\documentclass[international_finance_p2.tex]{subfiles}

\begin{document}
\setbeamercovered{transparent}
\section{International Parities}

\subsection{Interest Parity. Interest Rates and Inflation}
\begin{frame}{Interest Parity}{Interest Rates and Inflation}
\begin{itemize}[<+->]
\item
Interest rate parity is a no-arbitrage condition on the market under which investors will be indifferent to interest rates available on bank deposits in two different currencies.
\item
No-arbitrage condition exists when the market prices do not allow for profitable arbitrage.
\item
This condition does not always hold and this create potential opportunities for riskless profits from arbitrage deals.
\item
Two assumptions central to interest rate parity are capital mobility and perfect substitutability of domestic and foreign assets.
\end{itemize}
\end{frame}
\begin{frame}{Two forms of interest rate parities}
\begin{itemize}[<+->]
\item
uncovered interest rate parity (UIRP) exists when exposure to foreign exchange risk  is uninhibited;
\item
covered interest rate parity (CIRP) exists when a forward contract has been used to cover  exchange rate risk
\end{itemize}
\end{frame}
\begin{frame}{Uncovered interest rate parity (UIRP)}
\begin{align}
1+i_{USD}=\frac{E_t S_{t+k}}{S_t}(1+i_{EUR}),
\end{align}
where
$E_t S_{t+k}$ is the expected future spot exchange rate at time $t + k$;

$k$ is the number of periods into the future from time $t$;

$S_t$ is the current spot exchange rate at time $t$;

$i_{USD},\quad i_{EUR}$  are the interest rates in the domestic and foreign currencies, for example USD and EUR respectively.

The dollar return on dollar deposits, $1+i_{USD}$, is shown to be equal to the dollar return on euro deposits, $\frac{E_t S_{t+k}}{S_t}(1+i_{EUR})$.
\end{frame}

\begin{frame}{Covered interest rate parity (CIRP)}
\begin{align}
1+i_{USD}&=\frac{F_t}{S_t}(1+i_{EUR}) \quad \nonumber or\\
i_{USD}-i_{EUR}&=\frac{F_t-S_t}{S_t}
\end{align}

where

$F_t$ is the forward exchange rate at time $t$;

The dollar return on dollar deposits, $1+i_{USD}$, is shown to be equal to the dollar return on euro deposits, $\frac{F_t}{S_t}(1+i_{EUR})$.

Covered interest arbitrage is an arbitrage trading strategy whereby an investor capitalizes on the interest rate differential between two countries by using a forward contract to cover exchange rate risk.

\end{frame}
\subsection{The relation between Exchange Rates, Interest Rates, and Inflation}
\begin{frame}{The relation between Exchange Rates, Interest Rates, and Inflation}
\begin{itemize}[<+->]
\item
The nominal interest rate is the rate actually observed in the market.
\item
The real rate is a concept that measures the return after adjusting for inflation.
\end{itemize}
\end{frame}
\begin{frame}{The Fisher effect}
\begin{align}
i_{USD}&=r_{USD}+p^{USD}\\
i_{EUR}&=r_{EUR}+p^{EUR}
\end{align}

where

$i$ is the nominal interest rate;

$r$ is the real interest rate;

$p^{USD}$ is the expected rate of inflation (in this case in US dollars or EUR).
\end{frame}

\begin{frame}{Real interest rate parity (RIRP)}
\begin{align}
UIRP:\Delta E_t S_{t+k}&=E_t S_{t+k}-S_t\nonumber \\
&=i_{USD}-i_{EUR},\\
E_t S_{t+k}&= \Delta E_t (p_{t+k}^{USD} ) - \Delta E_t (p_{t+k}^{EUR} ),
\end{align}

where

$p_{t+k}^{USD}, \quad p_{t+k}^{EUR}$ represent expected rate of inflation for both currencies respectively (dollar and Euro in this example).

If the above conditions hold, then they can be combined and rearranged as the following:
\begin{align}
RIRP:i_{USD} - \Delta E_t (p_{t+k}^{USD})=i_{EUR}- \Delta E_t (p_{t+k}^{EUR} ),
\end{align}
\end{frame}
\begin{frame}{The link between interest rates, inflation, and exchange rates}
\begin{align}
RIRP:i_{USD}-i_{EUR}&=p_{t+k}^{USD}-p_{t+k}^{EUR}\nonumber\\
&=\frac{F_t-S_t}{S_t}.
\end{align}
The interest differential is equal to expected rates of inflation differential and is also equal to the forward premium.

