Sunday Times Teaser 3039 – Three Patterned Paving

by BRG on December 18, 2020

by Howard Williams

Published Sunday December 20 2020 (links)

James is laying foot-square stones in a rectangular block whose short side is less than 25 feet. He divides this area into three rectangles by drawing two parallel lines between the longest sides and into each of these three areas he lays a similar pattern.

The pattern consists of a band or bands of red stones laid concentrically around the outside of the rectangles with the centre filled with white stones. The number of red stone bands is different in each of the rectangles but in each of them the number of white stones used equals the number of outer {1] red stones used.

The total number of stones required for each colour is a triangular number (ie, one of the form 1+2+3+…).

What is the total area in square feet of the block?

[1] this word is unnecessary (it also creates an ambiguity).

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Brian Gladman permalink

This solution runs in under 100 microseconds.


from itertools import combinations
  
# Let m be the minimum block side, n be the sub-block
# side and b be the number of red bands so that:
#
#     m.n = 2.(m - 2.b).(n - 2.b)
#
# Solving for n in terms of m and b gives:
#
#     n = [m^2 / (m - 4.b) - (m - 4.b)] / 2

# the length of smaller side of the overall area
# (3 different b values requires that m > 12)
for m in range(13, 26):
  # collect (n, b) values
  nb = []
  for b in range(1, (m + 3) // 4):
    t = m - 4 * b
    n2 = m ** 2 // t - t
    if m ** 2 % t == 0 and n2 % 2 == 0:
      nb.append((n2 // 2, b))
      
  # if we find three or more (n, b) values
  for nb3 in combinations(nb, 3):
    n3, b3 = zip(*nb3)
    # the three sub-blocks form the longest side  
    if (long_side := sum(n3)) > m:
      stones = m * long_side
      # the overall number of stones of each colour
      # (i.e. stones / 2) must be a triangular number
      t = 4 * stones + 1
      if round(t ** 0.5) ** 2 == t:
        print(f"{stones} ({m} x {sum(n3)}) [{n3}, {b3}]")

from itertools import combinations

# Let m be the minimum block side, n be the sub-block

# side and b be the number of red bands so that:

# m.n = 2.(m - 2.b).(n - 2.b)

# Solving for n in terms of m and b gives:

# n = [m^2 / (m - 4.b) - (m - 4.b)] / 2

# the length of smaller side of the overall area

# (3 different b values requires that m > 12)

for m in range(13, 26):

# collect (n, b) values

nb = []

for b in range(1, (m + 3) // 4):

t = m - 4 * b

n2 = m ** 2 // t - t

if m ** 2 % t == 0 and n2 % 2 == 0:

nb.append((n2 // 2, b))

# if we find three or more (n, b) values

for nb3 in combinations(nb, 3):

n3, b3 = zip(*nb3)

# the three sub-blocks form the longest side

if (long_side := sum(n3)) > m:

stones = m * long_side

# the overall number of stones of each colour

# (i.e. stones / 2) must be a triangular number

t = 4 * stones + 1

if round(t ** 0.5) ** 2 == t:

print(f"{stones} ({m} x {sum(n3)}) [{n3}, {b3}]")

Sunday Times Teaser 3039 – Three Patterned Paving

by Howard Williams

Published Sunday December 20 2020 (links)

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