Questão 15 - 1ª Fase - IME 2025

Adotando-se um sistema de coordenadas cartesianas com E na origem, eixo x sobre $\overline{\mathrm{EH}}$, eixo y sobre $\overline{\mathrm{EF}}$ e eixo z sobre $\overline{\mathrm{EA}}$, tem-se:

$$\begin{gathered} {\mathrm{D}}_1 = \left(\frac{\mathrm{I}}{3}, 0, \mathrm{I}\right), {\mathrm{B}}_1 = \left(0, \frac{\mathrm{I}}{3}, \mathrm{I}\right) \mathrm{e} {\mathrm{H}}_1 = \left( \mathrm{l}, \frac{2\mathrm{l}}{3}, 0\right) \\ \overline{{\mathrm{D}}_1{\mathrm{B}}_1} = \sqrt{{\left(\frac{\mathrm{l}}{3} - 0\right)}^2 + {\left(0 - \frac{\mathrm{l}}{3}\right)}^2 + {\left(\mathrm{l}-\mathrm{l}\right)}^2} =\frac{ \mathrm{l}\sqrt{2}}{3} \\ \overline{{\mathrm{D}}_1{\mathrm{H}}_1} =\sqrt{{\left(\frac{\mathrm{l}}{3} - \mathrm{l}\right)}^2 + {\left(0 - \frac{2\mathrm{l}}{3}\right)}^2 + {\left(\mathrm{l}-0\right)}^2} =\frac{ \mathrm{l}\sqrt{17}}{3} \\ \overline{{\mathrm{B}}_1{\mathrm{H}}_1} =\sqrt{{\left(0 - \mathrm{l}\right)}^2 + {\left(\frac{\mathrm{l}}{3} - \frac{2\mathrm{l}}{3}\right)}^2 + {\left(\mathrm{l}-0\right)}^2} =\frac{ \mathrm{l}\sqrt{19}}{3} \end{gathered}$$

Observa-se que:

$$\begin{gathered} {\overline{{\mathrm{B}}_1{\mathrm{H}}_1}}^2 = {\left(\frac{ \mathrm{l}\sqrt{19}}{3}\right)}^2 = {\left(\frac{19\mathrm{l}}{9}\right)}^2 \\ {\overline{{\mathrm{D}}_1{\mathrm{B}}_1}}^2 + {\overline{{\mathrm{D}}_1{\mathrm{H}}_1}}^2 = {\left(\frac{ \mathrm{l}\sqrt{2}}{3}\right)}^2 + {\left(\frac{ \mathrm{l}\sqrt{17}}{3}\right)}^2 = \frac{{19\mathrm{l}}^2}{9} \end{gathered}$$

Portanto, o triângulo ${\mathrm{D}}_1{\mathrm{B}}_1{\mathrm{H}}_1$ é retângulo em ${\mathrm{D}}_1$ e tem área dada por:

$\frac{1}{2}\overline{{\mathrm{D}}_1{\mathrm{B}}_1} · \overline{{\mathrm{D}}_1{\mathrm{H}}_1} = \frac{1}{2} · \frac{ \mathrm{l}\sqrt{2}}{3} · \frac{ \mathrm{l}\sqrt{17}}{3} = \boxed{\frac{ {\mathrm{l}}^2\sqrt{34}}{18}}$