Math 55a Lecture Notes - Evan Chen

Math Olympiad Hardness Scale (MOHS) - Evan Chen

2 Speci cation Here is what each of the possible ratings means.1 Rating 0M: Sub-IMO. Problems rated 0 are too easy to use at IMO. I can often imagine such a problem could be solved by a strong student in an honors math class, even without olympiad training. Rating 5M: Very easy. This is the easiest rating which could actually appear while

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A Brief Introduction to Olympiad Inequalities

Evan Chen (April 30, 2014) A Brief Introduction to Olympiad Inequalities Example 1.2 Prove that a2 + b2 + c2 ab+ bc+ caand a4 + b4 + c4 a2bc+ b2ca+ c2ab. Proof. By AM-GM, a2 + b2 2 aband 2a4 + b4 + c4 4 a2bc: Similarly, b2 + c2 2 bcand

Barycentric Coordinates in Olympiad Geometry

problems with a formula sheet beside them. I’ve included Appendix B to facilitate these needs; I also have a version which includes just the formula sheet and the problems, which should be oating around. Happy bashing! 1For ABC counterclockwise, this is positive when P, Q and R are in counterclockwise order, and negative other-wise.

  Sheet, Coordinates, Formula, Formula sheet, Barycentric coordinates, Barycentric

The OTIS Excerpts - Evan Chen

Introduction The book is divided into algebra, combinatorics, and number theory. We do not cover geometry, for which Euclidean Geometry in Mathematical Olympiads [Che16] already serves the role of “comprehensive book”. The twelve main chapters in this book are structured in to four sections.

  Introduction, Excerpt, Otis, Combinatorics, The otis excerpts

Introduction to Functional Equations

Evan Chen (October 18, 2016) Introduction to Functional Equations Remark 2.4. There are of course other approaches. Here is an outline of another one. After showing f is an involution, one can simultaneously let x = f(t), y = f(u) and instead obtain f(t2 + u) = tf(t) + f(u) (check this!). This quickly becomes a \Cauchy equation", see below ...

  Equations

The Incenter/Excenter Lemma - Evan Chen

midpoint of arc BC. Show that L is the center of a circle through I, I A, B, C. A BC I L I A Proof. This is just angle chasing. Let A = \BAC, B = \CBA, C = \ACB, and note that A, I, L are collinear (as L is on the angle bisector). We are going to show that LB = LI, the other cases being similar. First, notice that \LBI = \LBC + \CBI = \LAC ...

  Lemma, Incenter, The incenter excenter lemma, Excenter

Exercises and Problems in Linear Algebra

16.3. Problems 110 16.4. Answers to Odd-Numbered Exercises111 Part 5. THE GEOMETRY OF INNER PRODUCT SPACES 113 Chapter 17. COMPLEX ARITHMETIC115 17.1. Background115 17.2. Exercises 116 17.3. Problems 118 17.4. Answers to Odd-Numbered Exercises119 Chapter 18. REAL AND COMPLEX INNER PRODUCT SPACES121 18.1. Background121 18.2. …

  Exercise, Linear, Problem, Algebra, Exercises and problems in linear algebra