two dissection problems (xpost from geometry.puzzles)
vanvera
but I just discovered this group, and I think the problem
can be interesting also here:
It is well known that it is possible to
cut a square in 7 different rectangles with the same
area. This is known as Blanche's Dissections.
see <http://mathworld.wolfram.com/BlanchesDissection.html>
(I was not able to get the original paper appeared on Eureka in 1971,
but I quite easily checked that the solution is unique
apart usual symmetries and scaling) and other solutions can be
obtained for 8 different rectangles. Unfortunately these solutions
have all non-integer sides.
My questions are similar, but I do not know the answers.
(1) Is it possible to cut a square in different rectangles with
the same perimeter ?
(2) Is it possible to cut an square in different rectangles with
the same area and INTEGRAL sides ?
bye,
g.
William Elliot
> (1) Is it possible to cut a square in different rectangles with
> the same perimeter ?
>
Huh? I'll read that as 'into different ...'
Sure, cut it up into smaller squares.
> (2) Is it possible to cut an square in different rectangles with
> the same area and INTEGRAL sides ?
>
Huh? I'll read that as 'into different ...'
No. What if square is 1/2 by 1/2 ?
vanvera
> On Sun, 19 Nov 2006, vanvera wrote:
>
>
>>(1) Is it possible to cut a square in different rectangles with
>>the same perimeter ?
>>
>
> Huh? I'll read that as 'into different ...'
> Sure, cut it up into smaller squares.
I beg your pardon for my bad use of english. It's not my language.
(I erroneously thought that the reference to Blanche's dissection was enough
to disambiguate my words).
What I mean is that the rectangles into which the square is divided
must be all different in size.
That is, if there are k rectangles with sides (x_1,y_1),
(x_2, y_2) .... (x_k, y_k) then we must NOT have
(x_i=x_j and y_i=y_j) or (x_i=y_j and y_i=x_j)
when i != j.
>>(2) Is it possible to cut an square in different rectangles with
>>the same area and INTEGRAL sides ?
>>
>
> Huh? I'll read that as 'into different ...'
> No. What if square is 1/2 by 1/2 ?
Again, I'll try to be more clear.
Does exist a square with integer sides which can be cutted into
rectangles (which have all different shapes, as above) but
have equal area and integer sides ?
g.
William Elliot
> William Elliot wrote:
> > On Sun, 19 Nov 2006, vanvera wrote:
> >
> >>(1) Is it possible to cut a square in different rectangles with
> >>the same perimeter ?
> >
> > Huh? I'll read that as 'into different ...'
> > Sure, cut it up into smaller squares.
>
> I beg your pardon for my bad use of english. It's not my language.
> (I erroneously thought that the reference to Blanche's dissection was enough
> to disambiguate my words).
>
post is better English than many speakers whose first language is
English. Indeed, yours is much preferable to theirs.
> What I mean is that the rectangles into which the square is divided
> must be all different in size.
> That is, if there are k rectangles with sides (x_1,y_1),
> (x_2, y_2) .... (x_k, y_k) then we must NOT have
>
> (x_i=x_j and y_i=y_j) or (x_i=y_j and y_i=x_j)
> when i != j.
>
Ok, you have square (0,0), (0,a), (a,0), (a,a)
Divide square into rectangles with corners
(0,0), (0,a), (a/3, 0), (a/3,a)
and
(a/3,0), (a/3,a), (a,0), (a,a).
> >>(2) Is it possible to cut an square in different rectangles with
> >>the same area and INTEGRAL sides ?
> >
> > Huh? I'll read that as 'into different ...'
> > No. What if square is 1/2 by 1/2 ?
>
> Again, I'll try to be more clear.
Excellent, you understand and appreciate the need for precision in
mathematics and geometry.
> Does exist a square with integer sides which can be cutted into
> rectangles (which have all different shapes, as above) but
> have equal area and integer sides ?
>
What if square is 1 x 1 ?
vanvera
>
> Ok, you have square (0,0), (0,a), (a,0), (a,a)
> Divide square into rectangles with corners
> (0,0), (0,a), (a/3, 0), (a/3,a)
> and
> (a/3,0), (a/3,a), (a,0), (a,a).
