Assume the puzzle can be solved. Then it must be solvable with a hole of any diameter, even zero. But if you drill a hole of zero diameter that is six inches long, you leave behind the volume of a six inch diameter sphere. Observe that triangles AOB and DOC are similar. L1 and L2 are given, and so L is a constant.
Every square tile covers an a-multiple of black squares.
Every square tile covers an a-multiple of black squares. This statement of how to position the colouring isn't good enough, I'm afraid. Suppose N columns are black with this positioning. The difference between these two numbers of black cells is H, which is not a multiple of a.