Team Selection Test

Lemma: Let $X$, $Y$, $Z$, $T$ be points on a plane. Then $$X{Y}^2+Y{Z}^2+Y{T}^2+T{X}^2\ge X{Z}^2+Y{T}^2.$$ Proof: Let $x=\overrightarrow{XY}$, $y=\overrightarrow{YZ}$, $z=\overrightarrow{ZT}$. Note that $\overrightarrow{XZ}=x+y$, $\overrightarrow{YT}=y+z$ and $\overrightarrow{XT}=x+y+z$. Then $X{Y}^2+Y{Z}^2+Y{T}^2+T{X}^2-X{Z}^2-Y{T}^2$ is equal to $$\begin{array}{rl}& x\cdot x+y\cdot y+z\cdot z+(x+y+z)\cdot (x+y+z)-(x+y)\cdot (x+y)-(y+z)\cdot (y+z)\\ & =x\cdot x+z\cdot z+2(x\cdot z)=(x+z)\cdot (x+z)\ge 0.\end{array}$$

Let the point $G^{\prime }$ be the centroid of the triangle $A^{\prime }{B}^{\prime }{C}^{\prime }$. Applying the lemma for $A$, $G^{\prime }$, $A^{\prime }$, $G$ gives $A{G}^{\prime 2}+G^{\prime }{A}^{\prime 2}+A^{\prime }{G}^2+G{A}^2\ge G^{\prime }{G}^2+A{A}^{\prime 2}$. As $A{A}^{\prime 2}=A^{\prime }{G}^2+A{A}_1^2$ we have $$A{A}_1^2\le A{G}^{\prime 2}+G^{\prime }{A}^{\prime 2}+G{A}^2-G^{\prime }{G}^2.$$

By similar inequalities for $B$ and $C$, we see that $A{A}_1^2+B{B}_1^2+C{C}_1^2$ is less than or equal to $$A{G}^{\prime 2}+G^{\prime }{A}^{\prime 2}+G{A}^2+B{G}^{\prime 2}+G^{\prime }{B}^{\prime 2}+G{B}^2+C{G}^{\prime 2}+G^{\prime }{C}^{\prime 2}+G{C}^2-3G^{\prime }{G}^2.(\ast )$$ By the Leibniz's theorem we obtain that $A{G}^{\prime 2}+B{G}^{\prime 2}+C{G}^{\prime 2}-3G^{\prime }{G}^2=G{A}^2+G{B}^2+G{C}^2$. It is well known that $G{A}^2+G{B}^2+G{C}^2=\frac{1}{3}(A{B}^2+B{C}^2+C{A}^2)$. As $ABC\cong A^{\prime }{B}^{\prime }{C}^{\prime }$, we also have $G^{\prime }{A}^{\prime 2}+G^{\prime }{B}^{\prime 2}+G^{\prime }{C}^{\prime 2}=\frac{1}{3}(A{B}^2+B{C}^2+C{A}^2)$. These three results conclude that $(\ast )$ is equal to $A{B}^2+B{C}^2+C{A}^2$ and we are done.