Fie $ABC-$un triunghi echilateral si punctele $P\in[AC], Q\in[AB],$ astfel incat dreapta $PQ$ sa fie tangenta la cercul inscris al triunghiul $ABC.$
Aratati ca cercurile $(P;|PB|), (Q;|QC|)$ si cercul $(O;R)-$circumscris triunghiului $ABC$ au un punct comun.
O problema data la Etapa a 3-a OM din RUSIA (2017)
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- Mesaje: 1493
- Membru din: Mar Oct 26, 2010 9:21 pm
- Localitate: ORADEA
- sunken rock
- Mesaje: 645
- Membru din: Joi Ian 06, 2011 2:49 pm
- Localitate: Constanta
Re: O problema data la Etapa a 3-a OM din RUSIA (2017)
Hint: After some analysis the problem is equivalent with:
On the smaller arc $BC$ of the circumcircle of the equilateral triangle $ABC$ take a point $S$. The perpendicular bisector of $BS$ intersects $AC$ at $P$ and the perpendicular bisector of $CS$ intersects $AB$ at $Q$, then $PQ$ is tangent to the incenter of $\triangle ABC$,
which seems easier to solve.
Best regards,
sunken rock
On the smaller arc $BC$ of the circumcircle of the equilateral triangle $ABC$ take a point $S$. The perpendicular bisector of $BS$ intersects $AC$ at $P$ and the perpendicular bisector of $CS$ intersects $AB$ at $Q$, then $PQ$ is tangent to the incenter of $\triangle ABC$,
which seems easier to solve.
Best regards,
sunken rock
A blind man sees the details better.