[ [ characteristics of perpendicularity]

The diagonal of a rectangular parallelepiped is equal to $a$ and forms an angle of $30^{\circ}$ with one face and $45^{\circ}$ with another. Find its volume.
Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a rectangular parallelepiped, in which $B D_{1}=a$. Then $A D_{1}$ and $C D_{1}$ are the orthogonal projections of the oblique lines $B D_{1}$ and $C D_{1}$ on the planes $A A_{1} D_{1} D$ and $D D_{1} C_{1} C$ respectively.

## Solution

Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a rectangular parallelepiped, in which $B D_{1}=a$. The line $A B$ is perpendicular to the intersecting lines $A D$ and $A A_{1}$ of the plane $A A_{1} D_{1} D$, so the line $A B$ is perpendicular to the plane $A A_{1} D_{1} D$. Therefore, $A B \perp A D_{1}$ and $A D_{1}$ is the orthogonal projection of the oblique line $B D_{1}$ on the plane $A A_{1} D_{1} D$. Similarly, we can prove that $C D_{1}$ is the orthogonal projection of the oblique line $B D_{1}$ on the plane $D D_{1} C_{1} C$. Thus, $A D_{1} B$ and $C D_{1} B$ are the angles of the line $B D_{1}$ with the planes $A A_{1} D_{1} D$ and $D D_{1} C_{1} C$ respectively.

Let $\angle A D_{1} B=30^{\circ}, \angle C D_{1} B=45^{\circ}$. From the right triangles $A B D_{1}, B C D_{1}$, and $C D D_{1}$, we find that

$$
\begin{gathered}
A B=B D_{1} \cdot \sin 30^{\circ}=\frac{1}{2} a, B C=B D_{1} \cdot \sin 45^{\circ}=\frac{1}{2} a \sqrt{2}, \\
C D_{1}=B D_{1} \cdot \cos 45^{\circ}=\frac{1}{2} a \sqrt{2} \\
D D_{1}=\sqrt{C D^{2}-C D^{2}}=\sqrt{C D^{2}-A B^{2}}=\sqrt{a^{2} / 2-a^{2} / 4}=\frac{1}{2} a
\end{gathered}
$$

Therefore,

$$
V\left(A B C D A_{1} B_{1} C_{1} D_{1}\right)=A B \cdot B C \cdot D D_{1}=\frac{1}{8} a^{3} \sqrt{2}
$$

## Answer

$$
\frac{1}{8} a^{3} \sqrt{2}
$$