Warning: this is an htmlized version!
The original is across this link,
and the conversion rules are here.
% (find-angg "LATEX/2017-1-GA-P1.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2017-1-GA-P1.tex"))
% (defun d () (interactive) (find-xpdfpage "~/LATEX/2017-1-GA-P1.pdf"))
% (defun e () (interactive) (find-LATEX "2017-1-GA-P1.tex"))
% (defun u () (interactive) (find-latex-upload-links "2017-1-GA-P1"))
% (find-xpdfpage "~/LATEX/2017-1-GA-P1.pdf")
% (find-sh0 "cp -v  ~/LATEX/2017-1-GA-P1.pdf /tmp/")
% (find-sh0 "cp -v  ~/LATEX/2017-1-GA-P1.pdf /tmp/pen/")
%   file:///home/edrx/LATEX/2017-1-GA-P1.pdf
%               file:///tmp/2017-1-GA-P1.pdf
%           file:///tmp/pen/2017-1-GA-P1.pdf
% http://angg.twu.net/LATEX/2017-1-GA-P1.pdf
\documentclass[oneside]{book}
\usepackage[colorlinks]{hyperref} % (find-es "tex" "hyperref")
%\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
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\usepackage{color}                % (find-LATEX "edrx15.sty" "colors")
\usepackage{colorweb}             % (find-es "tex" "colorweb")
%\usepackage{tikz}
%
% (find-dn6 "preamble6.lua" "preamble0")
%\usepackage{proof}   % For derivation trees ("%:" lines)
%\input diagxy        % For 2D diagrams ("%D" lines)
%\xyoption{curve}     % For the ".curve=" feature in 2D diagrams
%
\usepackage{edrx15}               % (find-angg "LATEX/edrx15.sty")
\input edrxaccents.tex            % (find-angg "LATEX/edrxaccents.tex")
\input edrxchars.tex              % (find-LATEX "edrxchars.tex")
\input edrxheadfoot.tex           % (find-dn4ex "edrxheadfoot.tex")
\input edrxgac2.tex               % (find-LATEX "edrxgac2.tex")
%
\begin{document}

\catcode`\^^J=10
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\def\eval#1{\directlua{#1}}
\def\pu{\directlua{pu()}}

\directlua{dofile "edrxtikz.lua"} % (find-LATEX "edrxtikz.lua")
\directlua{dofile "edrxpict.lua"} % (find-LATEX "edrxpict.lua")
%L V.__tostring = function (v) return format("(%.3f,%.3f)", v[1], v[2]) end




\def\V(#1){\VEC{#1}}





%   ____      _                    _ _           
%  / ___|__ _| |__   ___  ___ __ _| | |__   ___  
% | |   / _` | '_ \ / _ \/ __/ _` | | '_ \ / _ \ 
% | |__| (_| | |_) |  __/ (_| (_| | | | | | (_) |
%  \____\__,_|_.__/ \___|\___\__,_|_|_| |_|\___/ 
%                                                

{\setlength{\parindent}{0em}
\footnotesize
\par Geometria AnalÃtica
\par PURO-UFF - 2017.1
\par P1 - 7/jun/2017 - Eduardo Ochs
\par Respostas sem justificativas nÃo serÃo aceitas.
\par Proibido usar quaisquer aparelhos eletrÃ∧nicos.
% \par VersÃo: 14/mar/2016
% \par Links importantes:
% \par \url{http://angg.twu.net/2015.2-C2.html} (pÃgina do curso)
% \par \url{http://angg.twu.net/2015.2-C2/2015.2-C2.pdf} (quadros)
% \par \url{http://angg.twu.net/LATEX/2015-2-C2-material.pdf}
% \par {\tt eduardoochs@gmail.com} (meu e-mail)

}

\bsk
\bsk

\setlength{\parindent}{0em}
\def\T(Total: #1 pts){{\bf(Total: #1 pts)}}
\def\T(Total: #1 pts){{\bf(Total: #1)}}
\def\B       (#1 pts){{\bf(#1 pts)}}
% Usage:
% 1) \T(Total: 2.34 pts) Foo
% a) \B(0.45 pts) Bar



% (find-angg "LATEX/2015-2-GA-P2.tex")

1) \T(Total: 2.0 pts) Sejam $A=(2,0)$ e $B=(3,1)$.

a) \B(1.0 pts) Represente graficamente $r_0 = \setofst{Pâ\R^2}{\Vec{AP}Â\Vec{AB}=0}$.

b) \B(1.0 pts) Represente graficamente $r_2 = \setofst{Pâ\R^2}{\Vec{AP}Â\Vec{AB}=2}$.

\bsk
\bsk



2) \T(Total: 1.0 pts) Prove que se $\ww=a\uu+b\vv$ e $\uu$ e $\vv$ sÃo
nÃo-nulos entÃo

a) \B(0.5 pts) $\Pr_\uu \ww + \Pr_\vv \ww = \ww$ Ã falso em geral, 

b) \B(0.5 pts) $\Pr_\uu \ww + \Pr_\vv \ww = \ww$ Ã verdadeiro quando $\uuâ\vv$.

\bsk
\bsk



3) \T(Total: 2.5 pts) Seja $r:\frac x4 + \frac y3 = 1$.

a) \B(1.0 pts) Encontre um ponto $Pâ\setofxyst{y=2x}$ tal que $d(P,r)=1$.

b) \B(1.5 pts) Dê as equaÃÃes de duas retas diferentes, $s$ e $s'$,
paralelas a $r$ e tais que $d(r,s)=s(r,s')=2$.

