Advanced High-School Mathematics
Advanced High-School Mathematics David B. Surowski Shanghai American School Singapore American School January 29, 2011 ...
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i Preface/Acknowledgment The present expanded set of notes initially grew out of an attempt to flesh out the International Bacca ...
ii Preface/Acknowledgment intellectual rapport with the contents. I can only hope that the readers (if any) can find some someth ...
iii I would like to acknowledge the software used in the preparation of these notes. First of all, the typesetting itself made u ...
iv my 40-year career, these students pointed out countless errors in this document’s original draft. To them I owe an un-repayab ...
Contents ...
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1 Advanced Euclidean Geometry 1.1 Role of Euclidean Geometry in High-School Mathematics 1.2 Triangle Geometry 1.2.1 Basic notat ...
Chapter 1 Advanced Euclidean Geometry 1.1 Role of Euclidean Geometry in High-School Mathematics If only because in one’s “furthe ...
2 CHAPTER 1 Advanced Euclidean Geometry to Euclidean geometry. President James Garfield published a novel proof in 1876 of the P ...
SECTION 1.2 Triangle Geometry 3 1.2.2 The Pythagorean theorem One of the most fundamen- tal results is the well-known Pythagorea ...
4 CHAPTER 1 Advanced Euclidean Geometry In the diagram to the right, 4 ABC is a right triangle, segments [AB] and [AF] are perp ...
SECTION 1.2 Triangle Geometry 5 Proof. Note first that 4 AA′C′ and 4 CA′C′clearly have the same areas, which implies that 4 ABC′ ...
6 CHAPTER 1 Advanced Euclidean Geometry 4 ABCand 4 A′BC′are similar. Exercises Let 4 ABC and 4 A′B′C′ be given with ABĈ = A′̂B ...
SECTION 1.2 Triangle Geometry 7 In the figure to the right,ABCDis a parallelogram, and E is a point on the segment [AD]. The po ...
8 CHAPTER 1 Advanced Euclidean Geometry This implies in particular that for signed magnitudes, AB BA = − 1. Before proceeding fu ...
SECTION 1.2 Triangle Geometry 9 Lemma. Given the triangle 4 ABC, letXbe the intersection of a line throughAand meeting(BC). LetP ...
10 CHAPTER 1 Advanced Euclidean Geometry Proof. Assume that the lines in question are concurrent, meeting in the pointP. We then ...
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