n-Sectors of the Angles of a Triangle

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.