Regular pentagon sides are drawn as diameters of circles. Consider the circle that encloses these 5 circles and touches them. What fraction of the area of the large circle is covered by the 5 small circles?
Let the pentagon be denoted as $A B C D E$, the center of its inscribed circle as $O$, the radius of the inscribed circle as $\varrho$, and the midpoint of side $A B$ as $F$.

$1986-01-016-1$. eps

Since the pentagon is regular, $\angle B O F = \angle F O A = 36^{\circ}$. Our task is to determine the area ratios, so we can choose the side length of the pentagon to be 2 units, in which case the radius of the circle circumscribed around $A B$ is $r=1$, and $\varrho=\operatorname{ctg} 36^{\circ}$, so the radius of the large circle is $R=\varrho+r=1+\operatorname{ctg} 36^{\circ}$.

The circles circumscribed around the adjacent sides $C B$ and $B A$ of the pentagon intersect at a point different from $B$, let's denote this point as $P$, which is inside the pentagon.

The area $T$ of the covered region can be obtained by subtracting the area of the doubly covered parts from the total area of the circles. These parts are small circular segments, 10 in total, each being the difference between a sector of a circle and an isosceles triangle.

We can easily determine the central angle of the sector, for example, in the triangle $P F B$, $\angle O B F = 54^{\circ}$, and since $F B = F P$, $\angle P F B = 72^{\circ}$, so the area of the sector is 1/5 of the area of the small circle. The area of the isosceles triangle $P F B$ is, according to the known formula $t=\frac{a b \sin \gamma}{2}$, $\frac{r^{2} \sin 72^{\circ}}{2}$. Thus, $T=5 \cdot T_{\text {small circle }}-10\left(T_{\text {sector }}-T_{\text {triangle }}\right)=3 T_{\text {small circle }}+10 T_{\text {triangle }}=$ $3 \pi+5 \sin 72^{\circ}$.

Finally, the ratio of the areas covered by the 5 small circles and the area of the large circle is

$$
\frac{3 \pi+5 \sin 72^{\circ}}{\left(1+\operatorname{ctg} 36^{\circ}\right)^{2} \pi} \approx 0.8
$$