Edge Tessellations and The Stamp Folding Problem Ron Umble - - PowerPoint PPT Presentation

Edge Tessellations and The Stamp Folding Problem

Ron Umble Millersville University of Pennsylvania

Vassar College Math Chat

February 23, 2010

Ron Umble (Vassar College Math Chat) Edge Tessellations and Stamp Folding February 23, 2010 1 / 52

The Stamp Folding Problem

How many ways can one con…gure the perforation lines on a sheet of postage stamps so that it folds into a stack of single stamps?

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The Stamp Folding Problem

How many ways can one con…gure the perforation lines on a sheet of postage stamps so that it folds into a stack of single stamps? The solution will follow from our main result...

Ron Umble (Vassar College Math Chat) Edge Tessellations and Stamp Folding February 23, 2010 2 / 52

Tessellations

A tessellation (or tiling) of the plane is a collection of plane …gures that …lls the plane with no overlaps and no gaps

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Tessellations

A tessellation (or tiling) of the plane is a collection of plane …gures that …lls the plane with no overlaps and no gaps A regular tessellation has (non-trivial) translational symmetries in two independent directions

Ron Umble (Vassar College Math Chat) Edge Tessellations and Stamp Folding February 23, 2010 3 / 52

Tessellations

A tessellation (or tiling) of the plane is a collection of plane …gures that …lls the plane with no overlaps and no gaps A regular tessellation has (non-trivial) translational symmetries in two independent directions Examples include edge tessellations, which are generated by re‡ecting a polygon in its edges

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Main Result

Theorem 1. Exactly eight polygons generate edge tessellations:

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Main Result

Theorem 1. Exactly eight polygons generate edge tessellations: Most symmetric examples of Laves tilings (Grunbaum & Shephard, Tiling and Patterns, p. 96)

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Solving the Stamp Folding Problem

Perforation lines are lines of symmetry

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Solving the Stamp Folding Problem

Perforation lines are lines of symmetry Perforation lines miss stamp interiors

Ron Umble (Vassar College Math Chat) Edge Tessellations and Stamp Folding February 23, 2010 13 / 52

Solving the Stamp Folding Problem

Perforation lines are lines of symmetry Perforation lines miss stamp interiors Polygons in Thm 1 with interior angles 90 generate solutions

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Non-Solutions in Theorem 1

Polygons in Theorem 1 with an interior angle of 120 fail because...

Ron Umble (Vassar College Math Chat) Edge Tessellations and Stamp Folding February 23, 2010 14 / 52

Non-Solutions in Theorem 1

Polygons in Theorem 1 with an interior angle of 120 fail because... The bisector of a 120 interior angle contains an edge of some stamp

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The Edge Tesselation Problem

Which polygons generate an edge tessellation of the plane?

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Finding Edge Tesselations

Let G be a polygon that generates an edge tessellation

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Finding Edge Tesselations

Let G be a polygon that generates an edge tessellation If V is a vertex of G, then θ = m\V < 180

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Interior Angles at a Vertex

If G 0 is the re‡ection of G in an edge containing V , the interior angle

• f G 0 at V has measure θ; inductively, every interior angle at V has

measure θ

V θ

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Interior Angles at a Vertex

If G 0 is the re‡ection of G in an edge containing V , the interior angle

• f G 0 at V has measure θ; inductively, every interior angle at V has

measure θ

V θ θ

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Interior Angles at a Vertex

If G 0 is the re‡ection of G in an edge containing V , the interior angle

• f G 0 at V has measure θ; inductively, every interior angle at V has

measure θ

V θ θ θ

Ron Umble (Vassar College Math Chat) Edge Tessellations and Stamp Folding February 23, 2010 19 / 52

Interior Angles at a Vertex

If G 0 is the re‡ection of G in an edge containing V , the interior angle

• f G 0 at V has measure θ; inductively, every interior angle at V has

measure θ

V θ θ θ

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Crystallographic Restriction

An n-center of a tessellation is the center of a group of n rotational symmetries

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Crystallographic Restriction

An n-center of a tessellation is the center of a group of n rotational symmetries Crystallographic Restriction: If P is an n-center of a regular tessellation, then n = 2, 3, 4, 6

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Crystallographic Restriction

