Page 23 - RMGO 6

Rezolvarea problemelor din num˘arul anterior 23

−−−→ −−→ −−→ i −−→ j −→ −−→ −−→ −→
Solut ,ie. Avem P i Q j = P i A + AQ j = − AB + AC s , i P i G = P i A + AG =
2022 2022
i −−→ 2 1 −−→ −→ ä 2022 − 3i−−→ 1−→
Ä
− AB + · AB + AC = AB + AC, deci P i , G s , i Q j sunt
2022 3 2 3 · 2022 3
i j
coliniare dac˘ s , i numai dac˘ − = . Aceast˘ egalitate se poate rescrie,
a
a
a
2022 − 3i 2022
i j
2
2
a
succesiv: = , ij −674(i+j) = 0, (i−674)(j −674) = 2 ·337 . Rezult˘
i − 674 674
2 2 2
a
c˘ i − 674, j − 674 ∈ ±1, ±2, ±4, ±337, ±674, ±1348, ±337 , ±2 · 337 , ±4 · 337 .
Cum i, j ∈ {1, 2, . . . , 2021}, adic˘a i − 674, j − 674 ∈ {−673, −672, . . . , 1347},
rezult˘a c˘a i − 674 = 674 s , i j − 674 = 674, deci i = j = 1348. Astfel o singur˘a
a
a
dreapt˘ P i Q j trece prin G, s , i anume P 1348 Q 1348 (paralela dus˘ prin G la BC).
MGO 185. Demonstrat ,i c˘a ˆın orice triunghi ABC au loc identit˘at ,ile
X 2 3 X 2 X 2 2 9 X 2 2 X 4 9 X 4
m = · a , m m = · a b , m = · a .
a
a
a
b
4 16 16
Mih´aly Bencze, Bras , ov
2
2
P 2b + 2c − a 2
P 2 3 P 2
Solut ,ie. Folosind Formula medianei avem m = = · a ,
a
4 4
2
2
2
2
2
2
P (2b + 2c − a )(2c + 2a − b )
P 2 2 9 P 2 2 2 2 4 4
m m = = · (4b c +4b a −2b +4c +
a
b
4 16
2 2
2
2
P (2b + 2c − a )
2 2
2 2
4
2 2
2 2
4
2 2
9 P
P
4c a −2c b −2a c −2a +a b ) = · a b s , i m = =
a
16 4
1 P 9 P
4
2 2
4
2 2
4
4
2 2
· (4b + 4c + a + 8b c − 4b a − 4c a ) = · a .
16 16
Clasa a X-a
X 1
MGO 186. Fie x 1 , x 2 , . . . , x n > 0, n ∈ N, n ≥ 2 s , i m = . Ar˘atat ,i
x i x j
1≤i<j≤n
c˘a

3
X 1 4 m n(n − 1)
≤ .
» 3 3
x + x 8
1≤i<j≤n i j
Cˆand are loc egalitatea?
Daniel Jinga, Pites , ti
2
3
3
Solut ,ie. Avem (x i − x j ) (x i + x j ) ≥ 0, deci x + x ≥ x i x j (x i + x j ).
i j
Astfel, folosind s , i Inegalitatea Cauchy-Buniakowski-Schwarz, ipoteza s , i Ine-
1 1 1
P P
galitatea mediilor obt , inem » ≤ √ · √ ≤
3
1≤i<j≤n x + x 3 j 1≤i<j≤n x i x j x i + x j
i