The parity condition suggests that real interest rates will equalize between countries and that capital mobility will result in capital flows that eliminate opportunities for arbitrage.
\end{frame}
\end{document}


% !TEX root = RadCal_User_Guide.tex

\typeout{new file: Introduction_Chapter.tex}

\chapter{Introduction}

Thermal radiation fire plays a preponderant role in most fires as it is responsible for fire spread and fire growth due to thermal feedback from the hot upper layer in a compartment fire or the hot plume in an open pool fire, or controls the mass loss rate in large scale pool fire.
In all the aforementioned cases, it is crucial to accurately determinate the radiative heat transfer. The use of the Wien's displacement law quantifies the wavelength of the maximum radiative energy emitted by a blackbody. This law is expressed
by
\begin{equation}\label{eq:Wien}
 \rm \lambda_m T = 2897 \, \mu m.K
\end{equation}
with $\rm \lambda_m$ being the wavelength in units of m, corresponding to the peak intensity of blackbody emittance; T represents the temperature in units of Kelvin of the blackbody. Equation \ref{eq:Wien} says that in typical fire configurations, where the temperature ranges from 300~K to about 2500~K, the peak emittance wavelength varies from 10 $\rm \mu m$ to 1.1 $\rm \mu m$: almost all the radiative exchanges happen in the near to mid infrared range. Most of the components involved in fire are present in gas phase. The infrared spectrum of the gases is very discontinuous, with few narrow spectral areas that participate to radiative exchange. The rest of the spectrum is practically transparent. The participation propensity varies with the amount of the species present and the local temperature.

These intrinsics aspects of the radiative exchange between medium with important gradients of temperature, species amount, and non-homogeneity in species render this problem quite complex and some level of sophistication is needed in its treatment. The radiative properties of a gas species are dictated by quantum mechanics and are related to the composition and the structure of the component considered. The discrete nature of the energy levels (mostly vibrational and rotational modes) for a given molecule generate its infrared ``fingerprint''. While a thorough consideration of all the energy levels would give an exact assessment of the radiative exchange -- this is often referred to as line-by-line calculation in the literature -- this operating mode is still too computationally expensive for engineering applications and is only used for simple configurations. Moreover, the lack of data for both elevated temperatures and for most hydrocarbon species further restricts its applicability to fire scenarios. The development of narrow-band models constitutes a compromise between accuracy and efficiency. Narrow-band models divide the spectrum of interest into small spectral segments of uniform spectral properties and use statical representations of the energy levels over these segments. Narrow band models are computationally fast and accurate. In particular, they do not require detailed knowledge of all the active energy levels in a given molecule. They are easier to implement and use than Line-by-Line techniques.

RadCal was previously developed by Grosshandler \cite{Grosshandler1993} to predict radiative heat transfer from gases at elevated temperature using narrow-band models. RadCal is a computer program, originally written in FORTRAN 77, that computes the directional spectral intensity from a non-isothermal, non-uniform mixture of gases and soot, by spectrally solving the radiative transfer equation. In addition, RadCal returns the Planck mean absorption coefficient, an effective absorption coefficient, and other integrated quantities. Details about the different models used and the quantities printed by RadCal are provided in Chapter \ref{chap:SNB}.

The first version of RadCal was developed to predict the enhancement in radiation caused by the addition of pulverized coal to a 60 kW methanol-fired furnace~\cite{Grosshandler1976}. This first version considered the contributions of CO, $\rm CO_2$, $\rm H_2O$, and soot. Validation of first version of RadCal was documented in 1979 and can be found in Ref.~\cite{Grosshandler1979}. Predictions from RadCal for $\rm CO_2$, CO, and $\rm H_2O$ in individual or in mixtures were compared against published data. Good agreement was found except for some data. As Grosshandler states in Ref.~\cite{Grosshandler1993}:
\textit{``...The spectrum between 1.25 and 12.5 $\mu$m was satisfactorily reproduced, although some of the data at particular wavelengths differed from the prediction by as much as 17\%. Considerable disagreement occurred between the integrated emittance of $\rm CO_2$ as predicted from RADCAL and that computed from the charts of Hottel \cite{Hottel1954}. No one source for this disagreement was identified, but it was thought to be a combination of the difficulty in obtaining high accuracy spectral measurements under the full range of conditions investigated, the uncertainty associated with extrapolating total transmittance results beyond the measured temperature and pressure-pathlengths, and the approximations associated with the narrow-band models.''}

Methane was added to RadCal in 1985 \cite{Grosshandler1985}, along with an extension to 200 $\rm \mu m$ of the considered spectrum. The added methane data originates mostly from experiments performed by Brosmer \textit{et al.} \cite{Brosmer1985} and Lee \textit{et al.} \cite{Lee1964}. At this time, the code structure was updated and a new input file was created.

