Sunday Times Teaser 3318 – Squaring the Circle

by BRG on 26 April 2026

by Andrew Skidmore

Published Sunday April 26 2026 (link)

Liam is designing a logo as part of a school project. He has drawn a circle that touches all the sides of an octagon. The side lengths of the octagon are all different two-figure numbers of mm; in fact, they are all triangular numbers (ie, numbers of the type 1+2+3+4+…). Taken together these numbers use all the digits 0 to 9 (at least once each). If I told you the value of one of the lengths that is an odd number, then you would be able to answer the question below.

What (in mm) is the perimeter of the octagon?

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BRG permalink


from itertools import combinations

# if (but NOT only if) an octagon is tangential then 
# the sums of its alternate sides are equal

# possible two digit triangular numbers for the sides 
sides = tuple(n * (n + 1) // 2 for n in range(4, 14))

d = dict()
# select eight sides for the octagon
for c8 in combinations(sides, 8):
  # the length of the polygon perimeter must be even
  if not (p := sum(c8)) & 1:
    # all digits 0..9 must be present at least once
    if len(set("".join(str(n) for n in c8))) == 10:
      # pick four lengths for the even sides avoiding
      # duplicates
      for c3 in combinations(c8[1:], 3):
        if c8[0] + sum(c3) == p // 2:
          # collect solutions indexed on odd side lengths
          ce = (c8[0],) + c3
          co = tuple(c for c in c8 if c not in ce)
          tps = sum(((x,) for t in zip(ce, co) for x in t), ())
          print(tps, sum(tps))
          for n in c8:         
            if n & 1:
              # form and combine the even and odd sequences
              d[n] = d.get(n, []) + [tps]

# find a solution that is unique for an odd side length
for k, vs in d.items():
  if len(vs) == 1:
    print(f"perimeter: {sum(vs[0])} {vs[0]}")

from itertools import combinations

# if (but NOT only if) an octagon is tangential then

# the sums of its alternate sides are equal

# possible two digit triangular numbers for the sides

sides = tuple(n * (n + 1) // 2 for n in range(4, 14))

d = dict()

# select eight sides for the octagon

for c8 in combinations(sides, 8):

# the length of the polygon perimeter must be even

if not (p := sum(c8)) & 1:

# all digits 0..9 must be present at least once

if len(set("".join(str(n) for n in c8))) == 10:

# pick four lengths for the even sides avoiding

# duplicates

for c3 in combinations(c8[1:], 3):

if c8[0] + sum(c3) == p // 2:

# collect solutions indexed on odd side lengths

ce = (c8[0],) + c3

co = tuple(c for c in c8 if c not in ce)

tps = sum(((x,) for t in zip(ce, co) for x in t), ())

print(tps, sum(tps))

for n in c8:

if n & 1:

# form and combine the even and odd sequences

d[n] = d.get(n, []) + [tps]

# find a solution that is unique for an odd side length

for k, vs in d.items():

if len(vs) == 1:

print(f"perimeter: {sum(vs[0])} {vs[0]}")

John Z permalink


from itertools import combinations
from collections import defaultdict

# get the 2 digit triangular numbers
tr = tuple(i * (i + 1) // 2 for i in range(4 ,14))

sols = defaultdict(set)

# consider combinations of 8 triangular numbers
for trc in combinations(tr, 8):
  # get the set of digits used
  trstr = ''.join(str(tr) for tr in trc)
  # all ten digits 0-9 must be used and the sum of the triangular numbers even
  if len(set(trstr)) == 10 and (strc := sum(trc)) & 1 == 0:
    # for a circle to be inscribed in an octagon
    # the sums of alternate digits must be equal
    # consider combinations of alternate digits
    for trc4 in combinations(trc, 4):
      # do the two groups of four sum to the same value
      if sum(trc4) * 2 == strc:
        # save sums of 8 triangular digits indexed on odd tr numbers
        for a in trc:
          if a & 1:
            sols[a] |= {strc}

for x in sols.items():
  # there must be a unique solution for 1 triangular number
  if len(x[1]) == 1:
    print(x)

from itertools import combinations

from collections import defaultdict

# get the 2 digit triangular numbers

tr = tuple(i * (i + 1) // 2 for i in range(4 ,14))

sols = defaultdict(set)

# consider combinations of 8 triangular numbers

for trc in combinations(tr, 8):

# get the set of digits used

trstr = ''.join(str(tr) for tr in trc)

# all ten digits 0-9 must be used and the sum of the triangular numbers even

if len(set(trstr)) == 10 and (strc := sum(trc)) & 1 == 0:

# for a circle to be inscribed in an octagon

# the sums of alternate digits must be equal

# consider combinations of alternate digits

for trc4 in combinations(trc, 4):

# do the two groups of four sum to the same value

if sum(trc4) * 2 == strc:

# save sums of 8 triangular digits indexed on odd tr numbers

for a in trc:

if a & 1:

sols[a] |= {strc}

for x in sols.items():

# there must be a unique solution for 1 triangular number

if len(x[1]) == 1:

print(x)

John Z permalink

The above code will be more efficient by avoiding finding duplicates if

Python

break # no need to continue search once an alternate sides equivalence found for group of 8

1
2
3

break # no need to continue search once an alternate sides equivalence found for group of 8

is inserted after line 25 to terminate the ‘for’ loop of line 19.

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Sunday Times Teaser 3318 – Squaring the Circle

by Andrew Skidmore

Published Sunday April 26 2026 (link)

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