Sunday Times Teaser 2807 – Pentagons

by Nick MacKinnon

Published: 10 July 2016 (link)

I have taken five identical straight rods and joined each to the next by a hinge to make a flexible ring. With this ring I can make lots of different pentagonal shapes, and in particular I can make lots of pentagons with area equal to a whole number of square centimetres. The largest whole-numbered area achievable is a two-figure number, and the smallest whole-numbered area achievable is another two-figure number. In fact these two numbers use the same two digits but in the reverse order.

What is the largest whole-numbered area achievable?

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from math import ceil

# For side length L, the maximum area (achieved by a regular
# pentagon) is {(5.(5 + 2.5^(1/2)))^(1/2) / 4}.L^2
CP = (5 * (5 + 2 * 5 ** (1 / 2))) ** (1 / 2) / 4

# The minimum area (achieved when two sides are collapsed into
# one to leave an equilateral triangle) is {3^(1/2) / 4}.L^2 
CT = 3 ** (1/2) / 4

# the maximum area is a two digit number
for area_max in range(10, 100):
  
  # calculate the pentagon's side
  L = (area_max / CP) ** 0.5
  
  # and the minimum whole number area
  area_min = ceil(CT * L ** 2)
  
  # now check that the two areas are digit reversed
  if area_min == 10 * (area_max % 10) + area_max // 10:
    fs = 'The maximum area is {} cm^2 (rod length = ~{:.3f} cm).'
    print(fs.format(area_max, L))

from math import ceil

# For side length L, the maximum area (achieved by a regular

# pentagon) is {(5.(5 + 2.5^(1/2)))^(1/2) / 4}.L^2

CP = (5 * (5 + 2 * 5 ** (1 / 2))) ** (1 / 2) / 4

# The minimum area (achieved when two sides are collapsed into

# one to leave an equilateral triangle) is {3^(1/2) / 4}.L^2

CT = 3 ** (1/2) / 4

# the maximum area is a two digit number

for area_max in range(10, 100):

# calculate the pentagon's side

L = (area_max / CP) ** 0.5

# and the minimum whole number area

area_min = ceil(CT * L ** 2)

# now check that the two areas are digit reversed

if area_min == 10 * (area_max % 10) + area_max // 10:

fs = 'The maximum area is {} cm^2 (rod length = ~{:.3f} cm).'

print(fs.format(area_max, L))

This gives a maximum area of 61 square centimetres with a rod length of approximately 5.954 centimetres. Interestingly, since a rod length up to about 6.05 centimetres will give the same result, the rods could be of integer length.

Sunday Times Teaser 2807 – Pentagons

by Nick MacKinnon

Published: 10 July 2016 (link)

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