Squares and other polygons inscribed in curves

Squares and other polygonsinscribed in curves

Elizabeth [email protected]

Wellesley CollegeApril 10, 2012

Introduction Some Topology Rectangles Squares Other Polygons Thanks

A proof from “the book”

Theorem (Proof: H. Vaughan 1977)Let γ be a continuous closed curve in the plane withoutself-intersections, then there are four points on γ that are thevertices of a rectangle.

The proof is purely topological.

Principle

• Two objects are the “same" if there is a homeomorphismbetween them.

• Homeomorphism: 1-to-1, onto, continuous map with acontinuous inverse.

• Topological properties are preserved by homeomorphisms(ex. connectedness).

Setting up the proof — a topological detour

• Let γ be a continuous simple closed curve in the plane.• Let M(γ) = {{u, v} |u, v ∈ γ} (order doesn’t matter).• Since γ is homeomorphic to the standard unit circle S1,

we understand M(γ) by understanding M(S1).• Observe that S1 × S1 is represented by

•(u, v)

•(v ,u)

• (u,u)•

(x ,0)

• (0, x)

Visualizing a torus S1 × S1

Gluing a Torus - YouTube

http://www.youtube.com/watch?v=0H5_h-RB0T8

Structure of M(γ) = {{u, v} |u, v ∈ γ}

• S1 × S1 has both (u, v) and (v ,u).• Keep one half of S1 × S1. Cut this in half then reglue.

•x x•

FactM(γ) is homeomorphic to a Möbius band.Also, M(γ)’s boundary corresponds to pairs of points {u,u}.(This can be thought of as a copy of γ.)

Möbius bands are non-orientable surfaces

M.C. Escher

Real Projective PlaneDefinitionReal projective plane RP2 = {lines in R3 through the origin}

• Each line meets the unit sphere S2 in two antipodal points.• Identify a line ` with ` ∩ S2 = {x ,−x}.• Take half of S2 and glue together y and −y on the equator.• RP2 is a surface without boundary.

• 0•

−x•

−y•

Visualizing RP2

Learn more about surfaces:Beginning Topology,

Sue Goodman.

Boy’s surface

How can you visualize topological objects?A Topological Picture Book by George Francis

Real Projective Space and Möbius BandsFactRP2 − disc = Mb with boundary of disc as edge, orRP2 = Mb ∪ disc joined along their boundaries.

•a •b •c •a

disc•b •c

Möbiusband

•a•b •c•a

•e •d

A different view of the Möbius band...

... the Sudanese Möbius band.

Back to proof

• γ simple closed curve in the plane.• γ bounds a disc D (Schoenflies Theorem).• M(γ) = {{u, v} |u, v ∈ γ} homeomorphic to Möbius band.• Boundary of M(γ) is a copy of γ.• We know that gluing together a disc and Möbius band

along their boundary gives RP2.• RP2 = M(γ) ∪γ D

Back to proof

Proof continued. . .

Define F : RP2 → R3 by

F |disc is an inclusion of R2 ↪→ R3, and F |M(γ) isF ({u, v}) = (u+v

2 , ||u − v ||) (take midpoint in R2, length uv ).

• F agrees on γ and is a continuous function.• Fact: F cannot be an embedding (since RP2 is a closed

non-orientable surface).• F |disc is an embedding (by construction).• Problem points are on F |M(γ).

• There are pairs {u, v}, {a,b} with u+v2 = a+b

2 and||u − v || = ||a− b||.

• Thus uv and ab are the diagonals of a rectangle.

2 and||u − v || = ||a− b||.

The Square Peg problem

Question (Toeplitz, 1911)Let γ be a closed curve in the plane with no self-intersections.Are there four points on γ that are the vertices of a square?

Still an open question!Known results require γ to be “smooth enough”.

History of the Square Peg Problem

1. Emch, 1917, proved for piecewise analytic arcs2. Schnirelmann, 1929, proved for bounded curvature∗

3. Christiansen, 1950, for convex curves4. Ogilvy, 1950, for “nice enough” curves5. Jerrard, 1959, for analytic curves6. Guggenheim, 1965, fixes an error in Schnirelmann∗

7. Stromquist, 1989, for a class somewhat larger than C1

8. (Griffiths, 1991, contains serious errors)9. Nielsen, Wright, 2002, centrally symmetric C0 curves

10. Zivaljevik, 2008, preprint11. Cantarella, D, McCleary, joint work-in-progress.

Our results: Squares

Theorem (with Cantarella, McCleary)Let γ be a simple closed continuously differentiable curve in R2.Then there are an odd (or infinite) number of inscribed squaresabcd (in order) on γ.

What about space curves, or curves in Rk?

