Squares and other polygons inscribed in curves
Transcript of Squares and other polygons inscribed in curves
Squares and other polygonsinscribed in curves
Elizabeth [email protected]
Wellesley CollegeApril 10, 2012
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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).
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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
S1
S1
•(u, v)
•(v ,u)
• (u,u)•
(x ,0)
• (0, x)
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Visualizing a torus S1 × S1
Gluing a Torus - YouTube
http://www.youtube.com/watch?v=0H5_h-RB0T8
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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.
a
a
a
m
•x
x•
a
a
a
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mm
•x x•
a
a
mm
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.)
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Möbius bands are non-orientable surfaces
M.C. Escher
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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•
−x•
0•
x•
y•
−y•
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Visualizing RP2
Learn more about surfaces:Beginning Topology,
Sue Goodman.
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Boy’s surface
How can you visualize topological objects?A Topological Picture Book by George Francis
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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
•d
•e
disc•b •c
Möbiusband
•a•b •c•a
•e •d
•a
•c
•b
•a
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A different view of the Möbius band...
... the Sudanese Möbius band.
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Back to proof
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.
• 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
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Back to proof
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.
• 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
Introduction Some Topology Rectangles Squares Other Polygons Thanks
Back to proof
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.
• 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
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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.
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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.
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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.
Introduction Some Topology Rectangles Squares Other Polygons Thanks
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.
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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.
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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”.
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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.
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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?
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Our results: Squares
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.
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Our results: Squares
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.
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Our results: Squares
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.
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How would you prove something like this?
X X
Y
Y
α
β
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.
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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!
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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!
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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!
Introduction Some Topology Rectangles Squares Other Polygons Thanks
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!
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Why an odd number of squares?
An ellipse contains exactly one square:
Proof: Construct inscribed rectangles and use the IntermediateValue Theorem.
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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.
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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.
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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.
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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.
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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. . .
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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.
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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.
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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.
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Thank you!
Special thanks to
Jason CantarellaJohn McCleary
The movies and some pictures in this talkwere created by Jason Cantarella
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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.)
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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)
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Coda 2: A natural question
What happens to the loops of polygons for immersed curves?