Let $c(0, R)$ be a circle with diameter $AB$ and $C$ a point, on it different than $A$ and $B$ such that $\angle AOC > 90^o$. On the radius $OC$ we consider the point $K$ and the circle $(c_1)$ with center $K$ and radius $KC = R_1$. We draw the tangents $AD$ and $AE$ from $A$ to the circle $(c_1)$. Prove that the straight lines $AC, BK$ and $DE$ are concurrent
Problem
Source: 2010 Balkan Shortlist G8 BMO
Tags: concurrent, concurrency, Tangents, circles, geometry
04.04.2020 12:46
Already been posted.
04.04.2020 12:50
couldn't find using search, that's why I posted it, could you please give the link? edit: looks like this but not the same
04.04.2020 14:18
BMOSL 2010 G8 wrote: Let $c(0, R)$ be a circle with diameter $AB$ and $C$ a point, on it different than $A$ and $B$ such that $\angle AOC > 90^o$. On the radius $OC$ we consider the point $K$ and the circle $(c_1)$ with center $K$ and radius $KC = R_1$. We draw the tangents $AD$ and $AE$ from $A$ to the circle $(c_1)$. Prove that the straight lines $AC, BK$ and $DE$ are concurrent Let $AC\cap BK=\{N\}$ and $\mathcal C_1\cap CA=\{M\}$. Notice that $\angle KMC=\angle OAC\implies KM\|AB\implies \frac{BK}{KN}=\frac{AM}{AN}$. Now Menelaus on $\triangle COA$ on transversal $BN$ we get that $\frac{AC}{CN}\cdot\frac{NK}{KB}=1\implies \frac{AC}{CN}=\frac{BK}{NK}=\frac{MN}{MA}\implies (AN;MC)=-1\implies N$ lies on the Polar of $A$ WRT $\mathcal C_1$. But the Polar of $A$ is $DE$. Hence, $\{AC,BK,DE\}$ are concurrent. $\blacksquare$
12.05.2020 13:03
See also: Balkan MO 2013 G4