In a scalene triangle $\Delta ABC$, $AB<AC$, $PB$ and $PC$ are tangents of the circumcircle $(O)$ of $\Delta ABC$. A point $R$ lies on the arc $\widehat{AC}$(not containing $B$), $PR$ intersects $(O)$ again at $Q$. Suppose $I$ is the incenter of $\Delta ABC$, $ID \perp BC$ at $D$, $QD$ intersects $(O)$ again at $G$. A line passing through $I$ and perpendicular to $AI$ intersects $AB,AC$ at $M,N$, respectively. Prove that, if $AR \parallel BC$, then $A,G,M,N$ are concyclic.
Problem
Source: 2020 China Southeast 10.2
Tags: geometry, circumcircle, incenter
hectorraul
10.08.2020 11:48
Let $L$ be the midpoint of $BC$, $X,Y$ where the perpendicular bisector of $BC$ meets $(BAC)$, $Y$ in the same side of $A$ w.r.t $BC$, $Z=AX\cap BC$, $T$ the $A$-mixtilinear touching point, $K=MN\cap BC$. Take $G$ as the second intersection point of $(ABC)$ and $(AMN)$, We just need to prove that $G, D, Q$ are collinear. Claim 1: $A, L, Q$ are collinear. Not difficult to show. Claim 2: $T, I , Y$ are collinear. We just have to combine two facts, $TN, TM$ meet $(ABC)$ at the midpoints of arc $AC, AB$ and $TA$ is the $T-$symmedian in $MTN$. Claim 3: $X, T , K$ are collinear. $XL\cdot XY=XB^2=XI^2$, then $\angle IYX=\angle XIL$.Let $K'=XT\cap BC$. Now, $\angle XK'L=\angle TYX=\angle XIL=\angle XKL$, then $K=K'$. Claim 4: $Y, G , K$ are collinear. Let $H=TY\cap BC$. $\triangle NGM\sim\triangle CGB$, then $\angle GIK=\angle GLK$ meaning that $KGIL$ is cyclic. Let $\phi$ be the inversion centered at $K$ with radius $\sqrt{KB\cdot KC}$, it is easy to see that $Y=\phi(G)$ and the result follows. Claim 5: $G, D, Q$ are collinear. We know that $\phi(H)=L$ and $\phi(T)=X$. On the other hand $KGILX$ is cycli so its image is the line $TY$, then $\phi(I)=I$. Now, $KD\cdot KZ=KI^2$ then $\phi(D)=Z$. \[ \angle XGD=\angle KHG-\angle KDG=\angle KYX-\angle KYZ=\angle ZIL=\angle ZAL \]then $GD$ and $AL$ meet at $Q$ and we are done.
Steve12345
11.08.2020 14:44
Ok let $G'$ be the intersection of circumcircles $AMN$ and $ABC$. Proving $G',D,Q$ are collinear isn't that hard now. Obviously $AQ\cap BC=S$ where $S$ must be the midpoint of $BC$ (symmedian stuff and symmetry). Also let $D'$ be the A-excircle touch point with $BC$. It's wellknown that $DS=D'S$, $AD' || IS$ and $\triangle G'IS \sim \triangle G'MB$. From here we get $AD'$ and $G'S$ meet on the circumcircle of $ABC$. And we end by butterfly theorem to get $G',D,Q$ are collinear.
githjijjj
12.08.2020 09:09
https://artofproblemsolving.com/community/c6h1480699p8639270 Actually after you figure out that $Q$ is can be defined to be the intersection of $A$-median with the circumcircle, the problem becomes ISL G2 2016