Number 8 Spring 2018

Number 12 Spring 2019

Number 2 Spring 209 ROMANIAN MATHEMATICAL MAGAZINE Avilble online www.ssmrmh.ro Founding Editor DANIEL SITARU ISSN-L 250-0099 ROMANIAN MATHEMATICAL MAGAZINE PROBLEMS FOR JUNIORS JP.66. If, b [0; + ) nd

IOAN ŞERDEAN, DANIEL SITARU

Romanian Mathematical Magazine Web: http://www.ssmrmh.ro The Author: This article is published with open access. TRIGONOMETRIC SUBSTITUTIONS IN PROBLEM SOLVING PART IOAN ŞERDEAN, DANIEL SITARU Abstract.

On a class of three-variable inequalities. Vo Quoc Ba Can

On a class of three-variable inequalities Vo Quoc Ba Can 1 Theem Let a, b, c be real numbers satisfying a + b + c = 1 By the AM - GM inequality, we have ab + bc + ca 1, therefe setting ab + bc + ca = 1

Q.1 If a, b, c are distinct positive real in H.P., then the value of the expression, (A) 1 (B) 2 (C) 3 (D) 4. (A) 2 (B) 5/2 (C) 3 (D) none of these

Q. If a, b, c are distinct positive real in H.P., then the value of the expression, b a b c + is equal to b a b c () (C) (D) 4 Q. In a triangle BC, (b + c) = a bc where is the circumradius of the triangle.

Two geometric inequalities involved two triangles

OCTOGON MATHEMATICAL MAGAZINE Vol. 17, No.1, April 009, pp 193-198 ISSN 1-5657, ISBN 978-973-8855-5-0, www.hetfalu.ro/octogon 193 Two geometric inequalities involved two triangles Yu-Lin Wu 16 ABSTRACT.

ELEMENTARY PROBLEMS TREATED NON-ELEMENTARY

ELEMENTARY PROBLEMS TREATED NON-ELEMENTARY DANIEL SITARU Abstract. Witout entering into details regarding Fermat s teorem for functions aving multiple variables, te teorem of bilinear forms and Fubini

INMO-2001 Problems and Solutions

INMO-2001 Problems and Solutions 1. Let ABC be a triangle in which no angle is 90. For any point P in the plane of the triangle, let A 1,B 1,C 1 denote the reflections of P in the sides BC,CA,AB respectively.

1 Arithmetic calculations (calculator is not allowed)

1 ARITHMETIC CALCULATIONS (CALCULATOR IS NOT ALLOWED) 1 Arithmetic calculations (calculator is not allowed) 1.1 Check the result Problem 1.1. Problem 1.2. Problem 1.3. Problem 1.4. 78 5 6 + 24 3 4 99 1

A New Method About Using Polynomial Roots and Arithmetic-Geometric Mean Inequality to Solve Olympiad Problems

Polynomial Roots and Arithmetic-Geometric Mean Inequality 1 A New Method About Using Polynomial Roots and Arithmetic-Geometric Mean Inequality to Solve Olympiad Problems The purpose of this article is

Happy New Year 2008 Chuc Mung Nam Moi 2008

www.batdangthuc.net Happy New Year 008 Chuc Mung Nam Moi 008 Vietnam Inequality Forum - VIF - www.batdangthuc.net Ebook Written by: VIF Community User Group: All This product is created for educational

The Apollonius Circle as a Tucker Circle

Forum Geometricorum Volume 2 (2002) 175 182 FORUM GEOM ISSN 1534-1178 The Apollonius Circle as a Tucker Circle Darij Grinberg and Paul Yiu Abstract We give a simple construction of the circular hull of

A Proof of Gibson s and Rodgers Problem

A Proof of Gibson s and Rodgers Problem Nguyen Minh Ha February 4, 007 Abstract Let A 0 B 0 C 0 be a triangle with centroid G 0 inscribed in a circle Γ with center O. The lines A 0 G 0, B 0 G 0, C 0 G

