MMore about areas and centers of Poncelet polygons
Ana C. Chavez-CalizApril 14, 2020
Inspired by the work of R. Schwartz and S. Tabachnikov exposed in [13]about the loci of different centers of mass of Poncelet polygons, we studythe locus of another center of mass: the Circumcenter of Mass (studied indetail in [14]).Let P be an oriented n -gon P = ( V , . . . , V n ) where each V i ∈ R . Let A ( P ) denote the algebraic area of P (which is defined by formula (5)). Givenan appropriate triangulation of P , P = (cid:83) T i , let C i and A i be the circum-center and area of T i respectively.Assume A ( P ) is non-zero. The Circumcenter of Mass of P , CCM ( P ) isdefined as the weighted sum of the circumcenters: CCM ( P ) = (cid:88) i A i A ( P ) C i . (1)The Circumcenter of Mass does not depend on the triangulation and itcan be described in terms of the vertices V i = ( x i , y i ) by the formula: CCM ( P ) = 14 A ( P ) (cid:32) n (cid:88) i =1 y i ( x i − + y i − − x i +1 − y i +1 ) , n (cid:88) i =1 − x i ( x i − + y i − − x i +1 − y i +1 ) (cid:33) . (2)For more details and properties of the Circumcenter of Mass, consult [14]. Turn to [14] for more specifics about the conditions of the triangulation. a r X i v : . [ m a t h . M G ] A p r nother center discussed in this paper is CM . If P is a polygon, CM ( P ) is the center of mass of P , considering P as a “homogeneous lam-ina”. Similarly to CCM , if A ( P ) (cid:54) = 0, P = (cid:83) T i is a triangulation, and G i , A i denote the centroid and area of each T i respectively, then CM ( P )is defined as the weighted sum of the centroids: CM ( P ) = (cid:88) i A i A ( P ) G i . (3)If P has vertices V , . . . , V n , with V i = ( x i , y i ), the center of mass CM ( P ) can be expressed in terms of the coordinates of the vertices: CM ( P ) = 16 A ( P ) n (cid:88) i =1 ( x i y i +1 − x i +1 y i )( x i + x i +1 , y i + y i +1 ) . (4)showing that CM also does not depend on the triangulation.Paper [13] studies the locus of CM , however, there is a gap within theproof of the result. Proposition 2 of this paper recognizes this.Before continuing, let us recall Poncelet closure theorem: Theorem 1 ( Poncelet Porism ) . Let γ, Γ be two nested conics in R , with γ in the interior of Γ . Let W ∈ Γ , and (cid:96) be one of the tangents to γ passingthrough W : this line intersects Γ in a new point that we call W . Repeatthis construction, now starting from W , and considering the other tangent (cid:96) to γ passing through W to find W ; and so on.Suppose that for this particular choice of W , there is an n ∈ N such that W n +1 = W .Then, for any other V ∈ Γ if V , V , . . . ∈ Γ are obtained by the same con-struction described above, it happens that V n +1 = V . See Figure 1.The polygons with vertices ( V , . . . , V n ) inscribed in Γ and circumscribedabout γ are called Poncelet polygons . Significant references for this classic result are [2] and [5].Using the background set by Griffiths and Harris in [8], and Liouville’stheorem, we show the following result:2igure 1: Poncelet Theorem, with n = 5. Theorem 2.
Let γ, Γ be a pair of conics that admit a -parameter familyof Poncelet n -gons P t . The locus of the Circumcenter of Mass CCM ( P t ) isalso a conic. The same techniques and tools provide a proof of a result found previ-ously by Dan Reznik (shared via personal communication with S. Tabach-nikov):
Theorem 3.
Let γ , Γ be two concentric ellipses in general position admit-ting a -parameter family of Poncelet n -gons. Given P = ( V , . . . , V n ) oneof these Poncelet n -gons, let Q be a new polygon formed by the tangent linesto Γ at V i .If n is even, then A ( P ) · A ( Q ) stays constant within the Poncelet family. One can find several papers with results that share this flavor, like theones exposed in [1], [3], [4], [6], [7], [10], [11] and [12] to mention some.
