geg.: Punkte $A(1\mathrm{\mid }-2)A(1|-2)$, $C(-1,5\mathrm{\mid }3)C(-1\{,\}5|3)$, Pfeile $\overrightarrow{A{B}_n}(\phi )=\left(\begin{array}{c}4\cdot \sin \phi +3\\ {\displaystyle \frac{2}{\sin \phi }}\end{array}\right)\backslash overrightarrow\{AB\_n\}(\backslash varphi)=\backslash begin\{pmatrix\}\; 4\backslash cdot\; \backslash sin\backslash varphi+3\; \backslash \backslash \; \backslash dfrac\{2\}\{\backslash sin\backslash varphi\}\; \backslash end\{pmatrix\}$

ges.: Dreieck$A{B}_{1}CAB\_1C$ zum Viereck $A{B}_{1}C{D}_{1}AB\_1CD\_1$ ergänzen; Punkte $D_n(\phi )D\_n(\backslash varphi)$ berechnen

Ansatz und Rechnung:

1. Zeichnung im Koordinatensystem ergänzen:

Abbildung $B_{n}B\_n$ durch Punktspiegelung an $O(0\mathrm{\mid }0)O(0|0)$ auf $D_{n}D\_n$.

Ergänze das Dreieck $A{B}_{1}CAB\_1C$ zum Viereck $A{B}_{1}C{D}_{1}AB\_1CD\_1$.

Länge $O{B}_1=OB\_1\; =$Länge $O{D}_{1}OD\_1$

2. Rechnerischer Nachweis für $D_{n}D\_n$: Punkte $D_n(\phi )D\_n(\backslash varphi)$ berechnen

$\overrightarrow{O{B}_n}(\phi )=\overrightarrow{OA}\oplus \overrightarrow{A{B}_n}\backslash overrightarrow\{OB\_n\}(\backslash varphi)=\backslash overrightarrow\{OA\}\backslash oplus\backslash overrightarrow\{AB\_n\}$

$\overrightarrow{O{B}_n}=\left(\begin{array}{c}1\\ -2\end{array}\right)\oplus \left(\begin{array}{c}4\cdot \sin \phi +3\\ \frac{2}{\sin \phi }\end{array}\right)=\left(\begin{array}{c}4\cdot \sin \phi +4\\ -2+\frac{2}{\sin \phi }\end{array}\right)\phi \in ]0^{\circ };180^{\circ }[\backslash overrightarrow\{OB\_n\}=\backslash begin\{pmatrix\}\; 1\backslash \backslash -2\; \backslash end\{pmatrix\}\backslash oplus\; \backslash begin\{pmatrix\}\; 4\; \backslash cdot\; \backslash sin\backslash varphi\; +3\backslash \backslash \backslash frac\{2\}\{\backslash sin\backslash varphi\}\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\; 4\; \backslash cdot\; \backslash sin\backslash varphi\; +4\backslash \backslash -2+\backslash frac\{2\}\{\backslash sin\backslash varphi\}\; \backslash end\{pmatrix\}\backslash quad\; \backslash varphi\backslash in\; ]0^\backslash circ;180^\backslash circ[$

$\overrightarrow{O{D}_n}(\phi )\backslash overrightarrow\{OD\_n\}(\backslash varphi)$ ist dann punktgespiegelt an $OO$, also $-\overrightarrow{\mathbf{O}{\mathbf{B}}_{\mathbf{n}}}(\mathbf{\phi })\backslash color\; \{red\}\backslash mathbf\{-\backslash overrightarrow\{OB\_n\}(\backslash varphi)\}$

