For a regular triangle with side $a$, what fraction of the area of the circumscribed circle is covered by the circles of radius $\frac{a}{2}$ drawn around the vertices of the triangle?
The pairwise tangent circles with radius $a / 2$ can be rotated into each other by $120^{\circ}$, so their areas are equal. A part of the circles, consisting of a sector and two small segments, falls within the circumscribed circle. First, we will calculate the area of one of the sectors, specifically $A E D$. For this, we need to know the angle forming the sector, and due to axial symmetry, it is sufficient to determine half of it, denoted by $\beta$. In total, 6 such $\beta$-angled sectors fall within the circle. In addition, the covered area consists of 6 parts, each congruent to the segment $A D$ shown in the diagram. The area of these latter parts is calculated using the angle $\alpha$ of the sector $A O D$.

Consider the triangle $A O D$, where $O$ is the center of the circumscribed circle. Since the triangle is equilateral, the radius of its circumscribed circle is $2 / 3$ of its height, i.e., $r=\frac{2}{3} \frac{a \sqrt{3}}{2}=\frac{a \sqrt{3}}{3}$.

Let's apply the cosine rule to the angle $\beta$ in the triangle $A O D$:

$$
\begin{gathered}
\cos \beta=\frac{a^{2}+c^{2}-b^{2}}{2 a c}= \\
=\frac{\left(\frac{a \sqrt{3}}{3}\right)^{2}+\left(\frac{a}{2}\right)^{2}-\left(\frac{a \sqrt{3}}{3}\right)^{2}}{2 \cdot \frac{a \sqrt{3}}{3} \cdot \frac{a}{2}}=\frac{\sqrt{3}}{4}
\end{gathered}
$$

Thus, $\cos \beta=0.4330, \beta \approx 64^{\circ} 20'$, in radians:

$$
\begin{gathered}
360^{\circ}: 2 \pi=64.34^{\circ}: \beta \\
\beta=1.1230 \text { radians. }
\end{gathered}
$$

The area of one sector:

$$
\frac{\left(\frac{a}{2}\right)^{2} \cdot 1.1230}{2}=a^{2} \cdot \frac{1}{8} \cdot 1.1230
$$

The area of the segments is $\frac{1}{2} r^{2}(\widehat{\alpha}-\sin \alpha)$, where $\widehat{\alpha}$ is the value of the angle $\alpha$ in radians. (This formula can be found in function tables, but it can also be determined from the area of the sector and the triangle.)

$$
\alpha=180^{\circ}-2 \cdot 64.34^{\circ}=51.32^{\circ}
$$

Expressed in radians: 0.8957.

$$
T_{\text {segment }}=\frac{1}{2}\left(\frac{a \sqrt{3}}{3}\right)^{2}\left(0.8957-\sin 51.32^{\circ}\right)=\frac{1}{6} a^{2} \cdot 0.1150
$$

The area of the circumscribed circle: $\left(\frac{a \sqrt{3}}{3}\right)^{2} \pi$.

Finally, by dividing the sum of six times (1) and (2) by the area of the circumscribed circle, we get the desired ratio:

$$
\frac{6\left(a^{2} \cdot \frac{1}{8} \cdot 1.1230+\frac{1}{6} a^{2} \cdot 0.1150\right)}{\frac{a^{2}}{3} \cdot \pi}=\frac{\frac{3}{4} \cdot 1.1230+0.1150}{\frac{\pi}{3}} \approx 0.9140
$$

Thus, the circles cover more than 9/10 of the area of the circumscribed circle. (The calculations were performed with an accuracy of 4 decimal places.)

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