Problema (nível 3): A Adriana gosta de montar tapetes com os seguintes tipos de peças: Cantos: (image1) Laterais: (image2) Interiores: (image3) Ao lado está um tapete que a Adriana montou com $4$ cantos, $6$ laterais e $2$ interiores. De seguida, a Adriana decidiu montar um tapete com $30$ peças, das quais $4$ eram cantos, $14$ laterais e as restantes interiores. Quantas peças tem cada um dos lados do tapete? (image4)


Conteúdo referenciado:
[image1]
\begin{pspicture}(0,0)(1,1) \psarc[fillstyle=solid,fillcolor=azulclarologotipo,linewidth=0.3pt](0.1,0.9){0.2}{90}{180} \psline[fillstyle=solid,fillcolor=azulclarologotipo,linewidth=0.3pt](-0.1,0.9)(-0.1,-0.1)(0,-0.1)(0,0)(0.1,0)(0.1,-0.1)(0.2,-0.1)(0.2,0)(0.3,0)(0.3,-0.1)(0.4,-0.1)(0.4,0)(0.5,0)(0.5,-0.1)(0.6,-0.1)(0.6,0)(0.7,0)(0.7,-0.1)(0.8,-0.1)(0.8,0)(0.9,0)(0.9,-0.1)(1,-0.1)(1.05,-0.05)(1,0)(1,0.1)(1.1,0.1)(1.1,0.2)(1,0.2)(1,0.3)(1.1,0.3)(1.1,0.4)(1,0.4)(1,0.5)(1.1,0.5)(1.1,0.6)(1,0.6)(1,0.7)(1.1,0.7)(1.1,0.8)(1,0.8)(1,0.9)(1.1,0.9)(1.1,1)(1,1)(1,1.1)(0.1,1.1) \end{pspicture}
[image2]
\begin{pspicture}(0,0)(1,1) \pspolygon[fillstyle=solid,fillcolor=azulclarologotipo,linewidth=0.3pt](-0.1,0)(-0.05,-0.05)(0,0)(0.1,0)(0.1,-0.1)(0.2,-0.1)(0.2,0)(0.3,0)(0.3,-0.1)(0.4,-0.1)(0.4,0)(0.5,0)(0.5,-0.1)(0.6,-0.1)(0.6,0)(0.7,0)(0.7,-0.1)(0.8,-0.1)(0.8,0)(0.9,0)(0.9,-0.1)(1,-0.1)(1.05,-0.05)(1,0)(1,0.1)(1.1,0.1)(1.1,0.2)(1,0.2)(1,0.3)(1.1,0.3)(1.1,0.4)(1,0.4)(1,0.5)(1.1,0.5)(1.1,0.6)(1,0.6)(1,0.7)(1.1,0.7)(1.1,0.8)(1,0.8)(1,0.9)(1.1,0.9)(1.1,1)(1,1)(1,1.1)(-0.1,1.1)(-0.1,1)(0,1)(0,0.9)(-0.1,0.9)(-0.1,0.8)(0,0.8)(0,0.7)(-0.1,0.7)(-0.1,0.6)(0,0.6)(0,0.5)(-0.1,0.5)(-0.1,0.4)(0,0.4)(0,0.3)(-0.1,0.3)(-0.1,0.2)(0,0.2)(0,0.1)(-0.1,0.1) \end{pspicture}
[image3]
\begin{pspicture}(0,0)(1,1) \pspolygon[fillstyle=solid,fillcolor=azulclarologotipo,linewidth=0.3pt](-0.1,0)(-0.05,-0.05)(0,0)(0.1,0)(0.1,-0.1)(0.2,-0.1)(0.2,0)(0.3,0)(0.3,-0.1)(0.4,-0.1)(0.4,0)(0.5,0)(0.5,-0.1)(0.6,-0.1)(0.6,0)(0.7,0)(0.7,-0.1)(0.8,-0.1)(0.8,0)(0.9,0)(0.9,-0.1)(1,-0.1)(1.05,-0.05)(1,0)(1,0.1)(1.1,0.1)(1.1,0.2)(1,0.2)(1,0.3)(1.1,0.3)(1.1,0.4)(1,0.4)(1,0.5)(1.1,0.5)(1.1,0.6)(1,0.6)(1,0.7)(1.1,0.7)(1.1,0.8)(1,0.8)(1,0.9)(1.1,0.9)(1.1,1)(1.05,1.05)(1,1)(0.9,1)(0.9,1.1)(0.8,1.1)(0.8,1)(0.7,1)(0.7,1.1)(0.6,1.1)(0.6,1)(0.5,1)(0.5,1.1)(0.4,1.1)(0.4,1)(0.3,1)(0.3,1.1)(0.2,1.1)(0.2,1)(0.1,1)(0.1,1.1)(0,1.1)(-0.05,1.05)(0,1)(0,0.9)(-0.1,0.9)(-0.1,0.8)(0,0.8)(0,0.7)(-0.1,0.7)(-0.1,0.6)(0,0.6)(0,0.5)(-0.1,0.5)(-0.1,0.4)(0,0.4)(0,0.3)(-0.1,0.3)(-0.1,0.2)(0,0.2)(0,0.1)(-0.1,0.1) \end{pspicture}
[image4]
