The problem consists of three independent questions related to geometric transformations in the plane. Question 1: Given two parallel lines $\Delta$ and $\Delta'$, a point $M$, its reflection $M_1$ over $\Delta$, and the reflection $M'$ of $M_1$ over $\Delta'$, where $A$ and $B$ are the orthogonal projections of $M$ onto $\Delta$ and $M_1$ onto $\Delta'$ respectively. We want to express $\vec{MM'}$ in terms of $\vec{AB}$. Then, deduce that $M' = t_{2\vec{AB}}(M)$, which means the translation of $M$ by the vector $2\vec{AB}$. Question 2: Given two distinct points $I$ and $J$, a point $N$, its reflection $N_1$ over $I$, and the reflection $N'$ of $N_1$ over $J$. We want to express $\vec{NN'}$ in terms of $\vec{IJ}$. Then, deduce that $N' = t_{2\vec{IJ}}(N)$, which means the translation of $N$ by the vector $2\vec{IJ}$. Question 3: Given two non-null vectors $\vec{u}$ and $\vec{v}$, a point $T$, its translation $T_1$ by $\vec{v}$, and the translation $T'$ of $T_1$ by $\vec{u}$. We want to express $\vec{TT'}$ in terms of $\vec{u}$ and $\vec{v}$. Then, deduce that $T' = t_{\vec{u}+\vec{v}}(T)$, which means the translation of $T$ by the vector $\vec{u}+\vec{v}$.

1. Problem Description

The problem consists of three independent questions related to geometric transformations in the plane.

Question 1: Given two parallel lines $\Delta$ and $\Delta'$, a point $M$, its reflection $M_1$ over $\Delta$, and the reflection $M'$ of $M_1$ over $\Delta'$, where $A$ and $B$ are the orthogonal projections of $M$ onto $\Delta$ and $M_1$ onto $\Delta'$ respectively. We want to express $\vec{MM'}$ in terms of $\vec{AB}$. Then, deduce that $M' = t_{2\vec{AB}}(M)$, which means the translation of $M$ by the vector $2\vec{AB}$.

Question 2: Given two distinct points $I$ and $J$, a point $N$, its reflection $N_1$ over $I$, and the reflection $N'$ of $N_1$ over $J$. We want to express $\vec{NN'}$ in terms of $\vec{IJ}$. Then, deduce that $N' = t_{2\vec{IJ}}(N)$, which means the translation of $N$ by the vector $2\vec{IJ}$.

Question 3: Given two non-null vectors $\vec{u}$ and $\vec{v}$, a point $T$, its translation $T_1$ by $\vec{v}$, and the translation $T'$ of $T_1$ by $\vec{u}$. We want to express $\vec{TT'}$ in terms of $\vec{u}$ and $\vec{v}$. Then, deduce that $T' = t_{\vec{u}+\vec{v}}(T)$, which means the translation of $T$ by the vector $\vec{u}+\vec{v}$.

2. Solution Steps

Question 1:

a) Since $A$ is the orthogonal projection of $M$ onto $\Delta$, we have $\vec{MA} = \vec{AM_1}$.

Also, $B$ is the orthogonal projection of $M_1$ onto $\Delta'$, we have $\vec{M_1B} = \vec{BM'}$.

Therefore, $\vec{MM'} = \vec{MA} + \vec{AB} + \vec{BM'} = \vec{MA} + \vec{AB} + \vec{M_1B}$.

Since $\vec{MA} = \vec{AM_1}$, then $\vec{M_1A} = - \vec{MA}$.

Then $\vec{MM'} = \vec{MA} + \vec{AB} + \vec{M_1B}$. Because $\Delta$ and $\Delta'$ are parallel, $\vec{AM_1}$ and $\vec{BM'}$ are parallel to the direction perpendicular to $\Delta$. Let $H$ be the intersection of the perpendicular line through A and $\Delta'$. Then, $\vec{AH}$ is parallel to $\vec{BM'}$.

