Triangle ABC has incircle w centered as S that touches the sides BC,CA and AB at P,Q and R respectively. AB isn't equal AC, the lines QR and BC intersects at point M, the circle that passes through points B and C touches the circle w at point N, circumcircle of triangle MNP intersects with line AP at L (L isn't equal to P). Then prove that S,L and M lie on the same line
Problem
Source: Presendential Kazakhstan olympiad 1day 3 problem
Tags: geometry, circumcircle, perpendicular bisector, projective geometry, geometry proposed
Luis González
16.12.2009 16:43
Perpendicular bisector of $\overline{BC}$ cuts the circle $ \odot(BNC)$ at $ E$ lying on the same side of $ A$ with respect to $ BC.$ Since $ NE$ and $ NP$ are the external and internal bisector of $ \angle BNC,$ it follows that $ EN$ cuts $ BC$ at the harmonic conjugate of $ P$ with respect to $ BC,$ in other words, $ EN$ goes through $ M$ $ \Longrightarrow$ $ \angle MNP = 90^{\circ}.$ Then $ \angle MLP = \angle MNP = 90^{\circ}.$ Since $ LP$ is the polar of $ M$ with respect to the incircle $ (S),$ then $ PL$ goes through $ A.$
Ulanbek_Kyzylorda KTL
16.12.2009 17:14
can you explain me polarity please
Ulanbek_Kyzylorda KTL
17.12.2009 07:42
who has got another solution
nikolapavlovic
13.03.2016 18:41
Let the circle that touches $k$ and passes thru $BC$ be $\omega$ The homothety centered at $N$ taking $k$ to $\omega$.It takes $P$ to the midpoint of arc $BC$ thus $NP$ is the angle bisector of $\angle CNB$.$AP,BQ,CR$ are concurrent thus $(B,C;P,M)$ $\implies$ $\odot MNP$ is the Apollonius circle of $\triangle CNB$$\implies$ $\angle MNP=\frac{\pi}{2}$.All we need to prove now is : $$AP\perp MS$$$M$ is on the polar of $A$ wrt $k$ and $P$ wrt $k$ thus $AP$ is the polar of $M$ wrt $k$ so $MS\perp AP$ and we are done.