Without loss of generality, let's assume the side length of the biggest triangle equals $a$ and $BA_1=x$. Then $A_1C=a-x$ and $B_1C=x$, as shown below.

Let's also assume the side length of $\triangle{A_1B_1C_1}$ equals $a_1$. Consider $\triangle{A_1CB_1}$, we have $\angle{C}=60^\circ$ and $\angle{A_1B_1C}=120^\circ-\beta$. By the Law of Sine, we have $$\frac{a_1}{\sin{60^\circ}}=\frac{x}{\sin{\beta}}=\frac{a-x}{\sin{(120^\circ-\beta)}}$$ Canceling $x$ leads to $$a-\frac{\sin{\beta}}{\sin{60^\circ}}a_1 = \frac{\sin{(120^\circ-\beta)}}{\sin{60^\circ}}a_1$$ Or $$a_1=\frac{a\sin{60^\circ}}{\sin{\beta}}+\sin{(120^\circ-beta)}=\frac{a}{2\sin{(\beta+30^\circ)}}$$ Note that $\frac{1}{2\sin{(\beta+30^\circ)}}$ is a constant with respect to $\beta$. We find the side lengths of all these triangles form a geometric sequence whose common ratio $r=\frac{1}{2\sin{(\beta+30^\circ)}}$. It follows that their areas must also form a geometric sequence with a common ratio as $r^2 = \frac{1}{(2\sin{(\beta+30^\circ)})^2}$. By the sum of infinite geometric sequence formula, the following must hold: $$1=\frac{\dfrac{1}{(2\sin{(\beta+30^\circ)})^2}}{1-\dfrac{1}{(2\sin{(\beta+30^\circ)})^2}}$$ Solving the above equation yields $\sin{(\beta+30^\circ)}=\frac{\sqrt{2}}{2}$, where $0 < \beta < 60^\circ$, or, $\beta=\boxed{15^\circ}$.