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181.tex
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indent
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{\it George Weigt -- Advanced Calculus Homework \#4}
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$$\eqalign{
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I&=\int_0^{\pi/2}\ln(\cos x)\,dx\cr
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J&=\int_0^{\pi/2}\ln(\sin x)\,dx
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}$$
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\beginsection 1.
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First, rewrite the integral $I$ by using the substitution
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$x={\pi\over2}-y$. Show that
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$$I=J$$
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Solution: We have
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$$x={\pi\over2}-y,\qquad dx=-dy$$
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In addition, the limits of the integral must change from $x$ to $y$ according to
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$y(x)={\pi/2}-x$.
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$$\int_{y(0)}^{y(\pi/2)}=\int_{\pi/2}^0$$
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Therefore
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$$\eqalign{
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I&=\int_0^{\pi/2}\ln(\cos x)\,dx\cr
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&=-\int_{\pi/2}^0\ln(\cos({\pi\over2}-y))\,dy\cr
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&=\int_0^{\pi/2}\ln(\cos({\pi\over2}-y))\,dy
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}$$
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The last step removes the minus sign by flipping the limits of integration.
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Then from the hint, applying the substitution $\cos({\pi\over2}-y)=\sin y$ yields
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$$I=\int_0^{\pi/2}\ln(\cos({\pi\over2}-y))\,dy=
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\int_0^{\pi/2}\ln(\sin y)\,dy=J
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$$
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\beginsection 2.
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Next, add the two integrals together and show that
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$$I+J=-{\pi\over2}\ln2+{1\over2}\int_0^\pi\ln(\sin x)\,dx\eqno(4)$$
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Solution: We have
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$$\eqalign{
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I+J&=\int_0^{\pi/2}\ln(\cos x)\,dx+\int_0^{\pi/2}\ln(\sin x)\,dx\cr
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&=\int_0^{\pi/2}\big[\ln(\cos x)+\ln(\sin x)\big]\,dx\cr
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&=\int_0^{\pi/2}\ln(\cos x\sin x)\,dx\cr
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&=\int_0^{\pi/2}\ln(\hbox{$1\over2$}\sin 2x)\,dx\cr
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&=\int_0^{\pi/2}\big[\ln\hbox{$1\over2$}+\ln(\sin 2x)\big]\,dx\cr
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&=\int_0^{\pi/2}\ln\hbox{$1\over2$}\,dx+\int_0^{\pi/2}\ln(\sin2x)\,dx\cr
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}$$
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On the last line above, the left hand integral becomes
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$$\int_0^{\pi/2}\ln\hbox{$1\over2$}\,dx=(-\ln2)x\bigg|_0^{\pi/2}=-{\pi\over2}\ln2$$
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Using the substitution $y=2x$ and $dy=2dx$, the right hand integral becomes
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$$\int_0^{\pi/2}\ln(\sin2x)\,dx={1\over2}\int_0^\pi\ln(\sin y)\,dy$$
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Hence
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$$I+J=-{\pi\over2}\ln2+{1\over2}\int_0^\pi\ln(\sin x)\,dx$$
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\beginsection 3.
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Show that
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$$\int_0^\pi\ln(\sin x)\,dx=2J\eqno(5)$$
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Solution: Consider the following integral.
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$$\int_{\pi/2}^\pi\ln(\sin x)\,dx$$
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If we use the substitution
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$$y=x-{\pi/2},\qquad dy=dx,\qquad y(\pi/2)=0,\qquad y(\pi)=\pi/2$$
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we obtain
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$$\int_{\pi/2}^\pi\ln(\sin x)\,dx=\int_0^{\pi/2}\ln(\sin(y+\pi/2))\,dy
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=\int_0^{\pi/2}\ln(\cos y)\,dy=I$$
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Then by $I=J$ we have
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$$\int_{\pi/2}^\pi\ln(\sin x)\,dx=J$$
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Hence
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$$\int_0^\pi\ln(\sin x)\,dx
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=\int_0^{\pi/2}\ln(\sin x)\,dx+\int_{\pi/2}^\pi\ln(\sin x)\,dx
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=2J$$
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\beginsection 4.
