Difference between revisions of "Ptolemy's Inequality"
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− | <math> | + | <math>AB \cdot CD + BC \cdot DA \ge AC \cdot BD</math>, |
− | AB \cdot CD + BC \cdot DA \ge AC \cdot BD | ||
− | </math>, | ||
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− | with equality if and only if <math>ABCD </math> is a | + | with equality if and only if <math>ABCD</math> is a cyclic quadrilateral with diagonals <math>AC </math> and <math>BD </math> OR if <math>A, B, C, D</math> are collinear. |
This also holds if <math>A,B,C,D</math> are four points in space not in the same plane, but equality can't be achieved. | This also holds if <math>A,B,C,D</math> are four points in space not in the same plane, but equality can't be achieved. | ||
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− | BD = \frac{BA \cdot DC }{AP} \; ( | + | BD = \frac{BA \cdot DC }{AP} \; (1) |
</math>. | </math>. | ||
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− | BD = \frac{BC \cdot AD}{PC} \; ( | + | BD = \frac{BC \cdot AD}{PC} \; (2) |
</math>. | </math>. | ||
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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 09:01, 7 June 2023
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 if and only if is a cyclic quadrilateral with diagonals
and
OR if
are collinear.
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.