generalizing pascal's theorem - usna · generalizing pascal’s theorem will traves department...

Generalizing Pascal’s Theorem

Will Traves

Department of MathematicsUnited States Naval Academy

Bi-College ColloquiumBryn Mawr College and Haverford College

Philadelphia

07 OCT 2013

Traves (USNA) Generalizing Pascal’s Theorem Philadelphia, 07 OCT 2013 1 / 26

Reference

THE AMERICAN MATHEMATICAL

MONTHLYVOLUME 120, NO. 10 DECEMBER 2013

867Quick, Does 23 = 67 Equal 33 = 97?

A Mathematician’s Secret from Euclid to Today

David Pengelley

877Linear Algebra via Complex Analysis

Alexander P. Campbell and Daniel Daners

893Rigorous Computer Analysis of the Chow–Robbins Game

Olle Haggstrom and Johan Wastlund

901From Pascal’s Theorem to d-Constructible Curves

Will Traves

916The Cuoco Configuration

Roger Howe

NOTES

924A Generalization of the Leibniz Rule

Ulrich Abel

929Hamiltonian Cycles on Archimedean Solids Are Twisting Free

Richard Ehrenborg

933The Parbelos, a Parabolic Analog of the Arbelos

Jonathan Sondow

940Inequalities for Gamma Function Ratios

G. J. O. Jameson

945PROBLEMS AND SOLUTIONS

REVIEWS

953Real Analysis Through Modern InfinitesimalsBy Nader Vakil

James M. Henle with the assistance of Michael G. Henle

958EDITOR’S ENDNOTES

000INDEX TO VOLUME 120

An Official Publication of the Mathematical Association of America


Start at the Beginning

Line Arrangement due to Pappus of Alexandria (Synagogue; c. 340)

Richter-Gebert: 9 proofs in Perspectives on Projective Geometry, 2011


Pascal’s Mystic Hexagon Theorem

Pascal: placed the 6 intersection points on a conic (1639)

Converse: Braikenridge-Maclaurin Theorem


Why Mystic?


The Projective Plane

P2 is a compactification of R2

P2 = R2 ∪ line at∞

Parallel lines meet at infinity - one point at∞ for each slope.Line at infinity wraps twice around R2.


Topology of the Projective Plane

Thicken line at infinity:P2 = disk ∪ Mobius band

P2 can’t be embedded in R3

(Conway, Gordon, Sachs (1983):linked triangles in K6)

! 20 !

Projective space is also the union of a disc in R2 and a Möbius strip, and is equivalent to the sphere S2 with a blow up at one point.

The Relationship between RP2 and R3.

Earlier we claimed that RP2 is not a subset of R3, that it does not “fit” into R3. For this proof we call upon Conway, Gordon and Sachs’ 1983 result (ams.org) that 6 points all cannot be linked to one another such that the total linking number of all triangles formed is even, in R3. We will give an example of 6 points linked in RP2 , K6, such that the linking number of the set is even. First, to demonstrate Conway, Gordon and Sachs’ proof, an example of K6, 6 points linked in R3:

Figure 16

The linking numbers of triangles 124 and 356 is 0 because they are not linked; they could be pulled apart from each other without being caught like a chain link.

The linking numbers of triangles 246 and 135 is 1 because they are linked together.

If you add all of the linking numbers of all sets of triangles in this particular linking, you will

find the sum to be 1 (246 and 135 are the only linked triangles). This is consistent with Conway, Gordon and Sachs who claimed that as long as we are working in R3 then we will have an odd total linking number.


Bezout’s Theorem

Compactness of P2 allows us to count solutions:

Theorem (Bezout)Any two curves, without common components, defined by thevanishing of polynomials of degrees d1 and d2 meet in d1d2 points inP2, suitably interpreted.Lines meeting an Ellipse

y=0

4x2+9y2=36

y=1 y=2 y=3 Line meets curve in two

points (possibly imaginary).

Double point: tangency when y=2 4x2 + 9(4) = 36 so 4x2 = 0

It seems that such a curve of degree 1 always meets such a curve of degree 2 in (2)(1) points, if we count them properly.


Folklore

TheoremSuppose that k red lines meet k blue lines in a set Γ of k2 distinctpoints. If S = 0 is an irreducible curve of degree d through kd points ofΓ then the remaining points lie on a unique curve C of degreet = k − d.

Proof (Existence):R: deg k poly defining red lines.B: deg k poly defining blue lines.Pick P ∈ S \ Γ. Choose

Fa,b = aR + bBto vanish at P. Then S = 0 is acomponent of Fa,b = 0 (by Bezoutand S irred) and Fa,b/S defines C.�


Proof of Pascal’s Theorem

Theorem (Pascal)If 6 points A,B,C,a,b, c on an irreducible conic are joined by linesAb,Bc,Ca,aB,bC,aC, then the red lines meet the blue lines in 3 newcollinear points.

Proof: Since conic goesthrough 6 points ofΓ = R ∩ B, the line passesthrough the remaining 3points of Γ.


The 8⇒ 9 Theorem

Theorem (Cayley-Bacharach-Chasles)If two cubics C1 and C2 meet in 9 distinct points, then any other cubicC through 8 of these points goes through the ninth too.

Elementary proof on Terry Tao’s blog (July 15, 2011)


Inscribing an Octagon in a Cubic Curve

DefinitionAn n-gon P with n edges is inscribed in a curve C if every edge of Pmeets C only in 2-regular points, i.e. 2 edges of P pass through eachpoint.


