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== |
Construct a sphere passing through the points <math>B,C,D</math> and intersecting segments <math>AB,AC,AD</math> and <math>E,F,G</math>. We can now prove it through similar triangles, since the intersection of a sphere and a plane is always a circle. | Construct a sphere passing through the points <math>B,C,D</math> and intersecting segments <math>AB,AC,AD</math> and <math>E,F,G</math>. 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 <math>\bold a,\bold b,\bold c,\bold d</math>. | + | Let any four points be denoted by the vectors <math>\bold a,\bold b,\bold c,\bold d</math>. |
Note that | Note that | ||
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<math>\implies|\bold a-\bold b| |\bold c-\bold d|+|\bold a-\bold d| |\bold b-\bold c|\ge|\bold a-\bold c| |\bold b-\bold d|</math> | <math>\implies|\bold a-\bold b| |\bold c-\bold d|+|\bold a-\bold d| |\bold b-\bold c|\ge|\bold a-\bold c| |\bold b-\bold d|</math> | ||
| − | <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.
