Poncelet

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Mathematical Assoc. of America American Mathematical Monthly 121:1 September 1, 2014 3:29 p.m. poncelet AMM.tex page 1

A Simple Proof of Poncelets Theorem(on the occasion of its bicentennial)

Lorenz Halbeisen and Norbert Hungerbuhler

Abstract.We present a proof of Poncelets Theorem in the real projective plane which relies

only on Pascals Theorem.

1. INTRODUCTION In 1813, while Poncelet was in captivity as a war prisoner in

the Russian city of Saratov, he discovered the following theorem.

Theorem 1.1 (Poncelets Theorem). LetKandCbe nondegenerate conics in gen-eral position. Suppose there is ann-sided polygon inscribed inKand circumscribedaboutCsuch that none of its vertices belongs toC(in the case whenKandCinter-

sect or meet). Further suppose there is an(n 1)-sided polygonal chain with verticeson Ksuch that all its sides are tangent to Cand none of its vertices belongs to C.Then the side, which closes the polygonal chain, is also tangent to C.

An immediate consequence is the following more popular version of Poncelets

Theorem.

Corollary 1.2. LetKandCbe nondegenerate conics in general position which nei-ther meet nor intersect. Suppose there is an n-sided polygon inscribed in Kand cir-cumscribed aboutC. Then for any pointP ofK, there exists an n-sided polygon, alsoinscribed inKand circumscribed aboutC, which has Pas one of its vertices.

After his return to France, Poncelet published a proof in his book [17], which ap-

peared in 1822. He derived the version displayed above from a more general statement,where the sides of the polygon are tangent to conics Cifrom a pencil containing K. Hefirst proved the statement for a pencil of circles, thus generalizing theorems of Chap-

ple [5] and Euler [9] for triangles and of Fuss [10] for bicentric polygons. In order to

extend his main theorem from circles to conics, Poncelet then invoked a projection

theorem which states that every pair of conics with no more than two intersections can

be considered as the projective image of a pair of circles. Poncelet finished by arguing

that the case of more than two intersections followed by the principle of continuity.

Poncelets treatise was a milestone in the development of projective geometry, and his

theorem is widely considered the deepest and most beautiful result about conics.

Poncelets Theorem gained immediately the attention of the mathematical commu-

nity. Already in 1828, Jacobi gave in [13] an analytic proof for pairs of nested circles

by using the addition theorem for elliptic functions. In the sequel, Cayley investigated

algebraic conditions for two conics to be in Poncelet position. Using the theory of

Abelian integrals, he formulated a criterion in [4]. Cayley actually published a series

of papers dealing with Poncelets porism. In the early 20th century, Lebesgue revis-

ited Cayleys work and formulated the proof in the language of projective geometry

and algebra, see [15]. He used an observation by Hart who gave in [12] an elegant

argument for Poncelets Theorem for triangles. In recent times, Griffiths and Harris

MSC: Primary 51M04, Secondary 51A20;key words: Poncelets Porism, Pascals Theorem

January 2014] A SIMPLE PROOF OF PONCELETS THEOREM 1

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used Abels Theorem and the representation of elliptic curves by means of the Weier-

strass -function to establish the equivalence of Poncelets Theorem and the groupstructure on elliptic curves, see [11]. Poncelets Theorem has a surprising mechani-

cal interpretation for elliptic billiards in the language of dynamical systems: see [8]

or [7] for an overview of this facet. A common approach to all four classical closing

theorems (the Poncelet porism, Steiners Theorem, the Zigzag Theorem, and Emchs

Theorem) has recently been established by Protasov in [18]. King showed in [14], that

Poncelets porism is isomorphic to Tarskis plank problem (a problem about geometric

set-inclusion) and to Gelfands question (a number theoretic problem) via the con-

struction of an invariant measure. However, according to Berger [1, p. 203], all known

proofs of Poncelets Theorem are rather long and recondite.

