Pascal's theorem In projective geometry,

7:06 AM | BY ZeroDivide EDIT
In projective geometry, Pascal's theorem (also known as the Hexagrammum Mysticum Theorem) states that if six arbitrary points are chosen on a conic (i.e., ellipse, parabola or hyperbola) and joined by line segments in any order to form a hexagon, then the three pairs of opposite sides of the hexagon (extended if necessary) meet in three points which lie on a straight line, called the Pascal line of the hexagon. The theorem is valid in the Euclidean plane, but the statement needs to be adjusted to deal with the special cases when opposite sides are parallel.

Euclidean variants[edit]

The most natural setting for Pascal's theorem is in a projective plane since all lines meet and no exceptions need be made for parallel lines. However, with the correct interpretation of what happens when some opposite sides of the hexagon are parallel, the theorem remains valid in the Euclidean plane.
If exactly one pair of opposite sides of the hexagon are parallel, then the conclusion of the theorem is that the "Pascal line" determined by the two points of intersection is parallel to the parallel sides of the hexagon. If two pairs of opposite sides are parallel, then all three pairs of opposite sides form pairs of parallel lines and there is no Pascal line in the Euclidean plane (in this case, the line at infinity of the extended Euclidean plane is the Pascal line of the hexagon).

Related results[edit]


The intersections of the extended opposite sides of hexagon ABCDEF (right) lie on the Pascal line MNP (left).
This theorem is a generalization of Pappus's (hexagon) theorem – Pappus's theorem is the special case of a degenerate conic of two lines. Pascal's theorem is the polar reciprocal and projective dual ofBrianchon's theorem. It was formulated by Blaise Pascal in a note written in 1639 when he was 16 years old and published the following year as a broadside titled "Essay povr les coniqves. Par B. P.".[1]
A degenerate case of Pascal's Theorem (four points) is interesting; given points ABCD on a conic Γ, the intersection of alternate sides, AB ∩ CD, BC ∩ DA, together with the intersection of tangents at opposite vertices (A, C) and (B, D) are collinear in four points; the tangents being degenerate 'sides', taken at two possible positions on the 'hexagon' and the corresponding Pascal Line sharing either degenerate intersection. This can be proven independently using a property of pole-polar. If the conic is a circle, then another degenerate case tells us that for a triangle, the three points that appear as the intersection of a side line with the corresponding side line of the Gergonne triangle, are collinear.
Six is the minimum number of points on a conic about which special statements can be made, as five points determine a conic.
The converse is the Braikenridge–Maclaurin theorem, named for 18th century British mathematiciansWilliam Braikenridge and Colin Maclaurin (Mills 1984), which states that if the three intersection points of the three pairs of lines through opposite sides of a hexagon lie on a line, then the six vertices of the hexagon lie on a conic; the conic may be degenerate, as in Pappus's theorem.[2] The Braikenridge–Maclaurin theorem may be applied in the Braikenridge–Maclaurin construction, which is a synthetic construction of the conic defined by five points, by varying the sixth point.
The theorem was generalized by Möbius in 1847, as follows: suppose a polygon with 4n + 2 sides is inscribed in a conic section, and opposite pairs of sides are extended until they meet in 2n + 1 points. Then if 2n of those points lie on a common line, the last point will be on that line, too.

Hexagrammum Mysticum[edit]

If six unordered points are given on a conic section, they can be connected into a hexagon in 60 different ways, resulting in 60 different instances of Pascal's theorem and 60 different Pascal lines. This configuration of 60 lines is called the Hexagrammum Mysticum.[3]
As Thomas Kirkman proved in 1849, these 60 lines can be associated with 60 points in such a way that each point is on three lines and each line contains three points. The 60 points formed in this way are now known as the Kirkman points.[4] The pascal lines also pass, three at a time, through 20 Steiner points. There are 20 Cayley lines which consist of a Steiner point and three Kirkman points. The Steiner points also lie, four at a time, on 15 Plücker lines. Furthermore, the 20 Cayley lines pass four at a time through 15 points known as the Salmon points.[5]


Tangled-up hexagon ABCDEF, inscribed in a circle. Its sides are extended so that pairs of opposite sides intersect on Pascal's line. Each pair of extended opposite sides has its own color: one red, one yellow, one blue. Pascal's line is shown in white.

