From a point on the chord of a circle, we draw a secant. In which position of the secant will the area of the triangle be maximal, one vertex of which is the center of the circle, and the other two vertices are determined by the secant on the circumference of the circle? Construct this secant.
Given a circle with radius $r$, $C$ is the center of the circle, $A$ and $B$ are the two intersection points of a secant with the circle's circumference, and $a$ is the central angle corresponding to the chord $AB$: then the area of triangle $ABC$ is $t=\frac{r^{2}}{2} \sin \alpha$, which has a maximum value when $\alpha=90^{\circ}$ and $AB=r \sqrt{2}$.

Construction: Draw a circle with radius $\frac{r}{2} \sqrt{2}$ from point $C$ and construct a tangent from the given point $P$ to this circle; this tangent is the desired secant.

(Spitzer Vilmos, Pécs.)

Number of solutions: 42.