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In Euclidean geometry, a bicentric quadrilateral is a convex quadrilateral that has both an incircle and a circumcircle. The radii and center of these circles are called inradius and circumradius, and incenter and circumcenter respectively. From the definition it follows that bicentric quadrilaterals have all the properties of both tangential quadrilaterals and cyclic quadrilaterals. Other names for these quadrilaterals are chord-tangent quadrilateral and inscribed and circumscribed quadrilateral. It has also rarely been called a double circle quadrilateral and double scribed quadrilateral. 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. This is a corollary of Poncelet's porism, which was proved by the French mathematician Jean-Victor Poncelet (1788–1867).

quadrilateral
quadrilaterals
poncelet

Examples of bicentric quadrilaterals are squares, right kites, and isosceles tangential trapezoids.

A convex quadrilateral *ABCD* with sides *a*, *b*, *c*, *d* is bicentric if and only if opposite sides satisfy Pitot's theorem for tangential quadrilaterals *and* the cyclic quadrilateral property that opposite angles are supplementary; that is,

- [math]\displaystyle{ \begin{cases} a+c=b+d\\ A+C=B+D=\pi. \end{cases} }[/math]

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:^{[1]}

*WY*is perpendicular to*XZ*- [math]\displaystyle{ \frac{AW}{WB}=\frac{DY}{YC} }[/math]
- [math]\displaystyle{ \frac{AC}{BD}=\frac{AW+CY}{BX+DZ} }[/math]

The first of these three means that the *contact quadrilateral* *WXYZ* is an orthodiagonal quadrilateral.

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.^{[1]}

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.^{[1]}

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.)^{[1]}

There is a simple method for constructing a bicentric quadrilateral:

It starts with the incircle *C _{r}* around the centre

The validity of this construction is due to the characterization that, in a tangential quadrilateral *ABCD*, the contact quadrilateral *WXYZ* has perpendicular diagonals if and only if the tangential quadrilateral is also cyclic.

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^{[3]}^{[4]}^{[5]}^{[6]}^{[7]}

- [math]\displaystyle{ \displaystyle K = \sqrt{abcd}. }[/math]

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. Note that the converse does not hold: Some quadrilaterals that are not bicentric also have area [math]\displaystyle{ \displaystyle K = \sqrt{abcd}. }[/math]^{[8]} One example of such a quadrilateral is a non-square rectangle.

The area can also be expressed in terms of the tangent lengths *e*, *f*, *g*, *h* as^{[4]}^{:p.128}

- [math]\displaystyle{ K=\sqrt[4]{efgh}(e+f+g+h). }[/math]

A formula for the area of bicentric quadrilateral *ABCD* with incenter *I* is^{[5]}

- [math]\displaystyle{ K=AI\cdot CI+BI\cdot DI. }[/math]

If a bicentric quadrilateral has tangency chords *k*, *l* and diagonals *p*, *q*, then it has area^{[4]}^{:p.129}

- [math]\displaystyle{ K=\frac{klpq}{k^2+l^2}. }[/math]

If *k*, *l* are the tangency chords and *m*, *n* are the bimedians of the quadrilateral, then the area can be calculated using the formula^{[5]}

- [math]\displaystyle{ K=\left|\frac{m^2-n^2}{k^2-l^2}\right|kl }[/math]

This formula cannot be used if the quadrilateral is a right kite, since the denominator is zero in that case.

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

- [math]\displaystyle{ K=\frac{2MN\cdot EI\cdot FI}{EF} }[/math]

where *I* is the center of the incircle.^{[5]}

The area of a bicentric quadrilateral can be expressed in terms of two opposite sides and the angle *θ* between the diagonals according to^{[5]}

- [math]\displaystyle{ K=ac\tan{\frac{\theta}{2}}=bd\cot{\frac{\theta}{2}}. }[/math]

In terms of two adjacent angles and the radius *r* of the incircle, the area is given by^{[5]}

- [math]\displaystyle{ K=2r^2\left(\frac{1}{\sin{A}}+\frac{1}{\sin{B}}\right). }[/math]

The area is given in terms of the circumradius *R* and the inradius *r* as

- [math]\displaystyle{ K=r(r+\sqrt{4R^2+r^2})\sin \theta }[/math]

where *θ* is either angle between the diagonals.^{[9]}

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

- [math]\displaystyle{ K=2MN\sqrt{EQ\cdot FQ} }[/math]

where *Q* is the foot of the perpendicular to the line *EF* through the center of the incircle.^{[5]}

If *r* and *R* are the inradius and the circumradius respectively, then the area *K* satisfies the inequalities^{[10]}

- [math]\displaystyle{ \displaystyle 4r^2 \le K \le 2R^2. }[/math]

There is equality on either side only if the quadrilateral is a square.

