We are given a triangle $ABC$. $O$ is the circumcenter, $G$ is the centroid, and $H$ is the orthocenter of the triangle. $A'$, $B'$, and $C'$ are the midpoints of $BC$, $AC$, and $AB$ respectively. $h$ is a homothety with center $G$ and ratio $-\frac{1}{2}$. 1) a) Draw a figure. b) Prove that the image of the altitude $(AH)$ under the homothety $h$ is the perpendicular bisector of $(BC)$. c) Determine the images of the altitudes $(BH)$ and $(CH)$ under $h$. 2) Prove that the points $O$, $G$, and $H$ are collinear. The line containing $O$, $G$, and $H$ is called the Euler line of triangle $ABC$. 3) Let $\Omega$ be the circumcenter of triangle $A'B'C'$.

1. Problem Description

We are given a triangle $ABC$. $O$ is the circumcenter, $G$ is the centroid, and $H$ is the orthocenter of the triangle. $A'$, $B'$, and $C'$ are the midpoints of $BC$, $AC$, and $AB$ respectively. $h$ is a homothety with center $G$ and ratio $-\frac{1}{2}$.

1) a) Draw a figure.

b) Prove that the image of the altitude $(AH)$ under the homothety $h$ is the perpendicular bisector of $(BC)$.

c) Determine the images of the altitudes $(BH)$ and $(CH)$ under $h$.

2) Prove that the points $O$, $G$, and $H$ are collinear. The line containing $O$, $G$, and $H$ is called the Euler line of triangle $ABC$.

3) Let $\Omega$ be the circumcenter of triangle $A'B'C'$.

2. Solution Steps

1) a) Drawing the figure: This requires drawing a triangle ABC, locating the points O, G, H, A', B', and C' based on their definitions.

1) b) Proof that the image of the altitude $(AH)$ under $h$ is the perpendicular bisector of $(BC)$:

Since $h$ is a homothety with center $G$ and ratio $-\frac{1}{2}$, we have $\vec{GA'} = -\frac{1}{2}\vec{GA}$. Thus, $A'$ is the image of $A$ under the homothety $h$ if the altitude from $A$ is the line $(AH)$. Similarly, the image of $H$ under $h$ is $O$.

Since $AH$ is perpendicular to $BC$, the image of $AH$ under $h$ is also perpendicular to $BC$. Also $A'$ is the image of $A$ under $h$, then the image of line $(AH)$ must pass through $A'$. Since $A'$ is the midpoint of $BC$ and the image of $AH$ is perpendicular to $BC$ at $A'$, the image of $(AH)$ is the perpendicular bisector of $BC$.

1) c) Images of $(BH)$ and $(CH)$ under $h$:

Similarly, the image of the altitude $(BH)$ under $h$ is the perpendicular bisector of $AC$, which is the line perpendicular to $AC$ passing through $B'$. The image of the altitude $(CH)$ under $h$ is the perpendicular bisector of $AB$, which is the line perpendicular to $AB$ passing through $C'$.

2) Proof that $O$, $G$, and $H$ are collinear:

We know that $\vec{OH} = \vec{OA} + \vec{OB} + \vec{OC}$ and $\vec{OG} = \frac{1}{3}(\vec{OA} + \vec{OB} + \vec{OC})$.

Then $\vec{OH} = 3\vec{OG}$, which means $O$, $G$, and $H$ are collinear.

3) Location of $\Omega$:

Since $A'B'C'$ is the medial triangle, the circumcenter of $A'B'C'$ is the midpoint of $OH$. So $\Omega$ is the midpoint of the Euler line.

3. Final Answer

1) b) The image of the altitude (AH) by the homothety h is the perpendicular bisector of (BC).

1) c) The image of the altitude (BH) by the homothety h is the perpendicular bisector of (AC). The image of the altitude (CH) by the homothety h is the perpendicular bisector of (AB).

2) The points O, G, and H are collinear.

3) $\Omega$ is the midpoint of OH.