The Shortest Crease

What this might be about?

18 October 2015, Created with GeoGebra

Problem

An interesting optimization problem has been offered by Henry Ernest Dudeney:

Fold a page, so that the bottom outside corner touches the inside edge and the crease is the shortest possible. That is about as simple a question as we could put, but it will puzzle a good many readers to discover just where to make the fold. I give two examples of folding:

Dudeney's crease minimization - problem

It will be seen that the crease on the right is longer than that on the left, but the latter is not the shortest possible.

References

  1. H. E. Dudeney, 536 Puzzles & Curious Problems, Charles Scribner's Sons, 1967

Answer

Bisect $AB$ in $M.$ Bisect $AM$ at $E.$ Draw the line at $M$ perpendicular to $AB$ and the semicircle with diameter $BE.$ Let $N$ be the intersection of the two.

Dudeney's crease minimization - answer

The line $EN$ gives the direction of the shortest possible crease under the conditions.

Solution

The solution below is straightforward application of the Pythagorean theorem and the very beginning of the calculus.

Taking into account that Dudeney "folds a page," assume $AD\gt AB=4.$ Denote $x=BE.$

Dudeney's crease minimization - solution

Then successively: $AE=4-x;$ $EB'=BE=x;$ $AB'=\sqrt{x^{2}-(4-x)^{2}}.$

Further, triangles $AEB'$ and $FSB'$ are similar so that $\displaystyle FS=\frac{AE\cdot B'F}{AB'}=\frac{4(4-x)}{\sqrt{x^{2}-(4-x)^{2}}};$

$\begin{align} BS&=BF+FS\\ &=\sqrt{x^{2}-(4-x)^{2}}+\frac{4(4-x)}{\sqrt{x^{2}-(4-x)^{2}}}\\ &=\frac{x^{2}-(4-x)^{2}+4(4-x)}{\sqrt{x^{2}-(4-x)^{2}}}\\ &=\frac{4x}{\sqrt{x^{2}-(4-x)^{2}}}\\ &=\frac{2x}{\sqrt{2(x-2)}}. \end{align}$

Finally

$\begin{align} ES^{2}&=BE^2+BS^2\\ &=x^{2}+\frac{4x^{2}}{2(x-2)}\\ &=\frac{x^{3}}{x-2}\\ \end{align}$

Consider the function $\displaystyle f(x)=\frac{x^{3}}{\sqrt{x-2}}.$

$\displaystyle f'(x)=\frac{3x^{2}(x-2)-x^{3}}{(x-2)^{2}}=\frac{2x^{2}(x-3)}{(x-2)^2}.$

$x=0$ leads to no crease, but $x=3$ does. We next find that $f''(3)\gt 0$ so that $x=3$ is a local minimum.

This shows why Dudeney's answer is correct. With $x=3,$ $MN=\sqrt{2}$ and $BS=3\sqrt{2},$ implying that $ES$ passes through $N.$ Thus Dudeney's answer is a byproduct of the calculus solution. It would be exciting to arrive at his result without calculus.

Thus, according to Dudeney the minimum crease of a page with base $4$ equals $3\sqrt{3}.$ This crease can be obtained for any page whose vertical size is at least $3\sqrt{2}.$ However, it is obvious that, with the vertical side $b\lt 3\sqrt{3},$ the vertical crease through the midpoint $M$ will be shorter than $3\sqrt{3}.$ Obviously, Dudeney assumed that, for a "page," the ratio of the vertical to the horizontal side exceeds $3\sqrt{3}/4.$

 

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