New PDF release: Problem-Solving and Selected Topics in Euclidean Geometry:

By Sotirios E. Louridas

ISBN-10: 1461472733

ISBN-13: 9781461472735

"Problem-Solving and chosen themes in Euclidean Geometry: within the Spirit of the Mathematical Olympiads" includes theorems that are of specific price for the answer of geometrical difficulties. Emphasis is given within the dialogue of a number of tools, which play an important function for the answer of difficulties in Euclidean Geometry. ahead of the full resolution of each challenge, a key proposal is gifted in order that the reader can be capable of give you the resolution. functions of the elemental geometrical tools which come with research, synthesis, building and evidence are given. chosen difficulties that have been given in mathematical olympiads or proposed in brief lists in IMO's are mentioned. furthermore, a couple of difficulties proposed via best mathematicians within the topic are incorporated the following. The publication additionally includes new issues of their strategies. The scope of the ebook of the current e-book is to educate mathematical considering via Geometry and to supply idea for either scholars and lecturers to formulate "positive" conjectures and supply solutions.

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Additional resources for Problem-Solving and Selected Topics in Euclidean Geometry: In the Spirit of the Mathematical Olympiads

Example text

Let K, L be the intersections of AB with ED and EC, respectively. Show that AL2 + BK 2 = AB2 . 49) Proof We consider a rectangle KLNM, homothetic to the rectangle ABCD (see Fig. 26). Then the points A, M, E are collinear, and the points B, N, E are collinear. Let AK = a, KL = b, and BL = c. 38 3 Fundamentals on Geometric Transformations Fig. 50) b2 = 2ac. 51) which is equivalent to It is therefore enough to show that Eq. 51) holds true. Because of the homothety, we have that ABCD and MNLK are similar.

5 Homothety 31 Fig. 18 Homothety (Sect. 5) Fig. 19 Homothety (Sect. 5) 3. A characteristic criterion of homothety. A necessary and sufficient condition for a shape S to be homothetic to a shape S with ratio r = 0, 1 is that for each pair −−→ − → of points A, B of S there is a pair of points A , B of S such that A B = r AB. Proof Let S be homothetic to S with respect to O with ratio r = 0, 1. Let A, B be points of S and let A , B be their corresponding homologous points on S . 26) −→ − → OB = r OB.

The first homothety has center O and the second one has center O . 4 The common external tangents of two circles (if they exist) pass through the “external” center of homothety which is the intersection of the lines −−→ −→ KL and MM with KM and LM being parallel and having the same orientation. The common internal tangents of two circles pass through the “internal” center of −→ homothety which is the intersection of the lines KL and MM with the vectors KM −−→ and LM being parallel and having opposite orientations.

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Problem-Solving and Selected Topics in Euclidean Geometry: In the Spirit of the Mathematical Olympiads by Sotirios E. Louridas


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