Pascal's theorem: Difference between revisions

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{{short description|Theorem on the collinearity of three points generated from a hexagon inscribed on a conic}}

[[File:Pascaltheoremgenericwithlabels.svg|thumb|250px|Pascal line {{math|''GHK''}} of self-crossing hexagon {{math|''ABCDEF''}} inscribed in ellipse. Opposite sides of hexagon have the same color.]]

[[File:THPascal.svg|thumb|250px|right|The intersections of the extended opposite sides of simple [[cyclic polygon|cyclic]] hexagon {{math|''ABCDEF''}} (right) lie on the Pascal line MNP (left).]]

[[Image:Pascal'sTheoremLetteredColored.PNG|thumb|250px|Self-crossing hexagon {{math|''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.]]

In [[projective geometry]], '''Pascal's theorem''' (also known as the '''''hexagrammum mysticum theorem''''', [[Latin]] for mystical [[hexagram]]) states that if six arbitrary points are chosen on a [[conic section|conic]] (which may be an [[ellipse]], [[parabola]] or [[hyperbola]] in an appropriate [[affine plane]]) and joined by line segments in any order to form a [[hexagon]], then the three pairs of opposite [[Edge (geometry)|sides]] of the hexagon ([[extended side|extended]] if necessary) meet at three points which lie on a straight line, called the '''Pascal line''' of the hexagon. It is named after [[Blaise Pascal]].

The theorem is also valid in the [[Euclidean plane]], but the statement needs to be adjusted to deal with the special cases when opposite sides are parallel.

This theorem is a generalization of [[Pappus's hexagon theorem|Pappus's (hexagon) theorem]], which is the special case of a [[degenerate conic]] of two lines with three points on each line.

== Euclidean variants ==

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== Related results ==

This theorem is a generalization of [[Pappus's hexagon theorem|Pappus's (hexagon) theorem]] – Pappus's theorem is the special case of a [[degenerate conic]] of two lines. Pascal's theorem is the [[polar reciprocation|polar reciprocal]] and [[projective dual]] of [[Brianchon'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 (printing)|broadside]] titled "Essay pour les coniques. Par B. P."<ref name=orig>{{harvnb|Pascal|1640}}, translation {{harvnb|Smith|1959|p=326}}</ref>

Pascal's theorem is a special case of the [[Cayley–Bacharach theorem]].

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The theorem was generalized by [[August Ferdinand Möbius]] in 1847, as follows: suppose a polygon with {{math|4''n'' + 2}} sides is inscribed in a conic section, and opposite pairs of sides are extended until they meet in {{math|2''n'' + 1}} points. Then if {{math|2''n''}} of those points lie on a common line, the last point will be on that line, too.

==''Hexagrammum Mysticum''==

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 [[projective configuration|configuration]] of 60 lines is called the ''Hexagrammum Mysticum''.<ref>{{harvnb|Young|1930|p=67}} with a reference to Veblen and Young, ''Projective Geometry'', vol. I, p. 138, Ex. 19.</ref><ref>{{harvnb|Conway|Ryba|2012}}</ref>

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 [[projective configuration|configuration]] of 60 lines is called the ''Hexagrammum Mysticum''.<ref>{{harvnb|Young|1930|p=67}} with a reference to Veblen and Young, ''Projective Geometry'', vol. I, p. 138, Ex. 19.</ref><ref>{{harvnb|Conway|Ryba|2012}}</ref>

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'''.<ref>{{harvnb|Biggs|1981}}</ref> 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'''.<ref>{{harvnb|Wells|1991|p=172}}</ref>

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A short elementary proof of Pascal's theorem in the case of a circle was found by {{harvtxt|van Yzeren|1993}}, based on the proof in {{harv|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 by {{harvtxt|Stefanovic|2010}}.

We can infer the proof from existence of [[isogonal conjugate]] too. If we are to show that {{math|''X'' {{=}} ''AB'' ∩ ''DE''}}, {{math|''Y'' {{=}} ''BC'' ∩ ''EF''}}, {{math|''Z'' {{=}} ''CD'' ∩ ''FA''}} are collinear for concyclic {{math|''ABCDEF''}}, then notice that {{math|△''EYB''}} and {{math|△''CYF''}} are similar, and that {{math|''X''}} and {{math|''Z''}} will correspond to the isogonal conjugate if we overlap the similar triangles. This means that {{math|∠''BYXCYX'' {{=}} ∠''CYZ''}}, hence making {{math|''XYZ''}} collinear.

A short proof can be constructed using cross-ratio preservation. Projecting tetrad {{math|''ABCE''}} from {{math|''D''}} onto line {{math|''AB''}}, we obtain tetrad {{math|''ABPX''}}, and projecting tetrad {{math|''ABCE''}} from {{math|''F''}} onto line {{math|''BC''}}, we obtain tetrad {{math|''QBCY''}}. This therefore means that {{math|''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, {{math|''XYZ''}} are collinear.

