3L 3L + AP 3 AP + BP 3 BP 1 + CP 1 CP Hence, combining with ( ) gives δ 3 + δ 31 + δ 1 = 3L L = L 3 Proposed Problems. 1) Purser s theorem: ABC is a triangle with circumcircle (O) and ω is a circle in its plane. AX, BY, CZ are the tangent segments from A, B, C to ω. Show that ω is tangent to (O), if and only if ±AX BC ± BY CA ± CZ AB = 0 8
9 ) Circle ω touches the sides AB, AC of ABC at P, Q and its circumcircle (O). Show that the midpoint of P Q is either the incenter of ABC or the A-excenter of ABC, according to whether (O), ω are internally tangent or externally tangent. 3) ABC is A-right with circumcircle (O). Circle Ω B is tangent to the segments OB, OA and the arc AB of (O). Circle Ω C is tangent to the segments OC, OA and the arc AC of (O). Ω B, Ω C touch OA at P, Q, respectively. Show that: AB AC = AP AQ 4) Gumma, We are given a cirle (O, r) in the interior of a square ABCD with side length L. Let (O i, r i ) i = 1,, 3, 4 be the circles tangent to two sides of the square and (O, r) (externally). Find L as a fuction of r 1, r, r 3, r 4. 5) Two parallel lines τ 1, τ touch a circle Γ(R). Circle k 1 (r 1 ) touches Γ, τ 1 and a third circle k (r ) touches Γ, τ, k 1. We assume that all tangencies are external. Prove that R = r 1 r. 6)Victor Thébault ABC has incircle (I, r) and circumcircle (O). D is a point on AB. Circle Γ 1 (r 1 ) touches the segments DA, DC and the arc CA of (O). Circle Γ (r ) touches the segments DB, DC and the arc CB of (O). If ADC = ϕ, show that: r 1 cos ϕ + r sin ϕ = r References [1] I. Shariguin, Problemas de Geometrié (Planimetrié), Ed. Mir, Moscu, [] Vittasko, Sum of two tangents, equal to the distance of two points, AoPS, [3] My name is math, Tangent circles concurrent lines, AoPs, [4] Mathquark, Point [Feuerbach point of a triangle; FY + FZ = FX], AoPS, [5] Virgil Nicula, ABC and circle tangent to AB, AC and circumcircle, AoPS, [6] Shoki, Iran(3rd round)009, AoPS,
DEFINITIONS. Perpendicular Two lines are called perpendicular if they form a right angle.
DEFINITIONS Degree A degree is the 1 th part of a straight angle. 180 Right Angle A 90 angle is called a right angle. Perpendicular Two lines are called perpendicular if they form a right angle. Congruent
More informationProjective Geometry - Part 2
Projective Geometry - Part 2 Alexander Remorov alexanderrem@gmail.com Review Four collinear points A, B, C, D form a harmonic bundle (A, C; B, D) when CA : DA CB DB = 1. A pencil P (A, B, C, D) is the
More informationCollinearity and concurrence
Collinearity and concurrence Po-Shen Loh 23 June 2008 1 Warm-up 1. Let I be the incenter of ABC. Let A be the midpoint of the arc BC of the circumcircle of ABC which does not contain A. Prove that the
More informationSan Jose Math Circle April 25 - May 2, 2009 ANGLE BISECTORS
San Jose Math Circle April 25 - May 2, 2009 ANGLE BISECTORS Recall that the bisector of an angle is the ray that divides the angle into two congruent angles. The most important results about angle bisectors
More informationExercise Set 3. Similar triangles. Parallel lines
Exercise Set 3. Similar triangles Parallel lines Note: The exercises marked with are more difficult and go beyond the course/examination requirements. (1) Let ABC be a triangle with AB = AC. Let D be an
More informationInversion. Chapter 7. 7.1 Constructing The Inverse of a Point: If P is inside the circle of inversion: (See Figure 7.1)
Chapter 7 Inversion Goal: In this chapter we define inversion, give constructions for inverses of points both inside and outside the circle of inversion, and show how inversion could be done using Geometer
More informationSolutions to Practice Problems
Higher Geometry Final Exam Tues Dec 11, 5-7:30 pm Practice Problems (1) Know the following definitions, statements of theorems, properties from the notes: congruent, triangle, quadrilateral, isosceles
More informationChapter 6 Notes: Circles
Chapter 6 Notes: Circles IMPORTANT TERMS AND DEFINITIONS A circle is the set of all points in a plane that are at a fixed distance from a given point known as the center of the circle. Any line segment
More informationChapters 6 and 7 Notes: Circles, Locus and Concurrence
