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| [[File:PonceletQuad.png|thumb|Poncelet's closure theorem for bicentric quadrilaterals ABCD and EFGH]]
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| [[File:Bicentric kite 001.svg|thumb|right|A [[right kite]]]]
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| In [[Euclidean geometry]], a '''bicentric quadrilateral''' is a [[convex polygon|convex]] [[quadrilateral]] that has both an [[incircle]] and a [[circumcircle]]. The radii and center of these circles are called the ''inradius'' and ''circumradius'', and ''incenter'' and ''circumcenter'' respectively. From the definition it follows that bicentric quadrilaterals have all the properties of both [[tangential quadrilateral]]s and [[cyclic quadrilateral]]s. Other names for these quadrilaterals are '''chord-tangent quadrilateral'''<ref name=Dorrie>Dörrie, Heinrich, ''100 Great Problems of Elementary Mathematics: Their History and Solutions'', New York: Dover, 1965, pp. 188–193.</ref> and '''inscribed and circumscribed quadrilateral'''. It has also been called a '''double circle quadrilateral'''.<ref name=yun/>
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| If two circles, one within the other, are the incircle and the circumcircle of a bicentric quadrilateral, then every point on the circumcircle is the vertex of a bicentric quadrilateral having the same incircle and circumcircle.<ref>Weisstein, Eric W. "Poncelet Transverse." From ''MathWorld'' – A Wolfram Web Resource, [http://mathworld.wolfram.com/PonceletTransverse.html]</ref> This was proved by the French mathematician [[Jean-Victor Poncelet]] (1788–1867).
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| ==Special cases==
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| Examples of bicentric quadrilaterals are [[square (geometry)|squares]], [[right kite]]s, and [[Tangential trapezoid#Isosceles tangential trapezoid|isosceles tangential trapezoids]].
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| ==Characterizations==
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| [[File:Bicentric quadrilateral 2.png|thumb|A bicentric quadrilateral ABCD and its contact quadrilateral WXYZ]]
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| A convex quadrilateral ''ABCD'' with sides ''a'', ''b'', ''c'', ''d'' is bicentric [[if and only if]] opposite sides satisfy [[Pitot's theorem]] ''and'' opposite angles are [[Supplementary angles|supplementary]], that is
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| :<math>
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| \begin{cases}
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| a+c=b+d\\
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| A+C=B+D=\pi.
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| \end{cases}
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| </math>
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| Three other characterizations concern the points where the [[incircle]] in a [[tangential quadrilateral]] is tangent to the sides. If the incircle is tangent to the sides ''AB'', ''BC'', ''CD'', ''DA'' at ''W'', ''X'', ''Y'', ''Z'' respectively, then a tangential quadrilateral ''ABCD'' is also cyclic if and only if any one of the following three conditions holds:<ref name=Josefsson>{{citation |last=Josefsson |first=Martin |journal=Forum Geometricorum |pages=165–173 |title=Characterizations of Bicentric Quadrilaterals |url=http://forumgeom.fau.edu/FG2010volume10/FG201019.pdf |volume=10 |year=2010}}.</ref>
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| *''WY'' is [[perpendicular]] to ''XZ''
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| *<math>\frac{AW}{WB}=\frac{DY}{YC}</math>
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| *<math>\frac{AC}{BD}=\frac{AW+CY}{BX+DZ}</math>
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| The first of these three means that the ''contact quadrilateral'' ''WXYZ'' is an [[orthodiagonal quadrilateral]].
