Henry Ernest Dudeney/Modern Puzzles/139 - A Crease Problem/Solution

by : $139$

 * A Crease Problem
 * 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.
 * It will be seen that the crease $AB$ is considerably longer than $CD$, but the latter is not the shortest possible.


 * Dudeney-Modern-Puzzles-139.png

Solution

 * Dudeney-Modern-Puzzles-139-solution.png

Bisect $AB$ at $C$ and construct $CG$ parallel to $BH$.

Bisect $AC$ at $D$ and construct the semicircle between $B$ and $D$.

Let the semicircle $BD$ intersect $CG$ at $E$.

Let $DE$ be produced to intersect $BH$ at $F$.

Then $DF$ is the required crease.

Proof
First we assume that the length of the page is sufficiently long.

Otherwise simply folding the page in half could produce a shorter crease.

Suppose $t$ of the length of the bottom edge is folded over.

Then $1 - t$ of the length of the bottom edge is not folded over.

The angle adjacent to the bottom edge of the small triangle produced after folding is therefore:
 * $\cos^{-1} \paren {\dfrac {1 - t} t}$

The angle of the crease with respect to the bottom edge would be:
 * $\dfrac 1 2 \paren {180^\circ - \cos^{-1} \paren {\dfrac {1 - t} t} } = 90^\circ - \dfrac 1 2 \cos^{-1} \paren {\dfrac {1 - t} t}$

The length of the crease is therefore:

of the length of the bottom edge.

Let $\map f t = \dfrac {2 t^3} {2 t - 1}$.

Then $\map {f'} t = \dfrac {6 t^2 \paren {2 t - 1} - 2 \paren {2 t^3} } {\paren {2 t - 1}^2} = \dfrac {8 t^3 - 6 t^2} {\paren {2 t - 1}^2}$

$\map {f'} t = 0$ when $t = 0$ or $t = \dfrac 3 4$.

Hence the maximum length of the crease occurs when $t = \dfrac 3 4$, which is shown in the construction in the solution.