Proofs[edit]

Pascal's original note[1] has no proof, but there are various modern proofs of the theorem.
It is sufficient to prove the theorem when the conic is a circle, because any (non-degenerate) conic can be reduced to a circle by a projective transformation. This was realised by Pascal, whose first lemma states the theorem for a circle. His second lemma states that what is true in one plane remains true upon projection to another plane.[1] Degenerate conics follow by continuity (the theorem is true for non-degenerate conics, and thus holds in the limit of degenerate conic).
A short elementary proof of Pascal's theorem in the case of a circle was found by van Yzeren (1993), based on the proof in (Guggenheimer 1967). This proof proves the theorem for circle and then generalizes it to conics.
A short elementary computational proof in the case of the real projective plane was found byStefanovic (2010)
We can infer the proof from existence of isogonal conjugate too. If we are to show that X = AB ∩ DE,Y = BC ∩ EF, Z = CD ∩ FA are collinear for conconical ABCDEF, then notice that ADY and CYF are similar, and that X and Z will correspond to the isogonal conjugate if we overlap the similar triangles. This means that angle DYX = angle CYZ, hence making XYZ collinear.
A short proof can be constructed using cross-ratio preservation. Projecting tetrad ABCE from D onto line AB, we obtain tetrad ABPX, and projecting tetrad ABCE from F onto line BC, we obtain tetrad QBCY. This therefore means that R(AB; PX) = R(QB; CY), where one of the points in the two tetrads overlap, hence meaning that other lines connecting the other three pairs must coincide to preserve cross ratio. Therefore XYZare collinear.
Another proof for Pascal's theorem for a circle uses Menelaus' theorem repeatedly.
Dandelin, the geometer who discovered the celebrated Dandelin spheres, came up with a beautiful proof using "3D lifting" technique that is analogous to the 3D proof of Desargues' theorem. The proof makes use of the property that for every conic section we can find a one-sheet hyperboloid which passes through the conic. There also exists a simple proof for Pascal's theorem for a circle using Law of Sines and similarity.

Proof using cubic curves[edit]

Pascal's theorem has a short proof using the Cayley–Bacharach theorem that given any 8 points in general position, there is a unique ninth point such that all cubics through the first 8 also pass through the ninth point. In particular, if 2 general cubics intersect in 8 points then any other cubic through the same 8 points meets the ninth point of intersection of the first two cubics. Pascal's theorem follows by taking the 8 points as the 6 points on the hexagon and two of the points (say, M and N in the figure) on the would-be Pascal line, and the ninth point as the third point (P in the figure). The first two cubics are two sets of 3 lines through the 6 points on the hexagon (for instance, the set AB, CD, EF, and the set BC, DE, FA), and the third cubic is the union of the conic and the line MN. Here the "ninth intersection" P cannot lie on the conic by genericity, and hence it lies on MN.
The Cayley–Bacharach theorem is also used to prove that the group operation on cubic elliptic curves is associative. The same group operation can be applied on a cone if we choose a point E on the cone and a line MP in the plane. The sum of A and B is obtained by first finding the intersection point of line AB with MP, which is M. Next A and B add up to the second intersection point of the cone with line EM, which is D. Thus if Q is the second intersection point of the cone with line EN, then
(A + B) + C = D + C = Q = A + F = A + (B + C)
Thus the group operation is associative. On the other hand, Pascal's theorem follows from the above associativity formula, and thus from the associativity of the group operation of elliptic curves by way of continuity.

Proof using Bézout's theorem[edit]

Suppose f is the cubic polynomial vanishing on the three lines through AB, CD, EF and g is the cubic vanishing on the other three lines BC, DE, FA. Pick a generic point P on the conic and choose λ so that the cubic h = f + λg vanishes on P. Then h = 0 is a cubic that has 7 points A, B, C, D, E, F, P in common with the conic. But by Bézout's theorem a cubic and a conic have at most 3 × 2 = 6 points in common, unless they have a common component. So the cubic h = 0 has a component in common with the conic which must be the conic itself, so h = 0 is the union of the conic and a line. It is now easy to check that this line is the Pascal line.

A Property of Pascal's Hexagon[edit]

Given hexagon ABCDEF, let AC meet BD at G, BE meet CF at H, AE meets DF at I: Then, as well known, the six vertices of the hexagon lie on a conic if the points G, H, I are collinear. In addition, the two conditions are equivalent:[6]
\frac{\overline{GB}}{\overline{GD}} \times \frac{\overline{ID}}{\overline{IF}} \times \frac{\overline{HF}}{\overline{HC}} \times\frac{\overline{GC}}{\overline{GA}} \times \frac{\overline{IA}}{\overline{IF}} \times \frac{\overline{HE}}{\overline{HB}}=1

Degenerations of Pascals's theorem[edit]


Pascal's theorem: degenerations
There exist 5-point,4-point and 3-point degenerate cases of Pascal's theorem. In a degenerate case, two previously connected points of the figure will formally coincide and the connecting line becomes the tangent at the coalesced point. See the degenerate cases given in the added scheme and the external link on circle geometries. If one chooses suitable lines of the Pascal-figures as lines at infinity one gets many interesting figures on parabolas and hyperbolas (see the German sitesParabel and Hyperbel).

