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Minimize the crease

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MarkFL

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Feb 24, 2012
13,775
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:

creaselength.jpg

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

Wilmer

In Memoriam
Mar 19, 2012
376
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
....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)
 
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MarkFL

Administrator
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Feb 24, 2012
13,775
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

In Memoriam
Mar 19, 2012
376
Code:
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!
 
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MarkFL

Administrator
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Feb 24, 2012
13,775
...Can't wrap it up; take me out of my misery!
Okay, here is my solution:

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

creaselength2.jpg

By similarity, we may state:

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

Squaring, we obtain:

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

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

\(\displaystyle L^2=\frac{2x^3}{2x-W}\)

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

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

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

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

\(\displaystyle 4x-3W=0\)

\(\displaystyle x=\frac{3}{4}W\)

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:

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

Wilmer

In Memoriam
Mar 19, 2012
376
Thanks Mark.
Somewhat tougher than I thought...
Sure glad to see the SQRT(3)!
 

Wilmer

In Memoriam
Mar 19, 2012
376
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

In Memoriam
Mar 19, 2012
376
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