Crease

[cinelli]

|                                           |
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|                                           D
|                                      .  . .
|                                 .      .  .
|                             .         .   .
|                        .             .    .
|                    .               .      .
|                .                  .       .
|           .                      .        .
|       .                        .          .
|   .                           .           .
B'                             .            .
|.                            .             .
| .                         .               .
|  .                       .                .
|   .                     .                 .
|    .                   .                  .
|      .               .                    .
|       .             .                     .
|        .           .                      .
|         .        .                        .
|          .      .                         .
|            .   .                          .
|             . .                           .
A--------------C............................B

The diagram shows a sheet of paper, width W. The bottom corner B is folded to lie on the opposite edge, to B', making crease CD. This puzzle is to find the minimum length of the crease.

Cinelli

[servodude]

(2*W^2)^0.5

?

[mc2fool]

(2*W^2)^0.5

?

Yes, just eyeballing it surely CD is shortest when C is at A? Don't know how to prove that mathematically though....

[cinelli]

(2*W^2)^0.5

?

Yes, just eyeballing it surely CD is shortest when C is at A? Don't know how to prove that mathematically though....

No. Take a sheet of paper and make some exploratory folds. You will see that W*sqrt(2) is not the minimum.

Cinelli

[9873210]

(2*W^2)^0.5

?

Yes, just eyeballing it surely CD is shortest when C is at A? Don't know how to prove that mathematically though....

No. Take a sheet of paper and make some exploratory folds. You will see that W*sqrt(2) is not the minimum.

Cinelli

A sheet of paper or a strip of paper? Or a physically impossible infinite strip of paper? Or is this a question that cannot be answered with the given information?

[mc2fool]

(2*W^2)^0.5

?

Yes, just eyeballing it surely CD is shortest when C is at A? Don't know how to prove that mathematically though....

No. Take a sheet of paper and make some exploratory folds. You will see that W*sqrt(2) is not the minimum.

Cinelli

Not surprising as it couldn't possibly be that simple, eh?! But it does look like it, just eyeballing....

[cinelli]

A sheet of paper or a strip of paper? Or a physically impossible infinite strip of paper? Or is this a question that cannot be answered with the given information?

There is no trick. A sheet of A4 will do.

Cinelli

[9873210]

(2*W^2)^0.5

?

Yes, just eyeballing it surely CD is shortest when C is at A? Don't know how to prove that mathematically though....

No. Take a sheet of paper and make some exploratory folds. You will see that W*sqrt(2) is not the minimum.

Cinelli

Not surprising as it couldn't possibly be that simple, eh?! But it does look like it, just eyeballing....

Looking at the words and not the diagram

C may be on AB' rather than AB. This leads to a shorter crease. With a long enough strip, the crease length approaches the width.
end

[malkymoo]

I have given this problem a lot of thought, thinking that maybe there was an insight that would solve it, but when I realised that the solution involved quite a complicated differentiation I gave up. What little ability I once had with differentiation is now 50 years old.

There are solutions to this on the internet, search on "minimum crease length", once you get beyond the cricket references you can find solutions.

[mc2fool]

I have given this problem a lot of thought, thinking that maybe there was an insight that would solve it, but when I realised that the solution involved quite a complicated differentiation I gave up. What little ability I once had with differentiation is now 50 years old.

Yes, I got to that point too. I figured (although didn't work it through) that there were two (or was it three?) right angled triangles involved, and so it looked a simple-ish matter of combining Pythagoras on them and then ... ah, yes, calculus (shudder!) to iterate through to find the shortest.

It does strike me though, that it could be done by brute force in Excel, with a table of the pythag-calcs to get close, and then using goal seek ...

Not an acceptable way for this board though, methinks.

[Bubblesofearth]

Paper folded so that B meets B'

Then;

CD^2 = AB^2 + (AD-AC)^2

Maybe?

BoE

[mc2fool]

Paper folded so that B meets B'

Then;

CD^2 = AB^2 + (AD-AC)^2

Maybe?

BoE

Maybe I'm missing something but I don't see that. In fact, if you take the case of C being at A then AC is zero and you have CD^2 = CB^2 + CD^2, which obviously isn't right.

An observation, albeit maybe a dead obvious one: in the diagram as shown there are only two right angled triangles. BCD, with the right angle at B, and AB'C with the right angle at A. The third apparent right angled triangle, B'CD, is physically the same as BCD as it's just that bit of paper folded over.

In my earlier musings I thought it might be useful to pencil in another using a position E horizontally opposite to D (on the vertical above B') and so giving B'DE with the right angle at E, but now I can't remember why I thought so ....

[Bubblesofearth]

Maybe I'm missing something but I don't see that. In fact, if you take the case of C being at A then AC is zero and you have CD^2 = CB^2 + CD^2, which obviously isn't right.

An observation, albeit maybe a dead obvious one: in the diagram as shown there are only two right angled triangles. BCD, with the right angle at B, and AB'C with the right angle at A. The third apparent right angled triangle, B'CD, is physically the same as BCD as it's just that bit of paper folded over.

In my earlier musings I thought it might be useful to pencil in another using a position E horizontally opposite to D (on the vertical above B') and so giving B'DE with the right angle at E, but now I can't remember why I thought so ....

Apols, I wasn't at all clear on the first bit. B is taken up to the opposite top corner of the sheet and this becomes B'. You can't take B any higher as then it would no longer meet the paper's edge.