The aim of this paper is to give a simple proof of Poncelets Theorem in the real

projective plane. More precisely, we will show that Poncelets Theorem is a purely

combinatorial consequence of Pascals Theorem. Before we give several forms of the

latter, let us introduce some notation. For two pointsaand b, leta bdenote the linethrougha and b, and for two lines1 and2, let1 2 denote the intersection pointof these lines in the projective plane. In abuse of notation, we often write a b cinorder to emphasize that the points a,b,c are collinear. In the sequel, points are often

labeled with numbers, and lines with encircled numbers like .In this terminology, Pascals Theorem and its equivalent forms read as follows.

Pascals Theorem (cf. [16])

3

1

4

2

5

6

Any six points1, . . . ,6lie on a conic

if and only if

(1 2) (4 5)(2 3) (5 6)(3 4) (6 1)

are collinear.

Carnots Theorem (cf. [3, no. 396])

5

1

4

2

3

6

Any six points1, . . . ,6lie on a conic

if and only if

[(1 2) (3 4)] [(4 5) (6 1)](2 5)(3 6)

are concurrent.

2 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121

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Brianchons Theorem (cf. [2])

Any six lines , . . . , are tangent to a conic

if and only if

( ) ( )( ) ( )( ) ( )

are concurrent.

Carnots Theorem

Any six lines , . . . , are tangent to a conic

if and only if

[( ) ( )] [( ) ( )]( )( )

are collinear.

As a matter of fact, we would like to mention that if the conic is not degenerate,

then the collinear points in Pascals Theorem are always pairwise distinct (the same

applies to the concurrent lines in Brianchons Theorem).

Since the real projective plane is self-dual, Pascals Theorem and Brianchons The-

orem are equivalent. Moreover Carnots Theorem and its dual Carnots Theorem are

just reformulations of Pascals Theorem and Brianchons Theorem by exchanging the

points3and 5, and the lines and , respectively. Recall that if two adjacent points,say1 and 2, coincide, then the corresponding line1 2becomes a tangent with 1 ascontact point. Similarly, if two lines, say and , coincide, then becomes thecontact point of the tangent . As a last remark, we would like to mention that a conic

is in general determined by five points, by five tangents, or by a combination like threetangents and two contact points of these tangents.

The paper is organized as follows. In Section 2, we prove Poncelets Theorem for

the special case of triangles and at the same time we develop the kind of combinato-

rial arguments we shall use later. Section 3 contains the crucial tool which allows to

show that Poncelets Theorem holds for an arbitrary number of edges. Finally, in Sec-

tion 4, we use the developed combinatorial technics in order to prove some additional

symmetry properties of Poncelet-polygons.

2. PONCELETS THEOREM FOR TRIANGLES In order to prove Poncelets

Theorem for triangles, we will show that if the six vertices of two triangles lie on

a conicK, then the six sides of the triangles are tangents to some conic C.The crucial point in the proof of the following theorem (as well as in the proofs

of the other theorems of this paper) is to find the suitable numbering of points and

edges, and to apply some form of Pascals Theorem in order to find collinear points or

concurrent lines.

Theorem 2.1. If two triangles are inscribed in a conic and the two triangles do not

have a common vertex, then the six sides of the triangles are tangent to a conic.


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a1

a2

a3

b3

b2

b1

Proof. LetKbe a conic in which two triangles a1a2a3 and b1b2b3 are inscribedwhere the two triangles do not have a common vertex.

First, we introduce the following three intersection points:

I := (a1 a2) (b1 b2) ,

X := (a2 b3) (b2 a3) ,

I := (a3 a1) (b3 b1) .

In order to visualize the intersection points I,X, and I, we break up the conic Kanddraw it as two straight lines, one for each triangle as follows.

I

X

I

I

b1

a1

b2

a2

b3

a3

b1

a1

b2

a2

Now, we number the six points a1, a2, a3, b1, b2, b3 on the conicKas shown by thefollowing figure.

I

X

I

I

b14

a11

b25

a22

b33

a36

b14

a11

b25

a22


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By Pascals Theorem we get that the three intersection points

(1 2) (4 5) , (2 3) (5 6) , and (3 4) (6 1)

are collinear, which is the same as saying that the pointsIX I are collinear.In the next step, we label the sides of the triangles as shown in the following figure.