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| If ''E'', ''F'', ''G'', ''H'' are the midpoints of ''WX'', ''XY'', ''YZ'', ''ZW'' respectively, then the tangential quadrilateral ''ABCD'' is also cyclic [[if and only if]] the quadrilateral ''EFGH'' is a [[rectangle]].<ref name=Josefsson/>
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| According to another characterization, if ''I'' is the [[incenter]] in a [[tangential quadrilateral]] where the extensions of opposite sides intersect at ''J'' and ''K'', then the quadrilateral is also cyclic if and only if ''JIK'' is a [[right angle]].<ref name=Josefsson/>
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| Yet another [[necessary and sufficient condition]] is that a tangential quadrilateral ''ABCD'' is cyclic if and only if its [[Newton line]] is perpendicular to the Newton line of its contact quadrilateral ''WXYZ''. (The Newton line of a quadrilateral is the line defined by the midpoints of its diagonals.)<ref name=Josefsson/>
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| ==Construction==
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| There is a simple method for constructing a bicentric quadrilateral. Draw two [[perpendicular]] [[chord (geometry)|chords]] in a circle (it will be the incircle). At the endpoints of the chords, draw the [[tangent]]s to the circle. These intersect at four points, which are the [[vertex (geometry)|vertices]] of a bicentric quadrilateral.<ref>Alsina, Claudi and Nelsen, Roger, ''Icons of Mathematics. An exploration of twenty key images'', Mathematical Association of America, 2011, pp. 125-126.</ref> The validity of this construction is due to the characterization that, in a [[tangential quadrilateral]] ''ABCD'', the contact quadrilateral ''WXYZ'' has perpendicular [[diagonal]]s if and only if the tangential quadrilateral is also [[cyclic quadrilateral|cyclic]].
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| ==Area==
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| ===Formulas in terms of four quantities===
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| The [[area]] ''K'' of a bicentric quadrilateral can be expressed in terms of four quantities of the quadrilateral in several different ways. If the sides are ''a'', ''b'', ''c'', ''d'', then the area is given by<ref name=Weisstein/><ref name=Josefsson2/><ref name=Josefsson3/><ref name=Durell/><ref name=Yiu/>
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| :<math>\displaystyle K = \sqrt{abcd}.</math>
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| This is a special case of [[Brahmagupta's formula]]. It can also be derived directly from the trigonometric formula for the area of a [[tangential quadrilateral#Area|tangential quadrilateral]]. Note that the converse does not hold: Some quadrilaterals that are not bicentric also have area <math>\displaystyle K = \sqrt{abcd}.</math><ref>Lord, Nick, "Quadrilaterals with area formula <math>\displaystyle K = \sqrt{abcd}.</math>", ''Mathematical Gazette'' 96, July 2012, 345-347.</ref> One example of such a quadrilateral is a non-square [[rectangle]].
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| The area can also be expressed in terms of the [[Tangential quadrilateral#Special line segments|tangent lengths]] ''e'', ''f'', ''g'', ''h'' as<ref name=Josefsson2/>{{rp|p.128}}
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| :<math>K=\sqrt[4]{efgh}(e+f+g+h).</math>
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| A formula for the area of bicentric quadrilateral ''ABCD'' with incenter ''I'' is<ref name=Josefsson3/>
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| :<math>K=AI\cdot CI+BI\cdot DI.</math>
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| If a bicentric quadrilateral has [[Tangential quadrilateral#Special line segments|tangency chords]] ''k'', ''l'' and diagonals ''p'', ''q'', then it has the area<ref name=Josefsson2>{{citation
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| |last=Josefsson |first=Martin
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| |journal=Forum Geometricorum
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| |pages=119–130
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| |title=Calculations concerning the tangent lengths and tangency chords of a tangential quadrilateral
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| |url=http://forumgeom.fau.edu/FG2010volume10/FG201013.pdf
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| |volume=10
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| |year=2010}}.</ref>{{rp|p.129}}
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| :<math>K=\frac{klpq}{k^2+l^2}.</math>
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| If ''k'', ''l'' are the tangency chords and ''m'', ''n'' are the [[Quadrilateral#Special line segments|bimedians]] of the quadrilateral, then the area can be calculated using the formula<ref name=Josefsson3>{{citation
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| |last=Josefsson |first=Martin
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| |journal=Forum Geometricorum
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| |pages=155–164
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| |title=The Area of a Bicentric Quadrilateral
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| |url=http://forumgeom.fau.edu/FG2011volume11/FG201116.pdf
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| |volume=11
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| |year=2011}}.</ref>
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| :<math>K=\left|\frac{m^2-n^2}{k^2-l^2}\right|kl</math>
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| This formula cannot be used if the quadrilateral is a [[right kite]], since the denominator is zero in that case.