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==Proof using cubic curves==

[[File:THPascal.svg|thumb|250px|right|The intersections of the extended opposite sides of simple [[cyclic polygon|cyclic]] hexagon {{math|''ABCDEF''}} (right) lie on the Pascal line MNP (left).]]

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, {{math|''M''}} and {{math|''N''}} in the figure) on the would-be Pascal line, and the ninth point as the third point ({{math|''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 {{math|''AB, CD, EF''}}, and the set {{math|''BC, DE, FA''}}), and the third cubic is the union of the conic and the line {{math|''MN''}}. Here the "ninth intersection" {{math|''P''}} cannot lie on the conic by genericity, and hence it lies on {{math|''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 coneconic if we choose a point {{math|''E''}} on the coneconic and a line {{math|''MP''}} in the plane. The sum of {{math|''A''}} and {{math|''B''}} is obtained by first finding the intersection point of line {{math|''AB''}} with {{math|''MP''}}, which is {{math|''M''}}. Next {{math|''A''}} and {{math|''B''}} add up to the second intersection point of the coneconic with line {{math|''EM''}}, which is {{math|''D''}}. Thus if {{math|''Q''}} is the second intersection point of the coneconic with line {{math|''EN''}}, then

:<math>(A + B) + C = D + C = Q = A + F = A + (B + C)</math>

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Suppose {{math|''f''}} is the cubic polynomial vanishing on the three lines through {{math|''AB, CD, EF''}} and {{math|''g''}} is the cubic vanishing on the other three lines {{math|''BC, DE, FA''}}. Pick a generic point {{math|''P''}} on the conic and choose {{math|''λ''}} so that the cubic {{math|''h'' {{=}} ''f'' + ''λg''}} vanishes on {{math|''P''}}. Then {{math|''h'' {{=}} 0}} is a cubic that has 7 points {{math|''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&nbsp;&times;&nbsp;2&nbsp;=&nbsp;6 points in common, unless they have a common component. So the cubic {{math|''h'' {{=}} 0}} has a component in common with the conic which must be the conic itself, so {{math|''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 Propertyproperty of Pascal's Hexagonhexagon ==

Again given the hexagon on a conic of Pascal's theorem with the above notation for points (in the first figure), we have<ref>{{cite web|url=http://www.cut-the-knot.org/Generalization/OverlookedPascal.shtml|title=A Property of Pascal's Hexagon Pascal May Have Overlooked|date= 2014-02-03}}</ref>

:<math>\frac{\overline{GB}}{\overline{GA}} \times \frac{\overline{HA}}{\overline{HF}} \times \frac{\overline{KF}}{\overline{KE}} \times\frac{\overline{GE}}{\overline{GD}} \times \frac{\overline{HD}}{\overline{HC}} \times \frac{\overline{KC}}{\overline{KB}}=1.</math>

== Degenerations of PascalsPascal's theorem ==

[[File:Pascal-3456.png|450px|thumb|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 [[parabola#Properties of a parabola related to Pascal's theorem|parabolas]] and [[Hyperbola#Hyperbola asAs an affine image of the hyperbola y = 1/x|hyperbolas]].

==See also==

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* {{citation|last=Young|first=John Wesley|title=Projective Geometry|year=1930|publisher=The Mathematical Association of America|series=The Carus Mathematical Monographs, Number Four}}

* {{citation | last1=van Yzeren | first1=Jan | title=A simple proof of Pascal's hexagon theorem |mr=1252929 | year=1993 | journal=[[American Mathematical Monthly|The American Mathematical Monthly]] | issn=0002-9890 | volume=100 | issue=10 | pages=930–931 | doi=10.2307/2324214 | publisher=Mathematical Association of America | jstor=2324214}}

{{Commons category|Pascal's hexagram}}

==External links==

{{Commons category|Pascal's hexagram}}

* [http://www.cut-the-knot.org/Curriculum/Geometry/Pascal.shtml Interactive demo of Pascal's theorem (Java required)] at [[cut-the-knot]]

* [http://www.cut-the-knot.org/Curriculum/Geometry/PascalLines.shtml 60 Pascal Lines (Java required)] at [[cut-the-knot]]

* [https://archive.istoday/20121129152851/http://www.math.uregina.ca/~fisher/Norma/paper.html The Complete Pascal Figure Graphically Presented] by J. Chris Fisher and Norma Fuller (University of Regina)

* [http://www.mathematik.tu-darmstadt.de/~ehartmann/circlegeom.pdf ''Planar Circle Geometries, an Introduction to Moebius-, Laguerre- and Minkowski Planes''] (PDF; 891&nbsp;kB), Uni Darmstadt, S. 29–35.

* [http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.665.5892&rep=rep1&type=pdf How to Project Spherical Conics into the Plane] by Yoichi Maeda (Tokai University)

{{Blaise Pascal}}

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