Difference between revisions of "Ptolemy's Inequality"

Revision as of 13:52, 14 October 2007

Ptolemy's Inequality states that in for four points $\displaystyle A, B, C, D$ in the plane,

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

with equality iff. $\displaystyle ABCD$ is a cyclic quadrilateral with diagonals $\displaystyle AC$ and $\displaystyle BD$ .

Proof

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

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

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

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

Now, by the triangle inequality, we have $AP + PC \ge AC$ . Multiplying both sides of the inequality by $BD$ and using $(*)$ and $(**)$ 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 angles $BAP$ and $BDC$ are congruent, this would imply that the angles $BAC$ and $BPC$ are congruent, i.e., that $ABCD$ is a cyclic quadrilateral.

@@ Line 11: / Line 11: @@
 == Proof ==
-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>
@@ Line 19: / Line 19: @@
 </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>
@@ Line 27: / Line 27: @@
 </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 <math>(*) </math> and <math>(**) </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 angles <math>BAP </math> and <math>BDC </math> are congruent, this would imply that the angles <math>BAC </math> and <math>BPC </math> are [[congruent]], i.e., that <math>ABCD </math> is a cyclic quadrilateral.

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Difference between revisions of "Ptolemy's Inequality"

Revision as of 13:52, 14 October 2007

Proof