The problem consists of three separate geometry questions. Question 1: Given two parallel lines $(\Delta)$ and $(\Delta')$, a point $M$, $M_1 = S_{(\Delta)}(M)$ and $M' = S_{(\Delta')}(M_1)$, where $S$ denotes reflection across a line. Let $A$ be the orthogonal projection of $M$ onto $(\Delta)$, and $B$ be the orthogonal projection of $M_1$ onto $(\Delta')$. (a) Express the vector $\vec{MM'}$ in terms of $\vec{AB}$. (b) Deduce that $M' = t_{2\vec{AB}}(M)$, where $t$ denotes translation. Also deduce that $S_{(\Delta')} \circ S_{(\Delta)} = t_{2\vec{AB}}$. Question 2: Given two distinct points $I$ and $J$, a point $N$, $N_1 = S_I(N)$ and $N' = S_J(N_1)$, where $S$ denotes reflection across a point. (a) Express the vector $\vec{NN'}$ in terms of $\vec{IJ}$. (b) Deduce that $N' = t_{2\vec{IJ}}(N)$. Also deduce that $S_J \circ S_I = t_{2\vec{IJ}}$. Question 3: Given two non-null vectors $\vec{u}$ and $\vec{v}$, a point $T$, $T_1 = t_{\vec{v}}(T)$ and $T' = t_{\vec{u}}(T_1)$, where $t$ denotes translation by a vector. (a) Express the vector $\vec{TT'}$ in terms of $\vec{u}$ and $\vec{v}$. (b) Deduce that $T' = t_{\vec{u}+\vec{v}}(T)$. Also deduce that $t_{\vec{u}} \circ t_{\vec{v}} = t_{\vec{u}+\vec{v}}$.

1. Problem Description

The problem consists of three separate geometry questions.

Question 1: Given two parallel lines $(\Delta)$ and $(\Delta')$, a point $M$, $M_1 = S_{(\Delta)}(M)$ and $M' = S_{(\Delta')}(M_1)$, where $S$ denotes reflection across a line. Let $A$ be the orthogonal projection of $M$ onto $(\Delta)$, and $B$ be the orthogonal projection of $M_1$ onto $(\Delta')$.

(a) Express the vector $\vec{MM'}$ in terms of $\vec{AB}$.

(b) Deduce that $M' = t_{2\vec{AB}}(M)$, where $t$ denotes translation. Also deduce that $S_{(\Delta')} \circ S_{(\Delta)} = t_{2\vec{AB}}$.

Question 2: Given two distinct points $I$ and $J$, a point $N$, $N_1 = S_I(N)$ and $N' = S_J(N_1)$, where $S$ denotes reflection across a point.

(a) Express the vector $\vec{NN'}$ in terms of $\vec{IJ}$.

(b) Deduce that $N' = t_{2\vec{IJ}}(N)$. Also deduce that $S_J \circ S_I = t_{2\vec{IJ}}$.

Question 3: Given two non-null vectors $\vec{u}$ and $\vec{v}$, a point $T$, $T_1 = t_{\vec{v}}(T)$ and $T' = t_{\vec{u}}(T_1)$, where $t$ denotes translation by a vector.

(a) Express the vector $\vec{TT'}$ in terms of $\vec{u}$ and $\vec{v}$.

(b) Deduce that $T' = t_{\vec{u}+\vec{v}}(T)$. Also deduce that $t_{\vec{u}} \circ t_{\vec{v}} = t_{\vec{u}+\vec{v}}$.

2. Solution Steps

Question 1:

(a)

$\vec{MM_1} = 2\vec{MA}$

$\vec{M_1M'} = 2\vec{M_1B}$

$\vec{MM'} = \vec{MM_1} + \vec{M_1M'} = 2\vec{MA} + 2\vec{M_1B}$

Since $\Delta$ and $\Delta'$ are parallel, $\vec{AB} = \vec{AM_1}$. Therefore $\vec{AM_1} = \vec{AB}$.

$\vec{MM'} = 2(\vec{MA} + \vec{M_1B}) = 2(\vec{MA} + \vec{AM_1} + \vec{M_1B} - \vec{AM_1}) = 2(\vec{MM_1} + \vec{M_1B} - \vec{AM_1})= 2(\vec{MB} - \vec{AM_1})$.

Since MABM1 is a trapezoid, and A and B are projections of M and M

1. So the distance between the two lines is the same. Also MABM1 is rectangle.

So, $MA = M1B$. Then $\vec{MM'} = 2\vec{AB}$

(b)

Since $\vec{MM'} = 2\vec{AB}$, then $M' = t_{2\vec{AB}}(M)$.

$S_{(\Delta')} \circ S_{(\Delta)}(M) = S_{(\Delta')}(M_1) = M'$

Since $M' = t_{2\vec{AB}}(M)$, then $S_{(\Delta')} \circ S_{(\Delta)} = t_{2\vec{AB}}$.

Question 2:

(a)

$\vec{NN_1} = 2\vec{NI}$

$\vec{N_1N'} = 2\vec{N_1J}$

$\vec{NN'} = \vec{NN_1} + \vec{N_1N'} = 2\vec{NI} + 2\vec{N_1J} = 2\vec{NI} + 2(\vec{N_1I} + \vec{IJ}) = 2(\vec{NI} + \vec{N_1I}) + 2\vec{IJ} = 2(\vec{NI} - \vec{IN_1}) + 2\vec{IJ}$. Since N1 is the symmetric of N across I, then $\vec{IN} = - \vec{IN_1}$, meaning that $\vec{NI} = \vec{N_1I}$.

Therefore $\vec{NN'} = 2\vec{IJ}$.

(b)

Since $\vec{NN'} = 2\vec{IJ}$, then $N' = t_{2\vec{IJ}}(N)$.

$S_J \circ S_I(N) = S_J(N_1) = N'$

Since $N' = t_{2\vec{IJ}}(N)$, then $S_J \circ S_I = t_{2\vec{IJ}}$.

Question 3:

(a)

$\vec{TT_1} = \vec{v}$

$\vec{T_1T'} = \vec{u}$

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

(b)

Since $\vec{TT'} = \vec{u} + \vec{v}$, then $T' = t_{\vec{u}+\vec{v}}(T)$.

$t_{\vec{u}} \circ t_{\vec{v}}(T) = t_{\vec{u}}(T_1) = T'$

Since $T' = t_{\vec{u}+\vec{v}}(T)$, then $t_{\vec{u}} \circ t_{\vec{v}} = t_{\vec{u}+\vec{v}}$.

3. Final Answer

1)

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

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

2)

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

(b) $N' = t_{2\vec{IJ}}(N)$, $S_J \circ S_I = t_{2\vec{IJ}}$

3)

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

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