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| If ''M'' and ''N'' are the midpoints of the diagonals, and ''E'' and ''F'' are the intersection points of the extensions of opposite sides, then the area of a bicentric quadrilateral is given by
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| :<math>K=\frac{2MN\cdot EI\cdot FI}{EF}</math>
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| where ''I'' is the center of the incircle.<ref name=Josefsson3/>
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| ===Formulas in terms of three quantities===
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| The area of a bicentric quadrilateral can be expressed in terms of two opposite sides and the angle ''θ'' between the diagonals according to<ref name=Josefsson3/>
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| :<math>K=ac\tan{\frac{\theta}{2}}=bd\cot{\frac{\theta}{2}}.</math>
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| In terms of two adjacent angles and the radius ''r'' of the incircle, the area is given by<ref name=Josefsson3/>
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| :<math>K=2r^2\left(\frac{1}{\sin{A}}+\frac{1}{\sin{B}}\right).</math>
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| The area is given in terms of the circumradius ''R'' and the inradius ''r'' as
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| :<math>K=r(r+\sqrt{4R^2+r^2})\sin \theta</math>
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| where ''θ'' is either angle between the diagonals.<ref name=MJFG>{{citation
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| |last=Josefsson |first=Martin
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| |journal=Forum Geometricorum
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| |pages=237–241
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| |title=Maximal Area of a Bicentric Quadrilateral
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| |url=http://forumgeom.fau.edu/FG2012volume12/FG201222.pdf
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| |volume=12
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| |year=2012}}.</ref>
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| If ''M'' and ''N'' are the midpoints of the diagonals, and ''E'' and ''F'' are the intersection points of the extensions of opposite sides, then the area can also be expressed as
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| :<math>K=2MN\sqrt{EQ\cdot FQ}</math>
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| where ''Q'' is the foot of the normal to the line ''EF'' through the center of the incircle.<ref name=Josefsson3/>
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| ===Inequalities===
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| If ''r'' and ''R'' are the inradius and the circumradius respectively, then the [[area]] ''K'' satisfies the [[Inequality (mathematics)|inequalities]]<ref name=Alsina>Alsina, Claudi and Nelsen, Roger, ''When less is more: visualizing basic inequalities'', Mathematical Association of America, 2009, pp. 64-66.</ref>
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| :<math>\displaystyle 4r^2 \le K \le 2R^2.</math>
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| There is equality on either side only if the quadrilateral is a [[Square (geometry)|square]].
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| Another inequality for the area is<ref name=Crux>Inequalities proposed in ''Crux Mathematicorum'', 2007, Problem 1203, p. 39, [http://hydra.nat.uni-magdeburg.de/math4u/ineq.pdf]</ref>
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| :<math>K \le \tfrac{4}{3}r\sqrt{4R^2+r^2}</math>
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| where ''r'' and ''R'' are the inradius and the circumradius respectively.
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| A similar inequality giving a sharper upper bound for the area than the previous one is<ref name=MJFG/>
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| :<math>K \le r(r+\sqrt{4R^2+r^2})</math>
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| with equality holding if and only if the quadrilateral is a [[right kite]].
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| ==Angle formulas==
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| If ''a'', ''b'', ''c'', ''d'' are the length of the sides ''AB'', ''BC'', ''CD'', ''DA'' respectively in a bicentric quadrilateral ''ABCD'', then its vertex angles can be calculated with the [[Trigonometric functions|tangent function]]:<ref name=Josefsson3/>
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| :<math>\tan{\frac{A}{2}}=\sqrt{\frac{bc}{ad}}=\cot{\frac{C}{2}},</math>
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| :<math>\tan{\frac{B}{2}}=\sqrt{\frac{cd}{ab}}=\cot{\frac{D}{2}}.</math>
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| Using the same notations, for the [[Trigonometric functions|sine and cosine functions]] the following formulas holds:<ref name=Josefsson4>{{citation
