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We are making history again by presenting a new visual proof of the 2000 years old Ptolemy's theorem and Ptolemy's inequality. 00:00 Introduction 04:27 Geometry 101 08:19 Applications 14:46 Ptolemy's inequality 18:34 LIES 25:35 Animated proofs 28:57 Thank you! 30:53 Degenerate Easter Egg There are some other proofs of Ptolemy's theorem/inequality based on scaling and aligning suitable triangles. However, none of them is as slick, beautiful and powerful as Rainer's new proof. In particular, check out the animated scaling proof on the wiki page for Ptolemy's theorem (and this ) and check out the scaling proof by Claudi Asina and Roger Nelson: Proof Without Words: Ptolemy’s Inequality in Mathematics Magazine 87, (2014), p. 291. Rainer was inspired by a classic scaling based proof of Pythagoras theorem that I presented here You can find a couple of full text versions of the Almagest here #ptolemy For more background info check out the very comprehensive wiki pages on: Ptolemy’s theorem 's_theorem Ptolemy’s inequality 's_inequality Claudius Ptolemy The Almagest Trigonometric identities Cyclic quadrilateral The optic equation There are very interesting higher-dimensional versions of Ptolemy's theorem just like there are higher-dimensional versions of Pythagoras theorem. I did not get around to talking them today. Google ... Highly recommended: T. Brendan, How Ptolemy constructed trigonometry tables, The Mathematics Teacher 58 (1965), pp. 141-149 Tom M. Apostol, Ptolemy's Inequality and the Chordal Metric, Mathematics Magazine 40 (1967), pp. 233-235 an interactive exploration of Ptolemy's table of chords Ptolemy's theorem made a guest appearance in the the previous Mathologer video on the golden ratio: Here is a nice trick to make Ptolemy counterparts of Pythagorean triples. Take any two sets of Pythagorean triples: 5² = 3² 4², 13² = 12² 5², and combine them like this: 65² = 13² × 5²= 13²(4² 3²) = 52² 39²= 5²(12² 5²) = 60² 25². Now combining the two right angled triangles 52-39-65 and 25-60-65 along the common diagonal in any of four different ways gives a convex quadrilateral with all sides integers. Note that you automatically get 5 integer lengths and then Ptolemy's theorem guarantees that the remaining side is a fraction. Scaling up everything by the denominator of that fraction gives one of the special integer-everywhere quadrilaterals. See also Brahmagupta quadrilaterals. Here is a nice application of Ptolemy's theorem to a International Maths Olympiad problem In a cyclic quadrilateral the ratio of the diagonals equals the ratio of the sums of products of the sides that share the diagonals' end points: This extension of Ptolemy's theorem is part of the thumbnail for this video. T-shirt: cowsine :) Music: Floating branch by Muted and I promise by Ian Post. Enjoy, burkard
