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Jim Randell 7:03 am on 21 September 2025 Permalink | Reply
Tags: by: John Owen
Teaser 3287: Ferry route
From The Sunday Times, 21st September 2025 [link] [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?

[teaser3287]
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Alex.T.Sutherland, Frits, Jim Randell, and 1 other are discussing. Toggle Comments
- Jim Randell 7:18 am on 21 September 2025 Permalink | Reply
 See: Teaser 2549.
 
 If the diameter of the circle is x, and the lengths of the non-diameter legs are a, b, c, then we have:
 
 x³ − (a² + b² + c²)x − 2abc = 0
 
 The following Python program runs in 62ms. (Internal runtime is 643µs).
```
from enigma import (powers, inf, first, lt, subsets, pi, cb, sumsq, printf)

# squares less than 300
sqs = list(first(powers(1, inf, 2), lt(300)))

# choose 3 different lengths (x is the diameter)
for (a, b, x) in subsets(sqs, size=3):
 # duplicate one of the short lengths
 for c in (a, b):
 # check for a viable quadrilateral
 if not (x < a + b + c < pi * x): continue
 if not (cb(x) - sumsq([a, b, c]) * x - 2 * a * b * c == 0): continue
 # output solution
 printf("{a}, {b}, {x} [and {c}]")
```
 Solution: The leg lengths are 36, 49, 81 km.
 
 The duplicate leg length is 36 km.
 
 The next smallest solution is at 4× these lengths, which brings the diameter to 324 km.
 
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 - Jim Randell 8:49 am on 22 September 2025 Permalink | Reply
 We can suppose that the non-diameter sides are (a, b, a) (i.e. c = a, and it may be the case that a > b), and then the equation can be simplified to:
 
 2a² = x(x − b)
 
 So by choosing b and x we can then straightforwardly calculate the corresponding value for a, and check it is viable (i.e. a square):
```
from enigma import (powers, inf, first, lt, subsets, is_square, div, ordered, printf)

# squares less than 300
sqs = list(first(powers(1, inf, 2), lt(300)))

# choose 2 different lengths (x is the diameter)
for (b, x) in subsets(sqs, size=2):
 # a is the duplicated side
 a = is_square(div(x * (x - b), 2))
 if a is None or a not in sqs: continue
 # output solution
 (p, q) = ordered(a, b)
 printf("{p}, {q}, {x} [and {a}]")
```
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 - Frits 2:29 pm on 22 September 2025 Permalink | Reply
 
 I had come up with similar equation. You can also deduce the parity of “a”.
 
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- Ruud 10:35 am on 21 September 2025 Permalink | Reply
```
import math
import itertools

for ab, bs, sd, da in itertools.product((i * i for i in range(1, 18)), repeat=4):
 try:
 angle_ab = math.asin(ab / da) * 2
 angle_bs = math.asin(bs / da) * 2
 angle_sd = math.asin(sd / da) * 2

 if angle_ab + angle_bs + angle_sd == math.pi:
 print(sorted({ab, bs, sd, da}))
 except ValueError:
 ...
```
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 - Jim Randell 11:38 am on 21 September 2025 Permalink | Reply
 
 @Ruud: How does this code ensure that two of the leg lengths are the same?
 
 And it is probably better to use [[ math.isclose() ]] rather than testing floats for equality using ==.
 
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 - Ruud 12:55 pm on 21 September 2025 Permalink | Reply
 
 I just checked manually by checking that there are 3 different length. Could have added a simple test, of course.
 
 Yes, math.isclose would have been better, but as the equality check gave already an answer, I thought I could just refrain from isclose.
 
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 - Ruud 4:14 pm on 21 September 2025 Permalink | Reply
 @Jim
 I just checked manually that there are exactly three different lengths, but a test would be better.
 
 Yes, math.isclose would have beeen safer.
 
