, 2022

Solution:

First note that only triangles and quadrilaterals are possible.
There are $3$ possibilities:
- $1/2$ by $1/2$ right isosceles triangles
- $1$ by $1/8$ rectangles
- $1/2$ by $1/4$ rectangles

The first case has $16$ possibilities (there are $2$ choices for the orientation of each quadrant).
The second case has $2$ possibilities (either all horizontal or all vertical).

The third case is the trickiest. Label the quadrants $A, B, C, D$ where $A, B$ are at the top and $B, C$ are on the left. If each rectangle lies completely within a quadrant, there are $16$ ways. If rectangles span quadrants $A, B$ but not $C$ or $D$, there are $4$ ways. Similarly, there are $4$ ways each for [rectangles spanning $B, C$ but not $D, A$], [rectangles spanning $C, D$ but not $A, B$], and [rectangles spanning $D, A$ but not $B, C$]. Next, if rectangles span both $A, B$ and $C, D$, there is $1$ way, and if rectangles span both $B, C$ and $D, A$ there is $1$ way. Finally there are $2$ ways for each adjacent pair of quadrants to have a rectangle spanning them. This brings us to $16+4+4+4+4+1+1+2=36$ ways.

The final answer is $16+2+36=54$.