Sunday Times Teaser 3181 – “No slack, Alice!” says Grace

by BRG on 8 September 2023

by Stephen Hogg

Published Sunday September 10 2023 (link)

Grace mimicked Mr Dodgson: ‘’Take two different positive odd numbers with no shared prime factor. Multiply the larger by their sum, giving the perimeter of a distinct right-angled triangle with whole-number sides. Only such values and their multiples work.’’

Alice created a loop from part of a 1m thread. She was able to pull it tightly on a pinboard into a right-angled triangle with whole-number cm sides in two ways (with different-shaped triangles).

Alice then pulled the same loop tight over the thin polar spike of the classroom globe, down two meridians and along the equator. Thinking ‘’Logic’s dead!’’, she saw 90° equatorial angles and a non-zero polar angle, which obviously didn’t add to 180°.

Grace calculated the polar angle to the nearest degree. Curiously, transposing its two digits gave the globe’s whole-number centimetre radius.

Give this radius

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BRG permalink


from math import pi

# collect pythagorean triangles indexed on 
# their perimeters up 100 centimetres
d = dict()
for c in range(5, 51):
  for a in range(1, round(c / 2 ** (1 / 2))):
    b2 = (c * c - a * a)
    if (b := int(b2 ** (1/2))) ** 2 == b2:
      if (p := a + b + c) > 100:
        continue
      d[p] = d.get(p, set()) | {(a, b, c)}

# consider perimeters that allow two different
# shaped triangles
for p, s_abc in d.items():
  if len(s_abc) == 2:
    # now consider the two digit radius of the globe
    for r in range(10, 100):
      # calculate the polar angle in degrees
      # to the nearest integer
      th = round((180 / pi) * (p / r - pi))
      # does its digit reversed value match the radius?
      if 10 * (th % 10) + th // 10 == r:
        print(f"Radius {r}cm [perimeter {p}cm, triangles {s_abc}]")

from math import pi

# collect pythagorean triangles indexed on

# their perimeters up 100 centimetres

d = dict()

for c in range(5, 51):

for a in range(1, round(c / 2 ** (1 / 2))):

b2 = (c * c - a * a)

if (b := int(b2 ** (1/2))) ** 2 == b2:

if (p := a + b + c) > 100:

continue

d[p] = d.get(p, set()) | {(a, b, c)}

# consider perimeters that allow two different

# shaped triangles

for p, s_abc in d.items():

if len(s_abc) == 2:

# now consider the two digit radius of the globe

for r in range(10, 100):

# calculate the polar angle in degrees

# to the nearest integer

th = round((180 / pi) * (p / r - pi))

# does its digit reversed value match the radius?

if 10 * (th % 10) + th // 10 == r:

print(f"Radius {r}cm [perimeter {p}cm, triangles {s_abc}]")

Frits permalink

Using Dr Dodgson’s method:


from math import pi, ceil, gcd
from sys import argv

# return a Pythagorean triple
sides = lambda a, b, m: (m * b * a, m * (b * b - a * a) // 2, m * (b * b + a * a) // 2)

detail = 1
limit = 100 if len(argv) != 2 else int(argv[1])

# distance between the 2 equator points must be less than pi * r
if limit > 2 * pi * 99:
  limit = int(2 * pi * 99)
  print(f"resetting limit to {limit}")

# (a + 2)(2a + 2) <= limit
alim = int((9 + 2 * (limit - 4))**.5 / 2 - 1.5) + 1

if detail:
  print(f"{limit = }, {alim = }")

# take two different positive odd numbers with no shared prime factor. 
# multiply the larger by their sum, giving the perimeter of a distinct 
# right-angled triangle with whole-number sides

# generate possible perimeters (choose b boundary so that b(a + b) <= limit)
ps = [(p, (a, b), m) for a in range(1, alim, 2) 
      for b in range(a + 2, int((limit + 0.25 * a**2)**.5 - a / 2) + 1, 2)
      for m in range(1, limit // ((a + b) * b) + 1) 
      if (p := (a + b) * b * m) <= limit and gcd(a, b) == 1]
      
# only keep perimeters with two different-shaped triangles
ps = {p: lst for p, _, _ in ps 
     if len(lst := [sides(a, b, m) for p2, (a, b), m in ps if p2 == p]) == 2}

if detail:
  for k, vs in ps.items(): 
    print(f"perimeter: {k}, triangles: {vs}")
  print()

for p in ps.keys():
  # choose boundaries for radius <r> so angle <a> lies between 11 and 99
  for r in range(int(20 * p / (31 * pi)) + 1, ceil(180 * p / (191 * pi)) + 1):
    a = round(180 * (p - pi * r) / (pi * r))
    # reversed digits?
    if str(r) != str(a)[::-1]: continue
    print(f"answer: {r} cm for perimeter {p} cm")

from math import pi, ceil, gcd

from sys import argv

# return a Pythagorean triple

sides = lambda a, b, m: (m * b * a, m * (b * b - a * a) // 2, m * (b * b + a * a) // 2)

detail = 1

limit = 100 if len(argv) != 2 else int(argv[1])

