The centroid and circumcenter of $\Delta ABC$ are denoted by $G$ and $O$ respectively. If the perpendicular bisectors of $\overline{GA}$, $\overline{GB}$, $\overline{GC}$ intersect pairwise at points $A_1$, $B_1$, $C_1$ respectively, prove that $O$ is the centroid of $\Delta A_1B_1C_1$
Hello everybody. Please help me with the problem stated above which I could not solve. I guess this problem has not been asked before as I searched on Approach0.
My Progress:
Firstly, I let $D, E, F$ be the midpoints of the sides $BC$, $CA$ and $AB$. Now $A_1, B_1, C_1$ are the cicumcenters of $\Delta GBC$, $\Delta GCA$, $\Delta GAB$ respectively. Therefore $A_1D$, $B_1E$, $C_1F$ are the perpendicular bisectors of $\overline{BC}$, $\overline{CA}$, $\overline{CA}$ respectively, hence the lines $A_1D$, $B_1E$, $C_1F$ intersect at $O$. Therefore, it suffices to show that $\overline{A_1D}$, $\overline{B_1E}$, $\overline{C_1F}$ are the three medians of $\Delta A_1B_1C_1$.
Any help would be appreciated.
Thank You.
Please also help me with my previous problem. I can't rest without knowing where I went wrong in this question. Let $f$ be a bijection. Show that there exist infinitely many triples of nonnegative integers with $f(a) + f(c) = 2f(b)$.
Good point. Now, it is enough to prove that the midpoint of $B_1C_1$ lies on the perpendicular bisector of BC, since $O$ and $A_1$ lie there.
Let $X$ be the midpoint of $AG$ and $B_2, C_2$ midpoints of $AC$, $AB$ respectively.
Note that $A, X, C_2, C_1$ lie on the circle with diameter $AC_1$ and $A, X, B_2, B_1$ lie on the circle with diameter $AB_1$. Let $O_C$ and $O_B$ be the centers of these circles.
Let's do a homothety centered at $A$ with coefficient $0.5$. What we want to prove is thus equivalent to that the perpendicular bisector of $B_2C_2$ passes through midpoint of $O_B O_C$.
Let $Y$ be the midpoint of $B_2C_2$. We know that it lies on $AX$, that is, on the radical axis of the two circles. Let $Y_B$ and $Y_C$ be the other points of intersection of the line $B_2C_2$ with the circles.
Since $Y$ lies on the radical axis and $YB_2=YC_2$, by power of point we have $YY_B=YY_C$
Now, note that on the line $B_2C_2$ the order of points cannot be "$Y_B B_2 Y_C C_2$", since then $YB_2 < YY_B$ and $Y C_2 > YY_C$ and the LHS and RHS in both inequalities are equal. The order cannot be "$B_2 Y_B C_2 Y_C$" for the same reason.
So, the order is either "$B_2 Y_B Y_C C_2$" or "$Y_B B_2 C_2 Y_C$".
In both cases, $Y_BB_2 = Y_CC_2$. Project $O_B$ and $O_C$ onto $B_2C_2$. The images of $O_B$ and $O_C$ are the midpoints of $Y_BB_2$ and $YCC_2$ and since they are equal, the midpoint of $O_BO_C$ projects to the midpoint of $B_2C_2$.