Can it be proved by elementary geometry?
Suppose there are two circles $C_1$ and $C_2$ having respective radii $r_1$ and $r_2$, such that $r_1>r_2$, and respective centers $O_1$ and $O_2$.
Draw line $O_1O_2$ and a common external tangent which intersects circle 1 at $P_1$, circle 2 at $P_2$. These intersect at a point $Q$ such that $O_2$ lies between $O_1$ and $Q$. Draw radii $O_1P_1$ and $O_2O_P$ which are perpendicular to $P_1P_2$, thus parallel to one another. Thereby, triangles $O_1P_1Q$ and $O_2P_2Q$ are similar right triangles. From the proportionality of correspondung sides we then have:
where $s$ is the length of line segment $O_2Q$.
When we put $r_1=3, r_2=1$ this gives $s=2$ so that the right triangle $O_2P_2Q$ has a hypoteneuse twice as long as one leg. Then angle $O_2QP_2$ opposite this leg measures $30°$.
Now draw all three common tangents. They form an isosceles triangle enclosing the smaller circle, whose apex angle at $Q$ measures twice the angle $O_2QP_2$, or $60°$. The triangle is thereby certified equilateral.