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  • Unknown's avatar

    Jim Randell 8:07 am on 8 January 2025 Permalink | Reply
    Tags: by: Rex Lanham   

    Teaser 2595: Stylish fences 

    From The Sunday Times, 17th June 2012 [link] [link]

    Farmer Giles has a hexagonal field bounded by six straight fences. The six corners lie on a circle of diameter somewhere around 250 metres. At three alternate corners there are stiles. The angles of the hexagon at the corners without stiles are all the same. All six fences are different whole numbers of metres long. I have just walked in a straight line from one of the stiles to another and the distance walked was a whole number of metres.

    How many?

    [teaser2595]

     
    • Jim Randell's avatar

      Jim Randell 8:07 am on 8 January 2025 Permalink | Reply

      Suppose the field is:

      If the stiles are at A, C, E, then the angles at B, D, F are all equal, which means they are all 120°, and so the lines AC, AE, CE (= paths between stiles) are all equal length and so ACE is an equilateral triangle. (See: [ @wikipedia ]).

      If the diameter of the circle is d, then the length of the paths, p, is given by:

      p = d √(3/4)

      And d is around 250, let’s say in the range [245, 255], so the path is in the range [212.176, 220.836], so we can consider path lengths of 212 .. 221.

      For a given path length p (= AC) we can look for integer fence lengths (x, y) (= (AB, BC)) that complete the triangle ABC. By the cosine rule in ABC:

      p^2 = x^2 + y^2 + xy
      ⇒
      y^2 + xy + (x^2 − p^2) = 0

      which is a quadratic equation in y.

      So we can consider values for x and use the equation to find possible corresponding y values.

      And we need to find (at least) 3 different (x, y) pairs to complete the hexagon, so that all 6 fences are different lengths.

      The following Python program runs in 76ms. (Internal runtime is 7.5ms).

      from enigma import (irange, quadratic, sq, sqrt, printf)

r43 = sqrt(4, 3)

# consider possible path lengths
for p in irange(212, 221):
  # collect possible (x, y) pairs
  ss = list()
  for x in irange(1, p - 1):
    for y in quadratic(1, x, sq(x) - sq(p), domain='Z'):
      if x < y < p:
        ss.append((x, y))

  # we need (at least) 3 different pairs
  if len(ss) < 3: continue
  # output solution
  printf("p={p} d={d:.3f} -> {ss}", d=p * r43)

      Solution: The path is 217m long.

      There are actually 4 possible (x, y) pairs, and any 3 of them can be used to construct the hexagon:

      p=217 d=250.570 → [(17, 208), (77, 168), (87, 160), (93, 155)]

      If the puzzle had specified that the diameter of the circle was “around 400 m”, then we would have been able to give the lengths of the 6 fences:

      p=343 d=396.062 → [(37, 323), (112, 273), (147, 245)]

      Like

  • Unknown's avatar

    Jim Randell 10:12 am on 12 September 2023 Permalink | Reply
    Tags: by: Rex Lanham   

    Teaser 2651: Parsnip Farm 

    From The Sunday Times, 14th July 2013 [link] [link]

    On Parsnip Farm there was a triangular field with one of its boundaries running north to south. From that boundary’s southernmost point Farmer Giles built a fence in an easterly direction, thus partitioning the field into two smaller triangular fields, with the area of the northern field being forty per cent of the area of the southern field. Each side of each field was a whole number of furlongs in length, one of the sides of the southern field being twenty-five furlongs in length.

    What are the lengths of the other two sides of the southern field?

    [teaser2651]

     
    • Jim Randell's avatar

      Jim Randell 10:12 am on 12 September 2023 Permalink | Reply

      If the area of the original field is A, then it is split such that the northern field has an area N = (2/7)A and the southern field S = (5/7)A.

      The boundaries of N are all integers and so form a Pythagorean triple (x, y, z).

      And we can expand the northern field by a factor of f until its hypotenuse exactly covers the boundary of the original field, and we get this:

      The we can then write the area of the original field in two ways:

      A = (7/2)N = 7xy / 4
      A = fxy / 2
      ⇒
      f = 7/2

      And we can calculate the unknown sides of S:

      z′ = (f − 1)z = 5z/2
      y′ = hypot(7x/2, 5y/2) = hypot(7x, 5y) / 2

      This Python program runs in 58ms. (Internal runtime is 334µs).

      Run: [ @replit ]

      from enigma import (pythagorean_triples, div, ihypot, ordered, printf)

for (a, b, z) in pythagorean_triples(100):
  z_ = div(5 * z, 2)
  if z_ is None: continue
  N = ordered(a, b, z)
  for (x, y) in [(a, b), (b, a)]:
    y_ = div(ihypot(7 * x, 5 * y), 2)
    if y_ is None: continue
    S = ordered(x, z_, y_)
    if 25 in S:
      # output solution
      printf("N = {N}; S = {S}")

      Solution: The other two sides of the southern field are 6 and 29 furlongs.

      N is a (6, 8, 10) triangle, (area = 24 square furlongs ≈ 0.97 sq km).

      S is a (6, 25, 29) triangle, (area = 60 square furlongs ≈ 2.43 sq km).

      So the original field was a (8, 29, 35) triangle, (area = 84 square furlongs ≈ 3.40 sq km).

      Like

    • Frits's avatar

      Frits 5:44 pm on 12 September 2023 Permalink | Reply

      Avoiding imports.

       
is_square = lambda n: int(rt := n**.5) == rt

# generate Pythagoren triples for a fixed hypothenuse
def triples(h):
  sqs = {i * i for i in range(1, h)}
  return [(int(s1**.5), int(s2**.5)) for s1 in sqs 
           if (s2 := h**2 - s1) >= s1 and s2 in sqs]
  
side = 25

# can x be 25?  
# then f.x = 87.5, y' can never be a whole number  (f = 3.5)
# if other side (2.5y) is k.0 or k.5, so x != 25

# can y' be 25?
for c, d in triples(side): 
   # are these lenghts multiples of 2.5 and 3.5?
  x, y = -1, -1
  if (c % 2.5 == 0 and d % 3.5 == 0):
    y, x = c // 2.5, d // 3.5    
  if ((c % 3.5 == 0 and d % 2.5 == 0)):
    x, y = c // 3.5, d // 2.5    
  if x == -1: continue
  
  # y' can be 25
  if is_square(z2 := x**2 + y**2):
    z = int(z2**.5)
    if z % 2 == 0:
      print(f"answer {int(x)} and {int(2.5 * z)} furlongs")
  
# can z' be 25?
if side % 5: exit(0)
z = int(side * 2 / 5)

ts = triples(z)
# determine z'
for x, y in ts + [x[::-1] for x in ts]:
  if is_square(y1 := 3.5**2 * x**2 + 2.5**2 * y**2):
    print(f"answer {int(x)} and {int(y1**.5)} furlongs")

      Like

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