S2T2

Jim Randell 7:03 am on 21 September 2025 Permalink Reply
Tags: by: John Owen ( 32 )   

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  Jim Randell 7:18 am on 21 September 2025 Permalink | Reply

  See: Teaser 2549.

  If the diameter of the circle is x, and the lengths of the non-diameter legs are a, b, c, then we have:

  x³ − (a² + b² + c²)x − 2abc = 0

  The following Python program runs in 62ms. (Internal runtime is 643µs).

  from enigma import (powers, inf, first, lt, subsets, pi, cb, sumsq, printf)
  
  # squares less than 300
  sqs = list(first(powers(1, inf, 2), lt(300)))
  
  # choose 3 different lengths (x is the diameter)
  for (a, b, x) in subsets(sqs, size=3):
    # duplicate one of the short lengths
    for c in (a, b):
      # check for a viable quadrilateral
      if not (x < a + b + c < pi * x): continue
      if not (cb(x) - sumsq([a, b, c]) * x - 2 * a * b * c == 0): continue
      # output solution
      printf("{a}, {b}, {x} [and {c}]")
  

  Solution: The leg lengths are 36, 49, 81 km.

  The duplicate leg length is 36 km.

  

  The next smallest solution is at 4× these lengths, which brings the diameter to 324 km.

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    Jim Randell 8:49 am on 22 September 2025 Permalink | Reply

    We can suppose that the non-diameter sides are (a, b, a) (i.e. c = a, and it may be the case that a > b), and then the equation can be simplified to:

    2a² = x(x − b)

    So by choosing b and x we can then straightforwardly calculate the corresponding value for a, and check it is viable (i.e. a square):

    from enigma import (powers, inf, first, lt, subsets, is_square, div, ordered, printf)
    
    # squares less than 300
    sqs = list(first(powers(1, inf, 2), lt(300)))
    
    # choose 2 different lengths (x is the diameter)
    for (b, x) in subsets(sqs, size=2):
      # a is the duplicated side
      a = is_square(div(x * (x - b), 2))
      if a is None or a not in sqs: continue
      # output solution
      (p, q) = ordered(a, b)
      printf("{p}, {q}, {x} [and {a}]")
    

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      Frits 2:29 pm on 22 September 2025 Permalink | Reply

      I had come up with similar equation. You can also deduce the parity of “a”.

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  Ruud 10:35 am on 21 September 2025 Permalink | Reply

  import math
  import itertools
  
  for ab, bs, sd, da in itertools.product((i * i for i in range(1, 18)), repeat=4):
      try:
          angle_ab = math.asin(ab / da) * 2
          angle_bs = math.asin(bs / da) * 2
          angle_sd = math.asin(sd / da) * 2
  
          if angle_ab + angle_bs + angle_sd == math.pi:
              print(sorted({ab, bs, sd, da}))
      except ValueError:
          ...
  

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    Jim Randell 11:38 am on 21 September 2025 Permalink | Reply

    @Ruud: How does this code ensure that two of the leg lengths are the same?

    And it is probably better to use [[ math.isclose() ]] rather than testing floats for equality using ==.

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      Ruud 12:55 pm on 21 September 2025 Permalink | Reply

      I just checked manually by checking that there are 3 different length. Could have added a simple test, of course.

      Yes, math.isclose would have been better, but as the equality check gave already an answer, I thought I could just refrain from isclose.

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      Ruud 4:14 pm on 21 September 2025 Permalink | Reply

      @Jim
      I just checked manually that there are exactly three different lengths, but a test would be better.

      Yes, math.isclose would have beeen safer.

      Here’s my updated version:

      import math
      import itertools
      
      for ab, bs, sd, da in itertools.product((i * i for i in range(1, 18)), repeat=4):
          try:
              angle_ab = math.asin(ab / da) * 2
              angle_bs = math.asin(bs / da) * 2
              angle_sd = math.asin(sd / da) * 2
      
              if math.isclose(angle_ab + angle_bs + angle_sd, math.pi) and len(sol := {ab, bs, sd, da}) == 3:
                  print(sorted(sol))
          except ValueError:
              ...
      
      peek("done")
      

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  Alex.T.Sutherland 10:42 am on 28 September 2025 Permalink | Reply

  TEASER 3287.....FERRY ROUTE.
  Given:-		Cyclic Quadrilateral [A B C D] with lengths [a b c d], d = diameter; 
  		               All sides are of perfect square length.
  Method:-	Using Ptolemy's Theorem relating the diagonals to the sides
  		               of the quadrilateral and that right angles are created at B & C
  		              by the diagonals the following equation is derived:-
  
  		              (d^2 - a^2) * (d^2 - c^2) = (b*d + a*c)^2
  
  		              If 'c' is made equal to 'a' the following can be  derived:-
  
  Answer:-	2*a^2  = d*(d-b);
  
  		              Choose 3 squares from the given set to satisfy this equation.
  Side Line:-	The path could be 'abad' or'aabd' but the answer is the same for the lengths.
  		              (neglecting symmetry eg 'baa' is the same as 'aab')'
  		             For my answer the sum of the lengths 'aba' or 'aab' is a perfect square.
  

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