Outline • Examples of geometrical star designs • Angle sum of a triangle and other polygons • Properties of a pentagram • A method for constructing other stars • Connections with number theory and complex numbers • Angle sum of arbitrary stars • Turtle programming with Python
Examples – 5 points Somalia
Examples – 5 points Senegal
Examples – 5 points Timor-Leste
Examples – 5 points Saint Kitts and Nevis
Examples – 5 points Morocco
Examples – 5 points Ethiopia
Examples – 5 points
Examples – 6 points Israel
Examples – 6 points
Examples – 6 points 2 disconnected triangles
Examples – 7 points Australia
Examples – 7 points Sheriff badge, Suffolk County, NY
Examples – 7 points
Examples – 7 points 2 different types
Examples – 8 points Azerbaijan
Examples – 8 points 14 th century Iranian tile
Examples – 8 points
Examples – 8 points
Examples – 8 points connected disconnected 2 different types
Examples – 9 points Baha’i symbol Slipknot logo Baha’i symbol
Examples – 9 points connected disconnected connected 3 different types
Examples – other Aruba: 4 points Nauru: 12 points Malaysia: 14 points
Examples – other Aruba: 4 points Nauru: 12 points Malaysia: 2 x 7 points Marshall Islands: 24 pts
Angle sum of a triangle – Method 1 E • Consider triangle 𝐵𝐶𝐷 with angles 𝛽 , 𝛾 and 𝛿 𝛿 • Construct line-segment 𝐸𝐹 A C 𝛿 through 𝐵 parallel to 𝐶𝐷 𝛽 𝛾 • By alternate angles 𝛾 = ∠𝐵𝐶𝐷 = ∠𝐸𝐵𝐶 D 𝛿 = ∠𝐵𝐷𝐶 = ∠𝐹𝐵𝐷 • By adjacent angles in a 𝛾 straight angle 𝛽 + 𝛾 + 𝛿 = 180° B
Angle sum of a triangle – Method 2 • Consider triangle 𝐵𝐶𝐷 with 180° − 𝛿 angles 𝛽 , 𝛾 and 𝛿 A C 𝛿 • Extend sides to form exterior 𝛽 180° − 𝛽 angles 180° − 𝛽 etc. • Shrink 𝐵𝐶𝐷 to a single point with three exterior angles 180° − 𝛿 • By adjacent angles in a 𝛾 revolution 180° − 𝛾 180° − 𝛽 180° − 𝛽 + 180° − 𝛾 + B 180° − 𝛾 (180° − 𝛿) = 360° • Hence 𝛽 + 𝛾 + 𝛿 = 180°
Angle sum of a polygon – Method 1 • Consider an 𝑜 -sided polygon ( 𝑜 = 5 shown) • Divide the polygon into 𝑜 − 2 triangles by adding 𝑜 − 3 diagonals • By angle sum in each triangle, the total angle sum is 𝑜 − 2 × 180°
Angle sum of a polygon – Method 2 • Consider an 𝑜 -sided polygon with angles 𝛽 , 𝛾 , 𝛿 ,… 180° − 𝛾 180° − 𝛽 𝛽 • Extend sides to form exterior 𝛾 angles 180° − 𝛽 etc. • Shrink polygon to a single point with 𝑜 exterior angles 180° − 𝛿 • By adjacent angles in a 𝛿 revolution 180° − 𝛽 + 180° − 𝛾 + 180° − 𝛿 + ⋯ = 360° • Hence 𝛽 + 𝛾 + 𝛿 + ⋯ = 𝑜 − 2 × 180°
Properties of a regular pentagram • Inscribe inside a regular pentagon • Internal angles of pentagon are 36° 36° 3 × 180° ÷ 5 = 108° 36° 108° 72° • By isosceles triangles, angle at 36° 36° 108° each vertex of regular pentagram is 36° , with sum 180° 36° 36° • GeoGebra link: https://tinyurl.com/CMA2018pentagram
Properties of a regular pentagram • Inscribe inside a regular pentagon • Internal angles of pentagon are 3 × 180° ÷ 5 = 108° • By isosceles triangles, angle at each vertex of regular pentagram is 36° , with sum 180° 𝝔 • By similar triangles, the ratio of diagonal and side lengths is the 1+ 5 Golden Ratio 𝜚 = 2 , which is 𝟐 related to Fibonacci etc.