Cubics Admitting an n-gon

QuestionDoes every cubic admit an inscribed 8-gon? 10-gon? ...

Theorem (T-)Almost every cubic admits an inscribed 2n-gon with n ≥ 4. In fact,every elliptic curve admits such polygons.

Elliptic curves are smooth cubic curves- carries a group law: A,B,C collinear ⇐⇒ A + B + C = 0

- associativity depends on the 8⇒ 9 Theorem.- important for both theoretical (FLT) and practical reasons (ECM, DH)


Higher Degree Curves

2(t + 1) lines inscribed in a line → 2(t + 1)-gon in a degree t curve2(t + 2) lines inscribed in a conic→ 2(t + 2)-gon in a degree t curve...

QuestionDoes almost every degree t curve admit an inscribed 10-gon producedin this manner? 12-gon? . . .?

Theorem (T-, Roth)No (proof is a simple dimension count), but the question remains openfor degree 4 curves. Almost all quartics admit a 10-gon, a 12-gon anda 14-gon.


Inscribed 7-gons on an Elliptic Curve

Can we inscribe a 7-gon on an elliptic curve?- 21 (P,L) pairs with P ∈ C ∩ L- each point appears on 2 lines→ # pairs = 2 # points (Contradiction)What if each point was on 3 lines (a 3-regular inscribed 7-gon)?

No: Fano configuration cannot be embedded in P2.


3-regular Inscribed 9-gons

The Pappus configuration is a 3-regular 9-gon.

Theorem (T-)Every elliptic curve admits a 3-regular inscribed 9-gon. In fact, given apoint P ∈ P2 and an elliptic curve C there are 3 3-regular 9-gonsinscribed in C that pass through P.


4-regular Inscribed 12-gons

Maclaurin: the 9 points of inflection of an elliptic curve lie on 4 sets of 3lines.

Hesse (1844) studied the resulting 4-regular inscribed 12-gon.


Open Questions

QuestionFor which values of r and k do there always exist an inscribed r-regulark-gon in every elliptic curve? What about on every quartic curve?


10 points on a cubic curve

I want to study a simple question about points on cubic curves usingthe Cayley-Bacharach-Chasles Theorem.

QuestionWhen do 10 points lie on a cubic curve?

General equation for a cubic:

ax3 + bx2y + cx2 + dxy2 + exy + fx + gy3 + hy2 + iy + j = 0.

Each point Pi(xi , yi) imposes a linear constraint on the 10 coefficients:

ax3i + bx2

i yi + cx2i + dxiy2

i + exiyi + fxi + gy3i + hy2

i + iyi + j = 0.


Matrix Formulation

We gather these equations in a matrix formulation:

x31 x2

1 y1 x21 x1y2

1 x1y1 x1 y31 y2

1 y1 1x3

2 x22 y2 x2

2 x2y22 x2y2 x2 y3

2 y22 y2 1

x33 x2

3 y3 x23 x3y2

3 x3y3 x3 y33 y2

3 y3 1x3

4 x24 y4 x2

4 x4y24 x4y4 x4 y3

4 y24 y4 1

x35 x2

5 y5 x25 x5y2

5 x5y5 x5 y35 y2

5 y5 1x3

6 x26 y6 x2

6 x6y26 x6y6 x6 y3

6 y26 y6 1

x37 x2

7 y7 x27 x7y2

7 x7y7 x7 y37 y2

7 y7 1x3

8 x28 y8 x2

8 x8y28 x8y8 x8 y3

8 y28 y8 1

x39 x2

9 y9 x29 x9y2

9 x9y9 x9 y39 y2

9 y9 1x3

10 x210y10 x2

10 x10y210 x10y10 x10 y3

10 y210 y10 1

abcdefghij

=

0000000000

There is a non-zero solution when the determinant of the 10× 10matrix is zero.


A Geometric Approach

2 blue conics and 2 red conics define quartics meeting in 16 points.

Cayley-Bacharach-Chasles Theorem⇒ original 10 points lie on acubic iff the six new points lie on a conic.


Ruler and Compass Construction

This insight led David Wehlau and me to a ruler and compassconstruction that checks whether 10 points lie on a cubic (we’re stillnailing down the details).

The construction ought to be much faster than computing the 10× 10determinant.


Straightedge Construction

Meet and Join Algebra:- algebra of points and lines studied by Grassmann and Cayley- meet of two lines is their point of intersection- join of two points is the line through the points - allows algebraicformulation of straightedge-only constructions

Theorem (after Sturmfels and Whiteley)There exists a straightedge construction to determine if 10 points lie ona cubic (with about 100 million lines).


Redux: Cubics Admitting an n-gon

Theorem (T-)Every elliptic curve admits an inscribed 2n-gon with n ≥ 4.

Elliptic curve group law: A,B,C collinear ⇐⇒ A + B + C = 0.

Case n odd:

T = −A0 − Bn−1 = (P1 + A1) + (Pn−1 + Bn−2)= (P1 − P2 + · · · − Pn−1 − An−1) + (Pn−1 − Pn−2 + · · · − P1 − B0)= −An−1 − B0 = Q.


Case: n even

When n is even we get a pair of entwined n-gons:

This is reminiscent of a result due to Mobius (1848).


A result of A. F. Mobius

Theorem (Mobius)Inscribe a polygon with 4k + 2 sides into a conic and consider the2k + 1 points where opposite edges meet. If 2k of these points arecollinear, then so is the last one. Inscribe two 2k-gons in a conic andassociate edges cyclicly. The associated edges meet in 2k points andif 2k − 1 of these points are collinear, then so is the last one.


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