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Bulgarian Mathematical Olympiad

Geometry Difficulty 5.3 AIME, harder Prove it Bulgaria

Problem:
Let $O$ and $G$ be respectively the circumcenter and the centroid of $\triangle ABC$ and let $M$ be the midpoint of the side $AB$ . If $OG \perp CM$ , prove that $\triangle ABC$ is isosceles.

Solution

Solution:
Set $\vec{a}=\overrightarrow{OA}$ , $\vec{b}=\overrightarrow{OB}$ , $\vec{c}=\overrightarrow{OC}$ . We have that $\overrightarrow{OM}=\frac{1}{2}(\vec{a}+\vec{b})$ and hence
$\overrightarrow{OG}=\frac{1}{6}(3\vec{a}+\vec{b}+2\vec{c})$
On the other hand, $\overrightarrow{CM}=\frac{1}{2}(\vec{a}+\vec{b}-2\vec{c})$ . Then

$\begin{aligned} 0 &=\overrightarrow{OG} \cdot \overrightarrow{CM}=(3\vec{a}+\vec{b}+2\vec{c})(\vec{a}+\vec{b}-2\vec{c}) \\ &=3R^{2}+3\vec{a}\vec{b}-6\vec{a}\vec{c}+\vec{a}\vec{b}+R^{2}-2\vec{b}\vec{c}+2\vec{a}\vec{c}+2\vec{b}\vec{c}-4R^{2} \end{aligned}$
and therefore $0=4\vec{a}(\vec{b}-\vec{c})$ . Hence $OA \perp BC$ , i.e. $AB=AC$ .

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