Given a line $(D)$, a point $A$ on $(D)$, and a point $B$ on the line perpendicular to $(D)$ that passes through $A$. Construct the point $C$ such that $\vec{AC} = 3\vec{AB}$. 1) Construct the line $(D')$, which is the image of $(D)$ under the homothety centered at $B$ with a ratio of $-2$. 2) Find the locus of the centers of all homotheties with a ratio of $-2$ that transform $(D)$ into $(D')$.

1. Problem Description

Given a line $(D)$, a point $A$ on $(D)$, and a point $B$ on the line perpendicular to $(D)$ that passes through $A$. Construct the point $C$ such that $\vec{AC} = 3\vec{AB}$.

1) Construct the line $(D')$, which is the image of $(D)$ under the homothety centered at $B$ with a ratio of $-2$.

2) Find the locus of the centers of all homotheties with a ratio of $-2$ that transform $(D)$ into $(D')$.

2. Solution Steps

1) To construct $(D')$, the image of $(D)$ under the homothety with center $B$ and ratio $-2$, we need to find the image of at least two points on $(D)$. Let's find the image of $A$. Let $A'$ be the image of $A$ under this homothety. Then $\vec{BA'} = -2\vec{BA}$. This means that $A'$ lies on the line $BA$, and the distance from $B$ to $A'$ is twice the distance from $B$ to $A$, but in the opposite direction.

Since $(D')$ is the image of $(D)$ under a homothety with ratio $-2$, $(D')$ is parallel to $(D)$. Thus, $(D')$ is the line parallel to $(D)$ passing through $A'$.

2) Let $O$ be the center of a homothety with ratio $-2$ that transforms $(D)$ into $(D')$. Let $M$ be a point on $(D)$, and $M'$ be its image on $(D')$. Then $\vec{OM'} = -2\vec{OM}$. Let $I$ be the midpoint of $MM'$. Then

$\vec{OI} = \frac{\vec{OM} + \vec{OM'}}{2} = \frac{\vec{OM} - 2\vec{OM}}{2} = -\frac{1}{2}\vec{OM}$

This implies that $O$, $M$ and $I$ are collinear and $\vec{OI} = -\frac{1}{2}\vec{OM}$, so $\vec{MO} = -2\vec{OI}$ or $\vec{OM} = 2 \vec{IO}$. So $\vec{IO} = \frac{1}{2} \vec{OM}$. Since $M$ is a point on $(D)$, if the midpoint $I$ is fixed, the center $O$ will move along the line determined by $I$.

The locus of the midpoints of $MM'$, where $M \in (D)$ and $M' \in (D')$, is a line parallel to $(D)$ and $(D')$.

Specifically, if $A \in (D)$ and $A' \in (D')$, such that $\vec{BA'} = -2 \vec{BA}$, then $I_A$, the midpoint of $AA'$, is given by $\vec{BI_A} = \frac{\vec{BA} + \vec{BA'}}{2} = \frac{\vec{BA} - 2\vec{BA}}{2} = -\frac{1}{2}\vec{BA}$.

Thus, $I_A$ is on the line $BA$, and its distance from $B$ is half the distance from $B$ to $A$, and in the opposite direction.

The locus of midpoints is the line through $I_A$ and parallel to $(D)$.

Let $(L)$ be the locus of the centers of the homotheties. Let $O$ be a point on $(L)$.

Since $\vec{OM'} = -2 \vec{OM}$, $M' = h(M)$ where $h$ is the homothety.

Consider $A \in (D)$ and $A' \in (D')$. Then $\vec{OA'} = -2 \vec{OA}$. Also, $\vec{AC} = 3 \vec{AB}$. Let $h$ be a homothety such that $h((D)) = (D')$. If $O$ is the center, and $k=-2$, then we must have $\vec{OA'} = -2 \vec{OA}$ for some point $A' \in (D')$ corresponding to $A \in (D)$. Since $(D')$ is parallel to $(D)$, the locus is a line.

Let $M$ and $M'$ be points on $(D)$ and $(D')$ such that $\vec{OM'} = -2 \vec{OM}$. If $I$ is the midpoint of $M$ and $M'$, then $I = (M + M')/2$.

From the midpoint property, we have $\vec{OM} + \vec{OM'} = 2 \vec{OI}$, thus $\vec{OM} - 2\vec{OM} = 2 \vec{OI}$ and $\vec{OI} = -\frac{1}{2} \vec{OM}$.

So $O, I, M$ are collinear and since $M \in (D)$, $\vec{IM} = 3 \vec{IO}$. This implies that $\vec{OI} = -\frac{1}{3} \vec{IM}$.

Thus, $O$ describes a line such that $(D)$ can be obtained from this line by the homothety with center $I$ and ratio -

3. Consider the line through $C$, parallel to $(D)$. It transforms $(D)$ to $(D')$ with the ratio $-2$.

3. Final Answer

The locus of the centers is the line passing through $C$, parallel to $(D)$.