On the extended side $C D$ of the square $A B C D$, find the point $M$ for which the ratio $M A: M B$ is maximized.
Let one side of the square be $a$, $MC = x$, and $\frac{MA}{MB} = y$; then

$$
y^{2} = \frac{\overline{MA}^{2}}{\overline{MB}^{2}} = \frac{a^{2} + (a + x)^{2}}{a^{2} + x^{2}}
$$

or

$$
x^{2}(y^{2} - 1) - 2ax + a^{2}(y^{2} - 2) = 0
$$

For $x$ to be real, the discriminant of the equation must be non-negative, i.e.,

$$
a^{2} - a^{2}(y^{2} - 1)(y^{2} - 2) \geq 0,
$$

which implies the condition for the maximum:

$$
(y^{2} - 1)(y^{2} - 2) = 1
$$

or

$$
y^{4} - 3y^{2} + 1 = 0
$$

from which

$$
y^{2} = \frac{3 \pm \sqrt{5}}{2}
$$

Since $y > 1$ by the problem's meaning, the radical must be taken with a positive sign, so

$$
y^{2} = \frac{3 + \sqrt{5}}{2}
$$

thus

$$
x = \frac{2a}{1 + \sqrt{5}}
$$

or

$$
\frac{a}{2x} = \frac{1 + \sqrt{5}}{4} = \sin 54^{\circ}
$$

Therefore, $2MC = 2x$ is the hypotenuse of a right triangle, one of whose legs is $a$ and the angle opposite this leg is $54^{\circ}$, and thus this triangle can be constructed.

(Béla Bayer, Losoncz.)

The problem was also solved by: E. Baranyó, I. Bartók, A. Blau, A. Haar, Gy. Hirschfeld, F. Kertész, P. Pilczer, I. Pivnyik, K. Riesz, J. Sasvári, A. Schlesinger, H. Szmodics, L. Tóbiás.