Exercise 1319Geometry › Analytic geometryOlympiadAdded 5 Nov 2026

The distance between the circumcentre and the orthocentre

#vectors#orthocentre#circumcentre#Euler line

Statement

Let $ABC$ be a triangle with circumcentre $O$ , circumradius $R$ , orthocentre $H$ , centroid $G$ and sides $a,b,c$ .

Using vectors with origin $O$ , prove that $\vec{OH}=\vec{OA}+\vec{OB}+\vec{OC}$ , and deduce that $O,G,H$ are collinear with $OH=3\,OG$ .
Prove that $OH^2=9R^2-(a^2+b^2+c^2)$ .
Deduce $a^2+b^2+c^2\le9R^2$ , with equality iff $ABC$ is equilateral.

Hint

Set $\mathbf h=\mathbf a+\mathbf b+\mathbf c$ for the vectors $\mathbf a=\vec{OA}$ , etc., and show that $\mathbf h-\mathbf a$ is orthogonal to $\mathbf b-\mathbf c$ using $|\mathbf b|=|\mathbf c|=R$ .

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Solution

Write $\mathbf a,\mathbf b,\mathbf c$ for $\vec{OA},\vec{OB},\vec{OC}$ ; $|\mathbf a|=|\mathbf b|=|\mathbf c|=R$ .

1. Let $\mathbf h=\mathbf a+\mathbf b+\mathbf c$ and $K$ the point with $\vec{OK}=\mathbf h$ . Then
$(\mathbf h-\mathbf a)\cdot(\mathbf b-\mathbf c)=(\mathbf b+\mathbf c)\cdot(\mathbf b-\mathbf c)=|\mathbf b|^2-|\mathbf c|^2=0,$
so $AK\perp BC$ ; by symmetry $BK\perp CA$ and $CK\perp AB$ . So $K=H$ . The centroid is $\vec{OG}=\frac{\mathbf a+\mathbf b+\mathbf c}3$ , hence $\vec{OH}=3\vec{OG}$ .

2. Expanding,
$OH^2=3R^2+2(\mathbf a\cdot\mathbf b+\mathbf b\cdot\mathbf c+\mathbf c\cdot\mathbf a),$
and $a^2+b^2+c^2=6R^2-2(\mathbf a\cdot\mathbf b+\mathbf b\cdot\mathbf c+\mathbf c\cdot\mathbf a)$ , so $OH^2=9R^2-(a^2+b^2+c^2)$ .

3. $OH^2\ge0$ gives the inequality, with equality iff $H=O$ , i.e. iff the triangle is equilateral.

Check. Right triangle $3$ - $4$ - $5$ , $R=2.5$ : $OH=2.5$ and formula gives $9\cdot6.25-50=6.25=2.5^2$ ✓.

The distance between the circumcentre and the orthocentre

Statement

Solution

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