Trigonometrical identities and inequalities

Trigonometrical identities and inequalities Finbarr Holland January 1, 010 1 A review of the trigonometrical functions These are sin, cos, & tan. These are discussed in the Maynooth Olympiad Manual, which

Logarithmic Functions

Metropolitan Community College The Natural Logarithmic Function The natural logarithmic function is defined on (0, ) as ln x = x 1 1 t dt. Example 1. Evaluate ln 1. Example 1. Evaluate ln 1. Solution.

the mathscope All the best from Vietnamese Problem Solving Journals May 10, 2006 please download for free at our websites:

the mathscope All the best from Vietnamese Problem Solving Journals May 10, 2006 please download for free at our websites: www.math4u.de, & www.imo.org.yu translated by Phạm Văn Thuận, Eckard Specht Journal

CHAPTER 1 POLYNOMIALS

1 CHAPTER 1 POLYNOMIALS 1.1 Removing Nested Symbols of Grouping Simplify. 1. 4x + 3( x ) + 4( x + 1). ( ) 3x + 4 5 x 3 + x 3. 3 5( y 4) + 6 y ( y + 3) 4. 3 n ( n + 5) 4 ( n + 8) 5. ( x + 5) x + 3( x 6)

A NEW PROOF OF PTOLEMY S THEOREM

A NEW PROOF OF PTOLEMY S THEOREM DASARI NAGA VIJAY KRISHNA Abstract In this article we give a new proof of well-known Ptolemy s Theorem of a Cyclic Quadrilaterals 1 Introduction In the Euclidean geometry,

About the Japanese theorem

188/ ABOUT THE JAPANESE THEOREM About the Japanese theorem Nicuşor Minculete, Cătălin Barbu and Gheorghe Szöllősy Dedicated to the memory of the great professor, Laurenţiu Panaitopol Abstract The aim of

Maharashtra Board Class X Mathematics - Geometry Board Paper 2014 Solution. Time: 2 hours Total Marks: 40

Maharashtra Board Class X Mathematics - Geometry Board Paper 04 Solution Time: hours Total Marks: 40 Note: - () All questions are compulsory. () Use of calculator is not allowed.. i. Ratio of the areas

2010 Shortlist JBMO - Problems

Chapter 1 2010 Shortlist JBMO - Problems 1.1 Algebra A1 The real numbers a, b, c, d satisfy simultaneously the equations abc d = 1, bcd a = 2, cda b = 3, dab c = 6. Prove that a + b + c + d 0. A2 Determine

Algebraic Inequalities in Mathematical Olympiads: Problems and Solutions

Algebraic Inequalities in Mathematical Olympiads: Problems and Solutions Mohammad Mahdi Taheri July 0, 05 Abstract This is a collection of recent algebraic inequalities proposed in math Olympiads from

DISCUSSION CLASS OF DAX IS ON 22ND MARCH, TIME : 9-12 BRING ALL YOUR DOUBTS [STRAIGHT OBJECTIVE TYPE]

DISCUSSION CLASS OF DAX IS ON ND MARCH, TIME : 9- BRING ALL YOUR DOUBTS [STRAIGHT OBJECTIVE TYPE] Q. Let y = cos x (cos x cos x). Then y is (A) 0 only when x 0 (B) 0 for all real x (C) 0 for all real x

HANOI OPEN MATHEMATICS COMPETITON PROBLEMS

HANOI MATHEMATICAL SOCIETY NGUYEN VAN MAU HANOI OPEN MATHEMATICS COMPETITON PROBLEMS 2006-2013 HANOI - 2013 Contents 1 Hanoi Open Mathematics Competition 3 1.1 Hanoi Open Mathematics Competition 2006...

The circumcircle and the incircle

hapter 4 The circumcircle and the incircle 4.1 The Euler line 4.1.1 nferior and superior triangles G F E G D The inferior triangle of is the triangle DEF whose vertices are the midpoints of the sides,,.

Forum Geometricorum Volume 13 (2013) 1 6. FORUM GEOM ISSN Soddyian Triangles. Frank M. Jackson

Forum Geometricorum Volume 3 (203) 6. FORUM GEOM ISSN 534-78 Soddyian Triangles Frank M. Jackson bstract. Soddyian triangle is a triangle whose outer Soddy circle has degenerated into a straight line.

