Dissertativa 9 - 2ª Fase - Dia 1 - IME 2026

No triângulo ABM:  

$\mathrm{BM}=\mathrm{AB}·\mathrm{tg}15°=\mathrm{a}\left(2-\sqrt{3}\right)$

No triângulo CMN: 

$$\begin{gathered} \mathrm{CN}=\mathrm{CM}=\mathrm{a}-\mathrm{a}\left(2-\sqrt{3}\right)=\mathrm{a}\left(\sqrt{3}-1\right) \\ \mathrm{MN}=\mathrm{CM}\sqrt{2}=\mathrm{a}\left(\sqrt{3}-1\right)\sqrt{2} \\ \left[\mathrm{CMN}\right]=\frac{{\left(\mathrm{a}\left(\sqrt{3}-1\right)\right)}^2}{2}=\frac{{\mathrm{a}}^2{\left(\sqrt{3}-1\right)}^2}{2} \\ \mathrm{p}=\frac{2\mathrm{a}\left(\sqrt{3}-1\right)+\sqrt{2}\mathrm{a}\left(\sqrt{3}-1\right)}{2}=\frac{\left(2+\sqrt{2}\right)\mathrm{a}\left(\sqrt{3}-1\right)}{2} \end{gathered}$$

Sendo r o raio da circunferência inscrita no triângulo, tem-se: 

$\mathrm{r}=\frac{\left[\mathrm{CMN}\right]}{\mathrm{p}}=\frac{{\mathrm{a}}^2{\left(\sqrt{3}-1\right)}^2}{\left(2+\sqrt{2}\right)\mathrm{a}\left(\sqrt{3}-1\right)}=\boxed{\frac{\sqrt{3}-1}{2+\sqrt{2}}\mathrm{a}}$