# distance between the 2 equator points must be less than pi * r

if limit > 2 * pi * 99:

limit = int(2 * pi * 99)

print(f"resetting limit to {limit}")

# (a + 2)(2a + 2) <= limit

alim = int((9 + 2 * (limit - 4))**.5 / 2 - 1.5) + 1

if detail:

print(f"{limit = }, {alim = }")

# take two different positive odd numbers with no shared prime factor.

# multiply the larger by their sum, giving the perimeter of a distinct

# right-angled triangle with whole-number sides

# generate possible perimeters (choose b boundary so that b(a + b) <= limit)

ps = [(p, (a, b), m) for a in range(1, alim, 2)

for b in range(a + 2, int((limit + 0.25 * a**2)**.5 - a / 2) + 1, 2)

for m in range(1, limit // ((a + b) * b) + 1)

if (p := (a + b) * b * m) <= limit and gcd(a, b) == 1]

# only keep perimeters with two different-shaped triangles

ps = {p: lst for p, _, _ in ps

if len(lst := [sides(a, b, m) for p2, (a, b), m in ps if p2 == p]) == 2}

if detail:

for k, vs in ps.items():

print(f"perimeter: {k}, triangles: {vs}")

print()

for p in ps.keys():

# choose boundaries for radius <r> so angle <a> lies between 11 and 99

for r in range(int(20 * p / (31 * pi)) + 1, ceil(180 * p / (191 * pi)) + 1):

a = round(180 * (p - pi * r) / (pi * r))

# reversed digits?

if str(r) != str(a)[::-1]: continue

print(f"answer: {r} cm for perimeter {p} cm")

BRG permalink

I also produced a version using Dr Dodgson’s method:


from itertools import count, combinations
from math import pi, gcd
from sys import argv

# limit on the length of the string (2.pi.max(r) is
# the mmaximum sensible limit
limit = 100 if len(argv) != 2 else int(argv[1])
limit = min(limit, round(200 * pi))
# 'two ways' or 'only two ways'
only_two = False

# use Dr Dodgson's method to generate pythagorean triangles
d = dict()
for m in count(1, 2):
  if m * (m + 1) > limit:
    break
  for n in range(1, m, 2):
    if gcd(m, n) != 1:
      continue
    # form the triangle sides
    t = tuple(sorted((m * n, (m * m - n * n) // 2, (m * m + n * n) // 2)))
    if (p := m * (m + n)) > limit:
      break
    # add multiples of the base triangle 
    for r in range(1, limit // p + 1):
      d[r * p] = d.get(r * p, set()) | {tuple(r * n for n in t)}

cnt = 0
# connsider possible perimeters
for p in sorted(d.keys()):

  # do we want solutions for 'two' or 'only two' ways?
  if only_two and len(d[p]) > 2:
    continue

  # consider all pairings of triangles
  for t1, t2 in combinations(d[p], 2):
    
    # now consider the two digit radius of the globe
    for r in range(11, 100):
      srv = int(str(r)[::-1])
      # calculate the polar angle in degrees to nearest integer
      th = round((180 / pi) * (p / r - pi))
      # does it match the digit reversed value of the radius?
      if th == srv:
        f = 1 if limit == 100 else 15
        t1, t2 = f"{str(t1):>{f}}", f"{str(t2):>{f}}"
        cnt += 1
        print(f"{cnt:3} radius = {r:2} cm, p = {p:3} cm [{t1} {t2}]")

from itertools import count, combinations

from math import pi, gcd

from sys import argv

# limit on the length of the string (2.pi.max(r) is

# the mmaximum sensible limit

limit = 100 if len(argv) != 2 else int(argv[1])

limit = min(limit, round(200 * pi))

# 'two ways' or 'only two ways'

only_two = False

# use Dr Dodgson's method to generate pythagorean triangles

d = dict()

for m in count(1, 2):

if m * (m + 1) > limit:

break

for n in range(1, m, 2):

if gcd(m, n) != 1:

continue

# form the triangle sides

t = tuple(sorted((m * n, (m * m - n * n) // 2, (m * m + n * n) // 2)))

if (p := m * (m + n)) > limit:

break

# add multiples of the base triangle

for r in range(1, limit // p + 1):

d[r * p] = d.get(r * p, set()) | {tuple(r * n for n in t)}

cnt = 0

# connsider possible perimeters

for p in sorted(d.keys()):

# do we want solutions for 'two' or 'only two' ways?

if only_two and len(d[p]) > 2:

continue

# consider all pairings of triangles

for t1, t2 in combinations(d[p], 2):

# now consider the two digit radius of the globe

for r in range(11, 100):

srv = int(str(r)[::-1])

# calculate the polar angle in degrees to nearest integer

th = round((180 / pi) * (p / r - pi))

# does it match the digit reversed value of the radius?

if th == srv:

f = 1 if limit == 100 else 15

t1, t2 = f"{str(t1):>{f}}", f"{str(t2):>{f}}"

cnt += 1

print(f"{cnt:3} radius = {r:2} cm, p = {p:3} cm [{t1} {t2}]")

Sunday Times Teaser 3181 – “No slack, Alice!” says Grace

by Stephen Hogg

Published Sunday September 10 2023 (link)

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