It seems to me that those rectangles do not have the same
perimeter (this was a condition of problemi (1) ).
>>Does exist a square with integer sides which can be cutted into
>>rectangles (which have all different shapes, as above) but
>>have equal area and integer sides ?
>>
>
> What if square is 1 x 1 ?
Well, I'm looking for less trivial solutions (if any),
or a proof that the problem does not admit a non-trivial solution.
g.
Mark Spahn
"vanvera" <vanve...@CUTgmail.com> wrote in message
news:456055de$0$13765$4faf...@reader3.news.tin.it...
Here's a partial answer to your question, Is there a square of
integer side that can be partitioned into integer-side rectangles
that all have the same area but have all different shapes?
Express the area of each rectangle (all of them have the
same area) as a product of primes, such as 2^3*3*7^2 = 1176.
We can count how many integer-side rectangles of this area
there can be, by multiplying together the exponents plus 1:
(3+1)*(1+1)*(2+1) = 4*2*3 = 24 integer factors of 1176.
So there are 24 different rectangles of area 1176:
1x1176/1, 2x1176/2, 3x1176/3, 4x1176/4, 6x196, ..., 24x49.
How many such rectangles can be in the square?
With n such rectangles, their total area is 1176n, which
must be a square. The least such n is n = 2*3 = 6,
because 2*3 * 1176 = 2^4*3^2*7^2 = 7056 = 84^2.
Can we pick 6 of these 24 rectangles and fit them together
to form an 84-by-84 square?
Well, that's just some preliminary doodling...
-- Mark Spahn (West Seneca, NY, USA)
Patrick Hamlyn
>Ok, you have square (0,0), (0,a), (a,0), (a,a)
>Divide square into rectangles with corners
> (0,0), (0,a), (a/3, 0), (a/3,a)
>and
> (a/3,0), (a/3,a), (a,0), (a,a).
OP already posted that he has solutions for k=7 and k=8 for the area problem,
and now he is looking for solutions to the similar problem where perimeters are
equal, so clearly solutions to the area problem with smaller k are not of use.
His English, and his description, were plenty good enough for you to discern
this from his first post. The second post clarified the area problem and assumed
you were smart enough to extrapolate to the perimeter problem, clearly an
unfounded assumption.
>> >>(2) Is it possible to cut an square in different rectangles with
>> >>the same area and INTEGRAL sides ?
>> >
>> > Huh? I'll read that as 'into different ...'
>> > No. What if square is 1/2 by 1/2 ?
>>
>> Again, I'll try to be more clear.
>
>Excellent, you understand and appreciate the need for precision in
>mathematics and geometry.
>
>> Does exist a square with integer sides which can be cutted into
>> rectangles (which have all different shapes, as above) but
>> have equal area and integer sides ?
>>
>What if square is 1 x 1 ?
You answer
"Does [there] exist a square"
with
"What if [the] square is 1 x 1 ?"
Clearly you require much smaller, baby steps in logic spoonfed to you in order
to get this.
OP wants you to find an integer-sided square, you pick the size, such that you
can divide it into distinct rectangles with integer sides and equal perimeters.
To further clarify, for you alone (nobody else would need it): The rectangles
are to be distinct from each other. Their perimeters are to be equal, but they
don't need to be equal to the perimeter of the square you picked.
Patrick Hamlyn
Jeeez guys, he's looking for an answer to the *perimeter* problem. Read the
first post again.
This problem lends itself to solution by brute force, but it'll take an hour or
two to set up my tiling program with the requisite inputs, and if there's a
relatively simple solution out there someone will probably find it in minutes.
Patrick Hamlyn
I ran this through my tiler with the addition of a short front end which lists
rectangles of equal perimeter and finds all combinations of them with square
area, tried it out starting perimeter 20 and immediately it found:
CCCCCCCCC
CCCCCCCCC
DDDDDBBBB
DDDDDBBBB
DDDDDBBBB
DDDDDBBBB
DDDDDBBBB
DDDDDBBBB
AAAAAAAAA
And then I fixed the bug which didn't reset the rectangle count for successive
perimeters (the above has both perimeter 20 and 22 rectangles).
Now it finds nothing below perimeter 52, but I'll leave it running all day.
