Sunday Times Teaser 3330 – Bermuda Triangles

by BRG on 19 July 2026

by Peter Good

Published Sunday July 19 2026 (link)

My local estate agent was advertising a house named “Bermuda” which had a garden in the shape of a right-angled triangle containing a sundial. I went to view the property and it had nine straight and narrow footpaths. There was a path along each edge of the garden, one from each corner to the sundial and one from each edge to the sundial. Each of the latter paths was perpendicular to the edge it joined. Each path was a different whole number of feet in length, the longest being 500 feet and the two shortest being 36 and 77 feet. These two paths did not join the longest and were the only ones that were not a multiple of five feet.

What was the combined length of the nine footpaths?

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BRG permalink

# return the third side of a right angled triangle with:
#      c^2 = a^2 + b^2 if b is positive
#      c^2 = a^2 - b^2 if b is negative
def hypot(a, b):
  c2 = a * a + [-1, 1][b > 0] * b * b
  if c2 >= 0 and (c := round(c2 ** 0.5)) ** 2 == c2:
    return c

# consider the three paths along the triangle sides (a, b, c)
c = 500
for a in range(5, 355, 5):
  if (b := hypot(c, -a)) and b % 5 == 0:
    # consider the paths that don't connect with the triangle hypotenuse
    for p, q in ((36, 77), (77, 36)):
      # find the distances to the triangle vertexes (d, e, f)
      d, e, f = dv3 = hypot(p, q), hypot(a - p, q), hypot(b - q, p)
      if d and e and f and not (d % 5 or e % 5 or f % 5):
        # calculate the height of triangle (e, f, c) from the hypotenuse base (c)
        # using its area A = sqrt[s.(s - a).(s - b).(s - c)] where s is its semi-
        # perimiter
        s = (e + f + c) / 2
        r = 2 * (s * (s - e) * (s - f) * (s - c)) ** 0.5 / c
        abc, pqr = (a, b, c), (p, q, round(r))
        if len({*abc, *dv3, *pqr}) == 9 and not r % 5:
          sm = sum(abc) + sum(dv3) + sum(pqr)
          print(f"Total Footpath Length {sm} ft [{abc} {dv3} {pqr}]")

# return the third side of a right angled triangle with:

# c^2 = a^2 + b^2 if b is positive

# c^2 = a^2 - b^2 if b is negative

def hypot(a, b):

c2 = a * a + [-1, 1][b > 0] * b * b

if c2 >= 0 and (c := round(c2 ** 0.5)) ** 2 == c2:

return c

# consider the three paths along the triangle sides (a, b, c)

c = 500

for a in range(5, 355, 5):

if (b := hypot(c, -a)) and b % 5 == 0:

# consider the paths that don't connect with the triangle hypotenuse

for p, q in ((36, 77), (77, 36)):

# find the distances to the triangle vertexes (d, e, f)

d, e, f = dv3 = hypot(p, q), hypot(a - p, q), hypot(b - q, p)

if d and e and f and not (d % 5 or e % 5 or f % 5):

# calculate the height of triangle (e, f, c) from the hypotenuse base (c)

# using its area A = sqrt[s.(s - a).(s - b).(s - c)] where s is its semi-

# perimiter

s = (e + f + c) / 2

r = 2 * (s * (s - e) * (s - f) * (s - c)) ** 0.5 / c

abc, pqr = (a, b, c), (p, q, round(r))

if len({*abc, *dv3, *pqr}) == 9 and not r % 5:

sm = sum(abc) + sum(dv3) + sum(pqr)

print(f"Total Footpath Length {sm} ft [{abc} {dv3} {pqr}]")

Frits permalink

Another formula for r:

Python

# sqrt(e^4 - (e^2 - 500^2)^2 - (e^2 - f^2)^2 + 2 * f^2 * 500^2) / 1000

1
2
3

# sqrt(e^4 - (e^2 - 500^2)^2 - (e^2 - f^2)^2 + 2 * f^2 * 500^2) / 1000

Reply

Frits permalink

is_square = lambda n: None if (rt := round(n**.5))**2 != n else rt

# return 2 sides with hypothenuse h
def pyth_sides(h, mn, step=1):
  h2 = h * h
  rt = .5**.5
  for x in range(mn, int(rt * h) + 1, step):
    if (y := is_square(h2 - x * x)): 
      yield (x, y)
 
perp1, perp2 = 36, 77
edge3 = 500

# hypothenuse between the 2 shortest paths
if not (corner1 := is_square(perp1**2 + perp2**2)): exit(0)

# paths      
ps_fixed = [perp1, perp2, edge3, corner1]

sols = set()
# look for the 2 paths along the edge
for es in pyth_sides(edge3, 5, 5):
  # look for the 2 remaining paths from the corner 
  for i, e in enumerate(es):
    if not (corner2 := is_square((e - perp1)**2 + perp2**2)): continue
    if not (corner3 := is_square((es[1 - i] - perp2)**2 + perp1**2)): continue
    # look for the remaining path perpendicular to the edge
    for es2 in pyth_sides(corner2, 5, 5):
      for i2, e2 in enumerate(es2):
        if not (e3 := is_square(corner3**2 - e2**2)): continue 
        # double check if 2 variables add up to edge3
        if e3 + es2[1 - i2] != edge3: continue
        # we now know all the paths
        ps = ps_fixed + list(es) + [corner2, corner3, e2]
        # different path lengths
        if len(ps) != len(set(ps)): continue
        # check multiple of 5 requirement
        if sum(1 for p in ps if p % 5) != 2: continue
        sols.add(str(sum(ps)))
     
print("answer:", ' or '.join(sols))

is_square = lambda n: None if (rt := round(n**.5))**2 != n else rt

# return 2 sides with hypothenuse h

def pyth_sides(h, mn, step=1):

h2 = h * h

rt = .5**.5

for x in range(mn, int(rt * h) + 1, step):

if (y := is_square(h2 - x * x)):

yield (x, y)

perp1, perp2 = 36, 77

edge3 = 500

# hypothenuse between the 2 shortest paths

if not (corner1 := is_square(perp1**2 + perp2**2)): exit(0)

