Let the common area be $A$. Triangle: $\frac{\sqrt{3}}{4}a^2 = A$, so the perimeter is $3a = 2\sqrt{3\sqrt{3}}\sqrt{A}$. Square: perimeter $4\sqrt{A}$. Circle: $2\pi r = 2\sqrt{\pi}\sqrt{A}$. Dividing by $2\sqrt{A}$, the ratio is $\sqrt{3\sqrt{3}} : 2 : \sqrt{\pi}$.
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