Problema (nível 4): Uma roda com 8 $cm$ de raio, encostada a uma semicircunferência com 25 $cm$ de raio, como se mostra na figura, desloca-se apoiada no diâmetro da semicircunferência até que choca com esta. (image1) Qual é o comprimento da porção daquele diâmetro que não é tocada pela roda?


Conteúdo referenciado:
[image1]
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Resolução: Sejam $T$ o ponto de contacto entre a semicircunferência e a roda na posição inicial, $O_1$ o centro da semicircunferência, $O_2$ o centro da roda na posição inicial e $P$ o pé da perpendicular ao diâmetro traçada por $O_2$. (image2) Em $T$ a roda e a circunferência são tangentes, pelo que $T$, $O_1$ e $O_2$ são colineares. A porção de diâmetro que a roda toca é igual a $2\times\overline{PO_1}$. O triângulo $[PO_1O_2]$ é rectângulo, sendo $\overline{PO_2}=8$ $cm$ e $\overline{O_1O_2}=25-8=17$ $cm$. Pelo Teorema de Pitágoras, temos $17^2=8^2+\overline{PO_1}^2$ donde $\overline{PO_1}=15$ $cm$. Logo, a porção de diâmetro que é tocada pela roda mede $2\times 15=30\,cm$, pelo que a porção que não é tocada mede $50-30=20\,cm$. $\,$$\,$ Sociedade Portuguesa de Matemática $\;$ XVII ONM - Categoria B (10${\bf ^{\underline o}}$,11${\bf ^{\underline o}}$ e 12${\bf ^{\underline o}}$ anos) 2${\bf ^{\underline a}}$ Eliminatória $\;\;$ 27/01/99 (table1) $\,$$\,$


Conteúdo referenciado:
[image2]
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[table1]
\begin{tabular}{|p{7,7cm}|} \hline SUGEST\~{O}ES para a resolu\c{c}\~{a}o dos problemas\\ \hline \end{tabular}