 Here’s my updated version:
 
 import math import itertools for ab, bs, sd, da in itertools.product((i * i for i in range(1, 18)), repeat=4): try: angle_ab = math.asin(ab / da) * 2 angle_bs = math.asin(bs / da) * 2 angle_sd = math.asin(sd / da) * 2 if math.isclose(angle_ab + angle_bs + angle_sd, math.pi) and len(sol := {ab, bs, sd, da}) == 3: print(sorted(sol)) except ValueError: ... peek("done")
 
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- Alex.T.Sutherland 10:42 am on 28 September 2025 Permalink | Reply
```
TEASER 3287.....FERRY ROUTE.
Given:-		Cyclic Quadrilateral [A B C D] with lengths [a b c d], d = diameter; 
		 All sides are of perfect square length.
Method:-	Using Ptolemy's Theorem relating the diagonals to the sides
		 of the quadrilateral and that right angles are created at B & C
		 by the diagonals the following equation is derived:-

		 (d^2 - a^2) * (d^2 - c^2) = (b*d + a*c)^2

		 If 'c' is made equal to 'a' the following can be derived:-

Answer:-	2*a^2 = d*(d-b);

		 Choose 3 squares from the given set to satisfy this equation.
Side Line:-	The path could be 'abad' or'aabd' but the answer is the same for the lengths.
		 (neglecting symmetry eg 'baa' is the same as 'aab')'
		 For my answer the sum of the lengths 'aba' or 'aab' is a perfect square.
```
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Jim Randell 6:24 am on 13 July 2025 Permalink | Reply
Tags: by: John Owen, flawed ( 52 )
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 | Reply
 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:
 
 There were in fact two possible answers: 142 and 220 feet.
 Either was accepted as correct.
 
 .
 
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 - Frits 11:38 am on 13 July 2025 Permalink | Reply
 
 @Jim, shouldn’t we have: e + d + p = 2.x ?
 
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 - Jim Randell 11:56 am on 13 July 2025 Permalink | Reply
 
 @Frits: Why?
 
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 - Frits 12:18 pm on 13 July 2025 Permalink | Reply
 
 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 | Reply
 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}]")
```
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- Hugo 10:34 am on 21 July 2025 Permalink | Reply
 
 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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Jim Randell 2:42 am on 30 March 2025 Permalink | Reply
Tags: by: John Owen
Teaser 3262: Log cabins
From The Sunday Times, 30th March 2025 [link] [link]

Ann, Bert, Chloe and Dave own log cabins on the shore of a circular lake. Ann often rows across the lake (in a straight line) to visit Bert, and vice-versa. Chloe and Dave also visit each other in the same way.

The two routes cross in the lake at point X. Interestingly, the four distances from the log cabins to X are all whole numbers of metres (greater than 20m), but the list of distances contains no digit more than once. Chloe’s distance from her log cabin to X is twice Ann’s, while Bert’s distance is three times Ann’s.

In ascending order, what are the four distances from the log cabins to X?

[teaser3262]
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- Jim Randell 3:10 am on 30 March 2025 Permalink | Reply
  The cabins form a cyclic quadrilateral [ @wikipedia ].
  
  And if we denote the distances AX = a, BX = b, CX = c, DX = d. Then we have:
  
  a . b = c . d
  
  This is the intersecting chords theorem [ @wikipedia ].
  
  So:
  
  b = 3a
  c = 2a
  ⇒
  d = 3a/2
  
  Hence a must be even.
  
  The following Python program runs in 67ms. (Internal runtime is 106µs).
```
from enigma import (irange, inf, ediv, ordered, join, printf)

