Problema (nível 4): Se duas cordas paralelas de uma circunferência, de comprimento $10\;mm$ e $14\;mm$, distarem $6\;mm$ uma da outra, quanto mede a corda equidistante dessas duas?

Resolução: Caso 1: o centro da circunferência está entre as duas cordas (image1) Sejam $r$ o raio da circunferência e $x$ metade do comprimento da corda a determinar. É evidente que basta aplicar o Teorema de Pitágoras a três triângulos rectângulos com hipotenusa do mesmo comprimento $r$. Neste caso temos $$\left\{(image2)\right. $$ onde $a$ denota a distância do centro da circunferência ao ponto médio do segmento que separa as duas cordas (nos seus pontos médios). As duas primeiras equações dão-nos $a=2$ e $r=\sqrt{50}$. Então $x^2=46$ e, consequentemente, $2\,x=\sqrt{184}$. Caso 2: o centro da circunferência não está entre as duas cordas É evidente que o centro da circunferência nunca pode estar do lado da corda menor. Mas também não pode estar do lado da corda maior: (image3) Por um raciocínio perfeitamente análogo ao anterior obtemos o sistema de equações $$\left\{(image4)\right. $$ onde $a$ é a distância do centro da circunferência à corda maior. Daqui segue o absurdo $a=-1$!!!


Conteúdo referenciado:
[image1]
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[image2]
\begin{array}{l} 5^2+(3+a)^2=r^2\\ (3-a)^2+7^2=r^2\\ a^2+x^2=r^2 \end{array}
[image3]
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[image4]
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