Let $ABC$ be a scalene and acute triangle with circumcenter $O$. Let $\omega$ be the circle with center $A$, tangent to $BC$ at $D$. Suppose there are two points $F$ and $G$ on $\omega$ such that $FG \perp AO$, $\angle BFD = \angle DGC$ and the couples of points $(B,F)$ and $(C,G)$ are in different halfplanes with respect to the line $AD$. Show that the tangents to $\omega$ at $F$ and $G$ meet on the circumcircle of $ABC$.
Problem
Source: BMO SL 2019, G3
Tags: geometry, circumcircle, tangent
khanhnx
08.11.2020 05:27
Let $U, V$ be second intersections of $BF, CG$ and $\omega$. We have $\angle{DUV} = \angle{DGC} = \angle{DFB} = \angle{DVU},$ then $\triangle DUV$ is isosceles, hence $AD$ $\perp$ $UV$ or $UV$ $\parallel$ $BC$. From this, we have $\angle{BFG} = \angle{UVG} = 180^{\circ} - \angle{BCG}$ or $B, C, G, F$ lie on a circle. But $OB = OC, OF = OG$ then $F, G$ $\in$ $(ABC)$. Let $A'$ be reflection of $A$ in $O$ so $\angle{AFA'} = \angle{AGA'} = 90^{\circ},$ which means $A'F, A'G$ tangents $\omega$ at $F, G$
hydo2332
08.11.2020 05:39
What is BMO?
parmenides51
08.11.2020 05:49
hydo2332 wrote: What is BMO? In this case, Balkan MO, in other cases it might stand for British MO
Assassino9931
08.11.2020 23:30
Also Bulgaria IMO TST 2019, Day 2, Problem 1
Mahdi_Mashayekhi
26.02.2022 10:32
Claim1 : $BCGF$ is cyclic. Proof : $\angle FGC = \angle FGD + \angle DGC = \angle FDB + \angle BFD = \angle 180 - \angle FBD$. Claim2 : $O$ is center of $BCGF$. Proof : we have $FG \perp AO$ and $AF = AG$ so $O$ lies on perpendicular bisector of $FG$. $O$ also lies on perpendicular bisector of $BC$ so $O$ is the center. Now we have $OB = OC = OG = OF$ so $F,G$ are intersections of $\omega$ and the circle with center $A$. Now let tangents to $\omega$ at $F$ and $G$ meet at $S$. we know $S$ lies on perpendicular bisector of $FG$ witch is $AO$. $\angle FGS = \angle FAG/2 = \angle FAO = \angle FAS$ so $FAGS$ is cyclic so $S$ lies on circumcircle of $ABC$. we're Done.