Bar Billiards and Poncelet's Porism
WITOLDMOZGAWA(*)
ABSTRACT- Weprovethat for a given ovalCthere exist ovalsCinandCout, insideand outsideof C, such that thepairs (C, Cin) and (Cout,C) satisfy the Poncelet's porism for almost any number of reflections in their bar billiards.
1. Introduction.
Let us consider two ovalsC1 andC2, i.e., two smooth strictly convex closed plane curves. Let the ovalC2lieinsidethesecond ovalC1. From any point onC1, draw a tangent to C2 and extend it toC1 in theoppositedi- rection. From this point we draw another tangent, etc. For all tangents, the resulting Poncelet's transverse will be called a bar billiard since it is similar to a traditional game played years ago. In general in this game players scored points by knocking balls into the holes while avoiding toppling a skittlein themiddleof thetable. Heretheroleof theskittleis played by theovalC2 and it must be ``toppled'' by the tangent line. The general be- havior of such billiards is worth investigating but in this paper we will concentrate on the Poncelet's porism for this specific setting. Namely, we say that a bar billiard has a Poncelet's porism property if the following is true: if, on the oval C1, there is one point of origin for which a Poncelet transverse is closed, then the transverse will also close for any other point of origin on the oval. One can find an extensive bibliography on Poncelet's porism in [4]; a nice introduction to theory of billiards is provided in [3].
In this paper we prove that for a given ovalC, in a suitably smallC2- neighborhood there exist ovalsCinandCoutinsideand outsideofC, such that for a suitable large number of segments the pairs (C;Cin) and (Cout;C)
(*)Indirizzo dell'A.:InstytutMatematyki, UniwersytetMarii Curie-Sklodowskiej,6 pl. Marii Curie-Sklodowskiej 1, 20-031 Lublin, Poland6
E-mail: [email protected]
1991 Mathematics Subject Classification: 53A04, 51M15, 37E99.
satisfy Poncelet's porism in the above sense. In the course of the proof we bring into play the isoptics introduced and developed in [1] and [2].
2. Outer oval.
Let us begin with a simple algebraic lemma
LEMMA2.1. Let v;w2R2andhv;wiand[v;w]denote the dot product and the determinant of these vectors. Then
(1) vhv;wi whw;vi [v;w]J(w) (2) hv;wi [J(w);v],
where J is the positive rotation of anglep=2.
Next we consider an operator D
Daof theform Dp(t)
Da p(ta) p(t) cosa sina
acting on real functions defined on allR. From the elementary calculus it is clear that, for anyC1-functionp,
lima!0
Dp(t) Da p0(t) for anyt2R.
In what follows we give a preview of certain facts concerning isoptics as thefundamental tool in our constructions.
DEFINITION2.1. A plane, closed, simple, positively oriented, curve of positive curvature is called an oval.
Wetakea coordinatesystem with originOin theinterior ofC. Le tp(t), t2[0;2p], bethedistancefromOto thesupport linel(t) ofCperpendicular to thevectoreit costisint. It is well-known thatp(t) is of classC1and that theparametrization ofCin terms ofp(t) is given by the formula
z(t)p(t)eitp0(t)ieit; (2:1)
where ieit sinticost. Notethat thesupport functionp can beex- tended to a periodic function onRwith theperiod 2p.
DEFINITION 2.2. Let Ca be a locus of vertices of a fixed angle p a formed by two support lines of the oval C. The curve Cawill be called ana- isoptic of C.