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| |last=Josefsson |first=Martin
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| |journal=Forum Geometricorum
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| |pages=79–82
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| |title=A New Proof of Yun’s Inequality for Bicentric Quadrilaterals
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| |url=http://forumgeom.fau.edu/FG2012volume12/FG201208.pdf
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| |volume=12
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| |year=2012}}.</ref>
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| :<math>\sin{\frac{A}{2}}=\sqrt{\frac{bc}{ad+bc}}=\cos{\frac{C}{2}},</math>
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| :<math>\cos{\frac{A}{2}}=\sqrt{\frac{ad}{ad+bc}}=\sin{\frac{C}{2}},</math>
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| :<math>\sin{\frac{B}{2}}=\sqrt{\frac{cd}{ab+cd}}=\cos{\frac{D}{2}},</math>
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| :<math>\cos{\frac{B}{2}}=\sqrt{\frac{ab}{ab+cd}}=\sin{\frac{D}{2}}.</math>
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| The angle ''θ'' between the diagonals can be calculated from<ref name=Durell>Durell, C. V. and Robson, A., ''Advanced Trigonometry'', Dover, 2003, pp. 28, 30.</ref>
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| :<math>\displaystyle \tan{\frac{\theta}{2}}=\sqrt{\frac{bd}{ac}}.</math>
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| ==Inradius and circumradius==
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| The [[inradius]] ''r'' of a bicentric quadrilateral is determined by the sides ''a'', ''b'', ''c'', ''d'' according to<ref name=Weisstein>Weisstein, Eric, Bicentric Quadrilateral at ''MathWorld'', [http://mathworld.wolfram.com/BicentricQuadrilateral.html], Accessed on 2011-08-13.</ref>
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| :<math>\displaystyle r=\frac{\sqrt{abcd}}{a+c}=\frac{\sqrt{abcd}}{b+d}.</math>
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| The [[circumradius]] ''R'' is given as a special case of [[Parameshvara]]'s formula. It is<ref name=Weisstein/> | |
| :<math>\displaystyle R=\frac{1}{4}\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{abcd}}.</math>
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| The inradius can also be expressed in terms of the consecutive [[Tangential quadrilateral#Special line segments|tangent lengths]] ''e'', ''f'', ''g'', ''h'' according to<ref>M. Radic, Z. Kaliman, and V. Kadum, "A condition that a tangential quadrilateral is also a chordal one", ''Mathematical Communications'', 12 (2007) 33–52.</ref>{{rp|p. 41}}
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| :<math>\displaystyle r=\sqrt{eg}=\sqrt{fh}.</math>
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| These two formulas are in fact [[necessary and sufficient condition]]s for a [[tangential quadrilateral]] with inradius ''r'' to be [[cyclic quadrilateral|cyclic]]. | |
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| The four sides ''a'', ''b'', ''c'', ''d'' of a bicentric quadrilateral are the four solutions of the [[Quartic function|quartic equation]]
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| :<math>y^4-2sy^3+(s^2+2r^2+2r\sqrt{4R^2+r^2})y^2-2rs(\sqrt{4R^2+r^2}+r)y+r^2s^2=0</math>
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| where ''s'' is the semiperimeter, and ''r'' and ''R'' are the inradius and circumradius respectively.<ref name=Pop>Pop, Ovidiu T., "Identities and inequalities in a quadrilateral", ''Octogon Mathematical Magazine'', Vol. 17, No. 2, October 2009, pp 754-763.</ref>{{rp|p. 754}}
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| If there is a bicentric quadrilateral with inradius ''r'' whose [[Tangential quadrilateral#Special line segments|tangent lengths]] are ''e'', ''f'', ''g'', ''h'', then there exists a bicentric quadrilateral with inradius ''r<sup>v</sup>'' whose tangent lengths are ''e<sup>v</sup>'', ''f<sup>v</sup>'', ''g<sup>v</sup>'', ''h<sup>v</sup>'', where ''v'' may be any [[real number]].<ref name=Radic/>{{rp|pp.9-10}}
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| ===Inequalities===
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| The circumradius ''R'' and the inradius ''r'' satisfy the inequality
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| :<math>R\ge \sqrt{2}r</math>
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| which was proved by L. Fejes Tóth in 1948.<ref name=Radic>Radic, Mirko, "Certain inequalities concerning bicentric quadrilaterals, hexagons and octagons", ''Journal of Inequalities in Pure and Applied Mathematics'', Volume 6, Issue 1, 2005, [http://www.emis.de/journals/JIPAM/images/118_04_JIPAM/118_04.pdf]</ref> It holds with equality only when the two circles are [[concentric]] (have the same center as each other); then the quadrilateral is a [[square (geometry)|square]]. The inequality can be proved in several different ways, one is using the double inequality for the area above.
