\documentclass[%
version=last,%
paper=a5,%
fontsize=11pt,%
headings=small,%
%twoside,%
parskip=half,%  this option takes 2.5% more space than parskip
% draft=true,%
DIV=12,%
headinclude=false,%
pagesize]%
{scrartcl}
%\documentclass[a4paper,12pt]{article}
\usepackage[turkish]{babel}
\usepackage{hfoldsty}
\usepackage[neverdecrease]{paralist}
\usepackage{pstricks,pst-plot,pst-node}
\usepackage{amsmath,amsthm,amssymb,upgreek,bm}
\newcommand{\ydp}[1]{\overrightarrow{#1}} % "yönlü doğru parçası"
\newcommand{\comp}[4]{(#1:#2)\mathbin{\&}(#3:#4)}
\newcommand{\abs}[1]{\lvert#1\rvert}

\theoremstyle{definition}
\newtheorem{problem}{Problem}

\newtheorem*{sol}{\c C\"oz\"um}
%\newenvironment{solution}{\color{blue}\begin{sol}}{\end{sol}}
\newenvironment{solution}{\color{blue}\psset{linecolor=blue}\large\sffamily}{}
%\newenvironment{solution}{\color{white}\psset{linecolor=white}\large\sffamily}{}

\usepackage{verbatim}
%\let\solution=\comment
%\let\endsolution=\endcomment

\newcommand{\Q}{\mathbb Q}
%\usepackage{mathrsfs}
%\newcommand{\pow}[1]{\mathscr P(#1)}
\newcommand{\included}{\subseteq}
\renewcommand{\phi}{\varphi}
\newcommand{\lto}{\rightarrow}
\renewcommand{\leq}{\leqslant}
\renewcommand{\geq}{\geqslant}
\renewcommand{\setminus}{\smallsetminus}

%\renewcommand{\familydefault}{cmfib}
\pagestyle{empty}
\begin{document}

\title{Analitik Geometri (MAT 104)\\
Telafi S\i nav\i\ \color{blue}\c C\"oz\"umleri}
\date{14 Nisan 2014}
\author{David Pierce}
%\abstract{}
\maketitle
\thispagestyle{empty}

\begin{problem}
$ab:cd::ef:gh$ ise
  \begin{equation*}
ab:ef::cd:gh
  \end{equation*}
  orant\i s\i n\i\ kan\i tlay\i n.
(Bundan \"once kan\i tlad\i\u g\i m\i z teoremleri kullanabilirsiniz.)
\end{problem}

\begin{solution}
Baz\i\ $k$, $\ell$, ve $m$ i\c cin
\begin{align*}
  cd&=ak,&ef&=a\ell,&gh&=am,
\end{align*}
dolay\i s\i yla
\begin{equation*}
  b:k::\ell:m
\end{equation*}
\c c\"unk\"u, varsay\i mdan,
\begin{equation*}
  b:k::ab:ak::ab:cd::ef:gh::a\ell:am::\ell:m.
\end{equation*}
Sonu\c c olarak
\begin{gather*}
b:\ell::k:m,\\
ab:a\ell::ak:am,\\
ab:ef::cd:gh.  
\end{gather*}
\end{solution}

\newpage
\begin{problem}
$A$, $B$, ve $C$ noktalar\i\ birbirinden farkl\i\ ama ayn\i\ do\u gruda olsun,
ve bu do\u grunun pozitif y\"on\"u se\c cilmi\c s olsun.
$\ydp{AB}$ ve $\ydp{BC}$ 
y\"onl\"u do\u grular\i n\i n i\c saretlerinin ayn\i\ olmas\i\ i\c cin,
\begin{align*}
  AB&<AC,&BC&<AC
\end{align*}
ko\c sullar\i n\i n gerekli ve yeterli oldu\u gunu g\"osterin.
\end{problem}

\begin{solution}
Do\u gruda noktalar\i n s\i ras\i\ a\c sa\u g\i dakilerin biridir
(veya birinin tersidir):
\begin{align*}
  &ABC,&&CAB,&&BCA.
\end{align*}
\.Ilk durumda 
$\ydp{AB}$ ve $\ydp{BC}$ 
y\"onl\"u do\u grular\i n\i n i\c saretleri ayn\i d\i r
ve verilen ko\c sullar sa\u glan\i r.
Kalan durumlarda i\c saretler farkl\i d\i r,
ve ikinci durumda $BC>AC$,
\"u\c c\"unc\"u durumda $AB>AC$.
\end{solution}

\newpage

\begin{problem}
D\"uzlemde dik $xy$ eksenleri ve birim uzunlu\u gu se\c cilsin.
\begin{equation*}
  ax^2+by^2+cx+dy+e=0
\end{equation*}
denklemi,
$a$, $b$, $c$, $d$, ve $e$ sabitlerinin hangi ko\c sullar\i\ alt\i nda
parabol\"u tan\i mlar? hiperbol\"u? elipsi?
\end{problem}

\begin{solution}
Tan\i mlanan e\u gri parabol ise $ab=0$, ama $a\neq0$ veya $b\neq0$.
Bu ko\c sullar yeterlidir.

