geometry
The shape(Roughly to say, it a 4*5 rectangle – 2*2 rectangle):
You can draw a line(polygonal chain or a curve, and may violate the grid) to cut it into two shape, for example:
But that doesn't cut it into two congruent shapes.
So my question:
Is there any way to cut that shape into two congruent shapes?
It may be trivial, but I had try to solve it for hours…
You can and this generalizes to $n$-dimensional spaces as well. It's called the ham sandwich theorem and the two dimensional case is sometimes called the pancake theorem. In two dimensions you can use a "rotating knife" argument using the Intermediate Value theorem to ensure the existence of such a cut. Higher dimensional proofs require a bit more machinery but are essentially the same idea.
Aha, got it!
Reading this question gave me the idea to look at the perimeter of each piece; since these must be equal, and they use the same amount of perimeter for their border, the two polyominoes must each take up half of the boundary of the $L$-tromino. For topological regions, these two boundary portions must be contiguous. Also observe that the boundary split can't occur at a vertex exactly, because they're polyominoes; at each corner will be a square of the grid which will belong to one of the two pieces, giving it a portion of boundary a positive radius around that corner.
With this in mind, we can assume without loss of generality that one of the points $P$ of the border split lies strictly in the interior of the bottom edge (relative to the orientation of the image in the OP). This will force the other boundary point $Q$ to lie on the top edge or the edge immediately clockwise from it. So the "left" piece will span a length of $2$ units (assuming standard dimensions for the $L$-tromino) along the vertical axis, and thus the "right" piece must have those dimensions along either the vertical or the horizontal axis. But this is only possible if the $Q$ lies on the top edge:
This means that $P$ is on the right half of the bottom edge, so that the left piece contains two spokes of lengths $2$ and $1$ joined at a right angle. But such a subset cannot fit inside the right piece in an axis-aligned manner: the length $2$ component would have to be placed vertically, but then it would force the right piece to contain two points on the top and bottom edges with equal $x$-values, which can't happen if $P$ is beyond the midpoint of the bottom edge.
Similar reasoning to this seems like it might help in answering this question for other polyominoes, but I'm not sure it suffices to decide things in all cases.
Best Answer
Assuming the pieces do not have to be connected, we have the following solution: