Fold Square into Equilateral Triangle

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

We shall first establish the location of the largest equilateral triangle inscribed into a square and then demonstrate how this triangle can be obtained by folding the square. The argument for the former task is mainly intuitive.

First of all, Dudeney is right claiming that the triangle with sides equal to those of the square can be the largest possible. Indeed, moving slightly one of the corners of the triangle along the side of the square, while keeping two of their corners fixed, will increase the side and hence the area of the triangle and keep its third vertex within the square:

In a similar vein, we may claim that exactly on of triangle's vertices should coincide with one of the square's. Assume that this is not so:

Then the triangle can be slid to place one of its vertices into one of the square's corners:

This would allow - as before - to slide one of the other vertices of the triangle, thus, increasing its side length and the area:

Dudeney offers the following procedure:

From the above it is clear that the largest equilateral triangle inscribed into a square is bound to have one vertex at a corner of the square and the other two on the two sides not adjacent to that vertex. Further, for the triangle $AGK$ to be equilateral we need $AG=AK,$ implying $\angle BAK=DAG=15^{\circ}$ because $\angle GAK=60^{\circ}.$

What is need is to show that, if $\angle BAK=DAG=15^{\circ},$ then $GK=AK.$

As we know,

$\displaystyle\sin 15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4},$ $\displaystyle\cos 15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4},$ $\displaystyle\tan 15^{\circ}=2-\sqrt{3}.$

Assuming $AB=1,$ $\displaystyle AK=\frac{4}{\sqrt{6}+\sqrt{2}}=\sqrt{6}-\sqrt{2},$ $BK=2-\sqrt{3},$ $CK=CG=\sqrt{3}-1.$ It follows that $GK=\sqrt{2(\sqrt{3}-1)^{2}}=\sqrt{6}-\sqrt{2}=AK.$

It remains to be shown that Dudeney's folding really leads to the configuration where $AGK$ is equilateral.

When $B$ is folded onto $E\in IH,$ $AE=2AI,$ so that $\angle BAE=60^{\circ}$ and, therefore, $\angle BAF=30^{\circ}.$ Further, angles $BFA,$ $AFG,$ $CFG$ are all $60^{\circ}.$ $\displaystyle BF=\tan 30^{\circ}=\frac{\sqrt{3}}{3},\;$ $\displaystyle CF=1-\frac{\sqrt{3}}{3}=\frac{3-\sqrt{3}}{3}.\;$ $CG=CF\tan 60^{\circ}:$

$\displaystyle CG=\frac{3-\sqrt{3}}{3}\sqrt{3}=\sqrt{3}-1=CK,$

as required.