$X O Y 60^{\circ}$-angle $O X$ side, a point $A$ is taken such that $O A=a$. At point $A$, a perpendicular to $O X$ is drawn, intersecting $O Y$ at point $D$. A semicircle is constructed over $A D$, which intersects $O Y$ at point $B$. Questions:

1) What is the length of $B D$?
2) What is the surface area of the solid described by $B D$ when it rotates around $O X$?
3) What is the volume of the solid described by the triangle $A B D$ when it rotates around $O X$?
$1^{\circ}$. In the right-angled triangle $AB$ perpendicular to $OY$, we can write:

$$
\begin{aligned}
& B D: A D = A D: O D \\
& B D: a \sqrt{3} = a \sqrt{3}: 2 a
\end{aligned}
$$

from which

$$
B D = \frac{3 a}{2}
$$

$2^{\circ}$. From point $B$, we draw a perpendicular to $OX$, which intersects $OX$ at point $E$.

$$
\begin{gathered}
O B = O D - B D = \frac{a}{2}, \quad B E = O B \sin 60^{\circ} = \frac{a \sqrt{3}}{4} \\
O E = O B \cos 60^{\circ} = \frac{a}{4}, \quad A E = \frac{3 a}{4}
\end{gathered}
$$

The lateral surface area of the frustum described by $BD$:

$$
F = (B E + A D) \pi B D = \frac{15 a^{2} \pi \sqrt{3}}{8}
$$

$3^{\circ}$. The volume of the solid described by the triangle $ABD$ is obtained by subtracting the volume of the cone described by $AEB$ from the volume of the frustum described by $AEBD$; thus,

$$
\begin{gathered}
K = \frac{A E \pi}{3} \left( \overline{A D}^{2} + A D \cdot B E + \overline{B E}^{2} \right) - \frac{\overline{B E}^{2} \pi A E}{3} \\
K = \frac{A E \pi}{3} \left( \overline{A D}^{2} + A D \cdot B E \right) = \frac{15 a^{3} \pi}{16}
\end{gathered}
$$

(Szabó István, real gymnasium, VII. class, Debrecen.)

The problem was also solved by: Devecis Mihály, Feuer Mór, Freund Antal, Friedmann Bernát, Geist Emil, Goldstein Zsigmond, Grünhut Béla, Hofbauer Ervin, Kántor Nándor, Klein Mór, Riesz Frigyes, Schiffer Hugó, Simon Elek, Schwarz Endre, Szabó Károly.