Difference between revisions of "Ptolemy's Inequality"

Latest revision as of 08:01, 7 June 2023

Ptolemy's Inequality is a famous inequality attributed to the Greek mathematician Ptolemy.

Theorem

The inequality states that in for four points $A, B, C, D$ in the plane,

$AB \cdot CD + BC \cdot DA \ge AC \cdot BD$ ,

with equality if and only if $ABCD$ is a cyclic quadrilateral with diagonals $AC$ and $BD$ OR if $A, B, C, D$ are collinear.

This also holds if $A,B,C,D$ are four points in space not in the same plane, but equality can't be achieved.

Proof for Coplanar Case

We construct a point $P$ such that the triangles $APB, \; DCB$ are similar and have the same orientation. In particular, this means that

$BD = \frac{BA \cdot DC }{AP} \; (1)$ .

But since this is a spiral similarity, we also know that the triangles $ABD, \; PBC$ are also similar, which implies that

$BD = \frac{BC \cdot AD}{PC} \; (2)$ .

Now, by the triangle inequality, we have $AP + PC \ge AC$ . Multiplying both sides of the inequality by $BD$ and using equations $(1)$ and $(2)$ gives us

$BA \cdot DC + BC \cdot AD \ge AC \cdot BD$ ,

which is the desired inequality. Equality holds iff. $A$ , $P$ , and ${C}$ are collinear. But since the triangles $BAP$ and $BDC$ are similar, this would imply that the angles $BAC$ and $BDC$ are congruent, i.e., that $ABCD$ is a cyclic quadrilateral.

Outline for 3-D Case

Construct a sphere passing through the points $B,C,D$ and intersecting segments $AB,AC,AD$ and $E,F,G$ . 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 $\bold a,\bold b,\bold c,\bold d$ .

Note that

$(\bold a-\bold b)\cdot(\bold c-\bold d)+(\bold a-\bold d)\cdot(\bold b-\bold c)$

$=\bold a\cdot\bold c-\bold a\cdot\bold d-\bold b\cdot\bold c+\bold b\cdot\bold d+\bold a\cdot\bold b-\bold a\cdot\bold c-\bold d\cdot\bold b+\bold d\cdot\bold c$

$=\bold a\cdot\bold b-\bold a\cdot\bold d-\bold b\cdot\bold c+\bold c\cdot\bold d$

$=(\bold a-\bold c)\cdot(\bold b-\bold d)$ .

From the Triangle Inequality,

$|(\bold a-\bold b)\cdot(\bold c-\bold d)|+|(\bold a-\bold d)\cdot(\bold b-\bold c)|\ge|(\bold a-\bold c)\cdot(\bold b-\bold d)|$

$\implies AB\cdot CD+AD\cdot BC\ge AC\cdot BD$ .

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.

@@ Line 1: / Line 1: @@
-'''Ptolemy's Inequality''' states that in for four points <math> \displaystyle 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>
-<math>
+<math>AB \cdot CD + BC \cdot DA \ge AC \cdot BD</math>,
-\displaystyle AB \cdot CD + BC \cdot DA \ge AC \cdot BD
-</math>,
 </center>
-with equality [[iff]]. <math> \displaystyle ABCD </math> is a [[cyclic quadrilateral]] with [[diagonal]]s <math> \displaystyle AC </math> and <math> \displaystyle BD </math>.
+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.
-== Proof ==
+== Proof for Coplanar Case==
-We construct a point <math> \displaystyle P </math> such that the [[triangles]] <math> \displaystyle 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
 <center>
 <math>
-BD = \frac{BA \cdot DC }{AP} \; (*)
+BD = \frac{BA \cdot DC }{AP} \; (1)
 </math>.
 </center>
-But since this is a [[spiral similarity]], we also know that the triangles <math> \displaystyle ABD, \; PBC </math> are also similar, which implies that
+But since this is a [[spiral similarity]], we also know that the triangles <math>ABD, \; PBC </math> are also similar, which implies that
 <center>
 <math>
-BD = \frac{BC \cdot AD}{PC} \; (**)
+BD = \frac{BC \cdot AD}{PC} \; (2)
 </math>.
 </center>
-Now, by the [[triangle inequality]], we have <math> \displaystyle AP + PC \ge AC </math>.  Multiplying both sides of the inequality by <math> \displaystyle BC </math> and using <math> \displaystyle (*) </math> and <math> \displaystyle (**) </math> gives us
+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
 <center>
 <math>
-BA \cdot DC + BC \cdot AD \ge AC \cdot BC
+BA \cdot DC + BC \cdot AD \ge AC \cdot BD
 </math>,
 </center>
-which is the desired inequality.  Equality holds iff. <math> \displaystyle A </math>, <math> \displaystyle P </math>, and <math> \displaystyle {C} </math> are [[collinear]].  But since the angles <math> \displaystyle BAP </math> and <math> \displaystyle BDC </math> are congruent, this would imply that the angles <math> \displaystyle BAC </math> and <math> \displaystyle BPC </math> are [[congruent]], i.e., that <math> \displaystyle ABCD </math> is a cyclic quadrilateral.
+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:Inequalities]]
+[[Category:Geometric Inequalities]]

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