Sunday Times Teaser 3165 – Round the Bend

by BRG on May 19, 2023

by Howard Williams

Published Sunday May 21 2023 (link)

My new craft project involves card folding and requires a certain amount of precision and dexterity. For the penultimate stage a rectangular piece of card, with sides a whole number of centimetres (each less than 50cm), is carefully folded so that one corner coincides with that diagonally opposite to it. The resulting five-sided polygon also has sides of integer lengths in cm. The perimeter of the polygon is five twenty-eighths smaller than that of the perimeter of the original rectangular card.

As a final check I need to find the new area of the card.

What, in square centimetres, is the area of the polygon?

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BRG permalink

Here is the layout used in the program below:

The rectangle has sides of length a and b and its diagonal is of length c; the fold line, which crosses the diagonal at its mid-point and is perpendicular to it, is of length d. The relationships between the various lengths can be derived using similar triangles. Since the fold line has an integer length, we know that the diagonal must be rational which means that (a, b, c) a Pythagorean triple.


# the height of the rectangle
for a in range(1, 50):
  # the width of the rectangle
  for b in range(a, 50):
    # compute the length of the diagonal
    c = (a * a + b * b) ** (1 / 2)
    # the length of the fold line (an integer)
    d = a * c / b
    if round(d) == d:
      # the smallest side of the polygon (also an integer)
      s = (b * b - a * a) / (2 * b)
      if round(s) == s:
        # the perimeter condition (redundant)
        p1, p2 = 2 * (a + b), 2 * (a + s) + d
        if 23 * p1 == 28 * p2:
          area = a * b / 2 + a * (b * b - a * a) / (4 * b)
          print(f"Area = {area}cm^2 (rectangle: {a}cm x {b}cm).")

# the height of the rectangle

for a in range(1, 50):

# the width of the rectangle

for b in range(a, 50):

# compute the length of the diagonal

c = (a * a + b * b) ** (1 / 2)

# the length of the fold line (an integer)

d = a * c / b

if round(d) == d:

# the smallest side of the polygon (also an integer)

s = (b * b - a * a) / (2 * b)

if round(s) == s:

# the perimeter condition (redundant)

p1, p2 = 2 * (a + b), 2 * (a + s) + d

if 23 * p1 == 28 * p2:

area = a * b / 2 + a * (b * b - a * a) / (4 * b)

print(f"Area = {area}cm^2 (rectangle: {a}cm x {b}cm).")

Frits permalink

I ended up with more or less the same program as Brian.

Jim Randell already has used Pythagorean triples. Luckily I found a different approach (divisors).


# L1 = (-h/w)x + h
# line L1 passes through 2 corners (w, 0) and (0, h)
# L2 = (w/h)x + h/2 - w**2/(2h)
# both lines are perpendicular to each other and cross at (w/2, h/2)

# new triangle has two sides of length <w>, 2 sides of length <o>
# and one side of length <x>
for w in range(1, 49):                         # height is longer than width
  w2 = w * w
  # only check for divisors of w*w because of x2 formula
  for h in [x for x in range(w + 1, 50) if w2 % x == 0]:
    # consider triangle with sides x, w and (h - 2 * o)
    #x2 = (h - 2o)**2 + w2
    #x2 = ((h2 - 2oh) / h)**2 + w2
    # use that 2oh equals h2 - w2  (see below)
    #x2 = ((h2 - (h2 - w2)) / h)**2 + w2
    x2 = (w2 / h)**2 + w2
    # x2 has to be square
    if (x := x2**.5) % 1: continue

    # calculate side <o> which occurs twice
    # o equals L2(0) which is h/2 - w**2/(2h)
    o, r = divmod(h * h - w2, 2 * h)
    if r: continue

    # new perimeter is five twenty-eighths smaller than the old perimeter
    O = 2 * (h + w)
    N = 2 * (w + o) + x
    if 23 * O != 28 * N: continue

