A straight line can be drawn through a given point, intersecting a given line and forming an angle $\alpha$ with a given plane.
Solution. The lines inclined at an angle $\alpha$ to the given plane form the generators of a conical surface, whose vertex is the given point. The generators of this cone passing through the common points of the given line and the conical surface are the desired lines. There can be 2, 1, or 0 solutions depending on whether the line intersects, touches, or lies outside the cone.

(Ehrenfeld Nándor, Nyitra.)

Let the given point be denoted by $A$, the line by $a$, and the plane by $s$.

The axis of the mentioned cone is perpendicular to $s$; its intersection point with $s$ is $O$, the center of the base circle of the cone. The radius of this circle, $r$, can be determined as the leg of a right triangle, where the other leg is the real distance $d$ between $O$ and $A$, and the angle opposite this leg is $\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_02_bb7565d40ff379651decg-1.jpg?height=970&width=897&top_left_y=594&top_left_x=560)

By rotating the first trace of the plane $s$ around the first projection plane, we obtain the center $\left(O_{1}\right)$, around which the base circle of the cone can be drawn with radius $r$.

To find the common points of the line $a$ and the cone, we draw a plane ${ }_{1} s$ through $a$ and the vertex $A$ of the cone; this plane intersects the given plane at a line $b$. By rotating $b$, we obtain the rotated points $(X)_{1}$ and $(Y)_{1}$ of the points $X$ and $Y$, through which the desired generators pass. Finally, these generators $X A$ and $Y A$ intersect $a$ at the desired points $X_{1}$ and $Y_{1}$. (In our diagram, only the point $Y_{1}$ is shown, while the other point $X_{1}$ lies outside the paper frame.)

The problem was also solved by: Lengyel M. and Schlesinger K.