This report, a second edition of NIST Special Publication 1402, presents the latest enhancements brought to RadCal and aims to provide an exhaustive list of the mathematical and physical models used in RadCal, which was missing from the first edition. Chapter~\ref{chap:SNB} presents the fundamental mathematical models used in radiative heat transfer and presents the different narrow-band models used in RadCal. Chapter~\ref{chap:old_species} recalls the characteristics of the species that were present in the 1993 version of RadCal: $\rm H_2O$, $\rm CO_2$, CO, $\rm CH_4$, and soot. Additional hydrocarbons have been implemented into RadCal and the code has been rewritten, for its most part, into Fortran 2008. FTIR transmission measurements at the National Institute of Standards and Technology (NIST) were undertaken to provide highly resolved spectral absorption coefficients in the mid-IR and NIR as a function of temperature for many fuel species~\cite{Wakatsuki2005a,Wakatsuki2008,Yilmaz2008}. These measurements were performed over a range of temperatures from 300~K up to 1000~K for several fuel species, including paraffins (methane, ethane, propane, and n-heptane), olefins (ethylene and propylene), and other fuel-related species (methanol, toluene, and methyl-methacrylate). The uniform set of conditions and spectral resolution of these measurements have provided a set of data for developing calculation methodologies for absorption coefficients of these species in flame environments. New species and their associated narrow-band parameters are described in Chapter~\ref{chap:new_species}. The syntax of the input file was modified to make good use of native Fortran namelist that offers a more flexible way to input data. Chapter~\ref{chap::using_RADCAL} describes the new input file syntax, and how to compile and use RadCal. The code was modified following a modular approach and was translated into Fortran 2008 to benefit from recent advances in Fortran standard. The list of RadCal functions and subroutines is detailed in Chapter~\ref{chap:Code}.  The code has also been modified to account for any type of fuel mixture. Verification of the new species spectral data has been performed and results from these tests are reported in Chapter~\ref{sec:verification}. Finally, Chapter~\ref{chap::validation_tests} compares the new version of RadCal with the 1993 one for the various validation tests presented in the first edition of this report. This chapter also presents predicted quantities from an experimentally characterized small methanol pool fire.

Include here New RadCal features in a list:

\begin{itemize}
 \item New species spectral data between 700 and 4000~$\rm cm^{-1}$ and for temperature ranging from about 300~K to about 1000~K.
 \item New species data for: Ethylene, Ethane, Propylene, Propane, n-Heptane, Toluene, Methanol, Methane, MMA
 \item Code rewritten in Fortran 2008
 \item Code made modular
 \item More convenient input file
 \item Possibility to calculate the spectrum for any mixture of gas
 \item Updated and enhanced user guide
\end{itemize}


\documentclass[a4paper,12pt]{article}
\usepackage{epsfig,latexsym,amsmath,amssymb,epic,eepic,psfrag,subfigure,float,euscript,array}
\usepackage[latin1]{inputenc}
\usepackage{standalone}
\usepackage{tikz,pgf,pgfplots}

\newenvironment{exercise}[1][Uppgift]{\begin{trivlist} \item[\hskip
    \labelsep {\stepcounter{exerctr}\bfseries #1
      \arabic{exerctr}}]}{\end{trivlist}\vspace{10mm}}

\newcounter{exerctr}
\newcounter{abcctr}[exerctr]

