5. Sides $A B$ and $A D$ of parallelogram $A B C D$ are equal to 1 and $a$, respectively, $\angle B A D=\alpha$, and triangle $A B D$ is acute. For which $a$ and $\alpha$ will four circles of radius 1 with centers at the vertices of the parallelogram completely cover the parallelogram?
5. Let's describe .. around the acute-angled triangle $ABD \cdot$ circle $K$; let $O$ be the center of this circle (Fig. 34). We will show that if the radius $R$ of circle $K$ is greater than 1, then point $O$ will not be covered by any of the four circles with centers at $A, B, C, D$ and radius 1. Indeed, $AO = BO = DO = R > 1$, so the circles with centers at $A, B$, and $D$ will not cover point $O$. But the circle with center $C$ and radius 1 will also not cover it, because $CO > R > 1$: indeed, if point $C$ were located inside (or on the boundary) of circle $K$, then (acute!) $\angle BCD = \alpha$ would be no less than half the arc $BAD$, which is greater than $180^\circ$ because $\angle BDR$ is centered at point $A$. Similarly, it can be established that each of the triangles $APO, BMO, BNO, DNO$, and $DPO$ is entirely covered by a circle of radius 1 with a center at one of the points $A, B$, or $D$. Thus, three circles of radius 1 with centers at $A, B$, and $D$ cover the entire triangle $ABD$, from which, due to the central symmetry of the parallelogram, it follows that three circles of radius 1 with centers at $C, D$, and $B$ will completely cover the triangle $BCD$. Therefore, for four circles of radius 1 with centers at the vertices of the parallelogram to completely cover the parallelogram, it is necessary and sufficient that the radius $R$ of circle $K$ be no more than 1.

Now, by the Law of Sines applied to triangle $ABD$,

$$
2R = \frac{BD}{\sin \angle BAD} = \frac{BD}{\sin \alpha}
$$

But by the Law of Cosines, $BD = \sqrt{1 + a^2 - 2a \cos \alpha}$, and thus,

$$
R = \frac{\sqrt{1 + a^2 - 2a \cos \alpha}}{2 \sin \alpha}
$$

Thus, the condition that our four circles cover the parallelogram can be written as:

$$
\frac{\sqrt{1 + a^2 - 2a \cos \alpha}}{2 \sin \alpha} \leq 1
$$

or

$$
a^2 - 2a \cos \alpha + 1 - 4 \sin^2 \alpha \leq 0
$$

Since the equality $a^2 - 2a \cos \alpha + 1 - 4 \sin^2 \alpha = 0$ holds when

$$
a = \cos \alpha \pm \sqrt{\cos^2 \alpha - 1 + 4 \sin^2 \alpha} = \cos \alpha \pm \sqrt{3} \sin \alpha
$$

the solution to the last inequality is

$$
\cos \alpha - \sqrt{3} \sin \alpha \leq a \leq \cos \alpha + \sqrt{3} \sin \alpha
$$

But since $1 \cdot \cos \alpha$ is the projection of side $AB = 1$ of triangle $ABD$ onto side $AD = a$ and triangle $ABD$ is acute-angled, then $\cos \alpha < a$ and thus $\cos \alpha - \sqrt{3} \sin \alpha \leq a$.

Thus, four unit circles with centers at the vertices of parallelogram $ABCD$ will cover the parallelogram if and only if $a \leq \cos \alpha + \sqrt{3} \sin \alpha$.