Following the spirit of [8], [12], and [13], we complexify and projectivize thepicture. That is, even though the objects we work with are initially 2-realdimension objects, we allow the coordinates to take complex values and addthe points at infinity; so from now on we will be working in CP := C ∪ L ∞ .3igure 2: The conics γ and Γ are not in general position, since both tangents T p γ and T p Γ coincide.
Definition.
Let P be an oriented n − gon with vertices V , . . . , V n ∈ C ,where V i = ( x i , y i ) . The area of P is A ( P ) = 12 n (cid:88) i =1 ( x i y i +1 − x i +1 y i ) . (5) By convention, we set x n +1 := x , y n +1 := y , and in general, subindicesare taken modulo n . Notice the area of a polygon is invariant under equiaffine transformationsbut not under projective transformations.Recall that, if Γ ⊂ CP is a C curve and p ∈ Γ then T p Γ denotes theline tangent to Γ that passes through p . Similarly T Γ = { T p Γ : p ∈ Γ } .Given two conics γ , Γ ⊂ CP we say that γ, Γ are in general position ifthey intersect transversally, i.e. at every point in the intersection p ∈ γ ∩ Γ,we have T p γ ∩ T p Γ = { p } . See Figure 2 for a counterexample. Definition.
Let γ, Γ ⊂ CP be a pair of conics that admit a -parameterfamily of Poncelet n -gons. A Poncelet polygon of this family is called degen-erate if at least one of the vertices V i satisfy either V i ∈ γ ∩ Γ or T V i Γ ∈ T γ .This is illustrated in Figure 3.Otherwise, a Poncelet polygon is called non-degenerate . γ, Γ admit a 1-parameter family of Poncelet hexagons.The degenerate polygon P illustrates the case when one vertex belongs tothe intersection γ ∩ Γ, while the degenerate polygon P corresponds to thecase T V i Γ ∈ T γ .Notice that if P is a degenerate polygon then: • If V i ∈ γ ∩ Γ, then V i − k = V i + k for all k . Intuitively, the polygon“bends” at V i . • If T V i Γ ∈ T γ , then V i = V i +1 , and V i − k = V i +1+ k for all k . The vertices V i and V i +1 “glue together”.Figure 4 shows a picture of Poncelet polygons that “are about to collapse”.From this remark, we get the following observations regarding degeneratepolygons. Lemma 1. If γ , Γ ⊂ CP admit a -parameter family of Poncelet n -gons P t inscribed in Γ and circumscribed about γ , then1. If P is a degenerate Poncelet polygon, with finite vertices, then A ( P ) =0 .2. If γ, Γ are conics in general position, then there are n different de-generate polygons. Note:
The order of the label in the vertices matter: that is, if P =( V , . . . , V n ) and P (cid:48) = ( V (cid:48) , . . . , V (cid:48) n ) satisfy V i = V (cid:48) i + c for all i , with c con-5igure 4: A Poncelet polygon can degenerate if either two sides coincide orif two vertices coincide. P is “bending” and P has a pair of vertices thatare “gluing together”.stant, then we consider P and P (cid:48) to be equal if and only if c ≡ n ) . Proof.