An n-center of a tessellation is the center of a group of n rotational symmetries Crystallographic Restriction: If P is an n-center of a regular tessellation, then n = 2, 3, 4, 6 Re‡ecting in adjacent edges of G is a rotational symmetry about the common vertex V

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Crystallographic Restriction

An n-center of a tessellation is the center of a group of n rotational symmetries Crystallographic Restriction: If P is an n-center of a regular tessellation, then n = 2, 3, 4, 6 Re‡ecting in adjacent edges of G is a rotational symmetry about the common vertex V V is an n-center for some n = 2, 3, 4, 6

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Admissible Interior Angles

If G 0 is the image of G under a rotational symmetry at V , the number k

• f copies of G sharing vertex V is the order of the rotational subgroup at

V . Thus k = 3, 4, 6 and θ = 60, 90, 120.

G' G

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Admissible Interior Angles

Otherwise, G and G 0 rotate as a unit and the number k of copies of G sharing vertex V is twice the order of the rotational subgroup at V . Thus k = 4, 6, 8, 12 and θ = 30, 45, 60, 90.

G' G

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Admissible Polygons

Conclusion: Interior angles of G measure 30, 45, 60, 90, 120

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Admissible Polygons

Conclusion: Interior angles of G measure 30, 45, 60, 90, 120 Number of edges e 6 since the int angle sum 180 (e 2) 120e

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Admissible Polygons

Conclusion: Interior angles of G measure 30, 45, 60, 90, 120 Number of edges e 6 since the int angle sum 180 (e 2) 120e There are no regular tilings of the plane by pentagons

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Admissible Polygons

Conclusion: Interior angles of G measure 30, 45, 60, 90, 120 Number of edges e 6 since the int angle sum 180 (e 2) 120e There are no regular tilings of the plane by pentagons Therefore G is a triangle, a quadrilateral, or a hexagon

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Solutions with an Interior Angle of 120

If G has an interior angle of 120, then G is a 120-isosceles triangle, a 120-rhombus, a 60-90-120 kite, or a regular hexagon

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Solutions with Interior Angles Less Than 120

If the interior angles of G measure 90, the number of edges e 4 since the interior angle sum 180 (e 2) 90e

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Solutions with Interior Angles Less Than 120

If the interior angles of G measure 90, the number of edges e 4 since the interior angle sum 180 (e 2) 90e G is a rectangle, an equilateral, a 30-right, or an isosceles right triangle

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Periodic Orbits on Polygons: An Application

A periodic orbit of a billiard ball in motion on a frictionless table Ω bounded by a polygon G is a piecewise-linear constant speed curve α : R !Ω such that α (a) = α (b) for some a < b and α0 (a + t) = α0 (b + t) for almost all t 2 R.

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Periodic Orbits on Polygons

A periodic orbit α retraces the same path n 1 times

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Periodic Orbits on Polygons

A periodic orbit α retraces the same path n 1 times If n = 1, then α is primitive

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Periodic Orbits on Polygons

A periodic orbit α retraces the same path n 1 times If n = 1, then α is primitive If n > 1, then α is an n-fold iterate

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Periodic Orbits on Polygons

A periodic orbit α retraces the same path n 1 times If n = 1, then α is primitive If n > 1, then α is an n-fold iterate If α is primitive, αn denotes its n-fold iterate

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Periodic Orbits on Polygons

A periodic orbit α retraces the same path n 1 times If n = 1, then α is primitive If n > 1, then α is an n-fold iterate If α is primitive, αn denotes its n-fold iterate The period of a periodic orbit α is the number of times the ball strikes a bumper

Ron Umble (Vassar College Math Chat) Edge Tessellations and Stamp Folding February 23, 2010 28 / 52

Periodic Orbits on Polygons

A periodic orbit α retraces the same path n 1 times If n = 1, then α is primitive If n > 1, then α is an n-fold iterate If α is primitive, αn denotes its n-fold iterate The period of a periodic orbit α is the number of times the ball strikes a bumper If α is primitive of period k, then αn has period kn

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Fagnano’s Period 3 Orbit on An Acute Triangle

The orthic triangle is the inscribed triangle of least perimeter

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Fagnano’s Period 3 Orbit on An Acute Triangle

The orthic triangle is the inscribed triangle of least perimeter Vertices are feet of the altitudes

