Difference between revisions of "Circumcenter"

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[[Image:Circumcircle2.PNG|center]]
 
[[Image:Circumcircle2.PNG|center]]
  
== Proof that the perpendicular bisectors are concurrent ==
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== Proof that the perpendicular bisectors of a triangle are concurrent ==
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=== First Proof ===
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We consider a [[triangle]] <math>\displaystyle ABC</math>.  Let <math>\displaystyle O</math> be the intersection of the perpendicular bisectors of the [[side]]s <math>\displaystyle AB</math> and <math>\displaystyle BC</math>.  Since <math>\displaystyle O</math> lies on the perpendicular bisector of <math>\displaystyle AB</math>, it is [[equidistant]] from <math>\displaystyle A</math> and <math>\displaystyle B</math>; likewise, it is equidistant from <math>\displaystyle B</math> and <math>\displaystyle {C}</math>.  Hence <math>\displaystyle O</math> is equidistant from <math>\displaystyle A</math> and <math>\displaystyle {C}</math>; hence <math>\displaystyle O</math> also lies on the perpendicular bisector of <math>\displaystyle AC</math> (and is the circumcenter).
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=== Second Proof ===
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We start with a diagram:
 
We start with a diagram:
 
[[Image:Circumproof1.PNG|center]]
 
[[Image:Circumproof1.PNG|center]]
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By [[SAS Similarity]] <math>\triangle BFD\cong \triangle BAC</math>.  Thus <math>\angle BFD = \angle BAC</math> making <math>FD || AC</math>.  Since <math>EO\perp AC</math> and <math>AC\| FD, EO\perp FD</math> making <math>EH</math> an [[altitude]] of <math>DEF</math>.  Likewise, <math>DG</math> and <math>FI</math> are also altitudes.  Thus, the problem is reduced to proving that the altitudes of a triangle are concurrent.  This can be done using Ceva's Theorem (see [[orthocenter]] for more details).
 
By [[SAS Similarity]] <math>\triangle BFD\cong \triangle BAC</math>.  Thus <math>\angle BFD = \angle BAC</math> making <math>FD || AC</math>.  Since <math>EO\perp AC</math> and <math>AC\| FD, EO\perp FD</math> making <math>EH</math> an [[altitude]] of <math>DEF</math>.  Likewise, <math>DG</math> and <math>FI</math> are also altitudes.  Thus, the problem is reduced to proving that the altitudes of a triangle are concurrent.  This can be done using Ceva's Theorem (see [[orthocenter]] for more details).
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It is worth noting that the existance of the circumcenter is a much more fundamentally important theorem than it might seem, since it implies that three [[point]]s determine a circle.

Revision as of 18:21, 25 August 2006

The circumcenter is the center of the circumcircle of a polygon. However, it should be noted that only certain polygons can be circumscribed by a circle. All triangles have a circumcircle whose circumcenter is the intersection of the triangle's perpendicular bisectors. Quadrilaterals which have circumcircles are called cyclic quadrilaterals. Also, every regular polygon is cyclic.

Circumcircle2.PNG

Proof that the perpendicular bisectors of a triangle are concurrent

First Proof

We consider a triangle $\displaystyle ABC$. Let $\displaystyle O$ be the intersection of the perpendicular bisectors of the sides $\displaystyle AB$ and $\displaystyle BC$. Since $\displaystyle O$ lies on the perpendicular bisector of $\displaystyle AB$, it is equidistant from $\displaystyle A$ and $\displaystyle B$; likewise, it is equidistant from $\displaystyle B$ and $\displaystyle {C}$. Hence $\displaystyle O$ is equidistant from $\displaystyle A$ and $\displaystyle {C}$; hence $\displaystyle O$ also lies on the perpendicular bisector of $\displaystyle AC$ (and is the circumcenter).

Second Proof

We start with a diagram:

Circumproof1.PNG

One of the most common techniques for proving the concurrency of lines is Ceva's Theorem. However, there aren't any cevians in the diagram which would be needed for a direct application of Ceva's Theorem. Thus, we look for a way to make some by drawing in helpful lines. Drawing in $DE, EF$ and $FD$ (i.e. the medial triangle of $ABC$) does the trick.

Circumproof2.PNG

By SAS Similarity $\triangle BFD\cong \triangle BAC$. Thus $\angle BFD = \angle BAC$ making $FD || AC$. Since $EO\perp AC$ and $AC\| FD, EO\perp FD$ making $EH$ an altitude of $DEF$. Likewise, $DG$ and $FI$ are also altitudes. Thus, the problem is reduced to proving that the altitudes of a triangle are concurrent. This can be done using Ceva's Theorem (see orthocenter for more details).

It is worth noting that the existance of the circumcenter is a much more fundamentally important theorem than it might seem, since it implies that three points determine a circle.