Chapters 6 and 7 Notes: Circles, Locus and Concurrence IMPORTANT TERMS AND DEFINITIONS A circle is the set of all points in a plane that are at a fixed distance from a given point known as the center of
More informationIMO Geomety Problems. (IMO 1999/1) Determine all finite sets S of at least three points in the plane which satisfy the following condition:
IMO Geomety Problems (IMO 1999/1) Determine all finite sets S of at least three points in the plane which satisfy the following condition: for any two distinct points A and B in S, the perpendicular bisector
More informationChapter 3. Inversion and Applications to Ptolemy and Euler
Chapter 3. Inversion and Applications to Ptolemy and Euler 2 Power of a point with respect to a circle Let A be a point and C a circle (Figure 1). If A is outside C and T is a point of contact of a tangent
More information1. Find the length of BC in the following triangles. It will help to first find the length of the segment marked X.
1 Find the length of BC in the following triangles It will help to first find the length of the segment marked X a: b: Given: the diagonals of parallelogram ABCD meet at point O The altitude OE divides
More informationAdvanced Euclidean Geometry
dvanced Euclidean Geometry What is the center of a triangle? ut what if the triangle is not equilateral?? Circumcenter Equally far from the vertices? P P Points are on the perpendicular bisector of a line
More informationMath 531, Exam 1 Information.
Math 531, Exam 1 Information. 9/21/11, LC 310, 9:05-9:55. Exam 1 will be based on: Sections 1A - 1F. The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/531fa11/531.html)
More informationCIRCLE COORDINATE GEOMETRY
CIRCLE COORDINATE GEOMETRY (EXAM QUESTIONS) Question 1 (**) A circle has equation x + y = 2x + 8 Determine the radius and the coordinates of the centre of the circle. r = 3, ( 1,0 ) Question 2 (**) A circle
More informationCSU Fresno Problem Solving Session. Geometry, 17 March 2012
CSU Fresno Problem Solving Session Problem Solving Sessions website: http://zimmer.csufresno.edu/ mnogin/mfd-prep.html Math Field Day date: Saturday, April 21, 2012 Math Field Day website: http://www.csufresno.edu/math/news
More informationGeometry Regents Review
Name: Class: Date: Geometry Regents Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. If MNP VWX and PM is the shortest side of MNP, what is the shortest
More informationSection 9-1. Basic Terms: Tangents, Arcs and Chords Homework Pages 330-331: 1-18
Chapter 9 Circles Objectives A. Recognize and apply terms relating to circles. B. Properly use and interpret the symbols for the terms and concepts in this chapter. C. Appropriately apply the postulates,
More informationwww.pioneermathematics.com
Problems and Solutions: INMO-2012 1. Let ABCD be a quadrilateral inscribed in a circle. Suppose AB = 2+ 2 and AB subtends 135 at the centre of the circle. Find the maximum possible area of ABCD. Solution:
More informationThe University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Student Name:
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Wednesday, August 18, 2010 8:30 to 11:30 a.m., only Student Name: School Name: Print your name and the name of
More informationConjectures. Chapter 2. Chapter 3
Conjectures Chapter 2 C-1 Linear Pair Conjecture If two angles form a linear pair, then the measures of the angles add up to 180. (Lesson 2.5) C-2 Vertical Angles Conjecture If two angles are vertical
More information1 Solution of Homework
Math 3181 Dr. Franz Rothe February 4, 2011 Name: 1 Solution of Homework 10 Problem 1.1 (Common tangents of two circles). How many common tangents do two circles have. Informally draw all different cases,
More informationThree Lemmas in Geometry
Winter amp 2010 Three Lemmas in Geometry Yufei Zhao Three Lemmas in Geometry Yufei Zhao Massachusetts Institute of Technology yufei.zhao@gmail.com 1 iameter of incircle T Lemma 1. Let the incircle of triangle
More informationContents. 2 Lines and Circles 3 2.1 Cartesian Coordinates... 3 2.2 Distance and Midpoint Formulas... 3 2.3 Lines... 3 2.4 Circles...
