Sunday Times Teaser 3250 – Very Similar Triangles

by BRG on January 5, 2025

by Michael Fletcher

Published Sunday January 05 2025 (link)

I have a piece of A6 paper on which I draw two triangles. The triangles are very similar in two ways. First of all, they are both the same shape. Not only that, the lengths of two of the sides of one of the triangles are the same as the lengths of two of the sides of the other triangle, but one triangle is larger than the other.

If the sides of the triangles are whole numbers of millimetres and the triangles don’t have any obtuse angles.

What are the lengths of sides of the larger triangle?

From → Uncategorized

3 Comments Leave one →

BRG permalink


# A6 = 105mm x 148mm (181mm hypotenuse)
h_max = 181

# Triangles with sides (a, b, c) and (b, c, d)  with
# a / b = b / c = c / d == fraction(num / den) in its
# lowest terms. Hence d = a.(num / den)^3 which means
# that both a and d are multiples of cubes

# the cube root of d
for dr in range(2, 7):
  # the cube root of a
  for ar in range(1, dr):
    # consider possible multipliers  
    for m in range(1, h_max // dr ** 3 + 1):
      a, d = m * ar ** 3, m * dr ** 3
      b, c = a * dr // ar, d * ar // dr
      # check for valid triangles with no obtuse angles
      if c < a + b and c * c < a * a + b * b:
        print((b, c, d), [a, b, c])

# A6 = 105mm x 148mm (181mm hypotenuse)

h_max = 181

# Triangles with sides (a, b, c) and (b, c, d) with

# a / b = b / c = c / d == fraction(num / den) in its

# lowest terms. Hence d = a.(num / den)^3 which means

# that both a and d are multiples of cubes

# the cube root of d

for dr in range(2, 7):

# the cube root of a

for ar in range(1, dr):

# consider possible multipliers

for m in range(1, h_max // dr ** 3 + 1):

a, d = m * ar ** 3, m * dr ** 3

b, c = a * dr // ar, d * ar // dr

# check for valid triangles with no obtuse angles

if c < a + b and c * c < a * a + b * b:

print((b, c, d), [a, b, c])

John Z permalink


#A6 paper = 105mm x 148mm

# A / a = B / b = C / c = B / A = B / A = C / B 

for A in range(1, 106):
  for B in range(A + 1, 149):
    if B * B % A == 0 and A * A % B == 0:
      C = B * B // A
      if C * C < A * A + B * B: #no obtuse angle
        print('Large triangle:', A, B, C)
        a = A * A // B
        print('Small triangle:', a, A, B)

#A6 paper = 105mm x 148mm

# A / a = B / b = C / c = B / A = B / A = C / B

for A in range(1, 106):

for B in range(A + 1, 149):

if B * B % A == 0 and A * A % B == 0:

C = B * B // A

if C * C < A * A + B * B: #no obtuse angle

print('Large triangle:', A, B, C)

a = A * A // B

print('Small triangle:', a, A, B)

Frits permalink

I am not sure if the “fit” function is correct.


from collections import defaultdict

# A6 = 105mm x 148mm (181mm hypotenuse)

# the hypotenuse of an A6 piece of paper
h_max = int((148**2 + 105**2)**.5)

# does triangle (a, b, c) a <= b <= c fit into rectangle u by v with u <= v
def fit(a, b, c, u, v):
  if c <= v:
    # place side <c> on bottom side rectangular
    s = (a + b + c) / 2
    # altitude from base with length c
    h = (2 * (s * (s - a) * (s - b) * (s - c))**.5) / c
    return  h <= u
  else:
    # place side <c> from bottom left corner to top side rectangular
    w2 = v - (c**2 - u**2)**.5
    if b > v:
      h1 = (b**2 - v**2)**.5
      # place side <b> from bottom left corner to height <h1> at right side
      mx_a = (w2**2 + (u - h1)**2)**.5
    else:
      # place side <b> on bottom side rectangular
      mx_a = (u**2 + (w2 + b - v)**2)**.5

