Stamp Folding Puzzles: A Delightful Excursion in Recreational - - PowerPoint PPT Presentation

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Stamp Folding Puzzles: A Delightful Excursion in Recreational - - PowerPoint PPT Presentation

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Stamp Folding Puzzles: A Delightful Excursion in Recreational Geometry Ron Umble, speaker Millersville Univ of Pennsylvania MU/F&M Mathematics Colloquium April 7, 2011 (MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7,

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Stamp Folding Puzzles:

A Delightful Excursion in Recreational Geometry Ron Umble, speaker Millersville Univ of Pennsylvania

MU/F&M Mathematics Colloquium

April 7, 2011

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 1 / 28

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The Map Folding Problem

In the 1930s, Stanislav Ulam posed the following Map Folding Problem: How many ways can one fold a sheet of square "stamps" into a packet the size of one stamp? Easier problem: Suppose the "map" is a horizontal strip of stamps: Number stamps from left-to-right. Fold with stamp #1 face up & upright. Try this with a strip of three stamps.

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 2 / 28

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Solution for Three Stamps

A strip of three stamps can be folded six ways: (here the front of stamp #1 is marked)

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 3 / 28

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Folding a Strip of n Stamps

# Stamps: 2 3 4 5 6 7 8 9 10 · · · n # Foldings: 2 6 16 50 144 462 1392 4536 14060 · · · ? No formula for the nth term is known

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 4 / 28

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Folding a Strip of n Stamps

# Stamps: 2 3 4 5 6 7 8 9 10 · · · n # Foldings: 2 6 16 50 144 462 1392 4536 14060 · · · ? No formula for the nth term is known On-line Encyclopedia of Integer Sequences: A000136

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 4 / 28

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Folding a Strip of n Stamps

# Stamps: 2 3 4 5 6 7 8 9 10 · · · n # Foldings: 2 6 16 50 144 462 1392 4536 14060 · · · ? No formula for the nth term is known On-line Encyclopedia of Integer Sequences: A000136

OEIS was launched in 1996

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 4 / 28

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Folding a Strip of n Stamps

# Stamps: 2 3 4 5 6 7 8 9 10 · · · n # Foldings: 2 6 16 50 144 462 1392 4536 14060 · · · ? No formula for the nth term is known On-line Encyclopedia of Integer Sequences: A000136

OEIS was launched in 1996 Initiated in 1964 by Neil Sloane while a grad student at Cornell

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 4 / 28

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Folding a Strip of n Stamps

# Stamps: 2 3 4 5 6 7 8 9 10 · · · n # Foldings: 2 6 16 50 144 462 1392 4536 14060 · · · ? No formula for the nth term is known On-line Encyclopedia of Integer Sequences: A000136

OEIS was launched in 1996 Initiated in 1964 by Neil Sloane while a grad student at Cornell > 10, 000 new entries have been added each year

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 4 / 28

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Folding a Strip of n Stamps

# Stamps: 2 3 4 5 6 7 8 9 10 · · · n # Foldings: 2 6 16 50 144 462 1392 4536 14060 · · · ? No formula for the nth term is known On-line Encyclopedia of Integer Sequences: A000136

OEIS was launched in 1996 Initiated in 1964 by Neil Sloane while a grad student at Cornell > 10, 000 new entries have been added each year OEIS now has > 180, 500 entries

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 4 / 28

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Folding a Strip of n Stamps

# Stamps: 2 3 4 5 6 7 8 9 10 · · · n # Foldings: 2 6 16 50 144 462 1392 4536 14060 · · · ? No formula for the nth term is known On-line Encyclopedia of Integer Sequences: A000136

OEIS was launched in 1996 Initiated in 1964 by Neil Sloane while a grad student at Cornell > 10, 000 new entries have been added each year OEIS now has > 180, 500 entries

The Map Folding Problem is open and not well understood

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 4 / 28

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Stamp Folding Puzzles

Counting all possible foldings is a difficult problem...

