Sunday Times Teaser 3242 – Tynemouth Boating Lake

by BRG on 10 November 2024

by Phil Jackson

Published Sunday November 10 2024

When Tynemouth Boating Lake was being constructed, the builder hammered in stakes at two points in the ground, tied the ends of a rope to the stakes, and with a stick keeping the rope taut, marked out an ellipse as shown (not to scale). He was surprised to find that at some points on the circumference, the rope formed two different right-angled triangles, both with sides a whole number of feet long. The free length of the rope was 289 feet.

In increasing order, what were the five lengths of the sides of the two triangles?

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BRG permalink


# return the square root of a perfect square or None otherwise 
is_sqr = lambda n: w if n == (w := round(n ** 0.5)) ** 2 else None  
 
# the rope length
r_len = 289

# consider the triangle u^2 + v^2 = c^2 
# where c is the interfocal distance 
for u in range(1, r_len // 2 + 1):
  v = r_len - u
  if (c := is_sqr(u ** 2 + v ** 2)):

    # consider the triangle c^2 + w^2 = (x = r_len - w)^2 
    #       ==> c^2 = r_len.(r_len - 2.w) 
    #       ==> w = (r_len - c^2 / r_len) / 2
    if not (c * c) % r_len:
      w, r_ = divmod((r_len - c * c // r_len), 2)
      if not r_:
        x = r_len - w
        # print(f"{c = } {u = } {v = } {w = } {x = }")
        fs = ', '.join(str(n) for n in sorted([c, u, v, w, x])) 
        f, s, l = fs.rpartition(', ')
        print(f"The lengths are {f}, and {l}.")

# return the square root of a perfect square or None otherwise

is_sqr = lambda n: w if n == (w := round(n ** 0.5)) ** 2 else None

# the rope length

r_len = 289

# consider the triangle u^2 + v^2 = c^2

# where c is the interfocal distance

for u in range(1, r_len // 2 + 1):

v = r_len - u

if (c := is_sqr(u ** 2 + v ** 2)):

# consider the triangle c^2 + w^2 = (x = r_len - w)^2

# ==> c^2 = r_len.(r_len - 2.w)

# ==> w = (r_len - c^2 / r_len) / 2

if not (c * c) % r_len:

w, r_ = divmod((r_len - c * c // r_len), 2)

if not r_:

x = r_len - w

# print(f"{c = } {u = } {v = } {w = } {x = }")

fs = ', '.join(str(n) for n in sorted([c, u, v, w, x]))

f, s, l = fs.rpartition(', ')

print(f"The lengths are {f}, and {l}.")

Here is an analytic solution.

John Z permalink


sq = tuple(i * i for i in range(500))

nf = set() # potential values for distance between foci (stakes) of ellipse

# find side values for triangle with right angle on circumference of ellipse
for a in range(1, 289 // 2 + 1):
  b = 289 - a
  c2 = a * a + b * b
  if c2 in sq:
    c = sq.index(c2) # c is the distance between the stakes
    nf |= {(a, b, c)}

# consider possible distances between stakes 
for x, y, a in nf:
  for b in range(1, 289):
    if a * a + b * b == (289 - b) ** 2:
      
      print("The five sides are:", *sorted((x, y, a, b, 289 - b)))

sq = tuple(i * i for i in range(500))

nf = set() # potential values for distance between foci (stakes) of ellipse

# find side values for triangle with right angle on circumference of ellipse

for a in range(1, 289 // 2 + 1):

b = 289 - a

c2 = a * a + b * b

if c2 in sq:

c = sq.index(c2) # c is the distance between the stakes

nf |= {(a, b, c)}

# consider possible distances between stakes

for x, y, a in nf:

for b in range(1, 289):

if a * a + b * b == (289 - b) ** 2:

print("The five sides are:", *sorted((x, y, a, b, 289 - b)))

John Z permalink

The 7 lines of code in lines 4 -> 12 above can be rewritten as 2 lines of code using a comprehension:


# find side values for triangle with right angle on circumference of ellipse
# where c is the distance between the stakes
nf = [(a, 289 - a, sq.index(c2)) for a in range(1, 289 // 2 + 1)
      if (c2 := a * a + (289 - a) ** 2) in sq]

# find side values for triangle with right angle on circumference of ellipse

# where c is the distance between the stakes

nf = [(a, 289 - a, sq.index(c2)) for a in range(1, 289 // 2 + 1)

if (c2 := a * a + (289 - a) ** 2) in sq]

BRG permalink


from number_theory import pythag
from collections import defaultdict

# collect pythagorean triangles for which
# the length of two sides sum to 289
h2t = defaultdict(set)
s2t = defaultdict(set)
for t in pythag(h_max=289):
  if any(sum(t) - s == 289 for s in t):    
    # index on sides and hypotenuses
    s2t[t[0]].add(t)
    s2t[t[1]].add(t)
    h2t[t[2]].add(t)

# find matches between hypotenuses and sides
for k in h2t.keys() & s2t.keys():
  for a, b, c in h2t[k]:
    for d, e, f in s2t[k]:
      fs = ', '.join(str(n) for n in sorted({a, b, c, d, e, f})) 
      f, s, l = fs.rpartition(', ')
      print(f"The lengths are {f}, and {l}.")

from number_theory import pythag

from collections import defaultdict

# collect pythagorean triangles for which

# the length of two sides sum to 289

h2t = defaultdict(set)

s2t = defaultdict(set)

for t in pythag(h_max=289):

if any(sum(t) - s == 289 for s in t):

# index on sides and hypotenuses

s2t[t[0]].add(t)

s2t[t[1]].add(t)

h2t[t[2]].add(t)

# find matches between hypotenuses and sides

for k in h2t.keys() & s2t.keys():

for a, b, c in h2t[k]:

for d, e, f in s2t[k]:

fs = ', '.join(str(n) for n in sorted({a, b, c, d, e, f}))

f, s, l = fs.rpartition(', ')

print(f"The lengths are {f}, and {l}.")

Sunday Times Teaser 3242 – Tynemouth Boating Lake

by Phil Jackson

Published Sunday November 10 2024

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