Page 97 - RMGO 2

Rezolvarea problemelor din num˘arul anterior 97

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MGO 18. Determinat ,i x ∈ R astfel ˆıncˆat min |x| , 1 > x − 1 .

x
x
Aurel Chirit ,˘a, Slatina s , i Lucian Tut ,escu, Craiova

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Solut ,ie. x ∈ R verific˘a inegalitatea dat˘a dac˘a s , i numai dac˘a |x| > x − 1 s , i

x
2
1 1
1 1 1
> x − . Dar |x| > x − 2 x − 2 2 ⇔
⇔ x > ⇔ x > x − 2 +
x x x x x
2
1 − 2x 2 √ √ 1 1
1
< 0 ⇔ x ∈ −∞, − 2 ∪ 2 , +∞ = A, iar > x − ⇔ >

x
x 2 2 2 x x 2
2
1 1 1 √ √
2
2
x − ⇔ > x −2+ ⇔ x −2 < 0, x 6= 0 ⇔ x ∈ − 2, 2 \{0} = B.
x x 2 x 2
√ √ √
√
Solut , ia problemei este x ∈ A ∩ B, adic˘a x ∈ − 2, − 2 ∪ 2 , 2 .
2 2
0
0
0
0
0
0
MGO 19. Se consider˘a cubul ABCDA B C D s , i punctele M ∈ (A D ),
0
0
0
0
0
0
N ∈ (D C ) s , i P ∈ (BC) astfel ˆıncˆat A M = MD , ND = 2NC s , i PC = 3PB.
S , tiind c˘a AB = a, determinat ,i:
0
a) Distant ,a de la punctul D la planul (MNP);
b) Sinusul unghiului dintre planele (ABC) s , i (MNP).
Mihai Florea Dumitrescu, Potcoava
0
0
0
0
0
0
Solut ,ie. a) Fie PP kBB , P ∈ (B C ) s , i P Q ⊥ MN, Q ∈ MN. Avem A MNP =
0
a 2 √
2
2
0
0
A A B C D − A MD N − A NC P − A A MP B = , iar MN = D M + D N =
0
0
0
0
0
0
0
0
0
0
3 √
5a 2 · A MNP 0 4a p a 41
0
0
2
, deci P Q = = . Rezult˘a c˘a PQ = PP 02 + P Q = .
6 MN 5 5
0
A MNP · d(D , (MNP)) A MD N · PP 0
0
0
Avem V D MNP = = , deci d(D , (MNP)) =
0
3 √ 3
0
0
A MD N · PP 0 D M · D N · PP 0 2a 41
0
= = .
A MNP MN · PQ 41
0
0
0
0
b) sin (^ ((ABC), (MNP))) = sin (^ ((A B C ), (MNP))) = sin (^P QP) =
√
PP 0 5 41
= .
PQ 41
2
2
2
MGO 20. Dac˘a a, b, c > 0 s , i a + b + c = 1, demonstrat ,i c˘a are loc inegalitatea
1 1 1 1
+ + ≤ .
a + b a + c b + c 2abc
Florin St˘anescu, G˘aes , ti
2ab 2ac 2bc
Solut ,ie. Utilizˆand inegalit˘at , i uzuale avem · c + · b + · a ≤
a + b a + c b + c
a + b a + c b + c
2
2
2
· c + · b + · a = ab + ac + bc ≤ a + b + c = 1, de unde, ˆımp˘art , ind
2 2 2
prin 2abc, obt , inem inegalitatea din enunt , .