The first observation is straightforward.For the second part, we divide in two cases: • n even. Recall P = ( V , . . . , V n ) is degenerate if one of the followinghappens: – V i ∈ γ ∩ Γ for some index i . Since V i − k = V i + k , in particular one has V i + n − = V i − n +1 = V i + n +1 , (the last equality comes from the congruence mod n in the in-dices), which means V i + n has a unique tangent to γ , thus V i + n ∈ γ ∩ Γ.If P is one of these degenerate polygons, then it has n/ γ ∩ Γ and the rest repeatingtwice. In total, there are n different manners one can list thevertices of this polygon in the correct order: if one starts from avertex in the intersection, there is just one way to list the rest;but if one starts with any double vertex V i , then the next vertex6an be either V i +1 or V i − .Since γ, Γ are conics in general position, by B´ezout’s theorem, | γ ∩ Γ | = 4. Therefore we have 2 n of these degenerate polygons. – T V i Γ ∈ T γ for some i . Then V i = V i +1 , and V i − k = V i +1+ k . In particular V i − n = V i + n +1 = V i − n +1 , meaning T V i − n Γ ∈ T γ .As before, if P is one of these degenerate polygons, it has n/ T V i Γ ∈ T γ . Similarly, there are n differ-ent ways one can list the vertices of P in the correct order.In the dual space, (the space of lines) T γ and T Γ are conics aswell, so | T γ ∩ T Γ | = 4. There are then 2 n degenerate polygonsof this type.In the case where γ and Γ are in general position, these two casesare disjoint, and so we have 4 n degenerate Poncelet polygons intotal. • n odd. Applying a similar analysis to the previous case, a polygon P = ( V , . . . V n ) is degenerate if and only if exactly one vertex V i belongs to the intersection γ ∩ Γ and T V i + n − Γ ∈ T γ (see Figure 5).Then there are 4 n degenerate polygons. Theorem 4.
Let γ, Γ ⊂ CP be a pair of conics in general position thatadmit a -parameter family of Poncelet n − gons, with vertices in Γ and sidestangent to γ . Suppose there is a non-degenerate Poncelet polygon P such that A ( P ) = 0 . Then every Poncelet polygon of the family has area zero. Before going over the proof, let us examine the particular case when n = 4. Lemma 2.
A quadrilateral Q = ( V , V , V , V ) has area zero if and only ifthe diagonals V V and V V are parallel. n is odd (in this case n = 5), then each degenerate Ponceletpolygon has one vertex where the polygon “bends” (when V i ∈ γ ∩ Γ) andtwo vertices that “glue together” (if T V j Γ ∈ T γ , then V j = V j +1 ). Proof.
Notice A ( V , V , V , V ) = A ( V , V , V )+ A ( V , V , V ) = A ( V , V , V ) − A ( V , V , V ).The areas A ( V , V , V ) and A ( V , V , V ) are equal if and only if V V isparallel to V V . This proves the result. Proposition 1.
Let γ and Γ be conics admitting a -parameter family ofPoncelet quadrilaterals P t . If there is one non-degenerate Poncelet quadri-lateral with area zero, then all the Poncelet quadrilaterals have area zero.Moreover, the locus of each CCM ( P t ) and CM ( P t ) is a point at infinity,with CM ( P t ) = { [ a : b : 0] } and CCM ( P t ) = { [ − b : a : 0] } , for some a, b ∈ C not simultaneously .Proof. Recall that, if γ, Γ admit a 1-parameter family of Poncelet quadri-laterals P t , then the diagonals of P t for any t , intersect at a fixed point.This can be proved by taking a projective transformation that makes Γ and γ concentric and using the fact that the Poncelet map commutes with thereflection with respect to the center of γ and Γ.If one non-degenerate Poncelet quadrilateral has area zero, by Lemma 2the diagonals must be parallel and so they intersect at some point at infinity,8igure 6: If the diagonals of one non-degenerate Poncelet quadrilateral areparallel, then the diagonals of any other Poncelet quadrilateral in the familyare also parallel.say [ a : b : 0]. The diagonals of any other Poncelet quadrilateral then alsointersect at [ a : b : 0] and hence are parallel (see Figure 6).Recall that if P is a quadrilateral then CCM ( P ) lies in the intersection ofthe perpendicular bisectors of the diagonals. Therefore CCM ( P t ) = { [ − b : a : 0] } . Similarly CM ( P ) lies in the intersection of the line connectingthe centroids of V V V and V V V , and the line connecting the centroidsof V V V and V V V . In this case, these lines are both parallel to thediagonals, and so CM ( P t ) = { [ a : b : 0] } . Proof. (of Theorem 4)
