**Definition:** Two finite sets of polygons $A$ and $B$ are congruent if we can match polygons in $A$ in a one-one manner with polygons in $B$ with each matched pair of polygons mutually congruent.

**Question:** For every integer $n$, can every polygon be partitioned into $n$ sets of polygons such that the sets are congruent to one another?

**Remarks:** For $n =2$, the answer is yes. Indeed, one triangulates the input $m$-gon, then cuts each triangle into 3 kite polygons (https://en.wikipedia.org/wiki/Kite_(geometry)) that meet at a common vertex at the incenter of the triangle, thus resulting in a total of $\sim 3m$ kites. Then we cut each kite into 2 mutually congruent triangles and send each triangle into one of the output sets. This approach results in 2 congruent sets of polygons (indeed, triangles) with $\sim 3m$ elements each. For $n=4$, one can take the $n=2$ solution and divide each triangle piece in each set via kites into 2 congruent sets of 3 triangles each thus resulting in 4 mutually congruent sets with $\sim 9m$ triangles each. This approach should work for all powers of 2 values of $n$. *A natural question here is whether we can manage with less number of pieces in each congruent set.*

**Guess:** For other values of $n$, including even 3, the answer may be "not always". I have no proof for this.