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| [[Image:Equilateral triangle bicentric 001.svg|thumb|right|An [[equilateral triangle]]]]
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| [[Image:Bicentric kite 001.svg|thumb|right|A [[Bicentric quadrilateral|bicentric]] [[Kite (geometry)|kite]]]]
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| [[Image:Bicentric isosceles trapezoid 001.svg|thumb|right|A bicentric [[isosceles trapezoid]]]]
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| [[Image:Pentagon 001.svg|thumb|right|[[Regular polygon|A regular]] [[pentagon]]]]
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| In geometry, a '''bicentric polygon''' is a tangential [[polygon]] (a polygon all of whose sides are tangent to an inner [[incircle]]) which is also [[Cyclic polygon|cyclic]] — that is, [[inscribe]]d in an [[circumcircle|outer circle]] that passes through each vertex of the polygon. All [[triangle]]s and all [[regular polygon]]s are bicentric. On the other hand, a [[rectangle]] with unequal sides is not bicentric, because no circle can be tangent to all four sides.
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| ==Triangles==
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| {{main|Euler's theorem in geometry}}
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| Every triangle is bicentric.<ref>{{citation|title=The Facts on File Geometry Handbook|first=Catherine A.|last=Gorini|publisher=Infobase Publishing|year=2009|isbn=9780816073894|page=17|url=http://books.google.com/books?id=ZnkASIOYJWsC&pg=PA17}}.</ref> In a triangle, the radii ''r'' and ''R'' of the [[Incircle and excircles of a triangle|incircle]] and [[circumcircle]] respectively are related by the [[equation]]
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| :<math>\frac{1}{R-x}+\frac{1}{R+x}=\frac{1}{r}</math>
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| where ''x'' is the distance between the centers of the circles.<ref name="imo">{{citation|title=International Mathematical Olympiad: 1976-1990|first=István|last=Reiman|publisher=Anthem Press|year=2005|isbn=9781843312000|pages=170–171|url=http://books.google.com/books?id=xE_qYoJBpf4C&pg=PA170}}.</ref> This is one version of [[Euler's theorem in geometry|Euler's triangle formula]].
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| ==Bicentric quadrilaterals==
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| {{main|Bicentric quadrilateral}}
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| Not all [[quadrilateral]]s are bicentric (having both an incircle and a circumcircle). Given two circles (one within the other) with radii ''R'' and ''r'' where <math>R>r</math>, there exists a convex quadrilateral inscribed in one of them and tangent to the other [[if and only if]] their radii satisfy
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| :<math>\frac{1}{(R-x)^2}+\frac{1}{(R+x)^2}=\frac{1}{r^2}</math>
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| where ''x'' is the distance between their centers.<ref name="imo"/><ref>{{citation|title=Subjects for mathematical essays|first=Charles|last=Davison|publisher=Macmillan and co., limited|year=1915|page=98|url=http://books.google.com/books?id=Uz0_AQAAIAAJ&pg=PA98}}.</ref> This condition (and analogous conditions for higher order polygons) is known as [[Bicentric quadrilateral#Fuss' theorem and Carlitz' identity|Fuss' theorem]].<ref>{{citation|title=100 Great Problems of Elementary Mathematics: Their History and Solution|first=Heinrich|last=Dörrie|publisher=Courier Dover Publications|year=1965|isbn=9780486613482|page=192|url=http://books.google.com/books?id=i4SJwNrYuAUC&pg=PA192}}.</ref>
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| ==Regular polygons==
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| Every [[regular polygon]] is bicentric.<ref name="imo"/> In a regular polygon, the incircle and the circumcircle share a common center, which is also the center of the regular polygon, so the distance between the incenter and circumcenter is always zero.
