Bicentric quadrilateral

Poncelet's porism for bicentric quadrilaterals Template:Mvar and Template:Mvar

In Euclidean geometry, a bicentric quadrilateral is a convex quadrilateral that has both an incircle and a circumcircle. The radii and centers 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 special case of Poncelet's porism, which was proved by the French mathematician Jean-Victor Poncelet (1788–1867).

Special cases

A right kite

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

Characterizations

A bicentric quadrilateral ABCD and its contact quadrilateral WXYZ

A convex quadrilateral Template:Mvar with sides Template:Mvar 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,

$\{\begin{array}{c}a+c=b+d\\ A+C=B+D=\pi .\end{array}$

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 Template:Mvar at Template:Mvar respectively, then a tangential quadrilateral Template:Mvar is also cyclic if and only if any one of the following three conditions holds:

• Template:Mvar is perpendicular to Template:Mvar
• $\frac{\overline{AW}}{\overline{WB}}=\frac{\overline{DY}}{\overline{YC}}$
• $\frac{\overline{AC}}{\overline{BD}}=\frac{\overline{AW}+\overline{CY}}{\overline{BX}+\overline{DZ}}$

The first of these three means that the contact quadrilateral Template:Mvar is an orthodiagonal quadrilateral.

If Template:Mvar are the midpoints of Template:Mvar respectively, then the tangential quadrilateral Template:Mvar is also cyclic if and only if the quadrilateral Template:Mvar is a rectangle.

According to another characterization, if Template:Mvar is the incenter in a tangential quadrilateral where the extensions of opposite sides intersect at Template:Mvar and Template:Mvar, then the quadrilateral is also cyclic if and only if Template:Math is a right angle.

Yet another necessary and sufficient condition is that a tangential quadrilateral Template:Mvar is cyclic if and only if its Newton line is perpendicular to the Newton line of its contact quadrilateral Template:Mvar. (The Newton line of a quadrilateral is the line defined by the midpoints of its diagonals.)

Construction

A bicentric quadrilateral Template:Mvar with the contact quadrilateral Template:Mvar. Animation [https://commons.wikimedia.org/wiki/File:01-Bicentric_quadrilateral.gif see here

]

There is a simple method for constructing a bicentric quadrilateral:

It starts with the incircle Template:Mvar around the centre Template:Mvar with the radius Template:Mvar and then draw two to each other perpendicular chords Template:Mvar and Template:Mvar in the incircle Template:Mvar. At the endpoints of the chords draw the tangents Template:Mvar to the incircle. These intersect at four points Template:Mvar, which are the vertices of a bicentric quadrilateral. To draw the circumcircle, draw two perpendicular bisectors Template:Math on the sides of the bicentric quadrilateral Template:Mvar respectively Template:Mvar. The perpendicular bisectors Template:Math intersect in the centre Template:Mvar of the circumcircle Template:Mvar with the distance Template:Mvar to the centre Template:Mvar of the incircle Template:Mvar. The circumcircle can be drawn around the centre Template:Mvar.

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

Area

Formulas in terms of four quantities

The area Template:Mvar of a bicentric quadrilateral can be expressed in terms of four quantities of the quadrilateral in several different ways. If the sides are Template:Mvar, then the area is given by

$K=\sqrt{abcd}.$

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 $K=\sqrt{abcd}.$ One example of such a quadrilateral is a non-square rectangle.

The area can also be expressed in terms of the tangent lengths Template:Mvar asTemplate:Rp

$K=\sqrt[4]{efgh}(e+f+g+h).$

A formula for the area of bicentric quadrilateral Template:Mvar with incenter Template:Mvar is

$K=\overline{AI}\cdot \overline{CI}+\overline{BI}\cdot \overline{DI}.$

If a bicentric quadrilateral has tangency chords Template:Mvar and diagonals Template:Mvar, then it has areaTemplate:Rp

$K=\frac{klpq}{k^2+l^2}.$

If Template:Mvar are the tangency chords and Template:Mvar are the bimedians of the quadrilateral, then the area can be calculated using the formula

$K=\left|\frac{m^2-n^2}{k^2-l^2}\right|kl$

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

If Template:Mvar are the midpoints of the diagonals, and Template:Mvar are the intersection points of the extensions of opposite sides, then the area of a bicentric quadrilateral is given by

$K=\frac{2\overline{MN}\cdot \overline{EI}\cdot \overline{FI}}{\overline{EF}}$

where Template:Mvar is the center of the incircle.

Formulas in terms of three quantities

The area of a bicentric quadrilateral can be expressed in terms of two opposite sides and the angle Template:Mvar between the diagonals according to

$K=ac\tan \frac{\theta }{2}=bd\cot \frac{\theta }{2}.$

In terms of two adjacent angles and the radius Template:Mvar of the incircle, the area is given by

$K=2r^2(\frac{1}{\sin A}+\frac{1}{\sin B}).$

The area is given in terms of the circumradius Template:Mvar and the inradius Template:Mvar as

$K=r(r+\sqrt{4R^2+r^2})\sin \theta$

where Template:Mvar is either angle between the diagonals.

