S2T2

Jim Randell 7:49 am on 17 September 2019 Permalink Reply
Tags: by: Peter Harrison ( 6 ), flawed ( 53 )   

• 

  Jim Randell 7:50 am on 17 September 2019 Permalink | Reply

  As presented this puzzle has multiple solutions. However if the perimeter of the original rectangle is required to be a whole number of feet, then only one of the solutions remains.

  This Python program finds all the solutions to the puzzle numerically using the [[ find_zero() ]] function from the enigma.py library. It runs in 63ms.

  Run: [ @replit ]

  from enigma import (irange, cb, cbrt, sq, find_zero, catch, printf)
  
  # consider values for t, where T = t^3 is the total volume
  for t in irange(1, 100):
    T = cb(t)
  
    # calculate y (for x in [0, t])
    def Y(x):
      return cbrt(T - cb(x))
  
    # how close is x^2 + y^2 - xy to 243?
    def f(x):
      y = Y(x)
      return abs((sq(x) + sq(y) - x * y) - 243)
  
    # find a zero for f
    r = catch(find_zero, f, 0, cbrt(0.5 * T))
    if r is None: continue
    x = r.v
    y = Y(x)
    p = 2 * (x + y)
  
    # output solution
    printf("t={t}: x={x:.3f}, y={y:.3f}, p={p:.3f}")
  

  Solution: The (integer) perimeter of the original rectangle was 48 feet.

  Allowing non-integer solutions for the perimeter we get four solutions:

  t=16: x=0.857, y=15.999, p=33.712
  t=17: x=3.258, y=16.960, p=40.436
  t=18: x=6.255, y=17.745, p=48.000
  t=19: x=10.291, y=17.935, p=56.453

  Graphically we get a solution when the ellipse x² + y² − xy = 243 intersects the curve x³ + y³ = t³ for integer values of t.

  For positive x and y we get solutions for t = 16, 17, 18, 19, as shown in the graph below:

  

  The original rectangle has sides of length x and y, so the perimeter of the original rectangle is 2(x + y). Only t=18 gives an integer value for the perimeter (although the values for x and y are not integers). In this case the exact values are: x, y = 12 ± √33.

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  • 

    Jim Randell 8:39 am on 19 September 2019 Permalink | Reply

    Here is an analytical solution.

    Given the equations:

    x³ + y³ = t³
    x² + y² − xy = 243

    If we consider the product (x + y)(x² + y² − xy) we have:

    (x + y)(x² + y² − xy) = x³ + y³ = t³
    (x + y) = t³ / 243

    We also have:

    (x + y)² = (x² + y²) + 2xy
    (x + y)² = (243 + xy) + 2xy
    (x + y)² = 243 + 3xy
    xy = (x + y)² / 3 − 81

    So for a given value of t we can determine values for x + y and xy, say:

    x + y = a
    xy = b

    These have positive real solutions for x and y when a² ≥ 4b and b ≥ 0

    a² ≥ 4b
    (x + y)² ≥ 4xy
    (x + y)² ≥ (4/3)(x + y)² − 324
    (x + y)² ≤ 973
    (x + y) ≤ √973
    t³ ≤ 243 × √973
    t ≤ 19.643…

    b ≥ 0
    (x + y)² / 3 − 81 ≥ 0
    (x + y)² ≥ 243
    (x + y) ≥ √243
    t³ ≥ 243 × √243
    t ≥ 15.588…

    So t is in the range 16 to 19. And the required perimeter is given by p = 2(x + y) = 2t³ / 243.

    We can consider the 4 candidate values manually and look for an integer solution, or use a short program:

    Run: [ @replit ]

    from enigma import (irange, cb, div, printf)
    
    for t in irange(16, 19):
      p = div(2 * cb(t), 243)
      if p is None: continue
      printf("t={t}: p={p}")
    

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