Contents Lines and Circles 3.1 Cartesian Coordinates.......................... 3. Distance and Midpoint Formulas.................... 3.3 Lines.................................. 3.4 Circles..................................
More informationCircle Name: Radius: Diameter: Chord: Secant:
12.1: Tangent Lines Congruent Circles: circles that have the same radius length Diagram of Examples Center of Circle: Circle Name: Radius: Diameter: Chord: Secant: Tangent to A Circle: a line in the plane
More informationThe Euler Line in Hyperbolic Geometry
The Euler Line in Hyperbolic Geometry Jeffrey R. Klus Abstract- In Euclidean geometry, the most commonly known system of geometry, a very interesting property has been proven to be common among all triangles.
More informationTriangle Centers MOP 2007, Black Group
Triangle Centers MOP 2007, Black Group Zachary Abel June 21, 2007 1 A Few Useful Centers 1.1 Symmedian / Lemmoine Point The Symmedian point K is defined as the isogonal conjugate of the centroid G. Problem
More informationThe University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, January 24, 2013 9:15 a.m. to 12:15 p.m.
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Thursday, January 24, 2013 9:15 a.m. to 12:15 p.m., only Student Name: School Name: The possession or use of any
More informationIMO Training 2008 Circles Yufei Zhao. Circles. Yufei Zhao.
ircles Yufei Zhao yufeiz@mit.edu 1 Warm up problems 1. Let and be two segments, and let lines and meet at X. Let the circumcircles of X and X meet again at O. Prove that triangles O and O are similar.
More informationChapter 1. The Medial Triangle
Chapter 1. The Medial Triangle 2 The triangle formed by joining the midpoints of the sides of a given triangle is called the medial triangle. Let A 1 B 1 C 1 be the medial triangle of the triangle ABC
More informationGeometry Unit 5: Circles Part 1 Chords, Secants, and Tangents
Geometry Unit 5: Circles Part 1 Chords, Secants, and Tangents Name Chords and Circles: A chord is a segment that joins two points of the circle. A diameter is a chord that contains the center of the circle.
More informationTHREE DIMENSIONAL GEOMETRY
Chapter 8 THREE DIMENSIONAL GEOMETRY 8.1 Introduction In this chapter we present a vector algebra approach to three dimensional geometry. The aim is to present standard properties of lines and planes,
More informationDefinitions, Postulates and Theorems
Definitions, s and s Name: Definitions Complementary Angles Two angles whose measures have a sum of 90 o Supplementary Angles Two angles whose measures have a sum of 180 o A statement that can be proven
More informationCHAPTER 8 QUADRILATERALS. 8.1 Introduction
CHAPTER 8 QUADRILATERALS 8.1 Introduction You have studied many properties of a triangle in Chapters 6 and 7 and you know that on joining three non-collinear points in pairs, the figure so obtained is
More informationTest on Circle Geometry (Chapter 15)
Test on Circle Geometry (Chapter 15) Chord Properties of Circles A chord of a circle is any interval that joins two points on the curve. The largest chord of a circle is its diameter. 1. Chords of equal
More informationA Nice Theorem on Mixtilinear Incircles
A Nice Theorem on Mixtilinear Incircles Khakimboy Egamberganov Abstract There are three mixtilinear incircles and three mixtilinear excircles in an arbitrary triangle. In this paper, we will present many
More informationThe University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, August 16, 2012 8:30 to 11:30 a.m.
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Thursday, August 16, 2012 8:30 to 11:30 a.m., only Student Name: School Name: Print your name and the name of your
More informationGeometry Module 4 Unit 2 Practice Exam
Name: Class: Date: ID: A Geometry Module 4 Unit 2 Practice Exam Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which diagram shows the most useful positioning
More informationSIMSON S THEOREM MARY RIEGEL
SIMSON S THEOREM MARY RIEGEL Abstract. This paper is a presentation and discussion of several proofs of Simson s Theorem. Simson s Theorem is a statement about a specific type of line as related to a given
More informationThe University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, August 13, 2015 8:30 to 11:30 a.m., only.
