\documentclass[12pt, a4paper]{article}

%packages
\usepackage[ngerman]{babel}
\usepackage[utf8x]{inputenc}
%Formel packages
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsbsy}
\usepackage{amssymb}


\usepackage{enumerate}
\usepackage{geometry}
\geometry{a4paper, top=25mm, left=18mm, right=15mm, bottom=30mm,
headsep=10mm, footskip=12mm}


\begin{document}

\section*{Computer Graphics - Excercise 2}
\subsection*{3.1.1}
\subsection*{(a)}
The more obtuse the triangle, the more acute the angle of the corner of the voronoi area gets. Because the both sides of this corner has to be in a 90° angle to the sides of the traingle. 
\subsection*{(b)}
The formula of triangle is: $\frac{1}{2}*h*b$
As for the red triangle: $\frac{1}{2}*h*b$
with: $h=\frac{||p_i - p_j||}{2}$ and $b=$

\subsection*{(c)}

\subsection*{3.1.2}
\subsection*{(a)}
We know every vertex($v$) has three edges($e$) and 2 faces($f$). In addition every edge($e$) can separated into two half edges ($e_h$), this means $e_h = 2*e = 6*v$.\\
If we add the memory for every part it results in a formula:\\
\begin{center}
memory $= v*16\text{bytes} + e*4\text{bytes} + e_h*16\text{bytes} + f*4\text{bytes}$\\
$\rightarrow \text{through assumptions:}$\\
memory $= v*(16+3*4+6*16+2*4)\text{bytes}=v*132\text{bytes}$
\end{center}
\subsection*{(b)}
Because in a quad mash two triangles are combined into one quad, the resulting faces will be reduced by a half. The resulting ratio will be 1:3:1 (for v : e : f)
\subsection*{(c)}
We know every vertex($v$) has three edges($e$) and one face($f$). In addition every edge($e$) can separated into two half edges ($e_h$), this means $e_h = 2*e = 6*v$.\\
If we add the memory for every part it results in a formula:\\
\begin{center}
memory $= v*16\text{bytes} + e*4\text{bytes} + e_h*16\text{bytes} + f*4\text{bytes}$\\
$\rightarrow \text{through assumptions:}$\\
memory $= v*(16+3*4+6*16+1*4)\text{bytes}=v*132\text{bytes}$
\end{center}
\end{document}