I had never played this game before, so thanks for the interesting puzzle. I will note down here how I tackled this problem, briefly.
For convenience, I will lable the grid as below.
The first obvious thing is that,
if a cell lies next to the margin of the grid, it should be filled with a triangle such that one edge is lying on the margin, or it should be left empty.
Now, there are several places to start. (I'll explain them separately, while presenting them in one image.)
- Consider the cells containing number 3.
With aforementioned property, we can fill the two adjacent cells of each which lie on the margin. Now to create a tetromino, the two cells along the dotted line should contain ⬕s. And this will create our first square-tetromino in A10.
- Now move to region H5-J9.
The cells in the range J6-J8 and I7 cannot be filled with triangles, so they must be left empty to create an L-tetromino. Therefore, the marked edges (green) should contain the edge of the triangles outside.
- Consider 0.
J1, I2 and J3 cannot contain triangles, obviously.
Continuing from 0,
we see that I1 too cannot be filled with any type of triangle. So it should be left empty. Then if we leave I3 empty, we will get a pentomino. Thus I3 should be filled with a triangle, and the only possiblity is ⬔.
To extend J1-I1-I2 to a tetromino, there are two possibilities; T or skew-tetromino. To create a skew-tetromino, H1 should be filled with ◪, but that contradicts the earlier mentioned margin law. Therefore placing a ⬔ will give a T-tetromino as the only way.
And immediately, we can place a ⬔ in G1.
- Consider the four cells at bottom right corner.
Since J10 has to be a ◪, those four cells should present a square-tetromino.
- Then go to H8.
We can form an L-tetromino containing it, but we will get in trouble with G9. Similarly we cannot attach it to a skew-tetromino because of H10. Hence, we have to leave it isolated as a square-tetromino, and fill G9-H10 as same as I9-J10.
- Back to J3 again.
Currently, it cannot be a part of an L or a skew-tetromino. Therefore leave it as it is and after we can fill J4 with a ⬔.
Fill F10 with a ◪ and use the cells above E10 and F10 to complete square-tetromino.
Similarly, we can fill F1 with a ◩.
- Obviously, D9 should be a
◪. This will pave the way for a large square-teromino.
- In the same manner, we see I5 must be filled with a
⬕, predicting a large 1×4 tetromino.
- Now consider the three free sqaures in E7-F9.
Since G7 can be guaranteed to be filled with a ⬕, these three must along with F6 make an L-tetromino.
- And we can complete H6 and G6,
to create a square-tetromino.
- Due to marginal conditions of E6,
it must be a ◪.
- B6 can only be a
◪ which says that A5 should be a ◩.
We can't create a valid tetromino in the region marked in dotted lines below, therefore the two cells just filled should participate in a square-tetromino.
- Considering the currently empty cells in G4-H5,
we see that F5 should be a ⬕, and those cells should make an O-tetromino with it as the only possible way.
- We cannot do anything with D7,
as it cannot be a part of any other tetromino other than an O. Therefore it will be empty.
- Therefore D6 and E5,
must be ⬕ and ⬔ respectively.
- In the region, A2-E6,
a large straight-tetromino can be formed with several other square-tetrominoes.
- Now consider the last hint; the cell with number 2.
The adjacent unfilled cells in the range E2-F4 must create a T-tetromino.
- As the last step,
You know what to do : )
Here is our finalized grid!
Also thanks to OP for providing the Penpa link.