\newcommand{\abc}{\noindent\vspace{1mm}\\ {\bf
    \stepcounter{abcctr}(\alph{abcctr})\ }}
\newcommand{\bbm}{\begin{bmatrix}}
\newcommand{\ebm}{\end{bmatrix}}
\newcommand{\point}[1]{\hfill {\bf (#1p)}\\ \vspace{-5mm}}
\newcommand{\ctrb}{\EuScript{S}}
\newcommand{\Lap}{\mathcal{L}}
\newcommand{\obsv}{\EuScript{O}}
\newcommand{\realdel}[1]{\text{Re}\left\{#1\right\}}
\newcommand{\imagdel}{\text{Im}}
\newcommand{\bC}{\mathbb{C}}
\newcommand{\bR}{\mathbb{R}}
\newcommand{\bmpv}{\begin{minipage}[t]}
\newcommand{\bmps}{\begin{minipage}[t]{45mm}}
\newcommand{\bmpm}{\begin{minipage}[t]{90mm}}
\newcommand{\bmpl}{\begin{minipage}[t]{140mm}}
\newcommand{\emp}{\end{minipage}}
\newcommand*{\zethree}{\big(z - \mexp{-3h}\big)}
\newcommand*{\mexp}[1]{\ensuremath{\mathrm{e}^{#1}}}

\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
            \node[shape=circle,draw,inner sep=2pt] (char) {#1};}}

\addtolength{\topmargin}{-1cm}
\textheight 23.5cm
%\oddsidemargin 0.61cm
%\evensidemargin 0.61cm

\def\OctaveG{tf([0.5 1], [1 0 -1])}

\title{Computerized control partial exam 1 from fall semester 2016, modified}
\author{Kjartan Halvorsen}

\begin{document}

\maketitle

\begin{description}
\item[Time] September 13 17:30
\item[Place] 4101
\item[Permitted aids] The single colored page with your own notes, table of Laplace transforms, calculator
\end{description}

All answers should be readable and well motivated (if nothing else is written). Solutions/motivations should be written on the provided spaces in this exam. Use the last page if more space is needed.

\begin{center}
{\Large Good luck!} \\
\end{center}

\begin{tabular}{|l|l|}
\hline
\multicolumn{2}{|l|}{\bmpl
Matricula and name
\vspace*{18mm}
\emp}\\
\hline

\end{tabular}

\clearpage

%-----------------------------------------------------------------
\subsection*{The system}
The dynamic model of a ship with input $u$ being the rudder angle and the output $y$ being the heading (see figure \ref{fig:tanker}) can be described as a continuous-time second order system with a pole in the origin
\[ G(s) = \frac{K}{s(s + a)}. \]
For fully loaded, large tankers this dynamics is often unstable, meaning that $a<0$.
\begin{figure}[h]
\begin{center}
\includegraphics[width=0.8\linewidth]{tanker}
\caption{Heading of a ship controlled by rudder input.}
\label{fig:tanker}
\end{center}
\end{figure}

Consider for this exam the normalized continuous-time model of the tanker
\[ G(s) = \frac{1}{s(s - 1)}. \]

\clearpage
\subsection*{Problem 1 (50p)}

The system is sampled with sampling interval $h$ using step-invariant (zero-order hold) sampling. \textbf{Circle the correct pulse-transfer function below, and show your calculations}
\begin{enumerate}
 \item \( H(z) = \frac{(1-\mexp{h} -h)z - \big((1-h)\mexp{h}-1\big)}{(z-1)(z-\mexp{h})}\)
 \item \( H(z) = \frac{(-1+\mexp{h} -h)z - \big((1-h)\mexp{h}-1\big)}{(z-1)(z-\mexp{2h})}\)
 \item \( H(z) = \frac{(-1+\mexp{h} -h)z - \big((1-h)\mexp{h}-1\big)}{(z-1)(z-\mexp{h})}\)
 \end{enumerate}

\noindent
\fbox{
\bmpl
{\bf Derivation:}\\
\vspace*{150mm}
\emp}

\clearpage
\subsection*{Problem 2 (20p)}
Assume that the sampling period is $h=0.2$. In figure \ref{fig:complex-plane} draw the poles (crosses) and zero (circle) for both the continuous-time transfer function $G(s)$ and the  discretized pulse-transfer function $H(z)$ you determined in Problem 1.
   \begin{figure}[h]
   \begin{center}
   \includegraphics[]{complex-plane}
   \caption{Problem 2: Plot the poles of the continuous-time system (on the left) and the poles and zero of the discrete-time system (on the right). Indicate (with arrows and/or colors) corresponding pairs of continuous-time and discrete-time poles.}
   \label{fig:complex-plane}
   \end{center}
   \end{figure}

