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In paralelipipedul dreptunghic ABCDA'B'C'D', M si N sunt centrele fetelor A'B'C'D' si ADD'A'. Aratati ca daca \( AM\perp A^{\prime}C \) si \( C^{\prime}N\perp BD^{\prime} \) atunci paralelipipedul este cub.

Lema:

Daca \( A \) , \( B \) , \( C \) , \( D \) sunt patru puncte din plan atunci \( \overline{\underline{\left\|{\ AC\perp BD\ \Longleftrightarrow\ AB^2+CD^2=AD^2+BC^2\ \right\| \) .

Deoarece \( ACC^{\prime}A^{\prime} \) este dreptunghi si \( M\in [A^{\prime}C^{\prime}] \) inseamna ca punctele \( A\ ,\ C\ ,\ M\ ,\ A^{\prime} \) sunt coplanare .

Aplicand lema \( \Longrightarrow\ AM\perp A^{\prime}C\ \Longleftrightarrow\ AC^2+\underline{\underline{MA^{\prime}^2}}=AA^{\prime}^2+\underline{\underline{\underline{MC^2}}}\ \Longleftrightarrow\ AC^2+\underline{\underline{\frac{A^{\prime}C^{\prime}}{4}}}=AA^{\prime}^{2}+\underline{\underline{\underline{AA^{\prime}^{2}+\frac{A^{\prime}C^{\prime}}{4}}}} \)

\( \Longleftrightarrow\ AB^2+BC^2=2AA^{\prime}^2\ (1) \)

Analog aplicam lema in patrulaterul \( BND^{\prime}C^{\prime}\ :\ NC^{\prime}\perp BD^{\prime}\ \Longleftrightarrow\ BC^{\prime}^2+ND^{\prime}^2=C^{\prime}D^{\prime}^2+BN^2 \)

\( \Longleftrightarrow\ BC^2+AA^{\prime}^2=2AB^2\ (2) \)

Scazand relatiile \( (1) \) si \( (2) \) obtinem \( AB=AA^{\prime} \) si apoi \( AB=BC \), deci paralelipipedul este cub .