Problem 1614

1. **Choosing Points $B_k$:**
We claim that we can choose $B_k$ to be the midpoint of the arc $A_kA_{k+1}$ for all $1 \leq k \leq 30$ (where $A_{31} \equiv A_1$).

2. **Properties of $B_k$:**
Since $B_k$ is the midpoint of the arc $A_kA_{k+1}$, the line $OB_k$ (where $O$ is the center of the circle) is the perpendicular bisector of the chord $A_kA_{k+1}$. This implies that $OB_k \perp A_kA_{k+1}$.

3. Area Calculation:
We need to calculate the area of the $60$-gon $A_1B_1A_2B_2 \ldots A_{30}B_{30}$. We can break this area into smaller triangles and sum their areas.

4. Triangles Involved:
Each segment $A_kB_kA_{k+1}$ can be divided into two triangles: $\triangle A_kOB_k$ and $\triangle B_kOA_{k+1}$.

5. Area of Each Triangle:
Since $OB_k$ is perpendicular to $A_kA_{k+1}$, the area of $\triangle A_kOB_k$ and $\triangle B_kOA_{k+1}$ can be calculated using the formula for the area of a triangle:
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$
Here, the base is $A_kA_{k+1}$ and the height is the radius of the circle, which is $2$.

6. Summing the Areas:
The total area of the $60$-gon is the sum of the areas of all these triangles:
$$\text{Total Area} = \sum_{k=1}^{30} \left( \text{Area of } \triangle A_kOB_k + \text{Area of } \triangle B_kOA_{k+1} \right)$$
Since each triangle has an area of $\frac{1}{2} \times A_kA_{k+1} \times 2 = A_kA_{k+1}$, the total area is:
$$\text{Total Area} = \sum_{k=1}^{30} A_kA_{k+1}$$

7. **Perimeter of the Original $30$-gon:**
The perimeter of the original $30$-gon is:
$$\text{Perimeter} = \sum_{k=1}^{30} A_kA_{k+1}$$

8. Conclusion:
The numerical value of the area of the $60$-gon $A_1B_1A_2B_2 \ldots A_{30}B_{30}$ is equal to the numerical value of the perimeter of the original $30$-gon.

$\blacksquare$