[math-fun] "Share the Squares" dissection puzzle?
Has anyone devised a fun puzzle in which a 1-by-1, 4-by-4, 6-by-6, and 7-by-7 square are divided into smaller polyominoes which can then be reassembled to form a 2-by-2, 3-by-3, 5-by-5, and 8-by-8 square? (The title of the thread is a variation on Matt Parker's "Share the Power" puzzle, which is a generalization of this to higher powers, but without the embodiment via polyominoes.) My guess is that the best puzzle of this kind (i.e., the most challenging to solve) would be one that used a near-minimal number of pieces. I'm also seeking a polyomino implementation of the identity 0^2+3^2+5^2+6^2 = 1^2+2^2+4^2+7^2, though of course one of the squares has vanished! Jim Propp
Wonderful questions. The second question lends itself to an especially nice presentation: use all these pieces to make exactly three squares. Then use the same pieces to make exactly four squares. The best version would have a unique solution for each of these challenges. Presentation for the first question is siimilar: use all these pieces to make exactly four squares. Now find a different solution. If no one has a dissection for these challenges, then I'm going to look for a minimal one. On Mon, Jan 9, 2017 at 8:33 AM, James Propp <jamespropp@gmail.com> wrote:
Has anyone devised a fun puzzle in which a 1-by-1, 4-by-4, 6-by-6, and 7-by-7 square are divided into smaller polyominoes which can then be reassembled to form a 2-by-2, 3-by-3, 5-by-5, and 8-by-8 square?
(The title of the thread is a variation on Matt Parker's "Share the Power" puzzle, which is a generalization of this to higher powers, but without the embodiment via polyominoes.)
My guess is that the best puzzle of this kind (i.e., the most challenging to solve) would be one that used a near-minimal number of pieces.
I'm also seeking a polyomino implementation of the identity 0^2+3^2+5^2+6^2 = 1^2+2^2+4^2+7^2, though of course one of the squares has vanished!
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I hope you look for a minimal solution, Scott! (I wonder if anyone has already created software that automates the process?) I should mention that I asked Matt Parker if any of his readers had invented dissection puzzles based on the identity (n+0)^2+(n+3)^2+(n+5)^2+(n+6)^2=(n+1)^2+(n+2)^2+(n+4)^2+(n+7)^2 for specific values of n and he said no. In addition to n=0 and n=1 which I mentioned in my email, one might also consider negative values of n. n=-1: 2^2 + 4^2 + 5^2 = 3^2 + 6^2 (I like this one) n=-2: 3^2 + 4^2 = 5^2 (hello, Pythagoras!) n=-3: 3^2 + 3^2 = 1^2 + 1^2 + 4^2 (not so interesting) Here I've removed squares that occur on both sides of the equation. If one tries going beyond -3 to -4 etc., one finds that "reciprocity" sets in, and one doesn't get anything new. By the way, partitioning the set {1,2,3,4,5,6,7} into {1,2,4,7} and {3,5,6} reminds me of the numbers 24 (hours in a day), 7 (days in a week), and 365 (days in a year). I'm not sure if there's any way to exploit this coincidence, though. Jim On Mon, Jan 9, 2017 at 6:55 PM, Scott Kim <scottekim1@gmail.com> wrote:
Wonderful questions. The second question lends itself to an especially nice presentation: use all these pieces to make exactly three squares. Then use the same pieces to make exactly four squares. The best version would have a unique solution for each of these challenges. Presentation for the first question is siimilar: use all these pieces to make exactly four squares. Now find a different solution. If no one has a dissection for these challenges, then I'm going to look for a minimal one.
On Mon, Jan 9, 2017 at 8:33 AM, James Propp <jamespropp@gmail.com> wrote:
Has anyone devised a fun puzzle in which a 1-by-1, 4-by-4, 6-by-6, and 7-by-7 square are divided into smaller polyominoes which can then be reassembled to form a 2-by-2, 3-by-3, 5-by-5, and 8-by-8 square?
(The title of the thread is a variation on Matt Parker's "Share the Power" puzzle, which is a generalization of this to higher powers, but without the embodiment via polyominoes.)
