We are given a parallelogram $ABCD$ such that $AB = BD$. Let $f$ be an affine transformation such that $f(A) = B$, $f(B) = D$, $f(D) = C$. Let $t$ be the translation by vector $\overrightarrow{AB}$. 1. Determine the nature of the transformation $g$ defined by $g = t^{-1} \circ f$.

1. Problem Description

We are given a parallelogram $ABCD$ such that $AB = BD$. Let $f$ be an affine transformation such that $f(A) = B$, $f(B) = D$, $f(D) = C$. Let $t$ be the translation by vector $\overrightarrow{AB}$.

1. Determine the nature of the transformation $g$ defined by $g = t^{-1} \circ f$.

2. Construct the image of any point $M$ under the transformation $f$.

3. Show that $f$ has no invariant point.

2. Solution Steps

1. Nature of $g = t^{-1} \circ f$

$g(A) = t^{-1}(f(A)) = t^{-1}(B)$. Since $t$ is the translation by $\overrightarrow{AB}$, $t^{-1}$ is the translation by $\overrightarrow{BA}$. Therefore $t^{-1}(B) = A$. Thus $g(A) = A$.

$g(B) = t^{-1}(f(B)) = t^{-1}(D)$. Applying the translation $t^{-1}$ by vector $\overrightarrow{BA}$ to $D$, we get $g(B) = E$, where $\overrightarrow{BE} = \overrightarrow{AD}$. Since ABCD is a parallelogram, $\overrightarrow{AD} = \overrightarrow{BC}$. Since $AB=BD$, $ABCD$ is not a rhombus.

Since $AB = BD$, the parallelogram $ABDC$ is such that $AB=BD$.

$g(D) = t^{-1}(f(D)) = t^{-1}(C)$. Applying the translation $t^{-1}$ by vector $\overrightarrow{BA}$ to $C$, we get $g(D) = F$, where $\overrightarrow{BF} = \overrightarrow{AC}$. Since $ABCD$ is a parallelogram, $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AB} + \overrightarrow{AD}$. Also, $\overrightarrow{DF} = \overrightarrow{BF} - \overrightarrow{BD} = \overrightarrow{AC} - \overrightarrow{AB} = \overrightarrow{AD}$, so $DF=BC=AD$.

Since $g(A) = A$, $g$ is an affine transformation that fixes $A$.

We have $g(A) = A$, $g(B) = t^{-1}(D) = E$ such that $\overrightarrow{BE} = \overrightarrow{AD} = \overrightarrow{BC}$. Then $\overrightarrow{AE} = \overrightarrow{AB} + \overrightarrow{BE} = \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$. Thus $E = C$.

Therefore, $g(B) = C$.

$g(D) = t^{-1}(C)$. So $\overrightarrow{Ag(D)} = \overrightarrow{AD}$. We also have $\overrightarrow{AC}=\overrightarrow{AB} + \overrightarrow{AD}$

Then $g(D) = t^{-1}(C) = F$, such that $\overrightarrow{AF} = \overrightarrow{AD}$. So $F=D$. $g(D) = D$.

Therefore $g(A) = A$ and $g(D) = D$.

Since $g(A)=A$, $g(B)=C$, $g(D)=t^{-1}(C)$, if $ABCD$ is a parallelogram, $g(A)=A$, $g(B)=C$

then $g$ is a shear.

Since $g(A)=A$, $g(B)=C$, the line $AD$ is invariant.

$g$ is a shear with invariant line $(AD)$ and direction $\overrightarrow{AC}$.

2. Construct the image of any point $M$ under the transformation $f$.

To find the image of a point $M$ under $f$, consider $M$ as a linear combination of $A, B, D$. That is, $M = \alpha A + \beta B + \gamma D$ where $\alpha+\beta+\gamma = 1$.

Then $f(M) = \alpha f(A) + \beta f(B) + \gamma f(D) = \alpha B + \beta D + \gamma C$.

To construct $f(M)$, we can rewrite this as $f(M) = B + \beta \overrightarrow{BD} + \gamma \overrightarrow{BC}$.

Since ABCD is a parallelogram, $\overrightarrow{BC} = \overrightarrow{AD}$. So, $f(M) = B + \beta \overrightarrow{BD} + \gamma \overrightarrow{AD}$.

We can also write $f(M) = C + \alpha \overrightarrow{CB} + \beta \overrightarrow{CD}$ since $f(M) = \alpha B + \beta D + \gamma C = \alpha B + \beta D + (1-\alpha-\beta) C = \alpha (B-C) + \beta (D-C) + C = C - \alpha \overrightarrow{CB} - \beta \overrightarrow{CD}$

3. Show that $f$ has no invariant point.

Suppose $P$ is an invariant point, i.e., $f(P) = P$. Then $P = \alpha A + \beta B + \gamma D$ with $\alpha + \beta + \gamma = 1$. Applying $f$, we get

$f(P) = \alpha f(A) + \beta f(B) + \gamma f(D) = \alpha B + \beta D + \gamma C$.

If $f(P) = P$, then $\alpha A + \beta B + \gamma D = \alpha B + \beta D + \gamma C$.

$\alpha A + (\beta - \alpha) B + (\gamma - \beta) D - \gamma C = 0$

Since $ABCD$ is a parallelogram, $A-B-C+D = \vec{0}$. So we rewrite $C=A-B+D$,

Then $\alpha A + (\beta-\alpha) B + (\gamma-\beta)D - \gamma (A-B+D) = 0$

$(\alpha - \gamma)A + (\beta - \alpha + \gamma)B + (\gamma - \beta - \gamma)D = 0$

$(\alpha - \gamma)A + (\beta - \alpha + \gamma)B + (-\beta)D = 0$.

This means $\alpha - \gamma = 0$, $\beta - \alpha + \gamma = 0$, $-\beta = 0$.

Thus $\beta = 0$, $\alpha = \gamma$, so $\alpha + \beta + \gamma = 2\alpha = 1$, $\alpha = \gamma = 1/2$.

Then $P = \frac{1}{2}A + \frac{1}{2}D$, so $P$ is the midpoint of $AD$.

$f(P) = \frac{1}{2}f(A) + \frac{1}{2}f(D) = \frac{1}{2}B + \frac{1}{2}C$. So $f(P)$ is the midpoint of $BC$.

Since $ABCD$ is a parallelogram, the midpoint of $AD$ is not the same as the midpoint of $BC$.

Thus $f(P) \neq P$.

Hence $f$ has no invariant point.

3. Final Answer

1. The transformation $g$ is a shear with invariant line $(AD)$.

2. $f(M) = B + \beta \overrightarrow{BD} + \gamma \overrightarrow{AD}$, where $M = \alpha A + \beta B + \gamma D$ and $\alpha+\beta+\gamma = 1$.

3. $f$ has no invariant point.