Sunday Times Teaser 3191 – The Budgie’s Extra Ration

by BRG on November 17, 2023

by Paul Hughes

Published Friday November 17 2023 (link)

The budgie’s circular toy hung on a hook. Two equal legs suspended his budgerigar-seed dispensing chord from that hook. Parallel to the chord, a diameter crossed the middle. When budgie knocked his seed dispenser below the diameter, a triangle lit up, as shown, and he got an extra seed ration. In the design of the toy, the ratio of the length of the chord to the length of the smallest side of the lit triangle has been adjusted so that a right-angled triangle results, with the right angle on the diameter.

What is the square of that ratio?

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# Newton-Raphson root finder
def newton(f, f0, h):
  
  def der(f, x, dx):
    return (f(x + dx) - f(x - dx)) / (2 * dx)

cur, prv = f0, float("inf")
  while abs(cur - prv) >= h:
    prv, cur = cur, cur - f(cur) / der(f, cur, h)
  return cur

# return the canonical line constants <m> and <c>  
# in y = <m>.x + <c> given two points on the line
def line(p1, p2):
  (x1, y1), (x2, y2) = p1, p2
  if (d := x2 - x1):
    return (y2 - y1) / d, (x2 * y1 - x1 * y2) / d
  else:
    return float('inf'), float('inf')

#                     A (0,0)
#                    /|\
#                   / | \
#                  /  |  \
#               s /   |   \
#                /    |    \
#               /     | h   \
#    (d,1) D --/------|------\
#             /       |       \
#            /        |        \
#           /         |         \
# (-c,h) B /__________|__________\ C (c,h)
#                c         c
#
#  d = c / h;  c^2 = h.(2 - h) -> h = 1 + (1 - c^2)^(1/2)

# the lines AB and DC are perpendicular when 1 + m(AB).m(DC) is zero;  
# return this value and the ratio (2.c / s)^2 for the semi-chord <c>
def fn(c):
  h = 1 + (1 - c * c) ** (1 / 2)
  # line AB 
  mab = line((0, 0), (-c, h))[0]
  # line DC
  mdc = line((-c / h, 1), (c, h))[0]
  
  if mab and mdc:
    rsq = ((4 * c * c) / (1 + (c / h) ** 2))
    return 1 + mab * mdc, rsq

# find the <c> value for a right angle
c = newton(lambda x: fn(x)[0], 0.8, 1e-12)
print(f"sq(ratio) = {fn(c)[1]:.12g}")

# Newton-Raphson root finder

def newton(f, f0, h):

def der(f, x, dx):

return (f(x + dx) - f(x - dx)) / (2 * dx)

cur, prv = f0, float("inf")

while abs(cur - prv) >= h:

prv, cur = cur, cur - f(cur) / der(f, cur, h)

return cur

# return the canonical line constants <m> and <c>

# in y = <m>.x + <c> given two points on the line

def line(p1, p2):

(x1, y1), (x2, y2) = p1, p2

if (d := x2 - x1):

return (y2 - y1) / d, (x2 * y1 - x1 * y2) / d

else:

return float('inf'), float('inf')

# A (0,0)

# /|\

# / | \

# s / | \

# / | \

# / | h \

# (d,1) D --/------|------\

# / | \

# (-c,h) B /__________|__________\ C (c,h)

# c c

# d = c / h; c^2 = h.(2 - h) -> h = 1 + (1 - c^2)^(1/2)

# the lines AB and DC are perpendicular when 1 + m(AB).m(DC) is zero;

# return this value and the ratio (2.c / s)^2 for the semi-chord <c>

def fn(c):

h = 1 + (1 - c * c) ** (1 / 2)

# line AB

mab = line((0, 0), (-c, h))[0]

# line DC

mdc = line((-c / h, 1), (c, h))[0]

if mab and mdc:

rsq = ((4 * c * c) / (1 + (c / h) ** 2))

return 1 + mab * mdc, rsq

# find the <c> value for a right angle

c = newton(lambda x: fn(x)[0], 0.8, 1e-12)

print(f"sq(ratio) = {fn(c)[1]:.12g}")

Sunday Times Teaser 3191 – The Budgie’s Extra Ration

Published Friday November 17 2023 (link)

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