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| An extension of the previous inequality is<ref name=yun>Yun, Zhang, "Euler's Inequality Revisited", ''Mathematical Spectrum'', Volume 40, Number 3 (May 2008), pp. 119-121. First page available at [http://ms.appliedprobability.org/data/files/Abstracts%2040/40-3-6.pdf].</ref>
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| :<math>\frac{r\sqrt{2}}{R}\le \frac{1}{2}\left(\sin{\frac{A}{2}}\cos{\frac{B}{2}}+\sin{\frac{B}{2}}\cos{\frac{C}{2}}+\sin{\frac{C}{2}}\cos{\frac{D}{2}}+\sin{\frac{D}{2}}\cos{\frac{A}{2}}\right)\le 1</math>
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| where there is equality on either side if and only if the quadrilateral is a [[square (geometry)|square]].<ref name=Josefsson4/>{{rp|p. 81}}
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| The [[semiperimeter]] ''s'' of a bicentric quadrilateral satisfies<ref name=Radic/>{{rp|p.13}}
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| :<math>\sqrt{8r\left(\sqrt{4R^2+r^2}-r\right)}\le s \le \sqrt{4R^2+r^2}+r</math>
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| where ''r'' and ''R'' are the inradius and circumradius respectively.
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| ==Distance between the incenter and circumcenter==
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| [[File:Bicentric quadrilateral.png|thumb|A bicentric quadrilateral ABCD with incenter I and circumcenter O]]
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| ===Fuss' theorem===
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| Fuss' theorem gives a relation between the [[inradius]] ''r'', the [[circumradius]] ''R'' and the distance ''x'' between the [[incenter]] ''I'' and the [[circumcenter]] ''O'', for any bicentric quadrilateral. The relation is<ref name=Dorrie/><ref name=Yiu>Yiu, Paul, ''Euclidean Geometry'', [http://www.math.fau.edu/Yiu/EuclideanGeomeryNotes.pdf], 1998, pp. 158-164.</ref><ref>{{citation |last=Salazar |first=Juan Carlos |title=Fuss's Theorem |journal=Mathematical Gazette |volume=90 (July) |pages=306–307 |year=2006}}.</ref>
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| :<math>\frac{1}{(R-x)^2}+\frac{1}{(R+x)^2}=\frac{1}{r^2},</math>
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| or equivalently
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| :<math>\displaystyle 2r^2(R^2+x^2)=(R^2-x^2)^2.</math>
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| It was derived by [[Nicolaus Fuss]] (1755–1826) in 1792. Solving for ''x'' yields
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| :<math>x=\sqrt{R^2+r^2-r\sqrt{4R^2+r^2}}.</math>
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| Fuss's theorem says that if a quadrilateral is bicentric, then its two associated circles are related according to the above equations. In fact the converse also holds: given two circles (one within the other) with radii ''R'' and ''r'' and distance ''x'' between their centers satisfying the condition in Fuss' theorem, there exists a convex quadrilateral inscribed in one of them and tangent to the other.<ref>{{citation |last=Byerly |first=W. E. |title=The In- and-Circumscribed Quadrilateral |journal=The Annals of Mathematics |volume=10 |pages=123–128 |year=1909}}.</ref>
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| Applying <math>x^2 \ge 0</math> to the expression of Fuss's theorem for ''x'' in terms of ''r'' and ''R'' is another way to obtain the above-mentioned inequality <math>R \ge \sqrt{2}r.</math> A generalization is<ref name=Radic/>{{rp|p.5}}
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| :<math>2r^2+x^2\le R^2 \le 2r^2+x^2+2rx.</math>
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| ===Carlitz' identity===
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| Another formula for the distance ''x'' between the centers of the [[incircle]] and the [[circumcircle]] is due to the American mathematician [[Leonard Carlitz]] (1907–1999). It states that<ref>Calin, Ovidiu, ''Euclidean and Non-Euclidean Geometry a metric approach'', [http://math.emich.edu/~ocalin/Math341/Newpdf/driver13GeomB.pdf], pp. 153–158.</ref>
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| :<math>\displaystyle x^2=R^2-2Rr\cdot \mu</math>
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| where ''r'' and ''R'' are the [[inradius]] and the [[circumradius]] respectively, and
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| :<math>\displaystyle \mu=\sqrt{\frac{(ab+cd)(ad+bc)}{(a+c)^2(ac+bd)}} = \sqrt{\frac{(ab+cd)(ad+bc)}{(b+d)^2(ac+bd)}}</math>
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| where ''a'', ''b'', ''c'', ''d'' are the sides of the bicentric quadrilateral. Carlitz' identity is a generalization of [[Euler's theorem in geometry]] to a bicentric quadrilateral.
