When a coin is placed on a rotating table and is just about to slip, the centrifugal force acting on the coin equals the maximum static friction force. Let's denote the mass of the coin as $$m$$, the initial angular velocity as $$\omega_1$$, and the final angular velocity as $$\omega_2$$.

Initially, when the coin is placed at a distance of 1 cm from the center, the centrifugal force acting on the coin is:

$$F_1 = m r_1 \omega_1^2$$

where $$r_1 = 1 \mathrm{~cm}$$.

When the angular velocity is halved ($$\omega_2 = \frac{1}{2}\omega_1$$), the centrifugal force acting on the coin when it just slips is:

$$F_2 = m r_2 \omega_2^2 = m r_2 \left(\frac{1}{2}\omega_1\right)^2$$

Since the coin is just about to slip in both cases, the maximum static friction force remains the same. Therefore, we can equate the centrifugal forces:

$$m r_1 \omega_1^2 = m r_2 \left(\frac{1}{2}\omega_1\right)^2$$

Canceling the mass $$m$$ and the initial angular velocity $$\omega_1^2$$ from both sides, we get:

$$r_1 = r_2 \left(\frac{1}{2}\right)^2$$

Now, we can solve for $$r_2$$:

$$r_2 = \frac{r_1}{\left(\frac{1}{2}\right)^2} = \frac{1 \mathrm{~cm}}{\frac{1}{4}} = 4 \mathrm{~cm}$$

So, the coin will just slip when placed at a distance of 4 cm from the center when the angular velocity is halved.