New Invariants of Poncelet–Jacobi Bicentric Polygons

Abstract

The 1d family of Poncelet polygons interscribed between two circles is known as the Bicentric family. Using elliptic functions and Liouville’s theorem, we show (i) that this family has invariant sum of internal angle cosines and (ii) that the pedal polygons with respect to the family’s limiting points have invariant perimeter. Interestingly, both (i) and (ii) are also properties of elliptic billiard N-periodics. Furthermore, since the pedal polygons in (ii) are identical to inversions of elliptic billiard N-periodics with respect to a focus-centered circle, an important corollary is that (iii) elliptic billiard focus-inversive N-gons have constant perimeter. Interestingly, these also conserve their sum of cosines (except for the $N=4$ case).

Appendices

Appendix A: Polar Pedal Transformations

We review properties of polar and pedal transformations. A detailed treatment is found in [2, 11].

In the discussion that follows, all geometric objects are contained in a fixed plane. Let ${\mathcal {C}}$ be a circle centered at $f_1$. The polar transformation with respect to ${\mathcal {C}}$ maps each straight line not passing through $f_1$ into a point, and maps each point different from $f_1$ into a straight line. This is done in the following manner.

Let $p\ne f_{1}$ be a point and let ${p}^\dagger$ be the inversion of p with respect to ${\mathcal {C}}$. The straight line $L_{p}$ that passes through ${p}^\dagger$ and is orthogonal to the line joining p and ${p}^\dagger$ is the polar of p with respect to ${\mathcal {C}}$. Conversely, a line L not passing through $f_1$ has a point q as its pole with respect to ${\mathcal {C}}$ if $L=L_{q}$.

For a smooth curve $\gamma$ not passing through $f_1$, we can define the polar curve $\gamma ^{\star }$ in two equivalent ways. Let p be a point of $\gamma$ and $T_{p}\gamma$ the tangent line to $\gamma$ at p, we define $p^{\star }=(T_{p}\gamma )^{\star }$, and $\gamma ^{\star }$ is the curve generated by $p^{\star }$ as p varies along $\gamma$. We can also think of $\gamma ^{\star }$ as the curve that is the envelope of the 1-parameter family of lines $L_{p}$, where p is a point of $\gamma$.

The notion of a polar curve can be naturally extended to polygons in the following manner: let $L_{j}$, $j=1,2,\ldots ,N$ be the consecutive sides of a planar polygon $\mathcal {P}$, and let $q_{j}$ be the corresponding poles, then this indexed set of points are the vertices of what we call the polar polygon $\mathcal {P}^{\star }$. Alternatively, we can consider the polars of vertices of $\mathcal {P}$, and their consecutive intersections do define the vertices of $\mathcal {P}^{\star }$.

Although the next results are certainly classical, we could not find them explicitly in the literature, so we include them for the reader’s convenience.

Lemma 2

Let $\mathcal {E}$ be an ellipse and $f_1$ one its the foci. Then the polar curve $\mathcal {E}^{\star }$ with respect to a circle ${\mathcal {C}}$ centered at $f_1$ is a circle. Let $\mathcal {H}$ be a hyperbola and $f_1$ one of its foci. Then,  the polar curve $\mathcal {H}^{\star }$ with respect to a circle ${\mathcal {C}}$ centered at $f_1$ is a circle minus two points.

Proof

We will use polar coordinates for our computations. Without loss of generality, let $f_{1}=(0,0)$ and consider the parametrized conic given by:

$$\begin{aligned} \gamma (t)=\left[ \frac{a(1-e^2)}{1+e\cos {t}}\cos {t}, \frac{a(1-e^2)}{1+e\cos {t}}\sin {t}\right] , \end{aligned}$$

where if $e>1$, the trace of $\gamma$ is a hyperbola and if $e<1$, the trace of $\gamma$ is an ellipse. The expression for the polar curve $\gamma ^{\star }(t)$ is obtained by direct computation: compute the unit normal $\mathbf {n}(t)$ to $\gamma (t)$, and the distance d(t) from the tangent line through $\gamma (t)$ to $f_{1}$. This yields:

$$\begin{aligned} \gamma ^{\star }(t)=\left[ \frac{e+\cos {t}}{a(1-e^2)},\frac{\sin {t}}{a(1-e^2)}\right] , \end{aligned}$$

whose trace is clearly contained in a circle. For the hyperbola, the parameter t is such that $1+e\cos (t) \ne 0$, this is why $\mathcal {H}^{\star }$ is a circle minus two points. $\square$

