S2T2

Jim Randell 7:32 am on 13 May 2025 Permalink Reply
Tags: by: C Higgins ( 35 ), by: H Bradley ( 35 )   

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  Jim Randell 7:32 am on 13 May 2025 Permalink | Reply

  We are to construct the tangents to two non-touching circles. See: [ @wikipedia ]

  I am assuming that, as we are asked to find the radius of the smaller circle, then the two circles must be of different sizes. (And without this condition the Teaser puzzle itself is flawed).

  If the circles have radii r and R (where r < R), and the separation between their centres is d, then the tangents have length:

  outer tangent, length T: d^2 = T^2 + (R − r)^2
  inner tangent, length t: d^2 = t^2 + (R + r)^2

  This Python program runs in 59ms. (Internal runtime is 2.3ms).

  from enigma import (primes, irange, ircs, printf)
  
  # r = radius of smaller circle (a prime)
  for r in primes.between(2, 24):
    # R = radius of the larger circle (a prime)
    for R in primes.between(r + 1, 48 - r):
      # d = separation of centres (< 50)
      for d in irange(r + R + 1, 49):
        # calculate: T = length of outer tangent, t = length of inner tangent
        (T, t) = (ircs(d, -(R - r)), ircs(d, -(R + r)))
        if T is None or t is None or t == T: continue
        # output viable configurations
        printf("r={r} R={R} d={d} -> T={T} t={t}")
  

  Solution: The smaller circle has a radius of 7 cm.

  There are two possible scenarios that provide tangents with two different integer lengths:

  d=26, r=7, R=17 → T=24, t=10
  d=34, r=7, R=23 → T=30, t=16

  And so one of these must be the scenario presented in the magazine puzzle.

  The first of these solutions looks like this:

  

  ---

  If the circles were allowed to be the same size, then we can find the following additional viable configurations, that give two different integer tangent lengths:

  d=5, r=2, R=2 → T=5, t=3
  d=10, r=3, R=3 → T=10, t=8
  d=26, r=5, R=5 → T=26, t=24

  And the Teaser puzzle would not have a unique solution.

  These additional solutions can be seen by changing line 6 in the program to start the search for R from the value of r.

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    Frits 7:13 pm on 14 May 2025 Permalink | Reply

    In both scenarios we have (not a general rule):

    d^2 = T^2 + t^2 = 2.(r^2 + R^2)

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  Frits 11:02 am on 16 May 2025 Permalink | Reply

  from enigma import primes, pythagorean_triples, subsets
  from collections import defaultdict
  
  prms = set(primes.between(2, 48))
  
  # collect Pythagorean triples for each distance between the centres
  pt = defaultdict(list)
  for x, y, h in pythagorean_triples(49):
    pt[h] += [x]
    pt[h] += [y]
  
  for d, vs in pt.items():  
    for t, T in subsets(sorted(vs), 2):
      # make sure r and R are integers
      if t % 2 != T % 2: continue
      r, R = (T - t) // 2, (T + t) // 2
      if any(x not in prms for x in (r, R)): continue 
      print(f"{r=} {R=} {d=} -> {T=} {t=}")
  

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