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### The Shortest Crease

```
Date: 12/29/97 at 03:14:48
From: Kathy
Subject: The shortest crease

Q: A paper which is 6 units one side and 25 units another side. Put
the shorter side to face you. Fold the lower right corner to the left
side, making a crease. Our task is to find the minimum crease by
folding the right corner to the left side.

HINT: this question is related to derivative - you may use derivative
to solve it. (You are required to write down the equations that you
use and explain how you solve it.)

This question is a project given in my calculus class. We are
supposed to write equations to solve it. We have to write a bunch of
explanations and equations to complete a project paper. So far, I
have no idea how to solve it. Could you help me?

Thank you very much for your help.

Sincerely,
Kathy
```

```
Date: 12/29/97 at 12:22:00
From: Doctor Anthony
Subject: Re: The shortest crease

I have taken a general case with width equal to k, so you can see how
the working can be adjusted for different widths of paper.

If you take a length x along the width, and fold it, leaving (k-x)
unfolded, and let the corner meet the righthand edge somewhere along
its length, such that the side x makes an angle (theta) with the
length, then we can calculate the length of the fold and the area of
the folded triangle in terms of k and theta.

The other edge of the folded portion (at right angles to x) will make
an angle theta with the width, and it is easy to see that this length
is given by k/cos(theta)

We also have      x.sin(theta) = k-x

x(1+sin(theta)) = k

k
x = -------------
1 + sin(theta)

The length of the fold, by Pythagoras is given by

FOLD^2  = x^2 + [k/cos(theta)]^2

= [k/(1+sin(theta)]^2 + [k/cos(theta)]^2

= k^2[(1+sin(theta))^(-2) + sec^2(theta)]

Differentiate with respect to theta and we get

k^2[(-2)(1+sin(theta)^(-3)(cos(theta) +
2.sec(theta)sec(theta)tan(theta)]

k^2[-2.cos(theta)/(1+sin(theta))^3 + 2.sin(theta)/cos^3(theta)]

and equating this to zero, we get

cos^4(theta) = sin(theta)(1+sin(theta))^3

putting sin(theta) = 1/3 and then cos(theta) = sqrt(8)/3, this
equation is satisfied. So the minimum fold occurs with sin(theta)
= 1/3

x = k/(1+1/3) =  k/(4/3)  =  (3/4)k

FOLD^2 = (9/16)k^2 + (9/8)k^2  = (27/16)k^2

Fold = (3.Sqrt(3)/4)k

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Trigonometry

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