The framework of Griffiths and Harris exposed in [8]constitute the foundations of this proof. Papers like [13] make use of theirapproach to prove similar results.Recall that any conic in CP is isomorphic to ˆ C via stereographic projectionfrom any point.Consider the set of flags E = { ( p, (cid:96) ) : p ∈ Γ , p ∈ (cid:96), (cid:96) ∈ T γ } . The projection π : E →
Γ given by π ( p, (cid:96) ) = p is a two-to-one map, with fourbranch points: ( q, (cid:96) ) where q ∈ γ ∩ Γ. Computing the Euler characteristic,one obtains that E is topologically a torus. Moreover E ⊂ Γ × T γ is anelliptic curve. 9igure 7: σ ( p, (cid:96) ) = ( p (cid:48) , (cid:96) ), and τ ( p (cid:48) , (cid:96) ) = ( p (cid:48) , (cid:96) (cid:48) ).For the Poncelet map, consider the involutions σ ( p, (cid:96) ) = ( p (cid:48) , (cid:96) ) , τ ( p (cid:48) , (cid:96) ) = ( p (cid:48) , (cid:96) (cid:48) ) . See Figure 7.The Poncelet map is defined as T = τ ◦ σ . Since τ and σ are involutions, E admits a parameter t in which the Poncelet map is a translation of thetorus, T ( t ) = t + c , with c some constant.Now, let’s examine the area function for Poncelet polygons. Notice thateach flag ( p, (cid:96) ) gives rise to a unique Poncelet polygon with orientation.Also, the group generated by σ and τ is precisely the dihedral group D n .If Z n ⊂ D n is the cyclic group of order n , observe that A : E → ˆ C is a Z n -invariant meromorphic function, and since it is defined over a torus, A is an elliptic function.For conics γ, Γ in general position, the area has a simple pole exactlywhen one of the vertices goes to infinity. Then, A is an elliptic function oforder 4 n , with 2 n = | D n | simple poles coming from each one of the pointsof Γ at L ∞ . The poles are simple because γ and Γ are in general position.By Lemma 1, A has at least 4 n zeroes. If there was at least one non-degenerate Poncelet polygon with area zero, then this would contradict thefact that every non-constant elliptic function of order m has exactly m zeros.This implies that A must be constant.10 Limit of
C C M and
C M for Poncelet polygons From now on, we focus only on the case when not all Poncelet polygons havearea zero.Consider P = ( V , . . . , V n ) a degenerate polygon with finite vertices.Thanks to Theorem 4 we know we can find a 1-parameter family of non-zero-area Poncelet polygons P t = ( W ( t ) , . . . , W n ( t )) approaching to P as t goes to zero. Since the area of each P t is not zero, the Center of Mass CM ( P t ) and the Circumcenter of Mass CCM ( P t ) are well defined.Our goal for this section is to show that the limitslim t → CM ( P t ) and lim t → CCM ( P t ) , exist.The way the vertices W i ( t ) approach to V i depends on the parity of n and V i itself:(a) Around “bending” vertices: If V i ∈ γ ∩ Γ, then W i +1 ( t ) and W i − ( t )approach to V i +1 , as shown in Figure 8: • W i ( t ) = V i + −→ k i t + O ( t ), • W i +1 ( t ) = V i +1 + −→ k i +1 t + O ( t ) and • W i − ( t ) = V i +1 + λ i − −→ k i +1 t + O ( t ) with λ i − (cid:54) = 1 (since W i +1 ( t ) (cid:54) = W i − ( t ) for all t (cid:54) = 0).Figure 8: Behavior around V i ∈ γ ∩ Γ.11b)
Around “gluing” vertices: If T V i Γ ∈ T γ , and V i = V i +1 then thevertices W i ( t ) , W i +1 ( t ) are of the form: • W i ( t ) = V i + −→ k i t + O ( t ) and • W i +1 ( t ) = V i + λ i −→ k i t + O ( t ) with λ i (cid:54) = 1 (since W i ( t ) (cid:54) = W i +1 ( t )for all t (cid:54) = 0).Figure 9 illustrates the situation.Figure 9: Behavior around V i , when T V i Γ ∈ T γ .(c)
For the rest of the vertices: If V i = V j for some i (cid:54) = j non-consecutive,then V i +1 = V j − .Hence: • W i ( t ) = V i + −→ k i t + O ( t ), • W j ( t ) = V i + λ i −→ k i t + O ( t ) with λ i (cid:54) = 1, • W i +1 ( t ) = V i +1 + −→ k i +1 t + O ( t ) and • W j − ( t ) = V i +1 + λ i +1 −→ k i +1 t + O ( t ) with λ i +1 (cid:54) = 1.Figure 10: Behavior for the rest of the vertices of P t .12igure 11: Triangulation of P t . The figure in the left shows the triangulationaround vertices in the intersection γ ∩ Γ. The figure in the right shows thetriangulation for the rest of the vertices.
