Bicentric polygon

In geometry, a bicentric polygon is a tangential

Every triangle is bicentric.

where x is the distance between the centers of the circles.[2] This is one version of Euler's triangle formula.

Bicentric quadrilaterals

Not all quadrilaterals are bicentric (having both an incircle and a circumcircle). Given two circles (one within the other) with radii R and r where ${\displaystyle R>r}$, there exists a convex quadrilateral inscribed in one of them and tangent to the other if and only if their radii satisfy

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

where x is the distance between their centers.^[2][3] This condition (and analogous conditions for higher order polygons) is known as Fuss' theorem.^[4]

Polygons with n > 4

A complicated general formula is known for any number n of sides for the relation among the circumradius R, the inradius r, and the distance x between the circumcenter and the incenter.^[5] Some of these for specific n are:

${\displaystyle n=5:\quad r(R-x)=(R+x){\sqrt {(R-r+x)(R-r-x)}}+(R+x){\sqrt {2R(R-r-x)}},}$

${\displaystyle n=6:\quad 3(R^{2}-x^{2})^{4}=4r^{2}(R^{2}+x^{2})(R^{2}-x^{2})^{2}+16r^{4}x^{2}R^{2},}$

${\displaystyle n=8:\quad 16p^{4}q^{4}(p^{2}-1)(q^{2}-1)=(p^{2}+q^{2}-p^{2}q^{2})^{4},}$

where ${\displaystyle p={\tfrac {R+x}{r}}}$ and ${\displaystyle q={\tfrac {R-x}{r}}.}$

Regular polygons

Every

for these relations:

${\displaystyle n\!\,}$	${\displaystyle R\,{\text{and}}\,a\!\,}$	${\displaystyle r\,{\text{and}}\,a\!\,}$	${\displaystyle r\,{\text{and}}\,R\!\,}$
3	${\displaystyle R{\sqrt {3}}=a\!\,}$	${\displaystyle 2r={\frac {a}{3}}{\sqrt {3}}\!\,}$	${\displaystyle 2r=R\!\,}$
4	${\displaystyle R{\sqrt {2}}=a\!\,}$	${\displaystyle r={\frac {a}{2}}\!\,}$	${\displaystyle 2r=R{\sqrt {2}}\!\,}$
5	${\displaystyle R{\sqrt {\frac {5-{\sqrt {5}}}{2}}}=a\!\,}$	${\displaystyle r\left({\sqrt {5}}-1\right)={\frac {a}{10}}{\sqrt {50+10{\sqrt {5}}}}\!\,}$	${\displaystyle r({\sqrt {5}}-1)=R\!\,}$
6	${\displaystyle R=a\!\,}$	${\displaystyle {\frac {2r}{3}}{\sqrt {3}}=a\!\,}$	${\displaystyle {\frac {2r}{3}}{\sqrt {3}}=R\!\,}$
8	${\displaystyle R{\sqrt {2+{\sqrt {2}}}}=a\left({\sqrt {2}}+1\right)\!\,}$	${\displaystyle r{\sqrt {4-2{\sqrt {2}}}}={\frac {a}{2}}{\sqrt {4+2{\sqrt {2}}}}\!\,}$	${\displaystyle 2r\left({\sqrt {2}}-1\right)=R{\sqrt {2-{\sqrt {2}}}}\!\,}$
10	${\displaystyle ({\sqrt {5}}-1)R=2a\!\,}$	${\displaystyle 2r{\sqrt {25-10{\sqrt {5}}}}=5a\!\,}$	${\displaystyle {\frac {2r}{5}}{\sqrt {25-10{\sqrt {5}}}}={\frac {R}{2}}\left({\sqrt {5}}-1\right)\!\,}$

Thus we have the following decimal approximations:

${\displaystyle n\!\,}$		${\displaystyle R/a\!\,}$		${\displaystyle r/a\!\,}$		${\displaystyle R/r\!\,}$
${\displaystyle 3\,}$		${\displaystyle 0.577\,}$		${\displaystyle 0.289}$		${\displaystyle 2.000\,}$
${\displaystyle 4}$		${\displaystyle 0.707\,}$		${\displaystyle 0.500}$		${\displaystyle 1.414\,}$
${\displaystyle 5}$		${\displaystyle 0.851\,}$		${\displaystyle 0.688}$		${\displaystyle 1.236\,}$
${\displaystyle 6}$		${\displaystyle 1.000\,}$		${\displaystyle 0.866}$		${\displaystyle 1.155\,}$
${\displaystyle 8}$		${\displaystyle 1.307\,}$		${\displaystyle 1.207}$		${\displaystyle 1.082\,}$
${\displaystyle 10}$		${\displaystyle 1.618\,}$		${\displaystyle 1.539}$		${\displaystyle 1.051\,}$

Poncelet's porism

If two circles are the inscribed and circumscribed circles of a particular bicentric n-gon, then the same two circles are the inscribed and circumscribed circles of infinitely many bicentric n-gons. More precisely, every

Moreover, given a circumcircle and incircle, each diagonal of the variable polygon is tangent to a fixed circle. [7]

References