The base of the oblique parallelepiped ABCDA1B1C1D1 is a rhombus ABCD with side $a$ and an acute angle of $60^{\circ}$. The edge $A A1$ is also equal to $a$ and forms angles of $45^{\circ}$ with the edges $A B$ and $A D$. Find the volume of the parallelepiped.
Since the sum of two plane angles of a trihedral angle is greater than the third, the angle $B A D$ cannot be greater than $90^{\circ}$. Therefore, $\angle B A D=60^{\circ}$. Drop a perpendicular $A 1 H$ from the vertex $A 1$ of the parallelepiped $A B C D A 1 B 1 C 1 D 1$ to the plane of the base $A B C D$. Then from point $H$ drop perpendiculars $H P$ and $H Q$ to the lines $A B$ and $A D$ respectively. By the theorem of three perpendiculars, $A 1 P \perp A B$ and $A 1 Q \perp A D$. From the right triangles $A 1 A P$ and $A 1 A Q$ we find that

$$
A 1 P=A 1 Q=\frac{a}{\sqrt{2}}, A P=A Q=\frac{a}{\sqrt{2}}
$$

Since $H P=H Q$ (as orthogonal projections of equal oblique lines $A 1 P$ and $A 1 Q$), $A H$ is the bisector of angle $B A D$. Therefore,

$$
H P=A P \operatorname{tg} 30^{\circ}=\frac{a}{\sqrt{2}} \cdot \frac{1}{\sqrt{3}}=\frac{a}{\sqrt{6}} .
$$

From the right triangle $A 1 H P$ by the Pythagorean theorem we find that

$$
A 1 H=\sqrt{A_{1} P^{2}-H P^{2}}=\sqrt{\frac{a^{2}}{2}-\frac{a^{2}}{6}}=\frac{a}{\sqrt{3}}
$$

Therefore,

$$
V_{A B C D A} 1 B 1 C 1 D 1=S_{A B C D} \cdot A 1 H=a \cdot a \cdot \sin 60^{\circ} \cdot \frac{a}{\sqrt{3}}=\frac{\alpha^{2} \sqrt{3}}{2} \cdot \frac{a}{\sqrt{3}}=\frac{a^{2}}{2}
$$

## Answer

$\frac{a^{2}}{2}$.