10.337. Inside an equilateral triangle with side $a$, there are three equal circles, each of which touches two sides of the triangle and two other circles. Find the area of the part of the triangle that is outside these circles.
Solution.

Let $x$ be the radius of each of the circles. Then $A D=x \sqrt{3}, D E=2 x$, $E C=x \sqrt{3}$ (Fig. 10.122). Hence, $2 x \sqrt{3}+2 x=a, x=\frac{a}{2(\sqrt{3}+1)}=\frac{a(\sqrt{3}-1)}{4}$. The area of each of the circles is $\pi x^{2}=\frac{\pi a^{2}(2-\sqrt{3})}{8}$; therefore, the desired area $S=\frac{a^{2} \sqrt{3}}{4}-\frac{3 \pi a^{2}(2-\sqrt{3})}{8}=\frac{a^{2}(2 \sqrt{3}-6 \pi+3 \pi \sqrt{3})}{8}$.

Answer: $\frac{a^{2}(2 \sqrt{3}-6 \pi+3 \pi \sqrt{3})}{8}$.