Lemma 3

Let $\mathcal {E}_{1}$ and $\mathcal {E}_{2}$ be two confocal ellipses and $\mathcal {E}_{1}^{\star }$ and $\mathcal {E}_{2}^{\star }$ be the circles as in Lemma 2, then $f_{1}$ is a limiting point of the pencil of circles defined by $\mathcal {E}_{1}^{\star }$ and $\mathcal {E}_{2}^{\star }$. In a similar way,  let $\mathcal {E}$ and $\mathcal {H}$ be respectively an ellipse and hyperbola that are confocal,  and let $\mathcal {E}^{\star }$ and $\mathcal {H}^{\star }$ be the circle and the circle minus 2 points,  as in Lemma 2, then $f_{1}$ is a limiting point of the pencil of circles defined by $\mathcal {E}^{\star }$ and the circle that contains $\mathcal {H}^{\star }$.

Proof

Given two circles ${\mathcal {C}}_{1}$ and ${\mathcal {C}}_{2}$, a classical result states that the limiting points $\delta _{\pm }$ of the pencil of circles determined by ${\mathcal {C}}_{1}$ and ${\mathcal {C}}_{2}$ are such that the inversion of ${\mathcal {C}}_{1}$ and ${\mathcal {C}}_{2}$ with respect to circles centered on $\delta _{\pm }$ are concentric.

If we denote by $a_{i}$ and $e_{i}$, $i=1,2$, respectively, the semi-major diameter and eccentricity of the ellipses $\mathcal {E}_{1}$ and $\mathcal {E}_{2}$, then, by symmetry, we can define an unknown limiting point as $\delta _{p}=(x,0)$, and the concentric circle condition then becomes a quadratic equation in the variable x, where the coefficients depend on $a_{1}$, $a_{2}$, $e_{1}$ and $e_{2}$. Using the fact that $\mathcal {E}_{1}$ and $\mathcal {E}_{2}$ are confocal, which is equivalent to $e_{1}a_{1}=e_{2}a_{2}$, and with some algebraic manipulations, the quadratic equation can be written as:

$$\begin{aligned} (a_{1}-a_{2})(a_{1}+a_{2})(e_{2}a_{2}x+1)=0, \end{aligned}$$

so $f_{1}=(0,0)$ is indeed one of the limiting points of the pencil of circles. $\square$

Appendix B: Polar Image of Bicentric Pair

Consider the pair of nested circles:

$$\begin{aligned} {\mathcal {C}}_{{\mathrm{int}}}:\;x^2+y^2=r^2,\quad {\mathcal {C}}_{{\mathrm{ext}}}:\;(x+d)^2+y^2=R^2. \end{aligned}$$

Their limiting points $\ell _1$ and $\ell _2$ are given by [19, Limiting Points]:

$$\begin{aligned} \ell _1= (R^2 - d^2 - r^2 -\Delta )/(2d),\quad \ell _2=(R^2 - d^2 - r^2 + \Delta )/(2 d), \end{aligned}$$

where

$$\begin{aligned} \Delta =\sqrt{ \left( d+R+r \right) \left( R-d+r \right) \left( R+d-r \right) \left( R-d-r \right) }. \end{aligned}$$

Notice $\ell _1$ (resp. $\ell _2$) is internal (resp. external) to the circle pair. Below we show that the polar image of the ${\mathcal {C}}_{{\mathrm{int}}},{\mathcal {C}}_{{\mathrm{ext}}}$ pair with respect to a circle of radius $\rho$ centered on $\ell _1$ (resp. $\ell _2$) is a confocal pair of ellipses (resp. hyperbolas).

Lemma 4

The polar image of ${\mathcal {C}}_{{\mathrm{int}}}$ with respect to $\ell _1$ is the ellipse $\mathcal {E}$ centered at

$$\begin{aligned} \mathcal {O}_e=\left[ {\rho }^2\frac{d}{\Delta } +\frac{k-\Delta }{2d},0\right] , \end{aligned}$$

where $k=R^2 - d^2 - r^2$. Its semi-axes are given by : 

$$\begin{aligned} a^2={\rho }^4\left( \frac{2d^2r^2+ \Delta (k+\Delta )}{ 2 \Delta ^2 r^2}\right) ,\quad b^2= {\rho }^4\left( \frac{k+ \Delta }{2 \Delta r^2}\right) . \end{aligned}$$

Note that $c^2=a^2-b^2={\rho }^4d^2/{\Delta ^2}$.