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Thursday, August 13, 2015 8:30 to 11:30 a.m., only Student Name: School Name: The possession or use of any communications
More information3.1 Triangles, Congruence Relations, SAS Hypothesis
Chapter 3 Foundations of Geometry 2 3.1 Triangles, Congruence Relations, SAS Hypothesis Definition 3.1 A triangle is the union of three segments ( called its side), whose end points (called its vertices)
More informationConjectures for Geometry for Math 70 By I. L. Tse
Conjectures for Geometry for Math 70 By I. L. Tse Chapter Conjectures 1. Linear Pair Conjecture: If two angles form a linear pair, then the measure of the angles add up to 180. Vertical Angle Conjecture:
More informationThe Use of Dynamic Geometry Software in the Teaching and Learning of Geometry through Transformations
The Use of Dynamic Geometry Software in the Teaching and Learning of Geometry through Transformations Dynamic geometry technology should be used to maximize student learning in geometry. Such technology
More informationQUADRILATERALS CHAPTER 8. (A) Main Concepts and Results
CHAPTER 8 QUADRILATERALS (A) Main Concepts and Results Sides, Angles and diagonals of a quadrilateral; Different types of quadrilaterals: Trapezium, parallelogram, rectangle, rhombus and square. Sum of
More informationSelected practice exam solutions (part 5, item 2) (MAT 360)
Selected practice exam solutions (part 5, item ) (MAT 360) Harder 8,91,9,94(smaller should be replaced by greater )95,103,109,140,160,(178,179,180,181 this is really one problem),188,193,194,195 8. On
More informationThe University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Tuesday, August 13, 2013 8:30 to 11:30 a.m., only.
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Tuesday, August 13, 2013 8:30 to 11:30 a.m., only Student Name: School Name: The possession or use of any communications
More informationClass-10 th (X) Mathematics Chapter: Tangents to Circles
Class-10 th (X) Mathematics Chapter: Tangents to Circles 1. Q. AB is line segment of length 24 cm. C is its midpoint. On AB, AC and BC semicircles are described. Find the radius of the circle which touches
More informationAdditional Topics in Math
Chapter Additional Topics in Math In addition to the questions in Heart of Algebra, Problem Solving and Data Analysis, and Passport to Advanced Math, the SAT Math Test includes several questions that are
More informationBlue Pelican Geometry Theorem Proofs
Blue Pelican Geometry Theorem Proofs Copyright 2013 by Charles E. Cook; Refugio, Tx (All rights reserved) Table of contents Geometry Theorem Proofs The theorems listed here are but a few of the total in
More information4. How many integers between 2004 and 4002 are perfect squares?
5 is 0% of what number? What is the value of + 3 4 + 99 00? (alternating signs) 3 A frog is at the bottom of a well 0 feet deep It climbs up 3 feet every day, but slides back feet each night If it started
More informationThe common ratio in (ii) is called the scaled-factor. An example of two similar triangles is shown in Figure 47.1. Figure 47.1
47 Similar Triangles An overhead projector forms an image on the screen which has the same shape as the image on the transparency but with the size altered. Two figures that have the same shape but not
More informationCevians, Symmedians, and Excircles. MA 341 Topics in Geometry Lecture 16
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
More informationC relative to O being abc,, respectively, then b a c.
2 EP-Program - Strisuksa School - Roi-et Math : Vectors Dr.Wattana Toutip - Department of Mathematics Khon Kaen University 200 :Wattana Toutip wattou@kku.ac.th http://home.kku.ac.th/wattou 2. Vectors A
More informationBaltic Way 1995. Västerås (Sweden), November 12, 1995. Problems and solutions
Baltic Way 995 Västerås (Sweden), November, 995 Problems and solutions. Find all triples (x, y, z) of positive integers satisfying the system of equations { x = (y + z) x 6 = y 6 + z 6 + 3(y + z ). Solution.
