# Difference between revisions of "Ptolemy's Inequality"

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− | BD = \frac{BA \cdot DC }{AP} \; ( | + | BD = \frac{BA \cdot DC }{AP} \; (1) |

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− | BD = \frac{BC \cdot AD}{PC} \; ( | + | BD = \frac{BC \cdot AD}{PC} \; (2) |

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− | Now, by the [[triangle inequality]], we have <math>AP + PC \ge AC </math>. Multiplying both sides of the inequality by <math>BD</math> and using <math>( | + | Now, by the [[triangle inequality]], we have <math>AP + PC \ge AC </math>. Multiplying both sides of the inequality by <math>BD</math> and using equations <math>(1) </math> and <math>(2) </math> gives us |

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## Latest revision as of 19:10, 14 June 2020

**Ptolemy's Inequality** is a famous inequality attributed to the Greek mathematician Ptolemy.

## Contents

## Theorem

The inequality states that in for four points in the plane,

,

with equality for any cyclic quadrilateral with diagonals and .

This also holds if are four points in space not in the same plane, but equality can't be achieved.

## Proof for Coplanar Case

We construct a point such that the triangles are similar and have the same orientation. In particular, this means that

.

But since this is a spiral similarity, we also know that the triangles are also similar, which implies that

.

Now, by the triangle inequality, we have . Multiplying both sides of the inequality by and using equations and gives us

,

which is the desired inequality. Equality holds iff. , , and are collinear. But since the triangles and are similar, this would imply that the angles and are congruent, i.e., that is a cyclic quadrilateral.

## Outline for 3-D Case

Construct a sphere passing through the points and intersecting segments and . We can now prove it through similar triangles, since the intersection of a sphere and a plane is always a circle.

## Proof for All Dimensions?

Let any four points be denoted by the vectors .

Note that

.

From the Triangle Inequality,

.

## Note about Higher Dimensions

Similar to the fact that that there is a line through any two points and a plane through any three points, there is a three-dimensional "solid" or 3-plane through any four points. Thus in an n-dimensional space, one can construct a 3-plane through the four points and the theorem is trivial, assuming the case has already been proven for three dimensions.