Difference between revisions of "Ptolemy's Inequality"
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− | with equality | + | with equality for any cyclic quadrilateral <math>ABCD</math> with diagonals <math>AC </math> and <math>BD </math>. |
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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<math> | <math> | ||
− | BD = \frac{BA \cdot DC }{AP} \; ( | + | BD = \frac{BA \cdot DC }{AP} \; (1) |
</math>. | </math>. | ||
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<math> | <math> | ||
− | 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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− | 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== | ||
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<math>\implies AB\cdot CD+AD\cdot BC\ge AC\cdot BD</math>. | <math>\implies AB\cdot CD+AD\cdot BC\ge AC\cdot BD</math>. | ||
+ | |||
+ | ==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. | ||
==See Also== | ==See Also== | ||
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[[Category:Geometry]] | [[Category:Geometry]] | ||
− | [[Category: | + | [[Category:Inequalities]] |
− | [[Category: | + | [[Category:Geometric Inequalities]] |
Revision as of 17:08, 29 December 2021
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.