Angle sum of a cyclic pentagram • Consider a cyclic pentagram with angles 𝛽 , 𝛾 , 𝛿 , 𝜀 , 𝜁 at the vertices 𝛾 𝜀 • Construct radii through centre 𝑃 opposite angle 𝛽 • By circle geometry theorem, the 2𝛽 𝑃 associated angle at 𝑃 is 2𝛽 𝛽 𝜁 𝛿
Angle sum of a cyclic pentagram • Consider a cyclic pentagram with angles 𝛽 , 𝛾 , 𝛿 , 𝜀 , 𝜁 at the vertices 𝛾 𝜀 • Construct radii through centre 𝑃 opposite angle 𝛽 2𝛿 • By circle geometry theorem, the 2𝛽 2𝜁 associated angle at 𝑃 is 2𝛽 2𝛾 𝛽 2𝜀 • Similarly for all other angles 𝜁 • By adjacent angles in a revolution 𝛿 2𝛽 + 2𝛾 + 2𝛿 + 2𝜀 + 2𝜁 = 360° • Hence 𝛽 + 𝛾 + 𝛿 + 𝜀 + 𝜁 = 180°
Angle sum of any pentagram • Consider any pentagram with angles 𝛽 , 𝛾 , 𝛿 , 𝜀 , 𝜁 at the vertices 𝛾 𝜀 • Consider triangle with angles 𝛽 , 𝛾 and third angle 𝜄 = 180° − 𝛽 − 𝛾 𝛽 𝜄 𝜁 𝛿
Angle sum of any pentagram • Consider any pentagram with angles 𝛽 , 𝛾 , 𝛿 , 𝜀 , 𝜁 at the vertices 𝛾 𝜀 ∙∙ • Consider triangle with angles 𝛽 , 𝛾 and third angle 𝜄 = 180° − 𝛽 − 𝛾 ∙ • Similarly for the four other angles of the internal pentagon 𝛽 ∙ 𝜄 𝜁 • By angle sum of a pentagon 180° − 𝛽 − 𝛾 + 180° − 𝛾 − 𝛿 + 𝛿 180° − 𝛿 − 𝜀 + 180° − 𝜀 − 𝜁 + 180° − 𝜁 − 𝛽 = 540° • Hence… 𝛽 + 𝛾 + 𝛿 + 𝜀 + 𝜁 = 180°
Constructing regular stars 0 • Choose a number of vertices 𝑜 𝑜 = 5 𝑢 = 2 • Label 𝑜 vertices around a circle 0, 1, 2, … , 𝑜 − 1 4 1 • Choose a skipping number 𝑢 • For each 𝑙 = 0, 1, 2, … , 𝑜 − 1 , join vertex 𝑙 to vertex 𝑙 + 𝑢 (mod 𝑜) • Call this star 𝑇(𝑜, 𝑢) • The example shown is 𝑇 5,2 2 3 • Reversing the order, it is also 𝑇 5,3
Constructing regular stars 0 0 1 4 1 4 2 3 2 3 𝑇 5,1 𝑇 5,2 = 𝑇(5,3)
Constructing regular stars 0 0 0 1 4 1 4 1 4 2 3 2 3 2 3 𝑇 5,0 = 𝑇(5,5) 𝑇 5,1 = 𝑇(5,4) 𝑇 5,2 = 𝑇(5,3) 0 0 0 0 3 3 1 1 1 1 2 2 2 𝑇 2,1 𝑇 3,1 𝑇 4,1 𝑇 4,2
Constructing regular stars 0 0 0 1 4 1 4 1 4 2 3 2 3 2 3 𝑇 5,0 = 𝑇(5,5) 𝑇 5,1 = 𝑇(5,4) 𝑇 5,2 = 𝑇(5,3) 0 0 0 0 3 3 1 1 1 1 2 2 2 𝑇 2,1 𝑇 3,1 𝑇 4,1 𝑇 4,2 = 2 × S(2,1)
Constructing regular stars 𝑇(6,1) 𝑇 6,2 = 2 × 𝑇(3,1) 𝑇 6,3 = 3 × 𝑇(2,1) 𝑇(7,1) 𝑇(7,2) 𝑇(7,3)
Constructing regular stars 𝑇 8,2 = 2 × 𝑇 4,1 𝑇 8,4 = 4 × 𝑇(2,1) 𝑇 8,3 𝑇 9,2 𝑇 9,3 = 3 × 𝑇(3,1) 𝑇 9,4
