Fie ABCDA'B'C'D' un paralelipiped dreptunghic care are AB=a, BC=b, CC'=c. Punctele M, N, P, Q, R si S sunt situate pe muchiile [BC], [BB'], [A'B'], [A'D'], [DD'], respectiv [CD] astfel incat drumurile AX+XC', cu \( X\in\{\ M,\ N,\ P,\ Q,\ R,\ S\} \), au lungimi minime.
a) Aratati ca punctele M, N, P, Q, R, S sunt coplanare.
b) Calculati aria proiectiei poligonului MNPQRS pe planul (ABC).
,,Gh.Lazar'' 2005
Paralelipiped cu puncte pe muchii
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Marius Mainea
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a) Drumul \( AX+XC^{\prime} \) este minim cand \( A, X, C^{\prime} \) sunt coliniare (prin aducerea in acelasi plan a celor doua fete cu muchia comuna reprezentata de segmentul ce contine punctul \( X \)).
Atunci, vom avea ca \( \frac{BM}{MC}=\frac{AB}{CC^{\prime}}=\frac{a}{c}\Rightarrow \frac{BM}{BM+MC}=\frac{BM}{b}=\frac{a}{a+c}\Rightarrow BM=\frac{ab}{a+c} \) si \( MC=BC-BM=b-\frac{ab}{a+c}=\frac{ba+bc-ab}{a+c}=\frac{bc}{a+c} \).
Analog, \( QD^{\prime}=BM=\frac{ab}{a+c}; QA^{\prime}=MC=\frac{bc}{a+c}; BN=RD^{\prime}=\frac{ac}{a+b}; NB^{\prime}=RD=\frac{bc}{a+b}; SD=PB^{\prime}=\frac{ab}{b+c}; PA^{\prime}=SC=\frac{ac}{b+c} \). Fie \( E=AB\cap PN \).
\( PB^{\prime}\parallel BT\Rightarrow \bigtriangleup{PB^{\prime}N}\sim\bigtriangleup{EBN}\Rightarrow BE=\frac{BN}{B^{\prime}N}\cdot PB^{\prime}=\frac{\frac{ac}{a+b}}{\frac{bc}{a+b}}\cdot \frac{ab}{b+c}=\frac{a^2}{b+c} \).
Atunci \( \frac{BE}{SC}=\frac{\frac{a^2}{b+c}}{\frac{ac}{b+c}}=\frac{a}{c}=\frac{BM}{MC} \) si cum \( \angle{SCM}\equiv\angle{EBM}\Rightarrow \bigtriangleup{SCM}\sim \bigtriangleup{EBM}\Rightarrow \angle{BME}\equiv\angle{SMC}\Rightarrow S, M, E \) coliniare.
Asadar, \( PN\cap SM=\{E\}\Rightarrow \) punctele \( M, N, P, S \) sunt coliniare. Deci, \( S\in (MNP) \). Analog se demonstreaza ca \( R, Q\in (MNP) \).
b) Fie \( P^{\prime} \) si \( Q^{\prime} \) proiectiile punctelor \( P \) si \( Q \) pe planul \( (ABC) \).
Atunci proiectia poligonului \( MNPQRS \) pe planul \( ABC \) este \( BMSDQ^{\prime}P^{\prime} \).
\( S_{BMSDQ^{\prime}P^{\prime}}=S_{ABCD}-S_{AP^{\prime}Q^{\prime}}-S_{MCS}=S_{ABCD}-2\cdot S_{MCS}=ab-2\cdot \frac{MC\cdot CS}{2}=ab-\frac{bc}{a+c}\cdot \frac{ac}{b+c}=\frac{ab(a+c)(b+c)-bc\cdot ac}{(a+c)(b+c)}=\frac{ab(c^2+ab+ac+bc-c^2)}{(a+c)(b+c)}=\frac{ab(ab+ac+bc)}{(a+c)(b+c)} \).
Atunci, vom avea ca \( \frac{BM}{MC}=\frac{AB}{CC^{\prime}}=\frac{a}{c}\Rightarrow \frac{BM}{BM+MC}=\frac{BM}{b}=\frac{a}{a+c}\Rightarrow BM=\frac{ab}{a+c} \) si \( MC=BC-BM=b-\frac{ab}{a+c}=\frac{ba+bc-ab}{a+c}=\frac{bc}{a+c} \).
Analog, \( QD^{\prime}=BM=\frac{ab}{a+c}; QA^{\prime}=MC=\frac{bc}{a+c}; BN=RD^{\prime}=\frac{ac}{a+b}; NB^{\prime}=RD=\frac{bc}{a+b}; SD=PB^{\prime}=\frac{ab}{b+c}; PA^{\prime}=SC=\frac{ac}{b+c} \). Fie \( E=AB\cap PN \).
\( PB^{\prime}\parallel BT\Rightarrow \bigtriangleup{PB^{\prime}N}\sim\bigtriangleup{EBN}\Rightarrow BE=\frac{BN}{B^{\prime}N}\cdot PB^{\prime}=\frac{\frac{ac}{a+b}}{\frac{bc}{a+b}}\cdot \frac{ab}{b+c}=\frac{a^2}{b+c} \).
Atunci \( \frac{BE}{SC}=\frac{\frac{a^2}{b+c}}{\frac{ac}{b+c}}=\frac{a}{c}=\frac{BM}{MC} \) si cum \( \angle{SCM}\equiv\angle{EBM}\Rightarrow \bigtriangleup{SCM}\sim \bigtriangleup{EBM}\Rightarrow \angle{BME}\equiv\angle{SMC}\Rightarrow S, M, E \) coliniare.
Asadar, \( PN\cap SM=\{E\}\Rightarrow \) punctele \( M, N, P, S \) sunt coliniare. Deci, \( S\in (MNP) \). Analog se demonstreaza ca \( R, Q\in (MNP) \).
b) Fie \( P^{\prime} \) si \( Q^{\prime} \) proiectiile punctelor \( P \) si \( Q \) pe planul \( (ABC) \).
Atunci proiectia poligonului \( MNPQRS \) pe planul \( ABC \) este \( BMSDQ^{\prime}P^{\prime} \).
\( S_{BMSDQ^{\prime}P^{\prime}}=S_{ABCD}-S_{AP^{\prime}Q^{\prime}}-S_{MCS}=S_{ABCD}-2\cdot S_{MCS}=ab-2\cdot \frac{MC\cdot CS}{2}=ab-\frac{bc}{a+c}\cdot \frac{ac}{b+c}=\frac{ab(a+c)(b+c)-bc\cdot ac}{(a+c)(b+c)}=\frac{ab(c^2+ab+ac+bc-c^2)}{(a+c)(b+c)}=\frac{ab(ab+ac+bc)}{(a+c)(b+c)} \).