S2T2

Jim Randell 9:20 am on 3 August 2021 Permalink Reply
Tags: by: Nick MacKinnon ( 51 )   

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  Jim Randell 9:21 am on 3 August 2021 Permalink | Reply

  

  The maximal area is achieved with a cyclic pentagon. If the sides have length x the area is given by:

  Amax(x) = (x² / 4) √(5(5 + 2√5))

  If this value is not an integer, the maximum integer-valued area will be achieved by a slight deformation of the pentagon to give the floor of this value.

  I believe the minimal area is achieved with a degenerate pentagon where two of the sides are collapsed to give an equilateral triangle with a protruding spike.

  Amin(x) = (x² / 4) √3

  Again, if this value is not an integer, the minimum integer-valued area will be achieved by opening up the collapsed spike slightly (to give an actual pentagon), and increase the area to the ceiling of this value.

  This Python program considers possible Amax and Amin values (each 2 digits, one the reverse of the other), and calculates the range of x values that would give these values. If there are any common values then we have a viable solution. It runs in 50ms.

  Run: [ @replit ]

  from enigma import (irange, nreverse, sqrt, fdiv, printf)
  
  # area multipliers
  maxA = fdiv(sqrt(5 * (5 + 2 * sqrt(5))), 4)
  minA = fdiv(sqrt(3), 4)
  
  # consider 2 digit maximal area
  for M in irange(10, 99):
    m = nreverse(M)
    if not (9 < m < M): continue
  
    # calculate range of x values based on M
    xmin1 = sqrt(M, maxA)
    xmax1 = sqrt(M + 1, maxA)
  
    # calculate range of x value based on m
    xmin2 = sqrt(m - 1, minA)
    xmax2 = sqrt(m, minA)
  
    # intersect the intervals
    xmin = max(xmin1, xmin2)
    xmax = min(xmax1, xmax2)
    if not (xmin < xmax): continue
    printf("Amax={M} Amin={m} -> x = [{xmin}, {xmax}]")
  

  Solution: The largest whole-number area achievable is 61 cm².

  And the smallest whole-number area achievable is 16 cm².

  The length of the rods is between 5.995 cm and 6.003 cm, so (although not mentioned in the puzzle) an integer length (of 6 cm) for the rods could be intended.

  For 6 cm rods we have:

  Amax ≈ 61.937 cm²
  Amin ≈ 15.588 cm²

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  Frits 5:03 pm on 4 August 2021 Permalink | Reply

  Using Jim’s code as a base but avoiding square root calculations.
  The internal run time is 400 nanoseconds (with PyPy).

   
  # "x^2" related multipliers
  bigMult = 4 / (5 * (5 + 2 * 5**.5))**.5
  smallMult = 4 / 3**.5
  
  # consider 2 digit maximal area
  for t in range(2, 10):
    for u in range(1, t):
      M = 10 * t + u
      m = u * 10 + t
    
      # calculate range of x^2 values
      x2max1 = (M + 1) * bigMult
      x2min2 = (m - 1) * smallMult
  
      # leave inner loop if already a max is smaller than a min
      # as x2min2 grows faster than x2max1 for subsequent entries
      if x2max1 < x2min2: break
  
      x2min1 = x2max1 - bigMult
      x2max2 = x2min2 + smallMult
  
      # intersect the intervals
      x2min = max(x2min1, x2min2)
      x2max = min(x2max1, x2max2)
      if not(x2min < x2max): continue
      print(f"answer: {M} cm2")
  

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