Transweb Educational Services Pvt. Ltd Tel:

. An aeroplane flying at a constant speed, parallel to the horizontal ground, km above it, is observed at an elevation of 6º from a point on the ground. If, after five seconds, its elevation from the same

Calgary Math Circles: Triangles, Concurrency and Quadrilaterals 1

Calgary Math Circles: Triangles, Concurrency and Quadrilaterals 1 1 Triangles: Basics This section will cover all the basic properties you need to know about triangles and the important points of a triangle.

Nozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Ismailia Road Branch

Cairo Governorate Department : Maths Nozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Sheet Ismailia Road Branch Sheet ( 1) 1-Complete 1. in the parallelogram, each two opposite

SCTT The pqr-method august 2016

SCTT The pqr-method august 2016 A. Doledenok, M. Fadin, À. Menshchikov, A. Semchankau Almost all inequalities considered in our project are symmetric. Hence if plugging (a 0, b 0, c 0 ) into our inequality

Simple Proofs of Feuerbach s Theorem and Emelyanov s Theorem

Forum Geometricorum Volume 18 (2018) 353 359. FORUM GEOM ISSN 1534-1178 Simple Proofs of Feuerbach s Theorem and Emelyanov s Theorem Nikolaos Dergiades and Tran Quang Hung Abstract. We give simple proofs

Another Proof of van Lamoen s Theorem and Its Converse

Forum Geometricorum Volume 5 (2005) 127 132. FORUM GEOM ISSN 1534-1178 Another Proof of van Lamoen s Theorem and Its Converse Nguyen Minh Ha Abstract. We give a proof of Floor van Lamoen s theorem and

Answers. Chapter 9 A92. Angles Theorem (Thm. 5.6) then XZY. Base Angles Theorem (Thm. 5.6) 5, 2. then WV WZ;

9 9. M, 0. M ( 9, 4) 7. If WZ XZ, then ZWX ZXW ; Base Angles Theorem (Thm..6). M 9,. M ( 4, ) 74. If XZ XY, then XZY Y; Base Angles Theorem (Thm..6). M, 4. M ( 9, ) 7. If V WZV, then WV WZ; Converse of

ON THE GERGONNE AND NAGEL POINTS FOR A HYPERBOLIC TRIANGLE

INTERNATIONAL JOURNAL OF GEOMETRY Vol 6 07 No - ON THE GERGONNE AND NAGEL POINTS FOR A HYPERBOLIC TRIANGLE PAUL ABLAGA Abstract In this note we prove the existence of the analogous points of the Gergonne

New aspects of Ionescu Weitzenböck s inequality

New aspects of Ionescu Weitzenböck s inequality Emil Stoica, Nicuşor Minculete, Cătălin Barbu Abstract. The focus of this article is Ionescu-Weitzenböck s inequality using the circumcircle mid-arc triangle.

Let ABCD be a tetrahedron and let M be a point in its interior. Prove or disprove that [BCD] AM 2 = [ABD] CM 2 = [ABC] DM 2 = 2

SOLUTIONS / 343 In other words we have shown that if a face of our tetrahedron has a nonacute angle at one vertex the face of the tetrahedron opposite that vertex must be an acute triangle. It follows

HANOI OPEN MATHEMATICAL COMPETITON PROBLEMS

HANOI MATHEMATICAL SOCIETY NGUYEN VAN MAU HANOI OPEN MATHEMATICAL COMPETITON PROBLEMS HANOI - 2013 Contents 1 Hanoi Open Mathematical Competition 3 1.1 Hanoi Open Mathematical Competition 2006... 3 1.1.1

CBSE QUESTION PAPER CLASS-X MATHS

CBSE QUESTION PAPER CLASS-X MATHS SECTION - A Question 1:If sin α = 1 2, then the value of 4 cos3 α 3 cos α is (a)0 (b)1 (c) 1 (d)2 Question 2: If cos 2θ = sin(θ 12 ), where2θ and (θ 12 ) are both acute