If Template:Mvar are the midpoints of the diagonals, and Template:Mvar are the intersection points of the extensions of opposite sides, then the area can also be expressed as

$K=2\overline{MN}\sqrt{\overline{EQ}\cdot \overline{FQ}}$

where Template:Mvar is the foot of the perpendicular to the line Template:Mvar through the center of the incircle.

Inequalities

If Template:Mvar and Template:Mvar are the inradius and the circumradius respectively, then the area Template:Mvar satisfies the inequalities

$4r^2\le K\le 2R^2.$

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

Another inequality for the area isTemplate:Rp

$K\le \frac{4}{3}r\sqrt{4R^2+r^2}$

where Template:Mvar and Template:Mvar are the inradius and the circumradius respectively.

A similar inequality giving a sharper upper bound for the area than the previous one is

$K\le r(r+\sqrt{4R^2+r^2})$

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

In addition, with sides Template:Mvar and semiperimeter Template:Mvar:

$2\sqrt{K}\le s\le r+\sqrt{r^2+4R^2};$Template:Rp

$6K\le ab+ac+ad+bc+bd+cd\le 4r^2+4R^2+4r\sqrt{r^2+4R^2};$Template:Rp

$4K{r}^2\le abcd\le \frac{16}{9}{r}^2(r^2+4R^2).$Template:Rp

Angle formulas

If Template:Mvar are the length of the sides Template:Mvar respectively in a bicentric quadrilateral Template:Mvar, then its vertex angles can be calculated with the tangent function:

$\begin{array}{cc}\tan \frac{A}{2}& =\sqrt{\frac{bc}{ad}}=\cot \frac{C}{2},\\ \tan \frac{B}{2}& =\sqrt{\frac{cd}{ab}}=\cot \frac{D}{2}.\end{array}$

Using the same notations, for the sine and cosine functions the following formulas holds:

$\begin{array}{cc}\sin \frac{A}{2}& =\sqrt{\frac{bc}{ad+bc}}=\cos \frac{C}{2},\\ \cos \frac{A}{2}& =\sqrt{\frac{ad}{ad+bc}}=\sin \frac{C}{2},\\ \sin \frac{B}{2}& =\sqrt{\frac{cd}{ab+cd}}=\cos \frac{D}{2},\\ \cos \frac{B}{2}& =\sqrt{\frac{ab}{ab+cd}}=\sin \frac{D}{2}.\end{array}$

The angle Template:Mvar between the diagonals can be calculated from

$\tan \frac{\theta }{2}=\sqrt{\frac{bd}{ac}}.$

Inradius and circumradius

The inradius Template:Mvar of a bicentric quadrilateral is determined by the sides Template:Mvar according to

$r=\frac{\sqrt{abcd}}{a+c}=\frac{\sqrt{abcd}}{b+d}.$

The circumradius Template:Mvar is given as a special case of Parameshvara's formula. It is

$R=\frac{1}{4}\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{abcd}}.$

The inradius can also be expressed in terms of the consecutive tangent lengths Template:Mvar according toTemplate:Rp

$r=\sqrt{eg}=\sqrt{fh}.$

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

The four sides Template:Mvar of a bicentric quadrilateral are the four solutions of the quartic equation

$y^4-2s{y}^3+(s^2+2r^2+2r\sqrt{4R^2+r^2})y^2-2rs(\sqrt{4R^2+r^2}+r)y+r^2{s}^2=0$

where Template:Mvar is the semiperimeter, and Template:Mvar and Template:Mvar are the inradius and circumradius respectively.Template:Rp

If there is a bicentric quadrilateral with inradius Template:Mvar whose tangent lengths are Template:Mvar, then there exists a bicentric quadrilateral with inradius Template:Mvar whose tangent lengths are Template:Tmath where Template:Mvar may be any real number.Template:Rp

A bicentric quadrilateral has a greater inradius than does any other tangential quadrilateral having the same sequence of side lengths.Template:Rp

Inequalities

The circumradius Template:Mvar and the inradius Template:Mvar satisfy the inequality

$R\ge \sqrt{2}r$

which was proved by L. Fejes Tóth in 1948. 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 isTemplate:Rp

$\frac{r\sqrt{2}}{R}\le \frac{1}{2}(\sin \frac{A}{2}\cos \frac{B}{2}+\sin \frac{B}{2}\cos \frac{C}{2}+\sin \frac{C}{2}\cos \frac{D}{2}+\sin \frac{D}{2}\cos \frac{A}{2})\le 1$

where there is equality on either side if and only if the quadrilateral is a square.Template:Rp

The semiperimeter Template:Mvar of a bicentric quadrilateral satisfiesTemplate:Rp

$\sqrt{8r(\sqrt{4R^2+r^2}-r)}\le s\le \sqrt{4R^2+r^2}+r$

where Template:Mvar and Template:Mvar are the inradius and circumradius respectively.

