In a circle with radius $R$, by drawing a chord perpendicular to the diameter and connecting its endpoints with the endpoints of the diameter, under what condition will the difference in the areas of the two triangles formed with the chord as the common base be maximal?
I. solution. Let the sought chord be $2 x$, then according to the problem

$$
y=\left(R+\sqrt{\left.R^{2}-x^{2}\right)} x-\left(R-\sqrt{R^{2}-x^{2}}\right) x=2 x \sqrt{\left(R^{2}-x^{2}\right)}\right.
$$

or

$$
y=2 \sqrt{x^{2}\left(R^{2}-x^{2}\right)}
$$

Since $x^{2}+R^{2}-x^{2}=R^{2}$, in the case of maximum

$$
R^{2}-x^{2}=x^{2}
$$

and thus

$$
x=\frac{R}{2} \sqrt{2} \quad \text { and } \quad 2 x=R \sqrt{2}
$$

Therefore, the side of the square inscribed in the circle meets the condition.

II. solution. Let the chord perpendicular to the diameter be $2 x$, its central angle $2 \alpha$, then

$$
y=r \sin \alpha(r+r \cos \alpha)-r \sin \alpha(r-r \cos \alpha)=2 r^{2} \sin \alpha \cos \alpha
$$

or

$$
y=r^{2} \sin 2 \alpha
$$

This function is maximal when $2 \alpha=90^{\circ}$, in which case $y=r^{2}$.

(Béla Enyedi, Budapest.)

Number of solutions: 36.