My guess is that the best puzzle of this kind (i.e., the most challenging to solve) would be one that used a near-minimal number of pieces.
I'm also seeking a polyomino implementation of the identity 0^2+3^2+5^2+6^2 = 1^2+2^2+4^2+7^2, though of course one of the squares has vanished!
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I'm not sure that fewer pieces correlates with more challenging; it looks like there are some pretty simple solutions in which several of the polyominoes are just fairly large squares. On Mon, Jan 9, 2017 at 7:31 PM, James Propp <jamespropp@gmail.com> wrote:
I hope you look for a minimal solution, Scott! (I wonder if anyone has already created software that automates the process?)
I should mention that I asked Matt Parker if any of his readers had invented dissection puzzles based on the identity (n+0)^2+(n+3)^2+(n+5)^2+(n+6)^2=(n+1)^2+(n+2)^2+(n+4)^2+(n+7)^2 for specific values of n and he said no.
In addition to n=0 and n=1 which I mentioned in my email, one might also consider negative values of n.
n=-1: 2^2 + 4^2 + 5^2 = 3^2 + 6^2 (I like this one) n=-2: 3^2 + 4^2 = 5^2 (hello, Pythagoras!) n=-3: 3^2 + 3^2 = 1^2 + 1^2 + 4^2 (not so interesting)
Here I've removed squares that occur on both sides of the equation. If one tries going beyond -3 to -4 etc., one finds that "reciprocity" sets in, and one doesn't get anything new.
By the way, partitioning the set {1,2,3,4,5,6,7} into {1,2,4,7} and {3,5,6} reminds me of the numbers 24 (hours in a day), 7 (days in a week), and 365 (days in a year). I'm not sure if there's any way to exploit this coincidence, though.
Jim
On Mon, Jan 9, 2017 at 6:55 PM, Scott Kim <scottekim1@gmail.com> wrote:
Wonderful questions. The second question lends itself to an especially nice presentation: use all these pieces to make exactly three squares. Then use the same pieces to make exactly four squares. The best version would have a unique solution for each of these challenges. Presentation for the first question is siimilar: use all these pieces to make exactly four squares. Now find a different solution. If no one has a dissection for these challenges, then I'm going to look for a minimal one.
On Mon, Jan 9, 2017 at 8:33 AM, James Propp <jamespropp@gmail.com> wrote:
Has anyone devised a fun puzzle in which a 1-by-1, 4-by-4, 6-by-6, and 7-by-7 square are divided into smaller polyominoes which can then be reassembled to form a 2-by-2, 3-by-3, 5-by-5, and 8-by-8 square?
(The title of the thread is a variation on Matt Parker's "Share the Power" puzzle, which is a generalization of this to higher powers, but without the embodiment via polyominoes.)
My guess is that the best puzzle of this kind (i.e., the most challenging to solve) would be one that used a near-minimal number of pieces.
I'm also seeking a polyomino implementation of the identity 0^2+3^2+5^2+6^2 = 1^2+2^2+4^2+7^2, though of course one of the squares has vanished!
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I’m catching up - this is lovely. Has anyone posted a nice dissection yet? I tried briefly using both sets of squares to tile the plane - like the dissection proof of Pythagoras’ theorem - but it required cutting them up into smaller pieces. - Cris
On Jan 9, 2017, at 8:41 PM, Michael Collins <mjcollins10@gmail.com> wrote:
I'm not sure that fewer pieces correlates with more challenging; it looks like there are some pretty simple solutions in which several of the polyominoes are just fairly large squares.
On Mon, Jan 9, 2017 at 7:31 PM, James Propp <jamespropp@gmail.com> wrote:
I hope you look for a minimal solution, Scott! (I wonder if anyone has already created software that automates the process?)
I should mention that I asked Matt Parker if any of his readers had invented dissection puzzles based on the identity (n+0)^2+(n+3)^2+(n+5)^2+(n+6)^2=(n+1)^2+(n+2)^2+(n+4)^2+(n+7)^2 for specific values of n and he said no.
