Sunday Times Teaser 2617 – A Case of Ambiguity

by Des MacHale

Published: 18 November 2012 (link)

Every schoolchild knows that it is possible for two triangles to have two of their lengths-of-sides in common and one of their angles in common without the two triangles being identical or “congruent”. This can happen when the common angle is not included between the two common sides. I have found such a pair of non-congruent triangles with all the lengths of the sides being a whole number of centimetres and with the longest side of each triangle being 10cm in length.

What are the lengths of the other two sides of the smaller triangle?

2 Comments Leave one →

Brian Gladman permalink

Here is my solution:


from collections import defaultdict
from fractions import Fraction

d = defaultdict(list)
# c is the longest side with a <= b < c (there is a longest side)
c = 10
for a in range(1, c):
  # ensure that b > c - a so that the triangle is valid
  for b in range(max(a, c - a + 1), c):
    # for each side compute, the cosine of the angle opposite this side
    # and use this as a dictionary index to store a list of the sides
    # of all triangles having this angle as the non included angle
    top, bot, sides = a * a + b * b + c * c, 2 * a * b * c, (a, b, c)
    for x in sides:
      d[Fraction(top - 2 * x * x, bot // x)] += [sides]

fs = 'small triangle: {:s}, large triangle: {:s}, angle: cos^-1({:s})'
for f in d:
  # look for angles allowing two triangles
  if len(d[f]) > 1:
    # check that the triangles are different and that they share two sides
    if d[f][0] != d[f][1] and len(set(d[f][0]) & set(d[f][1])) == 2:
      print(fs.format(min(d[f][0], d[f][1]), max(d[f][0], d[f][1]), f))

from collections import defaultdict

from fractions import Fraction

d = defaultdict(list)

# c is the longest side with a <= b < c (there is a longest side)

c = 10

for a in range(1, c):

# ensure that b > c - a so that the triangle is valid

for b in range(max(a, c - a + 1), c):

# for each side compute, the cosine of the angle opposite this side

# and use this as a dictionary index to store a list of the sides

# of all triangles having this angle as the non included angle

top, bot, sides = a * a + b * b + c * c, 2 * a * b * c, (a, b, c)

for x in sides:

d[Fraction(top - 2 * x * x, bot // x)] += [sides]

fs = 'small triangle: {:s}, large triangle: {:s}, angle: cos^-1({:s})'

for f in d:

# look for angles allowing two triangles

if len(d[f]) > 1:

# check that the triangles are different and that they share two sides

if d[f][0] != d[f][1] and len(set(d[f][0]) & set(d[f][1])) == 2:

print(fs.format(min(d[f][0], d[f][1]), max(d[f][0], d[f][1]), f))

Brian Gladman permalink

Here is another version


#                   __
#                __/ / A
#             __/   /  
#        c __/     /    
#       __/       / b     b
#    __/         /        
# B /___________/__________ C
#        a1        a2 - a1
#
# From sine rule:
#
# (1)  b.sin(C) = c.sin(B)
#
# From cosine rule:
#
# (2)  b^2 = c^2 + a1^2 - 2.c.a1.cos(B) = c^2 + a2^2 - 2.c.a2.cos(B)
#
# From the right hand triangle:
#
# (3)  a2 - a1 = 2.b.cos(C)
#
# From equation (2):
#
# (4)  a2 + a1 = 2.c.cos(B)
#
# From (2) and (3):
#
# (5)  (a2 + a1)^2 - (a2 - a1)^2 = 4.a1.a2 =
#      = 4.c^2 - 4.c^2.sin^2(B) - 4.b^2 + 4.b^2.sin^2(C)
#
# Hence:
#
# (6)  a1.a2 = c^2 - b^2

ft = 'small triangle: {:s}, large triangle: {:s}'
c = 10
for b in range(1, c):
  p = c * c - b * b
  for f in range(1, min(c, int(p ** 0.5) + 1)):
    if not p % f and f < p // f < 11:
      print(ft.format(sorted([b, f, 10]), sorted([b, p // f, 10])))

# __

# __/ / A

# __/ /

# c __/ /

# __/ / b b

# __/ /

# B /___________/__________ C

# a1 a2 - a1

# From sine rule:

# (1) b.sin(C) = c.sin(B)

# From cosine rule:

# (2) b^2 = c^2 + a1^2 - 2.c.a1.cos(B) = c^2 + a2^2 - 2.c.a2.cos(B)

# From the right hand triangle:

# (3) a2 - a1 = 2.b.cos(C)

# From equation (2):

# (4) a2 + a1 = 2.c.cos(B)

# From (2) and (3):

# (5) (a2 + a1)^2 - (a2 - a1)^2 = 4.a1.a2 =

# = 4.c^2 - 4.c^2.sin^2(B) - 4.b^2 + 4.b^2.sin^2(C)

# Hence:

# (6) a1.a2 = c^2 - b^2

ft = 'small triangle: {:s}, large triangle: {:s}'

c = 10

for b in range(1, c):

p = c * c - b * b

for f in range(1, min(c, int(p ** 0.5) + 1)):

if not p % f and f < p // f < 11:

print(ft.format(sorted([b, f, 10]), sorted([b, p // f, 10])))

Sunday Times Teaser 2617 – A Case of Ambiguity

by Des MacHale

Published: 18 November 2012 (link)

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