We consider circular annuli with Poncelet’s porism property. We prove two identities which imply Chapple’s, Steiner’s and other formulas. All porisms can be expressed in the form in which elliptic functions are not used

Cieślak, Waldemar

Szczygielska, Elżbieta

English

University of Maria Curie-Skłodowska (UMCS): Scientific e-Journals / Uniwersytet Marii Curie-Skłodowskiej: e-czasopisma naukowe

doi: 10.2478/v10062-010-0011-0ANNALESU N I V E R S I T A T I S M A R I A E C U R I E - S K Ł O D O W S K AL U B L I N – P O L O N I AVOL. LXIV, NO. 2, 2010 SECTIO A 21–28WALDEMAR CIEŚLAK and ELŻBIETA SZCZYGIELSKAOn Poncelet’s porismAbstract. We consider circular annuli with Poncelet’s porism property. Weprove two identities which imply Chapple’s, Steiner’s and other formulas. Allporisms can be expressed in the form in which elliptic functions are not used.1. Introduction. An annulus formed by two circles will be called circularannulus. Poncelet [1] proved that if there exists one circuminscribed (simul-taneously inscribed in the outer and circumscribed on the inner) n-gon ina fixed circular annulus, then any point of the outer circle is the vertex ofsome circuminscribed n-gon. Circuminscribed polygon in a circular annuliwill be called a bicentric polygon. There is a general analytic expressionrelating the circumradius ρ, inradius r, and offset between the circumcenterand incenter a for a bicentric polygon. The condition for an n-gon to bebicentric is the following(1.1) sc(1nK (k) , k)=c√b2 − d2 + b√c2 − d2d (b+ c),whered =1ρ+ a, b =1ρ− a, c =1r,ω = cosh−1[1 +2c2(d2 − b2)d2 (b2 − c2)], k2 = 1− e−2ω,2000 Mathematics Subject Classification. 53A04.Key words and phrases. Porism, annulus, bicentric polygon.22 W. Cieślak and E. Szczygielskasc (x, k) is a Jacobi elliptic function and K (k) is a complete elliptic integralof the first kind, see [3], [1].Kerawala [3] established many porisms in a simple explicit form withoutresorting to the use of elliptic functions. In this paper we will show that itcan be done in any case.The paper [1] very precisely explains all details connected with Poncelet’sporism property.2. Main theorem. Let S ((u1, u2) , r) denote a circle with the center at(u1, u2) and the radius r. We consider an annulus A formed by two circlesS ((0, 0) , r) and S ((a, 0) , ρ) where a > 0 and a+ r < ρ. Let(2.1) z (t) = reit for t ∈ [0, 2pi]be a parametrization of S ((0, 0) , r). The circle S ((d, 0) , ρ) will be consid-ered in the form(2.2) w (t) = reit + λ (t) ieit for t ∈ [0, 2pi] .It is easy to verify that the positive-valued function λ is given by the formula(2.3) λ (t) =√a2 sin2 