Difference between revisions of "Ptolemy's Inequality"
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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 angles <math>BAP </math> and <math>BDC </math> are congruent, 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. | 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>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. | ||
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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>. |
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==See Also== | ==See Also== | ||
Revision as of 09:18, 11 June 2009
Ptolemy's Inequality is a famous inequality attributed to the Greek mathematician Ptolemy.
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 
.
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 
and 
gives us
,
which is the desired inequality. Equality holds iff. 
, , and 
are collinear. But since the angles 
and 
are congruent, 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.
General Proof
Let any four points be denoted by 
.
Note that
.
From the Triangle Inequality,
.
