Given a closed topological disk $D$ in the plane, is it always possible to find a straight line segment $L\subseteq D$ with endpoints on $\partial D$, and whose interior lies inside the interior of $D$, which divides $D$ into two parts of equal area?
I and a few friends at first believed this question to be intuitively true, but one of these friends eventually came up with the following counterexample: consider an equilateral triangle, and attach near each vertex a rectangle whose area is large compared to that of the triangle, oriented in a manner such that for an observer inside the triangle only a negligible portion of each rectangle can be seen. An approximation is illustrated below:
Despite my limited digital drawing abilities, suppose each of the rectangles have the same area and are long enough for the triangle's area to become negligible. Then, any line segment from the boundary of this disk to itself through the interior must have all but a negligible portion of two of the three rectangles on one side of the segment, hence the ratio between the larger and smaller areas is approximately $(2/3)/(1/3) = 2$, nowhere near the $1/1$ ratio demanded by the question.
And so I came up with this natural follow-up question: how high can the minimal ratio between areas on each side of a segment be for a given disk? For example, does there exist a topological disk $D$ such that for any line segment $L$ drawn as specified above, the smaller of the two parts created by this segment has area less than $1$% of the area of the larger part? $\epsilon $%?
The above counterexample, and perhaps my lack of imagination, lead me to hesitantly conjecture the following: for any closed topological disk $D$ in the plane, there exists a straight segment with endpoints on $\partial D$, and whose interior lies in the interior of $D$, which splits $D$ into two parts for which the ratio $L/S$ between the larger and smaller areas is less than $2$.
Is this true? If not, can the conjecture be made true by replacing $2$ by some sufficiently large constant?
My intuition agrees with your conjecture.
Let's try to prove it for polygons: There is a theorem stating that it is possible to triangulate polygons such that the triangles only use the vertices of the polygon. By adding vertices to the polygon we can make sure that the areas of all the triangles in the triangulation are arbitrarily close to each other, let's say they are all $\varepsilon$-close to $1$.
Now build a tree: The vertices of the trees are the triangles in the triangulation and two vertices are connected by an edge whenever the triangles share a common adge in the triangulation. Note that every vertex has at most $3$ edges. The graph is a tree, since our polygon is simply connected (homeomorphic to a disk).
Now the problem is to find an edge in the tree, such that removing the edge splits the tree into two connected components such that one contains at most twice as many vertices than the other. I think this can be done by starting on some edge in the tree and moving to an adjacent edge which better balances the tree (one needs to fill in the details here).
Now since all the vertices correspond to similarly sized triangles, the cut of the tree corresponds to cutting the polygon in such a way that the ratio of the two parts is at most 2:1 (up to some $\varepsilon$).
For the topological version of the problem we would need a theorem saying that every embedding of the sphere can be approximated by polygons, (which may not be true).