A question about chords that bisect the area of a plane convex set Let K be a compact convex subset of the Euclidean plane whose interior is non-empty. Does K always contain at least one point P that is the intersection point of at least 3 pairwise distinct chords of K, each of which bisects the area of K? I know that this is true when K is a triangle and suspect that it is true generally. But if it is, then I wonder whether there is a simple way to prove it.
 A: I think you can do better. You can choose any three angles that add to $\pi$ and find a point $K$ that has three lines separated by those angles going through it where each line bisects the area.  For an example, I will use $\frac \pi 2, \frac \pi 4, \frac \pi 4$, but the process will be the same for any set of angles.  
We can find horizontal and vertical lines that each divide the shape in half.  These lines meet in a point.  Now draw a line with slope $1$ that also divides the shape in half.  If it passes through the point where the other two lines meet, we are done.  Otherwise, suppose it crosses the vertical line above the point where the horizontal line does. Slowly rotate all three lines clockwise, moving them transversely to keep each dividing the shape in half.  The intersection points will be continuous functions of the rotation angle.  When we have rotated by $\pi$ radians, the intersection point of the slope $1$ line must again be above the intersection point of the other two.  To get there, it must have passed through the intersection point.  The rotation angle and intersection point where that happens is the point $K$ and set of directions we are looking for.
