Divide into 2 isosceles triangles

Python 3.8 (pre-release), 123 129 bytes

g=lambda*p:[[180-2*x,x]for x in p]
f=lambda a,b:((c:=180-a-b)==90or c>45>2in{a/b,b/a})*g(a,b)or(c/3in{a,b})*g(c/3,c/3*2)or f(b,c)

Try it online!

A function that takes in 2 angles, and returns the 2 isosceles triangles. If the given triangle cannot be divided, the function loops forever.

05AB1E, 39 38 bytes


Input as a pair of integers; output as a pair of pairs of integers.

Try it online or verify all test cases.


O                 # Take the sum of the (implicit) input-pair of angles
 ƵΔα              # Get the absolute difference with (compressed) 180
    ª             # Append that third angle to the (implicit) input-pair
     ©            # Store it in variable `®` (without popping)
90K               # Remove 90 from the triplet of angles
   D              # Duplicate it
®Qi               # If it's still equal to `®` (thus none were 90):
   ʒ              #  Filter the triplet by:
    45‹           #   Check that the angle is smaller than 45
               *  #   AND
        y2 S*     #   Check if the angle multiplied by 2
          3S* à   #   or multiplied by 3
       ®     å    #   is in the triplet of angles `®`
   }ß             #  After the filter: pop and push the minimum of the remaining angles
     x            #  Double it (without popping)
      ‚           #  Pair the non-doubled and doubled values together
  }ε              # After the if statement: map the angles in the pair to:
    90α           #  Get the absolute difference with 90
       ·          #  Double it
        y‚        #  And pair it with the non-mapped angle
                  # (after which the resulting pair of pairs is output implicitly)

See this 05AB1E tip of mine (section How to compress large integers?) to understand why ƵΔ is 180.

APL (Dyalog Unicode), 47 bytes

(⊢,⍨¨2×90-⊢)∘{90∊⍵:⍵~90⋄1 2×⌊/⍵∩∊⍵÷⊂2 3},,180-+

Try it online!

A tacit function that takes two angles as left and right arguments.

Uses the information found by Neil, modified to take care of xnor's test case:

A triangle can be divided into 2 isosceles triangles either if one of the angles is < 45° and is exactly one half or one third of one of the other angles, or if one of the angles is 90°.

Now, the base angles of the result can be found as follows:

  • If one of the angles is 90°, the bases are the other two angles.
  • Otherwise, one of the angles is < 45° and is exactly one half or one third of one of the other angles should hold, because the input is guaranteed to have a solution. In this case, the angle satisfying the condition becomes the base for one triangle, and the other triangle's base angle is twice the angle.

How it works: the code

(⊢,⍨¨2×90-⊢)∘{90∊⍵:⍵~90⋄1 2×⌊/⍵∩∊⍵÷⊂2 3},,180-+  ⍝ Left, Right: two angles
                                        ,,180-+  ⍝ Length-3 vector of three angles
             {                         }  ⍝ Find two base angles:
              90∊⍵:⍵~90⋄                  ⍝ If an angle is 90, the bases are the other two
                              ⍵∩∊⍵÷⊂2 3   ⍝ Otherwise, find the angles that are 1÷2 or 1÷3 of another
                        1 2×⌊/            ⍝ Take the minimum angle of that and attach its double
(          )∘  ⍝ Attach apex angles to two base angles
     2×90-⊢    ⍝ apex=180-2×base
 ⊢,⍨¨          ⍝ Attach each apex to the left of the base