JBTST I 2008, Problema 3

[Laurian Filip]

Fie \( ABC \) un triunghi ascutitunghic. Consideram triunghiul echilateral \( A^{\prime}UV, \) cu \( A^{\prime} \in (BC) \), \( U \in (AC) \), \( V \in (AB) \) astfel ca \( UV \parallel BC \). Definim analog punctele \( B^{\prime} \in (AC) \) si \( C^{\prime} \in (AB) \). Sa se demonstreze ca dreptele \( AA^{\prime} \), \( BB^{\prime} \) si \( CC^{\prime} \) sunt concurente.

[Ahiles]

Fie \( A^\prime U=UV=A^\prime V=a \). Atunci in triunghiurile \( BA^\prime V \) si \( UA^\prime C \) avem:
\( \frac{BA^\prime}{a}=\frac{\sin{(120-B)}}{\sin{B}} \)
\( \frac{CA^\prime }{a}=\frac{\sin{(120-C)}}{\sin{C}} \)
Impartind relatiile obtinem:
\( \frac{CA^\prime }{BA^\prime }=\frac{\sin{(120-C)}\cdot\sin{B} }{\sin{(120-B)}\cdot\sin{C}} \)
Analog obtinem:
\( \frac{BC^\prime }{AC^\prime }=\frac{\sin{(120-B)}\cdot\sin{A}}{\sin{(120-A)}\cdot\sin{B}} \)
\( \frac{AB^\prime }{CB^\prime }=\frac{\sin{(120-A)}\cdot\sin{C}}{\sin{(120-C)}\cdot\sin{A}} \)
Inmultind obtinem'
\( \frac{CA^\prime }{BA^\prime }.\frac{BC^\prime }{AC^\prime }.\frac{AB^\prime }{CB^\prime }=\frac{\sin{(120-C)}\cdot\sin{B} }{\sin{(120-B)}\cdot\sin{C}}.\frac{\sin{(120-B)}\cdot\sin{A}}{\sin{(120-A)}\cdot\sin{B}}.\frac{\sin{(120-A)}\cdot\sin{C}}{\sin{(120-C)}\cdot\sin{A}}=1 \)
Deci conform teoremei lui Ceva, drepetele \( AA^\prime \), \( BB^\prime \) si \( CC^\prime \) sunt concurente, q.e.d.