# paths

ps_fixed = [perp1, perp2, edge3, corner1]

sols = set()

# look for the 2 paths along the edge

for es in pyth_sides(edge3, 5, 5):

# look for the 2 remaining paths from the corner

for i, e in enumerate(es):

if not (corner2 := is_square((e - perp1)**2 + perp2**2)): continue

if not (corner3 := is_square((es[1 - i] - perp2)**2 + perp1**2)): continue

# look for the remaining path perpendicular to the edge

for es2 in pyth_sides(corner2, 5, 5):

for i2, e2 in enumerate(es2):

if not (e3 := is_square(corner3**2 - e2**2)): continue

# double check if 2 variables add up to edge3

if e3 + es2[1 - i2] != edge3: continue

# we now know all the paths

ps = ps_fixed + list(es) + [corner2, corner3, e2]

# different path lengths

if len(ps) != len(set(ps)): continue

# check multiple of 5 requirement

if sum(1 for p in ps if p % 5) != 2: continue

sols.add(str(sum(ps)))

print("answer:", ' or '.join(sols))

Frits permalink

I forgot the formula for one of the altitudes of a triangle when all the lengths of the sides are known. Today I didn’t feel like looking for it on the internet.

Reply

John Z. permalink

# longest side of right triangle is 500 feet (the hypotenuse)

# The two shortest paths are 36 and 77. They are perpendicular to the 2 legs
# and meet at the sundial so the diagonal = 85 = (36 ^ 2 + 77 ^ 2) ^ 0.5
p1, p2 = 36, 77

# first path from right-angle vertex to sundial
p3 = int((36 ** 2 + 77 ** 2) ** 0.5)

# find possible legs of the triangle: a ** 2 + b ** 2 = 500 ** 2
legs = [(a, int(b)) for a in range(5, 500, 5)
        if (b := (500 ** 2 - a ** 2) ** 0.5) == int(b)]

# consider the different triangles:
for a, b in legs:
  
  # second path from a vertex to sundial
  p4 = ((a - p1) ** 2 + p2 ** 2) ** 0.5
  if p4 != int(p4):
    continue
  p4 = int(p4)
  
  # third path from a vertex to sundial
  p5 = ((b - p2) ** 2 + p1 ** 2) ** 0.5
  if p5 != int(p5):
    continue
  p5 = int(p5)

# find the area of the triangle p4, p5, 500; start with the semi-perimeter
  s = (p4 + p5 + 500) / 2

area = (s * (s - p4) * (s - p5) * (s - 500)) ** 0.5
  
  # the perpendicular from the 500 foot hypotenuse:
  p6 = (2 * area) / 500
  if p6 != int(p6):
    continue
  p6 = int(p6)
  
  paths = (p1, p2, p3, p4, p5, p6, a, b, 500)

print('The paths are', ', '.join(str(s) for s in paths[:-1]), 'and 500 feet')
  print('They total:', sum(paths), 'feet')

# longest side of right triangle is 500 feet (the hypotenuse)

# The two shortest paths are 36 and 77. They are perpendicular to the 2 legs

# and meet at the sundial so the diagonal = 85 = (36 ^ 2 + 77 ^ 2) ^ 0.5

p1, p2 = 36, 77

# first path from right-angle vertex to sundial

p3 = int((36 ** 2 + 77 ** 2) ** 0.5)

# find possible legs of the triangle: a ** 2 + b ** 2 = 500 ** 2

legs = [(a, int(b)) for a in range(5, 500, 5)

if (b := (500 ** 2 - a ** 2) ** 0.5) == int(b)]

# consider the different triangles:

for a, b in legs:

# second path from a vertex to sundial

p4 = ((a - p1) ** 2 + p2 ** 2) ** 0.5

if p4 != int(p4):

continue

p4 = int(p4)

# third path from a vertex to sundial

p5 = ((b - p2) ** 2 + p1 ** 2) ** 0.5

if p5 != int(p5):

continue

p5 = int(p5)

# find the area of the triangle p4, p5, 500; start with the semi-perimeter

s = (p4 + p5 + 500) / 2

area = (s * (s - p4) * (s - p5) * (s - 500)) ** 0.5

# the perpendicular from the 500 foot hypotenuse:

p6 = (2 * area) / 500

if p6 != int(p6):

continue

p6 = int(p6)

paths = (p1, p2, p3, p4, p5, p6, a, b, 500)

print('The paths are', ', '.join(str(s) for s in paths[:-1]), 'and 500 feet')

print('They total:', sum(paths), 'feet')

Sunday Times Teaser 3330 – Bermuda Triangles

by Peter Good

Published Sunday July 19 2026 (link)

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