From Newsgroup: rec.puzzles

I have a sheet of stiff card and a sheet of newspaper. I cut a 100 mm
diameter circle out of the card and a square hole from the centre of the
sheet of newspaper. What is the smallest side length of the square hole
which will allow the circle to pass through the hole in the paper
without bending the card and without tearing the paper? For the purposes
of the puzzle, we can ignore the thickness of the card.

I hope that is clear. If not, can you suggest how to make the intention
clear?
--
David Entwistle

--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

In article <10cnn3t$3hgk4$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

I have a sheet of stiff card and a sheet of newspaper. I cut a 100 mm >diameter circle out of the card and a square hole from the centre of the >sheet of newspaper. What is the smallest side length of the square hole >which will allow the circle to pass through the hole in the paper
without bending the card and without tearing the paper?

But presumably not without "bending" the paper...

-- Richard
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles


richard@cogsci.ed.ac.uk (Richard Tobin) posted:

In article <10cnn3t$3hgk4$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

I have a sheet of stiff card and a sheet of newspaper. I cut a 100 mm >diameter circle out of the card and a square hole from the centre of the >sheet of newspaper. What is the smallest side length of the square hole >which will allow the circle to pass through the hole in the paper
without bending the card and without tearing the paper?

But presumably not without "bending" the paper...

-- Richard

Isn't bending the paper the whole trick?
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Wed, 15 Oct 2025 12:00:21 GMT, James Dow Allen <user4353@newsgrouper.org.invalid> wrote:

richard@cogsci.ed.ac.uk (Richard Tobin) posted:

In article <10cnn3t$3hgk4$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

I have a sheet of stiff card and a sheet of newspaper. I cut a 100 mm
diameter circle out of the card and a square hole from the centre of the >> >sheet of newspaper. What is the smallest side length of the square hole
which will allow the circle to pass through the hole in the paper
without bending the card and without tearing the paper?

But presumably not without "bending" the paper...

-- Richard

Isn't bending the paper the whole trick?

Yes!
--
This email has been checked for viruses by Avast antivirus software. www.avast.com
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On 15/10/2025 10:14, Richard Tobin wrote:

But presumably not without "bending" the paper...

Ah, no. My intention was to allow bending, or folding, of the paper. Not
the card, just the paper.

We could allow two answers, one with no folds and no bends to the paper
and one with paper folding.

The simple person that I am, I have to admit to having cut a disk out of
card and a square hole in a sheet of newspaper. I then spent longer than
you may imagine folding the paper and passing the disk through the hole.
Each time I, after looking at the flat sheet, was still slightly (not
sure of the word) "surprised(?)" that it actually fits through.
--
David Entwistle
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Thu, 16 Oct 2025 10:45:15 +0100, David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

On 15/10/2025 10:14, Richard Tobin wrote:

But presumably not without "bending" the paper...

Ah, no. My intention was to allow bending, or folding, of the paper. Not
the card, just the paper.

We could allow two answers, one with no folds and no bends to the paper
and one with paper folding.

The simple person that I am, I have to admit to having cut a disk out of >card and a square hole in a sheet of newspaper. I then spent longer than
you may imagine folding the paper and passing the disk through the hole. >Each time I, after looking at the flat sheet, was still slightly (not
sure of the word) "surprised(?)" that it actually fits through.

So, is the solution so well known that no one has bothered to post or
has it stumped too many?

At least James Dow Allen appears to know the answer ...

--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles


I'm fairly good with English (my first language)
and with logic, BUT for some reason sentences with two or
more negatives canceling each other often "throw" me.
(I dimly recall once citing a sentence with FIVE negatives
or so -- a sentence from a real news article, not designed
as a puzzle or parody.)

Sometimes I end up COUNTING the number of negatives in a
sentence to see if the total is odd or even.


James Dow Allen <user4353@newsgrouper.org.invalid> posted:

richard@cogsci.ed.ac.uk (Richard Tobin) posted:

But presumably not without "bending" the paper...

-- Richard

Isn't bending the paper the whole trick?

