Sunday Times Teaser 2782 – Spiders

by Graham Smithers

Published: 17 January 2016 (link)

Spiders Beth and Sam wake up in the bottom corner of a cuboidal barn (all of whose sides are whole numbers of metres). They want to reach the opposite bottom corner without actually walking across the floor. Beth decides to walk on one of five possible shortest routes, two of them being around the edge of the floor and the other three being over the walls and ceiling. Sam decides instead to spin a web directly to the point on the ceiling diagonally opposite the starting point and then to drop down into the corner. The total length of his journey is within five centimetres of a whole number of metres.

How high is the barn?

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Brian Gladman permalink


# the length and width of the barn
for a in range(1, 20):
  for b in range(1, a + 1):
    # there are two equal paths using the floor edges
    s1 = a + b
    # the height of the barn
    for c in range(1, 20):
      # there are two equal paths using the ceiling 
      # with a side and an end wall
      t = (a + c) ** 2 + (b + c) ** 2
      s2 = int(t ** 0.5 + 0.5)
      # these four paths must be equal
      if t != s2 ** 2 or s2 != a + b:
        continue
      # there are paths using the ceiling with either  
      # two side walls or two end walls but they are
      # not necessarily equal
      u = a ** 2 + (b + 2 * c) ** 2
      s3 = int(u ** 0.5 + 0.5)
      v = b ** 2 + (a + 2 * c) ** 2
      s4 = int(v ** 0.5 + 0.5)
      # if they are equal there are either four or six minimum paths
      if s3 == s4:
        continue
      # one of these paths must equal the other four
      if s3 == a + b and u == s3 ** 2 or s4 == a + b and v == s4 ** 2:
        # check that the other spider's path length is within 5cm
        # of an integer number of metres
        sp = (a * a + b * b + c * c) ** 0.5 + c
        d = (100 * sp) % 100
        if d < 5 or d > 95:
          print('The barn height is {} metres.'.format(c))

# the length and width of the barn

for a in range(1, 20):

for b in range(1, a + 1):

# there are two equal paths using the floor edges

s1 = a + b

# the height of the barn

for c in range(1, 20):

# there are two equal paths using the ceiling

# with a side and an end wall

t = (a + c) ** 2 + (b + c) ** 2

s2 = int(t ** 0.5 + 0.5)

# these four paths must be equal

if t != s2 ** 2 or s2 != a + b:

continue

# there are paths using the ceiling with either

# two side walls or two end walls but they are

# not necessarily equal

u = a ** 2 + (b + 2 * c) ** 2

s3 = int(u ** 0.5 + 0.5)

v = b ** 2 + (a + 2 * c) ** 2

s4 = int(v ** 0.5 + 0.5)

# if they are equal there are either four or six minimum paths

if s3 == s4:

continue

# one of these paths must equal the other four

if s3 == a + b and u == s3 ** 2 or s4 == a + b and v == s4 ** 2:

# check that the other spider's path length is within 5cm

# of an integer number of metres

sp = (a * a + b * b + c * c) ** 0.5 + c

d = (100 * sp) % 100

if d < 5 or d > 95:

print('The barn height is {} metres.'.format(c))

The shortest routes across the walls and ceilings can be seen by ‘flattening’ out the room as shown here:

The squares of the lengths of the different (equal) shortest paths in this puzzle are:[(a+b)^2] [(a+c)^2 + (b+c)^2] [b^2 + (a+2c)^2] [a^2+(b+2c)^2] The first two of these apply to two paths each, while the last two are different unless the room is square. If we equate the first three paths in pairs, we can show that [b(a-c)=c(a+c)] [b=a+c] [ab=2c(a+c)] And these in turn give (a=2c) and (b=3c) so the space diagonal becomes (sqrt(14)c). This shows that the length of the space diagonal is very close to (~374c) cm, from which the answer of 4 metres is quickly found

Jim Randell permalink

This Python program produces a series of solutions. (The number of solutions required can be specified on the command line). The smallest solution is the only barn with reasonable dimensions.


