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 $AG^2+BG^2+CG^2=\frac{a^2+b^2+c^2}{3}.$

$\blacksquare$

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2001 IMO Shortlist Problems/G3

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