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Fagnano’s Period 3 Orbit on An Acute Triangle

The orthic triangle is the inscribed triangle of least perimeter Vertices are feet of the altitudes Altitudes bisect the interior angles

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Unfolding Orbits on Equilateral Triangles

This technique is due to H. A. Schwarz

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Unfolding Orbits on Equilateral Triangles

This technique is due to H. A. Schwarz Let P be the initial position of the ball

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Unfolding Orbits on Equilateral Triangles

This technique is due to H. A. Schwarz Let P be the initial position of the ball Release the ball with velocity ! v

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Unfolding Orbits on Equilateral Triangles

This technique is due to H. A. Schwarz Let P be the initial position of the ball Release the ball with velocity ! v Re‡ect the triangle in the side the ball will strike next

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Unfolding Orbits on Equilateral Triangles

This technique is due to H. A. Schwarz Let P be the initial position of the ball Release the ball with velocity ! v Re‡ect the triangle in the side the ball will strike next Repeat until the ball arrives at some image Q of P with velocity ! v

Ron Umble (Vassar College Math Chat) Edge Tessellations and Stamp Folding February 23, 2010 30 / 52

Unfolding Orbits on Equilateral Triangles

This technique is due to H. A. Schwarz Let P be the initial position of the ball Release the ball with velocity ! v Re‡ect the triangle in the side the ball will strike next Repeat until the ball arrives at some image Q of P with velocity ! v PQ is an unfolding of the orbit

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A C C B B A

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Finding Orbits on Equilateral Triangles

Let T be the edge tessellation generated by an equilateral 4ABC

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Finding Orbits on Equilateral Triangles

Let H be the group generated by re‡ections in the lines of T . The action of H on an edge of a basic triangle generates a tessellation H by hexagons.

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Finding Orbits on Equilateral Triangles

Theorem 2. Let α be an orbit and let PQ be an unfolding. Then (i) α has even period i¤ Q lies on a horizontal edge of H. (ii) α has odd period i¤ α = γ2k1 for some k 1.

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Equivalent Orbits

Periodic orbits α and β are equivalent if there exist respective unfoldings PQ and RS and a horizontal translation τ such that RS = τ

• PQ
• .

S Q P R

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Equivalent Orbits

Periodic orbits α and β are equivalent if there exist respective unfoldings PQ and RS and a horizontal translation τ such that RS = τ

• PQ
• .

S Q P R

Two unfoldings that terminate on the same horizontal edge of H are equivalent.

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Equivalent Orbits

• Remark. Sequences of incidence angles in equivalent unfoldings di¤er

by a permutation. One could consider a …ner relation in which sequences di¤er by a cyclic permutation.

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Equivalent Orbits

• Remark. Sequences of incidence angles in equivalent unfoldings di¤er

by a permutation. One could consider a …ner relation in which sequences di¤er by a cyclic permutation. The period of a class is the period of its elements

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Equivalent Orbits

• Remark. Sequences of incidence angles in equivalent unfoldings di¤er

by a permutation. One could consider a …ner relation in which sequences di¤er by a cyclic permutation. The period of a class is the period of its elements Even (period) classes have cardinality c

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Equivalent Orbits

• Remark. Sequences of incidence angles in equivalent unfoldings di¤er

by a permutation. One could consider a …ner relation in which sequences di¤er by a cyclic permutation. The period of a class is the period of its elements Even (period) classes have cardinality c Odd classes are singletons

• γ2k1

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Equivalent Orbits

• Remark. Sequences of incidence angles in equivalent unfoldings di¤er

by a permutation. One could consider a …ner relation in which sequences di¤er by a cyclic permutation. The period of a class is the period of its elements Even (period) classes have cardinality c Odd classes are singletons

• γ2k1

Since H has countably many edges, there are countably many even classes

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Rhombic Coordinates

T is a tessellation of the plane with equilateral triangles

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Rhombic Coordinates

T is a tessellation of the plane with equilateral triangles P is a point on the horizontal edge AB of some basic triangle

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Rhombic Coordinates

T is a tessellation of the plane with equilateral triangles P is a point on the horizontal edge AB of some basic triangle s is the union of all line segments from P to its images

