Given a triangle ABC, I is the midpoint of [AC]. $f_m$ is a transformation such that for any point M, it maps to point M' where $\vec{MM'} = \vec{MA} - m\vec{MB} + \vec{MC}$, where $m$ is a real parameter. 1) Determine the nature and characteristic element of $f_2$. 2) Assume $m \ne 2$ and let $G_m$ be the point such that $\vec{G_mA} - m\vec{G_mB} + \vec{G_mC} = \vec{0}$. a) Determine and construct the set described by $G_m$ as $m$ varies. b) Determine the nature and characteristic elements of $f_m$ depending on the values of $m$.

1. Problem Description

Given a triangle ABC, I is the midpoint of [AC]. $f_m$ is a transformation such that for any point M, it maps to point M' where $\vec{MM'} = \vec{MA} - m\vec{MB} + \vec{MC}$, where $m$ is a real parameter.

1) Determine the nature and characteristic element of $f_2$.

2) Assume $m \ne 2$ and let $G_m$ be the point such that $\vec{G_mA} - m\vec{G_mB} + \vec{G_mC} = \vec{0}$.

a) Determine and construct the set described by $G_m$ as $m$ varies.

b) Determine the nature and characteristic elements of $f_m$ depending on the values of $m$.

2. Solution Steps

1) Determine the nature and characteristic element of $f_2$.

When $m=2$, we have $\vec{MM'} = \vec{MA} - 2\vec{MB} + \vec{MC}$.

Let's rewrite this using Chasles' relation by inserting B into $\vec{MA}$ and $\vec{MC}$:

$\vec{MM'} = (\vec{MB} + \vec{BA}) - 2\vec{MB} + (\vec{MB} + \vec{BC}) = \vec{MB} + \vec{BA} - 2\vec{MB} + \vec{MB} + \vec{BC} = \vec{BA} + \vec{BC}$.

Since $\vec{MM'} = \vec{BA} + \vec{BC} = \vec{u}$ (a constant vector), this is a translation by the vector $\vec{u} = \vec{BA} + \vec{BC}$.

2) a) Determine and construct the set described by $G_m$ as $m$ varies.

We have $\vec{G_mA} - m\vec{G_mB} + \vec{G_mC} = \vec{0}$.

Rearrange the equation:

$m\vec{G_mB} = \vec{G_mA} + \vec{G_mC}$.

Since I is the midpoint of AC, we have $\vec{IA} + \vec{IC} = \vec{0}$, or $\vec{GA} + \vec{GC} = 2\vec{GI}$ for any point G.

Therefore, $\vec{G_mA} + \vec{G_mC} = 2\vec{G_mI}$.

So, $m\vec{G_mB} = 2\vec{G_mI}$.

$\vec{G_mB} = \frac{2}{m}\vec{G_mI}$.

Since $m \ne 2$, $\vec{G_mI} = \frac{m}{2}\vec{G_mB}$.

This means that the points $G_m$, I, and B are collinear.

Also, $\vec{IG_m} = -\vec{G_mI} = -\frac{m}{2}\vec{G_mB} = \frac{m}{2}\vec{BG_m}$.

So, $G_m$ lies on the line (BI) and its position varies depending on the value of m. Since $m \ne 2$, $G_m \ne I$. The set of points $G_m$ is the line (BI) excluding the point I.

2) b) Determine the nature and characteristic elements of $f_m$ depending on the values of $m$.

$\vec{MM'} = \vec{MA} - m\vec{MB} + \vec{MC}$.

Rewrite the equation using Chasles' relation with point $G_m$:

$\vec{MM'} = (\vec{MG_m} + \vec{G_mA}) - m(\vec{MG_m} + \vec{G_mB}) + (\vec{MG_m} + \vec{G_mC})$

$\vec{MM'} = (1 - m + 1)\vec{MG_m} + \vec{G_mA} - m\vec{G_mB} + \vec{G_mC}$

Since $\vec{G_mA} - m\vec{G_mB} + \vec{G_mC} = \vec{0}$,

$\vec{MM'} = (2 - m)\vec{MG_m}$.

If $m=2$, we already know that $f_2$ is a translation. Now let's consider the cases when $m \ne 2$.

$\vec{M'M} = (m - 2)\vec{MG_m}$.

$\vec{M'M} = (m-2)\vec{MG_m} \implies \vec{M'G_m} + \vec{G_mM} = (m-2)\vec{MG_m}$.

$\vec{M'G_m} = (m-2)\vec{MG_m} - \vec{G_mM} = (m-2)\vec{MG_m} + \vec{MM}$.

$\vec{M'G_m} = (m-2)\vec{MG_m} + \vec{MG_m} + \vec{G_mM} = (m-1)\vec{MG_m}$.

Therefore, $\vec{G_mM'} = (1-m)\vec{G_mM}$.

This indicates that $G_m$ is a fixed point. $f_m$ is a homothety with center $G_m$ and ratio $1-m$.

3. Final Answer

1) $f_2$ is a translation by the vector $\vec{BA} + \vec{BC}$.

2) a) The set described by $G_m$ is the line (BI) excluding the point I.

2) b) If $m=2$, $f_m$ is a translation by the vector $\vec{BA} + \vec{BC}$. If $m \ne 2$, $f_m$ is a homothety with center $G_m$ and ratio $1-m$.