Problema (nível 4): Sejam ${\cal C}_1$ e ${\cal C}_2$ duas circunferências concêntricas de raios $r$ e $R$, respectivamente, com $r < R$. Os pontos $A$, $B$ e $C$, distintos dois a dois, pertencem a ${\cal C}_2$ e as cordas $[AB]$ e $[AC]$ são tangentes a ${\cal C}_1$. Sabendo que $R=5$ e $\overline{BC}=8$, determine o raio de ${\cal C}_1$.

Resolução: Sejam $M$ o ponto de tangência de $[AB]$ com ${\cal C}_1$ e $O$ o centro das duas circunferências. O segmento $[OM]$ é perpendicular a $[AB]$. Usando o teorema de Pitágoras nos triângulos rectângulos $[OMA]$ e $[BMO]$, conclui-se que $M$ é o ponto médio de $[AB]$. Analogamente, o ponto de tangência de $[AC]$ com ${\cal C}_1$, $M'$, é o ponto médio de $[AC]$. Como $\overline{OM}= \overline{OM'}$ e $[OA]$ é hipotenusa comum aos triângulos rectângulos $[OMA]$ e $[OAM']$, pelo teorema de Pitágoras, $\overline{MA}=\overline{AM'}$ e estes triângulos rectângulos são congruentes. Logo, os ângulos $\measuredangle MAO$ e $\measuredangle OAM'$ têm a mesma amplitude. A recta que passa por $A$ e $O$ intersecta a corda $[BC]$ num ponto $N$. Os triângulos $[BAN]$ e $[NAC]$ são congruentes, uma vez que $\measuredangle MAO$ e $\measuredangle OAM'$ têm a mesma amplitude, $[AN]$ é lado comum e $\overline{BA}=2\overline{MA}=2\overline{AM'}= \overline{AC}$. Então, $\overline{BN}= \overline{NC}$, isto é, $N$ é o ponto médio de $[BC]$. Solução 1 O triângulo $[ANB]$ é rectângulo em $N$ e semelhante ao triângulo $[OMA]$, porque têm os três ângulos congruentes. Assim, $\frac{\overline{OM}}{\overline{BN}}= \frac{\overline{OA}}{\overline{BA}}$. Sabe-se que $\overline{OM}= r$, $\overline{BN}=4$, $\overline{OA}=R=5$, $\overline{BA}=2\overline{MA}$, logo, tem-se $r= \frac{10}{\overline{MA}}$. Aplicando o teorema de Pitágoras ao triângulo $[OMA]$, obtém-se a equação $r^2 + (\frac{10}{r})^2 = 25$. As soluções desta equação são $r= 2 \sqrt{5}$ ou $r=\sqrt{5}$. Uma vez que $\frac{\overline{OA}}{\overline{BA}} <1$, $\overline{OM}=r < 4$. Conclui-se então que $r= \sqrt{5}$. (image1) Solução 2 O segmento $[BC]$ é uma corda da circunferência ${\cal C}_2$ de raio $R=5$. Uma vez que $\overline{BC}=8$, a amplitude do arco $BC$ é $\frac{8 \pi}{10} = \frac{4 \pi}{5}$. A amplitude do ângulo inscrito $\measuredangle BAC$ é assim $\frac{2 \pi}{5}$ e a amplitude de $\measuredangle BAO$ é $\frac{\pi}{5}$. O raio $r$ de ${\cal C}_1$ será igual a $\overline{OM}= R \mbox{ sen }\frac{\pi}{5}$. Resta então determinar $\mbox{ sen }\frac{\pi}{5}$. O triângulo $[BNO]$ é rectângulo em $N$ e a amplitude de $\measuredangle BON$ é $\frac{2\pi}{5}$. Como $\overline{BO}=5$ e $\overline{BN}=4$, obtém-se $\mbox{ sen }\frac{2\pi}{5}= \frac{4}{5}$, logo $\cos \frac{2\pi}{5}= \frac{3}{5}$. Usando agora a igualdade $\cos(2x)= 1 - 2 \mbox{ sen}^2x$, tem-se $\mbox{ sen }\frac{\pi}{5} = \frac{1}{\sqrt{5}}$. Finalmente, $r= \sqrt{5}$. (image2)


Conteúdo referenciado:
[image1]
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[image2]
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