The problem has two parts. a) Prove that $\Omega$ belongs to the line (OG) and that it is the midpoint of [OH]. Here, $\Omega$ is the center of the Euler circle, O is the circumcenter, G is the centroid, and H is the orthocenter of triangle ABC. b) Calculate the radius $r$ of the Euler circle as a function of the radius $R$ of the circumcircle of triangle ABC.

1. Problem Description

The problem has two parts.

a) Prove that $\Omega$ belongs to the line (OG) and that it is the midpoint of [OH]. Here, $\Omega$ is the center of the Euler circle, O is the circumcenter, G is the centroid, and H is the orthocenter of triangle ABC.

b) Calculate the radius $r$ of the Euler circle as a function of the radius $R$ of the circumcircle of triangle ABC.

2. Solution Steps

a) We use vector notation with O as the origin.

Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of vertices A, B, C respectively.

The position vector of the centroid G is given by:

$\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$

The position vector of the orthocenter H is given by:

$\vec{h} = \vec{a} + \vec{b} + \vec{c}$

Therefore, $\vec{h} = 3\vec{g}$.

This implies that O, G, and H are collinear, and this line is called the Euler line.

Let $\Omega$ be the center of the Euler circle. The Euler circle passes through the midpoints of the sides of the triangle, the feet of the altitudes, and the midpoints of the segments connecting the vertices to the orthocenter. The center of the Euler circle $\Omega$ is the midpoint of OH. Therefore,

$\vec{\omega} = \frac{\vec{o} + \vec{h}}{2} = \frac{\vec{0} + \vec{h}}{2} = \frac{\vec{h}}{2} = \frac{\vec{a} + \vec{b} + \vec{c}}{2}$

Since $\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$, we have $\vec{a} + \vec{b} + \vec{c} = 3\vec{g}$. Substituting this into the expression for $\vec{\omega}$, we get:

$\vec{\omega} = \frac{3\vec{g}}{2}$

This means that $\vec{\omega}$ is a scalar multiple of $\vec{g}$. Thus, O, G, and $\Omega$ are collinear. Therefore $\Omega$ belongs to the line (OG).

Since $\vec{\omega} = \frac{\vec{h}}{2}$, $\Omega$ is the midpoint of OH.

b) The radius $r$ of the Euler circle is half the radius $R$ of the circumcircle. This is a well-known property of the Euler circle.

The vertices of the medial triangle are the midpoints of the sides of triangle ABC. The medial triangle is similar to triangle ABC with a ratio of 1/

2. Thus, the radius of the circumcircle of the medial triangle (which is the Euler circle) is half the radius of the circumcircle of triangle ABC.

$r = \frac{R}{2}$

3. Final Answer

a) $\Omega$ belongs to the line (OG) and it is the midpoint of [OH].

b) $r = \frac{R}{2}$