S2T2

Jim Randell 8:08 am on 8 September 2020 Permalink Reply
Tags: by: J. S. P. Roberton ( 4 )   

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  Jim Randell 8:09 am on 8 September 2020 Permalink | Reply

  If we suppose the north and south boundaries of the field are separated by a distance of a, and the stiles divide the these boundaries into sections with length x and y, and the distances from the tree to the stiles are b and c, and the distance from the tree to the corner is z. Then we have a layout like this:

  

  When the children were younger they would run distances x and y.

  And when they were older the distances are b and c.

  Hence:

  b = x + 9
  c = y − 39
  ⇒ y − x = c − b + 48

  The distance between the two stiles, (b + c) forms the hypotenuse of a Pythagorean triangle with other sides of (y − x) = (c − b + 48) and a (triangle: S1 Y S2).

  So we can consider Pythagorean triples for this triangle. One of the non-hypotenuse sides corresponds to a, and the other when added to the hypotenuse gives:

  (b + c) + (c − b + 48) = 2c + 48

  So choosing one of the sides for a, allows us to calculate the value for c, and then b, x, y, z.

  The distance G S2 is the hypotenuse of the two triangles (G T S2) and (G X S2), and we can check that these hypotenuses match.

  The following Python program runs in 58ms.

  Run: [ @replit ]

  from enigma import (pythagorean_triples, div, sqrt, printf)
  
  # consider pythagorean triples corresponding to (a, c - b + 48, b + c)
  
  for (X, Y, Z) in pythagorean_triples(1000):
    # choose a side for a
    for (a, other) in [(X, Y), (Y, X)]:
      # and: other + Z = 2c + 48
      c = div(other + Z - 48, 2)
      if c is None or c < 1: continue
      b = Z - c
      x = b - 9
      y = c + 39
      if not (0 < b < c and 0 < x < y): continue
      # calculate z^2
      z2 = y * y - c * c
      if not (b * b + z2 == a * a + x * x): continue
      # output solution
      printf("({X}, {Y}, {Z}) -> a={a} b={b} c={c}, x={x} y={y} z={z:g}", z=sqrt(z2))
  

  Solution: The stiles are 200 yards apart. The oak tree is 117 yards from the gate.

  The short side of the field is 120 yards.

  The long side is 230 yards, and this is split into sections of 35 and 195 yards by the stiles.

  The oak tree is 44 yards from one stile and 156 yards from the other.

  ---

  This puzzle appeared in the 1986 book Microteasers (p 4) by Victor Bryant, in which he deals with solving puzzles using the BBC Micro. (See: [ @EnigmaticCode ]).

  Here is Victor Bryant’s BBC BASIC program (slightly modified by me to stop after the first solution is encountered, and print the run time for the program):

  >LIST
     10 REM Victor Bryant program for Teaser 1108
     15 start% = TIME
     20 FOR c = 26 TO 500  
     30   FOR b = 25 TO c - 1
     40     a = SQR((c + 39)^2 - c^2 + b^2 - (b - 9)^2)
     50     IF a <> INT(a) THEN 70
     60     IF a^2 <> (b + c)^2 - (c - b + 48)^2 THEN 70
     65     PRINT "Distances: stiles - tree = "; b; " and "; c; " & short side = "; a
     66     GOTO 85
     70   NEXT b
     80 NEXT c
     85 PRINT "(Time = "; (TIME - start%) / 100; "s)"
  
  >RUN
  Distances: stiles - tree = 44 and 156 & short side = 120
  (Time = 250.2s)
  
  >
  

  Here is a Python program using the same approach. It takes 65ms to find the first solution, or 137ms to run to completion. (On an emulated BBC Micro Victor’s program takes about an hour to run to completion).

  from enigma import (irange, subsets, is_square, printf)
  
  for (b, c) in subsets(irange(25, 500), size=2):
    a = is_square((c + 39)**2 - c**2 + b**2 - (b - 9)**2)
    if a is None: continue
    if a**2 + (c - b + 48)**2 != (b + c)**2 : continue
    printf("a={a} b={b} c={c}")
    break
  

  Comparing internal runtimes we find that my program runs in 2.4ms (to completion), and the port of Victor’s program in 17.3ms (to first solution; 219ms to completion). They search a similar solution space, (i.e. distance S1 S2 less than 1000 yards).