Another inequality for the area is^{[11]}^{:p.39,#1203}

- [math]\displaystyle{ K \le \tfrac{4}{3}r\sqrt{4R^2+r^2} }[/math]

where *r* and *R* are the inradius and the circumradius respectively.

A similar inequality giving a sharper upper bound for the area than the previous one is^{[9]}

- [math]\displaystyle{ K \le r(r+\sqrt{4R^2+r^2}) }[/math]

with equality holding if and only if the quadrilateral is a right kite.

In addition, with sides *a, b, c, d* and semiperimeter *s*:

- [math]\displaystyle{ 2\sqrt{K} \leq s \leq r+ \sqrt{r^2+4R^2}; }[/math]
^{[11]}^{:p.39,#1203}

- [math]\displaystyle{ 6K \leq ab+ac+ad+bc+bd+cd \leq 4r^2+4R^2+ 4r\sqrt{r^2+4R^2}; }[/math]
^{[11]}^{:p.39,#1203}

- [math]\displaystyle{ 4Kr^2\leq abcd \leq \frac{16}{9} r^2(r^2+4R^2). }[/math]
^{[11]}^{:p.39,#1203}

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 tangent function:^{[5]}

- [math]\displaystyle{ \tan{\frac{A}{2}}=\sqrt{\frac{bc}{ad}}=\cot{\frac{C}{2}}, }[/math]
- [math]\displaystyle{ \tan{\frac{B}{2}}=\sqrt{\frac{cd}{ab}}=\cot{\frac{D}{2}}. }[/math]

Using the same notations, for the sine and cosine functions the following formulas holds:^{[12]}

- [math]\displaystyle{ \sin{\frac{A}{2}}=\sqrt{\frac{bc}{ad+bc}}=\cos{\frac{C}{2}}, }[/math]
- [math]\displaystyle{ \cos{\frac{A}{2}}=\sqrt{\frac{ad}{ad+bc}}=\sin{\frac{C}{2}}, }[/math]
- [math]\displaystyle{ \sin{\frac{B}{2}}=\sqrt{\frac{cd}{ab+cd}}=\cos{\frac{D}{2}}, }[/math]
- [math]\displaystyle{ \cos{\frac{B}{2}}=\sqrt{\frac{ab}{ab+cd}}=\sin{\frac{D}{2}}. }[/math]

The angle *θ* between the diagonals can be calculated from^{[6]}

- [math]\displaystyle{ \displaystyle \tan{\frac{\theta}{2}}=\sqrt{\frac{bd}{ac}}. }[/math]

The inradius *r* of a bicentric quadrilateral is determined by the sides *a*, *b*, *c*, *d* according to^{[3]}

- [math]\displaystyle{ \displaystyle r=\frac{\sqrt{abcd}}{a+c}=\frac{\sqrt{abcd}}{b+d}. }[/math]

The circumradius *R* is given as a special case of Parameshvara's formula. It is^{[3]}

- [math]\displaystyle{ \displaystyle R=\frac{1}{4}\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{abcd}}. }[/math]

The inradius can also be expressed in terms of the consecutive tangent lengths *e*, *f*, *g*, *h* according to^{[13]}^{:p. 41}

- [math]\displaystyle{ \displaystyle r=\sqrt{eg}=\sqrt{fh}. }[/math]

These two formulas are in fact necessary and sufficient conditions for a tangential quadrilateral with inradius *r* to be cyclic.

The four sides *a*, *b*, *c*, *d* of a bicentric quadrilateral are the four solutions of the quartic equation

- [math]\displaystyle{ 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]

where *s* is the semiperimeter, and *r* and *R* are the inradius and circumradius respectively.^{[14]}^{:p. 754}

If there is a bicentric quadrilateral with inradius *r* whose tangent lengths are *e*, *f*, *g*, *h*, then there exists a bicentric quadrilateral with inradius *r ^{v}* whose tangent lengths are

A bicentric quadrilateral has a greater inradius than does any other tangential quadrilateral having the same sequence of side lengths.^{[16]}^{:pp.392–393}

The circumradius *R* and the inradius *r* satisfy the inequality

- [math]\displaystyle{ R\ge \sqrt{2}r }[/math]

which was proved by L. Fejes Tóth in 1948.^{[15]} It holds with equality only when the two circles are concentric (have the same center as each other); then the quadrilateral is a square. The inequality can be proved in several different ways, one using the double inequality for the area above.

An extension of the previous inequality is^{[17]}^{[18]}^{:p. 141}

- [math]\displaystyle{ \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]

where there is equality on either side if and only if the quadrilateral is a square.^{[12]}^{:p. 81}

The semiperimeter *s* of a bicentric quadrilateral satisfies^{[15]}^{:p.13}

- [math]\displaystyle{ \sqrt{8r\left(\sqrt{4R^2+r^2}-r\right)}\le s \le \sqrt{4R^2+r^2}+r }[/math]

where *r* and *R* are the inradius and circumradius respectively.