$\overrightarrow{O{D}_n}(\phi )=\left(\begin{array}{c}-(4\cdot \sin \phi +4)\\ -(-2+\frac{2}{\sin \phi })\end{array}\right)=\left(\begin{array}{c}-4\cdot \sin \phi -4)\\ 2-\frac{2}{\sin \phi })\end{array}\right)\backslash overrightarrow\{OD\_n\}(\backslash varphi)=\; \backslash begin\{pmatrix\}-(4\backslash cdot\; \backslash sin\backslash varphi\; +4)\backslash \backslash -(-2+\backslash frac\{2\}\{\backslash sin\backslash varphi\})\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-4\backslash cdot\; \backslash sin\backslash varphi\; -4)\backslash \backslash 2-\backslash frac\{2\}\{\backslash sin\backslash varphi\})\; \backslash end\{pmatrix\}$

$\Rightarrow {\mathbf{D}}_{\mathbf{n}}(-\mathbf{4}\cdot \sin \mathbf{\phi }-\mathbf{4}\mathbf{\mid }\mathbf{2}-\frac{\mathbf{2}}{\sin \mathbf{\phi }})\backslash mathbf\; \{\backslash Rightarrow\; D\_n(-4\backslash cdot\; \backslash sin\backslash varphi-4\backslash :\; |\backslash :\; 2-\backslash frac\{2\}\{\backslash sin\backslash varphi\})\}$

Check:

$\overrightarrow{O{B}_1}=\left(\begin{array}{c}4\cdot \sin (30\mathrm{°})+4\\ -2+\frac{2}{\sin (30\mathrm{°})}\end{array}\right)=\left(\begin{array}{c}4\cdot 0,5+4\\ -2+\frac{2}{0,5}\end{array}\right)=\left(\begin{array}{c}4\cdot 0,5+4\\ -2+\frac{2}{0,5}\end{array}\right)=\left(\begin{array}{c}6\\ 2\end{array}\right)\backslash overrightarrow\{OB\_1\}=\backslash begin\{pmatrix\}\; 4\; \backslash cdot\; \backslash sin(30°)\; +4\backslash \backslash -2+\backslash frac\{2\}\{\backslash sin(30°)\}\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\; 4\; \backslash cdot\; 0\{,\}5\; +4\backslash \backslash -2+\backslash frac\{2\}\{0\{,\}5\}\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\; 4\; \backslash cdot\; 0\{,\}5\; +4\backslash \backslash -2+\backslash frac\{2\}\{0\{,\}5\}\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\; 6\backslash \backslash 2\; \backslash end\{pmatrix\}$

$\overrightarrow{O{D}_n}(\phi )=\left(\begin{array}{c}-4\cdot \sin (30\mathrm{°})-4)\\ 2-\frac{2}{\sin (30\mathrm{°}})\end{array}\right)=\left(\begin{array}{c}-4\cdot 0,5-4)\\ 2-\frac{2}{0,5})\end{array}\right)=\left(\begin{array}{c}-6\\ -2\end{array}\right)\backslash overrightarrow\{OD\_n\}(\backslash varphi)\; =\backslash begin\{pmatrix\}-4\backslash cdot\; \backslash sin(30°)-4)\backslash \backslash 2-\backslash frac\{2\}\{\backslash sin(30°\})\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}-4\backslash cdot\; 0\{,\}5\; -4)\backslash \backslash 2-\backslash frac\{2\}\{0\{,\}5\})\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\; -6\backslash \backslash -2\; \backslash end\{pmatrix\}$

siehe Zeichnung!

Im Parallelogramm $A{B}_{n}C{D}_{n}AB\_nCD\_n$ ist der Punkt $O(0\mathrm{\mid }0)O(0|0)$ Mittelpunkt von $[B_n{D}_n][B\_nD\_n]$.

Damit das auch der Mittelpunkt von $[AC][AC]$ ist müssen $CC$ und $AA$ symmetrisch zum Ursprung liegen, also muss $CC$ die Koordinaten $(-1\mathrm{\mid }2)(-1|2)$ haben. Hier ist aber $C(-1,5\mathrm{\mid }3)C(-1\{,\}5|3)$. Daher ist unter den Vierecken $A{B}_{n}C{D}_{n}AB\_nCD\_n$ kein Parallelogramm!