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  • 

    Jim Randell 3:08 pm on 14 September 2020 Permalink | Reply

    Here’s my analytical solution:

    We have:

    x = b − 9
    y = c + 39

    and

    y² = c² + z²
    (c + 39)² = c² + z²
    2.39.c + 39.39 = z²
    c = (z² − 39.39) / 2.39
    c = (z²/39 − 39)/2

    Now, c is an integer, so z²/39 is an odd integer:

    z²/39 = 2k + 1
    z² = 39(2k + 1)

    and:

    c = (2k + 1 − 39)/2
    c = k − 19

    and: c > b > 9, so k > 29.

    Also, we have:

    a² = (b + c)² − (c − b + 48)²
    a² = (b² + 2bc + c²) − (b² + c² − 2bc − 96b + 96c + 2304)
    a² = 4bc − 96b + 96c + 2304
    a² = 4(b − 24)(c + 24)
    a² = 4(b − 24)(k + 5)

    So: b > 24 ⇒ k > 44.

    And:

    b² + z² = a² + (b − 9)²
    b² + 39(2k + 1) = 4(b − 24)(k + 5) + (b² − 18b + 81)
    78k + 39 = 4bk + 20b − 96k − 480 − 18b + 81
    2b(2k + 1) = 174k + 438
    b = (87k + 219) / (2k + 1)

    As k → ∞, b → 87/2 = 43.5, so: b ≥ 44 and k ≤ 175.

    This gives us a limited range of values for k, which we can check:

    from enigma import (irange, div, is_square, sqrt, printf)
    
    # consider possible values for k
    for k in irange(45, 175):
      # calculate a, b, c, x, y, z
      c = k - 19
      b = div(87 * k + 219, 2 * k + 1)
      if b is None or not (0 < b < c): continue
      a = is_square(4 * (b - 24) * (c + 24))
      if a is None: continue
      x = b - 9
      y = c + 39
      z = sqrt(78 * k + 39)
      if not (0 < x < y): continue
      # output solution
      printf("[k={k}] a={a} b={b} c={c}; x={x} y={y} z={z:g}")
    

    Or we can narrow down the values with further analysis:

    Now:

    (b − 24) = 39(k + 5) / (2k + 1)

    and so:

    a² = [2(k + 5)]² (39 / (2k + 1))

    The value of (2k + 1) is in the range 91 to 351, so the (39 / (2k + 1)) part, must contribute pairs of factors (which divide into the pairs of factors provided by the [2(k + 5)]² part).

    So (2k + 1) is some odd square multiple of 39 (which also means that z must be an integer).

    The only option in the required range is:

    (2k + 1) = 39×3² = 351
    k = 175
    a = 120, b = 44, c = 156
    x = 35, y = 195, z = 117

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  John Crabtree 6:11 am on 9 September 2020 Permalink | Reply

  (b + c)^2 = a^2 +(y – x)^2
  a^2 + x^2 = b^2 +z^2
  y^2 = c^2 + z^2
  Adding the three equations and simplifying gives bc + xy = z^2
  But z^2 = 39(2y – 39) = (39k)^2
  And so y = 39(k^2 + 1)/2
  Using b = x + 9 and c = y – 39, leads to x = 69/2 + 9/(2.k^2)
  For integral solutions, either k = 1 and then x = y = z = 39 which is invalid,
  or k = 3, x = 35, y = 195 and z = 117, etc with a = 120 as a check

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  • 

    John Crabtree 4:08 am on 10 September 2020 Permalink | Reply

    I should have included: a = 39k + 9/k = z + 9/k

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      Frits 10:37 am on 10 September 2020 Permalink | Reply

      Nice solution, I checked your equations.

      I think you have to proof k (and z?) is an integer
      before only only considering k=1 and k=3.

      fi, k = 3^0.5 would also lead to an integral solution for x.

      We only know for sure that a, b, c and x,y are natural numbers.

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        John Crabtree 4:02 pm on 11 September 2020 Permalink | Reply

        Frits, thank you. I had assumed that z was a whole number, which we are not told. We also need to consider a, =(2y -30)/k before determining possible values of k. y is a whole number, and so k must be rational, = n/m. But then for y to be a whole number, m= 1, and so k is an integer.
        I think that this makes the manual solution, which should exist, complete.

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    • 

      Frits 11:31 pm on 12 September 2020 Permalink | Reply

      Hi John,

      Unfortunately some of your text was lost during formatting. I can follow the logic if (2y -30)/k equals an integer expression. What formula or variable did you intend to use as left hand side of ” (2y -30)/k”?

      I now used opening and closing pre tags for this section.

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