Isn't it a bit too trivial? Simply consider the Apollonian Circles...

mecrazywong wrote: Isn't it a bit too trivial? Simply consider the Apollonian Circles... Just love the little things one finds are being claimed trivial from time to time Daniel

mecrazywong wrote: Isn't it a bit too trivial? Simply consider the Apollonian Circles... Remember this is for junior team selection, and it is supposed to be an easy problem on the test. It is not so trivial for a 14-year old

it's fairly easy with complex numbers also. let afix of point $X$ be $x$, and $m=0$ . From conditions above we have $|a|=|b+c|, |b|=|a+c|, |c|=|a+b|$, and if we rewrite $a$ as $x+iy$, $b$ as $m+ni$ and $c$ as $p+qi$ we obtain $x^2+y^2=(m+p)^2+(n+q)^2$, $m^2+n^2=(x+p)^2+(y+q)^2$ and $p^2+q^2=(m+x)^2+(y+n)^2$ . After suming this $(x+m+p)^2+(y+n+q)^2=0$, hence $a+b+c=0$ and $M$ is it's centroid.

Here is my solution:$BM=2MB_{1},CM=2MC_{1}$ and $C_{1}B_{1}=\frac{BC}2$ therefore $\frac{C_{1}M}{CM}=\frac{C_{1}B_{1}}{CB}=\frac{B_{1}M}{BC}=2 \iff \triangle C_{1}B_{1}M$ ~ $\triangle CBM$,thus $\angle{CBM}=\angle{C_{1}B_{1}M}$.But $C_{1}B_{1}$ is paralel to $BC$ therefore $B_{1},M,B$ are colinear.By the same manner we prove that $C_{1},M,C$ are colinear, hence M is the centroid of the triangle and the problem is solved!

We'll find after some computations that MG=0 (we'll use Leibinz theorem)

Valentin Vornicu wrote: Let $ABC$ be a triangle and $A_1$, $B_1$, $C_1$ the midpoints of the sides $BC$, $CA$ and $AB$ respectively. Prove that if $M$ is a point in the plane of the triangle such that \[ \frac{MA}{MA_1} = \frac{MB}{MB_1} = \frac{MC}{MC_1} = 2 , \]then $M$ is the centroid of the triangle. Let $G$ be the centroid of triangle $ABC$. Let $A'$ be the reflection of $A$ about $A_1$. See that $(A,A_1;G,A') = -1$ so $M \in \odot (GA')$ (Appolonius, as $\frac{MA}{MA_1} = 2$). Similarly. define $B',C'$. We see that $M$ is the point of concurrence of $\odot (GA'),\odot (GB'),\odot (GC')$. It is easy to see that these three circles cannot be coaxial. So, $M=G$. $\blacksquare$