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| For some regular polygons which can be [[Compass and straightedge constructions|constructed with compass and ruler]], we have the following formulas for the relation between the common [[Edge (geometry)|edge]] length ''a'', the radius ''r'' of the [[Incircle and excircles of a triangle|incircle]], and the radius ''R'' of the [[circumcircle]]:
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| {| class="wikitable"
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| | <math>n \!\, </math>
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| | <math>r \!\, </math>
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| | <math>R \!\, </math>
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| | <math>a \!\, </math>
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| |-
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| | [[Equilateral triangle|3]]
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| | <math> \frac{R}{2} = \frac{a}{6}\sqrt{3} \!\, </math>
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| | <math> 2r = \frac{a}{3}\sqrt{3} \!\, </math>
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| | <math> 2r\sqrt{3} = R\sqrt{3} \!\, </math>
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| |-
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| | [[Square (geometry)|4]]
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| | <math> \frac{R}{2}\sqrt{2} = \frac{a}{2} \!\, </math>
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| | <math> r\sqrt{2} = \frac{a}{2}\sqrt{2} \!\, </math>
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| | <math> 2r = R\sqrt{2} \!\, </math>
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| | [[Pentagon|5]]
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| | <math> \frac{R}{4}\left(\sqrt{5}+1\right) = \frac{a}{10}\sqrt{25+10\sqrt{5}} \!\, </math>
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| | <math> r\left(\sqrt{5}-1\right) = \frac{a}{10}\sqrt{50+10\sqrt{5}} \!\, </math>
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| | <math> 2r\sqrt{5-2\sqrt{5}} = \frac{R}{2}\sqrt{10-2\sqrt{5}} \!\, </math>
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| | [[Hexagon|6]]
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| | <math> \frac{R}{2}\sqrt{3} = \frac{a}{2}\sqrt{3} \!\, </math>
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| | <math> \frac{2r}{3}\sqrt{3} = a \!\, </math>
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| | <math> \frac{2r}{3}\sqrt{3} = R \!\, </math>
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| | [[Octagon|8]]
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| | <math> \frac{R}{2} \sqrt{2+\sqrt{2}} = \frac{a}{2}\left(\sqrt{2}+1\right) \!\, </math>
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| | <math> r \sqrt{4-2\sqrt{2}} = \frac{a}{2}\sqrt{4+2\sqrt{2}} \!\, </math>
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| | <math> 2r \left(\sqrt{2}-1\right) = R\sqrt{2-\sqrt{2}} \!\, </math>
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| | [[Decagon|10]]
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| | <math> \frac{R}{4} \sqrt{10+2\sqrt{5}} = \frac{a}{2}\sqrt{5+2\sqrt{5}} \!\, </math>
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| | <math> \frac{r}{5} \sqrt{50-10\sqrt{5}} = \frac{a}{2}\left(\sqrt{5}+1\right) \!\, </math>
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| | <math> \frac{2r}{5} \sqrt{25-10\sqrt{5}} = \frac{R}{2} \left(\sqrt{5}-1\right) \!\, </math>
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| |}
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| ==Poncelet's porism==
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| {{main|Poncelet's closure theorem}}
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| If two circles are the inscribed and circumscribed circles of a single bicentric ''n''-gon, then the same two circles are the inscribed and circumscribed circles of infinitely many bicentric ''n''-gons. More precisely,
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| every [[tangent line]] to the inner of the two circles can be extended to a bicentric ''n''-gon by placing vertices on the line at the points where it crosses the outer circle, continuing from each vertex along another tangent line, and continuing in the same way until the resulting [[polygonal chain]] closes up to an ''n''-gon. The fact that it will always do so is [[Poncelet's closure theorem]].<ref>{{citation|title=Poncelet's Theorem|first=Leopold|last=Flatto|publisher=American Mathematical Society|year=2009|isbn=9780821886267}}.</ref>
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| ==References==
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| {{reflist}}
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| == External links ==
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| * {{MathWorld|title=Bicentric polygon|urlname=BicentricPolygon}}
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| [[Category:Elementary geometry]]
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| [[Category:Polygons]]
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Wilber Berryhill is what his wife loves to call him and he totally loves this title. Alaska is the only location I've been residing in but now I'm contemplating other options. The preferred pastime for him and free psychic readings - fashionlinked.com, his children is fashion and he'll be starting something else alongside with it. Distributing production is where my main earnings comes from and it's some thing I truly enjoy.
Here is my homepage - best psychics (Afeen.Fbho.net)