Two chords of a circle intersect on the circle's circumference, with lengths $2a$ and $2b$ respectively; the perpendicular distance from the midpoint of one chord to the other is $d$. What is the radius of the circle?
Let the angle enclosed by the chords be $\alpha$, and the length of the straight line connecting the endpoints of the two chords be $c$. From the triangle thus formed:

$$
\begin{gathered}
c^{2}=4 a^{2}+4 b^{2}-8 a b \cos \alpha \\
\cos \alpha=\sqrt{1-\sin ^{2} \alpha} \\
\sin \alpha=\frac{d}{a}
\end{gathered}
$$

and thus

$$
\cos \alpha=\sqrt{\frac{a^{2}-d^{2}}{a^{2}}}
$$

Substituting this into (1):

$$
c^{2}=4 a^{2}+4 b^{2}-8 b \sqrt{a^{2}-d^{2}}
$$

The radius of the circle circumscribed around the triangle is

$$
r=\frac{c}{2 \sin \alpha}=\frac{a c}{2 d}
$$

Substituting (3) into (4):

$$
r=\frac{a}{d} \sqrt{a^{2}+b^{2}-2 b \sqrt{a^{2}-d^{2}}}
$$

If the angle enclosed by the two chords is obtuse, then:

$$
r=\frac{a}{d} \sqrt{a^{2}+b^{2}+2 b \sqrt{a^{2}-d^{2}}}
$$

If $a=d$, then

$$
r=\sqrt{a^{2}+b^{2}}
$$

(Szabó István, secondary school 6th grade, Debrecen.)

The problem was solved by: Friedmann Bernát, S.-A.-Ujhely; Galter János, Sz.-Udvarhely; Geiszt Emil, Győr; Grünhut Béla, Pécs; Kántor Nándor, Budapest; Oberle Károly, Budapest; Visnya Aladár, Pécs.