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Mark Spahn
"Patrick Hamlyn" <path@multipro.N_OcomSP_AM.au> wrote in message
news:int1m2563v7iajar3...@4ax.com...
Uh, sorry, but I was responding to the original poster's *second* problem:
(2) Is it possible to cut an square in different rectangles with
the same area and INTEGRAL sides ?
To make this clear, I quoted his second problem, reworded as:
Is there a square of integer side that can be partitioned into
integer-side rectangles that all have the same area but have
all different shapes?
Sorry I didn't make that as clear as I thought I did.
I see now that the ambiguity comes in my wording
"your question, Is there ... different shapes?"
Since the original poster asked two different questions,
I should have said
"your second question, Is there ... different shapes?"
-- Mark Spahn
Patrick Hamlyn
>Uh, sorry, but I was responding to the original poster's *second* problem:
> (2) Is it possible to cut an square in different rectangles with
> the same area and INTEGRAL sides ?
>To make this clear, I quoted his second problem, reworded as:
> Is there a square of integer side that can be partitioned into
> integer-side rectangles that all have the same area but have
> all different shapes?
>Sorry I didn't make that as clear as I thought I did.
>I see now that the ambiguity comes in my wording
>"your question, Is there ... different shapes?"
>Since the original poster asked two different questions,
>I should have said
>"your second question, Is there ... different shapes?"
>-- Mark Spahn
Your post was clear enough... my comprehension was lacking in this instance.
Looks like I was solving problem 3 "the unstated problem":
(3) Is it possible to cut a square into two or more distinct integral-sided
rectangles, all of which have the same perimeter.
(My answer so far: not with rectangles smaller than perimeter 62)
Patrick Hamlyn
>OP wants you to find an integer-sided square, you pick the size, such that you
>can divide it into distinct rectangles with integer sides and equal perimeters.
Doh... correction:
OP wants you to find an integer-sided square, you pick the size, such that you
can divide it into distinct rectangles with integer sides and equal areas.
Giovanni Resta
First of all, thank you all for the attention.
(I'm the OP (vanvera) writing from another account)
Yes, there are two problems.
Both regards the *nontrivial* dissection of integer-sided squares into
integer-sided rectangles, with the condition that the rectangles have
different shapes.
The two problems then differentiate asking
1) the rectangles have all the *same perimeter*,
2) the rectangles have all the *same area*.
The case (2), i.e, equal area, has been already solved with a minimum of
7 rectangles by Blanche Descartes, relaxing the condition on integer
sides.
Since I like integer numbers, this is not completely satisfactory...
I've a little of experience myself with tilers (nothing compared to
yours, I just wrote some programs for some specificic problems, like:
<http://www.imc.pi.cnr.it/resta/Poly/>
and <http://www.imc.pi.cnr.it/resta/rect/> )
but in this case, before writing an ad hoc tiler, I've looked
at the problem from an analytical point of view.
Using Mathematica I've fixed a tiling with 7 or 8 rectangles
and in both cases (perimeter and area) I've tried to solve
the equations arising from the conditions.
The process is quite tiresome, since I've not automatized it,
so, if for example I have 8 rectangles, then I have the 16
variables (the sides of the rectangles), and then, according to
the dwawing of the dissection I choose, I have to manually enter
the equations, and then help Mathematica a little.
In the case of the equal area I've confirmed (I think so, I do not have
the original paper) the result of Blanche Descartes, i.e., 7 rectangles
is the minimum. Unfortunately the solution (apart scaling and
symmetries) is unique and it is not rational. The same holds for the 3
different tilings with 8 rectangles that I've tried. In all the cases
I obtained a nonlinear system of equations which in pratice admits
an unique (irrational) solution, so there is no space to, say, vary a
parameter to obtain an infinite family of solutions among which
hopefully one can find a rational one.
Probably I'll try to solve the problem of equal area for some larger
tilings, but if the solution remains sustantially unique, then the
chance it is rational is quite scarce...
I just wonder if in case of tilings with a large enough number of
rectangles the problem start to have an infinite number of solutions
(of course discarting those that are just multiples).
Due to the difficulty of this problem I turned to the one
of rectangles of equal perimeter. In that case, since the equations
governing the problem are all linear, I did know that if a solution
exists then surely it exists also a rational solution and thus an
integer solution.
