Questão 14 - 1ª Fase - IME 2026

Na figura, $\mathrm{\alpha }$ e $\mathrm{\beta }$ são os ângulos agudos do triângulo $\mathrm{APB}$. Logo, $\mathrm{\alpha } + \mathrm{\beta } = 90°$.

$$\begin{gathered} \mathrm{AO} = \mathrm{BO} = \mathrm{r} \to \mathrm{O}\hat{\mathrm{A}}\mathrm{B} = \mathrm{O}\hat{\mathrm{B}}\mathrm{A} = \mathrm{\alpha } \\ \mathrm{O}\hat{\mathrm{B}}\mathrm{C} = \mathrm{A}\hat{\mathrm{B}}\mathrm{C} - \mathrm{O}\hat{\mathrm{B}}\mathrm{A} = 90° - \mathrm{\alpha } = \mathrm{\beta } \\ \mathrm{D}\hat{\mathrm{A}}\mathrm{O} = \mathrm{D}\hat{\mathrm{A}}\mathrm{B} - \mathrm{B}\hat{\mathrm{A}}\mathrm{P} = 90° - \mathrm{\alpha } = \mathrm{\beta } \end{gathered}$$

$∆\mathrm{ADO} \equiv ∆\mathrm{BCO} \left(\mathrm{ALA}\right) \to \mathrm{CO} = \mathrm{DO} = \frac{\mathrm{L}}{2}$

No triângulo $\mathrm{BCO}$:

$$\begin{gathered} \mathrm{r} = \mathrm{BO} = \sqrt{{\left(\mathrm{L}\right)}^2 + {\left(\frac{\mathrm{L}}{2}\right)}^2} = \frac{\mathrm{L}\sqrt{5}}{2} \\ ∆\mathrm{APB} ~ ∆\mathrm{BCO}: \\ \frac{\mathrm{AP}}{\mathrm{CO}} = \frac{\mathrm{AB}}{\mathrm{BO}} \to \frac{\mathrm{AP}}{{\displaystyle \frac{\mathrm{L}}{2}}} = \frac{\mathrm{L}}{{\displaystyle \frac{\mathrm{L}\sqrt{5}}{2}}}\to \mathrm{AP} = \frac{\mathrm{L}}{\sqrt{5}} \\ \frac{\mathrm{BP}}{\mathrm{BC}} = \frac{\mathrm{AB}}{\mathrm{BO}} \to \frac{\mathrm{BP}}{\mathrm{L}} = \frac{\mathrm{L}}{{\displaystyle \frac{\mathrm{L}\sqrt{5}}{2}}} \to \mathrm{BP} = \frac{2\mathrm{L}}{\sqrt{5}} \end{gathered}$$

Do enunciado: $\left[\mathrm{BPCO}\right] = 11$. Assim: 

 $$\begin{gathered} \left[\mathrm{ABCD}\right] - \left[\mathrm{ADO}\right] - \left[\mathrm{APB}\right] = \left[\mathrm{BPCO}\right] \to {\mathrm{L}}^2 - \frac{1}{2}\mathrm{L} · \frac{\mathrm{L}}{2} - \frac{1}{2} · \frac{\mathrm{L}}{\sqrt{5}} · \frac{2\mathrm{L}}{\sqrt{5}} = 11 \\ {\mathrm{L}}^2\left(1 - \frac{1}{4} - \frac{1}{5}\right) = 11 \to \frac{11}{20}{\mathrm{L}}^2 = 11 \to \mathrm{L} =\sqrt{20} = 2\sqrt{5} \end{gathered}$$

Logo:

$\mathrm{r} = \frac{\mathrm{L}\sqrt{5}}{2} = \frac{2\sqrt{5} · \sqrt{5}}{2} = \boxed{5}$