    # area is wo + (h-2o)w/2 + wo/2
    A = w * (h + o) / 2
    print("answer:", A, "cm^2")

# L1 = (-h/w)x + h

# line L1 passes through 2 corners (w, 0) and (0, h)

# L2 = (w/h)x + h/2 - w**2/(2h)

# both lines are perpendicular to each other and cross at (w/2, h/2)

# new triangle has two sides of length <w>, 2 sides of length <o>

# and one side of length <x>

for w in range(1, 49): # height is longer than width

w2 = w * w

# only check for divisors of w*w because of x2 formula

for h in [x for x in range(w + 1, 50) if w2 % x == 0]:

# consider triangle with sides x, w and (h - 2 * o)

#x2 = (h - 2o)**2 + w2

#x2 = ((h2 - 2oh) / h)**2 + w2

# use that 2oh equals h2 - w2 (see below)

#x2 = ((h2 - (h2 - w2)) / h)**2 + w2

x2 = (w2 / h)**2 + w2

# x2 has to be square

if (x := x2**.5) % 1: continue

# calculate side <o> which occurs twice

# o equals L2(0) which is h/2 - w**2/(2h)

o, r = divmod(h * h - w2, 2 * h)

if r: continue

# new perimeter is five twenty-eighths smaller than the old perimeter

O = 2 * (h + w)

N = 2 * (w + o) + x

if 23 * O != 28 * N: continue

# area is wo + (h-2o)w/2 + wo/2

A = w * (h + o) / 2

print("answer:", A, "cm^2")

@Brian, when I use (without even calling divisors()):


from number_theory import divisors

from number_theory import divisors

the program runs considerably slower . This doesn’t happen when I import divisors from enigma.

BRG permalink

@Frits, I believe Jim optimes factors and divisors in enigma for small values, whereas I optimise for large values by caching primes on import.

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John Z permalink

I asked ChatGPT to write the Python code to solve this Teaser. What it produced does not work. As a curiosity, here it is:


def solve_puzzle():
    original_perimeter = 0
    polygon_perimeter = 0
    polygon_area = 0

    # Iterate through all possible dimensions of the rectangular card
    for width in range(1, 50):
        for height in range(1, 50):
            original_perimeter = 2 * (width + height)

            # Calculate the lengths of the sides of the resulting polygon
            side1 = width
            side2 = height
            side3 = ((width ** 2) + (height ** 2)) ** 0.5

            # Calculate the perimeter of the resulting polygon
            polygon_perimeter = side1 + side2 + side3

            # Check if the perimeters satisfy the given condition
            if polygon_perimeter == original_perimeter - 5/28:
                # Calculate the area of the polygon
                semi_perimeter = polygon_perimeter / 2
                polygon_area = (semi_perimeter * (semi_perimeter - side1) * (semi_perimeter - side2) * (semi_perimeter - side3)) ** 0.5

                return polygon_area

    return None

# Call the function to solve the puzzle
result = solve_puzzle()

# Print the area of the polygon if a solution is found
if result:
    print(f"The area of the polygon is: {result} square centimeters")
else:
    print("No solution found.")

def solve_puzzle():

original_perimeter = 0

polygon_perimeter = 0

polygon_area = 0

# Iterate through all possible dimensions of the rectangular card

for width in range(1, 50):

for height in range(1, 50):

original_perimeter = 2 * (width + height)

# Calculate the lengths of the sides of the resulting polygon

side1 = width

side2 = height

side3 = ((width ** 2) + (height ** 2)) ** 0.5

# Calculate the perimeter of the resulting polygon

polygon_perimeter = side1 + side2 + side3

# Check if the perimeters satisfy the given condition

if polygon_perimeter == original_perimeter - 5/28:

# Calculate the area of the polygon

semi_perimeter = polygon_perimeter / 2

polygon_area = (semi_perimeter * (semi_perimeter - side1) * (semi_perimeter - side2) * (semi_perimeter - side3)) ** 0.5

return polygon_area

return None

# Call the function to solve the puzzle

result = solve_puzzle()

# Print the area of the polygon if a solution is found

if result:

print(f"The area of the polygon is: {result} square centimeters")

else:

print("No solution found.")

BRG permalink

Hi John, I have also tried this on some of the teasers with variable success. After seeing your attempt I tried to guide it to a correct solution by suggesting what was wrong with its efforts. I gave up after several attempts, none even close to being correct.

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Sunday Times Teaser 3165 – Round the Bend

by Howard Williams

Published Sunday May 21 2023 (link)

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