A regular hexagon $A B C D E F$ lying in the first picture plane is to be rotated about its side $A B$ until the first projection of the diagonal $B D$ is half of its true size. Construct the projections of the hexagon after the rotation and the first inclination angle of the raised plane.

 untranslated portion:  
请注意，原文中的数学符号和字母组合（如 $A B C D E F$、$A B$、$B D$）在翻译中保持不变，以确保数学表达的准确性。
First solution. $BD$ is the true size of a first trace line of the raised plane, whose first projection is $BD'$. Transform the hexagon to a new picture plane that is perpendicular to the first picture plane and $CF$; then the third image of the hexagon is a straight line, which we obtain if we draw a perpendicular from $D'$ to $t_{13}$ and intersect this perpendicular with a radius of length $BD$ from $A'''$.

![](https://cdn.mathpix.com/cropped/2024_05_02_8bdd37b1b3596237147eg-1.jpg?height=734&width=699&top_left_y=300&top_left_x=710)

This implies that the first inclination angle of the raised hexagon is $\alpha=60^{\circ}$.

From the third image, the first and second images of the hexagon can be easily reconstructed.

(Vince Szekeres, Pécs.)

Second solution. If the first projection of $BD$ is doubled, then it is the hypotenuse of a $60^{\circ}$ right triangle, whose smaller leg is $\frac{BD}{2}$, the first projection, and the other leg is $\frac{BD}{2} \sqrt{3}$, the first director of point $D$.

In this way, the two projections of point $D$ can be constructed, as well as those of point $E$. The first projections of points $C$ and $F$ lie on a line parallel to $AB$ through the midpoint of $BD'$, and their second projections lie on a line parallel to them through the midpoint of the distance $A''B''$ and $D''E''$.

(Pál Hajdu, Budapest.)