A $30$-gon $A_1A_2\cdots A_{30}$ is inscribed in a circle of radius $2$. Prove that one can choose a point $B_k$ on the arc $A_kA_{k+1}$ for $1 \leq k \leq 29$ and a point $B_{30}$ on the arc $A_{30}A_1$, such that the numerical value of the area of the $60$-gon $A_1B_1A_2B_2 \dots A_{30}B_{30}$ is equal to the numerical value of the perimeter of the original $30$-gon.
I claim this can be done by setting $B_k$ to be the midpoint of $A_kA_{k+1}$ for all $1\leq k\leq 30$ (where $A_{31}\equiv A_1$).
Let $OB_1$ intersect $A_1A_2$ at $H_1$, and define $H_2$, $H_3,\cdots, H_{30}$ similarly. Note that since $OA_1=OA_2$ and $BA_1=BA_2$, we have that $OB_1\perp A_1A_2$ (or more specifically that $OB_1$ is the perpendicular bisector of $A_1A_2$). Then using the fact that the radius of the circle is $2$, we have that \begin{align*}[A_1B_1A_2B_2\ldots A_{30}B_{30}]&=\sum_{k=1}^{30}\left([A_kOA_{k+1}]+[A_kB_kA_{k+1}]\right)=\sum_{k=1}^{30}\left(\dfrac12\cdot OH_k\cdot A_kA_{k+1}+\dfrac12\cdot B_kH_k\cdot A_kA_{k+1}\right)\\&=\sum_{k=1}^{30}\dfrac12\cdot A_kA_{k+1}\cdot OB_k=\sum_{k=1}^{30}A_kA_{k+1},\end{align*} as desired. $\blacksquare$
