Sunday Times Teaser 2663 – Missed the Plot

brian gladman permalink


# let the two smallest sides be a and b with a < b.  To make b the
# average of the other two sides requires that c = 2 * b - a.  Now
# using Heron's formula for the area and equating the area and the
# perimeter we can show that:
# 
#   (3 * b - 2 * a) * (2 * a - b) = 48
#
# so the bracketed expressions on the left are factors of 48. With
# these factors as f and g we obtain:
# 
#   a = (f + 3 * g) / 4 and b = (f + g) / 4

for f in range(1, 49):
  g, r = divmod(48, f)
  (a, s), b = divmod(f + 3 * g, 4), (f + g) // 2
  if not (r or s) and a < b:
    print('The sides are {:d}, {:d} and {:d} metres.'
            .format(a, b, 2 * b - a))

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# let the two smallest sides be a and b with a < b. To make b the

# average of the other two sides requires that c = 2 * b - a. Now

# using Heron's formula for the area and equating the area and the

# perimeter we can show that:

#

# (3 * b - 2 * a) * (2 * a - b) = 48

#

# so the bracketed expressions on the left are factors of 48. With

# these factors as f and g we obtain:

#

# a = (f + 3 * g) / 4 and b = (f + g) / 4

for f in range(1, 49):

g, r = divmod(48, f)

(a, s), b = divmod(f + 3 * g, 4), (f + g) // 2

if not (r or s) and a < b:

print('The sides are {:d}, {:d} and {:d} metres.'

.format(a, b, 2 * b - a))

Reply

Jim Randell permalink

My analysis gives the same equations as you found. I then consider the divisors of 48 (using the divisors() function from my enigma.py library. There are only 5 factorisations of 48 to consider.


# suppose the triangle has sides a, b, c
# where a < b, and c = b + (b - a) = 2b - a
#
# the semi-perimeter of the triangle s = (a + b + c)/2 = 3b/2
#
# and by Heron's formula the area of the triangle T is:
#
# T^2 = s(s - a)(s - b)(s - c)
#
# but T = 2s, hence:
#
# s(s - a)(s - b)(s - c) = 4s^2
# (s - a)(s - b)(s - c) = 4s
#
# substituting s = 3b/2 and eliminating c:
#
# (3b/2 - a)(3b/2 - b)(3b/2 - c) = 6b
# (3b - 2a)(2a - b) = 48
#
# both factors are positive integers, say f and g, f < g
# such that f.g = 48
#
# 2a - b = f
# 3b - 2a = g
#
# hence:
#
# b = (f + g)/2, a = (f + b)/2

from enigma import divisors, printf

n = 48
for f in divisors(n):
  g = n // f
  if g < f: break
  (b, r) = divmod(f + g, 2)
  if r: continue
  (a, r) = divmod(f + b, 2)
  if r: continue
  c = 2 * b - a
  printf("a={a} b={b} c={c}")

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# suppose the triangle has sides a, b, c

# where a < b, and c = b + (b - a) = 2b - a

#

# the semi-perimeter of the triangle s = (a + b + c)/2 = 3b/2

#

# and by Heron's formula the area of the triangle T is:

#

# T^2 = s(s - a)(s - b)(s - c)

#

# but T = 2s, hence:

#

# s(s - a)(s - b)(s - c) = 4s^2

# (s - a)(s - b)(s - c) = 4s

#

# substituting s = 3b/2 and eliminating c:

#

# (3b/2 - a)(3b/2 - b)(3b/2 - c) = 6b

# (3b - 2a)(2a - b) = 48

#

# both factors are positive integers, say f and g, f < g

# such that f.g = 48

#

# 2a - b = f

# 3b - 2a = g

#

# hence:

#

# b = (f + g)/2, a = (f + b)/2

from enigma import divisors, printf

n = 48

for f in divisors(n):

g = n // f

if g < f: break

(b, r) = divmod(f + g, 2)

if r: continue

(a, r) = divmod(f + b, 2)

if r: continue

c = 2 * b - a

printf("a={a} b={b} c={c}")

Reply

Jim Randell permalink

If you let Python do all the work it only finds the answer on the 11th (a, b, c) triple it tries.


# suppose the triangle has sides a, b, c
# where a < b, and c = b + (b - a) = 2b - a
#
# by Heron's formula the area of the triangle T is:
#
# (4T)^2 = (a + b + c)(b + c - a)(a + c - b)(a + b - c)
#
# and we need to find triangles such that T = a + b + c

from itertools import count
from enigma import irange, printf

def solve():
  for b in count(2):
    for a in irange(1, b - 1):
      c = 2 * b - a
      if not(c < a + b): continue
      # compute (4T)^2 both ways
      x1 = (a + b + c) * (b + c - a) * (a + c - b) * (a + b - c)
      x2 = (4 * (a + b + c)) ** 2
      if x1 == x2:
        printf("a={a} b={b} c={c}")
        return

solve()

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# suppose the triangle has sides a, b, c

# where a < b, and c = b + (b - a) = 2b - a

#

# by Heron's formula the area of the triangle T is:

#

# (4T)^2 = (a + b + c)(b + c - a)(a + c - b)(a + b - c)

#

# and we need to find triangles such that T = a + b + c

from itertools import count

from enigma import irange, printf

def solve():

for b in count(2):

for a in irange(1, b - 1):

c = 2 * b - a

if not(c < a + b): continue

# compute (4T)^2 both ways

x1 = (a + b + c) * (b + c - a) * (a + c - b) * (a + b - c)

x2 = (4 * (a + b + c)) ** 2

if x1 == x2:

printf("a={a} b={b} c={c}")

return

solve()

Reply

brian gladman permalink

Hi Jim,

I was too lazy to get a factor subroutine out although I would have done so had it been more of a challenge.

I also went on to solve for the triangle \((a, b, c=2b-a)\) with an area that is an integer multiple (\(k\)) of the semi-perimeter (\(s\)), which leads to the quadratic \[(3b – 2a)(2a – b) = 48k^2\] and the solution \[b = (2/3)\{2a\pm\sqrt{a^2-(3k)^2}\}\]

This equation also shows that the Pythagorean triple \((3k, 4k, 5k)\) provides a solution for any \(k\) value, although not all solutions are necessarily Pythagorean.

For example, the triangle \((13,14,15)\) provides another solution when \(k=4\).

Reply

ahmet cetinbudaklar permalink

If \(a\), \(b\) and \(c\) represent the sides of the plot with \(a=(b+c)/2\) and \(u=(a+b+c)/2\), then Heron’s formula for the area can be used to show that: \[a+b+c=\sqrt{u(u-a)(u-b)(u-c)}\] This soon provides a solution by trial and error.

Reply

Michael Broadey permalink

I generated Pythagorean Triples and found a result.

Reply

Sunday Times Teaser 2663 – Missed the Plot

by Alan Bergson

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