Perpendicularitate in prisma

[Claudiu Mindrila]

O prisma dreapta \( ABCDEFA_{1}B_{1}C_{1}D_{1}E_{1}F_{1} \) are baza hexagon regulat. Aratati ca:

a)\( AE_{1}\perp B_{1}E \) daca si numai daca \( AA_{1}=AB\sqrt{3} \);
b) In conditiile de mai sus avem \( \text{dist}\left(AE_{1},\ B_{1}E\right)=\frac{\sqrt{42}}{14}AB \)

Petru Braica, lista scurta, 2006

[Andi Brojbeanu]

a) Notam \( a=AB \) si \( b=AA_1 \). Atunci, \( AE=a\sqrt{3}, B_1E_1=2a, EE_1=b, AB_1=\sqrt{a^2+b^2} \).
Vom folosi lema:
Daca \( A \) , \( B \) , \( C \) , \( D \) sunt patru puncte din plan atunci \( AC\perp BD\ \Longleftrightarrow\ AB^2+CD^2=AD^2+BC^2\ \).
In cazul nostru, \( AE_1\perp B_1E\Leftrightarrow AE^2+B_1E_1^2=EE_1^2+AB_1^2\Leftrightarrow 3a^2+4a^2=b^2+a^2+b^2\Leftrightarrow 2b^2=6a^2\Leftrightarrow b=a\sqrt{3} \) sau \( AA_1=AB\sqrt{3} \).
b) In conditiile de la punctul a), avem: \( AE=a\sqrt{3}, AE_1=a\sqrt{6}, B_1E=a\sqrt{7}, AB_1=2a, B_1E_1=2a, EE_1=a\sqrt{3} \).
Atunci \( \bigtriangleup{AE_1B_1} \) si \( \bigtriangleup{AE_1E} \) sunt isoscele.
Fie \( M \) mijlocul segmentului \( [AE_1] \). Atunci \( EM\perp AE_1 \) si \( B_1M\perp AE_1 \).
Ducem \( MN\perp B_1E \). Deoarece \( AE_1\perp B_1M \) si \( AE_1\perp EM\Rightarrow AE_1\perp (B_1EE_1)\Rightarrow AE_1\perp MN\Rightarrow MN\perp AE_1 \).
Cum \( MN\perp AE_1 \) si \( MN\perp B_1E\Rightarrow d(AE_1, B_1E)=MN \).
Rezolvarea problemei se reduce la a arata ca \( MN=\frac{a\sqrt{42}}{14} \).
\( B_1M=\sqrt{B_1E_1^2-ME_1^2}=\sqrt{4a^2-\frac{6}{4}a^2}=\sqrt{\frac{10}{4}a^2}=\frac{a\sqrt{10}}{2}. \)
\( EM=\sqrt{EE_1^2-ME_1^2}=\sqrt{3a^2-\frac{6}{4}a^2}=\sqrt{\frac{6}{4}a^2}=\frac{a\sqrt{6}}{2}. \)
Din teorema cosinusului in \( \bigtriangleup{MB_1E} \), avem ca: \( cos(\angle{MEB_1})=\frac{ME^2+B_1E^2-B_1M^2}{2ME\cdot B_1E}=\frac{\frac{6}{4}a^2+7a^2-\frac{10}{4}a^2}{2\cdot \frac{a\sqrt{6}}{2}\cdot a\sqrt{7}}=\frac{6a^2}{a^2\sqrt{42}}=\frac{6}{\sqrt{42}} \).
\( sin(\angle{MEB_1})=\sqrt{1-cos^2(\angle{MEB_1})}=\sqrt{1-\frac{36}{42}}=\sqrt{\frac{6}{42}} \).
\( MN=ME\cdot sin(\angle{MEB_1})=\frac{a\sqrt{6}}{2}\cdot \sqrt{\frac{6}{42}}=\frac{3a}{\sqrt{42}}=\frac{3\sqrt{42}a}{42}=\frac{a\sqrt{42}}{14} \).