S2T2

Jim Randell 9:36 am on 7 July 2020 Permalink Reply
Tags: by: Alan Bergson ( 4 )   

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  Jim Randell 9:37 am on 7 July 2020 Permalink | Reply

  The triangle has sides of integer lengths (a, b, c) such that b = (a + c)/2.

  We can use Heron’s Formula [ @wikipedia ] to equate the area of the triangle with the perimeter, which gives us some restrictions on the sides of the triangle.

  However, we have encountered “equable” triangles before in Enigma 364, and generated a list of the 5 equable triangles, from which we can easily pick out the one that has one side that is an average of the other two.

  I also wrote code to generate k-equable triangles (integer sided triangles where the area is k times the perimeter), and that code can be reused here:

  Run: [ @repl.it ]

  from enigma import irange, isqrt, divc, divf, printf
  
  # find all triangles with integer sides where area = k . perimeter
  def triangles(k=1):
    K = 4 * k * k
    for p in irange(1, isqrt(3 * K)):
      for q in irange(max(p, divc(K + 1, p)), divf(3 * K, p)):
        (r, z) = divmod(K * (p + q), p * q - K)
        if r < q: break
        if z == 0: yield (p + q, p + r, q + r)
  
  # consider 1-equable triangles
  for (a, b, c) in triangles(1):
    if b - a == c - b:
      printf("a={a} b={b} c={c}")
  

  Solution: The lengths of the sides of the plot are: 6m, 8m, 10m.

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  GeoffR 10:54 am on 8 July 2020 Permalink | Reply

  
  for a in range (3, 25):
    for b in range (a+1, 25):
      for c in range(b+1, 25):
        # a < b < c and 2 * b = a + c
        if 2 * b != a + c: continue
        # Square of perimeter
        perim_sq = (a + b + c) ** 2
        # Area squared formula is based in Heron's formula
        area_sq = 1/16 *( 4*a*a*b*b - (a*a + b*b - c*c)**2)
        if perim_sq == area_sq:
          print(f"Length of sides are {a}, {b} and {c}m")
  
  # Length of sides are 6, 8 and 10m
  

  A (5,12,13) triangle has the area equal to the perimeter numerically, but does not have one side as the average of the other two sides.

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