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The Golden Ladder
- 1. The Golden Ladder ByWeixin Wu Advisor Professor Shablinsky May 2nd 2015
- 2. What is the golden ratio? ο Two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities.
- 3. Exact value of phi(Ο) and its properties ο Ο = (β5 + 1) / 2 β 1.618 ο Ο = 1 / Ο or Ο * Ο = 1 ο Ο = (β5 - 1) / 2 β 0.618 ο Ο2 = 1 + Ο ( x2 β x β 1 = 0, the positive root is Ο)
- 4. Some famous Euclidean geometric shapes
- 5. Apply Golden Ratio to tangent circles d = Ο d = 1 d = Ο d = Ο2 d = Ο3
- 6. Flip all triangles into the previous triangle d = Ο d = 1d = Ο2 d = Ο
- 7. Recursive property one ο All triangles are similar. ο The ratio between the corresponding sides of any triangle and the next smaller triangle is Ο.
- 8. Proof of recursive property one S1 S2 S3 S3 S4 S5
- 9. Proof of recursive property one S1 S2 S3 S3 S4 S5 π1 π3 = π2 π4 = π3 π5 =Ο οΌ
- 10. Proof of recursive property one S1 = (Ο + 1)/2 S2 = (Ο + Ο)/2 S3 = (1 + Ο)/2 S4 = (1 + Ο2)/2 S5 = (Ο + Ο2)/2 β¦.. Sn= (Οm + Οm+1)/2 Sn+1= (Οm + Οm+2)/2
- 11. Proof of recursive property one Sn = (Οm + Οm+1)/2 Sn+1 = (Οm + Οm+2)/2 Sn+2 = (Οm+1 + Οm+2)/2 Sn+3 = (Οm+1 + Οm+3)/2
- 12. Proof of recursive property one Case one: Sn Sn+2 = (Οm + Οm+1)/2 (Οm+1 + Οm+2)/2 = 1 + Ο Ο + Ο2 = 1 Ο = Ο
- 13. Proof of recursive property one Case two: Sn+1 Sn+3 = (Οm + Οm+2)/2 (Οm+1 + Οm+3)/2 = 1 + Ο2 Ο + Ο3 = 1 Ο = Ο
- 14. Proof of recursive property one Sn is the length of the nth side Sn Sn+2 = Ο
- 15. Recursive property two ο The area ratio between any triangle and its next smaller triangle is Ο2. A1 A2 S1 S2 S3 S3 S4 S5 A1 A2 = Ο2 οΌ
- 16. Proof of recursive property two π1 = π1 + π2 + π3 2 π΄1 = βπ1 β π1 β π1 β π1 β π2 β (π1 β π3) π2 = π3 + π4 + π5 2 π΄2 = βπ2 β π2 β π3 β π2 β π4 β (π2 β π5) By using Heronβs formula:
- 17. Proof of recursive property two π΄1 π΄2 = βπ1 π1 β π1 π1 β π2 (π1 β π3) βπ2 π2 β π3 π2 β π4 (π2 β π5) βπ1 βπ2 = β (π1 + π2 + π3)/2 (π3 + π4 + π5)/2 = β (Ο β π3 + Ο β π4 + Ο β π5) (π3 + π4 + π5) = β Ο(π3 + π4 + π5) π3 + π4 + π5 = βΟ β(π1 β π1) β(π2 β π3) = β (π2 + π3)/2 (π4 + π5)/2 = β (Ο β π4 + Ο β π5) (π4 + π5) = β Ο(π4 + π5 β π3) π4 + π5 β π3 = βΟ
- 18. Proof of recursive property two π΄1 π΄2 = βπ1 π1 β π1 π1 β π2 (π1 β π3) βπ2 π2 β π3 π2 β π4 (π2 β π5) = βΟ βΟ βΟ βΟ = Ο 2 π΄ π π΄ π + 1 = βππ π π β π π π π β π π + 1 (ππ β π π + 2 ) βππ + 1 π π + 1 β π π + 2 π π + 1 β π π + 3 (ππ + 1 β π π + 4 ) = βΟ βΟ βΟ βΟ = Ο 2
- 19. Proof of recursive property two An is the area of the nth triangle An An+1 = Ο 2
- 20. Recursive property three ο The ratio between the area of any triangle and the total area of all embedded triangles is Ο.
- 21. Proof of recursive property three A1 A2 A3 A4 An i=n+1 β Ai = Ο?
- 22. Proof of recursive property three π΄ π π΄ π+1 + π΄ π+2 + β― + π΄ π ππ π ππππ π‘π πππππππ‘π¦ An + An+1 + An+2 + β― + Am ππ π ππππ π‘π πππππππ‘π¦ π΄ π An+1 = Ο2
- 23. Proof of recursive property three An + A n π 2 + A n π 4 + β― + A n π 2π ππ π ππππ π‘π πππππππ‘π¦ π΄ π β 1 1 β 1 π2 By using geometric series, the above expression becomes:
- 24. Proof of recursive property three π΄ π π΄ π+1 + π΄ π+2 + β― + π΄ π = π΄ π π΄ π β 1 1 β 1 π2 β π΄ π π΄ π π΄ π β 1 1 β 1 π2 β π΄ π = π2 β 1 = π
- 25. Proof of recursive property three An is the area of the nth triangle An i=n+1 β Ai = Ο
- 26. More proofs in the future
- 27. The Golden Ladder in 3D
- 28. Thank you!