Moreover,^{[11]}^{:p.39,#1203}

- [math]\displaystyle{ 2sr^2\leq abc+abd+acd+bcd \leq 2r(r+\sqrt{r^2+4R^2})^2 }[/math]

and

- [math]\displaystyle{ abc+abd+acd+bcd \leq 2\sqrt{K}(K+2R^2). }[/math]
^{[11]}^{:p.62,#1599}

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^{[7]}^{[19]}^{[20]}

- [math]\displaystyle{ \frac{1}{(R-x)^2}+\frac{1}{(R+x)^2}=\frac{1}{r^2}, }[/math]

or equivalently

- [math]\displaystyle{ \displaystyle 2r^2(R^2+x^2)=(R^2-x^2)^2. }[/math]

It was derived by Nicolaus Fuss (1755–1826) in 1792. Solving for *x* yields

- [math]\displaystyle{ x=\sqrt{R^2+r^2-r\sqrt{4R^2+r^2}}. }[/math]

Fuss's theorem, which is the analog of Euler's theorem for triangles for bicentric quadrilaterals, 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^{[21]} (and then by Poncelet's closure theorem, there exist infinitely many of them).

Applying [math]\displaystyle{ 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]\displaystyle{ R \ge \sqrt{2}r. }[/math] A generalization is^{[15]}^{:p.5}

- [math]\displaystyle{ 2r^2+x^2\le R^2 \le 2r^2+x^2+2rx. }[/math]

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^{[22]}

- [math]\displaystyle{ \displaystyle x^2=R^2-2Rr\cdot \mu }[/math]

where *r* and *R* are the inradius and the circumradius respectively, and

- [math]\displaystyle{ \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]

where *a*, *b*, *c*, *d* are the sides of the bicentric quadrilateral.

For the tangent lengths *e*, *f*, *g*, *h* the following inequalities holds:^{[15]}^{:p.3}

- [math]\displaystyle{ 4r\le e+f+g+h \le 4r\cdot \frac{R^2+x^2}{R^2-x^2} }[/math]

and

- [math]\displaystyle{ 4r^2\le e^2+f^2+g^2+h^2 \le 4(R^2+x^2-r^2) }[/math]

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^{[15]}^{:p.5}

- [math]\displaystyle{ 8r\le a+b+c+d \le 8r\cdot \frac{R^2+x^2}{R^2-x^2} }[/math]

and

- [math]\displaystyle{ 4(R^2-x^2+2r^2)\le a^2+b^2+c^2+d^2 \le 4(3R^2-2r^2). }[/math]

The circumcenter, the incenter, and the intersection of the diagonals in a bicentric quadrilateral are collinear.^{[23]}

There is the following equality relating the four distances between the incenter *I* and the vertices of a bicentric quadrilateral *ABCD*:^{[24]}

- [math]\displaystyle{ \frac{1}{AI^2}+\frac{1}{CI^2}=\frac{1}{BI^2}+\frac{1}{DI^2}=\frac{1}{r^2} }[/math]

where *r* is the inradius.

If *P* is the intersection of the diagonals in a bicentric quadrilateral *ABCD* with incenter *I*, then^{[25]}

- [math]\displaystyle{ \frac{AP}{CP}=\frac{AI^2}{CI^2}. }[/math]

An inequality concerning the inradius *r* and circumradius *R* in a bicentric quadrilateral *ABCD* is^{[26]}

- [math]\displaystyle{ 4r^2 \le AI\cdot CI+BI\cdot DI \le 2R^2 }[/math]

where *I* is the incenter.

The lengths of the diagonals in a bicentric quadrilateral can be expressed in terms of the sides or the tangent lengths, which are formulas that holds in a cyclic quadrilateral and a tangential quadrilateral respectively.

In a bicentric quadrilateral with diagonals *p* and *q*, the following identity holds:^{[7]}

- [math]\displaystyle{ \displaystyle \frac{pq}{4r^2}-\frac{4R^2}{pq}=1 }[/math]

where *r* and *R* are the inradius and the circumradius respectively. This equality can be rewritten as^{[9]}

- [math]\displaystyle{ r=\frac{pq}{2\sqrt{pq+4R^2}} }[/math]

or, solving it as a quadratic equation for the product of the diagonals, in the form

- [math]\displaystyle{ pq=2r\left(r+\sqrt{4R^2+r^2}\right). }[/math]

An inequality for the product of the diagonals *p*, *q* in a bicentric quadrilateral is^{[10]}

- [math]\displaystyle{ \displaystyle 8pq\le (a+b+c+d)^2 }[/math]

where *a*, *b*, *c*, *d* are the sides. This was proved by Murray S. Klamkin in 1967.

Let ABCD be a bicentric quadrilateral, O is center of circle (ABCD). Then Incenters of four triangles OAB, OBC, OCD, ODA lie on a circle.^{[27]}

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