Let $ABC$ be an acute-angled triangle inscribed in a circle $k$. It is given that the tangent from $A$ to the circle meets the line $BC$ at point $P$. Let $M$ be the midpoint of the line segment $AP$ and $R$ be the second intersection point of the circle $k$ with the line $BM$. The line $PR$ meets again the circle $k$ at point $S$ different from $R$. Prove that the lines $AP$ and $CS$ are parallel.
Problem
Source: JBMO 2005, Problem 2
Tags: geometry, power of a point, radical axis, geometry proposed
pontios
30.10.2005 08:52
http://www.mathlinks.ro/Forum/viewtopic.php?t=58603 The power of $M$ w.r.t. $k$ is $MA^2$. Draw the circle $w$ passing through $P,B,R$ The point $M$ is on the radical axis of $w$ and $k$, so the power of $M$ w.r.t. the circle $w$ is $MA^2=MP^2$. This means that the segment $MP$ is tangent to $w$. So, the chord $PR$ is shown under equal angles from $P$ and $B$ : $\angle RPA = \angle PBR$ But $\angle PBR = \angle CBR = \angle CSR$ since $C,S,R,B$ are concyclic So we find that $\angle RPA = \angle CSR$, hence $CS \parallel AP$
Attachments:
Virgil Nicula
30.10.2005 09:56
$MP^2=MA^2=MR\cdot MB\Longrightarrow MP^2=MR\cdot MB\Longrightarrow MPR\sim MBP\ (s.a.s.)$ $\Longrightarrow \widehat {PSC}\equiv \widehat {RSC}\equiv \widehat {RBC}\equiv \widehat {MBP}\equiv \widehat {MPR}\equiv \widehat {APS}$ $\Longrightarrow \widehat {PSC}\equiv \widehat {APS}\Longrightarrow AP\parallel CS.$ Remark. I proposed this problem for J.B.M.O. 2005.
Sailor
08.02.2006 19:47
Consider an inversion of pole $P$ which invariates circle $k$. Consequently, $SC$ will transform into the circle circumscribed to $\triangle{BRP}$. Since $MA^2=MR\cdot{MB}=MP^2$ we get that the line $AP$ is tangent to the aforementioned circle and we are done!