Problema (nível 4): O octógono regular da figura está inscrito numa circunferência de raio 1 e $P$ é um ponto arbitrário desta circunferência. (image1) Calcula o valor de $\overline{PA}^{\;2}+\overline{PB}^{\;2}+\dots +\overline{PH}^{\;2}.$


Conteúdo referenciado:
[image1]
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Resolução: Como $[ABCDEFGH]$ é um trabalho regular, cada uma das diagonais $[AE]$, $[BF]$, $[CG]$ e $[DH]$ é um diâmetro da circunferência, pelo que os triângulos $[APE]$, $[BPF]$, $[CPG]$ e $[DPH]$ são rectângulos (em $P$). Portanto, pelo Teorema de Pitágoras, $$\overline{PA}^{\;2}+\overline{PE}^{\;2}=4,$$ $$\overline{PB}^{\;2}+ \overline{PF}^{\;2}=4,$$ $$\overline{PC}^{\;2}+\overline{PG}^{\;2}=4$$ e $$\overline{PD}^{\;2}+\overline{PH}^{\;2}=4.$$ Consequentemente, $$\overline{PA}^{\;2}+\overline{PB}^{\;2}+\dots +\overline{PH}^{\;2}=16$$