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.

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    $\begingroup$ Just to check, given a triangle, we would end up with three inner triangles $t_1, t_2, t_3$. Would the alternating triangle sets be $\{ t_1, t_3\}$ and $\{t_2\}$ in this case? $\endgroup$ – Guy Paterson-Jones Oct 19 '15 at 7:02
  • $\begingroup$ You are correct. $\endgroup$ – user277039 Oct 19 '15 at 7:30

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