Questão 12 - 1ª Fase - ITA 2026

A progressão geométrica ${\mathrm{r}}_{\mathrm{k}}$ é dada por ${\mathrm{r}}_{\mathrm{k}}: \mathrm{q}, {\mathrm{q}}^2, {\mathrm{q}}^3, {\mathrm{q}}^4 ... \Rightarrow {\mathrm{r}}_{\mathrm{k}}={\mathrm{q}}^{\mathrm{k}}, \mathrm{com} \mathrm{q} > 1.$

A circunferência ${\mathrm{C}}_{\mathrm{k}}: {\left(\mathrm{x}-{\mathrm{r}}_{\mathrm{k}}\right)}^2+{\left(\mathrm{y}-\mathrm{m}·{\mathrm{r}}_{\mathrm{k}}\right)}^2={\left({\mathrm{r}}_{\mathrm{k}}\right)}^2$ tem centro ${\mathrm{O}}_{\mathrm{k}}=\left({\mathrm{r}}_{\mathrm{k}}; {\mathrm{mr}}_{\mathrm{k}}\right)=\left({\mathrm{q}}^{\mathrm{k}}; \mathrm{m}·{\mathrm{q}}^{\mathrm{k}}\right)$ e raio ${\mathrm{r}}_{\mathrm{k}}={\mathrm{q}}^{\mathrm{k}}$.

Para que ${\mathrm{C}}_{\mathrm{k}}$ e ${\mathrm{C}}_{\mathrm{k}+1}$ sejam tangentes exteriormente, a distância entre seus centros deve ser igual à soma dos seus raios, ou seja:

$$\begin{gathered} \mathrm{d}\left({\mathrm{C}}_{\mathrm{k}}, {\mathrm{C}}_{\mathrm{k}+1}\right)={\mathrm{q}}^{\mathrm{k}}+{\mathrm{q}}^{\mathrm{k}+1}\iff {\left({\mathrm{q}}^{\mathrm{k}}+{\mathrm{q}}^{\mathrm{k}+1}\right)}^2={\left({\mathrm{q}}^{\mathrm{k}+1}-{\mathrm{q}}^{\mathrm{k}}\right)}^2+{\left({\mathrm{mq}}^{\mathrm{k}+1}-{\mathrm{mq}}^{\mathrm{k}}\right)}^2 \\ {\mathrm{q}}^{2\mathrm{k}}{\left(1+\mathrm{q}\right)}^2={\mathrm{q}}^{2\mathrm{k}}{\left(\mathrm{q}-1\right)}^2+{\mathrm{q}}^{2\mathrm{k}}·{\mathrm{m}}^2{\left(\mathrm{q}-1\right)}^2\iff {\left(1+\mathrm{q}\right)}^2={\left(\mathrm{q}-1\right)}^2+{\mathrm{m}}^2{\left(\mathrm{q}-1\right)}^2 \\ \iff 1+2\mathrm{q}+{\mathrm{q}}^2=1-2\mathrm{q}+{\mathrm{q}}^2+{\mathrm{m}}^2-2{\mathrm{m}}^2\mathrm{q}+{\mathrm{m}}^2{\mathrm{q}}^2 \\ \iff {\mathrm{m}}^2{\mathrm{q}}^2-\left(2{\mathrm{m}}^2+4\right)\mathrm{q}+{\mathrm{m}}^2=0 \left(*\right) \end{gathered}$$

Discriminante de (*):

$\mathrm{\Delta }={\left(2{\mathrm{m}}^2+4\right)}^2-4·{\mathrm{m}}^2·{\mathrm{m}}^2=16{\mathrm{m}}^2+16$

Cálculo de q:

$\mathrm{q}=\frac{2{\mathrm{m}}^2+4+\sqrt{16\left({\mathrm{m}}^2+1\right)}}{2{\mathrm{m}}^2}=\frac{{\mathrm{m}}^2+2+2\sqrt{{\mathrm{m}}^2+1}}{{\mathrm{m}}^2}=\boxed{{\left(\frac{\sqrt{{\mathrm{m}}^2+1}+1}{\mathrm{m}}\right)}^2}$

Observação: Se $\mathrm{q}=\frac{2{\mathrm{m}}^2+4-\sqrt{16\left({\mathrm{m}}^2+1\right)}}{2{\mathrm{m}}^2}={\left(\frac{\sqrt{{\mathrm{m}}^2+1}-1}{\mathrm{m}}\right)}^2$, tem-se a contradição:

$\mathrm{q}>1\Rightarrow {\left(\frac{\sqrt{{\mathrm{m}}^2+1}-1}{\mathrm{m}}\right)}^2>1\Rightarrow \sqrt{{\mathrm{m}}^2+1}>\mathrm{m}+1\Rightarrow {\mathrm{m}}^2+1>{\mathrm{m}}^2+2\mathrm{m}+1$