I

X

I

I

b1

a1

b2

a2

b3

a3

b1

a1

b2

a2

By Carnots Theorem we get that the six sides , . . . , of the two triangles are

tangents to a conic if and only if

[( ) ( )] [( ) ( )] ,

( ) , and

( )

are collinear. Now, this is the same as saying that the points X I I are collinear,which, as we have seen above, is equivalent to a1, a2, a3, b1, b2, b3 lying on a conic.

a1

a2

a3

b3

b2

b1

q.e.d.

As an immediate consequence we get Poncelets Theorem for triangles.

Corollary 2.2 (Poncelets Theorem for triangles). LetK andCbe nondegenerateconics. Suppose there is a triangle a1a2a3 inscribed inKand circumscribed aboutC. Then for any pointb1 ofKfor which two tangents to Cexist, there is a triangle

b1b2b3 which is also inscribed in Kand circumscribed aboutC.


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Proof. Let K and C be two conics with a triangle a1a2a3 inscribed in K andcircumscribed about C. Let b1 be an arbitrary point on K which is distinct froma1, a2, a3, and let b2 and b3 be distinct points on K such that b1 b2 and b1 b3are two tangents toC. By construction, we get that five sides of the trianglesa1a2a3and b1b2b3 are tangents toC. On the other hand, by Theorem 2.1, we know that allsix sides of these triangles are tangents to some conicC. Now, since a conic is deter-mined by five tangents, C andCcoincide, which implies that the triangle b1b2b3iscircumscribed aboutC. q.e.d.

As a special case of Brianchons Theorem we get the following.

Fact 2.3. LetCbe a conic and let the triangle a1a2a3 be circumscribed aboutC.Furthermore, lett1, t2, t3 be the contact points of the three tangents a2 a3, a3 a1, a1 a2. Then the three lines a1 t1,a2 t2, anda3 t3 meet in a point.

Proof. Label the three sides of the triangles as follows:

=a2 a3 = , =a3 a1= , =a1 a2= .

Then = t1, = t2, = t3, and by Brianchons Theorem we get

thata1 t1, a2 t2, a3 t3 meet in a point. q.e.d.

In general, for arbitraryn-gons tangent toC the analogous statement will be false.However, ifn is even and if the n-gon is at the same time inscribed in a conic K, asimilar phenomenon occurs (see Theorem 4.2).

3. THE GENERAL CASE LetK andCbe nondegenerate conics in general posi-tion. We assume that there is an n-sided polygona1, . . . , an which is inscribed inKsuch that all its n sides a1 a2, a2 a3, . . . , an a1are tangent to Cand none of itsvertices belongs toC. Let us assume that n is minimal with this property (thus, in par-ticular, the pointsa1, . . . , an are pairwise distinct). Further, let b1, . . . , bnbe an(n 1)-sided polygonal chain on Kwhere all n 1sides b1 b2, b2 b3, . . . , bn1 bnare tangent toCand none of its vertices is one ofa1, . . . , an or belongs toC. We do

not yet know thatbn b1 is tangent toCtoo. If we break up the conicKand draw itas two straight lines, one for the polygon and one for the polygonal chain, we get the

following situation.

bn1

bn

b1

b2

bn1

bn

b1

b2

an1

an

a1

a2

an1

an

a1

a2

In order to prove Poncelets Theorem, we have to show thatbn b1 is also tangent toC. This will follow easily from the following result.

Lemma 3.1. Forn 4, the three intersection points

I := (a1 a2) (b1 b2) ,

X := (a2 bn1) (b2 an1) , and

I := (an1 an) (bn1 bn) ,


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I

X

I

b1

b2

bn1

bn

a1

a2

an1

an

are pairwise distinct and collinear, which is visualized above by the dashed line.

Proof. Depending on the parity ofn, we have one of the following anchorings, fromwhich we will work step by step outwards.

neven, withk = n2

:

I

X

I

bk1

ak1

bk

ak

bk+1

ak+1

bk+2

ak+2

By Carnots Theorem we have that

I X I are collinear, whichproves the lemma forn= 4.

nodd, withk = n+12

:

I

X

I

bk1

ak1

bk

ak

bk+1

ak+1

1 2 3

4 5 6

By Pascals Theorem we have that

IX I are collinear.

