S2T2

Jim Randell 10:20 am on 28 July 2026 Permalink Reply
Tags: by: Andrew Skidmore ( 91 )   

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  Jim Randell 10:20 am on 28 July 2026 Permalink | Reply

  Starting with a triangle with sides a, b, c. Then the circular pond is concentric with the incircle of the triangle, so we are interested in the distances from the incentre (= I) to the vertices of the triangle (= u, v, w).

  

  The semi-perimeter (= s) of the triangle is given by:

  s = (a + b + c) / 2

  And the lengths of the sides of the triangle can be written:

  a = y + z
  b = x + z
  c = x + y
  ⇒
  s = x + y + z
  x = s − a
  y = s − b
  z = s − c

  And so the distances we are interested in are given by:

  u = hypot(r, s − a)
  v = hypot(r, s − b)
  w = hypot(r, s − c)

  We can calculate area (= A) of the triangle using Heron’s formula:

  A = √(s (s − a) (s − b) (s − c))

  And the inradius (= r) can then be calculated:

  r = A / s

  The following Python program considers possible sides of the triangle (a, b, c), and looks for scenarios where 2 of the distances from the incentre to the vertices are whole numbers.

  It runs in 76ms. (Internal runtime is 665µs).

  from enigma import (Rational, decompose, sq, is_square, sqrt, seq2str, printf)
  
  Q = Rational()
  
  # semi-perimeter
  s = 27
  # consider possible sides of the triangle (a, b, c); a + b + c = 2s
  for (a, b, c) in decompose(2*s, 3, increasing=1, sep=0):
    # if the area of the triangle is A, use Heron's formula to calculate A^2
    A2 = s * (s - a) * (s - b) * (s - c)
    # check for valid triangles
    if not (A2 > 0): continue
    # inradius = r, r = A/s, calculate r^2
    r2 = Q(A2, sq(s))
  
    # calculate the squares of the distances from the incentre to the vertices
    d2s = tuple(r2 + sq(s - v) for v in (a, b, c))
  
    # and look for situations where exactly 2 of these are perfect squares
    k = sum(1 for d2 in d2s if is_square(d2))
    if k != 2: continue
  
    # output solution
    ds = tuple(sqrt(d2) for d2 in d2s)
    printf("a={a} b={b} c={c} -> A2={A2} r2={r2} -> d2s={d2s} ds={ds}", d2s=seq2str(d2s), ds=seq2str(ds))
  

  Solution: The shortest side of the lawn is 6 m.

  And the other two sides are 24 m each. (So the triangle is isosceles).

  Giving distances from the incentre to the vertices of 4 m, 4 m, 21.166 m (= 8√7).

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  • 

    Jim Randell 11:48 am on 2 August 2026 Permalink | Reply

    If at least one of the (r² + (s − v)²) is a square number (to give a whole number root), then it follows that r² must be an integer, and so each of u², v², w² is also a whole numbers.

    And we can simplify the calculations of u², v², w² to be just in terms of a, b, c:

    u² = (s − a)bc / s
    v² = (s − b)ca / s
    w² = (s − c)ab / s

    The program can then be simplified slightly to:

    from enigma import (decompose, div, tuples, is_square, sqrt, seq2str, printf)
    
    # semi-perimeter
    s = 27
    # consider possible sides of the triangle (a, b, c); a + b + c = 2s
    for (a, b, c) in decompose(2*s, 3, increasing=1, sep=0):
      # check for valid triangles
      if not (a + b > c): continue
    
      # calculate the squares of the distances from the incentre to the vertices
      d2s = tuple(div((s - i) * j * k, s) for (i, j, k) in tuples((a, b, c), 3, circular=1))
      if None in d2s: continue
    
      # look for situations where exactly 2 of these are perfect squares
      k = sum(1 for d2 in d2s if is_square(d2))
      if k != 2: continue
    
      # output solution
      ds = tuple(sqrt(d2) for d2 in d2s)
      printf("a={a} b={b} c={c} -> d2s={d2s} ds={ds}", d2s=seq2str(d2s), ds=seq2str(ds))
    

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  Frits 1:54 pm on 28 July 2026 Permalink | Reply

  Doing most of the checks in decompose().

  from math import prod
  
  # decompose number <t> into <k> non-decreasing numbers and semi-perimeter logic
  def decompose(t, k, m=0, sp=0, s=[]):
    if k == 1:
      if not s or t >= s[-1]:
        s.append(t)
        # r^2 = (sp - a) * (sp - b) * (sp - c) / sp must be a positive integer
        p = prod(sp - x for x in s)
        if p > 0 and p % sp == 0:
          yield s, p // sp
    else:
      for n in range(m, (t // k) + 1):
        yield from decompose(t - n, k - 1, n, sp, s + [n])
  
  # semi-perimeter
  s = 27
  # consider possible sides of the triangle (a, b, c); a + b + c = 2s
  for (a, b, c), r2 in decompose(2 * s, 3, 1, s):
    # calculate the squares of the distances from the incentre to the vertices
    d2s = [r2 + (s - v)**2 for v in (a, b, c)]
  
    # and look for situations where exactly 2 of these are perfect squares
    if sum(round(rt := d2**.5) == rt for d2 in d2s) != 2: continue
  
    print("answer:", min(a, b, c))
  

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