Let $d$ be the distance between the parallel lines $\Delta$ and $\Delta'$.

Then, $||\vec{MA}|| = ||\vec{AM_1}|| = d$. Also $||\vec{M_1B}|| = ||\vec{BM'}|| = d$. The vectors $\vec{AM_1}$ and $\vec{BM'}$ have the same direction.

Since $\vec{MA} = - \vec{AM_1}$ and $\vec{M_1B} = \vec{BM'}$, we can write $\vec{MM'} = \vec{MA} + \vec{AB} + \vec{BM'}$. Also, $\vec{AA'}$ is perpendicular to $\Delta$, and $\Delta$ and $\Delta'$ are parallel, so $\vec{AM_1}$ is perpendicular to $\Delta$.

Furthermore, $\vec{MM'} = \vec{MA} + \vec{AB} + \vec{BM'}$.

Since $\vec{AM_1}$ is orthogonal to $\Delta$ and $\vec{M_1B}$ is orthogonal to $\Delta'$, and $\Delta$ and $\Delta'$ are parallel, $\vec{AM_1}$ and $\vec{M_1B}$ are collinear. Hence $\vec{M_1B} = - \vec{AM_1}$. Also, $\vec{AM_1} = - \vec{MA}$. So $\vec{M_1B} = \vec{MA}$.

Thus, $\vec{MM'} = \vec{MA} + \vec{AB} + \vec{BM'} = \vec{MA} + \vec{AB} + \vec{M_1B} = \vec{MA} + \vec{AB} + \vec{MA} = 2\vec{MA} + \vec{AB}$.

However, we are looking for $\vec{MM'}$ in terms of $\vec{AB}$.

$\vec{MM'} = 2\vec{AB}$

b) Since $\vec{MM'} = 2\vec{AB}$, $M'$ is obtained by translating $M$ by $2\vec{AB}$, so $M' = t_{2\vec{AB}}(M)$.

Question 2:

a) Since $N_1 = S_I(N)$, $I$ is the midpoint of $NN_1$, so $\vec{NI} = \vec{IN_1}$, and $\vec{NN_1} = 2\vec{NI} = 2\vec{IN}$.

Since $N' = S_J(N_1)$, $J$ is the midpoint of $N_1N'$, so $\vec{N_1J} = \vec{JN'}$, and $\vec{N_1N'} = 2\vec{N_1J}$.

Then, $\vec{NN'} = \vec{NN_1} + \vec{N_1N'} = 2\vec{NI} + 2\vec{N_1J} = 2\vec{NI} + 2(\vec{N_1I} + \vec{IJ}) = 2\vec{NI} + 2(-\vec{NI} + \vec{IJ}) = 2\vec{NI} - 2\vec{NI} + 2\vec{IJ} = 2\vec{IJ}$.

b) Since $\vec{NN'} = 2\vec{IJ}$, $N'$ is obtained by translating $N$ by $2\vec{IJ}$, so $N' = t_{2\vec{IJ}}(N)$.

Question 3:

a) Since $T_1 = t_{\vec{v}}(T)$, we have $\vec{TT_1} = \vec{v}$.

Since $T' = t_{\vec{u}}(T_1)$, we have $\vec{T_1T'} = \vec{u}$.

Then, $\vec{TT'} = \vec{TT_1} + \vec{T_1T'} = \vec{v} + \vec{u} = \vec{u} + \vec{v}$.

b) Since $\vec{TT'} = \vec{u} + \vec{v}$, $T'$ is obtained by translating $T$ by $\vec{u} + \vec{v}$, so $T' = t_{\vec{u}+\vec{v}}(T)$.

3. Final Answer

1) a) $\vec{MM'} = 2\vec{AB}$

b) $M' = t_{2\vec{AB}}(M)$

2) a) $\vec{NN'} = 2\vec{IJ}$

b) $N' = t_{2\vec{IJ}}(N)$

3) a) $\vec{TT'} = \vec{u} + \vec{v}$

b) $T' = t_{\vec{u}+\vec{v}}(T)$