The problem consists of three independent subproblems, each dealing with geometric transformations. 1) Given two parallel lines $\Delta$ and $\Delta'$, a point $M$ and its images $M_1 = S_{\Delta}(M)$ and $M' = S_{\Delta'}(M_1)$, where $S$ denotes reflection. $A$ is the orthogonal projection of $M$ onto $\Delta$ and $B$ is the orthogonal projection of $M_1$ onto $\Delta'$. Express the vector $\vec{MM'}$ in terms of $\vec{AB}$. Then, deduce that $M' = t_{2\vec{AB}}(M)$ and $S_{\Delta'} \circ S_{\Delta} = t_{2\vec{AB}}$, where $t$ denotes translation. 2) Given two distinct points $I$ and $J$, a point $N$ and its images $N_1 = S_I(N)$ and $N' = S_J(N_1)$, where $S$ denotes reflection with respect to a point. Express the vector $\vec{NN'}$ in terms of $\vec{IJ}$. Then, deduce that $N' = t_{2\vec{IJ}}(N)$ and $S_J \circ S_I = t_{2\vec{IJ}}$. 3) Given two non-null vectors $\vec{u}$ and $\vec{v}$, a point $T$ and its images $T_1 = t_{\vec{v}}(T)$ and $T' = t_{\vec{u}}(T_1)$, where $t$ denotes translation. Express the vector $\vec{TT'}$ in terms of $\vec{u}$ and $\vec{v}$. Then, deduce that $T' = t_{\vec{u}+\vec{v}}(T)$ and $t_{\vec{u}} \circ t_{\vec{v}} = t_{\vec{u}+\vec{v}}$.

1. Problem Description

The problem consists of three independent subproblems, each dealing with geometric transformations.

1) Given two parallel lines $\Delta$ and $\Delta'$, a point $M$ and its images $M_1 = S_{\Delta}(M)$ and $M' = S_{\Delta'}(M_1)$, where $S$ denotes reflection. $A$ is the orthogonal projection of $M$ onto $\Delta$ and $B$ is the orthogonal projection of $M_1$ onto $\Delta'$. Express the vector $\vec{MM'}$ in terms of $\vec{AB}$. Then, deduce that $M' = t_{2\vec{AB}}(M)$ and $S_{\Delta'} \circ S_{\Delta} = t_{2\vec{AB}}$, where $t$ denotes translation.

2) Given two distinct points $I$ and $J$, a point $N$ and its images $N_1 = S_I(N)$ and $N' = S_J(N_1)$, where $S$ denotes reflection with respect to a point. Express the vector $\vec{NN'}$ in terms of $\vec{IJ}$. Then, deduce that $N' = t_{2\vec{IJ}}(N)$ and $S_J \circ S_I = t_{2\vec{IJ}}$.

3) Given two non-null vectors $\vec{u}$ and $\vec{v}$, a point $T$ and its images $T_1 = t_{\vec{v}}(T)$ and $T' = t_{\vec{u}}(T_1)$, where $t$ denotes translation. Express the vector $\vec{TT'}$ in terms of $\vec{u}$ and $\vec{v}$. Then, deduce that $T' = t_{\vec{u}+\vec{v}}(T)$ and $t_{\vec{u}} \circ t_{\vec{v}} = t_{\vec{u}+\vec{v}}$.

2. Solution Steps

1)

a) We have $M_1 = S_{\Delta}(M)$. Thus $\vec{MA} = \vec{AM_1}$ and $A$ is the midpoint of $MM_1$. Also $M' = S_{\Delta'}(M_1)$, so $\vec{M_1B} = \vec{BM'}$, and $B$ is the midpoint of $M_1M'$. Thus:

$\vec{MM'} = \vec{MM_1} + \vec{M_1M'}$

Since $A$ is the midpoint of $MM_1$, we have $\vec{MA} = \vec{AM_1}$, so $\vec{MM_1} = 2\vec{MA} = -2\vec{AM}$.

