Let $ABC$ be a triangle with centroid $G$. Let the circumcircle of triangle $AGB$ intersect the line $BC$ in $X$ different from $B$; and the circucircle of triangle $AGC$ intersect the line $BC$ in $Y$ different from $C$. Prove that $G$ is the centroid of triangle $AXY$.
Problem
Source: RMO Mumbai 2016, P5
Tags: geometry, circumcircle
WizardMath
11.10.2016 10:29
By power of a point, if $M$ is the midpoint of $BC$, $MG.MA=MB.MX=MC.MY$ and so $XY$ has midpoint $M$, from which the conclusion follows due to $AG:GM=1:2$.
phymaths
11.10.2016 14:04
$POP$ straight away or trig bash..
mather007
26.10.2018 17:50
Same solution as WizardMath.... Too simple .
math_pi_rate
26.10.2018 18:18
You bumped a two year old topic just to tell that your solution is similar to another solution, and that you think its easy. Seriously, hats off to you.
Jishnu4414l
14.12.2023 19:14
Let $D=AG \cap BC$ Using Power of point on $D$, we have $DX.DB=DY.DC$ $\implies DX=DY$ (As $DB=DC$) Thus $D$ is the midpoint of $XY$. Now as $G$ is the centroid of $\triangle ABC$, we must have $AG:GD=2:1$. but $AD$ is the $A$ median in $\triangle AXY$ too. So $G$ must be the centroid of $\triangle AXY$ too. Our proof is thus complete.