S2T2

Jim Randell 10:27 am on 8 November 2020 Permalink Reply
Tags: by: Nick MacKinnon ( 43 )   

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  Jim Randell 10:27 am on 8 November 2020 Permalink | Reply

  We examined the general case of this problem in Enigma 34.

  And maximum width of the painting is given by:

  z = (a^(2/3) + b^(2/3))^(3/2)

  

  In particular this means that if we have a Pythagorean triple (x, y, z) (so: x² + y² = z²), then the maximum width painting for corridors of width x³ and y³ is z³.

  In this case: a = 27 = 3³ and b = 64 = 4³, so z = 5³ = 125.

  Solution: The painting is 125 ft wide.

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  • 

    Jim Randell 10:20 pm on 11 November 2020 Permalink | Reply

    Or a numerical solution:

    from math import cos, sin, pi
    from enigma import fdiv, find_min, printf
    
    # width of the legs
    (a, b) = (27, 64)
    
    # length of rod
    z = lambda theta: fdiv(a, cos(theta)) + fdiv(b, sin(theta))
    
    # find the minimum length
    r = find_min(z, 0, 0.5 * pi)
    
    # output solution
    printf("width = {r.fv:g} ft")
    

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• 

  Frits 4:13 pm on 10 November 2020 Permalink | Reply

  #    -----+-----------+
  #     27  .\          |
  #    -----+---\  W    |
  #           x |  \    |
  #             |y    \ |
  #             |.......+
  #             |       |
  #             |  64   |
  #
  # W =  Width painting
  #
  # W^2 = (x + 64)^2 + (y + 27)^2
  #
  # x / 27 = 64 / y --> xy = 27*64 --> y = 27*64/x
  #
  # W = ((x + 64)^2 + (27*64/x + 72)^2)^0.5
  #
  # W is minimal if W' = 0
  #
  # W' = 0.5(((x + 64)^2 + (27*64/x + 72)^2)^-0.5 * 
  #      (2 * (x + 64) + 2 * (27*64/x + 27)) * 
  #      -27*64/x^2  
  #
  # term 2, 3 = x + 64 + (27*64/x + 27)) * -27*64/x^2 
  #           = x + 64 - 27*27*64/x^2 - (27*64)^2/x^3
  #           = x^4 + 64 * x^3 - 27*27*64 * x - (27*64)^2
  #           = x^3 * (x + 64) - 27*27*64 * (x + 64)
  #           = (x^3 - 27*27*64) * (x + 64)
  #
  # W' = 0 if x = -64 or x = 36, disregard negative x
  #
  # x = 36 --> y = 48
  #
  # W^2 = (36 + 64)^2 + (48 + 27)^2 = 15625 --> W = 125
  

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• 

  Frits 7:58 pm on 10 November 2020 Permalink | Reply

  from enigma import SubstitutedExpression
  
  # the alphametic puzzle
  p = SubstitutedExpression(
    [
     # find the length of the shortest line segment from (UVW, 0) to (0, XYZ)
     # that passes through (27, 64).
     #
     # 64 / (UVW - 27) = (XYZ - 64) / 27
     
     "UVW*XYZ - 64*UVW == 27*XYZ",
     "UVW > 0",
     "XYX > 0",
    ],
    answer="((UVW)**2 + (XYZ)**2)**0.5, UVW, XYZ",
    d2i={},
    distinct="",
    accumulate=min,
    verbose=0)   
      
  # solve the puzzle
  r = p.run()
  
  ans = list(r.accumulate)
  print("answer =",ans[0])
  
  # answer =  125.0
  

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