Area of convex n-gon using triangles

Question

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. Is there a nice formula for the sum of the areas of alternating triangles? Is there a nice point such that the two different area values mentioned above are equivalent?

Answer 1

Suppose your point is $(x,y)$, and the vertices are $(x_1,y_1)\dots(x_n,y_n)$. Then the area of the first triangle is ${1\over2}\Big((x-x_1)(y-y_2)-(x-x_2)(y-y_1)\Big)$, which is in fact a linear function of $(x,y)$. A sum of several linear functions is again a linear function. So the problem in question basically means that we solve a single linear equation. Depending on the coefficients, the answer might be the whole plane (which, I believe, is the case for a regular even-sided n-gon), one particular line, or an empty set.

Upd. Come to think of it, all three situations are possible. The example of the first one is a regular square. The second occurs in a non-regular quadrangle, as explained by @joriki's comment. The third takes place in a hexagon with side lengths 1, 100, 1, 100, 1, 100 (in that order).