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Änderungsvorschläge von Jérôme Urhausen (Email vom 15.02.2014) umgesetzt.
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@ -341,10 +341,9 @@ schneiden sich.
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&= d(P, R)\\
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&= d(\varphi_2(P), \varphi_2(R))\\
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&= d(P', \varphi_2(R))\\
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&= d(Q', \varphi_2(R))
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\end{aligned}$\\
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und analog $d(Q', \varphi_1(R)) = d(Q', \varphi_2(R))$
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\item Seien $P$, $Q$ und $R$ Fixpunkte von $\varphi$, $R \in PG$
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\item Seien $P$, $Q$ und $R$ Fixpunkte von $\varphi$, $R \notin PQ$
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und $A \notin \overline{PQ} \cup \overline{PR} \cup \overline{QR}$.
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Sei $B \in \overline{PQ} \setminus \Set{P, Q}$. Dann ist
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$\varphi(B) = B$ wegen \cref{kor:beh2'}.
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@ -415,8 +414,8 @@ schneiden sich.
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\begin{beweis}
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Zeige $\angle PRQ < \angle RQP'$.
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Sei $M$ der Mittelpunkt der Strecke $\overline{QR}$. Sei
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$A \in MP^-$ mit $d(P,M) = d(M,A)$.
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Sei $M$ der Mittelpunkt der Strecke $\overline{QR}$ und $P' \in PQ^+ \setminus \overline{PQ}$.
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Sei $A \in MP^-$ mit $d(P,M) = d(M,A)$.
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\begin{figure}[ht]
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@ -460,19 +459,17 @@ schneiden sich.
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\label{fig:geometry-6}
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\end{figure}
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\begin{beweis}[von \cref{prop:14.7}]
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Seien $P, Q, R \in X$ mit $P, Q \in f \in G$.
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Sei $\varphi$ die Isometrie, die $Q$ auf $P$ und $P$ auf $P' \in f$
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\begin{beweis}
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Seien $P, Q \in f \in G$ und $\varphi$ die Isometrie, die $Q$ auf $P$ und $P$ auf $P' \in f$
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mit $d(P,P') = d(P, Q)$ abbildet und die Halbebenen bzgl. $f$ erhält.
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\underline{Annahme:} $\varphi(g) \cap g \neq \emptyset$\\
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$\Rightarrow$ Es gibt einen Schnittpunkt $\Set{R} = \varphi(g) \cap g$.\\
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Dann ist $\angle QPR < \angle RQP^-$ nach
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\cref{bem:14.9} und $\angle QPR = \angle RQP^-$, weil
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Dann ist $\angle QPR < \angle RQP'$ nach
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\cref{bem:14.9} und $\angle QPR = \angle RQP'$, weil
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$\varphi(\angle RQP') = \angle RPQ$.\\
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$\Rightarrow$ Widerspruch\\
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$\Rightarrow \varphi(g) \cap g = \emptyset$
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$\Rightarrow \varphi(g) \cap g = \emptyset \qed$
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\end{beweis}
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\begin{folgerung}\label{folgerung:14.10}%In Vorlesung: Folgerung 14.10
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@ -533,9 +530,19 @@ Sei im Folgenden \enquote{$\IWS$} die \enquote{Innenwinkelsumme}.
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\cref{folgerung:14.10}.
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\end{behauptung}
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\begin{beweis}[der Behauptung]
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Sei $M$ der Mittelpunkt $\overline{RC}$ und $A' \in MA^-$ mit
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$d(A', M) = d(A, M) \Rightarrow \triangle(MA'C)$ und
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\begin{beweis}
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Es seien $A, B, C \in X$ und $\triangle $ das Dreieck mit den
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Eckpunkten $A, B, C$ und $\alpha$ sei der Innenwinkel bei $A$,
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$\beta$ der Innenwinkel bei $B$ und $\gamma$ der Innenwinkel bei $C$.
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Sei $M$ der Mittelpunkt der Strecke $\overline{BC}$. Sei außerdem
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$\alpha_1 = \angle CAM$ und $\alpha_2 = \angle BAM$.
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Sei weiter $A' \in MA^-$ mit $d(A', M) = d(A, M)$.
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Die Situation ist in \cref{fig:prop14.11.2} skizziert.
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$ \Rightarrow \triangle(MA'C)$ und
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$\triangle(MAB)$ sind kongruent.
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$\Rightarrow \angle ABM = \angle A'CM$ und $\angle MA'C = \angle MAB$.