Proposition 2.
Let γ, Γ be a pair of conics that admit a -parameter familyof Poncelet polygons with area not constant zero. If P is a degenerate polygonwith finite vertices, and P t a -parameter family of non-degenerate Ponceletpolygons such that lim t → P t = P then lim t → CCM ( P t ) and lim t → CM ( P t ) exist.Proof. Regardless of the parity of n , we can triangulate the polygons P t intotriangles ( W j ( t ) , W j ( t ) , W j ( t )) such thatlim t → W j ( t ) = lim t → W j (cid:48) ( t ) = V i for exactly two indices j, j (cid:48) ∈ { j , j , j } and for some vertex V i of P .(a) If V i ∈ γ ∩ Γ, take ( W i − ( t ) , W i ( t ) , W i +1 ( t )) as part of the triangula-tion. This is shown in Figure 11 (left).(b) If V i = V j for some i ≤ j , add the triangles ( W i ( t ) , W i +1 ( t ) , W j ( t ))and ( W j − ( t ) , W j ( t ) , W i +1 ( t )) to the triangulation. This is illustratedin Figure 11 (right). 13 laim:
1. The area of each triangle tends to zero linearly with respect to t .2. The limit of the circumcenter of each triangle exists.3. The limit of the centroid of each triangle exists.Before proving the claim, notice that if it holds, then we are done. Con-sider the triangulation { T ( t ) , . . . T n − ( t ) } of P t described above. If the areaof each triangle A i ( t ) tends to zero linearly with respect to t and the area of P t is not zero, then the area A ( P t ) also tends to zero linearly with respectto t , and so lim t → A i ( t ) A ( P t ) exists.Denote by C i ( t ) the circumcenter of T i ( t ). Then the second part of the claimsecures that lim t → C i ( t ) exists for each subindex i . Hence,lim t → n − (cid:88) i =1 A i ( t ) A ( P t ) C i ( t )exists.Similarly, if G i ( t ) denotes the centroid of T i ( t ), then the third part ofthe claim ensures that lim t → G i ( t ) exists for all i . Thus,lim t → n − (cid:88) i =1 A i ( t ) A ( P t ) G i ( t )also exists. Proof of the claim:
Consider one triangle T ( t ) = ( W j ( t ) , W j ( t ) , W j ( t )) of this triangulation.Without lost of generality, supposelim t → W j ( t ) = lim t → W j ( t ) = V i . That is: • W j ( t ) = V i + −→ k i t + O ( t ) , • W j ( t ) = V i + λ −→ k i t + O ( t ) , • W j ( t ) = V i ± + −→ k i ± t + O ( t ) . λ (cid:54) = 1, and −→ k i is a vector tangent to Γ at V i , as shown in Figure12:Figure 12: The triangle ( W j ( t ) , W j ( t ) , W j ( t )) is part of the triangulation.Notice the vector V i − V i ± (cid:54) = 0 is not parallel to T V i Γ.Given V = ( x, y ) ∈ C , let V ⊥ := ( − y, x ). If (cid:104) , (cid:105) denotes the usual dotproduct, then the area of T ( t ) is: A ( T ( t )) = ( λ − (cid:104) V i − V i ± , −→ k ⊥ i (cid:105) t + O ( t ) . Recall λ (cid:54) = 1, and (cid:104) W, −→ k i ⊥ (cid:105) = 0 if and only if W ∈ T V i Γ, which is notthe case for V i − V i ± .For the second part, recall that the vertices of each triangle T i ( t ) alllie on Γ, which is a conic. Therefore none of the angles of any T i tend to π , and hence the circumcenter C i ( t ) of each triangle in the limit is also finite.Finally, for the third part we can even give precise coordinates of thecentroid of a degenerate triangle T i = ( V i , V i , V i ± ): CM ( T i ) = 2 V i + V i ± . This is because the centroid of a non-degenerate triangle is obtained by theintersection of the medians. Recall the centroid divides every median inratio 2 : 1. This provides the formula given above.15
Locus of the Circumcenter of Mass for PonceletPolygons
Theorem 5.