# consider distance AX = a (must be even)
for a in irange(22, inf, step=2):
  # BX = b = 3a, CX = c = 2a, DX = d = 3a/2
  (b, c, d) = (3 * a, 2 * a, ediv(3 * a, 2))
  ds = ordered(a, b, c, d)
  s = join(ds)
  # between them they use no digit more than once
  if len(s) > 10: break
  if len(s) != len(set(s)): continue
  # output solution
  printf("{ds} [a={a} b={b} c={c} d={d}]")
```
  Solution: The distances are: 36, 54, 72, 108 metres.
  
  a = 36
  b = 108
  c = 72
  d = 54
  and:
  a . b = c . d = 3888
  
  Manually we can follow a similar approach, considering even values for a starting at 22, until a solution is found:
  
  a = 22; repeated digits
  a = 24; b = 72; repeated digits
  a = 26; b = 78; c = 52; repeated digits
  a = 28; b = 84; repeated digits
  a = 30; b = 90; repeated digits
  a = 32; b = 96; c = 64; repeated digits
  a = 34; b = 102; c = 68; d = 51; repeated digits
  a = 36; b = 108; c = 72; d = 54; solution found
  
  For an exhaustive solution it is sufficient to check a ≤ 66 beyond which the distances use more than 10 digits, so there will always be duplication of digits. But there are no further solutions.
  
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Jim Randell 9:57 am on 3 December 2024 Permalink | Reply
Tags: by: John Owen
Teaser 2600: Mileometer
From The Sunday Times, 22nd July 2012 [link] [link]

My car’s digital mileometer has five digits, displaying whole miles. It also has a trip meter with five digits, displaying tenths of a mile (which resets to zero when reaching 10,000 miles). Soon after the car was new I reset the trip meter to zero (but never did that again). Not long after, I noticed that the two five-digit displays were the same ignoring the decimal point). The next time the displays were the same I wrote down the mileage and did so on each subsequent occasion until the mileage reached 100,000. The sum of all the mileages I wrote down was 500,000.

What were the five digits when the displays were first the same?

[teaser2600]
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- Jim Randell 9:57 am on 3 December 2024 Permalink | Reply
 Suppose the readings are the same at a distance ABCDE+F miles (the +F indicates + F tenths of a mile), then the readings are:
 
 milometer = ABCDE
 trip meter = ABCD+E
 
 The trip meter is actually reading XABCD+E miles, since it was reset, for some non-visible digit X.
 
 Which means the trip meter was reset at an actual distance of:
 
 ABCDE+F − XABCD+E
 
 And we want this to be soon after the car was new, maybe in the first 10K miles.
 
 This Python program looks at distances of possible duplicated readings, and collects them by the trip reset distance. It then looks for a distance where the sum of the milometer readings (except for the first one) is 500000.
 
 It runs in 530ms (using PyPy).
```
from enigma import (defaultdict, irange, subsets, nconcat, trim, printf)

# collect readings by reset distance
rs = defaultdict(list)

# consider possible 6-digit distances (in 0.1 mile units)
for (A, B, C, D, E, F) in subsets(irange(0, 9), size=6, select='M'):
 dist = nconcat(A, B, C, D, E, F)
 for X in irange(max(0, A - 1), A):
 # consider trip distances that match the milometer (in 0.1 mile units)
 trip = nconcat(X, A, B, C, D, E)
 # calculate trip reset point
 k = dist - trip
 if not (0 < k < 100000): continue
 rs[k].append(dist)

# output the milometer/trip readings
def output(d, k):
 (m, f) = divmod(d, 10)
 (t, g) = divmod(d - k, 10)
 t %= 10000
 printf("@ {m:05d}.{f} miles -> {m:05d} + {t:04d}.{g}")

