Given the shape below, the challenge is to bisect it by placing the two segments on the shape. All segments are length 6. The resulting two shapes need to have the same area but does not need to be the same. At least three solutions are required.
It appears that there are an infinite number of solutions in which:
- the outer ends of the two segments are positioned somewhere on the edges of your shape, and
- the inner ends of the two segments intersect.
The following diagram illustrates this.
At the top left is the solution found by @phenomist where area A is equal to area B.
If the intersection point of the top segment is moved a small distance p down the left edge of the top triangle, and intersection point of the bottom segment remains in place, then obviously area A is smaller than area B (see top right figure).
But if the bottom intersection is now moved a larger distance q up the bottom edge of the lower triangle it is equally obvious that a situation where area A is now larger than area B can be obtained (see bottom left figure).
As the areas A and B vary continuously with q, there will be some intermediate distance r (where 0 < r < q) where area A equals area B (bottom right figure).
This will work for any small p so there are an infinite number of solutions. If p is large enough it will reach the third solution provided by @Dr Xorile, where r = 6-p. For still larger p the solutions will just be diagonal mirrors of those already described.
But if the initial p displacement is down the other side of the triangle, there will be a further (infinite) set of solutions (in which the r displacement will still be up the bottom left edge of the lower triangle) that can be constructed in a similar manner.
Define X to be the point that makes AXB equilateral on the same side as everything else. Therefore, AX is parallel to DF and BX is parallel to FC. Let the bisection be FX and XB, which are both the same length as q=r. This satisfies the criterion: namely, each piece is composed of a congruent parallelogram and a congruent equilateral triangle. The left piece consists of ADFX+AXB and the right piece consists of BCFX+BCE.
@phenomist found the first two solutions. Here's a third:
We know that putting the lines at FC and CE would separate it into two parts with all on one side and none of the other. We also know that we can slide the two lines along CA (the dotted line) so that the ends are on FD and EB. Eventually we would get to DA and BA which would put all of the area onto the other side. So, by continuity, there must be a point which separates them into two equal parts somewhere between those two extremes.