The Shortest Crease

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

References

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

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 $x=BE.$

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 $3\sqrt{3}/4.$