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$\Rightarrow \alpha + \beta + \gamma =\IWS(\triangle ABC) = \IWS(\triangle AA'C)$
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@ -888,11 +895,11 @@ $\xRightarrow{\text{Strahlensatz}} \frac{a}{h_c} = \frac{c}{h_a} \rightarrow a \
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\item Ansatz: $\sigma = \begin{pmatrix}a & b\\c & d\end{pmatrix}$
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$\sigma(x_0) = \frac{ax_0 + b}{c x_0 + d} \overset{!}{=} 0$
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$\Rightarrow a x_0 + b = 0 \Rightarrow b = -a x_0$\\
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$\sigma(x_\infty) = \infty \Rightarrow c x_\infty + d = 0 \Rightarrow d = - x_\infty$\\
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$\sigma(x_\infty) = \infty \Rightarrow c x_\infty + d = 0 \Rightarrow d = - c x_\infty$\\
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$\sigma(x_1) = 1 \Rightarrow a x_1 + b = c x_1 + d$\\
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$a (x_1 - x_0) = c (x_1 - x_\infty) \Rightarrow c = a \frac{x_1 - x_0}{x_1 - x_\infty}$\\
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$\Rightarrow - a^2 \cdot x_\infty \frac{x_1 - x_0}{x_1 - x_\infty} + a^2 x_0 \frac{x_1 - x_0}{x_1 - x_\infty} = 1$\\
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$\Rightarrow a^2 \frac{x_1 - x_0}{x_1 - x_\infty} (x_0 - x_\infty) = 1$
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$\Rightarrow a^2 \frac{x_1 - x_0}{x_0 - x_\infty} (x_0 - x_\infty) = 1$
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$\Rightarrow a^2 = \frac{x_1 - x_\infty}{(x_1 - x_\infty) (x_1 - x_0)}$
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\item TODO d)
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\item Es genügt die Aussage für Matrizen aus \cref{prop:15.2d}
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@ -1001,7 +1008,8 @@ $\xRightarrow{\text{Strahlensatz}} \frac{a}{h_c} = \frac{c}{h_a} \rightarrow a \
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\item $\DV(0, 1, \infty, z_4) = \frac{(0- z_4) \cdot (\infty - 1)}{(0 -1) \cdot (\infty - z_4)} = \frac{z_4 \cdot (\infty - 1)}{\infty - z_4} = z_4$
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\item TODO
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\item Sei $\sigma \in \PSL_2(\mdc)$ mit $\sigma(z_1) = 0$, $\sigma(z_2) = 1$,
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$\sigma(z_3) = \infty$ (gibt es?)
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$\sigma(z_3) = \infty$. Ein solches $\sigma$ existiert, da man drei
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Parameter von $\sigma$ wählen darf.
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$\overset{\mathclap{\crefabbr{bem:15.4d}}}{\Rightarrow}\hspace{4mm} \DV(z_1, \dots, z_4) = \DV(0, 1, \infty, \sigma(z_4))$\\
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$\Rightarrow\hspace{4mm} \DV(z_1, \dots, z_4) \in \mdr \cup \Set{\infty}$\\
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@ -1052,18 +1060,18 @@ $\xRightarrow{\text{Strahlensatz}} \frac{a}{h_c} = \frac{c}{h_a} \rightarrow a \
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also gilt \obda $z_1 = \iu a$ und $z_2 = \iu b$ mit $a,b \in \mdr$ und $a < b$.
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\begin{align*}
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2d(\iu a, \iu b)&= \ln \mid \DV(0, \iu a, \infty, \iu b) \mid \\
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&= \ln \mid \frac{(0 - \iu b) (\infty - \iu a)}{(0 - \iu a)(\infty - \iu b)} \mid \\
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&= \ln \mid \frac{b}{a} \mid\\
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2d(\iu a, \iu b)&= \mid \ln \DV(0, \iu a, \infty, \iu b) \mid \\
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&= \mid \ln \frac{(0 - \iu b) (\infty - \iu a)}{(0 - \iu a)(\infty - \iu b)} \mid \\
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&= \mid \ln \frac{b}{a} \mid\\
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&= \ln b - \ln a
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\end{align*}
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Also: $d(z_1, z_2) \geq 0$, $d(z_1, z_2) = 0 \gdw z_1 = z_2$
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\begin{align*}
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2 d(z_2, z_1) &= \ln \DV(a_2, z_2, a_1, z_1)\\
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&= \ln \DV(\infty, \iu b, 0, \iu a)\\
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&\overset{\mathclap{\crefabbr{bem:15.4b.ii}}}{=}\hspace{5mm} \ln \DV(0, \iu b, \infty, \iu a)\\
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2 d(z_2, z_1) &= \mid \ln \DV(a_2, z_2, a_1, z_1) \mid\\
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&= \mid \ln \DV(\infty, \iu b, 0, \iu a) \mid\\
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&\overset{\mathclap{\crefabbr{bem:15.4b.ii}}}{=}\hspace{5mm} \mid \ln \DV(0, \iu b, \infty, \iu a) \mid \\
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&= 2 d(z_1, z_2)
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\end{align*}
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@ -1,15 +1,28 @@
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\begin{tikzpicture}
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\tkzSetUpPoint[shape=circle,size=10,color=black,fill=black]
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\tkzSetUpLine[line width=1]