Forn 5, the lemma will follow from the following two claims.

Claim 1. Letp andqbe integers with2 p < q n 1. Further, let

Ip1 := (ap1 ap) (bp1 bp) , Ip := (ap ap+1) (bp bp+1) ,

Iq := (aq1 aq) (bq1 bq) , Iq+1 := (aq aq+1) (bq bq+1) ,

and let

X := (ap bq) (bp aq) .

If Ip X Iq are pairwise distinct and collinear, then Ip1 X Iq+1 are alsopairwise distinct and collinear. This implication is visualized by the following figure.

Ip

X

Iq

bp1

bp

bp+1

bq1

bq

bq+1

ap1

ap

ap+1

aq1

aq

aq+1


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Ip1

X

Iq+1

bp1

bp

bp+1

bq1

bq

bq+1

ap1

ap

ap+1

aq1

aq

aq+1

Proof of Claim 1.

Ip

Iq

bp

bq

ap

aq

(a) By Brianchons Theorem, the lines,, are pairwise distinct and concurrent.

Ip

X

Iq

bp

bq

ap

aq

(b) The lines , , meet, by assumption, in X, and they are pairwise distinct.By (a), we have that and are distinct, and by symmetry also andare distinct.

Since a straight line meets a nondegenerate conic in at most two points, and arealso distinct.

Ip1

Iq+1

bp1

bp

bq1

bq

bq+1

ap1

ap

ap+1

aq

aq+1

(c) By Brianchons Theorem, the lines,,are pairwise distinct and concurrent.By (a) & (b) we get that ,, and meet inX, and by (c) we get that also ,, and

meet inX, which implies that Ip1 X Iq+1 are collinear and pairwise distinct.IfIp1= Iq+1, then the four lines ap1 ap, bp1 bp, aq aq+1, bq bq+1, whichare all tangent toC, would be concurrent. But then these four lines are not pairwisedistinct, and since the eight points ap1, ap, aq, aq+1, bp1, bp, bq, bq+1 are pairwisedistinct (recall that1 p 1< q+ 1 n), this contradicts our assumption that the


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conic Kis nondegenerate. By similar arguments it follows that both Ip1and Iq+1 aredistinct fromX. q.e.d.

Claim 2. LetIp1,Iq+1, andXbe as above, and let

X := (ap1 bq+1) (bp1 aq+1) .

IfIp1 X Iq+1 are pairwise distinct and collinear, then Ip1 X Iq+1 are

pairwise distinct and collinear too. This implication is visualized by the following

figure.

Ip1

X

Iq+1

bp1

bp

bp+1

bq1

bq

bq+1

ap1

ap

ap+1

aq1

aq

aq+1

Ip1

X Iq+1

bp1

bp

bp+1

bq1

bq

bq+1

ap1

ap

ap+1

aq1

aq

aq+1

Proof of Claim 2.

Ip1

X

Iq+1

bp

bq

ap

aq

(a) By assumption, the pointsIp1 X Iq+1 are pairwise distinct and collinear.

J

Ip1

X

bp

bq

ap

aq

1 2

34 5

6

(b) By Pascals Theorem, the points Ip1 X J are pairwise distinct andcollinear.


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J

X Iq+1

bp

bq

ap

aq

1 2

64 5

3

(c) By Pascals Theorem, the points X J Iq+1 are pairwise distinct andcollinear.

By (a) & (b) we get thatIp1 J Iq+1 are collinear, and by (c) we get that X

lies onJ Iq+1. Hence, Ip1 X Iq+1 are collinear. By (a), (c) and a symmetric

version of (c), the three pointsIp1, X, Iq+1 are pairwise distinct. q.e.d.

By an iterative application of Claim 1 & 2, we finally get the situation

I

X

I

b1

b2

bn1

bn

a1

a2

an1

an

in whichIX I are pairwise distinct and collinear. q.e.d.

