Sunday Times Teaser 2560 – Three Trees

by H Bradley and C Higgins

Published: 16 October 2011 (link)

Our front garden is circular, with a diameter less than 100 metres. Three trees grow on the perimeter: an ash, a beech and a cherry, each a different whole number of metres from each of the other two. A straight trellis 84 metres long joins the ash to the beech, and a honeysuckle grows at the point on the trellis nearest to the cherry tree. The distance from the cherry to the ash plus the distance from the ash to the honeysuckle equals the distance from the cherry to the beech.

What is the diameter of the garden?

2 Comments Leave one →

Brian Gladman permalink


from itertools import count

# Let the angle at triangle vertex A be alpha.  The
# diameter of the circumcircle for the triangle ABC
# is given by (where A is the trianngle's area):
#
# (1)    d = a.b.c / (2.A)
# 
# The area A = (1/2).b.c.sin(alpha) now gives:
#
# (2)    d = a / sin(alpha)
#
# THe given relationship BC = AC + AH is:
#
# (3)    a = b + b.cos(alpha)
#
# Using (2) and (3) to eliminate alpha gives:
#
# (4)    d^2 = a.b^2 / (2.b - a)
#
# Using the cosine rule for alpha:
#
# (5)   a^2 = b^2 + c^2 - 2.b.c.cos(alpha)
#
# with (3) gives:
#
# (6)   (a + c)^2 = (b + c)^2 + c^2

# the given value of c
c = 84
# try increasing b values until (6) gives a perfect square
# for a + c
for b in count(1):
  apc2 = (b + c) ** 2 + c ** 2
  apc = int(apc2 ** 0.5)
  a = apc - c
  # from (4) 2b - a is greater than zero
  if apc ** 2 == apc2 and 2 * b - a > 0:
    # now compute the square of the circumradius from (6)
    d2, r = divmod(a * b ** 2, 2 * b - a)
    d = int(d2 ** 0.5)
    if not r and d * d == d2:
      fs = 'Diameter = {} (a = {}, b = {}, c = {}).'
      print(fs.format(d, a, b, c))
      break

from itertools import count

# Let the angle at triangle vertex A be alpha. The

# diameter of the circumcircle for the triangle ABC

# is given by (where A is the trianngle's area):

# (1) d = a.b.c / (2.A)

# The area A = (1/2).b.c.sin(alpha) now gives:

# (2) d = a / sin(alpha)

# THe given relationship BC = AC + AH is:

# (3) a = b + b.cos(alpha)

# Using (2) and (3) to eliminate alpha gives:

# (4) d^2 = a.b^2 / (2.b - a)

# Using the cosine rule for alpha:

# (5) a^2 = b^2 + c^2 - 2.b.c.cos(alpha)

# with (3) gives:

# (6) (a + c)^2 = (b + c)^2 + c^2

# the given value of c

c = 84

# try increasing b values until (6) gives a perfect square

# for a + c

for b in count(1):

apc2 = (b + c) ** 2 + c ** 2

apc = int(apc2 ** 0.5)

a = apc - c

# from (4) 2b - a is greater than zero

if apc ** 2 == apc2 and 2 * b - a > 0:

# now compute the square of the circumradius from (6)

d2, r = divmod(a * b ** 2, 2 * b - a)

d = int(d2 ** 0.5)

if not r and d * d == d2:

fs = 'Diameter = {} (a = {}, b = {}, c = {}).'

print(fs.format(d, a, b, c))

break

BRG permalink

Looking at my original version, I thought it would be interesting to see what other solutions
are possible. It turns out that the number of solutions is finite.


# following on from the earlier analysis:
# 
# (7) (a - b).(a + b + 2.c) = c^2
# 
# (8) (a - b).(a + b + 2.c) =  c^2 = p.q with p < q
# 
# (9)  a - b = p        )  a = (p + q) / 2 - c
#      a + b + 2.c = q  )  b = a - p

from enigma import divisor_pairs
from fractions import Fraction as RF

c = 84
print('   a    b              d^2')
for p, q in divisor_pairs(c * c):
  t, r = divmod(p + q, 2)
  a = t - c
  b = a - p
  if not r and a > 0 and 2 * b > a:
    d2 = RF(a * b * b, b + b - a)
    u, d = d2.numerator, d2.denominator
    if (nd := round((u * d) ** 0.5)) ** 2 == u * d:
      d2 = f"({RF(nd, d)})^2" 
    print(f"{a:4} {b:4} {d2:>16}")

# following on from the earlier analysis:

# (7) (a - b).(a + b + 2.c) = c^2

# (8) (a - b).(a + b + 2.c) = c^2 = p.q with p < q

# (9) a - b = p ) a = (p + q) / 2 - c

# a + b + 2.c = q ) b = a - p

from enigma import divisor_pairs

from fractions import Fraction as RF

c = 84

print(' a b d^2')

for p, q in divisor_pairs(c * c):

t, r = divmod(p + q, 2)

a = t - c

b = a - p

if not r and a > 0 and 2 * b > a:

d2 = RF(a * b * b, b + b - a)

u, d = d2.numerator, d2.denominator

if (nd := round((u * d) ** 0.5)) ** 2 == u * d:

d2 = f"({RF(nd, d)})^2"

print(f"{a:4} {b:4} {d2:>16}")

Sunday Times Teaser 2560 – Three Trees

by H Bradley and C Higgins

Published: 16 October 2011 (link)

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