Lemma 5

The polar image of ${\mathcal {C}}_{{\mathrm{ext}}}$ with respect to $\ell _1$ is an ellipse $\mathcal {E}'$ confocal with $\mathcal {E}$ with semi-axes given by : 

$$\begin{aligned} a'^2= {\rho }^4\frac{(2R^2d^2 + \Delta (k'+\Delta ))}{ 2 \Delta ^2 R^2},\quad b'^2= {\rho }^4\left( \frac{k'+\Delta }{ 2 \Delta R^2}\right) , \end{aligned}$$

where $k'=R^2 + d^2 - r^2$.

Lemma 6

The polar image of ${\mathcal {C}}_{{\mathrm{int}}}$ with respect to $\ell _2$ is the hyperbola $\mathcal {H}$ centered at

$$\begin{aligned} \mathcal {O}_h=\left[ {-\rho }^2\frac{d}{\Delta } +\frac{k+\Delta }{2d},0\right] , \end{aligned}$$

with semi-axes given by : 

$$\begin{aligned} a_h^2={\rho }^4\left( \frac{ 2d^2r^2-\Delta (k - \Delta )}{2 \Delta ^2 r^2}\right) ,\quad b_h^2= {\rho }^4\left( \frac{k - \Delta }{2 \Delta r^2}\right) . \end{aligned}$$

Note that $c_h^2=a_h^2+b_h^2={{\rho }^4d^2}/{\Delta ^2}$. Note also that $c=c_h$.

Lemma 7

The polar image of ${\mathcal {C}}_{{\mathrm{out}}}$ with respect to $\ell _2$ is a hyperbola $\mathcal {H}'$ confocal with $\mathcal {H}$. Its semi-axes are given by : 

$$\begin{aligned} a_h'^2= {\rho }^4\left( \frac{2R^2d^2 - \Delta (k' - \Delta )}{ 2 \Delta ^2 R^2}\right) ,\quad b_h'^2= {\rho }^4 \left( \frac{k' - \Delta }{ 2 \Delta R^2}.\right) \end{aligned}$$

Appendix C: Polar Image of Confocal Pair

Consider a pair of origin-centered confocal ellipses $\mathcal {E}$ and $\mathcal {E}'$ with semi-axes a, b and $a',b'$, respectively. Their common foci $f_1,f_2$ lie at:

$$\begin{aligned} f_1=[-c,0],\quad f_2=[c,0], \end{aligned}$$

where $c^2=a^2-b^2$.

Below we show that the polar image of the $\mathcal {E},\mathcal {E}'$ pair with respect to a circle of radius $\rho$ centered on $f_1$ is a pair of nested circles ${\mathcal {C}}_{{\mathrm{int}}},{\mathcal {C}}_{{\mathrm{ext}}}$ with centers given by:

$$\begin{aligned} \mathcal {O}_{{\mathrm{int}}}=\left[ -c-\rho ^2\frac{c}{b^2},0\right] ,\quad \mathcal {O}_{{\mathrm{ext}}}=\left[ -c-\rho ^2\frac{c}{b'^2},0\right] . \end{aligned}$$

Note the distance d between said centers is given by:

$$\begin{aligned} d=\rho ^2\frac{ c\, (a^2 - {a'}^2)}{b^2\, {b'}^2}=\rho ^2 \frac{ca^2J^2}{b'^2}, \end{aligned}$$

where $J = \sqrt{a^2 - a'^2}/(ab).$

Their respective radii r, R are given by:

$$\begin{aligned} r=\rho ^2\frac{a}{b^2},\quad R=\rho ^2\frac{a'}{b'^2}. \end{aligned}$$

Let $\ell _1$ (resp. $\ell _2$) be the limiting point internal (resp. external) to ${\mathcal {C}}_{{\mathrm{int}}},{\mathcal {C}}_{{\mathrm{ext}}}$.

