Fie \( ABCD \) un tetraedru in care \( \triangle BCD \) este echilateral, iar \( \triangle ACD \) este dreptunghic isoscel cu \( m( \angle CAD)=90^\circ \). Notam cu \( d_{A},d_{B},d_{C},d_{D} \) distantele de la varfurile \( A,B,C,D \) la planele opuse in tetraedrul dat. Aratati ca:
\( 3d^2_{A}+d^2_{B}+d^2_{C}+d^2_{D}=2d_{A}d_{B}\sqrt{3}+2d_{C}d_{D} \)
Gizela Pascale, Revista Minus 1/2008
Identitate in tetraedru
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Claudiu Mindrila
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Identitate in tetraedru
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Andi Brojbeanu
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Se demonstreaza ca \( \mathcal{S}_{ABD}=\mathcal{S}_{ABC} \), deci \( d_C=d_D \). Notand \( BD=a \), obtinem \( \mathcal{S}_{BCD}=\frac{a^2\sqrt{3}}{2} \) si \( \mathcal{S}_{ACD}=\frac{a^2}{4} \). Atunci \( d_B=d_A\cdot \frac{\mathcal{S}_{BCD}}{\mathcal{S}_{ACD}}=d_A\sqrt{3} \).
Asadar, \( 3d_A^2+d_B^2=2\sqrt{3d_A^2d_B^2}=2d_Ad_B\sqrt{3} \) si \( d_C^2+d_D^2=2\sqrt{d_C^2d_D^2}=2d_Cd_D \) si prin insumare rezulta relatia ceruta.
Asadar, \( 3d_A^2+d_B^2=2\sqrt{3d_A^2d_B^2}=2d_Ad_B\sqrt{3} \) si \( d_C^2+d_D^2=2\sqrt{d_C^2d_D^2}=2d_Cd_D \) si prin insumare rezulta relatia ceruta.
Andi Brojbeanu
profesor, Liceul Teoretic "Lucian Blaga", Cluj-Napoca
profesor, Liceul Teoretic "Lucian Blaga", Cluj-Napoca