\bsk
\bsk



4) \T(Total: 2.0 pts) Sejam $A=(1,1)$, $B=(3,1)$, $C=(3,2)$, $A'=(4,1)$, $B'=(7,2)$.

a) \B(1.0 pts) Encontre um ponto $C'$ que faÃa os triÃngulos $ABC$ e
$A'B'C'$ serem semelhantes.

b) \B(1.0 pts) Verifique que $A\hat BC = A'\hat B'C'$ e que $B\hat AC
= B'\hat A'C'$.

\bsk
\bsk



5) \T(Total: 2.5 pts) Sejam $C$ o cÃrculo de centro $C_0=(0,2)$ e raio
$R=2$, $C'$ o cÃrculo de centro $C'_0=(1,0)$ e raio $R=1$, $r$ a reta
que passa por $C_0$ e $C'_0$; sejam $I$ e $I'$ os dois pontos de
$CâC'$, $s$ a reta que passa por $I$ e $i'$, $A$ o ponto de interseÃÃo
entre $r$ e $s$.

a) \B(0.1 pts) Dê as coordenadas de $I$.

b) \B(0.3 pts) Dê a equaÃÃo de $s$.

c) \B(0.4 pts) Dê as coordenadas de $A$.

d) \B(1.0 pts) Represente tudo graficamente e encontre $I'$ por simetria.

e) \B(0.7 pts) Encontre $I'$ pelo mÃtodo algÃbrico.


\newpage

Mini-gabarito:

(sem o desenvolvimento, sem desenhos, nÃo-revisado)

\msk

1a) $r_0 = \setofst{(x,y)â\R^2}{\Vec{(x-2,y)}Â\VEC{1,1}=0}
         = \setofst{(x,y)â\R^2}{y=2-x}$

1b) $r_2 = \setofst{(x,y)â\R^2}{\Vec{(x-2,y)}Â\VEC{1,1}=2}
         = \setofst{(x,y)â\R^2}{y=4-x}$

\bsk

2a) Contra-exemplo: se $\uu=\VEC{1,0}$, $\vv=\VEC{2,0}$, $a=1$, $b=1$
entÃo $\ww=\VEC{3,0}$ mas $\Pr_\uu \ww + \Pr_\vv \ww = \VEC{6,0}$.

2b) $\begin{array}[t]{rcl}
     \Pr_\uu \ww + \Pr_\vv \ww
        &=& \frac{\uuÂ(a\uu + b\vv)}{\uuÂ\uu} \uu + \frac{\vvÂ(a\uu + b\vv)}{\vvÂ\vv} \vv \\
        &=& \frac{a\uuÂ\uu + b\uuÂ\vv}{\uuÂ\uu} \uu + \frac{a\vvÂ\uu + b\vvÂ\vv}{\vvÂ\vv} \vv \\
        &=& \frac{a\uuÂ\uu}{\uuÂ\uu} \uu + \frac{b\vvÂ\vv}{\vvÂ\vv} \vv \\
        &=& a\uu + b\vv \\
        &=& \ww \\
     \end{array}
    $


\bsk

3) Sejam $A=(0,3)$, $B=(4,0)$. EntÃo $r =
\setofst{(x,y)â\R^2}{y=3-\frac 34 x}$ e $r$ passa por $A$ e $B$. AlÃm disso:
%
$$\begin{array}{l}
  \frac{1}{\sqrt{1+m^2}} =
  \frac{1}{\sqrt{1+\frac{9}{16}}} =
  \frac{1}{\sqrt{\frac{25}{16}}} =
  \frac 45,
  \\
  d((x,y),r) =
  \frac 45 d((x,y),(x,4-\frac 34 x)) =
  \frac 45 |(3-\frac 34 x) - y| =
  \frac {|12 - 3x - 4y|}{5},
  \\
  d((0,3+h),r) =
  \frac 45 d((0,3+h),(0,3)) =
  \frac 45 |h|,
  \\
  d((0,3+2\frac 54),r) = 2, \qquad (0,3+2\frac 54) = (0,5.5), \\
  d((0,3-2\frac 54),r) = 2, \qquad (0,3-2\frac 54) = (0,0.5). \\
  \end{array}
$$

% (find-es "ipython" "2017.1-GA-P1")


3a) Queremos $y=2x$ e $d((x,y),r) = \frac 45 |(3-\frac 34 x) - y| =
1$. As soluÃÃes sÃo $(7/11,14/11)$ e $(17/11, 34/11)$.

3b) $s: y=5.5-\frac 34 x$ e $s': y=0.5-\frac 34 x$.

\bsk

\def\ang{\operatorname{ang}}

4a) Seja $\vv = \Vec{AB} = \VEC{2,0}$; repare que $\ww = \Vec{BC} =
\VEC{0,1}$ à $\vv$ rodado $90°$ e multiplicado por $\frac 12$. Podemos
fazer a mesma coisa com $A'$ e $B'$: $\vv' = \Vec{A'B'} = \VEC{3,1}$,
e rodando isto $90°$ e multiplicando por $\frac 12$ obtemos $\ww' =
\VEC{-\frac 12, \frac 32}$. O ponto $C'$ pode ser $B'+\ww = (6.5,
3.5)$ ou $B'-\ww = (7.5, 0.5)$.

4b) $\cos \ang(\Vec{BA}, \Vec{BC}) = \cos \ang(\Vec{B'A'}, \Vec{B'C'}) = 0$;

$\cos \ang(\Vec{AB}, \Vec{AC}) = \cos \ang(\Vec{A'B'}, \Vec{A'C'}) =
\frac {2} {\sqrt 5}$.

\bsk


5a) $I=(0,0)$

5b) $y=x/2$

5c) $A=(0.8,0.4)$

5d) $I'=(1.6,0.8)$

5e) $I'=(1.6,0.8)$


\end{document}

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