Difference between revisions of "Ptolemy's Inequality"
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− | which is the desired inequality. Equality holds iff. <math>A </math>, <math>P </math>, and <math>{C} </math> are [[collinear]]. But since the | + | which is the desired inequality. Equality holds iff. <math>A </math>, <math>P </math>, and <math>{C} </math> are [[collinear]]. But since the triangles <math>BAP </math> and <math>BDC </math> are similar, this would imply that the angles <math>BAC </math> and <math>BDC </math> are [[congruent]], i.e., that <math>ABCD </math> is a cyclic quadrilateral. |
==Outline for 3-D Case== | ==Outline for 3-D Case== |
Revision as of 12:02, 17 June 2014
Ptolemy's Inequality is a famous inequality attributed to the Greek mathematician Ptolemy.
Contents
[hide]Theorem
The inequality states that in for four points in the plane,
,
with equality if and only if is a 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 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,
.