Let

,

, and

be the sets of

-sectors of the angles

,

, and

, respectively. The number of points of intersection between any two of the sets is

, hence, the number of points is at most

. For example, the maximum number of points for

is 3: one is the intersection of the bisectors of

and

, one is the intersection of the bisectors of

and

, and one is the intersection of the bisectors of

and

, but we know these three points coincide. The principle of inclusion-exclusion gives the number of distinct points:

.

When

, this gives

1=1+1+1-(1+1+1)+1.

Now we want to count the distinct points for any value of

. Some experimentation with this Demonstration shows that the result depends on the triangle's shape, especially its central or lateral symmetry. For example, for

, the number of lines is nine, which is odd, so that the bisectors of each angle are in each of

,

, and

. The number

|(A⋂B) ⋂ (B⋂C) ⋂ (C⋂A)| is 37 for an equilateral triangle with central symmetry, one for a scalene triangle with no symmetry, and nine for a isosceles with lateral symmetry; for

, both are zero, and so on.

By tabulating the numbers for each set and applying

*Mathematica*'s built-in function

FindSequenceFunction on the results, it appears (this is not a proof) that for a scalene triangle

. For an isosceles triangle without a right angle the formula appears to be

; for an equilateral or right isosceles triangle the formulas are complicated.