In trapezul dreptunghic ABCD \( AB\parallel CD \) , masura unghiului B este de \( 90^{\circ} \) si AB=2DC. In punctele A si D se ridica de aceeasi parte a planului (ABC) perpendiculare pe planul trapezului , pe care se iau punctele N si P, (AP si ND sunt perpendiculare pe plan) astfel incat DN=a si \( AP=\frac{a}{2} \). Stiind ca M este mijlocul laturii BC si triunghiul MNP este echilateral, determinati :
a) cosinusul unghiului dintre planele MNP si ABC.
b) distanta de la D la planul MNP.
G.Busuioc,N.Solomon,O.J.M.
Trapeze ,,perpendiculare''
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a) Facem notatiile \( BC=2b, AB=2c \). Evident, \( CD=c \).
Atunci, \( MN^2=ND^2+MD^2=ND^2+CD^2+MC^2=a^2+c^2+b^2, \)
\( MP^2=AP^2+AB^2+MB^2=\frac{a^2}{4}+4c^2+b^2, \)
\( NP^2=(DN-PA)^2+(BA-CD)^2+BC^2=\frac{a^2}{4}+c^2+4b^2 \).
Cum \( \bigtriangleup{MNP} \) echilateral, avem: \( MP^2=NP^2\Rightarrow \frac{a^2}{4}+4c^2+b^2=\frac{a^2}{4}+c^2+4b^2\Rightarrow 3c^2=3b^2\Rightarrow b=c \) si \( MN^2=NP^2\Rightarrow a^2+c^2+b^2=\frac{a^2}{4}+c^2+4b^2\Rightarrow \frac{3a^2}{4}=3b^2\Rightarrow b=c=\frac{a}{2} \).
Asadar, \( AB=a, BC=a, CD=\frac{a}{2}, AD=\sqrt{(AB-CD)^2+BC^2}=\sqrt{a^2+\frac{a^2}{4}}=\sqrt{\frac{5a^2}{4}}=\frac{a\sqrt{5}}{2} \).
\( S_{ADM}=S_{ABCD}-S_{ABM}-S_{CDM}=\frac{AB+CD}{2}\cdot BC-\frac{MC\cdot CD}{2}-\frac{AB\cdot BM}{2}=\frac{\frac{a}{2}+a}{2}\cdot a-\frac{\frac{a}{2}\cdot \frac{a}{2}}{2}-\frac{a\cdot\frac{a}{2}}{2}=\frac{3a^2}{4}-\frac{a^2}{8}-\frac{a^2}{4}=\frac{3a^2}{8} \).
Avem \( MN^2=a^2+c^2+b^2=a^2+\frac{a^2}{4}+\frac{a^2}{4}=\frac{3a^2}{2} \).
\( S_{MNP}=\frac{MN^2 \cdot \sqrt{3}}{4}=\frac{\frac{3a^2}{2}\cdot \sqrt{3}}{4}=\frac{3a^2\sqrt{3}}{8} \).
Atunci \( cos\angle((MNP);(ABC))=\frac{S_{ADM}}{S_{MNP}}=\frac{\frac{3a^2}{8}}{\frac{3a^2\sqrt{3}}{8}}=\frac{\sqrt{3}}{3} \).
b) \( V_{MDNP}=\frac{d(D;(MNP))\cdot S_{MNP}}{3}=\frac{d(M;(DNP))\cdot S_{DNP}}{3}\Rightarrow d(D;(MNP))=\frac{d(M;(DNP))\cdot S_{DNP}}{S_{MNP}} \).
Fie \( Q=pr_{M-AD} \). Cum \( DN\perp(ABC)\Rightarrow DN\perp MQ\Rightarrow MQ\perp DN \).
Din \( MQ\perp DN \) si \( MQ\perp DA\Rightarrow MQ\perp (DAN)\Rightarrow d(M;(DNP))=MQ \).
Avem \( S_{ADM}=\frac{MQ\cdot AD}{2}\Rightarrow MQ=\frac{2S_{ADM}}{AD}=\frac{2\cdot \frac{3a^2}{8}}{\frac{a\sqrt{5}}{2}}=\frac{3}{2\sqrt{5}}a \)
\( S_{DNP}=\frac{AD\cdot DN}{2}=\frac{\frac{a\sqrt{5}}{2}\cdot a}{2}=\frac{a^2\sqrt{5}}{4} \).
