Let $P=A_1{A}_2\cdots A_k$ be a convex polygon in the plane. The vertices $A_1,A_2,\dots ,A_k$ have integral coordinates and lie on a circle. Let $S$ be the area of $P$. An odd positive integer $n$ is given such that the squares of the side lengths of $P$ are all divisible by $n$. Prove that $2S$ is divisible by $n$.

Step 1 (Shoelace formula): By the Shoelace formula, $2S=|{\sum }_{i=1}^k(x_i{y}_{i+1}-x_{i+1}{y}_i)|$ where indices are cyclic. Each term $x_i{y}_{i+1}-x_{i+1}{y}_i$ is an integer since coordinates are integral.

Step 2 (Side lengths mod $n$): The squared side length $|A_i{A}_{i+1}{|}^2=(x_{i+1}-x_i{)}^2+(y_{i+1}-y_i{)}^2\equiv 0(\mathrm{mod}n)$.

Step 3 (Key lemma): Since all vertices lie on a circle of radius $R$, and $n$ is odd, we use the extended law of sines: $|A_i{A}_{i+1}|=2R\sin ({\theta }_i/2)$ where ${\theta }_i$ is the central angle. The Shoelace terms can be related to the central angles.

Step 4 (Working modulo $n$): Translate so that $A_1=(0,0)$. Then $x_1=y_1=0$ and each $|A_i{|}^2=x_i^2+y_i^2$ is constant (all on a circle) — actually after translation the center moves. Instead, use the circle equation: for center $(h,k)$ and radius $R$, $(x_i-h{)}^2+(y_i-k{)}^2=R^2$ for all $i$.

The difference $(x_{i+1}-x_i{)}^2+(y_{i+1}-y_i{)}^2\equiv 0(\mathrm{mod}n)$ combined with the cocircular condition gives congruence relations among the coordinates modulo $n$. Since $n$ is odd, we can use the factorization and properties of sums of two squares modulo odd numbers.

Step 5 (Conclusion): Combining the congruences from the side-length conditions with the Shoelace formula, every term in $2S$ picks up a factor of $n$, giving $n|2S$. $■$