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\tkzDefPoints{0/0/A, 4/0/B, 2/2/C, 6/2/D}
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\tkzDefPoints{0/0/A, 4/0/B, 2/2/C, 6/2/A'}
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\tkzMarkAngle[arc=l,size=0.8cm,color=green,fill=green!20](B,A,C)
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\path[draw] ++(25:.3) node[rotate=0] {$\alpha$};
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\node at (1,1.5) {$\beta$};
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\tkzDrawSegments(A,B A,C A,D B,C C,D)
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\tkzDrawPoints(A,B,C,D)
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\tkzMarkAngle[arc=l,size=1.15cm,color=green,fill=green!20,opacity=.8](A',A,C)
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\tkzLabelAngle[pos=0.9](A',A,C){$\alpha_1$}
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\tkzMarkAngle[arc=lll,size=1.15cm,color=green,fill=green!20,opacity=.8](B,A,A')
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\tkzLabelAngle[pos=0.9](B,A,A'){$\alpha_2$}
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\tkzMarkAngle[arc=l,size=0.7cm,color=red,fill=red!20,opacity=.8](C,B,A)
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\tkzLabelAngle[pos=0.5](C,B,A){$\beta$}
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\tkzMarkAngle[arc=l,size=0.6cm,color=blue,fill=blue!20,opacity=.8](A,C,B)
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\tkzLabelAngle[pos=0.4](A,C,B){$\gamma$}
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\tkzDrawSegments(A,B A,C A,A' B,C C,A')
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\tkzLabelPoint[below left](A){$A$}
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\tkzLabelPoint[below right](B){$B$}
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\tkzLabelPoint[above left](C){$C$}
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\tkzLabelPoint[above right](D){$D$}
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\end{tikzpicture}
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\tkzLabelPoint[above right](A'){$A'$}
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\tkzMarkAngle[arc=l,size=0.6cm,color=green,fill=green!20,opacity=.8](B,A,C)
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\path[draw] ++(25:.35) node[rotate=0] {$\alpha$};
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\tkzDrawSegments(A,B A,C)
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\tkzDrawPoints(A,B,C,A')
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\end{tikzpicture}
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@ -6,16 +6,29 @@
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\begin{tikzpicture}
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\tkzSetUpPoint[shape=circle,size=10,color=black,fill=black]
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\tkzSetUpLine[line width=1]
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\tkzDefPoints{0/0/A, 4/0/B, 2/2/C, 6/2/D}
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\tkzDefPoints{0/0/A, 4/0/B, 2/2/C, 6/2/A'}
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\tkzMarkAngle[arc=l,size=0.8cm,color=green,fill=green!20](B,A,C)
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\path[draw] ++(25:.3) node[rotate=0] {$\alpha$};
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\node at (1,1.5) {$\beta$};
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\tkzDrawSegments(A,B A,C A,D B,C C,D)
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\tkzDrawPoints(A,B,C,D)
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\tkzMarkAngle[arc=l,size=1.15cm,color=green,fill=green!20,opacity=.8](A',A,C)
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\tkzLabelAngle[pos=0.9](A',A,C){$\alpha_1$}
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\tkzMarkAngle[arc=lll,size=1.15cm,color=green,fill=green!20,opacity=.8](B,A,A')
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\tkzLabelAngle[pos=0.9](B,A,A'){$\alpha_2$}
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\tkzMarkAngle[arc=l,size=0.7cm,color=red,fill=red!20,opacity=.8](C,B,A)
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\tkzLabelAngle[pos=0.5](C,B,A){$\beta$}
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\tkzMarkAngle[arc=l,size=0.6cm,color=blue,fill=blue!20,opacity=.8](A,C,B)
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\tkzLabelAngle[pos=0.4](A,C,B){$\gamma$}
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\tkzDrawSegments(A,B A,C A,A' B,C C,A')
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\tkzLabelPoint[below left](A){$A$}
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\tkzLabelPoint[below right](B){$B$}
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\tkzLabelPoint[above left](C){$C$}
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\tkzLabelPoint[above right](D){$D$}
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\tkzLabelPoint[above right](A'){$A'$}
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\tkzMarkAngle[arc=l,size=0.6cm,color=green,fill=green!20,opacity=.8](B,A,C)
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\path[draw] ++(25:.35) node[rotate=0] {$\alpha$};
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\tkzDrawSegments(A,B A,C)
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\tkzDrawPoints(A,B,C,A')
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\end{tikzpicture}
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\end{document}
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