Difference between revisions of "Circumcenter"
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| − | The '''circumcenter''' is the [[center]] of the [[circumcircle]] of a [[polygon]]. | + | The '''circumcenter''' is the [[center]] of the [[circumcircle]] of a [[polygon]]. Only certain polygons can be circumscribed by a [[circle]]: all [[nondegenerate]] [[triangle]]s have a circumcircle whose circumcenter is the intersection of the [[perpendicular bisector]]s of the sides of the triangle. [[Quadrilateral]]s which have circumcircles are called [[cyclic quadrilateral]]s. Also, every [[regular polygon]] is [[cyclic]]. |
[[Image:Circumcircle2.PNG|center]] | [[Image:Circumcircle2.PNG|center]] | ||
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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 nondegenerate triangle <math>\triangle ABC</math>. Since the triangle is nondegenerate, <math>AB</math> and <math>BC</math> lie on different [[line]]s and so their perpendicular bisectors are not parallel and thus intersect. Let <math>O</math> be the intersection of these perpendicular bisectors. Since <math>O</math> lies on the perpendicular bisector of <math>AB</math>, it is [[equidistant]] from <math>A</math> and <math>B</math>; likewise, it is equidistant from <math>B</math> and <math>{C}</math>. Hence <math>O</math> is equidistant from <math>A</math> and <math>{C}</math>; hence <math>O</math> also lies on the perpendicular bisector of <math>AC</math> (and is the circumcenter). | ||
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| + | === Second Proof === | ||
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| + | We start with a diagram: | ||
| + | [[Image:Circumproof1.PNG|center]] | ||
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| + | One of the most common techniques for proving the [[concurrency]] of [[line]]s 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 <math>DE, EF</math> and <math>FD</math> (i.e. the [[medial triangle]] of <math>ABC</math>) does the trick. | ||
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| + | [[Image:Circumproof2.PNG|center]] | ||
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| + | By [[SAS Similarity]] <math>\triangle BFD\sim \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. | ||
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| + | It is worth noting that the existence of the circumcenter is a much more fundamentally important theorem than it might seem, since it implies that three [[point]]s determine a circle. | ||
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| + | [[Category:Geometry]] | ||
Latest revision as of 13:17, 4 August 2020
The circumcenter is the center of the circumcircle of a polygon. Only certain polygons can be circumscribed by a circle: all nondegenerate triangles have a circumcircle whose circumcenter is the intersection of the perpendicular bisectors of the sides of the triangle. Quadrilaterals which have circumcircles are called cyclic quadrilaterals. Also, every regular polygon is cyclic.
Proof that the perpendicular bisectors of a triangle are concurrent
First Proof
We consider a nondegenerate triangle 
. Since the triangle is nondegenerate, 
and 
lie on different lines and so their perpendicular bisectors are not parallel and thus intersect. Let 
be the intersection of these perpendicular bisectors. Since lies on the perpendicular bisector of , it is equidistant from 
and 
; likewise, it is equidistant from and 
. Hence is equidistant from and ; hence also lies on the perpendicular bisector of 
(and is the circumcenter).
Second Proof
We start with a diagram:
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 
and 
(i.e. the medial triangle of 
) does the trick.
By SAS Similarity 
. Thus 
making 
. Since 
and 
making 
an altitude of 
. Likewise, 
and 
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.
It is worth noting that the existence of the circumcenter is a much more fundamentally important theorem than it might seem, since it implies that three points determine a circle.