\noindent
\fbox{
\bmpl
{\bf Calculations:}\\
\vspace*{90mm}
\emp}

\clearpage

\subsection*{Problem 3 (40p)}
Assume, now, that the plant \(H(z) = \frac{1}{z-0.9}\) is controlled by feedback from the control error, as illustrated in figure~\ref{fig:feedback}, using the controller
\[ F(z) = K\frac{z}{z-1}. \]

\begin{figure}
\begin{center}
     \begin{tikzpicture}[scale = 0.8, node distance=25mm, block/.style={rectangle, draw, minimum width=15mm}, sumnode/.style={circle, draw, inner sep=2pt}]

     \node[coordinate] (refinput) {};
     \node[sumnode, right of=refinput, node distance=20mm] (sumerr) {\tiny $\sum$};
     \node[block, right of=sumerr] (controller) {$K\frac{z}{z-1}$};
     %\node[above of=controller, node distance=6mm] {controller};
     \node[block, right of=controller, node distance=35mm] (plant) {$\frac{1}{z-0.9}$};
     \node[sumnode, right of=plant, node distance=24mm] (sum) {\tiny $\sum$};
     %\node[above of=tank, node distance=6mm] {motor};
     \node[coordinate, right of=sum, node distance=20mm] (output) {};
     \node[coordinate, above of=sum, node distance=12mm] (disturbance) {};

     \draw[->] (refinput) -- node[above, pos=0.3] {$u_c(k)$} (sumerr);
     \draw[->] (sumerr) -- node[above] {$e(k)$} (controller);
     \draw[->] (controller) -- node[above] {$u(k)$} (plant);
     \draw[->] (plant) -- (sum);
     \draw[->] (sum) -- node [coordinate] (measure)  {} node [above, near end] {$y(k)$} (output);
     \draw[->] (disturbance) -- node[right, pos=0.2] {$d(k)$} (sum);
     \draw[->] (measure) -- ++(0,-14mm) -| node[right, pos=0.95] {$-$} (sumerr);
     \end{tikzpicture}
     \caption{Feedback control from the error signal.}
     \label{fig:feedback}
   \end{center}
 \end{figure}

\subsubsection*{(a) 20p}

Figure~\ref{fig:rlocus} shows the root locus for the closed-loop poles with respect to the gain $K$. In figure~\ref{fig:step}, four different step plots are shown for four different values of $K$. Identify (and circle) the corresponding step plot for each value of $K$ in the table below.

\begin{center}
\begin{tabular}{cl}
\(K\) & Step plot\\\hline
0.002 & A\hspace*{2mm} B\hspace*{2mm} C\hspace*{2mm} D\\
1.0 & A\hspace*{2mm}  B\hspace*{2mm}  C\hspace*{2mm} D\\
3.0 & A\hspace*{2mm} B\hspace*{2mm}  C\hspace*{2mm} D\\
4.0 & A\hspace*{2mm} B\hspace*{2mm}  C\hspace*{2mm} D\\ \hline
\end{tabular}
\end{center}

\begin{figure}[bp]
\begin{center}
\begin{tikzpicture}
    \node[anchor=south west,inner sep=0] at (0,0) {\includegraphics[width=0.5\linewidth]{p2_rlocus_rlocus-crop}};
    \node[coordinate, pin={[pin distance=20mm] 80:{$K=0.003$}}] at (7.17,3.47) {};
    \node[coordinate, pin={[pin distance=20mm] 115:{$K=3.8$}}] at (3.77,3.47) {};
\end{tikzpicture}

\caption{Root locus wrt the gain $K$.}
\label{fig:rlocus}
\end{center}
\end{figure}

\begin{figure}[tp]
\begin{center}
\begin{tabular}{cc}
A & B\\
\includegraphics[width=0.4\linewidth]{step-plot-3-crop}
&\includegraphics[width=0.4\linewidth]{step-plot-1-crop}\\
C & D\\
\includegraphics[width=0.4\linewidth]{step-plot-5-crop}
&\includegraphics[width=0.4\linewidth]{step-plot-2-crop}

\end{tabular}
\caption{Step responses for different values of $K$.}
\label{fig:step}
\end{center}
\end{figure}

\clearpage
\subsubsection*{(b) 20p}

Determine the gain $K$ so that the closed loop system has poles with realpart equal to $0.7$.

\noindent
\fbox{
\bmpl
{\bf Solution:}\\
\vspace*{150mm}
\emp}

\cleardoublepage

\noindent
{\bf If necessary,} you can continue your solutions on this page. Mark clearly which problem the solution corresponds to.