In addition to n=0 and n=1 which I mentioned in my email, one might also consider negative values of n.
n=-1: 2^2 + 4^2 + 5^2 = 3^2 + 6^2 (I like this one) n=-2: 3^2 + 4^2 = 5^2 (hello, Pythagoras!) n=-3: 3^2 + 3^2 = 1^2 + 1^2 + 4^2 (not so interesting)
Here I've removed squares that occur on both sides of the equation. If one tries going beyond -3 to -4 etc., one finds that "reciprocity" sets in, and one doesn't get anything new.
By the way, partitioning the set {1,2,3,4,5,6,7} into {1,2,4,7} and {3,5,6} reminds me of the numbers 24 (hours in a day), 7 (days in a week), and 365 (days in a year). I'm not sure if there's any way to exploit this coincidence, though.
Jim
On Mon, Jan 9, 2017 at 6:55 PM, Scott Kim <scottekim1@gmail.com> wrote:
Wonderful questions. The second question lends itself to an especially nice presentation: use all these pieces to make exactly three squares. Then use the same pieces to make exactly four squares. The best version would have a unique solution for each of these challenges. Presentation for the first question is siimilar: use all these pieces to make exactly four squares. Now find a different solution. If no one has a dissection for these challenges, then I'm going to look for a minimal one.
On Mon, Jan 9, 2017 at 8:33 AM, James Propp <jamespropp@gmail.com> wrote:
Has anyone devised a fun puzzle in which a 1-by-1, 4-by-4, 6-by-6, and 7-by-7 square are divided into smaller polyominoes which can then be reassembled to form a 2-by-2, 3-by-3, 5-by-5, and 8-by-8 square?
(The title of the thread is a variation on Matt Parker's "Share the Power" puzzle, which is a generalization of this to higher powers, but without the embodiment via polyominoes.)
My guess is that the best puzzle of this kind (i.e., the most challenging to solve) would be one that used a near-minimal number of pieces.
I'm also seeking a polyomino implementation of the identity 0^2+3^2+5^2+6^2 = 1^2+2^2+4^2+7^2, though of course one of the squares has vanished!
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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A k-tetrahedral number (k-simplicial would be better terminology) is just the analogue of a triangular or tetrahedral number but in k dimensions. The nth k-tetrahedral number is T_k(n) = (n+k-1)(n+k-2)...(n+1)n / k! or T_k(n) = (n+k-1)_C_k .* Question: Can T_k(n) be an exact kth power for k > 2 (and n > 1) ??? The question arose after a cursory check of the first few tetrahedral numbers, none of which seem to be perfect cubes. Of course, triangular numbers can be perfect squares, like 36, 1225, etc. —Dan _____________ * The formula is easily proved by induction, since e.g. the nth tetrahedral number is just a stack of the first n triangular numbers. But is there a geometrical reason that it should be the number of size-k subsets of a set of n+k-1 objects?
* The formula is easily proved by induction, since e.g. the nth tetrahedral number is just a stack of the first n triangular numbers. But is there a geometrical reason that it should be the number of size-k subsets of a set of n+k-1 objects?
Yes, the nth k-simplicial number counts the number of (k+1)-tuples: (a_0, a_1, a_2, ..., a_k) where each a_i is in {0, 1, 2, ...} and the sum is exactly n-1. [This is using the standard embedding of a regular k-simplex as the convex span of (k+1) orthonormal vectors.] Now, for instance, map the tuple (3, 2, 7, 4) to the string: ooo|oo|ooooooo|oooo which contains k copies of '|' and n-1 copies of 'o'. The result follows. Best wishes, Adam P. Goucher
One might also consider dissections in which the pieces are not polyominoes; some of these might make better puzzles than polyomino dissections. Jim On Mon, Jan 9, 2017 at 7:31 PM, James Propp <jamespropp@gmail.com> wrote:
I hope you look for a minimal solution, Scott! (I wonder if anyone has already created software that automates the process?)