It appears I'm not the only one who missed that Richard's
sentence contains TWO negatives ('not' and 'without') which cancel
each other. Richard's sentence is equivalent to

Presumably by "bending" the paper...

and thus equivalent to my reply.

Cheers,
James
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Tue, 21 Oct 2025 07:30:26 GMT, James Dow Allen wrote:

It appears I'm not the only one who missed that Richard's sentence
contains TWO negatives ('not' and 'without') which cancel each other. Richard's sentence is equivalent to

[quoted text muted]

and thus equivalent to my reply.

Thanks for pointing that out gently. Apologies to Richard for failing to
grasp the implications of his initial reply. I'll post what I believe to
be the solution, for anyone not following, later today.

Best wishes,
--
David Entwistle
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

In article <1761031826-4353@newsgrouper.org>,
James Dow Allen <user4353@newsgrouper.org.invalid> wrote:

It appears I'm not the only one who missed that Richard's
sentence contains TWO negatives ('not' and 'without') which cancel
each other.

But presumably not without "bending" the paper...

It wasn't meant to be obscure. I phrased it like that simply to
contrast with

without bending the card

in the original, and it was indeed a hint to the solution.

-- Richard
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Wed, 15 Oct 2025 09:47:57 +0100, David Entwistle wrote:

I have a sheet of stiff card and a sheet of newspaper. I cut a 100 mm diameter circle out of the card and a square hole from the centre of the sheet of newspaper. What is the smallest side length of the square hole
which will allow the circle to pass through the hole in the paper
without bending the card and without tearing the paper? For the purposes
of the puzzle, we can ignore the thickness of the card.

Card and Paper Puzzle Answer

Although it may seem surprising (it does to me), the minimum square side
is 50mm. The square is cut and then the newspaper is folded along the
diameter of the square hole and the circle offered up to the hole inside
the fold. By gently moving the opposite sides of the hole down the circle,
the paper forms a collapsed pyramid shape and the two cut edges form a straight line of total length 100mm. At this point the card circle will
slide through.
--
David Entwistle
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

In article <10d830n$3v56m$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

Card and Paper Puzzle Answer

Although it may seem surprising (it does to me), the minimum square side
is 50mm. The square is cut and then the newspaper is folded along the >diameter of the square hole and the circle offered up to the hole inside

Did you mean "diagonal"? Though "diameter" amounts to the same thing.

the fold. By gently moving the opposite sides of the hole down the circle, >the paper forms a collapsed pyramid shape and the two cut edges form a >straight line of total length 100mm. At this point the card circle will >slide through.

Now, what about a circular hole?

-- Richard
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Tue, 21 Oct 2025 16:27:36 -0000 (UTC), Richard Tobin wrote:

Did you mean "diagonal"? Though "diameter" amounts to the same thing.

Yes, I did mean diagonal. Thanks.

Now, what about a circular hole?

Good question. I think I know that one.

I guess there's also the obviously-related questions:

Card: equilateral triangle, square, regular pentagon, regular hexagon,
regular septagon, regular octagon, etc.

Hole: circular, equilateral triangle, square, regular pentagon, regular hexagon etc.

Some combinations may not warrant the extra effort involved in deriving an answer, but some could be interesting.
--
David Entwistle
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Tue, 21 Oct 2025 16:27:36 -0000 (UTC), Richard Tobin wrote:

Now, what about a circular hole?

I think I was on the right track, with a circular hole that passes through
the corners of the smallest square. Having tried it, the card disk passes through the hole, as you would expect - it does seem a little lose.

So, thinking it may be a circular hole with a circumference twice the card disk's diameter - that doesn't seem to fit.

I'm reasonably happy with my first thought, but I can't think of any way I
may be able to arrive at a definitive answer using mathematics.

Thanks,
--
David Entwistle
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

In article <10ddblc$1nubl$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

Now, what about a circular hole?

I think I was on the right track, with a circular hole that passes through >the corners of the smallest square. Having tried it, the card disk passes >through the hole, as you would expect - it does seem a little lose.

