A rectangular piece of paper has two opposite corners folded in such a way that the resulting fold is an equilateral hexagon. When is this task possible?

 untranslated portion:
Egy derékszögú négyszög alakú papírlapnak két szemközt fekvő csúcsát hajtsuk be úgy, hogy a keletkező idom egyenlő oldalú hatszög legyen. Mikor lehetséges a feladat. 

Translated:
A rectangular piece of paper has two opposite corners folded in such a way that the resulting fold is an equilateral hexagon. When is this task possible?
I. Solution. If the sides of the right-angled quadrilateral are $a, b$; the side of the question hexagon is $x$, then

$$
(a-x)^{2}+(b-x)^{2}=x^{2}
$$

from which

$$
x=a+b \pm \sqrt{2 a b}
$$

Only the negative value of the square root can be considered, because $x<\frac{1}{2}(a+b)$

II. Solution. Let the given rectangle be $A B C D$. According to the problem, points $E, F, G$, and $H$ must be found on the sides $A B, B C, C D$, and $D A$ respectively, such that $E B=B F=F G=D G$. The points $E$ and $H$ are obtained by the following construction:

From $A$, draw a circle with radius $A D$, which intersects $A B$ at $D_{1}$. From $D_{1}$, draw a circle with radius $D_{1} A$, which intersects the diagonal $B D$ at $A_{1}$. Complete the triangle $A D_{1} A_{1}$ to a rhombus, whose fourth vertex is $B_{1}$. The side $B_{1} D A B$ intersects at $E$ and the parallel to $A B_{1}$ from this point intersects the side $A D$ at $H$. The points $E$ and $H$ give the two points sought. Indeed, by the similarity theorem, we can write:

$$
\frac{A B_{1}}{E H}=\frac{A D}{H D}=\frac{B_{1} D}{E D}=\frac{B_{1} A_{1}}{E B}
$$

and thus

$$
E H=H D=E B
$$

(Kornis Ödön.)

The problem was also solved by: Filkorn J., Freibauer E., Glass M., Juvancz I., Kárf J., Kohn B., Krausz B., Krisztián Gy., Obláth R., Neumann J., Pollák N., Porkoláb J., Sasvári G., Sasvári J., Spitzer Ö., Tinyó J., Weisz J.