Page 135 - MATINF Nr. 13-14

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PROBLEME DE MATEMATICA PENTRU CONCURSURI 135

Solut ,ia 1 (Leonard Mihai Giugiuc, Drobeta Turnu Severin). Consider˘am A(−2u, 0), B(2v, 0) s , i
C(0, 2), unde u = ctg A s , i v = ctg B, deci u + v > 0. Avem:

mediana din B: x + y(u + 2v) − 2v = 0;
2
mediatoarea lui [BC]: xv − y + 1 − v = 0;

ˆın˘alt , imea din C: x = 0.

1 0 0

2
3
Cum ele sunt concurente, atunci v −1 1 − v 2 = 0, adic˘ 2v + uv − u = 0.
a

1 u + 2v −2v
Dac˘a v = 0, atunci u = 0, fals; dac˘a v = 1, atunci 2 = 0, fals; dac˘a v = −1, atunci −2 = 0,
2v 3
fals. Astfel v ∈ R − {−1, 0, 1} s , i u = .
1 − v 2
2v 3 1 + v 2
a) Cum u + v > 0, atunci > −v. Dac˘a v > 0, atunci > 0, deci v ∈ (0, 1).
1 − v 2 1 − v 2
1 + v 2
a
Dac˘ v < 0, atunci < 0, deci v ∈ (−∞, −1).
1 − v 2
π π Å 3π ã
ˆ , ∪ , π .
In concluzie, v ∈ (0, 1) ∪ (−∞, −1), adic˘a B ∈
4 2 4
3
2v 3 Å 2ctg B ã
b) Cum u = , atunci A = arcctg .
2
1 − v 2 1 − ctg B
Å 2 ã
1 − uv 1 − 2ctg B
Cum ctg C = , atunci C = arcctg .
u + v ctg B
Ä 1−2v 2 ä
A π − arcctg v − arcctg v π − arcctg v
c) Avem lim = lim Ä ä = −1 + lim Ä ä.
B→π C v→−∞ arcctg 1−2v 2 v→−∞ arcctg 1−2v 2
B<π v v
Folosind Regula lui L’Hˆopital avem

Ä 1−2v 2 ä 2 Ä 1−2v 2 ä 2
π − arcctg v 1 1 + v v 2 1 + v
lim Ä ä = lim · = lim · = 2.
v→−∞ arcctg 1−2v 2 v→−∞ 1 + v 2 1+2v 2 v→−∞ 1 + v 2 1 + 2v 2
v v 2

A
ˆ = 1.
In concluzie, lim
B→π C
B<π
ˆ
Solut ,ia 2 (Marius Damian, Br˘aila). a) Intr-un reper cartezian xOy consider˘am punctele A(a, b),
Å ã
a + c b
a ∈ R, b > 0, B(−c, 0), c > 0, C(c, 0), M , mijlocul lui [AC], D piciorul ˆın˘alt , imii din
2 2
C s , i E punctul de intersect , ie a dreptelor din enunt , ul problemei.
Ecuat , ia dreptei MB este

b bc
y − y B x − x B
a
= , adic˘ y = (x + c), deci y E = , (1).
y M − y B x M − x B a + 3c a + 3c