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 5 / 28

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Stamp Folding Puzzles

Counting all possible foldings is a difficult problem... So instead, let’s fold the stamps in some specified configuration

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 5 / 28

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Stamp Folding Puzzles

Counting all possible foldings is a difficult problem... So instead, let’s fold the stamps in some specified configuration Such probems are called Stamp Folding Puzzles

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 5 / 28

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Stamp Folding Puzzle #1

Fold this 4 × 4 sheet of stamps into a 2 × 2 square showing the four green squares

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 6 / 28

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Stamp Folding Puzzle #1

Fold this 4 × 4 sheet of stamps into a 2 × 2 square showing the four green squares yellow squares

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 6 / 28

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Stamp Folding Puzzle #1

Fold this 4 × 4 sheet of stamps into a 2 × 2 square showing the four green squares yellow squares blue squares

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 6 / 28

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Stamp Folding Puzzle #1

Fold this 4 × 4 sheet of stamps into a 2 × 2 square showing the four green squares yellow squares blue squares red squares

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 6 / 28

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Stamp Folding Puzzle #2

Fold this block of equilateral triangular stamps into a packet 9-deep with stamps in the following order: 2 6 7 5 9 3 4 1 8 (Hint: tuck 5 between 7 and 9.)

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 7 / 28

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Stamp Folding Puzzle #3

Fold this block of isosceles right triangular stamps into a packet 16-deep with stamps in the following order: 4 1 16 6 5 15 14 8 7 13 11 12 2 3 9 10

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 8 / 28

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Stamp Folding Puzzle #4

Fold this block of 60◦-right triangular stamps into a packet 12-deep with stamps in the following order: 5 2 8 9 7 3 4 11 12 1 6 10

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 9 / 28

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Fredrickson’s Conjecture*

Although triangular stamps have come in a variety of different triangular shapes, only three shapes seem suitable for [stamp] folding puzzles: equilateral, isosceles right triangles, and 60◦-right triangles.

*G. Fredrickson. "Piano-Hinged Dissections: Time to Fold!" A.K. Peters, Ltd., Wellesley, MA, 2006, p. 144.

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 10 / 28

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The HYKU Theorem

(Hall, York, Kirby, U - 2009) Exactly eight polygons generate edge tessellations of the plane:

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 11 / 28

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MU alum/students Matt Kirby, Josh York, Andrew Hall

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Corollary (Settling Fredrickson’s Conjecture)

Exactly four shapes are suitable for stamp folding puzzles: Rectangles; equilateral, isosceles right, 60◦-right triangles.

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Proof of Fredrickson’s Conjecture

Let G be a polygon generating a suitable edge tessellation

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Proof of Fredrickson’s Conjecture

Let G be a polygon generating a suitable edge tessellation Let V be a vertex of G

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 14 / 28

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Proof of Fredrickson’s Conjecture

Let G be a polygon generating a suitable edge tessellation Let V be a vertex of G The interior angle of G at V has measure θ < 180◦

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 14 / 28

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Proof of Fredrickson’s Conjecture

Let G be a polygon generating a suitable edge tessellation Let V be a vertex of G The interior angle of G at V has measure θ < 180◦ Let G be the reflection of G in an edge containing vertex V G' G V

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 14 / 28

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Admissible Interior Angles

The interior angle of G at V has measure θ; inductively, every interior angle at V has measure θ.

G' G V θ θ

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 15 / 28

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Admissible Interior Angles

The interior angle of G at V has measure θ; inductively, every interior angle at V has measure θ.

θ G G' θ V θ

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 16 / 28

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Admissible Interior Angles

The interior angle of G at V has measure θ; inductively, every interior angle at V has measure θ.

θ G G' θ V θ

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 17 / 28

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Admissible Interior Angles

A point P is an n-center of a tessellation if the group of rotational symmetries centered at P is generated by a rotation of φn =360/n◦

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 18 / 28

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Admissible Interior Angles

A point P is an n-center of a tessellation if the group of rotational symmetries centered at P is generated by a rotation of φn =360/n◦ An n-center is even if n is even

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 18 / 28

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Admissible Interior Angles

A point P is an n-center of a tessellation if the group of rotational symmetries centered at P is generated by a rotation of φn =360/n◦ An n-center is even if n is even Successively reflecting in the edges of G meeting at V is a rotational symmetry

