A hyperbolic proof of Pascal's Theorem
aa r X i v : . [ m a t h . HO ] D ec A HYPERBOLIC PROOF OF PASCAL’S THEOREM
MIGUEL ACOSTA AND JEAN-MARC SCHLENKER
Abstract.
We provide a simple proof of Pascal’s Theorem on cyclic hexagons, as wellas a generalization by Möbius, using hyperbolic geometry. Pascal’s Theorem
Blaise Pascal (1623–1662) is a towering intellectual figure of the XVII th century. Heis credited with inventing and building the first mechanical calculator, the Pascaline ,and with laying the foundations of probability theory, in particular in his correspondencewith Fermat – he came up for instance with Pascal’s triangle. He is also known for hiswork on hydrostatics and Pascal’s law (as well as the invention of the syringe), and fordiscovering the variation of air pressure with altitude – the SI unit of pressure is calledthe Pascal. However, he was most influential in his time as a philosopher and theologian,and well-known for “Pascal’s wager”.Pascal was raised and educated by his father, Etienne Pascal, who had a strong interestin the intellectual developments of his time and was an active member of a group ofscientists meeting around Martin Mersenne, including Desargues, Descartes and others.According to contemporary sources [7, p. 176], Blaise was extraordinarily precocious,and so passionate in studying mathematics that, when he was 11, his father forbid himto read any mathematics book before he turned 15 and knew latin and greek. Blaisetherefore continued studying geometry by himself and in secret and, at 16, publishedhis first treatise on projective geometry, the
Traité sur les coniques , which contained thefollowing remarkable statement.
Theorem A (Pascal’s Theorem) . Let
ABCDEF be a cyclic hexagon. Let X be theintersection point of AB and DE , Y the intersection point of BC and EF and Z theintersection point of CD and F A . Then X , Y and Z are aligned. The main goal of this note is to provide a simple proof, based on hyperbolic geometry,of Pascal’s Theorem. The statement has a natural setting in the projective plane. Insteadof considering a cyclic hexagon, one then considers a hexagon with vertices on a conic.Any non-degenerate conic is projectively equivalent to a circle, while the statements fordegenerate conics can be obtained by a limiting argument where a hexagon with verticeson a degenerate conic is obtained as a limit of hexagons with vertices on non-degenerateconics.By considering different permutations of the set of vertices, the configuration of Pas-cal’s theorem leads to a fascinating set of lines and points with a beautiful incidence Date : January 1, 2021.Both authors were partially supported by FNR grant AGoLoM OPEN/16/11405402.
A B C DEF Z XY
Figure 1.
The configuration of Pascal’s theorem (Theorem A) and Proposition B.structure. We recommend the articles of Conway and Ryba [1] and [2] describing thefull configuration, that they call mysticum hexagrammaticum . On the one hand, thisname describes their approach to label the lines and points by permutations of the set { A, B, C, D, E, F } , giving a natural way to state incidence relations and be able to iden-tify symmetries. On the other hand, it recalls the original name of Theorem A, givenby Pascal in : hexagrammum mysticum . The original work from Pascal did notsurvive, but there are many proofs of Theorem A, using a wide variety of tools. A proofusing algebraic geometry is sketched by Conway and Ryba in [1]. Other proofs involvecross-ratios and symmetries of the projective plane, as in [6] or Euclidean lengths andMenalaus’s theorem like in [3, p.77].The proof given here, based on hyperbolic geometry, is not really novel (it can bededuced easily from [6, page 436]) but it brings to light a striking link between Pascal’stheorem and elementary hyperbolic geometry. A recent and similar link is exhibitedby Drach and Schwarz in [4], where they revisit the seven-circles theorem, in Euclideangeometry, in terms of hyperbolic geometry. Even if those links between hyperbolic geom-etry and results on projective or Euclidean geometry do not lead to novel results, theygive a beautiful perspective to them.2. A hyperbolic statement of Pascal’s Theorem
We are going to use the Klein model of the hyperbolic plane. Consider an open disk ∆ on the projective plane, bounded by a circle Γ . The hyperbolic plane is defined as ∆ ,endowed with the Hilbert distance , defined as follows
Definition.