GIL Publishing House PO box 44, PO 3 ROMANIA

57 Title: Secrets in Inequalities volume - advanced inequalities - free chapter Author: Pham Kim Hung Publisher: GIL Publishing House ISBN: 978-606-500-00-5 Copyright c 008 GIL Publishing House. All rights

VKR Classes TIME BOUND TESTS 1-7 Target JEE ADVANCED For Class XI VKR Classes, C , Indra Vihar, Kota. Mob. No

VKR Classes TIME BOUND TESTS -7 Target JEE ADVANCED For Class XI VKR Classes, C-9-0, Indra Vihar, Kota. Mob. No. 9890605 Single Choice Question : PRACTICE TEST-. The smallest integer greater than log +

(b) g(x) = 4 + 6(x 3) (x 3) 2 (= x x 2 ) M1A1 Note: Accept any alternative form that is correct. Award M1A0 for a substitution of (x + 3).

Paper. Answers. (a) METHOD f (x) q x f () q 6 q 6 f() p + 8 9 5 p METHOD f(x) (x ) + 5 x + 6x q 6, p (b) g(x) + 6(x ) (x ) ( + x x ) Note: Accept any alternative form that is correct. Award A for a substitution

Chapter 6. Basic triangle centers. 6.1 The Euler line The centroid

hapter 6 asic triangle centers 6.1 The Euler line 6.1.1 The centroid Let E and F be the midpoints of and respectively, and G the intersection of the medians E and F. onstruct the parallel through to E,

2 Use the factor theorem to show that a + b c is a factor of. (a + b + c) 3 6(a + b + c)(a 2 + b 2 + c 2 )+8(a 3 + b 3 + c 3 ).

Section A: Pure Mathematics 1 The curve C 1 has parametric equations x = t 2, y = t 3, where

Nozha Directorate of Education Form : 2 nd Prep

Cairo Governorate Department : Maths Nozha Directorate of Education Form : 2 nd Prep Nozha Language Schools Geometry Revision Sheet Ismailia Road Branch Sheet ( 1) 1-Complete 1. In the parallelogram, each

1998 IMO Shortlist BM BN BA BC AB BC CD DE EF BC CA AE EF FD

IMO Shortlist 1998 Geometry 1 A convex quadrilateral ABCD has perpendicular diagonals. The perpendicular bisectors of the sides AB and CD meet at a unique point P inside ABCD. the quadrilateral ABCD is

Primitive Heronian Triangles With Integer Inradius and Exradii

Forum Geometricorum Volume 18 (2018) 71 77. FORUM GEOM ISSN 1534-1178 Primitive Heronian Triangles With Integer Inradius and Exradii Li Zhou Abstract. It is well known that primitive Pythagorean triangles

Note : This document might take a little longer time to print. more exam papers at : more exam papers at : more exam papers at : more exam papers at : more exam papers at : more exam papers at : more

CLASS 12 SUBJECT : MATHEMATICS

CLASS 2 SUBJECT : MATHEMATICS CBSE QUESTION PAPER 27(FOREIGN) General Instructions: (i) All questions are compulsory. (ii) Questions 4 in Section A carrying mark each (iii) Questions 5 2 in Section B carrying

A Brief Introduction to Inequalities

A Brief Introduction to Inequalities Anthony Erb Lugo Department of Mathematical Sciences University of Puerto Rico Mayaguez Campus Este Material Educativo es para ser distribuido de forma gratuita exclusivamente.

The uvw method - Tejs. The uvw method. by Mathias Bæk Tejs Knudsen

The uvw method - Tejs The uvw method by Mathias Bæk Tejs Knudsen The uvw method - Tejs BASIC CONCEPTS Basic Concepts The basic concept of the method is this: With an inequality in the numbers a, b, c R,

Solutions to the February problems.