It is convenient to parametrize thea-isopticCa by thesameangletso that theequation ofCatakes the form
za(t)p(t)eitDp(t) Da ieit: (2:2)
Since the functions involved are at least of classC1 weobservethat the given parametrization of an isoptic is of classC1. It can beshown that it is a regular curve. Note that
za(t)z(t)l(t;a)ieitz(ta)m(t;a)iei(ta) (2:3)
z0a(t) l(t;a)eitr(t;a)ieit (2:4)
for suitablefunctionsl,mandr. Moreover, we have l(t;a)Dp(t)
Da p0(t);
(2:5)
r(t;a)p(t)Dp0(t) Da : (2:6)
COROLLARY2.1. With the notations above we have 1) lim
a!0za(t)z(t);
2) lim
a!0l(t;a)0;
3) lim
a!0r(t;a)R(t), where R(t)p(t)p00(t) is the curvature ra- dius of the oval C at t.
LEMMA2.2. The function
ka(t) ka(t); 0<a<p k(t); a0
(2:7)
is continuous from the right ona, whereka(t)is the curvature of the isoptic za(t)at t.
PROOF. It follows immediately from the equation of an isoptic that z0a(t) l(t;a)eitr(t;a)ieit
(2:8)
and
jz0a(t)j2l2(t;a)r2(t;a):
(2:9)
The latter formula ensures that the isoptics of ovals are always regular curves. By the above considerations we get
a!0limz0a(t)lim
a!0 p0(t) Dp(t) Da
eit p(t)Dp0 Da
ieit
(2:10)
p(t)p00(t)ieitz0(t) and
lima!0jz0a(t)j2 lim
a!0 p0(t) Dp(t) Da
2
p(t)Dp0 Da
2
" #
(2:11)
p(t)p00(t)2 jz0(t)j2: Theformula for curvatureka(t) ([2]) of an isoptic curvecan betransformed to thefollowing
ka(t) 1 l2r2
32 2l22r2 R(t)rlR(t) cota lR(ta) sina
(2:12)
1
l2r2
32 2l22r2 R(t)r lDR(t) Da
:
Hence,
lima!0ka(t) 1
R(t)k(t):
(2:13)
p COROLLARY 2.2. There exists e>0 such that if 0<a<e then ka(t)>0for any t2R.
THEOREM2.1. For any oval C there exists an oval O containing oval C inside such that the corresponding bar billiard has the Poncelet's porism property.
PROOF. Le ta2p
n bean anglewith 0<a<efor theabovee. Then we can consider a bar billiard such that the starting oval isC2and itsa-isoptic isC1. In this case, for anytPoncelet's transverses close afternreflections and form ann-gon with or without self intersections. p
3. Inner oval.
In this section we consider the reverse problem: for a given ovalC, does there exist an oval Oinside Csuch that for a corresponding bar billiard Poncelet's porism holds?
First, weassumethat theovalCis parametrized as in formula (2.1) and that the origin of coordinates is placed at the Steiner point of C. Fur- thermore, the coordinate system is chosen in such a way that the tangent lineat z(0) is perpendicular to the x-axis. Then, in particular, we have z(0)(a;0) for some a>0 and p0(0)0. If weintroducethefollowing notation
q(t)(q1(t);q2(t))z(t) z(ta) (3:1)
then the linel(t) through thepointsz(t) andz(ta) is given by l(t): yq1(t)xq2(t) [z(t);q(t)]0:
(3:2)
Hence, we easily calculateh(t), its distancefrom theoriginO(0;0) h(t) [z(t);q(t)]
jq(t)j : (3:3)
LetW(t) be the angle contained between the positive direction of the axisOx and thevectorq(t), andx(t) be the angle between the positive direction ofOx andOO0, the shortest segment joiningOwith theline(3.2). Evidently, we havetherelation
x(t)p 2W(t):
(3:4)
Notethat thefollowing formulas aretrueforW(t) cosW(t)hz(0);q(t)i
jq(t)kz(0)j (3:5)
sinW(t)[z(0);q(t)]
jq(t)kz(0)j: (3:6)
Differentiating both sides of formula (3.5) we obtain W0(t)[q(t);q0(t)]
jq(t)j2 (3:7)
and, after some further manipulations we get W0(t)l(t;a)R(ta) m(t;a)R(t)