$ab\neq0$ olsun; o zaman
\begin{align*}
&\phantom{{}\iff{}}  ax^2+by^2+cx+dy+e=0\\
&\iff a\left(x^2+\frac cax+\frac{c^2}{4a^2}\right)
+b\left(y^2+\frac dby+\frac{d^2}{4b^2}\right)+e\\
&\phantom{MMMMMMMMMMMMMM}=\frac{c^2}{4a}+\frac{d^2}{4b}\\
&\iff a\left(x+\frac c{2a}\right)^2
+b\left(y+\frac d{2b}\right)^2
=\frac{c^2}{4a}+\frac{d^2}{4b}-e.
\end{align*}
\c Simdi
\begin{equation*}
\frac{c^2}{4a}+\frac{d^2}{4b}-e=f
\end{equation*}
olsun.
\begin{itemize}
\item 
E\u gri hiperbol ise $ab<0$ ve $f\neq0$, ve bu ko\c sullar yeterlidir;
\item
elips ise $a$, $b$, ve $f$'nin i\c saretleri ayn\i dir, ve bu yeterlidir.
\end{itemize}
\end{solution}

\newpage
\begin{problem}
\c Sekillerde
\begin{compactitem}
  \item
Bir do\u gru, $ABC$ \"u\c cgeninin kenarlar\i n\i\
$D$, $E$, ve $F$'de kessin;
\item
$AB$, \c cemberin \c cap\i\ olsun, ve $DG\perp AB$ olsun;
\item
$DH\perp DE$ ve $DH\cdot DE=DG^2$ olsun;
\item
$EK$, $DH$'ye paralel olsun ve $FH$'yi $K$'de kessin;
\item
$CL$, $DE$'ye paralel olsun ve $AB$'yi $L$'de kessin;
\item
$\ydp{EF}=a$, $\ydp{ED}=x$, $x>0$, $EK=b$, ve $DG=y$ olsun.
\end{compactitem}

  \psset{unit=1.2cm,labelsep=2pt}
\begin{pspicture}(-1,-0.6)(1.2,2)
\pspolygon(-1,0)(1,0)(0,1)
\uput[d](-1,0){$A$}
\uput[d](1,0){$B$}
\uput[ul](0,1){$C$}
\psline(0,0)(1,2)
\uput[d](0,0){$D$}
\uput[30](0.333,0.667){$E$}
\uput[r](1,2){$F$}
\psline(0,1)(1,2)
\psline(0,0)(1.2,-0.6)
\uput[r](1.2,-0.6){$H$}
\psline(1,2)(1.2,-0.6)
\psline(0.333,0.667)(1.133,0.267)
\uput[r](1.133,0.267){$K$}
\psline(0,1)(-0.5,0)
\uput[d](-0.5,0){$L$}
\end{pspicture}
\hfill
\begin{pspicture}(-1,-1.6)(1,1)
  \pscircle(0,0)1
\psline(-1,0)(1,0)
\psline(0,-1)(0,0)
\uput[l](-1,0){$A$}
\uput[r](1,0){$B$}
\uput[u](0,0){$D$}
\uput[d](0,-1){$G$}
\end{pspicture}
\hfill
\begin{pspicture}(-2,-1.6)(1.467,1)
\begin{solution}
\psline(0,0)(-2,-1)
\uput[d](-2,-1){$F$}
\psline(-1,0)(-2,-1)
\psline(0,0)(0.6,-1.2)
\uput[d](0.6,-1.2){$H$}
\psline(-2,-1)(1.467,-1.267)
\psline(0.667,0.333)(1.467,-1.267)
\uput[d](1.467,-1.267){$K$}
\psline(-1,0)(-2,0)
\psline(0,1)(-2,0)
\uput[d](-2,0){$L$}
\end{solution}
\pspolygon(-1,0)(1,0)(0,1)
\uput[ul](-1,0){$A$}
\uput[r](1,0){$B$}
\uput[30](0,1){$C$}
\psline(0,0)(0.667,0.333)
\uput[u](0,0){$D$}
\uput[ur](0.667,0.333){$E$}
\end{pspicture}

\begin{enumerate}[a)]
\item 
Sa\u gdaki \c sekli tamamlay\i n.
\item
Sadece $a$ ve $b$'yi kullanarak 
$AL\cdot LB:CL^2::\underline{\phantom{M}\text{
    \begin{solution}
      $b:\abs a$
    \end{solution}
}\phantom{M}}$?
\item
Sadece $a$, $b$, ve $x$'i kullanarak 
$y^2=\underline{\phantom{M}\text{
    \begin{solution}
      $bx-\displaystyle\frac ba\cdot x^2$
    \end{solution}
}\phantom{M}}$?
\end{enumerate}
\end{problem}

\begin{solution}
$  \begin{aligned}
AL\cdot LB:CL^2
&::\comp{AL}{CL}{LB}{CL}\\
&::\comp{AD}{FD}{DB}{ED}\\
&::AD\cdot DB:FD\cdot ED\\
&::DG^2:FD\cdot ED\\
&::DH\cdot DE:FD\cdot ED\\
&::DH:FD
::EK:FE
::b:\abs{a}.
  \end{aligned}$
  \begin{equation*}
y^2=DH\cdot x
=\frac{EK}{EF}\cdot DF\cdot x
=\frac b{\abs a}\cdot\abs{a-x}\cdot x
=\frac ba\cdot(a-x)\cdot x
  \end{equation*}
\c c\"unk\"u $a>0\iff a>x$.
\end{solution}

\end{document}