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Rhombic Coordinates

T is a tessellation of the plane with equilateral triangles P is a point on the horizontal edge AB of some basic triangle s is the union of all line segments from P to its images τ is a horizontal translation such that τ (s) misses the vertices of T

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Rhombic Coordinates

T is a tessellation of the plane with equilateral triangles P is a point on the horizontal edge AB of some basic triangle s is the union of all line segments from P to its images τ is a horizontal translation such that τ (s) misses the vertices of T O = τ (P) is the origin

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Rhombic Coordinates

T is a tessellation of the plane with equilateral triangles P is a point on the horizontal edge AB of some basic triangle s is the union of all line segments from P to its images τ is a horizontal translation such that τ (s) misses the vertices of T O = τ (P) is the origin AB is the unit of length

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Rhombic Coordinates

T is a tessellation of the plane with equilateral triangles P is a point on the horizontal edge AB of some basic triangle s is the union of all line segments from P to its images τ is a horizontal translation such that τ (s) misses the vertices of T O = τ (P) is the origin AB is the unit of length ! AB is the x-axis

Ron Umble (Vassar College Math Chat) Edge Tessellations and Stamp Folding February 23, 2010 43 / 52

Rhombic Coordinates

T is a tessellation of the plane with equilateral triangles P is a point on the horizontal edge AB of some basic triangle s is the union of all line segments from P to its images τ is a horizontal translation such that τ (s) misses the vertices of T O = τ (P) is the origin AB is the unit of length ! AB is the x-axis The line through O with inclination 60 is the y-axis

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Rhombic Coordinates

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Periodic Orbits and Rhombic Coordinates

Software for …nding periodic orbits: “Orbit Tracer 2” by Stephen Weaver at

http://www.millersville.edu/~rumble/StudentProjects/Weaver/Tessellation_v1.1.exe

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The Fundamental Region

Proposition 1. Every periodic orbit strikes some side of 4ABC with an incidence angle in the range 30 θ 60. Every periodic orbit has an unfolding in ΓO = f(x, y) j 0 x yg

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Classi…cation of Even Orbits

Let x, y 2 Z (x, y) lies on a horizontal edge of H i¤ x y (mod 3) .

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Classi…cation of Even Orbits

Let x, y 2 Z (x, y) lies on a horizontal edge of H i¤ x y (mod 3) . Theorem 3. There is a bijection f(x, y) 2 ΓO j x y (mod 3)g $ forbits with even periodg

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Classi…cation of Even Orbits

Let x, y 2 Z (x, y) lies on a horizontal edge of H i¤ x y (mod 3) . Theorem 3. There is a bijection f(x, y) 2 ΓO j x y (mod 3)g $ forbits with even periodg Theorem 4. (Classi…cation) (i) α has odd period i¤ α = γ2k1 for some k 1. (ii) α $ (x, y) has period 2n i¤ x y (mod 3) and x + y = n.

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Counting Classes of Periodic Orbits

Distinct classes may have the same period:

(1,10) (4,7) B C A O

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Counting Classes of Periodic Orbits

Theorem 5. (Counting Formulas) (i) The number of classes with period 2n is exactly O (n) = n + 2 2

• n + 2

3

• .

(ii) The number of primitive classes with period 2n is exactly P (n) = ∑

djn

µ (d) O (n/d) , where µ (d) = 8 < : 1, d = 1 (1)r , d = p1p2 pr for distinct primes pi 0,

• therwise.

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Conclusions

There are no orbits of period 2

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Conclusions

There are no orbits of period 2 There is at least one class of period 2n for each n 2

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Conclusions

There are no orbits of period 2 There is at least one class of period 2n for each n 2 The only primitive odd orbit is Fagnano’s period 3

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Conclusions

There are no orbits of period 2 There is at least one class of period 2n for each n 2 The only primitive odd orbit is Fagnano’s period 3 There are no primitive orbits of period 8, 12, or 20

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Conclusions

There are no orbits of period 2 There is at least one class of period 2n for each n 2 The only primitive odd orbit is Fagnano’s period 3 There are no primitive orbits of period 8, 12, or 20 All classes of period 2n are primitive i¤ n is prime

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Project Proposal

Find, classify and count the classes of periodic orbits on a polygon G that generates an edge tessellation

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The End

Thank you!

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