Teaser 3277: Croquet equipment
From The Sunday Times, 13th July 2025 [link] [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?
[teaser3277]
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Jim Randell 8:46 am on 13 July 2025 Permalink |
I get two solutions, so here is a straightforward solution to make sure I haven’t missed any thing.
But if we disallow distances of 60ft or more between any two positions on the walk (not just adjacent positions), then we can eliminate the larger of these solutions.
This Python program runs in 65ms. (Internal runtime is 5ms).
from enigma import (irange, ihypot, printf) M = 59 # max length of a path segment # let AB = EF = 2x for x in irange(1, M // 2): # e = distance E north from southern edge for e in irange(1, M): SE = ihypot(x, e) if SE is None or SE > M: continue # d = distance D is north of EF for d in irange(1, e - 1): FD = ihypot(x, d) if FD is None or FD > M: continue # p = distance P north of D for p in irange(1, M): SA = ihypot(x, e + d + p + p + d) if SA is None or SA > M: continue # total distance walked T = SE + SA + 2*FD + 4*x + 2*p printf("T={T} [SE={SE} FD={FD} SA={SA}; x={x} e={e} d={d} p={p}]")Solution: There are two viable solutions to the puzzle: 142 ft, and 220 ft.
Apparently the size of a croquet lawn is not fixed, but it should be a rectangle with sides in the ratio of 1.25.
The first of the solutions involves a playing area of 48 ft × 44 ft (ratio = 1.09).
And all positions on the route are within 60 ft of S (as indicated by the dotted line).
The second of the solutions involves a playing area of 96 ft × 42 ft (ratio = 2.29).
And B and F are further than 60 ft from S.
The published solution is: “142 feet”.
But a correction was issued with Teaser 3280:
.
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Frits 11:38 am on 13 July 2025 Permalink |
@Jim, shouldn’t we have: e + d + p = 2.x ?
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Jim Randell 11:56 am on 13 July 2025 Permalink |
@Frits: Why?
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Frits 12:18 pm on 13 July 2025 Permalink |
I assumed that the lawn was square (but that isn’t specified).
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Jim Randell 4:19 pm on 13 July 2025 Permalink |
Alternatively, using the [[
pythagorean_triples()]] function from the enigma.py library.The following Python program has an internal runtime of 122µs.
from enigma import (defaultdict, pythagorean_triples, subsets, div, printf) M = 59 # maximum length of a path segment # collect (<other>, <hypot>) sides for pythagorean triples t = defaultdict(list) for (x, y, z) in pythagorean_triples(M): t[x].append((y, z)) t[y].append((x, z)) # consider possible x values for x in sorted(t.keys()): if 2 * x > M: break # FD, SE, SA all have base x, and increasing vertical sides ts = sorted(t[x]) for ((d, FD), (e, SE), (a, SA)) in subsets(ts, size=3): p = div(a - e - 2*d, 2) if p is None or p < 1: continue # total distance walked T = SE + SA + 2*FD + 4*x + 2*p printf("T={T} [SE={SE} FD={FD} SA={SA}; x={x} e={e} d={d} p={p}]")LikeLike
Hugo 10:34 am on 21 July 2025 Permalink |
If hoop P has coordinates (0, 0) then it seems
A is at (12, -13), B is at (12, 13)
C is at (0, 8), D is at (0, -8)
E is at (-12, -13), F is at (-12, 13)
S is at (-24, -22).
It would have been kind of them to tell us that all the distances, including x and y separations, are integers.
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