More information11 th Annual Harvard-MIT Mathematics Tournament
11 th nnual Harvard-MIT Mathematics Tournament Saturday February 008 Individual Round: Geometry Test 1. [] How many different values can take, where,, are distinct vertices of a cube? nswer: 5. In a unit
More informationName Date Class. Lines and Segments That Intersect Circles. AB and CD are chords. Tangent Circles. Theorem Hypothesis Conclusion
Section. Lines That Intersect Circles Lines and Segments That Intersect Circles A chord is a segment whose endpoints lie on a circle. A secant is a line that intersects a circle at two points. A tangent
More informationThe Inversion Transformation
The Inversion Transformation A non-linear transformation The transformations of the Euclidean plane that we have studied so far have all had the property that lines have been mapped to lines. Transformations
More informationVisualizing Triangle Centers Using Geogebra
Visualizing Triangle Centers Using Geogebra Sanjay Gulati Shri Shankaracharya Vidyalaya, Hudco, Bhilai India http://mathematicsbhilai.blogspot.com/ sanjaybhil@gmail.com ABSTRACT. In this paper, we will
More informationLecture 24: Saccheri Quadrilaterals
Lecture 24: Saccheri Quadrilaterals 24.1 Saccheri Quadrilaterals Definition In a protractor geometry, we call a quadrilateral ABCD a Saccheri quadrilateral, denoted S ABCD, if A and D are right angles
More informationThree Lemmas in Geometry Solutions Yufei Zhao Massachusetts Institute of Technology
Three Lemmas in Geometry Solutions Yufei Zhao Massachusetts Institute of Technology yufei.zhao@gmail.com 1 Diameter of incircle 1. (IMO 1992) In the plane let C be a circle, l a line tangent to the circle
More information5.1 Midsegment Theorem and Coordinate Proof
5.1 Midsegment Theorem and Coordinate Proof Obj.: Use properties of midsegments and write coordinate proofs. Key Vocabulary Midsegment of a triangle - A midsegment of a triangle is a segment that connects
More informationGeometry Course Summary Department: Math. Semester 1
Geometry Course Summary Department: Math Semester 1 Learning Objective #1 Geometry Basics Targets to Meet Learning Objective #1 Use inductive reasoning to make conclusions about mathematical patterns Give
More informationCurriculum Map by Block Geometry Mapping for Math Block Testing 2007-2008. August 20 to August 24 Review concepts from previous grades.
Curriculum Map by Geometry Mapping for Math Testing 2007-2008 Pre- s 1 August 20 to August 24 Review concepts from previous grades. August 27 to September 28 (Assessment to be completed by September 28)
More informationAlgebra III. Lesson 33. Quadrilaterals Properties of Parallelograms Types of Parallelograms Conditions for Parallelograms - Trapezoids
Algebra III Lesson 33 Quadrilaterals Properties of Parallelograms Types of Parallelograms Conditions for Parallelograms - Trapezoids Quadrilaterals What is a quadrilateral? Quad means? 4 Lateral means?
More informationOn Mixtilinear Incircles and Excircles
Forum Geometricorum Volume 6 (2006) 1 16. FORUM GEOM ISSN 1534-1178 On Mixtilinear Incircles and Excircles Khoa Lu Nguyen and Juan arlos Salazar bstract. mixtilinear incircle (respectively excircle) of
More informationThe Geometry of Piles of Salt Thinking Deeply About Simple Things
The Geometry of Piles of Salt Thinking Deeply About Simple Things PCMI SSTP Tuesday, July 15 th, 2008 By Troy Jones Willowcreek Middle School Important Terms (the word line may be replaced by the word
More informationThe University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Wednesday, June 20, 2012 9:15 a.m. to 12:15 p.m., only Student Name: School Name: Print your name and the name
More informationMathematics (Project Maths Phase 3)
2014. M328 Coimisiún na Scrúduithe Stáit State Examinations Commission Leaving Certificate Examination 2014 Mathematics (Project Maths Phase 3) Paper 2 Ordinary Level Monday 9 June Morning 9:30 12:00 300
More informationUSA Mathematical Talent Search Solutions to Problem 5/2/16
5/2/16. Two circles of equal radius can tightly fit inside right triangle A, which has A = 13, = 12, and A = 5, in the three positions illustrated below. Determine the radii of the circles in each case.