Lemma 8

The limiting points $\ell _1,\ell _2$ are given by :  $[-c,0]$ and $[-c+\frac{\rho ^2}{c},0]$.

Appendix D: Bicentric Vertices: $N=3,4$

Consider a pair of circles

$$\begin{aligned} {\mathcal {C}}_1: x^2+y^2-r^2=0,\quad {\mathcal {C}}_2: (x+d)^2+y^2-R^2=0. \end{aligned}$$

1.1 D.1. $N=3$

Let $(x_0,y_0)=(r\cos t,r\sin t) \in {\mathcal {C}}_1$. Let $d^2=R(R-2r)$. Then, the 3-periodic orbit is parametrized by $\{P_1,P_2,P_3 \},$ where

$$\begin{aligned} P_1&=\left[ {\frac{\cos t(2\,d R \cos t + {R}^{2}- {d}^{2})+\Delta \,\sin t}{2\,R}}-d, {\frac{\sin t( 2d R\, \cos t + {R}^{2}-d^2) -\Delta \,\cos t }{2\,R}}\right] ,\\ P_2&=\left[ {\frac{ \cos t(2\,d R \cos t +{R}^{2}- {d}^{2})-\Delta \,\sin t }{2\,R}}-d,{\frac{\sin t(2d R\, \cos t + {R}^{2}- {d}^{2})+\Delta \,\cos t }{2\,R}}\right] ,\\ P_3&=\left[ -{\frac{ \left( R\cos t -d \right) \left( {R}^{2}-{d }^{2} \right) }{ R^2+d^2-2\,d R\cos t } }, {\frac{ -R\left( R^2-d^2 \right) \sin t }{ R^2+d^2-2\,d R \cos t }}\right] ,\\ \Delta&=\sqrt{ \left( R^2+d^2-2d R\,\cos t \right) \left( 3\, R^2-d^2 +2\,d R\cos t\right) }. \end{aligned}$$

Under the above pair of circles, the limiting points are at:

$$\begin{aligned} l_1&=\left[ \frac{ R^2-d^2 }{8\,d {R}^{2} } \left( \sqrt{ (9\,R^2-d^2) (R^2-d^2) }+3\, R^2+d^2 \right) ,0\right] ,\\ l_2&=l_1-\left[ \frac{\sqrt{9R^2 - d^2} (R^2-d^2)^{\frac{3}{2}}}{4R^2d}, 0\right] . \end{aligned}$$

1.2 D.2. $N=4$

Let $(x_0,y_0)\in {\mathcal {C}}_1$. The Cayley condition for a pair of circles to admit Poncelet 4-periodics due to Kerawala is [19, Poncelet’s Porism, Eq. 39]:

$$\begin{aligned} \frac{1}{(R-d)^2}+ \frac{1}{(R+d)^2}-\frac{1}{r^2}=0. \end{aligned}$$

Let $P_i=[x_i,y_i]$, $i=1,\ldots ,4$ denote the vertices of a bicentric 4-periodic. Let $\alpha =R^2+d^2$ and $\beta =R^2-d^2$. The vertices are parametrized as:

