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\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}

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\begin{solution}
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\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}