More informationCeva s Theorem. Ceva s Theorem. Ceva s Theorem 9/20/2011. MA 341 Topics in Geometry Lecture 11
MA 341 Topics in Geometry Lecture 11 The three lines containing the vertices A, B, and C of ABC and intersecting opposite sides at points L, M, and N, respectively, are concurrent if and only if 2 3 1
More informationCHAPTER 1. LINES AND PLANES IN SPACE
CHAPTER 1. LINES AND PLANES IN SPACE 1. Angles and distances between skew lines 1.1. Given cube ABCDA 1 B 1 C 1 D 1 with side a. Find the angle and the distance between lines A 1 B and AC 1. 1.2. Given
More informationMost popular response to
Class #33 Most popular response to What did the students want to prove? The angle bisectors of a square meet at a point. A square is a convex quadrilateral in which all sides are congruent and all angles
More informationThe University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, January 26, 2012 9:15 a.m. to 12:15 p.m.
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXMINTION GEOMETRY Thursday, January 26, 2012 9:15 a.m. to 12:15 p.m., only Student Name: School Name: Print your name and the name
More informationThe University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Student Name:
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Thursday, June 17, 2010 1:15 to 4:15 p.m., only Student Name: School Name: Print your name and the name of your
More informationSection 8.8. 1. The given line has equations. x = 3 + t(13 3) = 3 + 10t, y = 2 + t(3 + 2) = 2 + 5t, z = 7 + t( 8 7) = 7 15t.
. The given line has equations Section 8.8 x + t( ) + 0t, y + t( + ) + t, z 7 + t( 8 7) 7 t. The line meets the plane y 0 in the point (x, 0, z), where 0 + t, or t /. The corresponding values for x and
More informationCircle Theorems. This circle shown is described an OT. As always, when we introduce a new topic we have to define the things we wish to talk about.
Circle s circle is a set of points in a plane that are a given distance from a given point, called the center. The center is often used to name the circle. T This circle shown is described an OT. s always,
More informationMathematics Notes for Class 12 chapter 10. Vector Algebra
1 P a g e Mathematics Notes for Class 12 chapter 10. Vector Algebra A vector has direction and magnitude both but scalar has only magnitude. Magnitude of a vector a is denoted by a or a. It is non-negative
More informationAlgebraic Properties and Proofs
Algebraic Properties and Proofs Name You have solved algebraic equations for a couple years now, but now it is time to justify the steps you have practiced and now take without thinking and acting without
More informationwww.sakshieducation.com
LENGTH OF THE PERPENDICULAR FROM A POINT TO A STRAIGHT LINE AND DISTANCE BETWEEN TWO PAPALLEL LINES THEOREM The perpendicular distance from a point P(x 1, y 1 ) to the line ax + by + c 0 is ax1+ by1+ c
More informationGEOMETRY CONCEPT MAP. Suggested Sequence:
CONCEPT MAP GEOMETRY August 2011 Suggested Sequence: 1. Tools of Geometry 2. Reasoning and Proof 3. Parallel and Perpendicular Lines 4. Congruent Triangles 5. Relationships Within Triangles 6. Polygons
More informationEquation of a Line. Chapter H2. The Gradient of a Line. m AB = Exercise H2 1
Chapter H2 Equation of a Line The Gradient of a Line The gradient of a line is simpl a measure of how steep the line is. It is defined as follows :- gradient = vertical horizontal horizontal A B vertical
More informationPractice Test Answer and Alignment Document Mathematics: Geometry Performance Based Assessment - Paper
The following pages include the answer key for all machine-scored items, followed by the rubrics for the hand-scored items. - The rubrics show sample student responses. Other valid methods for solving
More informationHigh School Geometry Test Sampler Math Common Core Sampler Test
High School Geometry Test Sampler Math Common Core Sampler Test Our High School Geometry sampler covers the twenty most common questions that we see targeted for this level. For complete tests and break
More informationAngles in a Circle and Cyclic Quadrilateral
130 Mathematics 19 Angles in a Circle and Cyclic Quadrilateral 19.1 INTRODUCTION You must have measured the angles between two straight lines, let us now study the angles made by arcs and chords in a circle
More informationhttp://www.castlelearning.com/review/teacher/assignmentprinting.aspx 5. 2 6. 2 1. 10 3. 70 2. 55 4. 180 7. 2 8. 4
of 9 1/28/2013 8:32 PM Teacher: Mr. Sime Name: 2 What is the slope of the graph of the equation y = 2x? 5. 2 If the ratio of the measures of corresponding sides of two similar triangles is 4:9, then the
More information12. Parallels. Then there exists a line through P parallel to l.