Let
P, Q ∈ ∆ , and A, B be the intersection points of the line
P Q with Γ .Then d ( P, Q ) = 12 log (cid:18)
BPBQ · AQAP (cid:19) . This distance induces a complete Riemannian metric on ∆ , and thus notions of anglesand lengths. A key property of this Hilbert distance is that it is invariant under projective HYPERBOLIC PROOF OF PASCAL’S THEOREM 3 transformations that leave ∆ invariant. The geodesics are precisely the straight lines,but the angles are not the Euclidean ones. The circle Γ is then the boundary at infinityof the hyperbolic plane , and its points are called ideal points . The model is very rich, butwe focus here only on a few points related to the polarity in the projective plane andorthogonality in the hyperbolic plane.Given a circle in the projective plane, one can define a polarity relation between pointsand lines: there is a polar line for each point, and a pole for each line. By definition,given points P ∈ ∆ and Q ¯∆ , Q is in the polar line of P , and conversely, if and only if BPBQ · AQAP = − , where A and B are again the intersections of the line P Q with Γ . We will denote the polarline of P by P ∗ . The polarity relation is also invariant under projective transformationsthat leave ∆ invariant.It follows from the definition that if P, Q ∈ ∆ , then P and Q are both in the polarline of P ∗ ∩ Q ∗ , the intersection point of the polar lines P ∗ and Q ∗ . It follows that theline P Q is the polar line of P ∗ ∩ Q ∗ . As a consequence, the polarity relation preservesthe incidence: three points are aligned if and only if their polar lines are concurrent.This last fact allows to have dual statements. For example, the dual statement ofPascal’s theorem is Brianchon’s theorem: if a conic is inscribed in a hexagon, then thethree diagonals joining the opposite vertices of the hexagon are concurrent.Back to hyperbolic geometry, the link between polarity and the Klein model that weare going to use is the following: Proposition.
Two lines l and l in the hyperbolic plane are orthogonal if and only ifthe pole of l is contained in the extension of l to the projective plane. In particular, if l and l are two lines that do not intersect in the hyperbolic plane,they have a unique common perpendicular, that is the polar line of their intersectionpoint in the projective plane.The proposition follows from the projective invariance of the hyperbolic metric andof the polarity relation under projective transformations leaving ∆ invariant, since onecan always find such a projective transformation leaving ∆ invariant and bringing theintersection of l and l to the center of ∆ – in this case the proposition is easy to check.We can now restate Pascal’s theorem in terms of hyperbolic geometry. Consideringthe polar lines l , l and l of the points X , Y and Z of the statement of Theorem A, weobtain the following equivalent statement. Proposition B.
Let
ABCDEF be an ideal hyperbolic hexagon. Let l be the commonperpendicular to AB and DE , l the common perpendicular to BC and EF and l thecommon perpendicular to CD and F A . Then l , l and l are concurrent. Proof of Proposition B
The proof of Proposition B is based on the hyperbolic version of an elementary state-ment on triangles.
Theorem C.
Let
P QR be a triangle. Then the angle bisectors of
P QR are concurrent.
MIGUEL ACOSTA AND JEAN-MARC SCHLENKER
A B C DEF PQR
Figure 2.
The triangle
P QR and its hyperbolic angle bisectors.
Remark.
Theorem C holds for Euclidean, spherical and hyperbolic triangles. The proofis, in all three cases, elementary. A point of a triangle is in the bisector of an angleif and only if it is at equal distance from the two corresponding edges. Therefore, theintersection point of two angle bisectors is at equal distance from all three edges, and istherefore contained in the third bisector.
Lemma D.
