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soluţii şi barem

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Concursul interjudetean de matematica al Revistei SINUSEditia a II-a, Suceava, 18 noiembrie 2006Etapa finalaClasa a VIII-a2. Se considera dreptunghiul ABCD si E ( ABC)AM ⊥ EB,AN ⊥ECsi AP ED( M EB N EC P ED )a) CE ⊥ ( MNP);b) punctele A, M, N, si P sunt coplanare;c) MB + PD =NC .ME PE NE∉ astfel încât EA⊥( ABC).Fie⊥ ∈ ; ∈ ; ∈ . Aratati ca:BC ⊥ AE BC ⊥ AMRezolvare. a) ⇒ BC ⊥( ABE) ⇒ ⇒ AM ⊥( BCE)BC ⊥ AB AM ⊥ BEAM ⊥ ECAM ⊥ BCE ⇒ ⇒ EC ⊥ AMNAN ⊥ EC( ) ( ).................1p;.............................................................1p;( ***)AM ⊥ ECEC ⊥ AMN ⇒ ⇒ EC ⊥ MNPPN ⊥ EC( ) ( )( )CE ⊥ MNPb)CE ⊥ ANc)⇒ AM ⊂( MNP)AB = MB⋅BE MB ABAE = ME⋅BEME AE2 2⇒ = ;2 2..............................................................1p;...............................................................................1p;analog2PD AD= ................................................1p;2PE AE2 2 2 2MB PD AB + AD BD AC+ = = = .......................................................................1p;2 2 2ME PE AE AE AE2 2AC = NC⋅CE AC NC⇒ = ......................................................................................1p;22AE = NE⋅CEAE NE

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