Math 55a Lecture Notes - Evan Chen

Math 55a Lecture Notes Evan Chen Fall 2014 This is Harvard College’s famous Math 55a, instructed by Dennis Gaitsgory. The formal name for this …

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Math Olympiad Hardness Scale (MOHS) - Evan Chen

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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

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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

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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.

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The OTIS Excerpts - Evan Chen

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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.

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Introduction to Functional Equations

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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 ...

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The Incenter/Excenter Lemma - Evan Chen

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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 ...

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