Minimize Crease Length of Paper in Terms of Width

[MarkFL]

Hello MHB,

Consider a rectangular piece of paper of width $W$ laid on a flat surface. The lower left corner of the paper is bought over to the right edge of the paper, and the paper is smoothed flat creating a crease of length $L$, as in the diagram:

View attachment 1078

What is the minimal value of $L$ in terms of $W$?

 

[Wilmer]

Hello MHB,

Consider a rectangular piece of paper of width $W$ laid on a flat surface. The lower left corner of the paper is bought over to the right edge of the paper, and the paper is smoothed flat creating a crease of length $L$, as in the diagram:

https://www.physicsforums.com/attachments/1078
...A...... B

What is the minimal value of $L$ in terms of $W$?

When AB = 3W/4

a = AB
L^2 = (2a^3) / (2a - w)

Above is my contribution...(Coffee)

 

[MarkFL]

When AB = 3W/4

a = AB
L^2 = (2a^3) / (2a - w)

Above is my contribution...(Coffee)

This is correct...can you show how you arrived at this solution? (Sun)

 

[Wilmer]

A                       F 
                        D 
 
 
C                 B     E

a = BC = BD, b = AC = AD, c = AB, w = AF

Mark, that's a frustratingly delightful li'l problem!
Boils down to 2 congruent right triangles (ABC and ABD)
stuck together along the common hypotenuse AB, then
"completing the rectangle" (ACEF) thus forming
2 similar right triangles (ADF and BDE).

Simple enough it appears, but I'm kinda stuck.

I've been able to "see" that when c (your L) is at minimum,
then b = aSQRT(2) and c = aSQRT(3); so c = 3wSQRT(3) / 4,
using my previous observation: a = (3/4)w.

So I need to come up with some equation such that after
taking its 1st derivative, I'm left with 4c = 3wSQRT(3).

Can't wrap it up; take me out of my misery!

 

[MarkFL]

...Can't wrap it up; take me out of my misery!

Okay, here is my solution:

Spoiler

I have filled in the previous diagram with the information I need:

View attachment 1087

By similarity, we may state:

$$\frac{\sqrt{L^2-x^2}}{W}=\frac{x}{\sqrt{W(2x-W)}}$$

Squaring, we obtain:

$$\frac{L^2-x^2}{W^2}=\frac{x^2}{W(2x-W)}$$

$$L^2-x^2=\frac{Wx^2}{2x-W}$$

$$L^2=\frac{2x^3}{2x-W}$$

At this point we see that we require $$\frac{W}{2}<x\le W$$.

Minimizing $L^2$ will also minimize $L$, and so differentiating with respect to $x$ and equating to zero, we find:

$$\frac{d}{dx}\left(L^2 \right)=\frac{(2x-W)\left(6x^2 \right)-(2)\left(2x^3 \right)}{(2x-W)^2}=\frac{2x^2(4x-3W)}{(2x-W)^2}=0$$

Discarding the root outside of the meaningful domain, we are left with:

$$4x-3W=0$$

$$x=\frac{3}{4}W$$

The first derivative test easily shows that this is a minimum, as the linear factor in the numerator, the only factor which changes sign, goes from negative to positive across this critical value.

Thus, we may state:

$$L_{\min}=L\left(\frac{3}{4}W \right)=\sqrt{\frac{2\left(\frac{3}{4}W \right)^3}{2\left(\frac{3}{4}W \right)-W}}=\frac{3\sqrt{3}}{4}W$$

 

[Wilmer]

Thanks Mark.
Somewhat tougher than I thought...
Sure glad to see the SQRT(3)!

 

[Wilmer]

Damn! I was ok up to your L^2 = 2x^3 / (2x - W) , then to 2x^2(4x - 3W) = 0

Entirely forgot about this:
"Discarding the root outside of the meaningful domain, we are left with:"

Gotta good memory, but it's short (Emo)

(I don't have enough hair to do that!)

Again, NICE problem.

 

[Wilmer]

Of course, there will be no "all integer" solutions.
Interestingly, very few exist if "minimum L" condition is removed;
only 4 primitives keeping short leg of right triangle < 10000:
(right triangle sides, W):
75,100, 125, 96
845, 2028, 2197, 1440
2312, 4335, 4913, 3600
4375, 15000, 15625, 8064