#
# suppose the barn has (width, length, height) = (w, l, h)
#
# two of the shortest paths are along edges (RD + FW = p1, BD + LD = p2)
# so the length of the shortest paths is (l + w)
#
# the remaining shortest paths have lengths:
#
# hypot(l + h, w + h) [R + U + F = p3, B + U + L = p4]
# and hypot(l, w + 2h) [R + U + L = p5]
#
# so given l and w we can write quadratic equations for h.
#
# the fact there are exactly 5 shortest paths tells us w < l

from itertools import count
from enigma import irange, is_square, hypot, printf

# find n solutions
def solve(n=1):

  # consider increasing lengths
  for l in count(2):
    # and widths
    for w in irange(1, l - 1):
      # compute corresponding h (using p5)
      h = is_square(w * (w + 2 * l))
      if h is None or not(h > w): continue
      (h, r) = divmod(h - w, 2)
      if r > 0: continue
      # and check against p3, p4
      if not((l + w) ** 2 == (l + h) ** 2 + (w + h) ** 2): continue

      # calculate the length of diagonal through the cuboid
      d = hypot(l, w, h)
      x = abs(d - int(d + 0.5)) 
      # check it is within 5cm of a whole number of metres
      if x > 0.05: continue

      # lengths of paths for Beth and Sam
      (b, s) = (l + w, d + h)
      printf("h={h} w={w} l={l}, d={d:.2f} x={x:.2f}, beth={b} sam={s:.2f}")

      n -= 1
      if n == 0: return

from sys import argv
solve(1 if len(argv) < 2 else int(argv[1]))

# suppose the barn has (width, length, height) = (w, l, h)

# two of the shortest paths are along edges (RD + FW = p1, BD + LD = p2)

# so the length of the shortest paths is (l + w)

# the remaining shortest paths have lengths:

# hypot(l + h, w + h) [R + U + F = p3, B + U + L = p4]

# and hypot(l, w + 2h) [R + U + L = p5]

# so given l and w we can write quadratic equations for h.

# the fact there are exactly 5 shortest paths tells us w < l

from itertools import count

from enigma import irange, is_square, hypot, printf

# find n solutions

def solve(n=1):

# consider increasing lengths

for l in count(2):

# and widths

for w in irange(1, l - 1):

# compute corresponding h (using p5)

h = is_square(w * (w + 2 * l))

if h is None or not(h > w): continue

(h, r) = divmod(h - w, 2)

if r > 0: continue

# and check against p3, p4

if not((l + w) ** 2 == (l + h) ** 2 + (w + h) ** 2): continue

# calculate the length of diagonal through the cuboid

d = hypot(l, w, h)

x = abs(d - int(d + 0.5))

# check it is within 5cm of a whole number of metres

if x > 0.05: continue

# lengths of paths for Beth and Sam

(b, s) = (l + w, d + h)

printf("h={h} w={w} l={l}, d={d:.2f} x={x:.2f}, beth={b} sam={s:.2f}")

n -= 1

if n == 0: return

from sys import argv

solve(1 if len(argv) < 2 else int(argv[1]))

(The program uses routines from the enigma.py library).

It turns out that (h, w, l) = (k, 2k, 3k) for integers k, which leads to a simplified version of the program.

WB permalink

For the height of four metres, the room dimensions are 4, 8 and 12, then two shortest routes along edges of floor are 20. I cannot see how any other route with diagonals can also result in an exact length of 20.
What am I missing?

Reply
- Brian Gladman permalink
  
  Welcome to the site! The other routes go up a wall across the ceiling and down another wall to the final destination. You can see these routes if you think of the room as an upturned cardboard box and then ‘flatten’ it. This shows a number of ‘straight’ line paths between the two end points (I have now added a diagram above to show this).
  
  Reply
John Crabtree permalink

Looking at Brian’s diagram above, and looking at the red and blue line which meet at the top right corner, by inspection of the geometry, they can only be the same length if (b = a + c). Then eliminating a leads to a quadratic equation where (b = 0) (impossible) or (3c)

If (d) is the nearest integer to the diagonal distance, then it turns out that (d) and (c) are the base solution of the Pell equation (d^2 – 14c^2 = 1).

Reply

Sunday Times Teaser 2782 – Spiders

by Graham Smithers

Published: 17 January 2016 (link)

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