<p><strong>970.</strong> A regular heptadecagon \(P_1P_2P_3\ldots P_{17}\) is inscribed in a unit circle. Find \(\displaystyle\prod_{n=2}^{17} P_1P_n\).</p>
Step-by-Step Solution
Key Concept: The vertices of a regular heptadecagon on the unit circle are 17th roots of unity. The product of distances from one vertex to all others equals the derivative of the cyclotomic polynomial evaluated at that root.
<p><strong>Step 1:</strong> Place the heptadecagon on the unit circle with vertices at the 17th roots of unity: $P_k = e^{2\pi i(k-1)/17}$ for $k = 1, 2, \ldots, 17$.</p><p><strong>Step 2:</strong> Let $\omega = e^{2\pi i/17}$. Then $P_n = \omega^{n-1}$, so $P_1 = 1$ and we need $\displaystyle\prod_{n=2}^{17} |P_1 - P_n| = \prod_{n=2}^{17} |1 - \omega^{n-1}|$.</p><p><strong>Step 3:</strong> Reindex: $\displaystyle\prod_{n=2}^{17} |1 - \omega^{n-1}| = \prod_{j=1}^{16} |1 - \omega^j|$.</p><p><strong>Step 4:</strong> The polynomial $z^{17} - 1 = (z-1)(z^{16} + z^{15} + \cdots + z + 1)$. The roots of $\Phi_{17}(z) = z^{16} + z^{15} + \cdots + z + 1$ are exactly $\omega, \omega^2, \ldots, \omega^{16}$.</p><p><strong>Step 5:</strong> We need $\displaystyle\prod_{j=1}^{16}(1 - \omega^j)$. Since $\Phi_{17}(z) = \prod_{j=1}^{16}(z - \omega^j)$, evaluating at $z=1$: $\Phi_{17}(1) = \prod_{j=1}^{16}(1 - \omega^j)$.</p><p><strong>Step 6:</strong> For a prime $p$, $\Phi_p(1) = 1 + 1 + \cdots + 1$ ($p$ terms) $= p$. Thus $\Phi_{17}(1) = 17$.</p><p><strong>Step 7:</strong> Therefore $\displaystyle\prod_{n=2}^{17} P_1P_n = 17$.</p><p>∴ Answer: <strong>17</strong></p>
Correct Answer: 17