BoE

[mc2fool]

Apols, I wasn't at all clear on the first bit. B is taken up to the opposite top corner of the sheet and this becomes B'. You can't take B any higher as then it would no longer meet the paper's edge.

You mean like the rightmost one here?


Source (spoilers): https://datagenetics.com/blog/january22018/index.html

[Bubblesofearth]

You mean like the rightmost one here?


Source (spoilers): https://datagenetics.com/blog/january22018/index.html

Yes, nice pic btw.

BoE

[cinelli]

|                                           |
|                                           |
|                                           |
|                                           |
|                                           D
|                                      .  . .
|                                 .      .  .
|                             .         .   .
|                        .             .    .
|                    .               .      .
|                .                  .       .
|           .                      .        .
|       .                        .          .
|   .                           .           .
B'                             .            .
|.                            .             .
| .                         .               .
|  .                       .                .
|   .                     .                 .
|    .                   . c                .
|      .               .                    .
|       .             .                     .
|        . x         .                      .
|         .        .                        .
|          .      .                         .
|            .   .                          .
|             . .                           .
A--------------C............................B
    W-x                    x

Ingenious as these musings are, I have decided that allowing A to wander beyond the bounds of the paper wasn’t in the spirit of the diagram. My solution (not hidden) follows then we can move on to something else. This involves trigonometry and calculus, skills which some may have forgotten.

Let BC=x, CD=c, angle BCD=theta. AC = W-x. Note 0.5*W < x <= W. Then by the nature of paper folding triangles DCB and DCB' are congruent and angle B’CD = theta making angle ACB'=pi-2*theta.

In triangle BCD cos(theta)=x/c and in triangle AB’C cos(pi-2*theta)=(W-x)/x.
So -cos(2*theta)=W/x-1, cos(2*theta)=1-W/x, 2*cos(theta)^2-1 = 1-W/x
2*cos(theta)^2 = 2-W/x = 2*(x/c)^2
Hence 2*x*c^2-W*c^2 = 2*x^3, c^2 = 2*x^3 / (2*x-W) (1)
c will attain its minimum when c^2 does so we don’t need to extract the square root before differentiating wrt x: (d/dx)(c^2) = ((2*x-W)*6*x^2 – 4*x^3) / (2*x-W)^2

The minimum is attained when the numerator is zero:
8*x^3 – 6*W*x^2 = 0 or x^2 * (8*x 6*W) = 0

x can’t be zero so the only root is x = 3*W/4. Substituting this value into (1) gives
c^2 = 2*(3/4)^3 * W^3 / (0.5 * W)
giving the minimum c = (3*sqrt(3)/4) * W

Cinelli

[mc2fool]

Ingenious as these musings are, I have decided that allowing A to wander beyond the bounds of the paper wasn’t in the spirit of the diagram. My solution (not hidden) follows then we can move on to something else. This involves trigonometry and calculus, skills which some may have forgotten.

Calculus yes, trig very nearly. However, the spoiler link I posted earlier* has no trig, using only Pythagoras and (shudder) calculus.

I don't know if that makes it a "better" solution or not, I'll leave that to the maths boffs to argue, but there is an assertion in their solution that I don't quite see, being FH/EH = GA/EA. Could someone enlighten me on the why of that, please?

* Here again: https://datagenetics.com/blog/january22018/index.html

[9873210]

giving the minimum c = (3*sqrt(3)/4) * W

Cinelli

c = (3*sqrt(3)/4) * W = 1.299W

But if you fold the paper corner to opposite corner so that C is on AB', not AB, the length of the crease is W*sqrt(1+1/H^2)

For your A4 paper H=W*sqrt(2) so the crease length is W*sqrt(1.5) = 1.225W which is shorter, so you haven't solved the word problem because it has no constraint on the location of C.

[GoSeigen]

giving the minimum c = (3*sqrt(3)/4) * W

Cinelli

c = (3*sqrt(3)/4) * W = 1.299W

But if you fold the paper corner to opposite corner so that C is on AB', not AB, the length of the crease is W*sqrt(1+1/H^2)

For your A4 paper H=W*sqrt(2) so the crease length is W*sqrt(1.5) = 1.225W which is shorter, so you haven't solved the word problem because it has no constraint on the location of C.

But Cinelli also wrote:

I have decided that allowing A to wander beyond the bounds of the paper wasn’t in the spirit of the diagram.

so he can hardly be criticised for sticking to this constraint.

GS

[9873210]

giving the minimum c = (3*sqrt(3)/4) * W

Cinelli

c = (3*sqrt(3)/4) * W = 1.299W

But if you fold the paper corner to opposite corner so that C is on AB', not AB, the length of the crease is W*sqrt(1+1/H^2)

For your A4 paper H=W*sqrt(2) so the crease length is W*sqrt(1.5) = 1.225W which is shorter, so you haven't solved the word problem because it has no constraint on the location of C.

But Cinelli also wrote:

I have decided that allowing A to wander beyond the bounds of the paper wasn’t in the spirit of the diagram.

so he can hardly be criticised for sticking to this constraint.

GS

It's fair to say I elided this, sorry about that, but ...

Math problems do not have a spirit, they have explicit stated conditions.

It's not cheating to violate assumption. Finding loopholes is a major source of advancement.

(The same is true of engineering, except instead of "meeting explicit stated conditions" it "has to work", it does not matter if the old guard insist it can't work if it does work.)