Find the minimum length of the crease in a folded piece of paper

Consider a piece of paper with width 6 inches. The lower-right corner is folded over until the corner touches the left side of the paper. Find the minimum possible length of the crease in the paper. Find the angle the minimal crease makes with right edge of the paper. Assume the paper is sufficiently long such that the crease does not reach the top of the page.

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Let the angle the crease makes with the right edge of the paper be denoted by $\theta$ , the horizontal width of the fold by $x$ (i.e., the dotted line along the bottom of the paper length $x$ ), and let the length of the crease be $y$ . Then, from the diagram we see $x = y \sin \theta$ . Next, the triangle enclosed by the dotted black lines must be congruent to the triangle enclosed by blue lines since when we fold the paper over, the triangle enclosed by the black dotted lines is exactly the space the blue triangle was in before we made the fold. (Excuse the fact that the diagram is poorly draw and makes these two triangles look to be different sizes.) Keeping this in mind, consider the triangle in the lower left corner. This triangle has hypotenuse $x =y \sin \theta$ and the angle in the lower right corner must be $2 \theta$ (this is because the other two angles along the horizontal line on the bottom are both $90 - \theta$ , and $180 - (90 - \theta) - (90 - \theta) = 2 \theta$ ). Finally, the length of the leg of this triangle along the bottom of the piece of paper is $6 - x = 6 - y \sin \theta$ since we are given that the paper is 6 inches wide.

From trigonometry (specifically, $\cos$ equals adjacent over hypotenuse) we have

$\begin{align*} &&\cos (2 \theta) &= \frac{6 - y \sin \theta}{y \sin \theta} \\[9pt] \implies &&\cos (2 \theta) &= \frac{6}{y \sin \theta} - 1 \\[9pt] \implies && \cos (2 \theta) + 1 &= \frac{6}{y \sin \theta} \\[9pt] \implies && y \sin \theta &= \frac{6}{\cos (2 \theta) + 1} \\[9pt] \implies && y &= \frac{6}{\sin \theta (\cos (2 \theta) + 1)} \\[9pt] \implies && y &= \frac{6}{2 \sin \theta \cos^2 \theta}. \end{align*}$

The final equality follows from using the trig identity $\cos (2 \theta) = 2\cos^2 \theta - 1$ which implies $1 + \cos (2\theta) = 2 \cos^2 \theta$ .

Then, we want to minimize $y$ , which is the same as maximizing $2 \sin \theta \cos^2 \theta$ . Taking the derivative of this and setting it equal to 0 to find the critical points we have,

$\begin{align*} (2 \sin \theta \cos^2 \theta)' = 0 && \implies && 2 \cos^3 \theta - 4 \cos \theta \sin^2 \theta &= 0 \\ && \implies && \cos^2 \theta - 2 \sin^2 \theta &= 0 \\ && \implies && 1 - 3 \sin^2 \theta &= 0 \\ && \implies && \sin \theta &= \frac{1}{\sqrt{3}}. \end{align*}$

Since $1 - 3 \sin^2 \theta > 0$ for $\sin \theta < \frac{1}{\sqrt{3}}$ and $1 - 3 \sin^2 \theta < 0$ for $\sin \theta > \frac{1}{\sqrt{3}}$ we have that the $2 \sin \theta \cos^2 \theta$ has a maximum when $\sin \theta = \frac{1}{\sqrt{3}}$ . Therefore, plugging this into our expression for the length of the crease $y$ we have

$y = \frac{6}{\left( \frac{2}{\sqrt{3}} \right) \left( \frac{2}{3} \right)} = \left( \frac{3 \sqrt{3}}{4} \right)6 = \frac{9 \sqrt{3}}{2},$

and,

$\theta = \arcsin \left( \frac{1}{\sqrt{3}} \right) = \arctan \frac{\sqrt{2}}{2}.$

2 comments

November 14, 2018 at 8:52 pm

Molly says:

Very nice solution!
This is probably super simple, but I am missing how you got

$y = \frac{6}{\left( \frac{2}{\sqrt{3}} \right) \left( \frac{2}{3} \right)} = \left( \frac{3 \sqrt{3}}{4} \right)6 = \frac{9 \sqrt{3}}{2},$

when you plugged in $\sin\theta=\frac{1}{\sqrt3}$ . How do you know the value of cos?
Thank you!

- August 27, 2021 at 4:52 pm
  
  Anonymous says:
  
  Pythagoras: cos^2+sin^2=1

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Find the minimum length of the crease in a folded piece of paper

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