Page 76 - RMGO 3

76 Rezolvarea problemelor din num˘arul anterior

2 2
MGO 60. Aflat ,i a, b, c, d > 0, s , tiind c˘a a+b+c+d = 2 s , i + +
2a + b + c 2b + c + d
2 2 1 2 1
+ = + + .
2c + d + a 2d + a + b (a + b)(c + d) (a + c)(b + d) (a + d)(b + c)
Florin St˘anescu, G˘aes , ti
x 2 y 2 (x + y) 2
Solut ,ie. Utilizˆand Inegalitatea lui Bergstr¨om, + ≥ , unde z, t > 0,
z t z + t
x y 2 1 4
cu egalitate dac˘a s , i numai dac˘a = , avem = · ≤
z t 2a + b + c 2 a + b + a + c

1 1 1
+ , cu egalitate dac˘a s , i numai dac˘a b = c. Analog, obt , inem
2 a + b a + c

2 1 1 1 2 1 1 1 2
≤ + , ≤ + , ≤
2b + c + d 2 b + c b + d 2c + d + a 2 c + a c + d 2d + a + b

1 1 1
+ . Adunˆand membru cu membru aceste inegalit˘at , i s , i utilizˆand
2 d + a d + b
2 2 2
faptul c˘a a + b + c + d = 2 rezult˘a c˘a + + +
2a + b + c 2b + c + d 2c + d + a

2 1 1 1 1 1 1 1 1
≤ + + + + + =
2d + a + b 2 a + b c + d a + c b + d 2 a + d b + c
1 2 1
+ + . Dar, conform ipotezei, aceast˘a
(a + b)(c + d) (a + c)(b + d) (a + d)(b + c)
inegalitate devine egalitate, deci s , i inegalit˘at , ile adunate devin egalit˘at , i. Deducem
1
c˘a a = b = c = d = .
2

Clasa a IX-a

MGO 61. Se consider˘a triunghiul ABC s , i punctele M ∈ (AB), N ∈ (AC), P ∈
0
0
0
(BC) astfel ˆıncˆat AM = CP, AN = BP, BM = CN. Fie A , B , C mijloacele
0
0
0
segmentelor [BC], [AC], respectiv [AB], iar M , N , P mijloacele segmentelor
0
0
0
0
0
0
[NP], [MP], respectiv [MN]. Ar˘atat ,i c˘a A P · B N · C M = h a h b h c , unde
8
h a , h b , h c reprezint˘a lungimile ˆın˘alt ,imilor triunghiului ABC.
Mihai Florea Dumitrescu, Potcoava
Solut ,ie. Notˆand BM = CN = x, AM = CP = y s , i AN = BP = z, avem
y + z = a, x + z = b, x + y = c, deci x = p − a, y = p − b, z = p − c.
−−→ −−→ −−→ 1 −−→ −→ 1 −−→ −−→ 1 −−→ −→

0
0
0
0
Avem P A = AA − AP = AB + AC − AM + AN = AB + AC −
2 2
2
1 p − b −−→ p − c −→ p − a −−→ p − a −→
0
0
· AB + · AC = · AB + · AC. Rezult˘a c˘a A P =
2 c b 2c 2b
r
p − a 2 2 2 −−→ −→ p − a √
0
−−→ 1 −−→ 1 −→
0
P A = · · AB + · AC + · AB · AC = · 2 + 2 cos A =
2 c b bc 2