Unfortunately the same approach with Mathematica did not lead to
any solution for up to 8 rectangles.
Indeed I found some isoperimetral tilings, but unfortunately
the rectangles have not distinct shapes, like in the case
of a central square surrounded by 4 equal rectangles:
AAAB
CDDB
CDDB
CEEE
The fact that your tiler is running wihout finding solutions
(at least for the moment) confirms my suspect that this problem
if ever has a solution, it is not that trivial.
Thank you again for the attention,
giovanni
Patrick Hamlyn
The problem of equal areas looks very unlikely to yield integral tilings. There
are probably too few rectangles of a given area to give any solutions.
I tried up to area 3200, the next possible area (3456) defeated my program (out
of memory) since it is optimised for smaller pieces and is a bit wasteful of
memory.
My quick & dirty front-end program found these areas to pass to the tiler. Bugs
excluded, there shouldn't be any others under 9900.
Axx is area of rectangles
Sxx is square size (side, not area)
Nx is number of lists of rectangles which would be passed to the tiling program
last number on each line is the number of rectangles in each list
So the first line says for area 72 there are only two rectangles, and the square
would have side 12. (the rectangles would obviously be 8x9 and 6x12). Obviously
any list of only two rectangles is a non-starter, but I left them in for
completeness.
The line starting A720 indicates that for rectangles of area 720 there are six
sets of five rectangles each which will have the correct area for a square of
side 60 (ie there are six rectangles of area 720 with no side greater than 60,
so there are six ways of leaving out one of them).
This list is incredibly sparse compared to the equivalent problem for equal
perimeters.
A72, S12, N1 2
A144, S24, N1 4
A288, S24, N1 2
A300, S30, N1 3
A432, S36, N1 3
A450, S30, N1 2
A576, S48, N1 4
A648, S36, N1 2
A720, S60, N6 5
A800, S40, N1 2
A900, S60, N5 4
A1008, S84, N1 7
A1152, S48, N1 2
A1200, S60, N4 3
A1296, S72, N1 4
A1568, S56, N1 2
A1600, S80, N1 4
A1620, S90, N1 5
A1728, S72, N4 3
A1764, S84, N1 4
A1800, S120, N1 8
A1800, S60, N3 2
A2304, S96, N1 4
A2352, S84, N1 3
A2400, S120, N7 6
A2450, S70, N1 2
A2592, S72, N1 2
A2700, S90, N4 3
A2880, S120, N21 5
A3136, S112, N1 4
A3200, S80, N1 2
A3456, S144, N1 6
A3528, S84, N3 2
A3600, S180, N10 9
A3600, S120, N35 4
A3888, S108, N1 3
A3920, S140, N1 5
A4032, S168, N8 7
A4050, S90, N3 2
A4356, S132, N1 4
A4500, S150, N1 5
A4608, S96, N1 2
A4704, S168, N1 6
A4800, S120, N10 3
A4900, S140, N1 4
A5184, S144, N5 4
A5292, S126, N4 3
A5400, S180, N28 6
A5760, S240, N1 10
A5808, S132, N1 3
A5832, S108, N1 2
A6272, S112, N1 2
A6300, S210, N120 7
A6400, S160, N1 4
A6480, S180, N56 5
A6912, S144, N4 3
A7056, S252, N1 9
A7056, S168, N35 4
A7200, S240, N165 8
A7200, S120, N6 2
A7350, S210, N1 6
A7500, S150, N1 3
A7938, S126, N1 2
A8100, S270, N1 9
A8100, S180, N35 4
A8712, S132, N3 2
A8820, S210, N56 5
A9072, S252, N36 7
A9216, S192, N1 4
A9408, S168, N4 3
A9600, S240, N28 6
A9800, S140, N1 2
A9900, S330, N12 11
For comparison, here is the (complete) data for perimeter 64 and 66 rectangles.
In this case the lists of rectangles are of varying lengths due to their varying
areas. The actual rectangles are omitted here as before.
The first line tells us that 15 of the rectangles of perimeter 64 have the
correct area to make a square with side 51. The second line tells us there are
36 potential ways to put together perimeter-64 rectangles to make a square of
side 41, and there are from 8 to 11 rectangles in each list.
