52. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$; a plane is drawn through the edge $A A_{1}$, forming equal angles with the lines $B C$ and $B_{1} D$. Find these angles.
52. Let's take a point \( K \) on the extension of edge \( C C_{1} \) such that \( B_{1} K \| B C_{1} \), and through edge \( B B_{1} \) we draw a plane parallel to the given one (Fig. 1). This plane should pass either through the internal or the external bisector of angle \( D B_{1} K \). Since the ratio in which the plane passing through \( B B_{1} \) divides \( D K_{1} \),

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-051.jpg?height=503&width=574&top_left_y=485&top_left_x=335)

Fig. 1.

is equal to the ratio in which it divides \( D C \), there are two cases: the plane passes through a point \( N \) on edge \( D C \) such that \( |D N| / |N C| = \sqrt{3} / \sqrt{2} \), or it passes through a point \( M \) on its extension and again \( |D M| / |M C| = \sqrt{3} / \sqrt{2} \). Let's find the distance from point \( K \) to the first plane. This distance is equal to the distance from point \( C \) to line \( B N \). If this distance is \( x \), then

\[
x = \frac{2 S_{B N C}}{|B N|} = \frac{a \sqrt{2}}{(\sqrt{3} + \sqrt{2}) \sqrt{11 - 4 \sqrt{6}}} = \frac{a (\sqrt{6} - 1) \sqrt{2}}{5}
\]

and

\[
\sin \varphi = \frac{x}{|B_{1} K|} = \frac{\sqrt{6} - 1}{5}
\]

where \( \varphi \) is the angle between the plane \( B B_{1} N \) and the lines \( B_{1} D \) and \( B_{1} K \). The other angle is found in exactly the same way.

Answer: \( \arcsin \frac{\sqrt{6} \pm 1}{5} \).