With similar arguments as in the proof of Poncelets Theorem for triangles (Corol-

lary 2.2), we can now prove the general case of Poncelets Theorem (Theorem 1.1):

Proof of Poncelets Theorem. LetKandCbe nondegenerate conics in general posi-tion. We assume that there is an n-sided polygona1, . . . , an which is inscribed inKsuch that all its n sides a1 a2, a2 a3, . . . , an a1are tangent to Cand none of itsvertices belongs toC. Let us assume thatn is minimal with this property. Further weassume that there is an(n 1)-sided polygonal chain b1, . . . , bn whosen 1 sides

are tangent toCand none of its vertices is one of a1, . . . , anor belongs to C. We haveto show thatbn b1 is tangent toC.

! !

bn1

bn

b1

b2

bn1

bn

b1

b2

an1

an

a1

a2

an1

an

a1

a2

By Lemma 3.1 we know that I X I are pairwise distinct and collinear,

where I = (a1 a2) (b1 b2), I

= (an1 an) (bn1 bn), and X =(a2 bn1) (b2 an1). In order to show that bn b1 is tangent toC, we have tointroduce two more intersection points:

J := (an1 a1) (bn1 b1) ,

X := (an b1) (bn a1) .


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JI

X

I X I

b1

b2

b1

b2

bn1

bn

a1

a2

a1

a2

an1

an

We now apply Pascals Theorem twice as illustrated below.

JI

X

I

b1

b2

b1

b2

bn1

bn

a1

a2

a1

a2

an1

an1 12 2

34 45 5

6

(a) By Pascals Theorem, the points IX Jare pairwise distinct and collinear.

JI X

b1

b2

b1

b2

bn1

bn

a1

a2

a1

a2

an1

an1 2 6

34 5

(b) By Pascals Theorem, the points I JX are pairwise distinct and collinear.Since, by Lemma 3.1, I X I are pairwise distinct and collinear, by (a) we getthat I X J I are collinear, and by (b) we finally get that I X I arecollinear.

For the last step, we apply Carnots Theorem.

I

X

I

b1

b2

bn1

bn

a1

a2

an1

an

Since IX

I

are collinear, by Carnots Theorem

we get that the six lines , . . . , are tangent to some conic C. Now, since a conic is determined by five tangents, andthe five lines ,,,, are tangent toC,C andCcoincide. This implies that istangent toC, which is what we had to show. q.e.d.


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4. SYMMETRIES IN PONCELET-POLYGONS In this section we present some

symmetries in2n-sided polygons which are inscribed in some conic Kand circum-scribed about another conic C. To keep the terminology short, we shall call such apolygon a2n-Poncelet-polygonwith respect toK& C.

Theorem 4.1. LetKandCbe nondegenerate conics in general position which neither

meet nor intersect and leta1, . . . , a2n be the vertices of a2n-Poncelet-polygon withrespect to K& C. Further lett1, . . . , t2n be the contact points of the tangents a1 a2, . . . , a2n a1.

(a) All then diagonals a1 an+1, a2 an+2, . . . , an a2nmeet in a pointH0.

(b) All then linest1 tn+1, t2 tn+2, . . . , tn t2nmeet in the same pointH0.

Proof. (a) By the proof of Lemma 3.1, we get that the three points a1, an+1, and(a2 an+2) (an a2n)are collinear.

a1 an+1

a1

a2n

an+2

an+1

a1

a2

an

an+1

This is the same as saying that the three diagonalsa1 an+1,a2 an+2, andan a2n meet in a point, say H0. Now, by cyclic permutation we get that all n diagonalsmeet inH0.

(b) By the proof of Lemma 3.1, we get that the three points t1 H0 tn+1 arecollinear.

t1

H0

tn+1

a2

a1

a2n

an+3

an+2

an+1

a1

a2

a3

an

an+1

an+2

Thus, by cyclic permutation we get that allnlinest1 tn+1, t2 tn+2, . . . , tn t2npass throughH0, which implies that allnlines meet inH0. q.e.d.

In the last result, we show that the point H0 is independent of the particular2n-Poncelet-polygon (compare with Poncelets results no. 570 & 571 in [17]).

