$1^{\circ}$. A rectangle (e.g., made of paper) needs to be created that remains similar in shape when folded so that the shorter sides overlap each other.

$2^{\circ}$. a) A rectangle needs to be created that remains similar in shape when a square is cut off from the shorter side.

b) Starting from such a rectangle, imagine the procedure described in a) being repeated indefinitely and determine the sum of the areas of the rectangles that result from this process.
$1^{\circ}$. Let the sought right-angled quadrilateral be $A B C D$, where $A B < A D$; if the midpoint of $A D$ is denoted by $M$, and that of $B C$ by $N$, then

$$
\frac{A B}{A D}=\frac{A M}{A B}=\frac{A D}{2 A B}
$$

and thus

$$
\overline{A D}^{2}=2 \overline{A B}^{2}, \text { or } A D=A B \sqrt{2}
$$

The larger side of the sought right-angled quadrilateral is therefore the hypotenuse of a right-angled triangle, each leg of which is $A B$, the smaller side of the quadrilateral.

$2^{\circ}$. a) Let $A B C D$ again be the quadrilateral to be constructed, where $A B < A D$. The line $M N$ cuts off the square $A B N M$ from this quadrilateral; according to the problem:

$$
\frac{A B}{A D}=\frac{A D-A B}{A B} \quad \text { and thus } \quad \overline{A B}^{2}=A D(A C-A B)
$$

Therefore, the smaller side of the sought quadrilateral, $A B$, is the larger segment of the larger side, $A D$, divided according to the golden ratio.

b) Since $A B C D \sim M N C D$, we have

$$
\frac{A B C D}{M N C D}=\frac{\overline{A D}^{2}}{\overline{A B}^{2}} \quad \text { and thus } \quad M N C D=A B C D \cdot \frac{\overline{A B}^{2}}{\overline{A D}^{2}}
$$

If we cut off a square from the quadrilateral $M N C D$ in the manner indicated in the problem, and denote the area of the remaining rectangle by $t$, we get:

$$
\frac{M N C D}{t}=\frac{\overline{A D}^{2}}{\overline{A B}^{2}}
$$

from which, considering (1):

$$
t=A B C D \cdot \frac{\overline{A B}^{4}}{\overline{A D}^{4}}
$$

we see that the areas of these rectangles form a decreasing geometric progression, the common ratio of which is $\frac{\overline{A B}^{2}}{\overline{A D}^{2}}$; thus the sum of the areas is:

$$
T=\frac{A B C D}{1-\overline{A B}_{\overline{A D}^{2}}^{2}}=\overline{A D}^{2} \cdot \frac{A B \cdot A D}{\overline{A D}^{2}-\overline{A B}^{2}}=\overline{A D}^{2}
$$

Thus, the sum of the areas of the right-angled quadrilaterals is equal to the area of the square constructed on the larger side.

(Odön Kornis.)

The problem was also solved by: Barabás S., Breuer M., Czank K., Filkorn J., Freibauer E., Kohn B., Krausz B., Gy. Krisztián, Gy. Lupta, R. Obláth, Gy. Perl, J. Porkoláb, G. Sasvári, J. Weisz.