Let
ABDE be an ideal hyperbolic quadrilateral. Then, the common orthog-onal to AB and DE is the angle bisector of the lines AD and BE .Proof. Let l be the angle bisector of the lines AD and BE . The hyperbolic reflectionon l exchanges A and B , so the line AB is preserved by the reflection. Thus, the anglebisector of the two lines is orthogonal to AB . Similarly, l is also perpendicular to DE ,so it is precisely the common perpendicular to AB and DE . (cid:3) Proof of Proposition B.
If the diagonals AD , BE and CF are concurrent at a point P ,then Lemma D implies that l , l and l contain P , and are therefore concurrent.Now, suppose that the diagonals AD , BE and CF are not concurrent. We call P theintersection point of BE and CF , Q the intersection point of AD and CF , and R theintersection point of AD and BE .It follows from Lemma D that the angle bisector of P QR at P is the common per-pendicular l to AB and DE , while the angle bisector of P QR at Q is the commonperpendicular l to BC and EF , and the angle bisector of P QR at R is the commonperpendicular l to CD and F A .By Theorem C applied to the triangle
P QR , the lines l , l and l are concurrent, andthe result follows. (cid:3) The Möbius generalization
In 1847, Möbius proved a generalization of Pascal’s Theorem for (4 n + 2) -gons [5]. HYPERBOLIC PROOF OF PASCAL’S THEOREM 5 A i A i +1 A i +2 n +1 A i +2 n +2 m i +1 m i l i Figure 3.
The region R i . Theorem E (Möbius, 1847) . Let A A · · · A n +2 be a cyclic (4 n +2) -gon. Let X , . . . , X n +1 be the intersection points of the pairs of opposite sides of A A · · · A n +2 . If X , . . . , X n are aligned, then X n +1 lies in the same line as X , . . . , X n . By considering the polar lines l , . . . , l n +1 of X , . . . , X n +1 , we obtain the correspond-ing hyperbolic statement, for which the proof of Proposition B extends easily. Proposition F.
Let A A · · · A n +2 be a hyperbolic ideal (4 n + 2) -gon. Let l , . . . , l n +1 be the common perpendicular to the pairs of opposite sides of A A · · · A n +2 . If l , . . . , l n are concurrent, then the common intersection point belongs also to l n +1 . For each i ∈ { , . . . , n + 1 } , let m i be the line joining A i and A i +2 n +1 . Let R i be thebe the union of the two quarters defined by m i and m i +1 that contain the sides A i A i +1 and A i +2 n +1 A i +2 n +2 , as in Figure 3, where the indices are taken modulo n + 1 . Observethat, by Lemma D, the line l i is the set of points of R i that is at the same distance from m i and m i +1 . If l , . . . , l n are concurrent at a point P , then P is at the same distancefrom all the lines m i , so, in particular, it is at the same distance from m n +1 and m .The only remaining point to complete the proof (and the reason the statement is falsefor n -gons) is given by the following lemma. Lemma G. If P ∈ R ∩ · · · ∩ R n , then P ∈ R ∩ · · · ∩ R n +1 Proof.
Consider Cartesian equations for the lines m i , that we still denote by m i , so m i ( A i ) = m i ( A i +2 n +1 ) = 0 . Up to changing signs, we can suppose that m i ( A i +1 ) > .Thus, for each i ∈ { , . . . , n } the region R i is defined by the inequality m i m i +1 < ,but the region R n +1 , bounded by m n +1 and m , is defined by m n +1 m > . Now, if P ∈ R ∩ · · · ∩ R n , then m ( P ) m ( P ) < , . . . , m n ( P ) m n +1 ( P ) < . By multiplyingthis even number of inequalities, we obtain m n +1 ( P ) m ( P ) > , so P ∈ R n +1 . (cid:3) Acknowledgement.
The authors are grateful to Arseniy Akopyan for pointing out ref-erence [6].
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Department of mathematics, Université du Luxembourg, Maison du Nombre, 6, Avenuede la Fonte, L-4364 Esch-sur-Alzette, Luxembourg.
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