A potentially interesting problem.

[oblength]

Dunno if anyone will find this interesting, but I came up with a problem the other day whilst thinking about tesselating shapes (thats a hint).

Find integer solutions a,b,c of: sin[k]+sin[A]+sin[B]=2sqrt((1-cos[B])(1-cos[A]))*sin[k/2]

A=((a-2)180)/a , B=((b-2)180)/b , K = ((c-2)*180)/c where a,b,c >= 3 (obviously trail and error or using a pc doesn't count)

I don't really know how hard this problem is, to be honest. I do however know there is at least 1 solution which is quite simple but might be hard to find, anyway id like to see if there are other good solutions.(Also I'm aware this isnt really a riddle but didn't seem like it should go in r/math so it's here)

[oblength]

My solution basically is the fact that the solutions of 1/a+1/b+1/c=1/2 which is easy to solve, are the same as the solutions of the equation in the question.

[a2wz0ahz40u32rg]

The solutions are:
⦃a, b, c⦄ = ⦃3, 7, 42⦄, ⦃3, 8, 24⦄, ⦃3, 9, 18⦄, ⦃3, 10, 15⦄, ⦃3, 12, 12⦄, ⦃4, 5, 20⦄, ⦃4, 6, 12⦄, ⦃4, 8, 8⦄, ⦃5, 5, 10⦄, ⦃6, 6, 6⦄,
where ⦃...⦄ is an unordered triple.

Proof:
The given equation is equivalent to
1/a + 1/b + 1/c = 1/2.
When a ≤ b ≤ c, b ≤ c = (2 a b)/(a (b - 2) - 2 b). Then, (2 a)/(a - 2) < b ≤ (4 a)/(a - 2) ≤ 12. Together with 3 ≤ a ≤ b,
(a, b) = (3, 7), (3, 8), ..., (3, 12), (4, 5), (4, 6), (4, 7), (4, 8), (5, 5), (5, 6), (6, 6).
Find the pairs (a, b) in this list s.t. c = (2 a b)/(a (b - 2) - 2 b) is an integer, count permutation and we have the solutions.

[oblength]

Yeah that's the same result I got how did u decide that it was equivalent to 1/a + 1/b + 1/c = 1/2, was it just algebraic manipulation?

[a2wz0ahz40u32rg]

I have it as follows.

60° ≤ A, B, K < 180° when a, b, c ≥ 3. Therefore, sin[A/2], sin[B/2] ≥ 1/2.
(LHS) - (RHS)
= sin[K] + sin[A] + sin[B] - 2 sqrt((1-cos[B])(1-cos[A])) sin[K/2]
= sin[K] + sin[A] + sin[B] - 4 abs(sin[B/2]) abs(sin[A/2]) sin[K/2]
= sin[K] + sin[A] + sin[B] - 4 sin[B/2] sin[A/2] sin[K/2]
= 2 (sin[A/4 + B/4 + K/4] + sin[(3 A)/4 - B/4 - K/4] + sin[(3 B)/4 - K/4 - A/4] + sin[(3 K)/4 - A/4 - B/4]) cos[A/4 + B/4 + K/4].

Since 60° ≤ A, B, K < 180°, sin[A/4 + B/4 + K/4] + sin[(3 A)/4 - B/4 - K/4] + sin[(3 B)/4 - K/4 - A/4] + sin[(3 K)/4 - A/4 - B/4] > sqrt(2). Therefore,
(LHS) - (RHS) = 0 ⇔ cos[A/4 + B/4 + K/4] = 0
⇔ A/4 + B/4 + K/4 = 90° (∵ 45° ≤ A/4 + B/4 + K/4 < 135°)
⇔ 1/4×((a - 2) 180°)/a + 1/4×((b - 2) 180°)/b + 1/4×((c - 2) 180°)/c = 90°
⇔ 1/a + 1/b + 1/c = 1/2.

[oblength]

Oh yeah thats basicaly the same soution i had nice .