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Finally, by combining your answers in (4) and (5), evaluate $I$ and $J$.
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\medskip\noindent
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Solution: From $I=J$ we have $I+J=2J$. Hence from (4) and (5) we have
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$$-{\pi\over2}\ln2+{1\over2}\int_0^\pi\ln(\sin x)\,dx
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=\int_0^\pi\ln(\sin x)\,dx$$
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Therefore
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$$\int_0^\pi\ln(\sin x)\,dx=-\pi\ln2$$
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Combining the above result with (5) we have
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$$\int_0^\pi\ln(\sin x)\,dx=-\pi\ln2=2J$$
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Therefore
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$$I=J=-{\pi\over2}\ln2$$
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\end
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\noindent
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{\it George Weigt -- Geometry Homework \#5}
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\beginsection Page 58, problem 7.
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In the Taxicab Plane show that if $A=(-{5\over2},2)$,
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$B=({1\over2},2)$, $C=(2,2)$, $P=(0,0)$, $Q=(2,1)$
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and $R=(3,{3\over2})$ then $A{-}B{-}C$ and $P{-}Q{-}R$.
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Show that $\overline{AB}\simeq\overline{PQ}$,
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$\overline{BC}\simeq\overline{QR}$ and
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$\overline{AC}\simeq\overline{PR}$.
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Sketch an appropriate picture.
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\medskip\noindent
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Solution: First, compute all the distances.
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$$
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\eqalign{
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AB&=\left|-{5\over2}-{1\over2}\right|+|2-2|=3\cr
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BC&=\left|{1\over2}-2\right|+|2-2|={3\over2}\cr
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AC&=\left|-{5\over2}-2\right|+|2-2|={9\over2}\cr
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}
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$$
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$$
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\eqalign{
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PQ&=|0-2|+|0-1|=3\cr
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QR&=|2-3|+\left|1-{3\over2}\right|={3\over2}\cr
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PR&=|0-3|+\left|0-{3\over2}\right|={9\over2}\cr
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}
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$$
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We have $AB+BC=AC$ therefore $A{-}B{-}C$.
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We have $PQ+QR=PR$ therefore $P{-}Q{-}R$.
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We have $AB=PQ$ therefore $\overline{AB}\simeq\overline{PQ}$.
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We have $BC=QR$ therefore $\overline{BC}\simeq\overline{QR}$.
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We have $AC=PR$ therefore $\overline{AC}\simeq\overline{PR}$.
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\beginsection Page 58, problem 8.
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Let $A=(0,0)$, $B=({1\over10},1$) and $C=(1,1)$ be points in $R^2$
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with the max distance $d_S(P,Q)=max\{|x_1-x_2|,|y_1-y_2|\}$.
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Prove that $\overline{AB}\simeq\overline{AC}$.
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Sketch the two segments. Do they look congruent?
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\medskip\noindent
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Solution: First, compute the distances.
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$$\eqalign{
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AB=d_S(A,B)&=max\left\{\left|0-{1\over10}\right|,|0-1|\right\}=1\cr
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AC=d_S(A,C)&=max\{|0-1|,|0-1\}=1\cr
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}$$
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We have $AB=AC$ therefore $\overline{AB}\simeq\overline{AC}$.
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They look congruent when projected onto the $y$-axis.
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\beginsection Page 58, problem 9.
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In the Poincare Plane let $P=(1,2)$ and $Q=(1,4)$.
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If $A=(0,2)$ and $B=(1,\sqrt3)$, find $C\in\overrightarrow{AB}$ with
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$\overline{AC}\simeq\overline{PQ}$.
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\medskip\noindent
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Solution: First, compute $PQ$, the distance from $P$ to $Q$.
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For $P$ and $Q$ we have $x_1=x_2$ therefore $P$ and $Q$ are on a type I line.
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We have
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$$PQ=\left|\ln{4\over2}\right|=\ln2$$
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Next, $A$ and $B$ are on a type II line. We have
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$$c={y_2^2-y_1^2+x_2^2-x_1^2\over2(x_2-x_1)}
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={3-4+1-0\over2(1-0)}=0,\qquad
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r=\sqrt{(x_1-c)^2+y_1^2}=2$$
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Next, apply the standard ruler to $A$ and $B$.
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$$\eqalign{
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f(A)&=\ln\left({0-0+2\over 2}\right)=\ln 1=0\cr
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f(B)&=\ln\left({1-0+2\over\sqrt3}\right)=\ln\sqrt3\cr
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}$$
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Next, we want to find $C$. We have
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$$AC=|f(A)-f(C)|=\ln2$$
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therefore
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$$f(C)=f(A)\pm\ln2$$
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Note that $f(B)>f(A)$.
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In order to have $C\in\overrightarrow{AB}$ we must choose $f(C)>f(A)$ therefore
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$$f(C)=f(A)+\ln2=\ln2$$
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hence
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$$t=\ln2,\qquad C=(r\tanh t+c,r\mathop{\rm sech} t)=(1.2,1.6)$$
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Check:
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$$AC=\left|\ln\left({{0-0+2\over2}\over{1.2-0+2\over1.6}}\right)\right|
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=\left|\ln{1\over2}\right|=\ln2$$
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\beginsection Page 58, problem 10.
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In the Taxicab Plane let $P=(1,-2)$, $Q=(2,5)$,
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$A=(4,-1)$ and $B=(3,2)$.
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Find $C\in\overrightarrow{AB}$ with $\overline{AC}\simeq\overline{PQ}$.
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\medskip\noindent
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Solution:
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$$PQ=|1-2|+|-2-5|=8$$
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$$m={y_2-y_1\over x_2-x_1}={-1-2\over4-3}=-3,
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\qquad b=y_1-mx_1=11$$
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$$\eqalign{
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f(A)&=x(1+|m|)=4(1+|-3|)=16\cr
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f(B)&=x(1+|m|)=3(1+|-3|)=12\cr
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}$$
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We have
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$$AC=|f(A)-f(C)|=PQ=8$$
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therefore
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$$f(C)=f(A)\pm8$$
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Since $f(B)<f(A)$ we must have $f(C)<f(A)$. Therefore
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$$f(C)=f(A)-8=8$$
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hence $x=2$, $y=mx+b=5$, i.e.
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$$C=(2,5)$$
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Check:
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$$AC=|4-2|+|-1-5|=8$$
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\beginsection Page 62, problem 1.
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Prove that $\angle ABC=\angle CBA$ in a metric geometry.
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\medskip\noindent
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Solution:
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The objects $\angle ABC$ and $\angle CBA$ are sets of points.
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Their equivalance is proven by showing that
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each one is a subset of the other.
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\item{i.} Let $P\in\angle ABC$.
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%Then either $P\in\overrightarrow{BA}$ or $P\in\overrightarrow{BC}$.
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Then $P\in(\overrightarrow{BA}\cup\overrightarrow{BC})$
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which implies $P\in\angle CBA$.
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Therefore $\angle ABC\subset\angle CBA$.
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\item{ii.} Let $P\in\angle CBA$.
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%Then either $P\in\overrightarrow{BC}$ or $P\in\overrightarrow{BA}$.
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Then $P\in(\overrightarrow{BC}\cup\overrightarrow{BA})$
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which implies $P\in\angle ABC$.
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Therefore $\angle CBA\subset\angle ABC$.
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\par\noindent
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By (i) and (ii) above, $\angle ABC=\angle CBA$.
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Note that asserting membership in a ray
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requires a betweeness property that is fulfilled by the ruler
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of a metric geometry.
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\beginsection Page 62, problem 2.
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Let $D$, $E$, and $F$ be three noncollinear points of a metric geometry
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and let $\ell$ be a line that contains at most one of $D$, $E$, and $F$.
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Prove that each of
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$\overleftarrow D\overrightarrow E$,
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$\overleftarrow D\overrightarrow F$ and
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$\overleftarrow E\overrightarrow F$ intersects $\ell$ in at most one point.