Solutions to the February problems. 348. (b) Suppose that f(x) is a real-valued function defined for real values of x. Suppose that both f(x) 3x and f(x) x 3 are increasing functions. Must f(x) x x also

ISI, CMI, Olympiads Study Material

B. Stat. (Hons.) Admission Test : 2009 Group A (Each of the following questions has exactly one correct option and you have to identify it) 1. If k times the sum of the first n natural numbers is equal

A = (a + 1) 2 = a 2 + 2a + 1

A = (a + 1) 2 = a 2 + 2a + 1 1 A = ( (a + b) + 1 ) 2 = (a + b) 2 + 2(a + b) + 1 = a 2 + 2ab + b 2 + 2a + 2b + 1 A = ( (a + b) + 1 ) 2 = (a + b) 2 + 2(a + b) + 1 = a 2 + 2ab + b 2 + 2a + 2b + 1 3 A = (

The Law of Averages. MARK FLANAGAN School of Electrical, Electronic and Communications Engineering University College Dublin

The Law of Averages MARK FLANAGAN School of Electrical, Electronic and Communications Engineering University College Dublin Basic Principle of Inequalities: For any real number x, we have 3 x 2 0, with

Answers: ( HKMO Final Events) Created by Mr. Francis Hung Last updated: 13 December Group Events

Individual Events SI a 900 I1 P 100 I k 4 I3 h 3 I4 a 18 I5 a 495 b 7 Q 8 m 58 k 6 r 3 b p R 50 a m 4 M 9 4 x 99 q 9 S 3 b 3 p Group Events 15 16 w 1 Y 109 SG p 75 G6 x 15 G7 M 5 G8 S 7 G9 p 60 G10 n 18

UNC Charlotte 2005 Comprehensive March 7, 2005

March 7, 2005 1. The numbers x and y satisfy 2 x = 15 and 15 y = 32. What is the value xy? (A) 3 (B) 4 (C) 5 (D) 6 (E) none of A, B, C or D Solution: C. Note that (2 x ) y = 15 y = 32 so 2 xy = 2 5 and

1 Solution to Homework 4

Solution to Homework Section. 5. The characteristic equation is r r + = (r )(r ) = 0 r = or r =. y(t) = c e t + c e t y = c e t + c e t. y(0) =, y (0) = c + c =, c + c = c =, c =. To find the maximum value

Objective Mathematics

. In BC, if angles, B, C are in geometric seq- uence with common ratio, then is : b c a (a) (c) 0 (d) 6. If the angles of a triangle are in the ratio 4 : :, then the ratio of the longest side to the perimeter

x n+1 = ( x n + ) converges, then it converges to α. [2]

1 A Level - Mathematics P 3 ITERATION ( With references and answers) [ Numerical Solution of Equation] Q1. The equation x 3 - x 2 6 = 0 has one real root, denoted by α. i) Find by calculation the pair

2013 Sharygin Geometry Olympiad

Sharygin Geometry Olympiad 2013 First Round 1 Let ABC be an isosceles triangle with AB = BC. Point E lies on the side AB, and ED is the perpendicular from E to BC. It is known that AE = DE. Find DAC. 2

Cambridge International Examinations Cambridge Ordinary Level

Cambridge International Examinations Cambridge Ordinary Level *8790810596* ADDITIONAL MATHEMATICS 4037/13 Paper 1 October/November 2017 2 hours Candidates answer on the Question Paper. No Additional Materials

MATH 31BH Homework 5 Solutions

MATH 3BH Homework 5 Solutions February 4, 204 Problem.8.2 (a) Let x t f y = x 2 + y 2 + 2z 2 and g(t) = t 2. z t 3 Then by the chain rule a a a D(g f) b = Dg f b Df b c c c = [Dg(a 2 + b 2 + 2c 2 )] [

Hanoi Open Mathematical Competition 2017

Hanoi Open Mathematical Competition 2017 Junior Section Saturday, 4 March 2017 08h30-11h30 Important: Answer to all 15 questions. Write your answers on the answer sheets provided. For the multiple choice

[STRAIGHT OBJECTIVE TYPE] log 4 2 x 4 log. x x log 2 x 1

[STRAIGHT OBJECTIVE TYPE] Q. The equation, log (x ) + log x. log x x log x + log x log + log / x (A) exactly one real solution (B) two real solutions (C) real solutions (D) no solution. = has : Q. The

