S2T2

Jim Randell 10:35 am on 27 May 2026 Permalink Reply
Tags: by: C Higgins ( 39 ), by: H Bradley ( 39 )   

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  Jim Randell 10:35 am on 27 May 2026 Permalink | Reply

  This Python program considers possible n-gons for n = 3 .. 10, and possible 2-digit side lengths. And looks for viable combinations where the maximum volume is a recognisable 6-digit date.

  It runs in 94ms. (Internal runtime is 25ms).

  from math import (tan, pi)
  from enigma import (irange, fdiv, polygon_area_regular, cproduct, find_max, snapf, nsplit, printf)
  
  # calculate volume of a box, starting with a regular n-gon
  # with side <x>, and then make a box by measuring <y> from
  # each corner and cutting off a quadrilateral from each corner
  def vol(n, x, y):
    # the area of the base = A
    A = polygon_area_regular(n, x - 2*y)
    # height of the sides = z
    t = pi * fdiv(n - 2, 2 * n)
    z = y * tan(t)
    # volume
    return A * z
  
  # consider possible n-gons and possible 2-digit sides (= x)
  for (n, x) in cproduct([irange(3, 10), irange(10, 99)]):
  
    # maximise the volume (= V)
    r = find_max((lambda y: vol(n, x, y)), 0, 0.5 * x)
    # volume is close to a 6-digit integer
    V = snapf(r.fv, t=1e-4, default=None)
    # V should be recognisable as a 6-digit date (dd/mm/yy)
    if V is None or V < 100000: continue
    (dd, mm, yy) = nsplit(V, k=3, base=100)
    if not (0 < dd < 32 and 0 < mm < 13): continue
    y = r.v
  
    # output solution
    printf("n={n} x={x} -> y={y:.4f} V={V} ({dd:02d}/{mm:02d}/{yy:02d})")
  

  Solution: The volume of the box is 140625 cm³ .

  So the grandfather’s DOB is 14/06/25 (14th June 1925).

  The original sheet of metal was a hexagon (6-gon) with a side length of 75 cm.

  Cuts perpendicular to the sides at a distance 12.5 cm from each corner are made to produce a hexagonal base with side length 50 cm (area = 3750 √3 cm²) and six rectangular sides with height 12.5 √3 cm.

  The volume of the box is then: (3750 √3) × (12.5 √3) = 3750 × 12.5 × 3 = 140625.

  

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  • 

    Jim Randell 3:30 pm on 28 May 2026 Permalink | Reply

    Using some analysis:

    The equation for the volume is:

    V(n, x, y) = (1/4) n y (x − 2y)² cot²(𝝅/n)

    And, for a fixed n and x this is at a maximum when:

    y = x/6

    hence:

    V = (1/54) n x³ cot²(𝝅 / n)

    And for an integer volume the cot²(𝝅/n) part (= c) must be rational, restricting the value of n to one of the following:

    n = 3 → c = 1/3
    n = 4 → c = 1
    n = 6 → c = 3

    And so we can simplify the program:

    from enigma import (irange, cproduct, div, cb, nsplit, printf)
    
    # rational (n, cot^2(pi/n)) pairs as (n, p/q)
    ncs = {
      3: (1, 3),
      4: (1, 1),
      6: (3, 1),
    }
    
    for ((n, (p, q)), x) in cproduct([ncs.items(), irange(10, 99)]):
    
      # calculate volume V
      V = div(n * cb(x) * p, 54 * q)
      # must be recognisable as a 6-digit date (dd/mm/yy)
      if V is None or V < 100000: continue
      (dd, mm, yy) = nsplit(V, k=3, base=100)
      if not (0 < dd < 32 and 0 < mm < 13): continue
    
      # output solution
      printf("n={n} x={x} -> V={V} ({dd:02d}/{mm:02d}/{yy:02d})")
    

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