Ptolemy's Inequality · 1 Theorem · 2 Proof for Coplanar Case · 3 Outline for 3-D Case · 4 Proof for All Dimensions? · 5 Note about Higher Dimensions · 6 See Also ... Proof for Coplanar Case · Proof for All Dimensions?

In the inversion-based proof of Ptolemy's inequality, transforming four co-circular points by an inversion centered at one of them causes the other three to ...

In (*) and its proof, 3 can be replaced by any positive integer n and that gives Ptolemy's inequality in n-dimensional space. The following are some exercises.

24 мая 2015 г. · Here we collect some proofs of the following nice geometric result. If ABCD is a quadrilateral, then AB.CD + BC.DA \geq AC.BD with equality if ABCD is cyclic.

4 сент. 2021 г. · Theorem 6.4.1 Ptolemy's inequality. In any quadrangle, the product of diagonals cannot exceed the sum of the products of its opposite sides; ...

The theorem says that if the quadrilateral can be inscribed in a circle then Ptolemy's condition is satisfied.

Ptolemy's theorem states the relationship between the diagonals and the sides of a cyclic quadrilateral. It is a powerful tool to apply to problems about ...

Ptolemy's theorem gives a relationship between the side lengths and the diagonals of a cyclic quadrilateral; it is the equality case of Ptolemy's Inequality. ...

There are elegant proofs to Ptolemy, as well as instructive ones that explain the method being used very well. Exercise 2.1. Prove Ptolemy's Inequality with ...

16 мая 2014 г. · ABCD is a convex quadrilateral in which AB2+CD2=BC2+AD2. Prove by vectors that diagonals AC and BD are orthogonal to each other. linear algebra - Explanation for the proof of Ptolemy-inequality geometry - Proof of Ptolemy's inequality using $z \mapsto \frac1z Does the Ptolemy's inequality hold for concave quadrilaterals? geometry - Ptolemy's inequality and Pompeiu's theorem Другие результаты с сайта math.stackexchange.com