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YCor
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Partition of Polygonspolygons into 'Congruent Sets'congruent sets of Polygons'polygons'

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

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

Remarks: For n =2$n =2$, the answer is yes. Indeed, one triangulates the input m$m$-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 ~ 3m$\sim 3m$ 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 ~3m$\sim 3m$ elements each. For n=4$n=4$, one can take the n=2$n=2$ 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 ~9m$\sim 9m$ triangles each. This approach should work for all powers of 2 values of n$n$. A natural question here is whether we can manage with less number of pieces in each congruent set.

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

Partition of Polygons into 'Congruent Sets of Polygons'

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 ~ 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 ~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 ~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.

Partition of polygons into 'congruent sets of polygons'

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.

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Martin Sleziak
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Nandakumar R
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Partition of Polygons into 'Congruent Sets of Polygons'

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 ~ 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 ~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 ~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.