$$\begin{aligned} x_{1}&=\Delta \,y_0-{\frac{ \left( \beta +2\,d x_0 \right) \left( d \alpha -\beta x_0 \right) }{2\,\alpha }},\\ y_{1}&=-\Delta \,x_0+{\frac{ \left( 2\,d \beta x_0+ \alpha ^{2} \right) y_0}{2\,\alpha }},\\ x_{2}&=-\Delta \,y_0-{\frac{ \left( \beta +2\,dx_0 \right) \left( d \beta - \beta x_0 \right) }{2\,\alpha }},\\ y_{2}&=\Delta \,x_0+{\frac{2\,d \beta y_0\,x_0+ \alpha ^{2}y_0}{2\,\alpha }},\\ x_3&= \frac{ ((( x_0^2 \alpha \beta - 4 x_0^2 \alpha ^2 + 3/2 \beta ^3 - 2 \beta ^2 \alpha ) \sqrt{2 \alpha - 2 \beta } + \beta (2 \Delta \alpha y_0 + 8 \alpha ^2 x_0 - 8 \alpha \beta x_0 + \beta ^2 x_0)) \beta )}{(4 (\sqrt{2 \alpha - 2 \beta } \alpha x_0 + \beta (\beta - 2 \alpha )/2)^2)},\\ y_3&=\frac{\alpha \beta (\alpha (2 x_0 y_0 \alpha + \beta ( x_0 y_0 - 2 \Delta )) \sqrt{2 \alpha - 2 \beta } + (4 \Delta x_0 - 2 \beta y_0) \alpha ^2 - 2 \Delta \alpha \beta x_0 + \beta ^3 y_0)}{(2 \sqrt{2 \alpha - 2 \beta } \alpha x_0 - 2 \alpha \beta + \beta ^2)^2},\\ x_4&=-\frac{((( x_0^2 \alpha \beta - 4 x_0^2 \alpha ^2 + 3/2 \beta ^3 - 2 \beta ^2 \alpha ) \sqrt{2 \alpha - 2 \beta } + \beta (-2 \Delta \alpha y_0 + 8 \alpha ^2 x_0 - 8 \alpha \beta x_0 + \beta ^2 x_0)) \beta )}{(4 (\sqrt{2 \alpha - 2 \beta } \alpha x_0 + \beta (\beta - 2 \alpha )/2)^2)},\\ y_4&=\frac{(\alpha (2 x_0 y_0 \alpha + \beta ( x_0 y_0 + 2 \Delta )) \sqrt{2 \alpha - 2 \beta } + (-4 \Delta x_0 - 2 \beta y_0) \alpha ^2 + 2 \Delta \alpha \beta x_0 + \beta ^3 y_0) \beta }{(2 \sqrt{2 \alpha - 2 \beta } \alpha x_0 - 2 \alpha \beta + \beta ^2)^2},\\ \Delta&=\frac{\sqrt{-2\sqrt{2\alpha - 2\beta }\; \alpha x_0\beta ^2 - 2 x_0^2\alpha ^3 + 2 x_0^2\alpha ^2\beta + \alpha ^2\beta ^2 + \alpha \beta ^3 - \beta ^4}}{ (2\alpha }. \end{aligned}$$

Under the above pair of circles, the limiting points are at:

$$\begin{aligned} l_1=\left[ \frac{R^2 - d^2}{2d}, 0\right] ,\quad l_2= \left[ \frac{d(R^2-d^2)}{R^2 + d^2}, 0\right] . \end{aligned}$$

Appendix E: Limiting Pedal Perimeters for $N=3$ and $N=4$

Below we consider 3- and 4-periodics in the confocal pair where a, b are the semi-axes of the outer ellipse has axes (a, b). Below, set $\delta =\sqrt{a^4-a^2 b^2+b^4}$ and $c^2=a^2-b^2$.

Fig. 5

$N=3$ case: the bicentric family (solid orange) is the poristic family [8]. Its sum of cosines is invariant and equal to those of the two limit point pedals (pink and purple). The Gergonne points $X_{7,1}$ and $X_{7,2}$ of each pedal are stationary. (colour figure online) live

1.1 E.1. $N=3$ case

Referring to Fig. 5, the perimeter $L^\dagger$ of the inversive polygon for the $N=3$ family, originally derived in [15, Prop. 4] is given by:

$$\begin{aligned} L^\dagger =L_+=\rho ^2 \frac{\sqrt{ \left( 8\,{a}^{4}+4\,{a}^{2}{b}^{2}+2\,{b}^{4} \right) \delta +8\,{a}^{6}+3\,{a}^{2}{b}^{4}+2\,{b}^{6}}}{{a}^{2}{b}^{2}}. \end{aligned}$$

By Corollary 1, this is equal to the perimeter $L_-$ of the bicentric pedal with respect to the focal limiting point.

$$\begin{aligned} L_{-}&= {\frac{ \left( 9\,{R}^{2}-{d}^{2} \right) \left( {R}^{2}-{d}^{2} \right) \sqrt{2}\rho ^2}{16\,{R}^{4}d}\sqrt{- \left( {R}^{2}-{d}^{2} \right) ^{{\frac{3}{2}}}\sqrt{9\,{R}^{2}-{d}^{2}}+ 3R^4 + 6R^2d^2 - d^4}},\\ R&=(2a^4 - 2a^2b^2 + b^4 + (2a^2 -b^2)\delta )a\rho ^2/b^6,\quad d=(2a^2 - b^2 + 2\delta )c\rho ^2a^2/b^6. \end{aligned}$$

