an optimisation problem folding paper

Question

A standard $8.5$ inches by $11$ inches piece of paper is folded so that one corner touches the opposite long side and the crease runs from the adjacent short side to the other long side, as shown in the picture below. What is the minimum length of the crease?

So I put the shorter leg of the folded portion (since it's a right triangle) as $x$ and the longer leg $y$. Then the crease is $\sqrt{x^2+y^2}$ and deriving it (which is what I'm supposed to be doing for optimisation) in terms of x to minimise the crease length, I got $\frac{x+yy'}{\sqrt{x^2+y^2}} = 0$, so $x+yy' = 0$. Then I was able to find some similar triangles - one with hypotenuse $x$ and one leg $8.5-x$ and the other with hypotenuse $y$ and the not-corresponding leg $8.5$. From there I was able to get the equation $y=\sqrt{y^2-8.5^2} + \sqrt{x^2-(8.5-x)^2}$. Then I started squaring the equation but it got quite messy, and I'm not even sure if the problem is supposed to be so messy... Can someone help me solve this?

Disclaimer: I do know that there is a post with the same problem on this site, but unfortunately the answers there didn't really help me, and I don't get what I am doing wrong.

Answer 1

Let me denote the corners in your figure clockwise (starting from the top left corner) by $A$, $B$, $C$, $D$, $E$, and $F$. This means $|AB|=b=8.5$, $|AE|=a=11$, $|DF|=x$, $|CF|=y$, ... Furthermore, I denote the point on the left edge of the paper which has the same height as $C$ by $G$.

I define $z=|EF|$. From the two right triangles $DEF$ and $CFG$ we obtain $z^2+(b-x)^2=x^2$ and $(y-z)^2+b^2=y^2$. Solving these equations for $x$ and $y$ yields $$x=\frac{b^2+z^2}{2b}\qquad\mbox{ and }\qquad y=\frac{b^2+z^2}{2z}.$$ Hence, $$\sqrt{x^2+y^2}=\frac{(b^2+z^2)^{3/2}}{2bz}.$$ This is what needs to be minimized with respect to $z$. The first-order optimality condition yields $z=b/\sqrt{2}$. Hence, the solution is $$x=\frac{3b}{4}\qquad\mbox{ and }\qquad y=\frac{3b}{2\sqrt{2}}.$$

Answer 2

$|EG|=c(t)\rightarrow \displaystyle\min_t$

\begin{align} |BF|&=\sqrt2\,\sqrt{h\,t\,(h\,t-\sqrt{(h^2\,t^2-w^2})} ,\\ \sin\phi&=\tfrac{\sqrt2}2\,\sqrt{1-\sqrt{1-\frac{w^2}{(h\,t)^2}}} ,\\ \cos\phi&= \tfrac12\,\sqrt{2+2\,\sqrt{1-\frac{w^2}{(h\,t)^2}}} ,\\ c(t)&=\frac{h\,t}{\cos\phi} =\dots \end{align}

And the optimum can be found as

\begin{align} t_{\min}&=\frac{3\sqrt2\,w}{4\,h}= \frac{51\sqrt2}{88} \approx 0.8196 ,\\ c(t_{\min})&= \frac{51\,\sqrt3}{8} \approx 11.0418 . \end{align}

Answer 3

Perhaps, a simpler approach would be to express $|EG|$ in terms of $\angle ABF=\phi$.

In this case \begin{align} |BF|&=\frac{w}{\cos\phi} ,\\ |BK|&=\tfrac12\,\frac{w}{\cos\phi} ,\\ |BE|&=\frac{|BK|}{\cos(\tfrac\pi2-\phi)} =\frac{w}{2\,\cos\phi\sin\phi} ,\\ |EG|&=\frac{|BE|}{\cos\phi} =\tfrac12\,\frac{w}{\cos^2\phi\sin\phi} \\ &=\tfrac12\,\frac{w}{(1-\sin^2\phi)\,\sin\phi} \\ &=\tfrac12\,\frac{w}{(1-x^2)\,x} . \end{align}

Now we can use $x=\sin\phi$ as a control parameter:

\begin{align} |EG|&=f(x)= \tfrac12\,\frac{w}{(1-x)(1+x)x} = \frac{w}4\left( \frac{1}{1-x} -\frac{1}{1+x} +\frac{2}{x} \right) ,\\ f'(x)&= \frac{w}4\left( \frac{1}{(1-x)^2} +\frac{1}{(1+x)^2} -\frac{2}{x^2} \right) \\ &= \frac{w}2\cdot \frac{3\,x^2-1}{(1-x)^2(1+x)^2\,x^2} . \end{align}

Solution to \begin{align} f'(x)&=0\quad \text{gives } x=\pm\tfrac{\sqrt3}3, \\ |EG|_{\min}&=\tfrac34\,w\,\sqrt3 =\tfrac{51}8\,\sqrt3 \approx 11.0418 . \end{align}