From Newsgroup: rec.puzzles

Taken from 'Amusements in Mathematics' by Henry Ernest Dudeney.

Here is a little cutting out poser. I take a strip of paper, measuring
five inches by one inch, and, by cutting it into five pieces, the parts
fit together and form a square, as shown in this illustration [*]. Now
it is quite an interesting puzzle to discover how we can do this in only
four pieces.

[*] The illustration is available at the following link.

https://archive.org/details/amusementsinmath00dude/page/37/mode/1up

If you don't have access to this, I shall describe the arrangement...

The illustration shows the 5 x 1 rectangle cut into five parts: two 2 x
1 rectangles and one 1 x 1 square. The rectangles are further divided
into two parts, each, along the diagonal. This forms one square and four triangles. The four triangles are arranged such that the hypotenuse, of
each, forms the outer perimeter of a square. The remaining 1 x 1 square
then fits neatly into the space left in the middle.
--
David Entwistle

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From Newsgroup: rec.puzzles

On Sun, 21 Sep 2025 09:13:52 +0100, David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

Taken from 'Amusements in Mathematics' by Henry Ernest Dudeney.

Here is a little cutting out poser. I take a strip of paper, measuring
five inches by one inch, and, by cutting it into five pieces, the parts
fit together and form a square, as shown in this illustration [*]. Now
it is quite an interesting puzzle to discover how we can do this in only >four pieces.

[*] The illustration is available at the following link.

https://archive.org/details/amusementsinmath00dude/page/37/mode/1up

If you don't have access to this, I shall describe the arrangement...

The illustration shows the 5 x 1 rectangle cut into five parts: two 2 x
1 rectangles and one 1 x 1 square. The rectangles are further divided
into two parts, each, along the diagonal. This forms one square and four >triangles. The four triangles are arranged such that the hypotenuse, of >each, forms the outer perimeter of a square. The remaining 1 x 1 square
then fits neatly into the space left in the middle.

Did anyone get this? Two days on, and no responses.

The five piece solution kind of begs to be discovered as we are
looking for a square of side sqrt(5) and

sqrt(5) = sqrt(1^2 + 2^2)

and all that is screaming to be seen in the diagram.

Never being good at this kind of puzzle, the four piece solution
is beyond me. Will keep staring at the square of side sqrt(5),
though!
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From Newsgroup: rec.puzzles

On 23/09/2025 17:07, Charlie Roberts wrote:

Did anyone get this? Two days on, and no responses.

I didn't find a solution. I was looking at strips cut, aligned with the diagonal, and that didn't work.

I've looked at the solution and Dudeney says: "First find the side of
the square (the mean proportional between the length and height of the rectangle), and the solution is obvious"...

He then runs through a general solution for strips of various
proportions. It looks quite ingenious - perhaps not something we all
should know, but something of use to some. I'll post the solution, if no
one has it, at the weekend.
--
David Entwistle
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From Newsgroup: rec.puzzles

On Thu, 25 Sep 2025 09:56:43 +0100, David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

On 23/09/2025 17:07, Charlie Roberts wrote:

Did anyone get this? Two days on, and no responses.

I didn't find a solution. I was looking at strips cut, aligned with the >diagonal, and that didn't work.

I've looked at the solution and Dudeney says: "First find the side of
the square (the mean proportional between the length and height of the >rectangle), and the solution is obvious"...

Well, the first part *is* obivous, but after that .... is it still
obvious? Certainly, not ot me!

He then runs through a general solution for strips of various
proportions. It looks quite ingenious - perhaps not something we all
should know, but something of use to some. I'll post the solution, if no
one has it, at the weekend.

Would love to see the general cases. I guess there has to be
some restriction on the sides of the rectangle.
--
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From Newsgroup: rec.puzzles

On 21/09/2025 09:13, David Entwistle wrote:

Taken from 'Amusements in Mathematics' by Henry Ernest Dudeney.

Here is a little cutting out poser. I take a strip of paper, measuring
five inches by one inch, and, by cutting it into five pieces, the parts
fit together and form a square, as shown in this illustration [*]. Now
it is quite an interesting puzzle to discover how we can do this in only four pieces.

SPOILER.
POILER.
OILER.
ILER.
LER.
ER.
R.
.

Dudeney's solution says:

The illustration[*] shows how to cut the four pieces and form with them
a square. First find the side of the square (the mean proportional
between the length and height of the rectangle), and the method is
obvious. If our strip is exactly in the proportions 9 x 1, or 16 x 1, or
25 x 1, we can clearly cut it in 3, 4, or 5 rectangular pieces
respectively to form a square. Excluding these special cases, the
general law is that for a strip in length more than n^2 times the
breadth, and not more than (n + 1)^2 times the breadth, it may be cut in
n+2 pieces to form a square, and there will be n - 1 rectangular pieces
like piece 4 in the diagram. Thus, for example, with a strip 24 x 1, the length is more than 16 and less than 25 times the breadth. Therefore it
can be done in 6 pieces (n here being 4), 3 of which will be
rectangular. In the case where n equals 1, the rectangle disappears and
we get a solution in three pieces. Within these limits, of course, the
sides need not be rational: the solution is purely geometrical.

[*] An image of the solution is available on-line at the following links:

http://puzzles.50webs.org/a153.html

and

https://archive.org/details/amusementsinmath00dude/page/172/mode/1up
--
David Entwistle
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From Newsgroup: rec.puzzles

On Sun, 28 Sep 2025 10:46:56 +0100, David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

Dudeney's solution says:

The illustration[*] shows how to cut the four pieces and form with them
a square. First find the side of the square (the mean proportional
between the length and height of the rectangle), and the method is
obvious. If our strip is exactly in the proportions 9 x 1, or 16 x 1, or
25 x 1, we can clearly cut it in 3, 4, or 5 rectangular pieces
respectively to form a square. Excluding these special cases, the
general law is that for a strip in length more than n^2 times the
breadth, and not more than (n + 1)^2 times the breadth, it may be cut in
n+2 pieces to form a square, and there will be n - 1 rectangular pieces
like piece 4 in the diagram. Thus, for example, with a strip 24 x 1, the >length is more than 16 and less than 25 times the breadth. Therefore it
can be done in 6 pieces (n here being 4), 3 of which will be
rectangular. In the case where n equals 1, the rectangle disappears and
we get a solution in three pieces. Within these limits, of course, the
sides need not be rational: the solution is purely geometrical.

[*] An image of the solution is available on-line at the following links:

http://puzzles.50webs.org/a153.html

and

https://archive.org/details/amusementsinmath00dude/page/172/mode/1up

Amazing! Thanks for this general solutuion.

As I said earlier, I am absolutely terrible at dissection puzzles.
Always thought of them as one-off wonders. But, now, here
is a general solution for this class of problems.

I do not remember seeing anything like this solution in Martin
Gardner's writings, but I could have missed it.

One to salt away.

Thanks.
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