I was recently re-reading my question here (Area of convex n-gon using triangles) and I did not want to rebump it, but I had another question.
Here it is for reference: Suppose we have a convex $n$-gon and a point inside the $n$-gon or on the sides of the $n$-gon, and suppose one extended lines from all the vertices of the $n$-gon to make $n$ triangles with two of its vertices on the $n$-gon and the third vertex being the point itself.
Now, my new question: suppose we have only one point such that the sums of the areas of alternating triangles are equivalent. Is there anything special about it? (spoiler: yes, there is, but what is it?) My professor proposed it to me and I can't figure out how to solve it.
By alternating triangles, I mean if you took the triangles, and went clockwise, and alternated triangles and took the sum of those areas (area 1=triangle 1+triangle 3+triangle 5+..., area2=triangle2+triangle4+triangle6+...), and this point should cause area1=area2.