\begin{pspicture}(0,0)(1,3) \def\interior{\pspolygon[fillstyle=solid,fillcolor=azulclarologotipo,linewidth=0.3pt](-0.1,0)(-0.05,-0.05)(0,0)(0.1,0)(0.1,-0.1)(0.2,-0.1)(0.2,0)(0.3,0)(0.3,-0.1)(0.4,-0.1)(0.4,0)(0.5,0)(0.5,-0.1)(0.6,-0.1)(0.6,0)(0.7,0)(0.7,-0.1)(0.8,-0.1)(0.8,0)(0.9,0)(0.9,-0.1)(1,-0.1)(1.05,-0.05)(1,0)(1,0.1)(1.1,0.1)(1.1,0.2)(1,0.2)(1,0.3)(1.1,0.3)(1.1,0.4)(1,0.4)(1,0.5)(1.1,0.5)(1.1,0.6)(1,0.6)(1,0.7)(1.1,0.7)(1.1,0.8)(1,0.8)(1,0.9)(1.1,0.9)(1.1,1)(1.05,1.05)(1,1)(0.9,1)(0.9,1.1)(0.8,1.1)(0.8,1)(0.7,1)(0.7,1.1)(0.6,1.1)(0.6,1)(0.5,1)(0.5,1.1)(0.4,1.1)(0.4,1)(0.3,1)(0.3,1.1)(0.2,1.1)(0.2,1)(0.1,1)(0.1,1.1)(0,1.1)(-0.05,1.05)(0,1)(0,0.9)(-0.1,0.9)(-0.1,0.8)(0,0.8)(0,0.7)(-0.1,0.7)(-0.1,0.6)(0,0.6)(0,0.5)(-0.1,0.5)(-0.1,0.4)(0,0.4)(0,0.3)(-0.1,0.3)(-0.1,0.2)(0,0.2)(0,0.1)(-0.1,0.1)} \def\lateral{\pspolygon[fillstyle=solid,fillcolor=azulclarologotipo,linewidth=0.3pt](-0.1,0)(-0.05,-0.05)(0,0)(0.1,0)(0.1,-0.1)(0.2,-0.1)(0.2,0)(0.3,0)(0.3,-0.1)(0.4,-0.1)(0.4,0)(0.5,0)(0.5,-0.1)(0.6,-0.1)(0.6,0)(0.7,0)(0.7,-0.1)(0.8,-0.1)(0.8,0)(0.9,0)(0.9,-0.1)(1,-0.1)(1.05,-0.05)(1,0)(1,0.1)(1.1,0.1)(1.1,0.2)(1,0.2)(1,0.3)(1.1,0.3)(1.1,0.4)(1,0.4)(1,0.5)(1.1,0.5)(1.1,0.6)(1,0.6)(1,0.7)(1.1,0.7)(1.1,0.8)(1,0.8)(1,0.9)(1.1,0.9)(1.1,1)(1,1)(1,1.1)(-0.1,1.1)(-0.1,1)(0,1)(0,0.9)(-0.1,0.9)(-0.1,0.8)(0,0.8)(0,0.7)(-0.1,0.7)(-0.1,0.6)(0,0.6)(0,0.5)(-0.1,0.5)(-0.1,0.4)(0,0.4)(0,0.3)(-0.1,0.3)(-0.1,0.2)(0,0.2)(0,0.1)(-0.1,0.1)} \def\canto{\psarc[fillstyle=solid,fillcolor=azulclarologotipo,linewidth=0.3pt](0.1,0.9){0.2}{90}{180} \psline[fillstyle=solid,fillcolor=azulclarologotipo,linewidth=0.3pt](-0.1,0.9)(-0.1,-0.1)(0,-0.1)(0,0)(0.1,0)(0.1,-0.1)(0.2,-0.1)(0.2,0)(0.3,0)(0.3,-0.1)(0.4,-0.1)(0.4,0)(0.5,0)(0.5,-0.1)(0.6,-0.1)(0.6,0)(0.7,0)(0.7,-0.1)(0.8,-0.1)(0.8,0)(0.9,0)(0.9,-0.1)(1,-0.1)(1.05,-0.05)(1,0)(1,0.1)(1.1,0.1)(1.1,0.2)(1,0.2)(1,0.3)(1.1,0.3)(1.1,0.4)(1,0.4)(1,0.5)(1.1,0.5)(1.1,0.6)(1,0.6)(1,0.7)(1.1,0.7)(1.1,0.8)(1,0.8)(1,0.9)(1.1,0.9)(1.1,1)(1,1)(1,1.1)(0.1,1.1)} \rput(1,-1.1){\rotatebox{90}{\canto}} \rput(2.1,-0.1){\rotatebox{180}{\lateral}} \rput(3.2,-0.1){\rotatebox{180}{\lateral}} \rput(4.3,-0.1){\rotatebox{180}{\canto}} \rput(1,0){\rotatebox{90}{\lateral}} \rput(1.1,0){\interior} \rput(2.2,0){\interior} \rput(3.3,1){\rotatebox{270}{\lateral}} \rput(0,1.1){\canto} \rput(1.1,1.1){\lateral} \rput(2.2,1.1){\lateral} \rput(3.3,2.1){\rotatebox{270}{\canto}} \end{pspicture}

Resolução: Solução 1: Como há $30$ peças, então as dimensões do tapete podem ser $1\times 30$, $2\times 15$, $3\times 10$ ou $5\times 6$. No primeiros dois casos, não haveria peças interiores; no terceiro caso, haveria apenas $8$ peças interiores; no último caso, de facto, existem $12$ peças interiores. Logo as dimensões do tapete são $5\times 6$. Solução 2: Como há $30-14-4=12$ peças interiores, então as dimensões do interior podem ser $1\times 12$, $2\times 6$ ou $3\times 4$. No primeiro caso, o tapete teria dimensões $3\times 14$, ou seja, $42$ peças; no segundo caso, o tapete teria dimensões $4\times 8$, ou seja, $32$ peças; no terceiro caso, de facto, o tapete tem dimensões $5\times 6$, ou seja, $30$ peças. Logo as dimensões do tapete são $5\times 6$.