# 2001 IMO Shortlist Problems/G3

## Problem

Let $ABC$ be a triangle with centroid $G$. Determine, with proof, the position of the point $P$ in the plane of $ABC$ such that $AP{\cdot}AG + BP{\cdot}BG + CP{\cdot}CG$ is a minimum, and express this minimum value in terms of the side lengths of $ABC$.

## Solution

We claim that the expression is minimized at $P=G$, resulting it having a value of $(a^2+b^2+c^2)/3$ ( $a,b,c$ being the side lengths of $ABC$).

We will use vectors, with $G=\vec{0}$ (meaning that $\vec A+\vec B+\vec C=\vec 0$). Note that by Cauchy-Schwarz, $$\|\vec A\| \|\vec A-\vec P \|+\|\vec B\|\|\vec B-\vec P \|+\|\vec C\| \|\vec C-\vec P \|$$ $$\ge \vec A\cdot (\vec A-\vec P) +\vec B\cdot (\vec B-\vec P)+\vec C\cdot (\vec C-\vec P)$$ $$=\|\vec A\|^2+\|\vec B\|^2+\|\vec C\|^2-\vec P\cdot (\vec A+\vec B+\vec C)=\|\vec A\|^2+\|\vec B\|^2+\|\vec C\|^2,$$ and this bound is clearly reached by $\vec P=\vec 0$. Furthermore, equality is only reached when $\vec A$, $\vec B$, $\vec C$ are scalar multiples of $\vec A-\vec P$, $\vec B-\vec P$, $\vec C-\vec P$, respectively. This means that $\vec P$ is a scalar multiple of $\vec A$, $\vec B$, and $\vec C$, so $\vec P=\vec 0$. (Note that $\vec A$ and $\vec B$ are linearly independent, since the centroid is not on $AB$.)

Now all that remains is to calculate $\|\vec A\|^2+\|\vec B\|^2+\|\vec C\|^2=AG^2+BG^2+CG^2$. To calculate $AG$, first let $D$ be the midpoint of $BC$. Then by Stewart's theorem, $$AD^2\cdot a+\frac{a^3}{4}=\frac{b^2a}{2}+\frac{c^2a}{2}$$ $$AD^2=\frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4}=\frac{2b^2+2c^2-a^2}{4}.$$

Furthermore, $AG=2/3\cdot AD$, so $$AG^2=\frac{2b^2+2c^2-a^2}{9}.$$ By similar reasoning, we can calculate $BG^2=(2c^2+2a^2-b^2)/9$ and $CG^2=(2a^2+2b^2-c^2)/9$, so $$AG^2+BG^2+CG^2=\frac{a^2+b^2+c^2}{3}.$$ $\blacksquare$