Convex, 4-sided shape with an incircle and a circumcircle
In
Euclidean geometry, a bicentric quadrilateral is a
convexquadrilateral 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[1] and inscribed and circumscribed quadrilateral. It has also rarely been called a double circle quadrilateral[2] and double scribed quadrilateral.[3]
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.[4] This is a special case of
Poncelet's porism, which was proved by the French mathematician
Jean-Victor Poncelet (1788–1867).
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,
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:[5]
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.[5]
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.[5]
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.)[5]
Construction
There is a simple method for constructing a bicentric quadrilateral:
It starts with the incircle Cr around the
centreI with the radius r and then draw two to each other
perpendicularchordsWY and XZ in the incircle Cr. At the endpoints of the chords draw the
tangentsa, b, c, d to the incircle. These intersect at four points A, B, C, D, which are the
vertices of a bicentric quadrilateral.[6]
To draw the circumcircle, draw two
perpendicular bisectorsp1, p2 on the sides of the bicentric quadrilateral a respectively b. The perpendicular bisectors p1, p2 intersect in the centre O of the circumcircle CR with the distance x to the centre I of the incircle Cr. The circumcircle can be drawn around the centre O.
The validity of this construction is due to the characterization that, in a
tangential quadrilateralABCD, the contact quadrilateral WXYZ has perpendicular
diagonals if and only if the tangential quadrilateral is also
cyclic.
Area
Formulas in terms of four quantities
The
areaK 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[7][8][9][10][11]
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 [12] One example of such a quadrilateral is a non-square
rectangle.
The area can also be expressed in terms of the
tangent lengthse, f, g, h as[8]: p.128
A formula for the area of bicentric quadrilateral ABCD with incenter I is[9]
If a bicentric quadrilateral has
tangency chordsk, l and diagonals p, q, then it has area[8]: p.129
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[9]
This formula cannot be used if the quadrilateral is a
right kite, since the denominator is zero in that case.
If M, N are the midpoints of the diagonals, and E, F are the intersection points of the extensions of opposite sides, then the area of a bicentric quadrilateral is given by
The area of a bicentric quadrilateral can be expressed in terms of two opposite sides and the angle θ between the diagonals according to[9]
In terms of two adjacent angles and the radius r of the incircle, the area is given by[9]
The area is given in terms of the circumradius R and the inradius r as
where θ is either angle between the diagonals.[13]
If M, N are the midpoints of the diagonals, and E, F are the intersection points of the extensions of opposite sides, then the area can also be expressed as
where Q is the foot of the perpendicular to the line EF through the center of the incircle.[9]
Inequalities
If r and R are the inradius and the circumradius respectively, then the
areaK satisfies the
inequalities[14]
There is equality on either side only if the quadrilateral is a
square.
Another inequality for the area is[15]: p.39, #1203
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[13]
with equality holding if and only if the quadrilateral is a
right kite.
In addition, with sides a, b, c, d and
semiperimeters:
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:[9]
The four sides a, b, c, d of a bicentric quadrilateral are the four solutions of the
quartic equation
where s is the semiperimeter, and r and R are the inradius and circumradius respectively.[18]: 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 rv whose tangent lengths are where v may be any
real number.[19]: pp.9–10
A bicentric quadrilateral has a greater inradius than does any other tangential quadrilateral having the same sequence of side lengths.[20]: pp.392–393
Inequalities
The circumradius R and the inradius r satisfy the inequality
which was proved by L. Fejes Tóth in 1948.[19] 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[2][21]: p. 141
where there is equality on either side if and only if the quadrilateral is a
square.[16]: p. 81
It was derived by
Nicolaus Fuss (1755–1826) in 1792. Solving for x yields
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[23] (and then by
Poncelet's closure theorem, there exist infinitely many of them).
Applying to the expression of Fuss's theorem for x in terms of r and R is another way to obtain the above-mentioned inequality A generalization is[19]: p.5
Carlitz' identity
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[24]
where a, b, c, d are the sides of the bicentric quadrilateral.
Inequalities for the tangent lengths and sides
For the
tangent lengthse, f, g, h the following inequalities holds:[19]: p.3
and
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[19]: p.5
In a bicentric quadrilateral with
diagonalsp, q, the following identity holds:[11]
where r and R are the
inradius and the
circumradius respectively. This equality can be rewritten as[13]
or, solving it as a
quadratic equation for the product of the diagonals, in the form
An inequality for the product of the diagonals p, q in a bicentric quadrilateral is[14]
where a, b, c, d are the sides. This was proved by
Murray S. Klamkin in 1967.
Four incenters lie on a circle
Let ABCD be a bicentric quadrilateral and O the center of its circumcircle. Then the incenters of the four triangles △OAB, △OBC, △OCD, △ODA lie on a circle.[28]
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