%\end{document}

%*****************************************************************
%*****************************************************************
\newpage
\setcounter{page}{1}

\section*{Solutions}
\subsection*{Problem 1}
   First calculate the step-response of the continous-time system
   \[Y(s) = G(s)\frac{1}{s} = \frac{1}{s^2(s-1)} = \frac{1}{s-1} - \frac{1}{s} - \frac{1}{s^2}.\]
   The inverse Laplace-transform gives
   \[ y(t) = \mexp{t} - 1 - t\]
   Sampling this function gives
   \[ y(kh) = \mexp{kh} -1 - kh\]
   which has the Z-transform
   \[Y(z) = \frac{z}{z - \mexp{h}} - \frac{z}{z-1} - \frac{hz}{(z-1)^2}\]
   Dividing the z-transform of the system response to that of the input (the step) gives
   \begin{align*}
   H(z) &= \frac{Y(z)}{U(z)} = \frac{z-1}{z}Y(z) = \frac{z-1}{z-\mexp{h}} - 1 - \frac{h}{z-1}\\
        &= \frac{(z-1)^2 - (z-1)(z-\mexp{h}) - h(z-\mexp{h})}{(z-1)(z-\mexp{h})}\\
	&= \frac{(z-1)(z-1 -z + \mexp{h}) - hz + h\mexp{h}}{(z-1)(z-\mexp{h})}\\
        &= \frac{ (\mexp{h} - 1 -h)z - (\mexp{h}-1-h\mexp{h})}{(z-1)(z-\mexp{h})}\\
        &= \frac{ (\mexp{h} - 1 -h)z - \big( (1-h)\mexp{h}-1\big)}{(z-1)(z-\mexp{h})}\\
   \end{align*}

   The correct pulse transfer function is the third.

\subsection*{Problem 2}
The discrete-time poles are in $z=1$ and $z=\mexp{0.2} \approx 1.22$. The zero is  in
\[ z = - \frac{(1-h)\mexp{0.2} - 1}{\mexp{0.2} - 1 -h} \approx -1.07. \]
\begin{center}
\includegraphics[width=0.8\linewidth]{complex-plane-facit}
\end{center}

\subsection*{Problem 3}

\subsubsection*{(a)}

The root locus starts with two poles at $1$ and $0.9$. For small values of $K$ we will have poles that are slow and completely damped. The only such response is \textbf{B}. As $K$ increases we will have closed-loop poles that follows the unit circle a bit inside it. The poles will have little damping and will increase in speed with $K$. Finally, one pole move outside the unit circle, and the system becomes unstable. It is easy to see the response that is slow (B) and unstable (C). It is a bit difficult to read off the difference in osciallation period between the other two responses. In summary, we get

\begin{center}
\begin{tabular}{cl}
\(K\) & Step plot\\\hline
0.002 & A\hspace*{2mm} \circled{B}\hspace*{2mm} C\hspace*{2mm} D\\
1.0 & A\hspace*{2mm}  B\hspace*{2mm}  C\hspace*{2mm} \circled{D}\\
3.0 & \circled{A}\hspace*{2mm} B\hspace*{2mm}  C\hspace*{2mm} D\\
4.0 & A\hspace*{2mm} B\hspace*{2mm}  \circled{C}\hspace*{2mm} D\\ \hline
\end{tabular}
\end{center}

\subsubsection*{(b)}

The closed-loop system from command signal to the output is
\[ H_c(z) = \frac{K \frac{z}{z-1}\frac{1}{z-0.9}}{1 + K \frac{z}{z-1}\frac{1}{z-0.9}} = \frac{Kz}{(z-1)(z-0.9) + Kz}. \]
The characteristic equation is
\[ (z-1)(z-0.9) + Kz = z^2 - (1.9-K)z + 0.9 = 0 \]
with solution
\[ z = \frac{1.9-K}{2} \pm \frac{1}{2}\sqrt{(1.9-K)^2 - 5.6}. \]
We know from the root locus that for a value of  $K$ that gives poles with real part $0.7$, then the poles are complex-conjugated and so the expression under the root sign must be negative. The real part is given by the first term in the solution to the quadratic equation. We get
\[ \frac{1.9-K}{2} = 0.7 \quad \Rightarrow \quad K = 1.9-1.4 = 0.5 \]

