Subfield: Number Theory


Question: Let $n$ be an odd positive integer. In the Cartesian plane, a cyclic polygon $P$ with area $S$ is chosen. All its vertices have integral coordinates, and the squares of its side lengths are all divisible by $n$. Prove that $2 S$ is an integer divisible by $n$.

Solution:
Let $P=A_{1} A_{2} \ldots A_{k}$ and let $A_{k+i}=A_{i}$ for $i \geqslant 1$. By the Shoelace Formula, the area of any convex polygon with integral coordinates is half an integer. Therefore, $2 S$ is an integer. We shall prove by induction on $k \geqslant 3$ that $2 S$ is divisible by $n$. Clearly, it suffices to consider $n=p^{t}$ where $p$ is an odd prime and $t \geqslant 1$.

For the base case $k=3$, let the side lengths of $P$ be $\sqrt{n a}, \sqrt{n b}, \sqrt{n c}$ where $a, b, c$ are positive integers. By Heron's Formula,

$$
16 S^{2}=n^{2}\left(2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2}\right) .
$$

This shows $16 S^{2}$ is divisible by $n^{2}$. Since $n$ is odd, $2 S$ is divisible by $n$.

Assume $k \geqslant 4$. If the square of length of one of the diagonals is divisible by $n$, then that diagonal divides $P$ into two smaller polygons, to which the induction hypothesis applies. Hence we may assume that none of the squares of diagonal lengths is divisible by $n$. As usual, we denote by $\nu_{p}(r)$ the exponent of $p$ in the prime decomposition of $r$. We claim the following.

- Claim. $\nu_{p}\left(A_{1} A_{m}^{2}\right)>\nu_{p}\left(A_{1} A_{m+1}^{2}\right)$ for $2 \leqslant m \leqslant k-1$.

Proof. The case $m=2$ is obvious since $\nu_{p}\left(A_{1} A_{2}^{2}\right) \geqslant p^{t}>\nu_{p}\left(A_{1} A_{3}^{2}\right)$ by the condition and the above assumption.

Suppose $\nu_{p}\left(A_{1} A_{2}^{2}\right)>\nu_{p}\left(A_{1} A_{3}^{2}\right)>\cdots>\nu_{p}\left(A_{1} A_{m}^{2}\right)$ where $3 \leqslant m \leqslant k-1$. For the induction step, we apply Ptolemy's Theorem to the cyclic quadrilateral $A_{1} A_{m-1} A_{m} A_{m+1}$ to get

$$
A_{1} A_{m+1} \times A_{m-1} A_{m}+A_{1} A_{m-1} \times A_{m} A_{m+1}=A_{1} A_{m} \times A_{m-1} A_{m+1}
$$

which can be rewritten as

$$
\begin{aligned}
A_{1} A_{m+1}^{2} \times A_{m-1} A_{m}^{2}= & A_{1} A_{m-1}^{2} \times A_{m} A_{m+1}^{2}+A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2} \\
& -2 A_{1} A_{m-1} \times A_{m} A_{m+1} \times A_{1} A_{m} \times A_{m-1} A_{m+1}
\end{aligned}
\tag{1}
$$

From this, $2 A_{1} A_{m-1} \times A_{m} A_{m+1} \times A_{1} A_{m} \times A_{m-1} A_{m+1}$ is an integer. We consider the component of $p$ of each term in (1). By the inductive hypothesis, we have $\nu_{p}\left(A_{1} A_{m-1}^{2}\right)>\nu_{p}\left(A_{1} A_{m}^{2}\right)$. Also, we have $\nu_{p}\left(A_{m} A_{m+1}^{2}\right) \geqslant p^{t}>\nu_{p}\left(A_{m-1} A_{m+1}^{2}\right)$. These give

$$
\nu_{p}\left(A_{1} A_{m-1}^{2} \times A_{m} A_{m+1}^{2}\right)>\nu_{p}\left(A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2}\right)
\tag{2}
$$

Next, we have $\nu_{p}\left(4 A_{1} A_{m-1}^{2} \times A_{m} A_{m+1}^{2} \times A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2}\right)=\nu_{p}\left(A_{1} A_{m-1}^{2} \times A_{m} A_{m+1}^{2}\right)+$ $\nu_{p}\left(A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2}\right)>2 \nu_{p}\left(A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2}\right)$ from (2). This implies

$$
\nu_{p}\left(2 A_{1} A_{m-1} \times A_{m} A_{m+1} \times A_{1} A_{m} \times A_{m-1} A_{m+1}\right)>\nu_{p}\left(A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2}\right) .
\tag{3}
$$

Combining (1), (2) and (3), we conclude that

$$
\nu_{p}\left(A_{1} A_{m+1}^{2} \times A_{m-1} A_{m}^{2}\right)=\nu_{p}\left(A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2}\right) .
$$

By $\nu_{p}\left(A_{m-1} A_{m}^{2}\right) \geqslant p^{t}>\nu_{p}\left(A_{m-1} A_{m+1}^{2}\right)$, we get $\nu_{p}\left(A_{1} A_{m+1}^{2}\right)<\nu_{p}\left(A_{1} A_{m}^{2}\right)$. The Claim follows by induction.



From the Claim, we get a chain of inequalities

$$
p^{t}>\nu_{p}\left(A_{1} A_{3}^{2}\right)>\nu_{p}\left(A_{1} A_{4}^{2}\right)>\cdots>\nu_{p}\left(A_{1} A_{k}^{2}\right) \geqslant p^{t}
$$

which yields a contradiction. Therefore, we can show by induction that $2 S$ is divisible by $n$.