I should mention that I asked Matt Parker if any of his readers had invented dissection puzzles based on the identity (n+0)^2+(n+3)^2+(n+5)^2+(n+6)^2=(n+1)^2+(n+2)^2+(n+4)^2+(n+7)^2 for specific values of n and he said no.
In addition to n=0 and n=1 which I mentioned in my email, one might also consider negative values of n.
n=-1: 2^2 + 4^2 + 5^2 = 3^2 + 6^2 (I like this one) n=-2: 3^2 + 4^2 = 5^2 (hello, Pythagoras!) n=-3: 3^2 + 3^2 = 1^2 + 1^2 + 4^2 (not so interesting)
Here I've removed squares that occur on both sides of the equation. If one tries going beyond -3 to -4 etc., one finds that "reciprocity" sets in, and one doesn't get anything new.
By the way, partitioning the set {1,2,3,4,5,6,7} into {1,2,4,7} and {3,5,6} reminds me of the numbers 24 (hours in a day), 7 (days in a week), and 365 (days in a year). I'm not sure if there's any way to exploit this coincidence, though.
Jim
On Mon, Jan 9, 2017 at 6:55 PM, Scott Kim <scottekim1@gmail.com> wrote:
Wonderful questions. The second question lends itself to an especially nice presentation: use all these pieces to make exactly three squares. Then use the same pieces to make exactly four squares. The best version would have a unique solution for each of these challenges. Presentation for the first question is siimilar: use all these pieces to make exactly four squares. Now find a different solution. If no one has a dissection for these challenges, then I'm going to look for a minimal one.
On Mon, Jan 9, 2017 at 8:33 AM, James Propp <jamespropp@gmail.com> wrote:
Has anyone devised a fun puzzle in which a 1-by-1, 4-by-4, 6-by-6, and 7-by-7 square are divided into smaller polyominoes which can then be reassembled to form a 2-by-2, 3-by-3, 5-by-5, and 8-by-8 square?
(The title of the thread is a variation on Matt Parker's "Share the Power" puzzle, which is a generalization of this to higher powers, but without the embodiment via polyominoes.)
My guess is that the best puzzle of this kind (i.e., the most challenging to solve) would be one that used a near-minimal number of pieces.
I'm also seeking a polyomino implementation of the identity 0^2+3^2+5^2+6^2 = 1^2+2^2+4^2+7^2, though of course one of the squares has vanished!
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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
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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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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-- Forewarned is worth an octopus in the bush.
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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-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Nice puzzle. I'll bite with a solution: the integers with this property are exactly all the numbers that are the sum of two squares. I have a proof that for all sums of two squares one can construct the dissection, but I'm missing the other way around. Best, Sébastien On 10 January 2017 at 20:26, Dan Asimov <asimov@msri.org> wrote:
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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-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I like this, too. As usual, I'd find it even more elegant to see a minimal geometric dissection of a small collection of integer-sided square *tori* into another collection having the same total area. For equal sums of nth powers, the same question for cubical n-tori. E.g., ----- What is the minimal dissection of cubical 3-tori of sides 3, 4, 5 into one of side 6 ? ----- —Dan
On Jan 9, 2017, at 8:33 AM, James Propp <jamespropp@gmail.com> wrote:
Has anyone devised a fun puzzle in which a 1-by-1, 4-by-4, 6-by-6, and 7-by-7 square are divided into smaller polyominoes which can then be reassembled to form a 2-by-2, 3-by-3, 5-by-5, and 8-by-8 square?
(The title of the thread is a variation on Matt Parker's "Share the Power" puzzle, which is a generalization of this to higher powers, but without the embodiment via polyominoes.)
My guess is that the best puzzle of this kind (i.e., the most challenging to solve) would be one that used a near-minimal number of pieces.
I'm also seeking a polyomino implementation of the identity 0^2+3^2+5^2+6^2 = 1^2+2^2+4^2+7^2, though of course one of the squares has vanished!
participants (10)
-
Adam P. Goucher -
Cris Moore -
Dan Asimov -
Fred Lunnon -
James Propp -
Michael Collins -
Michael Kleber -
Scott Kim -
Seb Perez-D -
Veit Elser