Theorem 4.2. LetK andC be nondegenerate conics in general position which nei-ther meet nor intersect and leta1, . . . , a2n andb1, . . . , b2n be the vertices of two2n-Poncelet-polygons with respect to K& C. Further lett1, . . . , t2n andt

1, . . . , t

2n

be the contact points of the Poncelet-polygons. Then all 4n lines a1 an+1, . . .,t1 tn+1, . . .,b1 bn+1, . . .,t

1 t

n+1, . . . meet in a pointH0. Moreover, oppositesides of the Poncelet-polygons meet on a fixed lineh, wherehis the polar ofH0, bothwith respect toCandK.

Proof. By Theorem 4.1 we know that the2n lines a1 an+1, . . ., t1 tn+1, . . . meetin a pointH0. First, we show that the polar h of the poleH0 with respect to C is thesame as the polarh ofH0 with respect toK, and then we show that the point H0 isindependent of the choice of the2n-Poncelet-polygon.


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ai+1

ak+1

ai+n

ak+n

ai

ak+n+1

ak

ai+n+1

tk

tk+n

ti

ti+n

H0

P

Q

First notice that in the figure above, H0 is on the polarp ofPwith respect to theconicCand thatH0 is also on the polar p

ofPwith respect to the conic K(see forexample Coxeter and Greitzer [6, Theorem 6.51]). Thus,Plies on the polar h ofH0with respect toC, as well as on the polarh ofH0 with respect otK. Since the sameapplies to the pointQ, the polars handh coincide, which shows that the poleH0hasthe same polar with respect to both conics.

The fact thatH0 is independent of the choice of the 2n-Poncelet-polygon is just aconsequence of the following.

Claim. LetH0 be as above and leth be the polar ofH0 (with respect to K orC).Choose an arbitrary pointP onh. Lets1& s2 be the two tangents from P toC andletA & A andB & B be the intersection points ofs1 ands2 withK.

H0

P

BA

B

A

s1 s2

ThenH0= (AB) (B A).

Proof of Claim. By a projective transformation, we may assume that his the line atinfinity. Then, the pole H0 becomes the common center of both conics and the claimfollows by symmetry. q.e.d.


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Now, leta1, . . . , a2n andb1, . . . , b2n be the vertices of two2n-Poncelet-polygonswith respect to K& C. Furthermore, let H0= (a1 an+1) (a2 an+2)and H

0=

(b1 bn+1) (b2 bn+2), and let h and h be their respective polars. Choose any

pointPwhich lies on bothh andh, and draw the two tangents fromP toC whichintersect K in the points A, A, B , B . If the conics K and C do not meet (whatwe assume), then these points are pairwise distinct and by the Claim we get H0 =(AB) (B A) =H0. q.e.d.

Notice, that for n= 3, H0is the Brianchon point with respect toCof the Poncelet-hexagon, andh its Pascal line with respect to K. So, for n >3, the pointH0 is thegeneralized Brianchon pointwith respect toCof the2n-Poncelet-polygon, andh itsgeneralized Pascal linewith respect toK.

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Elementargeometrie: Die Relation zwischen der Distanz der Mittelpuncte und den Radien zweier Kreise

zu finden, von denen der eine einem unregelmaigen Polygon eingeschrieben, der andere demselben

umgeschrieben ist,J. Reine Angew. Math.3 (1828) 376389.

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LORENZ HALBEISEN ([email protected]) received his PhD from the ETH Zurich in 1994.

After research positions in France, Spain, and California, he became lecturer at Queens University Belfast andis now senior scientist at the ETH Zurich. His interests range across set theory, finite and infinite combinatorics,

geometry, and number theory.

Department of Mathematics, ETH Zentrum, R amistrasse 101, 8092 Zurich, Switzerland

NORBERT HUNGERBUHLER ([email protected]) received his PhD at ETH Zurich in

1994. After research positions in Germany and the US he became assistant professor at the University of

Alabama at Birmingham and later professor at the University of Fribourg. Currently he is professor at ETH.

His interests range across analysis, geometry, discrete mathematics and number theory.

Department of Mathematics, ETH Zentrum, R amistrasse 101, 8092 Zurich, Switzerland


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