= Solution
Let the <equilateral triangle> have side <length> $a$. Its altitude is $H=\sqrt3a/2$, and its <incircle> has <radius> $r=H/3=a/(2\sqrt3)$: the <centre> lies on a median, one third of the altitude above the base. The <area> of the triangle is $aH/2=\sqrt3a^2/4$, while the <circle> has <area> $\pi r^2=\pi a^2/12$. Therefore
$$
\boxed{\frac{\mathcal A_{\rm circle}}{\mathcal A_{\rm triangle}}
=\frac{\pi}{3\sqrt3}=\frac{\pi\sqrt3}{9}\approx0.604600.}
$$
The side <length> cancels, as required for a dimensionless ratio between similar configurations.
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