Theorem (with Cantarella, McCleary)Let γ be a simple closed continuously differentiable curve in Rk .Then there are an odd (or infinite) number of square-likequadrilaterals abcd inscribed in order on γ with equal sides|ab| = |bc| = |cd | = |da| and equal diagonals |ac| = |bd |.

CorollaryFor a simple closed C1 plane curve, there are an odd (orinfinite) number of inscribed squares.

Theorem (corners don’t matter)For a curve of finite total curvature without cusps, there is atleast one inscribed square.

How would you prove something like this?

IdeaIf a pair of “surfaces” α, β ofdimension d1, d2 intersect in aspace of dimension k, then theintersection has dimensiond1 + d2 − k.On the left, d1 = d2 = 2 andk = 3, so the intersection has di-mension 1.

PrincipleAs long as the boundary of β stays in Y and the boundary of αstays in X, we can deform the tube β and the sheet α howeverwe want, and the intersection will still contain a curve that goesall the way around β.

Dimension count for squares

• A point in R2 has two coordinates (x , y). Four points in R2

have eight coordinates (x1, y1, x2, y2, x3, y3, x4, y4).The space of quadruples of points in R2 is eight dim’l.

• A point on a curve γ is described by one coordinate (howfar it is from the start).The space of quadruples of points on γ is four dim’l.

• A square in R2 is described by four coordinates: thecoordinates of the center (x , y), an angle θ by which thesquare is rotated, and the length ` of a side of the square.The space of squares in R2 is four dim’l.

• The space of squares which are inscribed in a curve is4 + 4− 8 = 0 dimensional!

Why an odd number of squares?

An ellipse contains exactly one square:

Proof: Construct inscribed rectangles and use the IntermediateValue Theorem.

Odd number of squares, continued. . .

• If we move the ellipse to some other curve, we deform the4-dimensional space of quadruples of points on the curve.

• We can show that during the deformation, the boundary ofthis space stays away from the boundary of the space ofsquares, and. . .

• . . . the intersection will contain at least one point.• More intersections may appear, but they appear in pairs.• Hence # of squares is 1+ an even number, so it is odd.

Arguments like this are found inDifferential Topology by Victor Guillemin, Alan Pollack.

Our results: Loops of polygons

The next result doesn’t have a parallel in the literature. . .

Theorem (with Cantarella, McCleary)For a generic simple closed C1-smooth curve in Rk , there is asmooth curve of inscribed equilateral n-gons so that one pointon the n-gon travels all of the way around the curve.

This is best understood by watching a movie. . .

Equilateral Triangles on the Jellybean Curve

The set of equilateral triangles on this curve fragments into 7distinct loops. One goes all the way around, the other twogroups of three don’t.

Why is there a loop of polygons?

A circle contains such a loop for any n:

and the rest of the proof (dimension count, deform the circle toyour curve and watch the loop of n-gons deform) goes alongthe same lines as before.

In fact our Loop of Polygons Theorem holds for n-gons(x1, x2, . . . , xn) whose side-lengths are in any constructible ratioR = r1 : · · · : rn. That is |x2 − x1| : · · · : |x1 − xn| = r1 : · · · : rn.

Constructible: we assume each ri > 0 and R obeys the triangleinequalities: ri ≤ r1 + · · ·+ ri−1 + ri+1 + · · ·+ rn.

Theorem (with Cantarella, McCleary)Given a constructible ratio R and a C1-smooth curve in Rk ,there is an arbitrarily C1-close smooth curve γ such thatPolR ∩ Cn[γ] is a collection of smooth, disjoint orientedcircles C.

Moreover the collection [C] represents +1 in H1(Cn[γ];Z) ∼= Z.

Thank you!

Special thanks to

Jason CantarellaJohn McCleary

The movies and some pictures in this talkwere created by Jason Cantarella

Coda 1: A bold conjecture

Among our equilateral quadrilaterals are four squares. Theseare the unique concyclic equilateral quadrilaterals.

ConjectureIn any simple closed plane curve, there are four inscribedconcyclic quadrilaterals with any admissible edgelength ratior1 : · · · : r4.(Partial results by Makeev.)

Why is this so important?

This would give a purely topological proof of the

Theorem (Four-Vertex Theorem)There are at least four critical points for curvature (vertices) onany simple closed C2 curve in the plane.

Proof.The limit of 4 concyclic points coming together on a curve is avertex (Mukhopadyaya). This is one of the degenerateconcyclic 4-tuples (p1, . . . ,p4) with

x1 : · · · : x4 → 1 : 1 : 1 : 3. (1)

Coda 2: A natural question

What happens to the loops of polygons for immersed curves?

Squares and other polygons inscribed in curves

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