# consider possible reset distances
for (k, vs) in rs.items():
 vs = sorted(vs)
 # sum the milometer readings (except the first one)
 t = sum(v // 10 for v in trim(vs, head=1))
 if t == 500000:
 # output solution
 output(k, k)
 for v in vs: output(v, k)
 printf("t = {t}")
 printf()
```
 Solution: When the displays were first the same they were showing 01235 (or 0123.5 for the trip meter).
 
 The trip meter was reset as a distance of 1111.6 miles, and so the readings were:
 
 @ 01111.6 miles → 01111 + 0000.0 [trip reset]
 @ 01235.1 miles → 01235 + 0123.5 [1st equal display]
 @ 12346.2 miles → 12346 + 1234.6 [2nd equal display – noted down]
 @ 23457.3 miles → 23457 + 2345.7 [3rd equal display – noted down]
 @ 34568.4 miles → 34568 + 3456.8 [4th equal display – noted down]
 @ 45679.5 miles → 45679 + 4567.9 [5th equal display – noted down]
 @ 56790.6 miles → 56790 + 5679.0 [6th equal display – noted down]
 @ 67901.7 miles → 67901 + 6790.1 [7th equal display – noted down]
 @ 79012.8 miles → 79012 + 7901.2 [8th equal display – noted down]
 @ 90123.9 miles → 90123 + 9012.3 [9th equal display – noted down]
 @ 90124.0 miles → 90124 + 9012.4 [10th equal display – noted down]
 
 And the sum of the values noted down is:
 
 12346 + 23457 + 34568 + 45679 + 56790 + 67901 + 79012 + 90123 + 90124
 = 500000
 
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Jim Randell 7:58 am on 5 November 2024 Permalink | Reply
Tags: by: John Owen

Teaser 2608: Football logic

From The Sunday Times, 16th September 2012 [link] [link]

In the logicians’ football tournament, each of three teams (captained by Alf, Bert and Charlie) plays each other once, with three points for a win and one for a draw. In working out their order at the end of the tournament, “goals scored” are used to determine the order of teams with the same points total. Each captain only knows the scores in his own team’s games. At the end I asked the captains in turn whether they knew which position they had finished in. The replies were:

Alf: “no”;
Bert: “no”;
Charlie: “no”;
Alf: “yes”.

In which position did Charlie’s team finish? And what was the result in the game between the other two teams?

[teaser2608]

Jim Randell 8:01 am on 5 November 2024 Permalink | Reply

If a captain knows that they are tied on points with another team, then they cannot know which way the tie breaking will go, as they do not know how many goals the other team scored in their remaining match, so any situation where teams are tied on points will necessarily require the captain to respond that they do not know their final position.

This Python program uses the [[ Football() ]] helper class, and the [[ filter_unique() ]] function from the enigma.py library to solve the puzzle.

It runs in 69ms. (Internal runtime is 838µs).

from enigma import (Football, filter_unique, printf)

# scoring system
football = Football(games='wdl', points=dict(w=3, d=1))

# calculate positions from points, ties are indicated with '='
def pos(pts):
  rs = [None] * len(pts)
  p = 1
  for x in sorted(set(pts), reverse=1):
    k = pts.count(x)
    v = str(p)
    if k > 1: v += '='
    for (i, y) in enumerate(pts):
      if y == x:
        rs[i] = v
    p += k
  return tuple(rs)

is_tie = lambda x: x.endswith('=')

# generate possible situations
def generate():
  # consider possible game outcomes
  for (AB, AC, BC) in football.games(repeat=3):
    # table for each team
    A = football.table([AB, AC], [0, 0])
    B = football.table([AB, BC], [1, 0])
    C = football.table([AC, BC], [1, 1])
    # return (<match-outcomes>, <positions>)
    yield ((AB, AC, BC), pos([A.points, B.points, C.points]))

# indices for the teams, and the matches
(A, B, C) = (AB, AC, BC) = (0, 1, 2)
matches = { A: (AB, AC), B: (AB, BC), C: (AC, BC) }

# can team <j> deduce their finishing position knowing the outcomes of their own matches?
# <f> = 0 if the answer is "no"; = 1 if it is "yes"
def answer(ss, j, f):
  # which matches is team <j> involved in?
  ms = matches[j]
  fnf = lambda t: tuple(t[0][i] for i in ms)
  fng = lambda t: t[1][j]
  # find results with unique/non-unique positions
  r = filter_unique(ss, fnf, fng)
  if f == 0:
    # return results that are non-unique, or end with a tie on points
    return r.non_unique + [t for t in r.unique if is_tie(fng(t))]
  else:
    # return results that are unique, but don't end with a tie on points
    return [t for t in r.unique if not is_tie(fng(t))]

# start with all possible outcomes and positions
ss = list(generate())

# A cannot deduce his finishing position
ss = answer(ss, A, 0)

# B cannot deduce his finishing position
ss = answer(ss, B, 0)

# C cannot deduce his finishing position
ss = answer(ss, C, 0)

# A can now deduce his finishing position
ss = answer(ss, A, 1)

# output solutions
for ((AB, AC, BC), (posA, posB, posC)) in ss:
  printf("posC = {posC}; AB = {AB} [AB={AB} AC={AC} BC={BC}; posA={posA} posB={posB} posC={posC}]")

Solution: Charile finished in second place. The game between Alf and Bert was a draw.

The match outcomes are either:

(A v B = draw; C beat A; B beat C)
B = 1w 1d = 4 pts (1st)
C = 1w = 3 pts (2nd)
A = 1d = 1 pt (3rd)

or:

(A v B = draw; A beat C; C beat B)
A = 1w 1d = 4 pts (1st)
C = 1w = 3pts (2nd)
B = 1d = 1pt (3rd)

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Jim Randell 4:38 pm on 28 July 2023 Permalink | Reply
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Jim Randell 4:55 pm on 8 January 2023 Permalink | Reply
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Jim Randell 4:29 pm on 6 January 2023 Permalink | Reply
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Jim Randell 4:00 pm on 7 October 2022 Permalink | Reply
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Jim Randell 9:14 am on 5 June 2022 Permalink | Reply
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Jim Randell 8:57 am on 21 September 2021 Permalink | Reply
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