In fine, \( d(D;(MNP))=\frac{\frac{3}{2\sqrt{5}}a\cdot \frac{a^2\sqrt{5}}{4}}{\frac{3a^2\sqrt{3}}{8}}=\frac{a\sqrt{3}}{3} \).
Atunci, \( MN^2=ND^2+MD^2=ND^2+CD^2+MC^2=a^2+c^2+b^2, \)
\( MP^2=AP^2+AB^2+MB^2=\frac{a^2}{4}+4c^2+b^2, \)
\( NP^2=(DN-PA)^2+(BA-CD)^2+BC^2=\frac{a^2}{4}+c^2+4b^2 \).
Cum \( \bigtriangleup{MNP} \) echilateral, avem: \( MP^2=NP^2\Rightarrow \frac{a^2}{4}+4c^2+b^2=\frac{a^2}{4}+c^2+4b^2\Rightarrow 3c^2=3b^2\Rightarrow b=c \) si \( MN^2=NP^2\Rightarrow a^2+c^2+b^2=\frac{a^2}{4}+c^2+4b^2\Rightarrow \frac{3a^2}{4}=3b^2\Rightarrow b=c=\frac{a}{2} \).
Asadar, \( AB=a, BC=a, CD=\frac{a}{2}, AD=\sqrt{(AB-CD)^2+BC^2}=\sqrt{a^2+\frac{a^2}{4}}=\sqrt{\frac{5a^2}{4}}=\frac{a\sqrt{5}}{2} \).
\( S_{ADM}=S_{ABCD}-S_{ABM}-S_{CDM}=\frac{AB+CD}{2}\cdot BC-\frac{MC\cdot CD}{2}-\frac{AB\cdot BM}{2}=\frac{\frac{a}{2}+a}{2}\cdot a-\frac{\frac{a}{2}\cdot \frac{a}{2}}{2}-\frac{a\cdot\frac{a}{2}}{2}=\frac{3a^2}{4}-\frac{a^2}{8}-\frac{a^2}{4}=\frac{3a^2}{8} \).
Avem \( MN^2=a^2+c^2+b^2=a^2+\frac{a^2}{4}+\frac{a^2}{4}=\frac{3a^2}{2} \).
\( S_{MNP}=\frac{MN^2 \cdot \sqrt{3}}{4}=\frac{\frac{3a^2}{2}\cdot \sqrt{3}}{4}=\frac{3a^2\sqrt{3}}{8} \).
Atunci \( cos\angle((MNP);(ABC))=\frac{S_{ADM}}{S_{MNP}}=\frac{\frac{3a^2}{8}}{\frac{3a^2\sqrt{3}}{8}}=\frac{\sqrt{3}}{3} \).
b) \( V_{MDNP}=\frac{d(D;(MNP))\cdot S_{MNP}}{3}=\frac{d(M;(DNP))\cdot S_{DNP}}{3}\Rightarrow d(D;(MNP))=\frac{d(M;(DNP))\cdot S_{DNP}}{S_{MNP}} \).
Fie \( Q=pr_{M-AD} \). Cum \( DN\perp(ABC)\Rightarrow DN\perp MQ\Rightarrow MQ\perp DN \).
Din \( MQ\perp DN \) si \( MQ\perp DA\Rightarrow MQ\perp (DAN)\Rightarrow d(M;(DNP))=MQ \).
Avem \( S_{ADM}=\frac{MQ\cdot AD}{2}\Rightarrow MQ=\frac{2S_{ADM}}{AD}=\frac{2\cdot \frac{3a^2}{8}}{\frac{a\sqrt{5}}{2}}=\frac{3}{2\sqrt{5}}a \)
\( S_{DNP}=\frac{AD\cdot DN}{2}=\frac{\frac{a\sqrt{5}}{2}\cdot a}{2}=\frac{a^2\sqrt{5}}{4} \).
In fine, \( d(D;(MNP))=\frac{\frac{3}{2\sqrt{5}}a\cdot \frac{a^2\sqrt{5}}{4}}{\frac{3a^2\sqrt{3}}{8}}=\frac{a\sqrt{3}}{3} \).