Consideram triunghiurile \( ABC \) si \( A_1B_1C_1 \) cu \( AB=A_1B_1 \), \( m(\angle BAC)=m(\angle B_1A_1C_1)=60^o \) si \( m(\angle ABC) +m(\angle A_1B_1C_1)=180^o \). Sa se arate ca:
\( \frac{1}{AB}=\frac{1}{AC} + \frac{1}{A_1C_1} \).
ONM Problema 1
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Problema propusa este o consecinta imediata a urmatoarei leme :
relatia Ptolemeu obtinem \( AS=AB+AC \) . Se stie (sau se arata usor din asemanarea \( ABS\sim ADC \) )
ca \( AD\cdot AS=AB\cdot AC \) . In concluzie, \( AD\cdot (AB+AC)=AB\cdot AC \) , adica \( \frac {1}{AD}=\frac {1}{AB}+\frac {1}{AC} \) .
Dem. Daca \( w \) este cercul circumscris al \( \triangle ABC \) si \( \{A,S\}=AD\cap w \) , atunci \( \triangle BCS \) este echilateral . DinLema. Fie \( \triangle ABC \) pentru care \( A=120^{\circ} \) . Notam \( D\in (BC)\ ,\ \angle DAB\equiv\angle DAC \) . Atunci \( \frac {1}{AD}=\frac {1}{AB}+\frac {1}{AC} \) .
relatia Ptolemeu obtinem \( AS=AB+AC \) . Se stie (sau se arata usor din asemanarea \( ABS\sim ADC \) )
ca \( AD\cdot AS=AB\cdot AC \) . In concluzie, \( AD\cdot (AB+AC)=AB\cdot AC \) , adica \( \frac {1}{AD}=\frac {1}{AB}+\frac {1}{AC} \) .