Unless there's an analytical reason waiting to be discovered which prevents an
integer solution, this looks a lot more promising than the equal-area problem,
although progress is now fairly slow, taking hours for each perimeter-size.
P64, S51, N1 15
P64, S41, N36 11 11 10 10 10 10 10 9 10 10 10 10 10 10 9 9 9 9 9 9 9 9 9 9 9 9 9
9 8 9 9 8 8 8 8 8
P64, S38, N94 10 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 8 8 8 8 8 9 9 9 9 9 8 8 8
8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 7 9 9 9 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8
7 7 8 8 8 8 8 8 8 8 7 7 7 7 8 7 7 7 7 7 7
P64, S44, N36 12 11 11 11 11 11 10 11 11 10 10 10 11 11 11 10 10 10 10 10 10 10
10 10 10 9 10 10 10 10 10 10 9 9 9 9
P64, S47, N7 13 12 12 12 12 12 11
P64, S39, N60 10 10 10 9 9 9 9 8 9 9 8 8 8 8 9 9 9 9 9 9 9 8 8 8 9 8 8 8 8 8 8 8
8 8 8 8 8 8 8 8 7 8 8 8 8 8 8 8 8 8 8 7 7 7 7 7 7 7 7 7
P64, S36, N89 9 9 9 9 9 9 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 7 7 8 8 8 8 8 8
8 7 7 7 8 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 7 7 8 7 7 7 7 7 7 7 7 7 7 7
7 7 7 7 7 7 7 7 6 7 7 7 6 6 6 6
P64, S33, N62 8 8 8 7 7 7 7 7 7 7 7 7 7 7 6 7 7 7 7 7 6 7 6 6 6 6 6 6 6 6 6 6 6
7 7 7 7 7 7 7 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 5 5 5 5 5 5 5
P64, S45, N11 12 12 11 11 11 11 11 10 10 10 9
P64, S40, N79 10 10 10 10 10 10 10 10 9 9 10 10 10 10 9 9 9 9 9 9 9 9 9 9 10 9 9
9 9 9 9 9 9 9 9 9 9 9 8 8 8 8 10 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 8 8 8 8 9 9
8 8 8 8 8 8 8 8 8 8 8 7
P64, S34, N61 8 8 8 8 8 8 8 7 7 8 8 8 7 7 7 7 7 7 7 8 7 7 7 7 7 7 7 7 7 7 7 7 7
7 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 6 6 6 6 6 6 6 6 6 6 6 6
P64, S37, N60 9 9 9 9 9 9 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 7 7 7 8 8 8 8 8 8
8 8 8 8 8 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 6 6 6
P64, S43, N28 11 11 11 11 11 11 10 10 11 11 10 10 10 10 10 10 9 10 10 10 10 9 9
9 9 9 9 9
P64, S31, N48 7 7 7 7 7 6 6 6 6 6 6 6 6 6 6 6 6 6 5 5 5 5 6 6 6 6 6 6 6 6 5 5 6
5 5 5 5 5 5 5 5 5 5 5 5 5 5 4
P64, S35, N63 8 8 8 8 8 8 8 8 8 8 7 7 7 7 8 8 7 7 7 7 7 7 7 7 7 6 8 7 7 7 7 7 7
7 7 7 7 7 6 6 7 7 7 7 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 5
P64, S32, N30 7 7 7 7 7 6 6 7 7 7 7 6 6 6 6 6 6 7 6 6 6 6 6 6 6 6 5 5 5 5
P64, S46, N14 12 12 11 11 12 11 11 11 11 11 11 10 10 10
P64, S42, N62 10 10 10 10 10 10 10 10 9 10 10 9 10 9 9 9 9 10 10 10 10 10 10 10
10 10 9 9 9 10 10 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 8 9 9 9 9 9 9 8 8 8 8 8 8 8
P64, S49, N3 13 13 12
P64, S48, N1 13
P64, S30, N7 6 5 5 5 5 5 5
P64, S29, N12 5 5 5 5 5 5 5 4 5 4 4 4
P64, S27, N1 4
P66, S42, N114 11 11 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 9 9 10 10
10 10 10 10 10 10 10 9 9 10 10 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 8 8 10 10
10 10 10 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 8 8 8 8 9 9 9 9 9 9 9 9
9 9 9 9 8 8 9 8 8 8 8 8 8 8 8 8 8 8 8
P66, S48, N12 13 13 13 12 12 12 12 12 12 12 12 11
P66, S36, N98 9 9 9 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 7 7 7 7 8 8 8