The perimeter $L_+$ of the bicentric pair with respect to the non-focal limiting point is given by:

$$\begin{aligned} L_{+}&= {\frac{ \left( 9\,{R}^{2}-{d}^{2} \right) \left( {R}^{2}-{d}^{2} \right) \sqrt{2}\rho ^2}{16\,{R}^{4}d}\sqrt{ \left( {R}^{2}-{d}^{2} \right) ^{{\frac{3}{2}}}\sqrt{9\,{R}^{2}-{d}^{2}}+ 3R^4 + 6R^2d^2 - d^4}},\\ R&=(2a^4 - 2a^2b^2 + b^4 + (2a^2 -b^2)\delta )a\rho ^2/b^6,\quad d=(2a^2 - b^2 + 2\delta )c\rho ^2a^2/b^6. \end{aligned}$$

The sum of cosines of a triangle is given by $1+r/R$ and is therefore constant for the $N=3$ bicentric family. Let $\theta '_i$ denote the angles of the bicentric polygon. The sum of its cosines can be derived as:

$$\begin{aligned} \sum \cos \theta ' = 1+\frac{r}{R} = \frac{3R^2 - d^2}{2R^2}. \end{aligned}$$

(10)

Proposition 1

The sum of cosines for the first and second $N=3$ bicentric pedals are constant and identical to (10).

Note: In terms of the associated elliptic billiard parameters, this is given by [15, Prop. 6]:

$$\begin{aligned} \sum \cos {\theta ^\dagger }_{(N=3)}=\frac{\delta (a^2+c^2-\delta )}{a^2c^2}. \end{aligned}$$

Proof

Using CAS, it follows from straightforward calculations with the orbit parametrized in Appendix D. $\square$

The two limiting pedals have stationary Gergonne points $X_7$. The first one was derived in [9, Proposition 1]:

$$\begin{aligned} X_{7,1}= & {} \left[ c\left( 1-\frac{\rho ^2}{\delta +c^2}\right) ,0\right] , \\ X_{7,1}= & {} \frac{ (R^2 - d^2)((R^2 - d^2)^{3/2}\sqrt{9R^2 - d^2} + 3R^4 + 6R^2d^2 - d^4)}{16R^4d}. \end{aligned}$$

Fig. 6

In the $N=4$ case, remarkable things happen: (i) the sum of cosines of the $f_1$-pedal (pink) is not constant; (iii) its perimeter is the same as the corresponding billiard 4-periodic; (iv) the vertices of the $l_2$-pedal (purple) are collinear and (v) the sum of its cosines is 4. (colour figure online) Video

1.2 E.2. $N=4$ case

Referring to Fig. 6, the perimeter $L^\dagger$ of the inversive polygon for billiard 4-periodics was originally derived in [9, Prop. 18]. It is identical to the perimeter of 4-periodics themselves and given by:

$$\begin{aligned} L^\dagger = L_{+,N=4}= \,{\frac{4\sqrt{{a}^{2}+{b}^{2}}}{{b}^{2}}}. \end{aligned}$$

(11)

Proposition 2

In the $N=4$ family,  the vertices of the bicentric pedal with respect to the non-focal limiting point are collinear.

Proof

The polar image of the bicentric family with respect to $\ell _2$ is a pair of confocal hyperbolas, see Appendix B, i.e., the polar image of bicentric 4-periodics is a billiard family. It can be shown its vertices are concyclic with the two hyperbolic foci $f_1',f_2'$, one of which coincides with $\ell _2$. Therefore, the inversion of said vertices with respect to $\ell _2$ is a set of collinear points. $\square$

As before, Eq. (11) is the same as the perimeter of the first bicentric pedal. The perimeter $L_+$ of the non-focal bicentric pedal is given by:

$$\begin{aligned} L_{-,N=4} = \frac{4a^2}{b^2 c}. \end{aligned}$$

Regarding the sum of cosines, it is well known a circle-inscribed quadrilateral has supplementary opposing angles, i.e.:

Observation 1

The sum of cosines of a bicentric $N=4$ family is null.

Since the second bicentric pedal is a degenerate polygon:

Observation 2

The sum of cosines of the second limiting pedal to the $N=4$ bicentric family is equal to 4.