Let γ, Γ be a pair of conics in general position that admit a -parameter family of Poncelet n -gons P t , with not-constant zero area. Thenthe locus CCM ( P t ) is also a conic.Proof. Recall that Γ intersects the line at infinity in two different points,say L and M .Observe that the x and y coordinates of CCM on E are meromorphic D n -invariant functions, with 4 n simple poles: 2 n = | D n | of them comingfrom each one of the flags that give rise to the polygon with a vertex in L ,and 2 n more coming from the polygon with one of the vertices in M .Say P is a Poncelet polygon with a vertex at infinity. Denote by L + , L − and M + , M − the adjacent vertices to L and M respectively. Then CCM ( P )coincides with the circumcenter of ( L − , L, L + ) (in case L is a vertex of P )or the circumcenter of ( M − , M, M + ) (if M is a vertex of P ), since this isthe term that overpowers in formula (1). The circumcenter of ( L − , L, L + )coincides the intersection of the perpendicular bisector of L − L + at infinity,and similarly for the circumcenter of ( M − , M, M + ).Let z be a local holomorphic parameter of E at L . Then x ( z ) = u z + u + . . . , y ( z ) = u z + u + . . . , where [ u : u : 0] is the intersection of the perpendicular bisector of L − L + at infinity.Similarly, if w is a local holomorphic parameter of E at M , then x ( w ) = v w + v + . . . , y ( w ) = v w + v + . . . , where [ v : v : 0] is the intersection of the perpendicular bisector of M − M + at infinity.Consider f ( x, y ) = Ax + By + Cxy + Dx + Ey . We want to find A, B, C, D, E ∈ C such that f has no poles at L and M .Expanding f in the local parameters z and w for L and M :16 ( x ( z ) , y ( z )) = 1 z (cid:0) u A + u B + u u C (cid:1) +1 z [2 u u A + 2 u u B + ( u u + u u ) C + u D + u E ] + . . .f ( x ( w ) , y ( w )) = 1 w (cid:0) v A + v B + v v C (cid:1) +1 w [2 v v A + 2 v v B + ( v v + v v ) C + v D + v E ] + . . . So we are looking for
A, B, C, D, E non-trivial solution to the system oflinear equations u A + u B + u u C = 0 . u u A + 2 u u B + ( u u + u u ) C + u D + u E = 0 .v A + v B + v v C = 0 . v v A + 2 v v B + ( v v + v v ) C + v D + v E = 0 . (6)Recall u i , v i are constants that depend on γ , Γ and the relative positionof γ with respect to Γ.Since we have more variables than equations, the system is consistentand the solution set is at least 1-dimensional.If A, B, C, D, E is a non-trivial solution of (6), then f ( x, y ) = Ax + By + Cxy + Dx + Ey has no poles on E , meaning this function is constant, and so the coordinates x ( t ) , y ( t ) of CCM ( P t ) satisfy Ax ( t ) + By ( t ) + Cx ( t ) y ( t ) + Dx ( t ) + Ey ( t ) = F, for some F ∈ C . Hence, the locus of CCM ( P t ) is a conic.17igure 13: The ellipses γ , Γ admit in this case a 1-parameter family ofPoncelet hexagons. P has vertices in Γ and sides tangent to γ , while Q hassides tangent to Γ. Through numerical experiments, Dan Reznik has found several invariantsrelated to billiard trajectories. There is one of them with its proof:
Theorem 6.