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 18 / 28

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Admissible Interior Angles

A point P is an n-center of a tessellation if the group of rotational symmetries centered at P is generated by a rotation of φn =360/n◦ An n-center is even if n is even Successively reflecting in the edges of G meeting at V is a rotational symmetry V is an n-center for some n > 1

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 18 / 28

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Admissible Interior Angles

A point P is an n-center of a tessellation if the group of rotational symmetries centered at P is generated by a rotation of φn =360/n◦ An n-center is even if n is even Successively reflecting in the edges of G meeting at V is a rotational symmetry V is an n-center for some n > 1 θ =    φn if the bisector of ∠V is a line of symmetry

1 2φn

therwise

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 18 / 28

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Admissible Interior Angles

Case n = 2 : φ2 = 180◦

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 19 / 28

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Admissible Interior Angles

Case n = 2 : φ2 = 180◦ θ < 180◦ ⇒ θ = 90◦

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 19 / 28

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Admissible Interior Angles

Case n = 2 : φ2 = 180◦ θ < 180◦ ⇒ θ = 90◦ Case n = 3 : φ3 = 120◦ ⇒ θ ∈ {60◦, 120◦}

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 19 / 28

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Admissible Interior Angles

Case n = 2 : φ2 = 180◦ θ < 180◦ ⇒ θ = 90◦ Case n = 3 : φ3 = 120◦ ⇒ θ ∈ {60◦, 120◦} Case n ≥ 4 : φn ≤ 90◦ ⇒ θ ∈

90◦, 72◦, 60◦, 513

7

, 45◦, . . .
(MU/F&M Mathematics Colloquium )

Stamp Folding Puzzles April 7, 2011 19 / 28

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Admissible Interior Angles

Case n = 2 : φ2 = 180◦ θ < 180◦ ⇒ θ = 90◦ Case n = 3 : φ3 = 120◦ ⇒ θ ∈ {60◦, 120◦} Case n ≥ 4 : φn ≤ 90◦ ⇒ θ ∈

90◦, 72◦, 60◦, 513

7

, 45◦, . . .
Conclude: θ ∈ {x | nx = 360◦, n ≥ 3}

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 19 / 28

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Admissible Interior Angles

Case n = 2 : φ2 = 180◦ θ < 180◦ ⇒ θ = 90◦ Case n = 3 : φ3 = 120◦ ⇒ θ ∈ {60◦, 120◦} Case n ≥ 4 : φn ≤ 90◦ ⇒ θ ∈

90◦, 72◦, 60◦, 513

7

, 45◦, . . .
Conclude: θ ∈ {x | nx = 360◦, n ≥ 3}

Obtuse interior angles measure 120◦

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 19 / 28

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Angle Bisectors and Lines of Symmetry

If θ = 120◦, three copies of G share vertex V

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 20 / 28

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Angle Bisectors and Lines of Symmetry

If m∠V = 120◦, three copies of G share vertex V Let e and e be the edges of G that meet at V , and labeled so that the angle from e to e measures 120◦

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 21 / 28

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Angle Bisectors and Lines of Symmetry

Let e and e be the images of e and e under a 120◦ rotation

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 22 / 28

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Angle Bisectors and Lines of Symmetry

Let e and e be the images of e and e under a 120◦ rotation Then e and bisector s of ∠V are collinear

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 22 / 28

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Angle Bisectors and Lines of Symmetry

Let e and e be the images of e and e under a 120◦ rotation Then e and bisector s of ∠V are collinear e is the reflection of e in bisector s

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 22 / 28

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Angle Bisectors and Lines of Symmetry

Let e and e be the images of e and e under a 120◦ rotation Then e and bisector s of ∠V are collinear e is the reflection of e in bisector s Bisector s is on a line of symmetry

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 22 / 28

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Angle Bisectors and Lines of Symmetry

Argument above depends only on the parity of n

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 23 / 28

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Angle Bisectors and Lines of Symmetry

Argument above depends only on the parity of n Conclusion: If n is odd, the bisector s of ∠V is a line of symmetry

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 23 / 28

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Angle Bisectors and Lines of Symmetry

Argument above depends only on the parity of n Conclusion: If n is odd, the bisector s of ∠V is a line of symmetry A suitable edge tessellation has strictly even n-centers

