Problema (nível 4): Na figura $\overline{AB}=9$ e $\overline{AD}=8$. As duas circunferências, tangentes entre si, têm centros $E$ e $F$ e são tangentes aos lados do rectângulo $[ABCD]$ nos pontos $M$, $N$, $X$ e $Y$. Sabendo que o raio da circunferência de centro $F$ mede $2$, quanto mede o raio da circunferência de centro $E$? (image1)


Conteúdo referenciado:
[image1]
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Resolução: Sejam $I$ o ponto de intersecção das rectas definidas por $[FN]$ e $[EM]$ e $r=\overline{EM}=\overline{XE}$ a medida do raio da circunferência de centro $E$. Sendo $M$ e $N$ pontos de tangência entre as circunferências e os lados do rectângulo, tem-se $[FN]\perp [BC]$ e $[ME]\perp [AB]$, pelo que $[FN] \: / \!\! / \: [DC]$ e $[ME]\: / \!\! / \: [AD]$. Assim $$\overline{IF}=\overline{DC}-\overline{FN}-\overline{EX}=7-r \quad \mbox{ e } \quad \overline{IE}=\overline{AD}-\overline{EM}-\overline{FY}=6-r.$$ Por outro lado, $\overline{EF}=r+2$ e o triângulo $[EFI]$ é rectângulo. Pelo Teorema de Pitágoras tem-se $\overline{IF}^2+\overline{IE}^2=\overline{EF}^2$, ou seja $(7-r)^2+(6-r)^2=(r+2)^2$. As soluções desta equação são $r=3$ e $r=27$, mas como $r<8$ conclui-se que $r=3$.