Stars and number theory 𝑜 • 𝑇 𝑜, 𝑢 = 𝑇(𝑜, 𝑜 − 𝑢) , so we can assume 0 ≤ 𝑢 ≤ 2 • 𝑇 𝑜, 1 is an 𝑜 -sided polygon 𝑜 𝑢 • If 𝑒 = gcd 𝑜, 𝑢 , then 𝑇 𝑜, 𝑢 is 𝑒 rotated copies of 𝑇 𝑒 , 𝑒 • Therefore 𝑇 𝑜, 𝑢 is connected if and only if gcd 𝑜, 𝑢 = 1 • For 𝑜 ≥ 3 , the number of different connected stars with 𝜒 𝑜 𝑜 vertices is 2 , where 𝜒 is the Euler-totient function
Stars and complex numbers 2𝜌 • It is well known that if 𝜕 = cis is the principal 𝑜 th 𝑜 root of unity, the powers 1, 𝜕, 𝜕 2 , 𝜕 3 , … , 𝜕 𝑜−1 form a regular 𝑜 -gon in the complex plane • In general, if 𝑨 = 𝜕 𝑢 is another 𝑜 th root of unity, then the powers 1, 𝑨, 𝑨 2 , 𝑨 3 , … , 𝑨 𝑜−1 form 𝑇(𝑜, 𝑢) • This follows from index laws and the fact that 𝑨 𝑜 = 1 so indices can be reduced modulo 𝑜 • Adjusting the magnitude of 𝑨 introduces spiralling
Angle sum of any pentagram • Consider any pentagram with angles 𝛽 , 𝛾 , 𝛿 , 𝜀 , 𝜁 at the vertices 𝛾 𝜀 • Extend sides to form exterior angles 180° − 𝛽 etc. • Shrink pentagram to a single point with 𝑜 exterior angles 𝛽 𝜁 𝛿
Angle sum of any pentagram • Consider any pentagram with angles 𝛽 , 𝛾 , 𝛿 , 𝜀 , 𝜁 at the vertices 𝛾 𝜀 • Extend sides to form exterior angles 180° − 𝛽 etc. • Shrink pentagram to a single point with 𝑜 exterior angles 𝛽 𝜁 𝛿
Angle sum of any pentagram • Consider any pentagram with angles 𝛽 , 𝛾 , 𝛿 , 𝜀 , 𝜁 at the vertices 𝑇(5,2) 𝛾 𝜀 • Extend sides to form exterior angles 180° − 𝛽 etc. • Shrink pentagram to a single point 𝑢 = 2 with 𝑜 exterior angles which account 𝛽 𝜁 for 𝟑 full revolutions • Then 𝛿 180° − 𝛽 + 180° − 𝛾 + 180° − 𝛿 + 180° − 𝜀 + 180° − 𝜁 = 2 × 360° • Hence 𝛽 + 𝛾 + 𝛿 + 𝜀 + 𝜁 = 180°
Angle sum of any star • Consider the star 𝑇(𝑜, 𝑢) with angles 𝛽 1 , 𝛽 1 , …, 𝛽 𝑜 at the vertices (note that we do not need to assume they are equal!) • The exterior angle at vertex 𝑙 is 180° − 𝛽 𝑙 • Since 𝑢 vertices are skipped at each step, it takes 𝑢 revolutions to trace around the full star • Hence the sum of exterior angles is 𝑜 𝑜 = 180° − 𝛽 𝑙 = 𝑜 × 180° − 𝛽 𝑙 2𝑢 × 180° = 𝑢 × 360° 𝑙=1 𝑙=1 • Therefore the sum of angles at the vertices is (𝑜 − 2𝑢) × 180°
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