Questão 3 - IME - 1ª Fase - IME 2025

${\mathrm{z}}_1=\sqrt{2}\left(\cos \frac{\mathrm{\pi }}{4}+\mathrm{i} \mathrm{sen}\frac{\mathrm{\pi }}{4}\right) \mathrm{e} {\mathrm{z}}_{\mathrm{n}}={\left(\sqrt{2}\right)}^{\mathrm{n}}\left(\cos \frac{\mathrm{n\pi }}{4}+\mathrm{i} \mathrm{sen}\frac{\mathrm{n\pi }}{4}\right)$

A área dos triângulos formados pelos afixos de três termos consecutivos será:

$$\begin{gathered} 2\mathrm{A}=\left|\begin{array}{ccc}{\left(\sqrt{2}\right)}^{\mathrm{n}}\cos \frac{\mathrm{n\pi }}{4}& {\left(\sqrt{2}\right)}^{\mathrm{n}}\mathrm{sen}\frac{\mathrm{n\pi }}{4}& 1\\ {\left(\sqrt{2}\right)}^{\mathrm{n}+1}\cos \frac{\left(\mathrm{n}+1\right)\mathrm{\pi }}{4}& {\left(\sqrt{2}\right)}^{\mathrm{n}+1}\mathrm{sen}\frac{\left(\mathrm{n}+1\right)\mathrm{\pi }}{4}& 1\\ {\left(\sqrt{2}\right)}^{\mathrm{n}+2}\cos \frac{\left(\mathrm{n}+2\right)\mathrm{\pi }}{4}& {\left(\sqrt{2}\right)}^{\mathrm{n}+2}\mathrm{sen}\frac{\left(\mathrm{n}+2\right)\mathrm{\pi }}{4}& 1\end{array}\right|= \\ =2^{\mathrm{n}} \left|\begin{array}{ccc}\cos \frac{\mathrm{n\pi }}{4}& \mathrm{sen}\frac{\mathrm{n\pi }}{4}& 1\\ \sqrt{2} \cos \frac{\left(\mathrm{n}+1\right)\mathrm{\pi }}{4}& \sqrt{2} \mathrm{sen}\frac{\left(\mathrm{n}+1\right)\mathrm{\pi }}{4}& 1\\ 2 \cos \frac{\left(\mathrm{n}+2\right)\mathrm{\pi }}{4}& 2 \mathrm{sen}\frac{\left(\mathrm{n}+2\right)\mathrm{\pi }}{4}& 1\end{array}\right|= \\ =2^{\mathrm{n}-1 }\overline{)\sqrt{2}\left(\cos \frac{\mathrm{n\pi }}{4}\mathrm{sen}\frac{\left(\mathrm{n}+1\right)\mathrm{\pi }}{4}-\mathrm{sen}\frac{\mathrm{n\pi }}{4}\cos \frac{\left(\mathrm{n}+1\right)\mathrm{\pi }}{4}\right)} \\ -2\left(\mathrm{sen}\frac{\left(\mathrm{n}+2\right)\mathrm{\pi }}{4}\cos \frac{\mathrm{n\pi }}{4}-\cos \frac{\left(\mathrm{n}+2\right)\mathrm{\pi }}{4}\mathrm{sen}\frac{\mathrm{n\pi }}{4}\right) \\ \overline{)+2\sqrt{2}\left(\mathrm{sen}\frac{\left(\mathrm{n}+2\right)\mathrm{\pi }}{4}\cos \frac{\left(\mathrm{n}+1\right)\mathrm{\pi }}{4}-\cos \frac{\left(\mathrm{n}+2\right)\mathrm{\pi }}{4}\mathrm{sen}\frac{\left(\mathrm{n}+1\right)\mathrm{\pi }}{4}\right)}= \\ =2^{\mathrm{n}-1}\left|\sqrt{2}\mathrm{sen}\left(\frac{\mathrm{\pi }}{4}\right)-2\mathrm{sen} \left(\frac{\mathrm{\pi }}{2}\right)+2\sqrt{2} \mathrm{sen}\left(\frac{\mathrm{\pi }}{4}\right)\right|=2^{\mathrm{n}-1} \end{gathered}$$

Note que a maior área será para o valor de n = 18, logo a área será $\boxed{\mathrm{A}=2^{17}}$.