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\medskip\noindent
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Solution:
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By hypothesis
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$$\ell=\overleftarrow P\overrightarrow Q,\qquad P\ne D,E,F$$
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In an incidence geometry, two points determine a unique line.
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Therefore
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$$
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\overleftarrow P\overrightarrow Q\ne\overleftarrow D\overrightarrow E,\qquad
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\overleftarrow P\overrightarrow Q\ne\overleftarrow D\overrightarrow F,\qquad
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\overleftarrow P\overrightarrow Q\ne\overleftarrow E\overrightarrow F
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$$
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In an incidence geometry two lines are either parallel, or they intersect in
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exactly one point.
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If $\overleftarrow P\overrightarrow Q$ is parallel to
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$\overleftarrow D\overrightarrow E$ then the fact that
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$\overleftarrow P\overrightarrow Q\ne\overleftarrow D\overrightarrow E$
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implies that
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$$\overleftarrow P\overrightarrow Q\cap\overleftarrow D\overrightarrow E=\emptyset$$
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Therefore $\overleftarrow P\overrightarrow Q$ intersects
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$\overleftarrow D\overrightarrow E$ at one point or not at all.
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Repeat for $\overleftarrow D\overrightarrow F$ and
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$\overleftarrow E\overrightarrow F$.
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\beginsection Page 62, problem 3.
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Prove that if $\triangle ABC=\triangle DEF$ in a metric geometry then
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$\overleftarrow A\overrightarrow B$ contains exactly two of the points
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$D$, $E$ and $F$.
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\medskip\noindent
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Solution:
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By $\triangle ABC=\triangle DEF$ we have
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$$\overline{AB}\subset\overline{DE}\cup\overline{EF}\cup\overline{FD}$$
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Consequently, since $D$, $E$ and $F$
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are not collinear, exactly one of the following is true.
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$$
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\overline{AB}\subset\overline{DE}
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\quad\hbox{or}\quad
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\overline{AB}\subset\overline{EF}
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\quad\hbox{or}\quad
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\overline{AB}\subset\overline{FD}
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$$
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%Hence, $A$, $B$ and exactly two of the points $D$, $E$ and $F$ are collinear.
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Therefore $\overleftarrow A\overrightarrow B$ contains exactly two of the points
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$D$, $E$ and $F$.
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\beginsection Page 62, problem 5.
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In a metric geometry, prove that if $A$, $B$ and $C$ are not collinear
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then $\overline{AB}=\overleftarrow A\overrightarrow B\cap\triangle ABC$.
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\medskip\noindent
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Solution:
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We have
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$$\eqalign{
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\overleftarrow A\overrightarrow B\cap\triangle ABC&=
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(\overleftarrow A\overrightarrow B\cap\overline{AB})\cup
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(\overleftarrow A\overrightarrow B\cap\overline{BC})\cup
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(\overleftarrow A\overrightarrow B\cap\overline{CA})\cr
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&=\overline{AB}\cup\{B\}\cup\{A\}\cr
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&=\overline{AB}
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}$$
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\beginsection Page 68, problem 1.
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If $S_1$ and $S_2$ are convex subsets of a metric geometry, prove that
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$S_1\cap S_2$ is convex.
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\medskip\noindent
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Solution:
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Assume $S_1\cap S_2$ contains at least one line segment.
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Let $P,Q\in S_1\cap S_2$.
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Then $P,Q\in S_1$ and $P,Q\in S_2$.
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Hence by hypothesis we have $\overline{PQ}\subset S_1$ and $\overline{PQ}\subset S_2$.
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This implies that $\overline{PQ}\subset S_1\cap S_2$.
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Therefore $S_1\cap S_2$ is convex.
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\beginsection Page 68, problem 3.
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If $H_1$ is a half plane determined by $\ell$ prove that $H_1\cup\ell$ is convex.
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\medskip\noindent
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Solution:
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We want to show that for every $P,Q\in(H_1\cup\ell)$ we have
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$\overline{PQ}\subset(H_1\cup\ell)$.
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There are three cases.
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\item{i.} $P,Q\in\ell$ implies $\overline{PQ}\subset\ell$.