t+ 2ar cos t+ ρ2 − r2 − a2 − a sin t for t ∈ [0, 2pi] .A fixed point w (t) of S ((a, 0) , ρ) belongs to some α-isoptic Cα.Figure 1.We have(2.5) tanα (t)2=λ (t)r.Let(2.6) ϕ (t) = t+ α (t) ,On Poncelet’s porism 23and ϕ[0] (t) = t, ϕ[1] = ϕ, ϕ[n] = ϕ(ϕ[n−1])for n = 2, 3, . . . (see [2, (4.4)]).The formula (2.5) implies thateiα(t) =r2 − λ2 (t)r2 + λ2 (t)+ i2λ (t) rr2 + λ2 (t),i.e.,(2.7) eiϕ(t) =(r + λ (t) i|r + λ (t) i|)2eit.We consider, for each fixed t ∈ [0, 2pi], the following sequence of complexnumbers(2.8) ωm (t) =r + λ(ϕ[m] (t))i∣∣r + λ (ϕ[m] (t)) i∣∣ for m = 0, 1, 2, . . . .The formula (2.7) can be rewritten in the form(2.9) eiϕ(t) = ω20 (t) eit.Moreover, we haveeiϕ[2](t) = eiϕ(ϕ(t)) = ω20 (ϕ (t)) eiϕ(t) = ω21 (t)ω20 (t) eit = (ω0 (t)ω1 (t))2 eitand(2.10) eiϕ[m](t) = (ω0 (t)ω1 (t) . . . ωm−1 (t))2 eit.According to [2, Theorem 4.2], a convex annulus possesses Poncelet’s porismproperty if and only if for some natural number n ≥ 3 the function ϕ[n] (t)−t− 2pi vanishes.From this statement and the formula (2.10) we easily get the main theo-rem.Theorem 2.1. Let a circular annulus A possess Poncelet’s porism propertyfor a fixed natural number n ≥ 3. Then the following identities hold(2.11) (ω0ω1 . . . ωn−1)2 ≡ 1and(2.12) Imω0ω1 . . . ωn−1 ≡ 0.In the next sections we will show that Chapple’s, Steiner’s and otherformulas (see [1], [4]) are consequences of the identity (2.12).24 W. Cieślak and E. Szczygielska3. Bicentric triangle. If a triangle can be circuminscribed in a circularannulus A, then the so-called Euler triangle formula holds(3.1) a2 = ρ2 − 2rρ,see [1], [4].Let λm = λ(ϕ[m] (0))for m = 0, 1, 2.Figure 2.Clearly, we see that(3.2) λ0 = λ2 =√ρ2 − (r − a)2 , λ1 =√(ρ− a)2 − r2.From (2.12) we get0 = Imω0 (0)ω1 (0)ω2 (0) = Imω20 (0)ω1 (0)= Im (r + λ0i)2 (r + λ1i) =(r2 − λ20)λ1 + 2r2λ0,i.e., (λ20 − r2)λ21 = 4r4λ20.Hence we have the following formulas, each of them being a consequence ofthe previous one.[ρ2 − (r − a)2 − r2]2 [(ρ− a)2 − r2]= 4r4[ρ2 − (r − a)2],[ρ2 − (r − a)2 − r2]2(ρ− a)2−[(ρ2 − (r − a)2)2 − 2r2 (ρ2 − (r − a)2)+ r4] r2= 4r4[ρ2 − (r − a)2],On Poncelet’s porism 25[ρ2 − (r − a)2 − r2]2(ρ− a)2= r2[(ρ2 − (r − a)2)2+ 2r2(ρ2 − (r − a)2)+ r4],[ρ2 − (r − a)2 − r2]2(ρ− a)2 = r2[ρ2 − (r − a)2 + r2]2,[ρ2 − (r − a)2 − r2](ρ− a) = r[ρ2 − (r − a)2 + r2],(ρ2 − (r − a)2)(ρ− a)− r2 (ρ− a) = r(ρ2 − (r − a)2)+ r3,(ρ2 − (r − a)2)(ρ− a− r) = r2 (ρ− a+ r) ,a3 − (ρ+ r) a2 − (ρ2 − 2rρ) a+ (ρ+ r) (ρ2 − 2rρ) = 0,(a− ρ− r) (a2 − ρ2 + 2rρ) = 0and the Euler triangle formula is proved.4. Bicentric quadrileteral. If a 4-gon can be circuminscribed in a circu-lar annulus A, then the following formula