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 23 / 28

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Angle Bisectors and Lines of Symmetry

Argument above depends only on the parity of n Conclusion: If n is odd, the bisector s of ∠V is a line of symmetry A suitable edge tessellation has strictly even n-centers Even n-centers ⇒ θ ∈ S = {x | nx = 180◦, n ≥ 2} = {90◦, 60◦, 45◦, 36◦, 30◦, . . .}

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 23 / 28

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Angle Bisectors and Lines of Symmetry

Argument above depends only on the parity of n Conclusion: If n is odd, the bisector s of ∠V is a line of symmetry A suitable edge tessellation has strictly even n-centers Even n-centers ⇒ θ ∈ S = {x | nx = 180◦, n ≥ 2} = {90◦, 60◦, 45◦, 36◦, 30◦, . . .} Only non-obtuse polygons generate suitable edge tessellations

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 23 / 28

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Admissible Polygons

Let g be the number of edges of G

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 24 / 28

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Admissible Polygons

Let g be the number of edges of G Interior angle sum: 180◦ (g − 2) ≤ 90◦g ⇒ g ≤ 4

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 24 / 28

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Admissible Polygons

Let g be the number of edges of G Interior angle sum: 180◦ (g − 2) ≤ 90◦g ⇒ g ≤ 4 Case g = 4 : Each interior angle θi ≤ 90◦

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 24 / 28

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Admissible Polygons

Let g be the number of edges of G Interior angle sum: 180◦ (g − 2) ≤ 90◦g ⇒ g ≤ 4 Case g = 4 : Each interior angle θi ≤ 90◦ Interior angle sum of a quadrilateral: 360◦ ⇒ θi = 90◦

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 24 / 28

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Admissible Polygons

Let g be the number of edges of G Interior angle sum: 180◦ (g − 2) ≤ 90◦g ⇒ g ≤ 4 Case g = 4 : Each interior angle θi ≤ 90◦ Interior angle sum of a quadrilateral: 360◦ ⇒ θi = 90◦ Conclusion: G is a rectangle

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 24 / 28

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Admissible Right Triangles

Case g = 3 : G = ∆ABC

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 25 / 28

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Admissible Right Triangles

Case g = 3 : G = ∆ABC m∠A ≤ m∠B and m∠C = 90◦

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 25 / 28

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Admissible Right Triangles

Case g = 3 : G = ∆ABC m∠A ≤ m∠B and m∠C = 90◦ m∠A + m∠B = 90◦

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 25 / 28

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Admissible Right Triangles

Case g = 3 : G = ∆ABC m∠A ≤ m∠B and m∠C = 90◦ m∠A + m∠B = 90◦ (m∠A, m∠B) ∈ S × S ⇒ (m∠A, m∠B) ∈ {(30◦, 60◦) , (45◦, 45◦)}

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 25 / 28

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Admissible Right Triangles

Case g = 3 : G = ∆ABC m∠A ≤ m∠B and m∠C = 90◦ m∠A + m∠B = 90◦ (m∠A, m∠B) ∈ S × S ⇒ (m∠A, m∠B) ∈ {(30◦, 60◦) , (45◦, 45◦)} Conclusion: G is a 60◦-right or an isosceles-right triangle

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Admissible Acute Triangles

Each interior angle θi ≤ 60◦

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Admissible Acute Triangles

Each interior angle θi ≤ 60◦ Interior angle sum of a triangle: 180◦ ⇒ θi = 60◦

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 26 / 28

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Admissible Acute Triangles

Each interior angle θi ≤ 60◦ Interior angle sum of a triangle: 180◦ ⇒ θi = 60◦ G is a equilateral triangle

(MU/F&M Mathematics Colloquium ) Stamp Folding Puzzles April 7, 2011 26 / 28

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Admissible Acute Triangles

Each interior angle θi ≤ 60◦ Interior angle sum of a triangle: 180◦ ⇒ θi = 60◦ G is a equilateral triangle The proof is complete

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Recap:

Exactly four shapes are suitable for stamp folding puzzles: Rectangles; equilateral, isosceles right, 60◦-right triangles.

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SLIDE 69

The End

Thank you!

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