Let ${AB}$ be a chord of a circle ${(c)}$ centered at ${O}$, and let ${K}$ be a point on the segment ${AB}$ such that ${AK<BK}$. Two circles through ${K}$, internally tangent to ${(c)}$ at ${A}$ and ${B}$, respectively, meet again at ${L}$. Let ${P}$ be one of the points of intersection of the line ${KL}$ and the circle ${(c)}$, and let the lines ${AB}$ and ${LO}$ meet at ${M}$. Prove that the line ${MP}$ is tangent to the circle ${(c)}$. Theoklitos Paragyiou (Cyprus)
Problem
Source: 2016 JBMO Shortlist G7
Tags: geometry, JBMO
FabrizioFelen
08.10.2017 13:04
Let $\omega_{A}=\odot (AKL)$ and $\omega_{B}=\odot (BKL)$, since $\omega_A$ and $\omega _B$ are tangent to $\odot (ABC)$ it's so easy note that: $$2\measuredangle BLK=2\measuredangle ALK =\overarc{AK}=\overarc{KB}=\overarc{AB}=\measuredangle AOB$$$\Longrightarrow$ $\measuredangle AOB=\measuredangle ALK+\measuredangle BLK=\measuredangle ABL$ $\Longrightarrow$ $ABOL$ is cyclic, by simple angle chasing we get $LK$ and $LO$ are the internal bisector and external bisector of $\measuredangle ALB$ respectively. $\Longrightarrow$ $(A,B,K,M)=-1$ and $MO\perp LK$, combining both results we get $\overline{KL}$ is the polar of $M$ wrt $\odot (ABC)$. Finally, so from $P$ is in the polar of $M$, we get $MP$ is tangent to $\odot (ABC)$.
navi_09220114
08.10.2017 14:25
Observe that ALK=arc AB=1/2AOB=BLK=> ALB=AOB, ALOB cyclic and LK bisect angle ALB. But OA=OB hence OL is external angle bisector of ALB => (A,B;K,M)=-1. Together with OLK=90 means LK is the polar line of M, hence MP is tangent to the circle.
bever209
08.08.2020 20:17
Do the same thing as @above to get (A,B;K,M)=-1 Then taking perspectivity from P gives (A,B;Q,P)=-1, (where Q is the tangent to the (c) that is not equal to P). So since MQ is a tangent, then MP is a tangent, and we r done.