Lorenz Halbein and Norbert Hungerb¨uhler
MSC:Primary51M04,Secondary51A20;key words:Poncelet’s Porism,Pascal’s Theorem
January2014]A SIMPLE PROOF OF PONCELET’S THEOREM1
ud Abel’s Theorem and the reprentation of elliptic curves by means of the Weier-strass ℘-function to establish the equivalence of Poncelet’s Theorem and the group
structure on elliptic curves,e [11].Poncelet’s Theorem has a surprising mechani-
cal interpretation for elliptic billiards in the language of dynamical systems:e [8]
or [7]for an overview of this facet.A common approach to all four classical closing
theorems (the Poncelet porism,Steiner’s Theorem,the Zigzag Theorem,and Emch’s
Theorem)has recently been established by Protasov in [18].King showed in [14],that
Poncelet’s porism is isomorphic to Tarski’s plank problem (a problem about geometric
t-inclusion)and to Gelfand’s question (a number theoretic problem)via the con-
struction of an invariant measure.However,according to Berger [1,p.203],all known
proofs of Poncelet’s Theorem are rather long and recondite.
The aim of this paper is to give a simple proof of Poncelet’s Theorem in the real
projective plane.More precily,we will show that Poncelet’s Theorem is a purely
combinatorial conquence of Pascal’s Theorem.Before we give veral forms of the
latter,let us introduce some notation.For two points a and b ,let a −b denote the line
through a and b ,and for two lines ℓ1and ℓ2,let ℓ1∧ℓ2denote the interction point
of the lines in the projective plane.In abu of notation,we often write a −b −c in
order to emphasize that the points a ,b ,c are collinear.In the quel,points are often
labeled with numbers,and lines with encircled numbers like ③.
In this terminology,Pascal’s Theorem and its equivalent forms read as follows.
养胃粥食谱
Pascal’s Theorem (cf.[16])314256Any six points 1,...,6lie on a conic if and only if (1−2)∧(4−5)(2−3)∧(5−6)(3−4)∧(6−1)are collinear.Carnot’s Theorem (cf.[3,no.396])
5怎么腌牛排
1
4
2
3
6
Any six points 1,...,6lie on a conic
if and only if
[(1−2)∧(3−4)]−[(4−5)∧(6−1)]
(2−5)
(3−6)
are concurrent.
2c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
Brianchon’s Theorem (cf.[2])①②③④⑤⑥
Any six lines ①,...,⑥are tangent to a conic if and only if (①∧②)−(④∧⑤)(②∧③)−(⑤∧⑥)(③∧④)−(⑥∧①)are concurrent.Carnot’s Theorem ∗
①②
⑤
④③
⑥忘不了前任
Any six lines ①,...,⑥are tangent to a conic
if and only if
[(①∧②)−(③∧④)]∧[(④∧⑤)−(⑥∧①)]
(②∧⑤)
(③∧⑥)
are collinear.
As a matter of fact,we would like to mention that if the conic is not degenerate,
then the collinear points in Pascal’s Theorem are always pairwi distinct (the same
英式发音applies to the concurrent lines in Brianchon’s Theorem).
Since the real projective plane is lf-dual,Pascal’s Theorem and Brianchon’s The-
orem are equivalent.Moreover Carnot’s Theorem and its dual Carnot’s Theorem ∗are
just reformulations of Pascal’s Theorem and Brianchon’s Theorem by exchanging the
points 3and 5,and the lines ③and ⑤,respectively.Recall that if two adjacent points,
换屏幕
say 1and 2,coincide,then the corresponding line 1−2becomes a tangent with 1as生化危机6好玩吗
contact point.Similarly,if two lines,say ①and ②,coincide,then ①∧②becomes the
contact point of the tangent ①.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 three
tangents and two contact points of the tangents.
The paper is organized as follows.In Section 2,we prove Poncelet’s Theorem for
the special ca of triangles and at the same time we develop the kind of combinato-
rial arguments we shall u later.Section 3contains the crucial tool which allows to
show that Poncelet’s Theorem holds for an arbitrary number of edges.Finally,in Sec-
tion 4,we u the developed combinatorial technics in order to prove some additional
symmetry properties of Poncelet-polygons.
2.PONCELET’S THEOREM FOR TRIANGLES In order to prove Poncelet’s
Theorem for triangles,we will show that if the six vertices of two triangles lie on
a conic K ,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 Pascal’s Theorem in order to find collinear points or
concurrent lines.5820
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.
January 2014]A SIMPLE PROOF OF PONCELET’S THEOREM 3
a 1
a 2
a 3
b 3
b 2
b 1
Proof.Let K be a conic in which two triangles △a 1a 2a 3and △b 1b 2b 3are inscribed
where the two triangles do not have a common vertex.
写信给监狱的亲人怎么写First,we introduce the following three interction points:
I :=(a 1−a 2)∧(b 1−b 2),
X :=(a 2−b 3)∧(b 2−a 3),
I ′:=(a 3−a 1)∧(b 3−b 1).
In order to visualize the interction points I ,X ,and I ′,we break up the conic K and
draw it as two straight lines,one for each triangle as follows.I X I ′I
b 1a 1b 2a 2b 3a 3b 1a 1b 2
a 2
Now,we number the six points a 1,a 2,a 3,b 1,b 2,b 3on the conic K as shown by the
following figure.I X I ′I
b 14a 11b 25
a 22
b 33a 36b 14
a 11
b 2
5
a 22
4c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 121
By Pascal’s Theorem we get that the three interction points
(1−2)∧(4−5),(2−3)∧(5−6),and (3−4)∧(6−1)
are collinear,which is the same as saying that the points I −X −I ′are collinear.
In the next step,we label the sides of the triangles as shown in the following figure.
I X I ′I
b 1a 1b 2a 2b 3a 3b 1a 1b 2
a 2
②①⑥②⑤④③⑤By Carnot’s Theorem ∗we get that the six sides ①,...,⑥of the two triangles are
tangents to a conic if and only if
[(①∧②)−(③∧④)]∧[(④∧⑤)−(⑥∧①)],
(②∧⑤),and
(③∧⑥)
are collinear.Now,this is the same as saying that the points X −I −I ′are collinear,
which,as we have en above,is equivalent to a 1,a 2,a 3,b 1,b 2,b 3lying on a conic.
a 1
a 2
a 3
b 3
b 2
b 1
As an immediate conquence we get Poncelet’s Theorem for triangles.
Corollary 2.2(Poncelet’s Theorem for triangles).Let K and C be nondegenerate
conics.Suppo there is a triangle
△a 1a 2a 3inscribed in K and circumscribed about
C .Then for any point b 1of K for which two tangents to C exist,there is a triangle
△b 1b 2b 3which is also inscribed in K and circumscribed about C .
January 2014]A SIMPLE PROOF OF PONCELET’S THEOREM 5
