Given three non-collinear points $A$, $B$, and $C$. For any point $M$ in the plane, we associate a point $M'$ defined by $3\vec{AM'} - 2\vec{AM} = \vec{BC}$. 1) Explain why each point $M$ corresponds to a unique point $M'$. 2) We define a function $f$ from the plane into itself. Determine the images of $A$, $B$, and $C$ by $f$. 3) Find a point invariant by $f$. 4) Demonstrate that $f$ is a homothety and give its center and ratio.

1. Problem Description

Given three non-collinear points $A$, $B$, and $C$. For any point $M$ in the plane, we associate a point $M'$ defined by $3\vec{AM'} - 2\vec{AM} = \vec{BC}$.

1) Explain why each point $M$ corresponds to a unique point $M'$.

2) We define a function $f$ from the plane into itself. Determine the images of $A$, $B$, and $C$ by $f$.

3) Find a point invariant by $f$.

4) Demonstrate that $f$ is a homothety and give its center and ratio.

2. Solution Steps

1) We have $3\vec{AM'} - 2\vec{AM} = \vec{BC}$. Then

$3\vec{AM'} = 2\vec{AM} + \vec{BC}$

$\vec{AM'} = \frac{2}{3}\vec{AM} + \frac{1}{3}\vec{BC}$.

Since the right side is uniquely determined by $M$, $\vec{AM'}$ is uniquely determined. Therefore $M'$ is unique.

2) We need to find $f(A), f(B), f(C)$. Let $A', B', C'$ be the images of $A, B, C$ respectively.

For $M=A$, we have $3\vec{AA'} - 2\vec{AA} = \vec{BC}$, which means $3\vec{AA'} = \vec{BC}$, so $\vec{AA'} = \frac{1}{3}\vec{BC}$. Thus, $A' = A + \frac{1}{3}\vec{BC}$.

For $M=B$, we have $3\vec{AB'} - 2\vec{AB} = \vec{BC}$, which means $3\vec{AB'} = 2\vec{AB} + \vec{BC}$. Thus, $\vec{AB'} = \frac{2}{3}\vec{AB} + \frac{1}{3}\vec{BC}$.

For $M=C$, we have $3\vec{AC'} - 2\vec{AC} = \vec{BC}$, which means $3\vec{AC'} = 2\vec{AC} + \vec{BC}$. Thus, $\vec{AC'} = \frac{2}{3}\vec{AC} + \frac{1}{3}\vec{BC}$.

3) We are looking for a point $I$ such that $f(I) = I$, i.e., $3\vec{AI} - 2\vec{AI} = \vec{BC}$, so $\vec{AI} = \vec{BC}$. Then $I = A + \vec{BC}$.

4) We have $\vec{AM'} = \frac{2}{3}\vec{AM} + \frac{1}{3}\vec{BC}$. Also, $\vec{AI} = \vec{BC}$.

Then $\vec{AM'} = \frac{2}{3}\vec{AM} + \frac{1}{3}\vec{AI}$.

Subtracting $\vec{AI}$ from both sides,

$\vec{AM'} - \vec{AI} = \frac{2}{3}\vec{AM} + \frac{1}{3}\vec{AI} - \vec{AI} = \frac{2}{3}\vec{AM} - \frac{2}{3}\vec{AI}$

$\vec{IM'} = \frac{2}{3}(\vec{AM} - \vec{AI}) = \frac{2}{3}\vec{IM}$.

Therefore, $f$ is a homothety with center $I$ and ratio $\frac{2}{3}$.

The center is $I = A + \vec{BC}$, such that $\vec{AI} = \vec{BC}$.

3. Final Answer

1) Each point $M$ corresponds to a unique point $M'$ because $\vec{AM'} = \frac{2}{3}\vec{AM} + \frac{1}{3}\vec{BC}$ is uniquely determined by $M$.

2) $f(A) = A + \frac{1}{3}\vec{BC}$, $\vec{AB'} = \frac{2}{3}\vec{AB} + \frac{1}{3}\vec{BC}$, $\vec{AC'} = \frac{2}{3}\vec{AC} + \frac{1}{3}\vec{BC}$.

3) $I = A + \vec{BC}$.

4) $f$ is a homothety with center $I = A + \vec{BC}$ and ratio $\frac{2}{3}$.