S2T2

Jim Randell 4:45 pm on 13 August 2021 Permalink Reply
Tags: by: Nick MacKinnon ( 51 )   

• 

  Jim Randell 5:07 pm on 13 August 2021 Permalink | Reply

  (See also: Teaser 1917).

  With a suitable diagram this puzzle reduces to solving a quadratic equation.

  By reflecting the table along its sides we can convert the paths of the ball into straight lines, whose length can be easily calculated. (For the purposes of this puzzle the ball and the pockets can be considered to be points of negligible size).

  

  We see that the 1st distance A is given by:

  A² = (12)² + (12 + x)²

  And the second distance B by:

  B² = (24)² + (6 + x)²

  And they are related by:

  B = (13/10)A
  10B = 13A
  100B² = 169A²

  This gives us a quadratic equation in x, which we can solve to find the required answer (= B).

  Run: [ @replit ]

  from enigma import (quadratic, sq, hypot, printf)
  
  # dimensions of the table
  (X, Y) = (6, 12)
  
  # B = (P/Q) A
  # (P^2)(A^2) = (Q^2)(B^2)
  (P2, Q2) = (sq(13), sq(10))
  
  # calculate the coefficients of the quadratic: a.x^2 + b.x + c = 0
  a = P2 - Q2
  b = P2 * 2 * Y - Q2 * 2 * X
  c = P2 * 2 * sq(Y) - Q2 * (sq(2 * Y) + sq(X))
  printf("[({a:+d})x^2 ({b:+d})x ({c:+d})]")
  
  # and solve the equation to determine x
  for x in quadratic(a, b, c, domain='F'):
    if 0 < x < X:
      # calculate the paths
      A = hypot(12, 12 + x)
      B = hypot(24, 6 + x)
      printf("A={A:g} B={B:g} [x={x:g}]")
  

  Solution: On the second shot the ball travelled 26 feet.

  The required equation is:

  69x² + 2856x − 12528 = 0

  This factorises as:

  3(x − 4)(23x + 1044) = 0

  From which the value of x is obvious, and the value of B can then be calculated.

  We find:

  x = 4
  A = 20
  B = 26

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  • 

    Frits 6:25 pm on 13 August 2021 Permalink | Reply

    @Jim, I verified your quadratic solution. It directly follows from the diagram.

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  • 

    Jim Randell 11:51 am on 16 August 2021 Permalink | Reply

    Or finding the answer numerically:

    Run: [ @replit ]

    from enigma import (hypot, fdiv, find_value, printf)
    
    # calculate paths A and B
    A = lambda x: hypot(12, 12 + x)
    B = lambda x: hypot(24, 6 + x)
    
    # find an x value that gives B/A = 1.3
    x = find_value((lambda x: fdiv(B(x), A(x))), 1.3, 0, 6).v
    printf("A={A:g} B={B:g} [x={x:g}]", A=A(x), B=B(x))
    

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