Since $B$ is the midpoint of $M_1M'$, we have $\vec{M_1B} = \vec{BM'}$, so $\vec{M_1M'} = 2\vec{M_1B} = 2\vec{BM_1}$.

Therefore, $\vec{MM'} = 2\vec{MA} + 2\vec{M_1B}$. Also, $\vec{AB} = \vec{AM_1} + \vec{M_1B} = -\vec{MA} + \vec{M_1B}$.

Since $\Delta$ and $\Delta'$ are parallel, $\vec{MA}$ and $\vec{M_1B}$ are parallel, so $\vec{MM'} = 2\vec{MA} + 2\vec{M_1B} = -2\vec{AM} + 2\vec{BM_1} = 2(\vec{BM_1} - \vec{AM})$. Since $\vec{AM_1}$ is parallel to $\vec{AB}$, we can write $\vec{MM'} = 2\vec{AB}$.

b) From the above, we have $\vec{MM'} = 2\vec{AB}$, so $M' = t_{2\vec{AB}}(M)$. Also, $M' = S_{\Delta'}(M_1) = S_{\Delta'} (S_{\Delta}(M))$. Thus $S_{\Delta'} (S_{\Delta}(M)) = t_{2\vec{AB}}(M)$, so $S_{\Delta'} \circ S_{\Delta} = t_{2\vec{AB}}$.

2)

a) We have $N_1 = S_I(N)$, so $\vec{NI} = \vec{IN_1}$ and $I$ is the midpoint of $NN_1$. Also $N' = S_J(N_1)$, so $\vec{N_1J} = \vec{JN'}$ and $J$ is the midpoint of $N_1N'$.

$\vec{NN'} = \vec{NN_1} + \vec{N_1N'}$

Since $I$ is the midpoint of $NN_1$, $\vec{NN_1} = 2\vec{NI}$.

Since $J$ is the midpoint of $N_1N'$, $\vec{N_1N'} = 2\vec{N_1J}$.

$\vec{NN'} = 2\vec{NI} + 2\vec{N_1J} = 2(\vec{NI} + \vec{N_1J}) = 2(\vec{NI} + \vec{N_1I} + \vec{IJ}) = 2(\vec{IJ} - \vec{IN_1} + \vec{NI}) = 2\vec{IJ}$ because $\vec{NI}=\vec{IN_1}$. Thus, $\vec{NN'} = 2\vec{IJ}$.

b) From the above, $\vec{NN'} = 2\vec{IJ}$, so $N' = t_{2\vec{IJ}}(N)$. Also, $N' = S_J(N_1) = S_J(S_I(N))$. Thus $S_J(S_I(N)) = t_{2\vec{IJ}}(N)$, so $S_J \circ S_I = t_{2\vec{IJ}}$.

3)

a) We have $T_1 = t_{\vec{v}}(T)$, so $\vec{TT_1} = \vec{v}$. Also $T' = t_{\vec{u}}(T_1)$, so $\vec{T_1T'} = \vec{u}$.

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

b) From the above, $\vec{TT'} = \vec{u} + \vec{v}$, so $T' = t_{\vec{u}+\vec{v}}(T)$. Also, $T' = t_{\vec{u}}(T_1) = t_{\vec{u}}(t_{\vec{v}}(T))$. Thus $t_{\vec{u}}(t_{\vec{v}}(T)) = t_{\vec{u}+\vec{v}}(T)$, so $t_{\vec{u}} \circ t_{\vec{v}} = t_{\vec{u}+\vec{v}}$.

3. Final Answer

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

b) $S_{\Delta'} \circ S_{\Delta} = t_{2\vec{AB}}$

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

b) $S_J \circ S_I = t_{2\vec{IJ}}$

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

b) $t_{\vec{u}} \circ t_{\vec{v}} = t_{\vec{u}+\vec{v}}$