In a given circle, we draw two diameters that are perpendicular to each other, and from an arbitrary point $P$ on the circumference of the circle, we drop perpendiculars to these diameters. Determine the position of $P$ such that the sum of the lengths of these perpendiculars is maximized.
If we denote the angle formed by the radius $OP = r$ of point $P$ with one of the perpendiculars as $\alpha$, then the sum of the perpendiculars dropped from the diameter is

$$
s=r(\sin \alpha+\cos \alpha)
$$

from which

$$
s^{2}=r^{2}\left(\sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha\right)
$$

or

$$
s^{2}=r^{2}(1+\sin 2 \alpha)
$$

and

$$
s=r \sqrt{1+\sin 2 \alpha}
$$

$s$ is maximum when $\sin 2 \alpha$ is the largest, i.e., when $\alpha=45^{\circ}$. Therefore, point $P$ is given by the intersection of the circle's circumference with the angle bisector of the right angle formed by the diameters.

(Miletits Ernö, Györ.)

Number of solutions: 29.