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\item{ii.} $P,Q\in H_1$ implies $\overline{PQ}\subset H_1$.
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\item{iii.} The remaining case is $P\in\ell$ and $Q\in H_1$.
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Let $X$ be defined such that $P{-}X{-}Q$.
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Note that $P\not\in\overline{XQ}$ therefore $X$ and $Q$ are in the same
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half plane, hence $X\in H_1$.
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Because $H_1$ is convex we have $\overline{XQ}\subset H_1$.
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Hence $\overline{PQ}\subset(H_1\cup\ell)$.
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\par\noindent
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Therefore $H_1\cup\ell$ is convex.
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\beginsection Page 68, problem 4.
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If $H_1$ and $H_2$ are the half planes determined by the line $\ell$,
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prove that neither $H_1$ nor $H_2$ is empty.
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\medskip\noindent
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Solution:
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By definition we have $S-\ell=H_1\cup H_2$.
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An incidence geometry has at least three points
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that are not collinear therefore $H_1\cup H_2\ne\emptyset$.
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Let $P\in\ell$ and $Q\in(H_1\cup H_2)$.
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Since $Q\not\in\ell$ there is a unique line $\overleftarrow P\overrightarrow Q\ne\ell$.
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Let $X$ be defined such that $X{-}P{-}Q$. Then $X$ is in one half plane
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and $Q$ is in the other.
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Therefore $H_1\ne\emptyset$ and $H_2\ne\emptyset$.
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\beginsection Page 68, problem 5.
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If $H_1$ is a half plane determined by the line $\ell$, prove that
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$H_1$ has at least three noncollinear points.
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\medskip\noindent
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Solution:
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Let $A$ and $B$ be defined such that $A,B\in\ell$, $A\ne B$.
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Let $C$ be defined such that $C\in H_1$.
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%Since $A\ne B$ we have
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%$\overleftarrow A\overrightarrow C\ne\overleftarrow B\overrightarrow C$.
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%Furthermore, we have
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%$\overleftarrow A\overrightarrow C\cap\overleftarrow B\overrightarrow C=\{C\}$
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%since two distinct lines can intersect in at most one point.
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Let $X$ and $Y$ be defined such that $A{-}X{-}C$ and $B{-}Y{-}C$.
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Then we have $X,Y\in H_1$
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along with $Y\not\in\overleftarrow A\overrightarrow C$.
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Therefore we have $X,Y,C\in H_1$ and $X$, $Y$ and $C$ are not collinear.
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\medskip\noindent
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More explanation about $Y\not\in\overleftarrow A\overrightarrow C$.
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Because $A\ne B$ we have $\overleftarrow A\overrightarrow C\ne\overleftarrow B\overrightarrow C$.
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Since two distinct lines can intersect in at most one point, and that point is $C$,
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we have $\overleftarrow A\overrightarrow C\cap\overleftarrow B\overrightarrow C=\{C\}$.
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Since $Y\in\overleftarrow B\overrightarrow C$ and $Y\not\in\{C\}$,
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we must have $Y\not\in\overleftarrow A\overrightarrow C$.
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\beginsection Page 68, problem 13.
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If $A$, $B$, $C$ are noncollinear in a metric geometry,
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prove that $\triangle ABC$ is convex.
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\medskip\noindent
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Solution:
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We have
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$$\triangle ABC=\overline{AB}\cup\overline{BC}\cup\overline{CA}$$
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Let $P$ be defined such that $B{-}P{-}C$.
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Since distinct lines can intersect in at most one point we have
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$$\eqalign{
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\overleftarrow A\overrightarrow P\cap\overleftarrow A\overrightarrow B=\{A\}\cr
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\overleftarrow A\overrightarrow P\cap\overleftarrow A\overrightarrow C=\{A\}\cr
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\overleftarrow A\overrightarrow P\cap\overleftarrow B\overrightarrow C=\{P\}\cr
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}$$
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Therefore $\overleftarrow A\overrightarrow P\cap\triangle ABC=\{A,P\}$.
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Hence $\overline{AP}\not\subset\triangle ABC$.
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Therefore $\triangle ABC$ is not convex.
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\end
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