Difference between revisions of "Ptolemy's Inequality"
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− | '''Ptolemy's Inequality''' states that in for four points <math>A, B, C, D </math> in the plane, | + | '''Ptolemy's Inequality''' is a famous inequality attributed to the Greek mathematician Ptolemy. |
+ | |||
+ | ==Theorem== | ||
+ | The inequality states that in for four points <math>A, B, C, D </math> in the plane, | ||
<center> | <center> | ||
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</center> | </center> | ||
− | with equality | + | with equality for any cyclic quadrilateral <math>ABCD</math> with diagonals <math>AC </math> and <math>BD </math>. |
− | == Proof == | + | 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. |
+ | |||
+ | == Proof for Coplanar Case== | ||
We construct a point <math>P </math> such that the [[triangles]] <math>APB, \; DCB </math> are [[similar]] and have the same [[orientation]]. In particular, this means that | We construct a point <math>P </math> such that the [[triangles]] <math>APB, \; DCB </math> are [[similar]] and have the same [[orientation]]. In particular, this means that | ||
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<center> | <center> | ||
<math> | <math> | ||
− | BD = \frac{BA \cdot DC }{AP} \; ( | + | BD = \frac{BA \cdot DC }{AP} \; (1) |
</math>. | </math>. | ||
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<center> | <center> | ||
<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. | ||
+ | |||
+ | ==Proof for All Dimensions?== | ||
+ | |||
+ | Let any four points be denoted by the vectors <math>\bold a,\bold b,\bold c,\bold d</math>. | ||
+ | |||
+ | Note that | ||
+ | |||
+ | <math>(\bold a-\bold b)\cdot(\bold c-\bold d)+(\bold a-\bold d)\cdot(\bold b-\bold c)</math> | ||
+ | |||
+ | <math>=\bold a\cdot\bold c-\bold a\cdot\bold d-\bold b\cdot\bold c+\bold b\cdot\bold d+\bold a\cdot\bold b-\bold a\cdot\bold c-\bold d\cdot\bold b+\bold d\cdot\bold c</math> | ||
+ | |||
+ | <math>=\bold a\cdot\bold b-\bold a\cdot\bold d-\bold b\cdot\bold c+\bold c\cdot\bold d</math> | ||
+ | |||
+ | <math>=(\bold a-\bold c)\cdot(\bold b-\bold d)</math>. | ||
+ | |||
+ | From the Triangle Inequality, | ||
+ | |||
+ | <math>|(\bold a-\bold b)\cdot(\bold c-\bold d)|+|(\bold a-\bold d)\cdot(\bold b-\bold c)|\ge|(\bold a-\bold c)\cdot(\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>. | ||
+ | |||
+ | ==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== | ||
+ | *[[Ptolemy's Theorem]] | ||
[[Category:Geometry]] | [[Category:Geometry]] | ||
[[Category:Inequality]] | [[Category:Inequality]] | ||
[[Category:Theorems]] | [[Category:Theorems]] |
Revision as of 20:10, 14 June 2020
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.