Let γ , Γ be two concentric ellipses in general position admit-ting a -parameter family of Poncelet n -gons. Given P = ( V , . . . , V n ) oneof these Poncelet n -gons, let Q be a new polygon formed by the tangent linesto Γ at V i (see Figure 13). If n is even, then A ( P ) · A ( Q ) stays constantwithin the Poncelet family.Proof. By applying an equiaffine transformation, we can assume γ is a cir-cle, with both γ and Γ centered at the origin. As before, the product of theareas can be thought of as a meromorphic function g : E → ˆ C . It would beenough to check that g has no poles to conclude that g must be constant.Notice g may have a pole if one of the polygons P or Q has a vertex atinfinity.If Q had a vertex at infinity, this would imply that T V i Γ and T V i +1 Γare parallel. This happens only when − V i = V i +1 . Since n is even, andthe Poncelet map commutes with the reflection in the origin, one has that − V i = V i + n . So, V i +1 = V i + n : this is possible only when n = 4, V = V , V = V . Notice that V i ∈ γ ∩ Γ for all i , thus P is a degenerate polygon,18ith finite vertices. In this case, A has a simple pole at Q , but a simple zeroat P . Then g has no pole in this case.If Q has only finite vertices and P has a point at infinity (say V n ), thenthe lines connecting V n with V and V n with V n − are parallel and by sym-metry with respect to the reflection in the origin, V = − V n − .By symmetry with respect to the reflection in the origin, one also hasthat V i = − V n − i , for all i , giving a pairing between finite vertices. Since n iseven, the only possibility is that P is a degenerate polygon, with V n/ = V n and • V n/ , V n/ ∈ γ ∩ Γ if n ≡ • V n − = V n +24 , V n +24 = V n − and hence their tangent to Γ is alsotangent to γ , if n ≡ A has a simple pole at P , but because P is degenerate, Q is alsodegenerate (with finite vertices) and so, A has a simple zero at Q . Hence g has no pole at P and therefore is constant. The centers CM , CM and CCM still make sense in different geometries(for instance, [14] goes over the definition and properties of
CCM in spher-ical and hyperbolic geometry). One may feel tempted to look into sphericaland hyperbolic geometry and see if any of these theorems still hold. Unfor-tunately, the scenario doesn’t look very promising.For instance, consider S . As described in [9], a spherical conic is theintersection of the sphere S with a quadratic cone. It doesn’t make anyharm to assume that all our configuration of spherical conics and polygonsare contained in the northern hemisphere: S := { ( x, y, z ) : x + y + z = 1 , z > } . If π : S → { ( x, y,
1) : x, y ∈ R } ∼ = R is the central projection, notice π andits inverse π − preserve conics and geodesics. Hence the Poncelet porismstill holds in the spherical case. 19igure 14: The locus of the center of mass of Poncelet triangles is not aconic.Suppose γ, Γ ⊂ S are two conics that admit a 1-parameter family ofPoncelet triangles. For the specific case of triangles, it happens that theaverage of the vertices (projected into the sphere) coincides with the inter-section of the medians. That is, if T = ( V , V , V ) with V i ∈ S is a sphericaltriangle, and CM ( T ) ∈ S is the intersection of the medians of T , then: CM ( T ) = V + V + V || V + V + V || = CM ( T ) . For this specific case of triangles, denote CM ( T ) := CM ( T ) = CM ( T ).Figure 14 shows the projection of a configuration of conics γ, Γ ⊂ S that admit a 1-parameter family of Poncelet triangles T t ⊂ S . One canobserve π [ CM ( T t )] is not a conic, so CM ( T t ) ⊂ S is also not a conic. This paper wouldn’t be possible without the guidance received from S.Tabachnikov, who suggested the topic in the first place. I’m deeply gratefulfor all the stimulating discussions and all the insightful comments and cor-rections.Many thanks to A. Akopyan, D. Rudenko and M. Ghomi for the valuablesuggestions and to R. Schwartz, whose work and ideas inspired this paper.I also acknowledge the support received from the U.S. National ScienceFoundation through the grant DMS-1510055.20 eferences [1] A. Akopyan, R. Schwartz, and S. Tabachnikov. Billiards in ellipsesrevisited. arXiv preprint arXiv:2001.02934 , 2020.[2] M. Berger.
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