SUMMATIVE ASSESSMENT I, IX / Class IX

I, 0 SUMMATIVE ASSESSMENT I, 0 0 MATHEMATICS / MATHEMATICS MATHEMATICS CLASS CLASS - IX - IX IX / Class IX MA-0 90 Time allowed : hours Maximum Marks : 90 (i) (ii) 8 6 0 0 (iii) 8 (iv) (v) General Instructions:

(e) (i) Prove that C(x) = C( x) for all x. (2)

Revision - chapters and 3 part two. (a) Sketch the graph of f (x) = sin 3x + sin 6x, 0 x. Write down the exact period of the function f. (Total 3 marks). (a) Sketch the graph of the function C ( x) cos

The Apollonius Circle and Related Triangle Centers

Forum Geometricorum Qolume 3 (2003) 187 195. FORUM GEOM ISSN 1534-1178 The Apollonius Circle and Related Triangle Centers Milorad R. Stevanović Abstract. We give a simple construction of the Apollonius

Hanoi Open Mathematical Olympiad

HEXAGON inspiring minds always Let x = 6+2 + Hanoi Mathematical Olympiad 202 6 2 Senior Section 20 Find the value of + x x 7 202 2 Arrange the numbers p = 2 2, q =, t = 2 + 2 in increasing order Let ABCD

Senior Team Maths Challenge Regional Final 2008 Mini Relay Round

A1 Solve the simultaneous equations: 2x 5y 2z 4 7x y 2z 13 9x 4 y 1 Write down the value of. 2 z yx 2 A2 T is the number that you will receive The expression T x 4 1 2x 2 Tx 7x 5 can be written in the

CBSE SAMPLE PAPER Class IX Mathematics Paper 1 (Answers)

CBSE SAMPLE PAPER Class IX Mathematics Paper 1 (Answers) 1. Solution: We have, 81 36 x y 5 Answers & Explanations Section A = ( 9 6 x) ( y 5 ) = ( 9 6 x + y 5 ) (9 6 x y 5 ) [ a b = (a + b)(a b)] Hence,

Geometry. A. Right Triangle. Legs of a right triangle : a, b. Hypotenuse : c. Altitude : h. Medians : m a, m b, m c. Angles :,

Geometry A. Right Triangle Legs of a right triangle : a, b Hypotenuse : c Altitude : h Medians : m a, m b, m c Angles :, Radius of circumscribed circle : R Radius of inscribed circle : r Area : S 1. +

Bounds for Elements of a Triangle Expressed by R, r, and s

Forum Geometricorum Volume 5 05) 99 03. FOUM GEOM ISSN 534-78 Bounds for Elements of a Triangle Expressed by, r, and s Temistocle Bîrsan Abstract. Assume that a triangle is defined by the triple, r, s)

(This type of questions may be asked in the examination )

34. The slope of a straight line parallel to the line 2x + 4y + 5 = 0 is... a) 2 b) 1 / 2 c) - 1 / 2 d) - 2 35. The angle of inclination of a straight line whose slope is is... a) 0 0 b) 30 0 c) 60 0 d)

SUMMATIVE ASSESSMENT-1 SAMPLE PAPER (SET-2) MATHEMATICS CLASS IX

SUMMATIVE ASSESSMENT-1 SAMPLE PAPER (SET-) MATHEMATICS CLASS IX Time: 3 to 3 1 hours Maximum Marks: 80 GENERAL INSTRUCTIONS: 1. All questions are compulsory.. The question paper is divided into four sections

DISCRETE INEQUALITIES

Vasile Cîrtoaje DISCRETE INEQUALITIES VOLUME 1 SYMMETRIC POLYNOMIAL INEQUALITIES ART OF PROBLEM SOLVING 2015 About the author Vasile Cîrtoaje is a Professor at the Department of Automatic Control and

Vectors - Applications to Problem Solving

BERKELEY MATH CIRCLE 00-003 Vectors - Applications to Problem Solving Zvezdelina Stankova Mills College& UC Berkeley 1. Well-known Facts (1) Let A 1 and B 1 be the midpoints of the sides BC and AC of ABC.

