The problem is based on an equilateral triangle $ABC$ with side length 4cm and center $O$. $\Gamma$ is the circumcircle of $ABC$. $I$ is the midpoint of $[AB]$ and $J$ is the midpoint of $[OI]$. The lines $(OA)$ and $(OC)$ intersect $\Gamma$ at $D$ and $E$ respectively. The problem asks us to find the point $G$ such that $\vec{GA} + \vec{GB} + \vec{GC} + \vec{GD} + \vec{GE} = \vec{0}$. We need to express $\vec{OG}$ in terms of $\vec{OB}$ and then in terms of $\vec{OJ}$ and $\vec{OD}$. We need to show that lines $(OB)$ and $(DJ)$ intersect at $G$. Finally, we are given a transformation $f$ such that $4\vec{MM'} = \vec{MA} + \vec{MB} + \vec{MC} + \vec{MD} + \vec{ME}$ and $M' = f(M)$. We must show that $f$ is a homothety, specifying its center and ratio. We also need to find the images of points $B$ and $D$ under the transformation $f$.

1. Problem Description

The problem is based on an equilateral triangle $ABC$ with side length 4cm and center $O$. $\Gamma$ is the circumcircle of $ABC$. $I$ is the midpoint of $[AB]$ and $J$ is the midpoint of $[OI]$. The lines $(OA)$ and $(OC)$ intersect $\Gamma$ at $D$ and $E$ respectively.

The problem asks us to find the point $G$ such that $\vec{GA} + \vec{GB} + \vec{GC} + \vec{GD} + \vec{GE} = \vec{0}$. We need to express $\vec{OG}$ in terms of $\vec{OB}$ and then in terms of $\vec{OJ}$ and $\vec{OD}$. We need to show that lines $(OB)$ and $(DJ)$ intersect at $G$.

Finally, we are given a transformation $f$ such that $4\vec{MM'} = \vec{MA} + \vec{MB} + \vec{MC} + \vec{MD} + \vec{ME}$ and $M' = f(M)$. We must show that $f$ is a homothety, specifying its center and ratio. We also need to find the images of points $B$ and $D$ under the transformation $f$.

2. Solution Steps

2a) Express $\vec{OG}$ in function of $\vec{OB}$

Since $\vec{GA} + \vec{GB} + \vec{GC} + \vec{GD} + \vec{GE} = \vec{0}$, we have:

$\vec{OA} - \vec{OG} + \vec{OB} - \vec{OG} + \vec{OC} - \vec{OG} + \vec{OD} - \vec{OG} + \vec{OE} - \vec{OG} = \vec{0}$

$\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} + \vec{OE} - 5\vec{OG} = \vec{0}$

$5\vec{OG} = \vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} + \vec{OE}$

Since $ABC$ is an equilateral triangle centered at $O$, and $D$ and $E$ are on the circumcircle, $\vec{OA} + \vec{OC} = \vec{OD} + \vec{OE}$.

Then $5\vec{OG} = \vec{OB} + 2(\vec{OA} + \vec{OC})$

Since $O$ is the center of the equilateral triangle $ABC$, $\vec{OA} + \vec{OB} + \vec{OC} = \vec{0}$, so $\vec{OA} + \vec{OC} = -\vec{OB}$.

$5\vec{OG} = \vec{OB} + 2(-\vec{OB}) = -\vec{OB}$

$\vec{OG} = -\frac{1}{5}\vec{OB}$

2b) Express $\vec{OG}$ in function of $\vec{OJ}$ and $\vec{OD}$

We know that $J$ is the midpoint of $[OI]$, so $\vec{OJ} = \frac{1}{2}\vec{OI}$. Since $I$ is the midpoint of $[AB]$, $\vec{OI} = \frac{1}{2}(\vec{OA}+\vec{OB})$, and $\vec{OA}+\vec{OB}+\vec{OC} = \vec{0}$ thus $\vec{OI} = \frac{1}{2}(-\vec{OC} + \vec{OB})$. Therefore $\vec{OJ} = \frac{1}{4}(-\vec{OC}+\vec{OB})$.

From the previous result, we have $\vec{OB} = -5\vec{OG}$. Also, $D$ is on $(OA)$, so $\vec{OA} = k\vec{OD}$ for some constant $k$.

Since $\vec{OA} + \vec{OC} = -\vec{OB}$, $\vec{OC} = -\vec{OB} - \vec{OA} = 5\vec{OG} - k\vec{OD}$.