Moreover,Template:Rp

$2s{r}^2\le abc+abd+acd+bcd\le 2r(r+\sqrt{r^2+4R^2}{)}^2$

and

$abc+abd+acd+bcd\le 2\sqrt{K}(K+2R^2).$ Template:Rp

Distance between the incenter and circumcenter

A bicentric quadrilateral Template:Mvar with incenter I and circumcenter O

Fuss' theorem

Fuss' theorem gives a relation between the inradius Template:Mvar, the circumradius Template:Mvar and the distance Template:Mvar between the incenter Template:Mvar and the circumcenter Template:Mvar, for any bicentric quadrilateral. The relation is

$\frac{1}{(R-x{)}^2}+\frac{1}{(R+x{)}^2}=\frac{1}{r^2},$

or equivalently

$2r^2(R^2+x^2)=(R^2-x^2{)}^2.$

It was derived by Nicolaus Fuss (1755–1826) in 1792. Solving for Template:Mvar yields

$x=\sqrt{R^2+r^2-r\sqrt{4R^2+r^2}}.$

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 Template:Mvar and Template:Mvar and distance Template:Mvar between their centers satisfying the condition in Fuss' theorem, there exists a convex quadrilateral inscribed in one of them and tangent to the other (and then by Poncelet's closure theorem, there exist infinitely many of them).

Applying $x^2\ge 0$ to the expression of Fuss's theorem for Template:Mvar in terms of Template:Mvar and Template:Mvar is another way to obtain the above-mentioned inequality $R\ge \sqrt{2}r.$ A generalization isTemplate:Rp

$2r^2+x^2\le R^2\le 2r^2+x^2+2rx.$

Carlitz' identity

Another formula for the distance Template:Mvar between the centers of the incircle and the circumcircle is due to the American mathematician Leonard Carlitz (1907–1999). It states that

$x^2=R^2-2Rr\cdot \mu$

where Template:Mvar and Template:Mvar are the inradius and the circumradius respectively, and

$\mu =\sqrt{\frac{(ab+cd)(ad+bc)}{(a+c{)}^2(ac+bd)}}=\sqrt{\frac{(ab+cd)(ad+bc)}{(b+d{)}^2(ac+bd)}}$

where Template:Mvar are the sides of the bicentric quadrilateral.

Inequalities for the tangent lengths and sides

For the tangent lengths Template:Mvar the following inequalities holds:Template:Rp

$4r\le e+f+g+h\le 4r\cdot \frac{R^2+x^2}{R^2-x^2}$

and

$4r^2\le e^2+f^2+g^2+h^2\le 4(R^2+x^2-r^2)$

where Template:Mvar is the inradius, Template:Mvar is the circumradius, and Template:Mvar is the distance between the incenter and circumcenter. The sides Template:Mvar satisfy the inequalitiesTemplate:Rp

$8r\le a+b+c+d\le 8r\cdot \frac{R^2+x^2}{R^2-x^2}$

and

$4(R^2-x^2+2r^2)\le a^2+b^2+c^2+d^2\le 4(3R^2-2r^2).$

Other properties of the incenter

The circumcenter, the incenter, and the intersection of the diagonals in a bicentric quadrilateral are collinear.

There is the following equality relating the four distances between the incenter Template:Mvar and the vertices of a bicentric quadrilateral Template:Mvar:

$\frac{1}{\stackrel{2}{\stackrel{‾}{AI}}}+\frac{1}{\stackrel{2}{\stackrel{‾}{CI}}}=\frac{1}{\stackrel{2}{\stackrel{‾}{BI}}}+\frac{1}{\stackrel{2}{\stackrel{‾}{DI}}}=\frac{1}{r^2}$

where Template:Mvar is the inradius.

If Template:Mvar is the intersection of the diagonals in a bicentric quadrilateral Template:Mvar with incenter Template:Mvar, then

$\frac{\overline{AP}}{\overline{CP}}=\frac{\stackrel{2}{\stackrel{‾}{AI}}}{\stackrel{2}{\stackrel{‾}{CI}}}.$

Properties of the diagonals

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 Template:Mvar, the following identity holds:

$\frac{pq}{4r^2}-\frac{4R^2}{pq}=1$

where Template:Mvar and Template:Mvar are the inradius and the circumradius respectively. This equality can be rewritten as

$r=\frac{pq}{2\sqrt{pq+4R^2}}$

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

$pq=2r(r+\sqrt{4R^2+r^2}).$

An inequality for the product of the diagonals Template:Mvar in a bicentric quadrilateral is

$8pq\le (a+b+c+d{)}^2$

where Template:Mvar are the sides. This was proved by Murray S. Klamkin in 1967.

Four incenters lie on a circle

Let Template:Mvar be a bicentric quadrilateral and Template:Mvar the center of its circumcircle. Then the incenters of the four triangles Template:Math lie on a circle.

See also

• Bicentric polygon
• Ex-tangential quadrilateral