THE INVERSE TRIGONOMETRIC FUNCTIONS

THE INVERSE TRIGONOMETRIC FUNCTIONS Question 1 (**+) Solve the following trigonometric equation ( x ) π + 3arccos + 1 = 0. 1 x = Question (***) It is given that arcsin x = arccos y. Show, by a clear method,

So, eqn. to the bisector containing (-1, 4) is = x + 27y = 0

Q.No. The bisector of the acute angle between the lines x - 4y + 7 = 0 and x + 5y - = 0, is: Option x + y - 9 = 0 Option x + 77y - 0 = 0 Option x - y + 9 = 0 Correct Answer L : x - 4y + 7 = 0 L :-x- 5y

Distances Among the Feuerbach Points

Forum Geometricorum Volume 16 016) 373 379 FORUM GEOM ISSN 153-1178 Distances mong the Feuerbach Points Sándor Nagydobai Kiss bstract We find simple formulas for the distances from the Feuerbach points

2007 Mathematical Olympiad Summer Program Tests

2007 Mathematical Olympiad Summer Program Tests Edited by Zuming Feng Mathematics Olympiad Summer Program 2007 Tests 1 Practice Test 1 1.1. In triangle ABC three distinct triangles are inscribed, similar

A SYNTHETIC PROOF OF DAO S GENERALIZATION OF GOORMAGHTIGH S THEOREM

Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 84-5569, Vol3, (014), Issue, pp15-19 A SYNTHETIC PROOF OF DAO S GENERALIZATION OF GOORMAGHTIGH S THEOREM TRAN HOANG SON ABSTRACT

OLYMON. Produced by the Canadian Mathematical Society and the Department of Mathematics of the University of Toronto. Issue 6:4.

OLYMON Produced by the Canadian Mathematical Society and the Department of Mathematics of the University of Toronto. Issue 6:4 April, 005 Please send your solution to Prof. Edward J. Barbeau Department

Cambridge International Examinations CambridgeOrdinaryLevel

Cambridge International Examinations CambridgeOrdinaryLevel * 2 5 4 0 0 0 9 5 8 5 * ADDITIONAL MATHEMATICS 4037/12 Paper1 May/June 2015 2 hours CandidatesanswerontheQuestionPaper. NoAdditionalMaterialsarerequired.

ASSIGNMENT ON MATRICES AND DETERMINANTS (CBSE/NCERT/OTHERSTATE BOARDS). Write the orders of AB and BA. x y 2z w 5 3

1 If A = [a ij ] = is a matrix given by 4 1 3 A [a ] 5 7 9 6 ij 1 15 18 5 Write the order of A and find the elements a 4, a 34 Also, show that a 3 = a 3 + a 4 If A = 1 4 3 1 4 1 5 and B = Write the orders

PROVING SOME GEOMETRIC IDENTITIES BY USING THE DETERMINANTS

Journal of cience and Arts Year No () pp 385-39 OIGINAL PAPE POVING OME GEOMETIC IDENTITIE BY UING THE DETEMINANT DAM VAN NHI Manuscript received: 7; Accepted paper: 79; Published online: Abstract In this

State Precalculus/Trigonometry Contest 2008

State Precalculus/Trigonometry Contest 008 Select the best answer for each of the following questions and mark it on the answer sheet provided. Be sure to read all the answer choices before making your

41st International Mathematical Olympiad

41st International Mathematical Olympiad Taejon, Korea, July 2000 1 Two circles Γ 1 and Γ 2 intersect at M and N Let l be the common tangent to Γ 1 and Γ 2 so that M is closer to l than N is Let l touch

Sec 4 Maths SET D PAPER 2

S4MA Set D Paper Sec 4 Maths Exam papers with worked solutions SET D PAPER Compiled by THE MATHS CAFE P a g e Answer all questions. Write your answers and working on the separate Answer Paper provided.

