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>, | ||
</center> | </center> | ||
| − | 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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<math> | <math> | ||
| − | BD = \frac{BA \cdot DC }{AP} \; ( | + | BD = \frac{BA \cdot DC }{AP} \; (1) |
</math>. | </math>. | ||
</center> | </center> | ||
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<math> | <math> | ||
| − | BD = \frac{BC \cdot AD}{PC} \; ( | + | BD = \frac{BC \cdot AD}{PC} \; (2) |
</math>. | </math>. | ||
</center> | </center> | ||
| − | 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]] |
Latest revision as of 08:01, 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.