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| ===Inequalities for the tangent lengths and sides===
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| For the [[Tangential quadrilateral#Special line segments|tangent lengths]] ''e'', ''f'', ''g'', ''h'' the following inequalities holds:<ref name=Radic/>{{rp|p.3}}
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| :<math>4r\le e+f+g+h \le 4r\cdot \frac{R^2+x^2}{R^2-x^2}</math>
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| and
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| :<math>4r^2\le e^2+f^2+g^2+h^2 \le 4(R^2+x^2-r^2)</math>
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| where ''r'' is the inradius, ''R'' is the circumradius, and ''x'' is the distance between the incenter and circumcenter. The sides ''a'', ''b'', ''c'', ''d'' satisfy the inequalities<ref name=Radic/>{{rp|p.5}}
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| :<math>8r\le a+b+c+d \le 8r\cdot \frac{R^2+x^2}{R^2-x^2}</math>
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| and | |
| :<math>4(R^2-x^2+2r^2)\le a^2+b^2+c^2+d^2 \le 4(3R^2-2r^2).</math>
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| ==Other properties of the incenter==
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| The [[circumcenter]], the [[incenter]], and the intersection of the [[diagonal]]s in a bicentric quadrilateral are [[collinear]].<ref>Bogomolny, Alex, ''Collinearity in Bicentric Quadrilaterals'' [http://www.maa.org/editorial/knot/BicentricQuadri.html], 2004.</ref>
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| There is the following equality relating the four distances between the incenter ''I'' and the vertices of a bicentric quadrilateral ''ABCD'':<ref>Juan Carlos Salazar, ''Fuss Theorem for Bicentric Quadrilateral'', 2003, [http://tech.groups.yahoo.com/group/Hyacinthos/message/6768].</ref>
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| :<math>\frac{1}{IA^2}+\frac{1}{IC^2}=\frac{1}{IB^2}+\frac{1}{ID^2}=\frac{1}{r^2}</math>
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| where ''r'' is the inradius.
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| An inequality concerning the inradius ''r'' and circumradius ''R'' in a bicentric quadrilateral ''ABCD'' is<ref>Post at ''Art of Problem Solving'', 2009, [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=260539]</ref>
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| :<math>4r^2 \le IA\cdot IC+IB\cdot ID \le 2R^2</math>
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| where ''I'' is the incenter.
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| ==Properties of the diagonals==
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| The lengths of the diagonals in a bicentric quadrilateral can be expressed in terms of [[Cyclic quadrilateral#Diagonals|the sides]] or [[Tangential quadrilateral#Diagonals and tangency chords|the tangent lengths]], which are formulas that holds in a [[cyclic quadrilateral]] and a [[tangential quadrilateral]] respectively.
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| In a bicentric quadrilateral with [[diagonal]]s ''p'' and ''q'', the following identity holds:<ref name=Yiu/>
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| :<math>\displaystyle \frac{pq}{4r^2}-\frac{4R^2}{pq}=1</math>
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| where ''r'' and ''R'' are the [[inradius]] and the [[circumradius]] respectively. This equality can be rewritten as<ref name=MJFG/>
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| :<math>r=\frac{pq}{2\sqrt{pq+4R^2}}</math>
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| or, solving it as a [[quadratic equation]] for the product of the diagonals, in the form
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| :<math>pq=2r\left(r+\sqrt{4R^2+r^2}\right).</math>
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| An inequality for the product of the diagonals ''p'', ''q'' in a bicentric quadrilateral is<ref name=Alsina/> | |
| :<math>\displaystyle 8pq\le (a+b+c+d)^2</math>
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| where ''a'', ''b'', ''c'', ''d'' are the sides. This was proved by [[Murray S. Klamkin]] in 1967.
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| ==See also==
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| *[[Bicentric polygon]]
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| *[[Ex-tangential quadrilateral]]
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| *[[Poncelet's closure theorem]]
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| ==References==
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| {{reflist}}
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| [[Category:Quadrilaterals]]
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| [[Category:Polygons]]
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