Dissertativas 9 - 2ª Fase Dia 1 - IME 2025

Seja R o raio das circunferências, conforme a figura:

Nos triângulos ${\mathrm{AC}}_1\mathrm{B}$ e ${\mathrm{BC}}_2\mathrm{C}$, tem-se:

$$\begin{gathered} \mathrm{A}\hat{\mathrm{B}}{\mathrm{C}}_1=\mathrm{B}\hat{\mathrm{A}}{\mathrm{C}}_1=\mathrm{\gamma } \mathrm{e} \mathrm{A}\widehat{{\mathrm{C}}_1}\mathrm{B}=180°-2\mathrm{\gamma } \\ \mathrm{C}\hat{\mathrm{B}}{\mathrm{C}}_2=\mathrm{B}\hat{\mathrm{C}}{\mathrm{C}}_2=\mathrm{\alpha } \mathrm{e} \mathrm{B}\widehat{{\mathrm{C}}_2}\mathrm{C}=180°-2\mathrm{\alpha } \end{gathered}$$

O quadrilátero ${\mathrm{BC}}_1{\mathrm{MC}}_2$ e as diagonais $\overline{\mathrm{BM}} \mathrm{e} \overline{{\mathrm{C}}_1{\mathrm{C}}_2}$ também são bissetrizes. Assim:

$$\begin{gathered} {\mathrm{C}}_1\hat{\mathrm{B}}\mathrm{M}={\mathrm{C}}_2\hat{\mathrm{B}}\mathrm{M}=\frac{180°-\mathrm{\alpha }-\mathrm{\gamma }}{2}=90°-\frac{\left(\mathrm{\alpha }+\mathrm{\gamma }\right)}{2} \\ {\mathrm{C}}_1\widehat{{\mathrm{C}}_2}\mathrm{B}={\mathrm{C}}_1\widehat{{\mathrm{C}}_2}\mathrm{M}={\mathrm{C}}_2\widehat{{\mathrm{C}}_1}\mathrm{B}={\mathrm{C}}_2\widehat{{\mathrm{C}}_1}\mathrm{M}=90°-\left(90°-\frac{\left(\mathrm{\alpha }+\mathrm{\gamma }\right)}{2}\right)=\frac{\mathrm{\alpha }+\mathrm{\gamma }}{2} \end{gathered}$$

No triângulo ${\mathrm{AC}}_1\mathrm{M}$, tem-se:

$$\begin{gathered} \overline{{\mathrm{C}}_1\mathrm{A}}=\overline{{\mathrm{C}}_1\mathrm{M}}=\mathrm{R} \\ \mathrm{A}\widehat{{\mathrm{C}}_1}\mathrm{M}=180°-2\mathrm{\gamma }+2·\frac{\mathrm{\alpha }+\mathrm{\gamma }}{2}=180°-\mathrm{\gamma }+\mathrm{\alpha } \end{gathered}$$

No triângulo ${\mathrm{CC}}_2\mathrm{M}$, tem-se:

$$\begin{gathered} \overline{{\mathrm{C}}_1\mathrm{A}}=\overline{{\mathrm{C}}_1\mathrm{M}}=\mathrm{R} \\ \mathrm{C}\widehat{{\mathrm{C}}_2}\mathrm{M}=360°-\left(180°-2\mathrm{\alpha }+2·\frac{\mathrm{\alpha }+\mathrm{\gamma }}{2}\right)=180°-\mathrm{\gamma }+\mathrm{\alpha } \end{gathered}$$

Segue que os triângulos ${\mathrm{AC}}_1\mathrm{M} \mathrm{e} {\mathrm{CC}}_2\mathrm{M}$ são congruentes pelo caso LAL; dessa forma, $\overline{\mathrm{AM}}=\overline{\mathrm{CM}}$. Essa igualdade implica que M pertence à mediatriz de A e C, fato que independe do tamanho do raio das circunferências.

O lugar geométrico procurado, portanto, é a mediatriz do segmento $\overline{\mathrm{AC}}.$