These dissection problems remind me of the lovely puzzle: For which integers n in Z+ does there exist a right triangle R(n) that can be dissected into n congruent copies of a triangle similar to R(n) ? —Dan
On Jan 10, 2017, at 7:51 AM, Michael Kleber <michael.kleber@gmail.com> wrote:
Bob Wainwright's "Partridge Problem" in general asks for a dissection-based proof of 1^3 + 2^3 + 3^3 + ... + n^3 = (1+2+...+n)^2, using appropriately scaled copies of a single polygon for whatever n you can find a solution for. As with Jim's original dissection question, the squaring comes from scaling up; the cubing comes from both scaling and by increasing multiplicity (one thing of size 1, two things of size 2, ..., n things of size n).
Veit is referring to Patrick Hamlyn's discovery that the 30-60-90 triangle has a Partridge dissection with n=4. You can see a picture of it on Eric Friedman's page http://www2.stetson.edu/~efriedma/mathmagic/0802.html if you search for "30-60-90".
I'll take this opportunity to remind everyone that there's a great open question here: Does there exist any polygon with partridge number 2? That is, is there any polygon P such that one copy of P and two double-sized copies 2P can be used to tile a triple-sized copy 3P?
--Michael
On Tue, Jan 10, 2017 at 10:00 AM, Cris Moore <moore@santafe.edu> wrote:
But is there a nice picture of this dissection somewhere?
- Cris
On Jan 10, 2017, at 7:34 AM, Veit Elser <ve10@cornell.edu> wrote:
My favorite dissection of this kind is the proof of
1^3 + 2^3 + 3^3 + 4^3 = (1+2+3+4)^2
that I learned about in this forum (all tiles are 30-60-90 triangles).
-Veit
On Jan 9, 2017, at 11:19 PM, James Propp <jamespropp@gmail.com> wrote:
One might also consider dissections in which the pieces are not polyominoes; some of these might make better puzzles than polyomino dissections.
Jim
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