12. Parallels Given one rail of a railroad track, is there always a second rail whose (perpendicular) distance from the first rail is exactly the width across the tires of a train, so that the two rails
More informationWarm-up Theorems about triangles. Geometry. Theorems about triangles. Misha Lavrov. ARML Practice 12/15/2013
ARML Practice 12/15/2013 Problem Solution Warm-up problem Lunes of Hippocrates In the diagram below, the blue triangle is a right triangle with side lengths 3, 4, and 5. What is the total area of the green
More informationUnit 2 - Triangles. Equilateral Triangles
Equilateral Triangles Unit 2 - Triangles Equilateral Triangles Overview: Objective: In this activity participants discover properties of equilateral triangles using properties of symmetry. TExES Mathematics
More informationIncenter Circumcenter
TRIANGLE: Centers: Incenter Incenter is the center of the inscribed circle (incircle) of the triangle, it is the point of intersection of the angle bisectors of the triangle. The radius of incircle is
More informationShape, Space and Measure
Name: Shape, Space and Measure Prep for Paper 2 Including Pythagoras Trigonometry: SOHCAHTOA Sine Rule Cosine Rule Area using 1-2 ab sin C Transforming Trig Graphs 3D Pythag-Trig Plans and Elevations Area
More informationWarm-up Tangent circles Angles inside circles Power of a point. Geometry. Circles. Misha Lavrov. ARML Practice 12/08/2013
Circles ARML Practice 12/08/2013 Solutions Warm-up problems 1 A circular arc with radius 1 inch is rocking back and forth on a flat table. Describe the path traced out by the tip. 2 A circle of radius
More informationTangent Properties. Line m is a tangent to circle O. Point T is the point of tangency.
CONDENSED LESSON 6.1 Tangent Properties In this lesson you will Review terms associated with circles Discover how a tangent to a circle and the radius to the point of tangency are related Make a conjecture
More informationGeometry Handout 2 ~ Page 1
1. Given: a b, b c a c Guidance: Draw a line which intersects with all three lines. 2. Given: a b, c a a. c b b. Given: d b d c 3. Given: a c, b d a. α = β b. Given: e and f bisect angles α and β respectively.
More information15. Appendix 1: List of Definitions
page 321 15. Appendix 1: List of Definitions Definition 1: Interpretation of an axiom system (page 12) Suppose that an axiom system consists of the following four things an undefined object of one type,
More informationThe University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Wednesday, January 28, 2015 9:15 a.m. to 12:15 p.m.
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Wednesday, January 28, 2015 9:15 a.m. to 12:15 p.m., only Student Name: School Name: The possession or use of any
More informationOn Triangles with Vertices on the Angle Bisectors
Forum Geometricorum Volume 6 (2006) 247 253. FORUM GEOM SSN 1534-1178 On Triangles with Vertices on the ngle isectors Eric Danneels bstract. We study interesting properties of triangles whose vertices
More informationThe University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, August 13, 2009 8:30 to 11:30 a.m., only.
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Thursday, August 13, 2009 8:30 to 11:30 a.m., only Student Name: School Name: Print your name and the name of your
More informationx R 2 x = (x 1, x 2 ) or p = (x, y) R 3
Euclidean space 1 Chapter 1 Euclidean space A. The basic vector space We shall denote by R the field of real numbers. Then we shall use the Cartesian product R n = R R... R of ordered n-tuples of real
More informationAngles that are between parallel lines, but on opposite sides of a transversal.
GLOSSARY Appendix A Appendix A: Glossary Acute Angle An angle that measures less than 90. Acute Triangle Alternate Angles A triangle that has three acute angles. Angles that are between parallel lines,
More information/27 Intro to Geometry Review
/27 Intro to Geometry Review 1. An acute has a measure of. 2. A right has a measure of. 3. An obtuse has a measure of. 13. Two supplementary angles are in ratio 11:7. Find the measure of each. 14. In the
More information