A plane passes through a point $P$ in space. Three lines also pass through $P$. Construct a plane through $P$ that makes the same angle with each of the three lines.
First, we construct a plane $S$ that makes the same angle with each line but does not pass through the point $P$. If we translate this plane so that it passes through $P$, we get the desired plane.

Assume that the three lines are not in the same plane. Measure a given segment $d$ from $P$ in some direction on each line (see diagram). Let the endpoint of these segments different from $P$ be $A$, $B$, and $C$. We will show that the plane $S$ determined by the points $A$, $B$, and $C$ makes the same angle with each line. Let the orthogonal projection of $P$ onto the plane $S$ be $Q$. The triangles $P Q A$, $P Q B$, and $P Q C$ are congruent because they have two equal sides and the larger angle opposite to these sides. Therefore, the lines $P A$, $P B$, and $P C$ make the same angle with the plane $S$. The segment $d$ can be measured on the three lines in $2 \cdot 2 \cdot 2 = 8$ different ways; the 8 planes obtained on $s$ are symmetric with respect to $P$, so the problem generally has $\frac{8}{2} = 4$ solutions.

If the three lines are in the same plane, then this is the only plane that makes the same (zero-degree) angle with each line.

András Máthé (Budapest, Apáczai Cs. János Gymnasium, 10th grade)

![](https://cdn.mathpix.com/cropped/2024_05_02_7fc2c4b01332454be5b2g-1.jpg?height=545&width=672&top_left_y=722&top_left_x=716)