Since $\vec{OJ} = \frac{1}{4}(-\vec{OC}+\vec{OB}) = \frac{1}{4}(-5\vec{OG}+k\vec{OD} - 5\vec{OG}) = \frac{1}{4}(-10\vec{OG}+k\vec{OD})$

Then $4\vec{OJ} = -10\vec{OG}+k\vec{OD}$, so $10\vec{OG} = k\vec{OD}-4\vec{OJ}$

$\vec{OG} = \frac{k}{10}\vec{OD} - \frac{2}{5}\vec{OJ}$

Since $\vec{OA} + \vec{OB} + \vec{OC} = 0$, $\vec{OA} = \vec{OD}$ (since $D$ is on $\Gamma$), $\vec{OB} = -5 \vec{OG}$

$5\vec{OG} = \vec{OD} + \vec{OB} + \vec{OC} + \vec{OD} + \vec{OE} = \vec{OD} - 5\vec{OG} + \vec{OC} + \vec{OD} + \vec{OE}$.

$10\vec{OG} = 2\vec{OD} + \vec{OC} + \vec{OE} - 5\vec{OG}$.

$\vec{OG} = \frac{1}{5} (-\vec{OB})$

2c) Show that $(OB)$ and $(DJ)$ are secant at $G$.

$\vec{OG} = -\frac{1}{5}\vec{OB}$, so $O, G, B$ are collinear. We also have $\vec{OJ} = \frac{1}{2} \vec{OI} = \frac{1}{4} (\vec{OB} - \vec{OC})$.

We also know that $\vec{OA} = \vec{OD}$ because $OA=OD$. Let $x = \frac{1}{5}$. Then $\vec{OG} = -x\vec{OB}$ or $\vec{OG} = x\vec{BO}$. Thus $\vec{OG}$ is on line $(OB)$.

$\vec{DJ} = \vec{OJ} - \vec{OD} = \frac{1}{4}(\vec{OB}-\vec{OC}) - \vec{OD}$.

Also, $5\vec{OG} = \vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}+\vec{OE}$ and $\vec{OA}+\vec{OB}+\vec{OC} = \vec{0}$, thus $5\vec{OG} = \vec{OD}+\vec{OE} = \vec{OD}+\vec{OC}$. Since $\vec{OC}=\vec{OE}$.

2d)

$4\vec{MM'} = \vec{MA} + \vec{MB} + \vec{MC} + \vec{MD} + \vec{ME}$

Let $M = G$, $4\vec{GG'} = \vec{GA} + \vec{GB} + \vec{GC} + \vec{GD} + \vec{GE} = \vec{0}$, thus $\vec{GG'} = \vec{0}$ and $G' = G$. Therefore $G$ is the center of the homothety.

$4\vec{GM'} = \vec{GA} + \vec{GB} + \vec{GC} + \vec{GD} + \vec{GE} - 5\vec{GM}$

$4\vec{GM'} = - 5\vec{MG}$, thus $4\vec{GM'} = 5\vec{GM}$, thus $\vec{GM'} = \frac{5}{4}\vec{GM}$.

Therefore $f$ is a homothety with center $G$ and ratio $5/4$.

$f(B)$: $4\vec{BB'} = \vec{BA} + \vec{BB} + \vec{BC} + \vec{BD} + \vec{BE} = \vec{BA} + \vec{BC} + \vec{BD} + \vec{BE}$

$f(D)$: $4\vec{DD'} = \vec{DA} + \vec{DB} + \vec{DC} + \vec{DD} + \vec{DE} = \vec{DA} + \vec{DB} + \vec{DC} + \vec{DE}$

Since $G$ is the center of the homothety, $f(B) = B'$, such that $\vec{GB'} = \frac{5}{4}\vec{GB}$. Similarly, for $f(D) = D'$, $\vec{GD'} = \frac{5}{4}\vec{GD}$.

3. Final Answer

2a) $\vec{OG} = -\frac{1}{5}\vec{OB}$

2b) $\vec{OG} = \frac{k}{10}\vec{OD} - \frac{2}{5}\vec{OJ}$, where $\vec{OA} = k\vec{OD}$.

3a) $f$ is a homothety with center $G$ and ratio $\frac{5}{4}$.

3b) $\vec{GB'} = \frac{5}{4}\vec{GB}$ and $\vec{GD'} = \frac{5}{4}\vec{GD}$