sina(l2(t;a)r2(t;a)) >0;
(3:8)
wherel(t;a) andm(t;a) are the same as in formulas (2.3) and (2.4). Moreover from the geometric interpretation of these functions, we see that they are nonzero and thatm<0. Thus, it follows from elementary calculus that there exists an inverse functionc(x)W 1 x p
2
t. This function allows us to define a new support function
P(x)h(c(x)):
(3:9)
NotethatjOO0j P(x) and recall thatxis the angle between the positive direction of axisOxandOO!0
. This construction will be used to find an en- velopezizoof thefamily of lines (3.2) in theform
zizo(x)P(x)eixdP(x) dx ieix: (3:10)
After some computations we have dP(x)
dx dh
dt(c(x)) dc(x)
dx [z0;q]hq;qi hq;zi[q;q0] jqj[q;q0] (c(x)):
(3:11)
In this frameworktis a generic parameter and we may consider an equation of the envelope in the form
zizo(t)
P p 2W(t)
ei p2W(t) dP dx
p 2W(t)
iei p2W(t) (3:12)
P(t)EitP0x(t)iEit; where
Eit [z(0);q(t)]
jq(t)kz(0)j;hz(0);q(t)i jq(t)kz(0)j
; (3:13)
iEit hz(0);q(t)i
jq(t)kz(0)j;[z(0);q(t)]
jq(t)kz(0)j
: (3:14)
Ultimately, the above considerations lead to zizo(t) [z(t);q(t)]
jq(t)j Eit (3:15)
[z0(t);q(t)]hq(t);q(t)i hq(t);z(t)i[q(t);q0(t)]
jq(t)j[q(t);q0(t)] iEit; wherez0(t)dz
dt(t) andq0(t)dq
dt(t). Let us make the convention that in the reminder of this paper prime will denote differentiation with respect tot.
By taking the derivative, we obtain
LEMMA3.1.
Eit0
iEit[q(t);q0(t)]
jq(t)j2 ; (3:16)
iEit0
Eit[q(t);q0(t)]
jq(t)j2 : (3:17)
THEOREM3.1. The curve zizo(t)is the closed envelope of the family of linesfl(t): t2Rg.
PROOF. By lemma (3.1) it follows that z0izo(t) P(t)[q(t);q0(t)]
jq(t)j2 dP dx(t) 0!
iEit (3:18)
and from (3.14) weobtain that
iEita q(t) jq(t)j: (3:19)
This means that the tangent vector toz0izo(t) is linearly dependent onq(t).
Note, that we don't claim that it is a nonzero vector. p We do not yet know whetherzizo(t)!z(t) whena!0. To this end we prove
LEMMA3.2.
a!0limzizo(t)z(t):
(3:20)
PROOF. Write [z(0);q(t)]
jq(t)kz(0)j(rsina lcosa) sint(lsinarcosa) cost l2r2
q ;
(3:21)
hz(0);q(t)i
jq(t)kz(0)j (rsina lcosa) cost(lsinarcosa) cost l2r2
q ;
(3:22)
wherell(t;a) andrr(t;a). By passing to thelimit, weobtain lima!0Eiteit;
(3:23)
lima!0iEitieit: (3:24)
Similarly, since q(t)
jq(t)j(rsina lcosa)eit(lsinarcosa)ieit l2r2
q ;
(3:25)
weget
lima!0P(t) lim
a!0
[z(t);q(t)]
jq(t)j p(t):
(3:26)
When we analyze formula (3.25) we easily see that
a!0lim q(t)
sina R(t)ieit; (3:27)
a!0lim q0(t)
sinaR(t)eit R0(t)ieit (3:28)
Theaboveformulas and theformula (3.15) give
a!0lim dP
dx(t)p0(t);
(3:29)
which implies that
lima!0zizo(t)z(t):
(3:30)
p Our next task is to prove
a!0limz0izo(t)z0(t):
(3:31) Recall that
dP
dx(t)h0(t) jq(t)j2 [q(t);q0(t)]; (3:32)
so that
z0izo(t) h(t)[q(t);q0(t)]
jq(t)j2 h0(t) jq(t)j2 [q(t);q0(t)]
!0! iEit: (3:33)
In order to determine lim
a!0z0izo(t) we need only two facts:
lima!0
[q(t);q0(t)]
jq(t)j2 1;
(3:34)
lima!0
q0(t)
sina2R0(t)eit(R(t) R00(t))ieit: (3:35)
Hence, after some tedious computations we get
a!0lim h0(t) jq(t)j2 [q(t);q0(t)]
!0
p00(t);
(3:36) that implies
LEMMA3.3. We have
lima!0z0izo(t)z0(t):
(3:37)
In particular, the envelope which can be called an inner isoptic is a regular curve ifais sufficiently small.
Notethat for somevalues ofathe inner isoptic can develop cusps.
Our last problem before formulating the final theorem is to find lima!0z00izo(t).
LEMMA3.4. Under the above notations
lima!0z00izo(t)R0(t)ieit R(t)eitz00(t):
(3:38)
In particular, the function
kizo(t) kizo(t); 0<a<p;
k(t); a0;
(3:39)
wherekizo(t)denotes the curvature of the inner isoptic zizo(t), is continuous from the right.
PROOF. Write
z00izo(t) h0(t)[q(t);q0(t)]
jq(t)j2 h(t) [q(t);q0(t)]
jq(t)j2
!0 (
(3:40)
h0(t) jq(t)j2 [q(t);q0(t)]
!00) iEit
h(t)[q(t);q0(t)]
jq(t)j2 h0(t) jq(t)j2 [q(t);q0(t)]
!0
( )
[q(t);q0(t)]
jq(t)j2 Eit: Note that in the above representation the second term tends to R(t)eitand h0(t)[q(t);q0(t)]
jq(t)j2 !p0(t). It remains to find two other limits in the first term.
This is a very long and cumbersome job, and after computations similar to those we have done earlier, we obtain
lima!0
[q(t);q0(t)]
jq(t)j2
!0
0;
(3:41)
lima!0 h0(t) jq(t)j2 [q(t);q0(t)]
!00
p000(t):
(3:42)
Hence, the lemma holds. p
COROLLARY 3.1. There exists e>0 such that if 0<a<e then kizo(t)>0for anyt2R.
THEOREM3.2. For any oval C there exists an oval O, contained in C such that corresponding bar billiard has Poncelet's porism property.
PROOF. Le t a2p
n bean anglesuch that 0<a<e for theabovee.
Then we can consider a bar billiard with the starting ovalC1and its inner isoptic zizo(t) take n forC2. In this case, for any t, the Poncelet's trans- verses close after n reflections and form an n-gon with or without self
intersections. p
REFERENCES
[1] K. BENKO - W. CIESÂLAK - S. GOÂZÂDZÂ - W. MOZGAWA, On isoptic curves, An.
Stiint. Univ. Al. I. Cuza Iasi, Ser. Noua, Mat.,36, No. 1 (1990), pp. 47±54.
[2] W. CIESÂLAK - A. MIERNOWSKI - W. MOZGAWA, Isoptics of a closed strictly convexcurve, Global differential geometry and global analysis, Proc. Conf., Berlin/Ger. 1990, Lect. Notes Math.,1481(1991), pp. 28±35.
[3] S. TABACHNIKOV,Geometry and billiards, Student Mathematical Library 30.
Providence, RI: American Mathematical Society, (2005).
[4] H. BOS- C. KERS- F. OORT- D. RAVEN,Poncelet's closure theorem, Expos.
Math.,5(1987), pp. 289±364
[5] E. W. WEISSTEIN, Poncelet's Porism, From MathWorld±A Wolfram Web Resource. http://mathworld.wolfram.com/PonceletsPorism.html
Manoscritto pervenuto in redazione il 12 maggio 2007.