Questão 10 - 1ª Fase - IME 2025

Pelas informações do enunciado, tem-se:

$$\begin{gathered} \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{2 + 2\sqrt{2} \sqrt{3 + \sqrt{\sqrt{2}- \sqrt{12}+ \sqrt{18 - \sqrt{128}}}}}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{2 + 2\sqrt{2} \sqrt{3 + \sqrt{\sqrt{2}- \sqrt{12}+ \sqrt{18 - 8\sqrt{2}}}}}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{2 + 2\sqrt{2}\sqrt{3 + \sqrt{\sqrt{2}- \sqrt{12}+ \sqrt{{\left(4 - \sqrt{2}\right)}^2}}}}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{2 + 2\sqrt{2} \sqrt{3 + \sqrt{\sqrt{2}- \sqrt{12}+ 4 - \sqrt{2}}}}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{2 + 2\sqrt{2}\sqrt{3 + \sqrt{4- \sqrt{12}}}}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{2 + 2\sqrt{2} \sqrt{3 + \sqrt{4- 2\sqrt{3}}}}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{2 + 2\sqrt{2}\sqrt{3 + \sqrt{{\left(\sqrt{3} - 1\right)}^2}}}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{2 + 2\sqrt{2} \sqrt{3 + \sqrt{3} - 1}}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{2 + 2\sqrt{2} \sqrt{2 + \sqrt{3}}}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{ 2 + 2\sqrt{2} \sqrt{{\left(\sqrt{\frac{3}{2}}+ \sqrt{\frac{1}{2}}\right)}^2}}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{ 2 + 2\sqrt{2} \left(\sqrt{\frac{3}{2}}+ \sqrt{\frac{1}{2}}\right)}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{2 + 2\sqrt{3}+2}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{4 + 2\sqrt{3}}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{{\left(\sqrt{3} + 1\right)}^2}} \\ \mathrm{r} =\sqrt{3 - \sqrt{3} + \sqrt{3} + 1} \\ \mathrm{r} =\sqrt{4} \Rightarrow \mathrm{r} = 2 \end{gathered}$$

Desse modo, a inequação é dada por:

$$\begin{gathered} \sqrt{2025 + \sqrt{\mathrm{t}}} + \sqrt{2025 - \sqrt{\mathrm{t}}} \le \sqrt{2025 · 2} \\ \sqrt{2025 + \sqrt{\mathrm{t}}} + \sqrt{2025 - \sqrt{\mathrm{t}}} \le \sqrt{4050} \end{gathered}$$

Como ambos os lados são positivos:

$$\begin{gathered} {\left(\sqrt{2025 + \sqrt{\mathrm{t}}} + \sqrt{2025 - \sqrt{\mathrm{t}}}\right)}^2 \le {\sqrt{4050}}^2 \\ 2025 + \sqrt{\mathrm{t}} + 2025 - \sqrt{\mathrm{t}} + 2\sqrt{{2025}^2 - \mathrm{t}} \le 4050 \\ 4050 + 2\sqrt{{2025}^2 - }\mathrm{t} \le 4050 \\ \sqrt{{2025}^2 - }\mathrm{t} \le 0 \end{gathered}$$

Como também $\sqrt{{2025}^2 - \mathrm{t}} \ge 0$, então: 

$\sqrt{{2025}^2 - \mathrm{t}} = 0 \Rightarrow \mathrm{t} ={ 2025}^2.$