Sunday Times Teaser 3277 – Croquet Equipment

by BRG on July 13, 2025

by John Owen

Published Sunday July 13 2025 (link)

Our non-standard croquet lawn has six hoops, at positions A to F, and a central peg at P. The equipment storage box is in the southwest corner at S, and the lawn is symmetrical in both the east-west and north-south directions. The diagram is not to scale, but D is south of the projected line from S to E. Also AB is half the width of the lawn. When setting out the equipment, I walk along the route SEFDPCBAS. All of the distances on my route between positions are whole numbers of feet (less than 60), and both D and E are whole numbers of feet north of the south boundary.

What is the total distance of my route?

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BRG permalink

Sadly I get two answers.


from itertools import combinations

def ptr(x, y):
  z = round((t := x * x + y * y) ** (1 / 2))
  return z if z * z == t else None

# consider the half-width of AB             
for w in range(1, 30):
  # collect possible distances of E and A north
  # of S and the related diagonals SE and SA
  xhs = []
  for x in range(1, 60):
    if (h := ptr(w, x)) and h < 60:
      xhs.append((x, h))
  
  # consider possible pairings for these values 
  for (x, SE), (y, SA) in combinations(xhs, 2):
    # consider the half distance <z> between C and D
    # and the distance <r> of D north of the line EF
    for z in range(1, 60):
      r, _r = divmod(y - x - 2 * z, 2)
      # r must be less than x (D is south of the extended line SE)
      if not _r and 0 < r < x:
        if (DF := ptr(w, r)) and DF < 60:
          d = SE + SA + 4 * w + 2 * (DF + z)
          print(f"{d = } ({SE = } {DF = } {SA = }) "
                f"({w = :2} {x = :2} {y = } {z = :2} {r = :2})")

from itertools import combinations

def ptr(x, y):

z = round((t := x * x + y * y) ** (1 / 2))

return z if z * z == t else None

# consider the half-width of AB

for w in range(1, 30):

# collect possible distances of E and A north

# of S and the related diagonals SE and SA

xhs = []

for x in range(1, 60):

if (h := ptr(w, x)) and h < 60:

xhs.append((x, h))

# consider possible pairings for these values

for (x, SE), (y, SA) in combinations(xhs, 2):

# consider the half distance <z> between C and D

# and the distance <r> of D north of the line EF

for z in range(1, 60):

r, _r = divmod(y - x - 2 * z, 2)

# r must be less than x (D is south of the extended line SE)

if not _r and 0 < r < x:

if (DF := ptr(w, r)) and DF < 60:

d = SE + SA + 4 * w + 2 * (DF + z)

print(f"{d = } ({SE = } {DF = } {SA = }) "

f"({w = :2} {x = :2} {y = } {z = :2} {r = :2})")

Both Jim Randell (at S2T2) and Frits (below) realised that ALL three pythagorean triangles share a common side before I did. This allows a very compact solution:


from itertools import combinations        

# there are three pythagorean triangles all of which
# share a common side, <w>, the half-width of AB
for w in range(1, 30):
  # list (in order) the other sides and hypotenuses
  # of all such triangles for hypotenuses below 60
  xhs = [(x, h) for h in range(w + 1, 60) if
    (x := round((t := (h * h - w * w)) ** (1/2))) ** 2 == t]
  
  # consider possible triples created from these values 
  for (r, DF), (x, SE), (y, SA) in combinations(xhs, 3):
    # ensure that CP and PD (<z> here) are positive integers 
    z, _r = divmod(y - x - 2 * r, 2)
    if not _r and z > 0:
      d = SA + SE + 4 * w + 2 * (z + DF)
      print(f"{d = } ({SE = } {DF = } {SA = }) "
            f"({w = :2} {x = :2} {y = } {z = :2} {r = :2})")

from itertools import combinations

# there are three pythagorean triangles all of which

# share a common side, <w>, the half-width of AB

for w in range(1, 30):

# list (in order) the other sides and hypotenuses

# of all such triangles for hypotenuses below 60

xhs = [(x, h) for h in range(w + 1, 60) if

(x := round((t := (h * h - w * w)) ** (1/2))) ** 2 == t]

# consider possible triples created from these values

for (r, DF), (x, SE), (y, SA) in combinations(xhs, 3):

# ensure that CP and PD (<z> here) are positive integers

z, _r = divmod(y - x - 2 * r, 2)

if not _r and z > 0:

d = SA + SE + 4 * w + 2 * (z + DF)

print(f"{d = } ({SE = } {DF = } {SA = }) "

f"({w = :2} {x = :2} {y = } {z = :2} {r = :2})")

Frits permalink

Shorter and a little bit more efficient.


from itertools import combinations

# we have 3 different Pythagorean formulas with side w^2 (AB = 2.w)

# dictionary of Pythagorean triples below 60 with one side <w> below 30
d = {w: [(round(b), c) for c in range(w + 1, 60) 
         if round(b := (c**2 - w**2)**.5) == b] for w in range(1, 30)}

sols = set()
for w, vs in d.items():
  # choose three different Pythagorean triples with hypothenusas FD, SE and AS
  for (o1, FD), (hE, SE), (o3, AS) in combinations(vs, 3):
    # distances from south boundary for  D
    hD = o1 + hE
    DP, r = divmod(o3 - 2 * hD + hE, 2)
    if r or DP < 1: continue
    
    # calculate total distance
    tot = SE + AS + 4 * w + 2 * (FD + DP)
    sols.add(str(int(tot)))

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

from itertools import combinations

# we have 3 different Pythagorean formulas with side w^2 (AB = 2.w)

# dictionary of Pythagorean triples below 60 with one side <w> below 30

d = {w: [(round(b), c) for c in range(w + 1, 60)

if round(b := (c**2 - w**2)**.5) == b] for w in range(1, 30)}

sols = set()

for w, vs in d.items():

# choose three different Pythagorean triples with hypothenusas FD, SE and AS

for (o1, FD), (hE, SE), (o3, AS) in combinations(vs, 3):

# distances from south boundary for D

hD = o1 + hE

DP, r = divmod(o3 - 2 * hD + hE, 2)

if r or DP < 1: continue

# calculate total distance

tot = SE + AS + 4 * w + 2 * (FD + DP)

sols.add(str(int(tot)))

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

Sunday Times Teaser 3277 – Croquet Equipment

by John Owen

Published Sunday July 13 2025 (link)

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