Problem A. 911. (September 2025)
A. 911. For which integers \(\displaystyle n \ge 4\) is it true that the area of the regular \(\displaystyle n\)-gon is equal to the product of two of its diagonals?
Based on the idea of Mihály Hujter, Budapest
(7 pont)
Deadline expired on October 10, 2025.
Solution. Let \(\displaystyle \omega=e^{2\pi i/n}\), the condition is satisfied if and only if there are some integers \(\displaystyle 2\le k,m\le n-2\) such that
| \(\displaystyle |\omega^k-1|\cdot|\omega^m-1| = \frac{n}{4}|\omega^2-1| \) | \(\displaystyle (*) \) |
\(\displaystyle |\omega^k-1|^2|\omega^m-1|^2-\bigg(\frac{n}{4}\bigg)^2|\omega^2-1|^2 = 0 \)
| \(\displaystyle (\omega^k-1)(\omega^{n-k}-1)\cdot(\omega^m-1)(\omega^{n-m}-1) -\bigg(\frac{n}4\bigg)^2(\omega^2-1)(\omega^{n-2}-1) = 0 \) | \(\displaystyle (**) \) |
so \(\displaystyle \omega\) is a root of the polynomial
\(\displaystyle P(x) = (x^k-1)(x^{n-k}-1)\cdot(x^m-1)(x^{n-m}-1) -\bigg(\frac{n}4\bigg)^2(x^2-1)(x^{n-2}-1). \)
At the same time, \(\displaystyle \omega\) is a root of the cyclotomic polynomial \(\displaystyle \Phi_n(x)\), which is irreducible over \(\displaystyle \mathbb{Q}\), so \(\displaystyle P(x)\) must be divisible by \(\displaystyle \Phi_n(x)\), and therefore \(\displaystyle (**)\) and \(\displaystyle (*)\) must hold true for all primitive \(\displaystyle n\)th complex roots of unity.
Taking product of \(\displaystyle (*)\) on all primitive roots,
\(\displaystyle 4^{\varphi(n)} \ge \prod_{\mathrm{gcd}(r,n)=1} |\omega^{kr}-1|\cdot|\omega^{m r}-1| =\bigg(\frac{n}4\bigg)^{\varphi(n)} \underbrace{\left|\prod_{\mathrm{gcd}(r,n)=1}(\omega^{2r}-1)\right|}_{\text{some positive integer}} \ge\bigg(\frac{n}4\bigg)^{\varphi(n)}, \)
therefore \(\displaystyle n\le 16\).
By checking all cases, it can be found that for \(\displaystyle 8\) and \(\displaystyle 12\) there are suitable integers \(\displaystyle 2,4\) for \(\displaystyle n=8\), and \(\displaystyle 4,8\) for \(\displaystyle n=12\).
Statistics:
23 students sent a solution. 7 points: Diaconescu Tashi, Forrai Boldizsár, Rajtik Sándor Barnabás, Xiaoyi Mo. 6 points: Tianyue DAI. 5 points: 1 student. 4 points: 1 student. 1 point: 10 students. 0 point: 5 students. Unfair, not evaluated: 1 solutions.
Problems in Mathematics of KöMaL, September 2025