8 8 8 7 7 7 7 7 7 7 7 7 7 7 7 7 6 8 8 8 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 6 7 7 7 7
7 7 7 6 6 6 6 6 6 7 6 6 6 6 6 6 6 6 6 6 6 6 6 6 5
P66, S40, N132 10 10 10 10 10 10 10 10 10 9 9 9 9 9 9 10 9 9 9 9 9 9 9 9 9 9 9 9
9 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 8 8 8 9 9 9 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 9
9 9 9 9 9 9 9 9 9 8 8 8 8 8 9 9 8 8 8 8 8 8 8 8 8 8 8 8 8 9 8 8 8 8 8 8 8 8 8 8
8 8 8 8 8 8 8 7 8 8 8 8 8 7 7 7 7 7 7 7 7 7 7 7
P66, S46, N41 12 12 12 12 12 12 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11
10 10 11 11 11 11 11 11 11 11 11 11 10 10 10 11 10 10 10
P66, S44, N71 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 10 10 10 10 10
10 11 11 11 11 10 10 10 10 10 10 10 10 10 10 10 10 10 9 9 11 11 11 10 10 10 10
10 10 10 10 10 10 10 10 10 9 10 10 9 9 9 9 9 9 9 9 9 8
P66, S34, N94 8 8 8 8 8 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7
7 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 6 6 7 7 6 6 6 6 6 6 6 6 6 6 6 6
7 6 6 6 6 6 6 6 6 6 6 6 6 6 5 6 5 5 5 5 5
P66, S38, N132 9 9 9 9 9 9 9 9 9 9 9 9 8 8 8 9 9 8 8 8 8 8 8 8 8 8 8 9 8 8 8 8 8
8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 7 8 8 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8
8 8 8 8 8 8 8 8 8 7 8 8 8 8 7 7 7 8 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 7 7 7 7 7 7 7
7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 6 6
P66, S32, N58 7 7 7 7 7 7 7 7 7 7 7 7 7 6 6 6 6 6 6 7 7 6 6 6 6 6 6 6 6 7 6 6 6
6 6 6 6 6 6 5 5 6 5 5 5 5 5 5 5 5 5 5 5 5 5 4 4 4
P66, S50, N6 13 12 13 12 12 12
P66, S52, N1 13
P66, S30, N20 5 5 5 5 5 5 5 5 5 4 4 5 4 4 4 4 4 4 4 4
P66, S28, N4 4 4 3 3
P66, S26, N2 3 3
Patrick Hamlyn
Here is the most promising set of rectangles I've found for the equal-area
problem:
420 x 30
360 x 35
350 x 36
315 x 40
300 x 42
280 x 45
252 x 50
225 x 56
210 x 60
200 x 63
180 x 70
175 x 72
168 x 75
150 x 84
140 x 90
126 x 100
120 x 105
All have area 12600
Any 14 of these 17 rectangles have the same area as a 420 x 420 square.
However I believe a tiling is unlikely.
Patrick Hamlyn
>Both regards the *nontrivial* dissection of integer-sided squares into
>integer-sided rectangles, with the condition that the rectangles have
>different shapes.
>
>The two problems then differentiate asking
>1) the rectangles have all the *same perimeter*,
For this one, if there is a solution the perimeter can't be less than 72.
vanvera
>>The two problems then differentiate asking
>>1) the rectangles have all the *same perimeter*,
>
>
> For this one, if there is a solution the perimeter can't be less than 72.
Thanks, you saved me a lot of programming and computing time...
I should propose easier problems next time...
ciao,
g.
jasen
> On Sun, 19 Nov 2006, vanvera wrote:
>
>> (1) Is it possible to cut a square in different rectangles with
>> the same perimeter ?
>>
> Huh? I'll read that as 'into different ...'
> Sure, cut it up into smaller squares.
I think He means "into dissimilar ..."
Bye.
Jasen