    return a <= mx_a

# consider sides (b / f, b, b * f, b * f * f) with fraction f = e / r
# a^2 + b^2 >= c^2  or  1 / f^2 + 1 >= f^2  or  r^4 + r^2 * e^2 >= e^4

r_e = defaultdict(list)
# b must be divisible by a square as d = b * f * f
for r in range(2, 14):
  r2 = r * r
  r4 = r2 * r2
  # numerator of f
  for e in range(r + 1, 2 * r):
    # no obtuse angles
    if r4 + r2 * e * e < e * e * e * e: break
    r_e[r] += [e]

for b in range(next(iter(r_e.keys()))**2, h_max - 1):
  for r in r_e.keys():
    if b % r**2: continue
    # d = b * (e / r)^2 <= h_max and also f^2 < 2 so e < r * sqrt(2)
    mx_e = min(int(r * 2**.5), int((h_max * r**2 / b)**.5))
    for e in r_e[r]:   # f = e / r
      if e > mx_e: break

      # (a, b, c, d) = (r * b / e, b, b * e / r, b * (e / r)^2)
      if ((r * b) % e == 0 and
          (fc := (b * e) / r) == (c := int(fc)) and
          (fd := (e * c) / r) == (d := int(fd))):
        if fit((a := int((r * b) / e)), b, c, 105, 148):
          print("answer:", (b, c, d))

from collections import defaultdict

# A6 = 105mm x 148mm (181mm hypotenuse)

# the hypotenuse of an A6 piece of paper

h_max = int((148**2 + 105**2)**.5)

# does triangle (a, b, c) a <= b <= c fit into rectangle u by v with u <= v

def fit(a, b, c, u, v):

if c <= v:

# place side <c> on bottom side rectangular

s = (a + b + c) / 2

# altitude from base with length c

h = (2 * (s * (s - a) * (s - b) * (s - c))**.5) / c

return h <= u

else:

# place side <c> from bottom left corner to top side rectangular

w2 = v - (c**2 - u**2)**.5

if b > v:

h1 = (b**2 - v**2)**.5

# place side <b> from bottom left corner to height <h1> at right side

mx_a = (w2**2 + (u - h1)**2)**.5

else:

# place side <b> on bottom side rectangular

mx_a = (u**2 + (w2 + b - v)**2)**.5

return a <= mx_a

# consider sides (b / f, b, b * f, b * f * f) with fraction f = e / r

# a^2 + b^2 >= c^2 or 1 / f^2 + 1 >= f^2 or r^4 + r^2 * e^2 >= e^4

r_e = defaultdict(list)

# b must be divisible by a square as d = b * f * f

for r in range(2, 14):

r2 = r * r

r4 = r2 * r2

# numerator of f

for e in range(r + 1, 2 * r):

# no obtuse angles

if r4 + r2 * e * e < e * e * e * e: break

r_e[r] += [e]

for b in range(next(iter(r_e.keys()))**2, h_max - 1):

for r in r_e.keys():

if b % r**2: continue

# d = b * (e / r)^2 <= h_max and also f^2 < 2 so e < r * sqrt(2)

mx_e = min(int(r * 2**.5), int((h_max * r**2 / b)**.5))

for e in r_e[r]: # f = e / r

if e > mx_e: break

# (a, b, c, d) = (r * b / e, b, b * e / r, b * (e / r)^2)

if ((r * b) % e == 0 and

(fc := (b * e) / r) == (c := int(fc)) and

(fd := (e * c) / r) == (d := int(fd))):

if fit((a := int((r * b) / e)), b, c, 105, 148):

print("answer:", (b, c, d))

Sunday Times Teaser 3250 – Very Similar Triangles

by Michael Fletcher

Published Sunday January 05 2025 (link)

Leave a Reply Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta