Sunday Times Teaser 2922 – Inscribed

by Andrew Skidmore

Published September 23 2018 (link)

The Republic of Mathematica has an unusual rectangular flag. It measures 120 cm by 60 cm and has a red background. It features a green triangle and a white circle. All the lengths of the sides of the triangle and the radius of the circle (which touches all three sides of the triangle) are whole numbers of cm. Also, the distances from the vertices of the triangle to the centre of the circle are whole numbers of cm. The flag has a line of symmetry.

What is the length of the shortest side of the triangle?

One Comment Leave one →

Brian Gladman permalink


from collections import defaultdict
from itertools import permutations
from number_theory import pythag

# the triangle is isoceles (b, c, c) with base b,
# height h and incircle radius r = b.h / (b + 2.c)
#
# the height h is given by h^2 = c^2 - (b / 2)^2
# with the distance from the top vertex to the 
# centre of the circle being h - r
#
# the distance from a lower vertex to the circle
# centre is given by l^2 = r^2 + (b / 2)^2
#
# hence we need two pythagorean triangles with
# the same side (b / 2):
#
#    (b / 2, h, c) and (b / 2, r, l)

# store pythagorean triangles indexed on twice
# their non-hypotenuse sides (i.e. b)
s2p = defaultdict(list)
for x, y, z in pythag(h_max = 120):
  s2p[2 * x].append((y, z))
  s2p[2 * y].append((x, z))

# consider sides (b) that give two or more triangles
for b, tl in s2p.items():
  if len(tl) >= 2:

    # consider pairs of triangles for (h, c) and (r, l)
    for (h, c), (r, l) in permutations(tl, 2):
      
      # compute the incircle radius and check it against
      # the value from the pythagorean triangle 
      if divmod(h * b, b + 2 * c) == (r, 0):
        print(f'b = {b}, c = {c}, r = {r}, l = {l}, h = {h}')

from collections import defaultdict

from itertools import permutations

from number_theory import pythag

# the triangle is isoceles (b, c, c) with base b,

# height h and incircle radius r = b.h / (b + 2.c)

# the height h is given by h^2 = c^2 - (b / 2)^2

# with the distance from the top vertex to the

# centre of the circle being h - r

# the distance from a lower vertex to the circle

# centre is given by l^2 = r^2 + (b / 2)^2

# hence we need two pythagorean triangles with

# the same side (b / 2):

# (b / 2, h, c) and (b / 2, r, l)

# store pythagorean triangles indexed on twice

# their non-hypotenuse sides (i.e. b)

s2p = defaultdict(list)

for x, y, z in pythag(h_max = 120):

s2p[2 * x].append((y, z))

s2p[2 * y].append((x, z))

# consider sides (b) that give two or more triangles

for b, tl in s2p.items():

if len(tl) >= 2:

# consider pairs of triangles for (h, c) and (r, l)

for (h, c), (r, l) in permutations(tl, 2):

# compute the incircle radius and check it against

# the value from the pythagorean triangle

if divmod(h * b, b + 2 * c) == (r, 0):

print(f'b = {b}, c = {c}, r = {r}, l = {l}, h = {h}')

Sunday Times Teaser 2922 – Inscribed

by Andrew Skidmore

Published September 23 2018 (link)

Leave a comment to Brian Gladman Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta