S2T2

Jim Randell 8:23 am on 19 February 2019 Permalink Reply
Tags: by: W G E Jackson ( 2 )   

• 

  Jim Randell 8:40 am on 19 February 2019 Permalink | Reply

  Let’s suppose that the field is a unit square, and that Ali’s age is a (a whole number).

  Then the total of all the ages is d = a + 2(a + 11) + 74 = 3a + 96, so the corresponding areas of the triangular plots are:

  Hassan = 74 / d
  Rashid, Riad = (a + 11) / d
  Ali = a / d

  Splitting two sides of the square at x, y ∈ (0, 1) we have:

  For Rashid and Riad:

  (a + 11) / d = x / 2
  (a + 11) / d = (1 − x)(1 − y) / 2

  and for Ali:

  a / d = y / 2

  So, given a value for a we have:

  d = 3a + 96
  x = 2(a + 11) / d
  y = 2a / d
  (1 − x)(1 − y) = x

  So we can consider whole number values for a that give a solution to these equations:

  from enigma import (Rational, irange, printf)
  
  Q = Rational()
  
  # choose a value for a
  for a in irange(1, 50):
    # total age
    d = 3 * a + 96
    # calculate x and y
    (x, y) = (Q(2 * a + 22, d), Q(2 * a, d))
    # do they satisfy the third equation?
    if x == (1 - x) * (1 - y):
      printf("a={a} [x={x}, y={y}]")
  

  Solution: Ali is 24.

  A viable solution to the equations is:

  a = 24, x = 5/12, y = 2/7

  so the fields look like this:

  

  Hassan has 37/84 (≈ 44.1%) of the field. The twins have 5/24 (≈ 20.8%) each. And Ali has the remaining 1/7 (≈ 14.3%).

  There is a further solution to the equations at:

  a = −208/5, x = 17/8, y = 26/9

  but this doesn’t give a viable answer.

  LikeLike

• 

  John Crabtree 4:24 pm on 16 April 2021 Permalink | Reply

  The four equations involving a, d, x and y can be reduced to:
  5a^2 + 88a -4992 = 0. Then a = -8.8 +/- 32.8

  LikeLike