It seems clear (?) that when passing a circular disk through a hole,
at some point the full diameter of the circle must between two
extremes of the hole. And the distance between those two extremes
can't be more half the perimeter of the hole, since the perimeter must
run from one extreme to the other on each side of the disk. So that
gives us an upper bound - the diameter of the disk can't be more than
half the perimeter of the hole.

Can we achieve that limit?

For the square hole, it's easy to make folds to extend the perimeter
into two coincident straight lines and get a slot twice the length
of the sides.

For the circular hole, obviously no finite sequence of folds will make
the circumference into straight lines, but we can in theory approach arbitrarily close to it. To get the full length will require a
continuous deformation of the paper.

-- Richard
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Thu, 23 Oct 2025 14:21:17 -0000 (UTC), Richard Tobin wrote:

For the circular hole, obviously no finite sequence of folds will make
the circumference into straight lines, but we can in theory approach arbitrarily close to it. To get the full length will require a
continuous deformation of the paper.

Thanks. I agree with the theory. But, it feels tricky to get anywhere
close to the limit. If it continues raining, I'll have a look at an
octagonal hole, with a limiting perimeter tomorrow.
--
David Entwistle
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On 23/10/2025 15:21, Richard Tobin wrote:

In article <10ddblc$1nubl$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

Now, what about a circular hole?

I think I was on the right track, with a circular hole that passes through >> the corners of the smallest square. Having tried it, the card disk passes
through the hole, as you would expect - it does seem a little lose.

It seems clear (?) that when passing a circular disk through a hole,
at some point the full diameter of the circle must between two
extremes of the hole. And the distance between those two extremes
can't be more half the perimeter of the hole, since the perimeter must
run from one extreme to the other on each side of the disk. So that
gives us an upper bound - the diameter of the disk can't be more than
half the perimeter of the hole.

Can we achieve that limit?

For the square hole, it's easy to make folds to extend the perimeter
into two coincident straight lines and get a slot twice the length
of the sides.

For the circular hole, obviously no finite sequence of folds will make
the circumference into straight lines, but we can in theory approach arbitrarily close to it. To get the full length will require a
continuous deformation of the paper.

But then we have to decide if such a continuous deformation is possible. The good thing about
simple folds is that it is clear they exist and how the paper behaves. Maybe we can create a
sequence of approximating solutions with folds, and consider some kind of limit process, and finally
show the paper limit is still locally flat (has not been stretched at all). I'm not convinced that
will work...

Mike.

-- Richard

--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Thu, 23 Oct 2025 15:33:24 -0000 (UTC), David Entwistle wrote:

Thanks. I agree with the theory. But, it feels tricky to get anywhere
close to the limit. If it continues raining, I'll have a look at an
octagonal hole, with a limiting perimeter tomorrow.

With a sheet of paper with an octagonal hole cut in it, it naturally folds along eight axes, aligned with the vertices of the octagon, alternatively
in and out, when viewed from one side.

The edges of the octagonal-hole won't form a straight line, as this would require the inward folds, of the remainder of the paper, to move in and through the fold on the opposite side of the hole. It is hard to explain,
but clear when you try it.

The disk won't pass through the hole because the edges of the octagon
won't align along a straight line, but favour a three dimensional star
shape.
--
David Entwistle
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

In article <10dfl8i$2h2fj$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

The edges of the octagonal-hole won't form a straight line

https://www.cogsci.ed.ac.uk/~richard/unfolded.jpg

https://www.cogsci.ed.ac.uk/~richard/folded.jpg

-- Richard
--- Synchronet 3.21a-Linux NewsLink 1.2

From Newsgroup: rec.puzzles

On Fri, 24 Oct 2025 11:28:00 -0000 (UTC), Richard Tobin wrote:

https://www.cogsci.ed.ac.uk/~richard/unfolded.jpg

https://www.cogsci.ed.ac.uk/~richard/folded.jpg

Well blow me down. That's very impressive.

I can confirm performing those folds, on an octagon of side 25 mm, the 100
mm diameter disk just slides through.

That's a whole puzzle on its own.
--
David Entwistle
--- Synchronet 3.21a-Linux NewsLink 1.2