holds(4.1)(ρ2 − a2)2 = 2r2 (ρ2 + a2) ,see [1], [4].Let λm = λ(ϕ[m] (0))for m = 0, 1, 2, 3.Figure 3.It is easy to see that(4.2) λ0 = λ3 =√ρ2 − (r − a)2, λ1 = λ2 =√ρ2 − (r + a)2.26 W. Cieślak and E. SzczygielskaFrom (2.12) we have0 = Imω0 (0)ω1 (0)ω2 (0)ω3 (0) = Im (ω0 (0)ω1 (0))2= Im ((r + λ0i) (r + λ1i))2 = Im(r2 − λ0λ1 + r (λ0 + λ1) i)= 2r (λ0 + λ1)(r2 − λ0λ1)i.e.r2 = λ0λ1.Hence we getr4 =[ρ2 − (r − a)2] [ρ2 − (r + a)2],r4 = ρ4 − 2 (r2 + a2) ρ2 + r4 − 2r2a2 + a4,ρ4 − 2a2ρ2 + a4 = 2r2ρ2 + 2r2a2.This is exactly the formula (4.1).5. Bicentric pentagon. If a 5-gon can be circuminscribed in a circularannulus A, then eight equivalent formulas are known (see [4, (43)–(50)]).Let λm = λ(ϕ[m] (0))for m = 0, 1, 2, 3, 4.Figure 4.It is easy to see that(5.1) λ0 = λ4 =√ρ2 − (r − a)2, λ2 =√(ρ− a)2 − r2, λ1 = λ3.We note thata2 = λ20 + r2 + ρ2 − 2ρ√λ20 + r2 cos γ,sinβ =r√λ20 + r2, cosβ =λ0√λ20 + r2,λ0 + λ12ρ= cos (β + γ) .On Poncelet’s porism 27Hence we obtainλ1 = 2ρ[cosβ cos γ − sinβ√1− cos2 γ]− λ0= 2ρ λ0√λ20 + r2λ20 + r2 + ρ2 − a22ρ√λ20 + r2− r√λ20 + r2√√√√1−(λ20 + r2 + ρ2 − a22ρ√λ20 + r2)2− λ0=1λ20 + r2[λ0(λ20 + r2 + ρ2 − a2)− r√4ρ2(λ20 + r2)− (λ20 + r2 + ρ2 − a2)2]− λ0=1λ20 + r2[λ0(ρ2 − a2)− r√4ρ2 (λ20 + r2)− (λ20 + r2 + ρ2 − a2)2]=1λ20 + r2[λ0(ρ2 − a2)− 2arλ0] ,i.e.,(5.2) λ1 =1λ20 + r2λ0(ρ2 − a2 − 2ar) = λ0 ρ2 − a2 − 2arρ2 − a2 + 2ar .From (2.12) we get0 = Imω20 (0)ω21 (0)ω2 (0) = Im ((r + λ0i) (r + λ1i))2 (r + λ2i)=[(r2 − λ0λ1)2 − r2 (λ0 + λ1)2]λ2 + 2r2 (r2 − λ0λ1) (λ0 + λ1)or0 = Imω20 (0)ω21 (0)ω2 (0) = Im (r + λ0i)2 (r + λ1i)2 (r + λ2i)=[(r2 − λ20) (r2 − λ21)− 4r2λ0λ1]λ2 + 2r2 [(r2 − λ20)λ1 + (r2 − λ21)λ0] .Now, it suffices to substitute λ0, λ2 and λ1 given by (5.1) and (5.2),respectively. For n ≥ 6 all calculations are similar.References[1] Bos, H. J . M., Kers, C., Dort, F. and Raven, D. W., Poncelet’s closure theorem, Expo.Math. 5 (1987) 289–364.[2] Cieślak, W., Szczygielska, E., Circuminscribed polygons in a plane annulus, Ann. Univ.Mariae Curie-Skłodowska Sect. A 62 (2008), 49–53.[3] Kerawala, S. M., Poncelet porism in two circles, Bull. Calcutta Math. Soc. 39 (1947),85–105.[4] Weisstein, E. W., Poncelet’s Porism, From Math World — A Wolfram Web Resource.http://mathworld.wolfram.com/PonceletsPorism.html28 W. Cieślak and E. SzczygielskaW. Cieślak E. SzczygielskaZakład Matematyki Państwowa Wyższa Szkoła ZawodowaPolitechnika Lubelska w Białej Podlaskiejul. Nadbystrzycka 40 ul. Sidorska 95/9720-618 Lublin 21-500 Biała PodlaskaPoland Polande-mail: izacieslak@wp.pl e-mail: eszczygielska@o2.plReceived December 10, 2009

On Poncelet’s porism

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