Problema (nível 4): Considera um cone de altura $1\;cm$ e base cujo raio mede $\sqrt 2\;cm$. No interior deste cone encontra-se um cubo assente na base. Sabendo que os vértices superiores do cubo tocam na superfície lateral do cone, determina a medida do lado do cubo.

Resolução: Representemos por $l$ o lado do cubo. Na figura seguinte $A$ é o vértice do cone, $B$ é um ponto de contacto entre o cubo e o cone, $O$ é o centro da secção do cone que contém o lado superior do cubo e $O'$ e $B'$ são os pontos da base que estão sobre as rectas $AO$ e $AB$, respectivamente. (image1) (image2) Como os triângulos $[A B O]$ e $[AB'O']$ são semelhantes, tem-se $$ \frac{\overline{OA}}{\overline{O'A}}=\frac{\overline{OB}}{\overline{O'B'}}. $$ Como $\overline{O'A}=1$, $\overline{O'B'}=\sqrt{2}$ e $\overline{OA}=\overline{O'A}- \overline{O'O}=1-l$, tem-se $\overline{O B}=\sqrt2(1-l)$. Consideremos agora a figura seguinte, onde está representada a secção do cone que contém a face superior do cubo. (image3) Como $[B O C]$ é um triângulo rectângulo em $O$, tem-se, pelo teorema de Pitágoras, $\overline{O B}^2=\overline{B C}^2-\overline{O C}^2$. Como $\overline{B C}=l$ e $\overline{O B}=\overline{O C}$, vem $\overline{O B}=\frac{\sqrt2}{2}\,l.$ Igualando as duas expressões obtidas para $\overline{O B}$, conclui-se que $\sqrt2\,(1-l)=\frac{\sqrt2}{2}\,l$, pelo que $l=\frac23\,cm$.


Conteúdo referenciado:
[image1]
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[image2]
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[image3]
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