Sunday Times Teaser 3287 – Ferry Route

by BRG on September 21, 2025

by John Owen

Published Sunday September 21 2025 (link)

A large circular sea, whose diameter is less than 300km, is served by a ferry that makes a clockwise route around half of the sea, serving the ports of Ayton, Beaton, Seaton and Deaton in that order, then back to Ayton. Deaton is diametrically opposite Ayton. Each of the four legs of its route is a straight line, two of them being the same length. The lengths of all of the legs are whole numbers of km, and they all happen to be square numbers.

In increasing order, what are the three different leg lengths?

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BRG permalink

See teaser 2549 for the derivation of the formulae used below.


from itertools import combinations
from math import cos, acos

# see Teaser 2549 for derivation of formulae

# squares up to 300
sq = tuple(x * x for x in range (1, 18))

for p, q in combinations(sq, 2):
  for a, b, c in ((p, p, q), (p, q, q)):      
    # s = ((a^2 + b^2 + c^2) / 3)^(1 / 2)
    s = ((a * a + b * b + c * c) / 3) ** (1 / 2)
    d = 2 * s * cos(acos(a * b * c / s ** 3) / 3)
    if d < 300 and abs(d - round(d)) < 1e-14:
      print(f"{a} {b} {c} -> {d:.0f} (km)")

from itertools import combinations

from math import cos, acos

# see Teaser 2549 for derivation of formulae

# squares up to 300

sq = tuple(x * x for x in range (1, 18))

for p, q in combinations(sq, 2):

for a, b, c in ((p, p, q), (p, q, q)):

# s = ((a^2 + b^2 + c^2) / 3)^(1 / 2)

s = ((a * a + b * b + c * c) / 3) ** (1 / 2)

d = 2 * s * cos(acos(a * b * c / s ** 3) / 3)

if d < 300 and abs(d - round(d)) < 1e-14:

print(f"{a} {b} {c} -> {d:.0f} (km)")

and


from itertools import combinations
from number_theory import divisors

# squares up to 300
sq = tuple(x * x for x in range (1, 18))

# consider three squares, two the same
for p, q in combinations(sq, 2):
  # note: order of (a,b,c) irrelevant
  for a, b, c in ((q, p, p),):
    for d in divisors(2 * a * b * c):
      if d >= 300:
        break
      s = a * a + b * b + c * c
      # d^3 == (a^2 + b^2 + c^2).d + 2.a.b.c      
      if d ** 3 == s * d + 2 * a * b * c:
        print(f"{a} {b} {c} -> {d:.0f} (km)")

from itertools import combinations

from number_theory import divisors

# squares up to 300

sq = tuple(x * x for x in range (1, 18))

# consider three squares, two the same

for p, q in combinations(sq, 2):

# note: order of (a,b,c) irrelevant

for a, b, c in ((q, p, p),):

for d in divisors(2 * a * b * c):

if d >= 300:

break

s = a * a + b * b + c * c

# d^3 == (a^2 + b^2 + c^2).d + 2.a.b.c

if d ** 3 == s * d + 2 * a * b * c:

print(f"{a} {b} {c} -> {d:.0f} (km)")

John Z permalink


# use Ptolemy's theorem for a circumscribed quadrilateral: ef=ac+bd:
# product of the diagonals = sum of the products of opposite sides
# as d is a diameter we also have
# d ** 2 == a ** 2 + e ** 2 and
# d ** 2 == c ** 2 + f ** 2

# a = AB, b = BC, c = CD and d = DA
# a chord of a given length always subtends the same angle so their order is
# irrelevant. We chose to have a repeated chords a = AB = c = CD which
# simplifies the application of Ptolemy below

from itertools import combinations

# squares up to 300
sqs = tuple(i * i for i in range(1, 18))

for x, y, d in combinations(sqs, 3):
  for a, b in((y, x), (x, y)):
    if d * d == 2 * a * a + b * d:
      print(x, y, d,'km')

# use Ptolemy's theorem for a circumscribed quadrilateral: ef=ac+bd:

# product of the diagonals = sum of the products of opposite sides

# as d is a diameter we also have

# d ** 2 == a ** 2 + e ** 2 and

# d ** 2 == c ** 2 + f ** 2

# a = AB, b = BC, c = CD and d = DA

# a chord of a given length always subtends the same angle so their order is

# irrelevant. We chose to have a repeated chords a = AB = c = CD which

# simplifies the application of Ptolemy below

from itertools import combinations

# squares up to 300

sqs = tuple(i * i for i in range(1, 18))

for x, y, d in combinations(sqs, 3):

for a, b in((y, x), (x, y)):

if d * d == 2 * a * a + b * d:

print(x, y, d,'km')

Sunday Times Teaser 3287 – Ferry Route

by John Owen

Published Sunday September 21 2025 (link)

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