Another Question on congruent partitions
A few days back, I put up a question at mathoverflow.
Definition: A perfect congruent partition of a planar region C is a partition of it into some finite number n of pieces that are all mutually congruent (any piece can be transformed into another piece by an isometry. We consider only cases where each piece is connected and is bounded by a simple curve) and with no part of C left out (not part of any piece).
Question: If from a convex planar region C, 2 connected, mutually congruent pieces (pieces need not be convex) can be cut such that no point of the boundary of C is left out, then will C always allow a perfect congruent partition into 2 pieces (with no point on the entire C left out)? I can't think of a counterexample or proof.
Note: The question can be generalized from 2 pieces to n pieces.
Partial Answer: For n >=5, the answer is "not necessarily".
Example for n = 5 (Obviously generalizable to higher values of n). Consider a square S and a slightly smaller square S' got by insetting S. C is a convex region got by replacing each edge of S with an outward bulging arc of very large radius - ie very close to a line segment.
Now, consider cutting C into 5 mutually congruent pieces. It is pretty much obvious that no perfect congruent partition of C into 5 pieces is possible. It is also easily seen that the entire boundary of C can be 'used' by dividing that part of C between its boundary and S' (S' obviously is in the interior of C) into 4 mutually congruent thin pieces. What remains of C after these 4 pieces are cut is the interior of S'. There is more than enough space there to cut another thin piece congruent to the 4 already cut. So, we can have 5 mutually congruent pieces that can be cut from C that together use the entire boundary of C (in fact, only 4 of the 5 can do it) but we cannot cut C entirely into 5 mutually congruent pieces.
This simple construction generalizes to n =6 (just start with a regular pentagon instead of square) but does not seem to give a clue for n =4,3,2.
Definition: A perfect congruent partition of a planar region C is a partition of it into some finite number n of pieces that are all mutually congruent (any piece can be transformed into another piece by an isometry. We consider only cases where each piece is connected and is bounded by a simple curve) and with no part of C left out (not part of any piece).
Question: If from a convex planar region C, 2 connected, mutually congruent pieces (pieces need not be convex) can be cut such that no point of the boundary of C is left out, then will C always allow a perfect congruent partition into 2 pieces (with no point on the entire C left out)? I can't think of a counterexample or proof.
Note: The question can be generalized from 2 pieces to n pieces.
Partial Answer: For n >=5, the answer is "not necessarily".
Example for n = 5 (Obviously generalizable to higher values of n). Consider a square S and a slightly smaller square S' got by insetting S. C is a convex region got by replacing each edge of S with an outward bulging arc of very large radius - ie very close to a line segment.
Now, consider cutting C into 5 mutually congruent pieces. It is pretty much obvious that no perfect congruent partition of C into 5 pieces is possible. It is also easily seen that the entire boundary of C can be 'used' by dividing that part of C between its boundary and S' (S' obviously is in the interior of C) into 4 mutually congruent thin pieces. What remains of C after these 4 pieces are cut is the interior of S'. There is more than enough space there to cut another thin piece congruent to the 4 already cut. So, we can have 5 mutually congruent pieces that can be cut from C that together use the entire boundary of C (in fact, only 4 of the 5 can do it) but we cannot cut C entirely into 5 mutually congruent pieces.
This simple construction generalizes to n =6 (just start with a regular pentagon instead of square) but does not seem to give a clue for n =4,3,2.
posted by R.Nandakumar @ 9:43 AM
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