Page 37 - RMGO 5

O extindere a unei probleme din Mathematical Reflections nr. 6/2021 37

BA 0 CB 0 AC 0
Demonstrat¸ie. Not˘am x = , y = , z = (coordonatele baricentrice ale
0
0
0
A C B A C B
punctului P), unde xyz = 1. Fie C 1 s , i B 1 mijloacele segmentelor CP, respectiv BP.
Not˘am {β} = BC 1 ∩ AC, {γ} = CB 1 ∩ AB, {D} = AG a ∩ BC, {E} = BG b ∩ CA,
{F} = CG c ∩ AB,
0
Folosind Relat ,ia lui Van Aubel s , i Teorema lui Menelaus ˆın 4ACC cu transver-
sala B − C 1 − β, obt , inem

CP 1 CC 1 xy + 1 C 1 C xy + 1
= + y ⇒ = ⇒ = ;
PC 0 x PC 0 2x C 1 C 0 xy + 2x + 1
BC 0 βA C 1 C βA (z + 1)(xy + 2x + 1)
· · = 1 ⇒ = .
BA βC C 1 C 0 βC xy + 1
γA (y + 1)(xz + 2z + 1)
Analog, = .
γB y(xz + 1)
Aplicˆand Teorema lui Ceva ˆın triunghiul ABC cu cevianele AD, Bβ, Cγ
DB βC γA
deducem c˘ · · = 1, deci
a
DC βA γB
DB (z + 1)(xy + 2x + 1)y(xz + 1) (xy + 2x + 1)(z + 1)(y + 1)
= =
DC (y + 1)(xz + 2z + 1)(xy + 1) (y + 1)(xz + 2z + 1)(xy + 1)
z(xy + 2x + 1)(z + 1) z(xy + 2x + 1)
= = .
(xz + 2z + 1)(xyz + z) xz + 2z + 1

EC FA
Scriem dou˘ relat , ii similare pentru rapoartele s , i :
a
EA FB
EC x(yz + 2y + 1) FA y(xz + 2z + 1)
= , = .
EA xy + 2x + 1 FB yz + 2y + 1

a
a
Cum xyz = 1, aplicˆand reciproca Teoremei lui Ceva rezult˘ c˘ dreptele AG a , BG b ,
CG c sunt concurente.

Bibliografie

[1] https://faculty.evansville.edu/ck6/encyclopedia/etc.html