See also[edit]

Notes[edit]

  1. ^ Jump up to:a b c Pascal 1640, translation Smith 1959, p. 326
  2. Jump up^ Coxeter & Greitzer 1967, p. 76
  3. Jump up^ Young 1930, p. 67 with a reference to Veblen and Young, Projective Geometry, vol. I, p. 138, Ex. 19.
  4. Jump up^ Biggs 1981
  5. Jump up^ Wells 1991, p. 172
  6. Jump up^ http://www.cut-the-knot.org/Generalization/OverlookedPascal.shtml

References[edit]

  • Guggenheimer, Heinrich W. (1967), Plane geometry and its groups, San Francisco, Calif.: Holden–Day Inc., MR 0213943
  • Mills, Stella (March 1984), "Note on the Braikenridge–Maclaurin Theorem", Notes and Records of the Royal Society of London (The Royal Society) 38 (2): 235–240, doi:10.1098/rsnr.1984.0014, JSTOR 531819
  • Pascal, Blaise (1640). "Essay povr les coniqves" (facsimile). Niedersächsiche Landesbibliothek, Gottfried Wilhelm Leibniz Bibliothek. Retrieved 21 June 2013.
  • Wells, David (1991), The Penguin Dictionary of Curious and Interesting Geometry, London: Penguin Books, ISBN 0-14-011813-6
  • Young, John Wesley (1930), Projective Geometry, The Carus Mathematical Monographs, Number Four, The Mathematical Association of America
  • van Yzeren, Jan (1993), "A simple proof of Pascal's hexagon theorem", The American Mathematical Monthly (Mathematical Association of America) 100 (10): 930–931, doi:10.2307/2324214, ISSN 0002-9890, JSTOR 2324214, MR 1252929

  • Hexagrammum Mysticum Theorem

    The mathematician Pascal was most famous for Pascal's triangle, a number pattern that used in the binomial expansion formula. Today, we will introduce a theorem in geometry that bears his name. The Pascal's Hexagrammum Mysticum Theorem states that if we draw a hexagon inscribed in a circle then the three pairs of opposite sides of the hexagon intersect at three points which lie on a straight line.


    Pascal discovered this amazing hexagon theorem when he was only 16 years old and published it in a poster with the title "Essay pour les coniques" (Essay on Conics). Below is a picture of a rare copy kept in the National Library of France.



    We can recognize Pascal's hexagon theorem on the figure appeared on the top of the poster. The hexagon is PQVONK.The first pair of opposite sides, PK and VO, intersect at the point M. The second pair of opposite sides, KN and QV, intersect at the point S. Therefore, by Pascal's theorem, the third pair of opposite sides, PQ and NO, must intersect at a point on the line MS. In other words, the three lines MS, NO and PQ must intersect at a common point.

    An equivalent statement of Pascal's theorem: three lines MS, NO, PQ must intersect at a single point


    Different configurations of Pascal's theorem

    Pascal's theorem has many configurations. The six vertices of the hexagon can be arranged on the circle in any particular order. With each arrangement of the vertices, we have an instance of the Pascal's theorem. Thanks to this variety, Pascal deduced hundreds of corollaries from this theorem.

    Now you are invited to draw different pictures of Pascal's theorem. You can draw a circle and then pick out six points: 1, 2, 3, 4, 5, 6, on the circle of any particular order. Draw the intersections of three pairs of lines {12,45}, {23,56}, {34,61}, and then connect the three intersecting points. After that you can choose your own favorite picture of Pascal's theorem.

    Here is an example:


    My favorite picture of Pascal's theorem is the following.


    Pascal's theorem for the conics

    So far, we have only stated the Pascal's theorem for the circle. What makes Pascal's theorem so interesting is that it also holds for all types of conics.  So instead of choosing six points on a circle, we can choose six points on an ellipse, or on a parabola, or on a hyperbola, the theorem is still true! The three intersecting points still lie on a straight line! How amazing is that!


    Here is a picture of an ellipse:

    Here is a parabola:

    And here is a hyperbola:

    When the conic degenerated into two straight lines, Pascal's theorem becomes Pappus' Theorem. This is the logo of Math Garden Blog.



    Let us stop here for now, hope to see you again in the next post.




    Homework.

    1. Draw different configurations for Pascal's theorem.

    2. Find a proof of Pascal's theorem.
↑