Cambridge International Examinations Cambridge Ordinary Level

Cambridge International Examinations Cambridge Ordinary Level * 3 9 5 8 3 8 6 4 7 * ADDITIONAL MATHEMATICS 4037/13 Paper1 October/November 016 hours CandidatesanswerontheQuestionPaper. NoAdditionalMaterialsarerequired.

International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL EXAMINATIONS PAPER 2 MAY/JUNE SESSION 2002

International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL EXAMINATIONS ADDITIONAL MATHEMATICS 0606/2 PAPER 2 MAY/JUNE SESSION 2002 2 hours Additional materials: Answer paper Electronic

Add Math (4047) Paper 2

1. Solve the simultaneous equations 5 and 1. [5]. (i) Sketch the graph of, showing the coordinates of the points where our graph meets the coordinate aes. [] Solve the equation 10, giving our answer correct

4. The G.C.D of 15x 4 y 3 z 5,12x 2 y 7 z 2... a) 15x 4 y 7 z 5 b)3x 2 y 3 z 2 c)12x 2 y 3 z 2 d)3x 4 y 7 z 5

4. The slope of a straight line parallel to the line x + 4y + 5 = 0 is... a) b) 1 / c) - 1 / d) - 5. The angle of inclination of a straight line whose slope is is... a) 0 0 b) 0 0 c) 60 0 d) 90 0 6. The

Introduction to Differential Equations

Chapter 1 Introduction to Differential Equations 1.1 Basic Terminology Most of the phenomena studied in the sciences and engineering involve processes that change with time. For example, it is well known

Geometry. Class Examples (July 29) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014

Geometry lass Examples (July 29) Paul Yiu Department of Mathematics Florida tlantic University c a Summer 2014 1 The Pythagorean Theorem Theorem (Pythagoras). The lengths a

Problems in Algebra. 2 4ac. 2a

Problems in Algebra Section Polynomial Equations For a polynomial equation P (x) = c 0 x n + c 1 x n 1 + + c n = 0, where c 0, c 1,, c n are complex numbers, we know from the Fundamental Theorem of Algebra

( ) = ( ) ( ) = ( ) = + = = = ( ) Therefore: , where t. Note: If we start with the condition BM = tab, we will have BM = ( x + 2, y + 3, z 5)

Chapter Exercise a) AB OB OA ( xb xa, yb ya, zb za),,, 0, b) AB OB OA ( xb xa, yb ya, zb za) ( ), ( ),, 0, c) AB OB OA x x, y y, z z (, ( ), ) (,, ) ( ) B A B A B A ( ) d) AB OB OA ( xb xa, yb ya, zb za)

Cambridge International Examinations Cambridge Ordinary Level

Cambridge International Examinations Cambridge Ordinary Level * 9 4 5 1 6 2 0 9 2 4 * ADDITIONAL MATHEMATICS 4037/12 Paper 1 October/November 2015 2 hours Candidates answer on the Question Paper. No Additional

STRAIGHT LINES EXERCISE - 3

STRAIGHT LINES EXERCISE - 3 Q. D C (3,4) E A(, ) Mid point of A, C is B 3 E, Point D rotation of point C(3, 4) by angle 90 o about E. 3 o 3 3 i4 cis90 i 5i 3 i i 5 i 5 D, point E mid point of B & D. So

Geometry. Class Examples (July 1) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014

Geometry lass Examples (July 1) Paul Yiu Department of Mathematics Florida tlantic University c b a Summer 2014 21 Example 11: Three congruent circles in a circle. The three small circles are congruent.

Section 4.8 Anti Derivative and Indefinite Integrals 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

Section 4.8 Anti Derivative and Indefinite Integrals 2 Lectures College of Science MATHS 101: Calculus I (University of Bahrain) 1 / 28 Indefinite Integral Given a function f, if F is a function such that

(b + c) 2 + at(b + c) 1 4 (2 + t)(b + c)2 a (2 t)(b + c)2 + at(b + c) a 2.

1. Determine all real constants t such that whenever a, b, c are the lengths of the sides of a triangle, then so are a 2 + bct, b 2 + cat, c 2 + abt. Proposed by S. Khan, UNK The answer is the interval