In the figure shown $$ABC$$ is a uniform wire. If the center of mass of the wire lies vertically below point $$A$$, then $$\frac{BC}{AB}$$ is close to

Place point $$B$$ at the origin $$(0, 0)$$ with segment $$BC$$ along the x-axis.

Segment $$AB$$ (Length $$L_1$$): Center of mass is at its midpoint. $$x_1 = \frac{L_1 \cos 60^\circ}{2} = \frac{L_1}{4}$$

Segment $$BC$$ (Length $$L_2$$): Center of mass is at its midpoint. $$x_2 = \frac{L_2}{2}$$

Point $$A$$: The x-coordinate is $$L_1 \cos 60^\circ = \frac{L_1}{2}$$.

The problem states the system COM lies vertically below $$A$$, so $$X_{com} = x_A$$:

$$\frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{L_1}{2}$$

$$\frac{L_1(\frac{L_1}{4}) + L_2(\frac{L_2}{2})}{L_1 + L_2} = \frac{L_1}{2}$$

$$L_1^2 + 2L_2^2 = 2L_1(L_1 + L_2)$$

$$2L_2^2 - 2L_1L_2 - L_1^2 = 0$$

$$r = \frac{L_2}{L_1}$$: $$2r^2 - 2r - 1 = 0$$

$$r = \frac{2 + \sqrt{4 - 4(2)(-1)}}{2(2)} = \frac{2 + \sqrt{12}}{4} = \frac{1 + \sqrt{3}}{2}$$

$$r \approx \frac{1 + 1.732}{2} = \mathbf{1.366}$$