\end{document}


\documentclass{memoir}
\usepackage{notestemplate}

%\logo{~/School-Work/Auxiliary-Files/resources/png/logo.png}
%\institute{Rice University}
%\faculty{Faculty of Whatever Sciences}
%\department{Department of Mathematics}
%\title{Class Notes}
%\subtitle{Based on MATH xxx}
%\author{\textit{Author}\\Gabriel \textsc{Gress}}
%\supervisor{Linus \textsc{Torvalds}}
%\context{Well, I was bored...}
%\date{\today}

%\makeindex

\begin{document}

% \maketitle

% Notes taken on

\subsection{Free Homotopy of Curves}
\label{sub:free_homotopy_of_curves}

In order to state Cauchy's theorem in complete generality, we need to construct a notion of homotopy for curves.

\begin{defn}[Free Homotopy]
	A \textbf{free homotopy} of closed curves in \(\Omega \) is a continuous map \(\gamma(\tau,t)\) from \([0,1]\times [t_0,t_1]\) to \(\Omega \) such that
\begin{align*}
	\gamma(\tau,t_0) = \gamma(\tau,t_1)
\end{align*}
for every \(\tau\in [0,1]\). We can denote \(\gamma_\tau(t) := \gamma(\tau,t)\).\\

We say \(\gamma_0,\gamma_1\) are \textbf{homotopic} if there exists a free homotopy with \(\gamma(0,t) = \gamma_0\) and \(\gamma(1,t) = \gamma_1\).
\end{defn}
Of course, if \(\Omega \) is convex, then any two curves are automatically freely homotopic.

% \printindex
\end{document}


\graphicspath{{Pics/}}

\newpage\section{Centers of inside and outside}

\impden{Incenter and Co.}{
    Let $\triangle ABC$ be an ordinary triangle, $I$ is
    its incenter, $D, E, F$ are the touch points of the incenter with $BC, CA,
    AB$ and $ D', E', F' $ are the reflections of $D, E, F$ wrt $ I $.
    Let the $ I_a, I_b, I_c $ excircles touch $BC, CA, AB$ at $D_1, E_1,
    F_1$.\\

    Let $M_a, M_b, M_c$ be the midpoints of the smaller arcs $BC, CA, AB$, and
    $M_A, M_B, M_C$ be the midpoints of the major arcs $BC, CA, AB$. $ M $ are
    the midpoint of $ BC $.  Let $ A' $ be the antipode of $ A $ wrt $ \odot
    ABC $.\\

    Let $(I_a)$ touch $BC, CA, AB$ at $D_A, E_A, F_A$. So, $D_A \equiv D_1$.\\

    Call $ EF $, `$ A $-tangent line', and $ DE, DF $ similarly. And call $
    E_AF_A $ `$ A_A $-tangent line.\\

    \figdf{.8}{incenter_main}{All the primary points related to the incenter
    and the excenters}
}

\newpage

\begin{minipage}{.59\linewidth}
    \lem{Antipode and Incenter}{
        $ A'I, \odot ABC, \odot AEIF $ are concurrent at $ Y_A $. And $ Y_A,
        D, M_a $ are collinear.
    }

    \lem{}{
        $ DD_H \perp EF $, then $ D_H, I, A' $ are collinear.
    }

    \lem{}{
        \[\frac{FD_H}{D_HE} = \frac{BD}{DC}\]
    }

    \vspace{2em}

    \lem{Arc Midpoint as Centers}{
        \[M_AE_1 = M_AF_1\quad M_BF_1 = M_BD_1\quad M_CD_1 = M_CE_1\]
    Moreover, $I, O$ are the orthocenter and the circumcenter of $\triangle
    I_aI_bI_c$
    }

    \vspace{4em}

    \lem{Incircle Touchpoint and Cevian}{
        Let a cevian be $ AX $ and let $ I_1, I_2 $ be the incirlces of $
        \triangle ABX, \triangle ACX $. Then $ D, I_1, I_2, X $ are concyclic.
        And the other common tangent of $ \odot I_1 $ and $ \odot I_2 $ goes
        through $ D $.
    }

\end{minipage}\hfill%
\begin{minipage}{.39\linewidth}
    \figdf{.9}{antipode_incenter}{\autoref{lemma:Antipode and Incenter}}

    \figdf{.9}{excenter_touchpoint_bigarc-midpoints}{\autoref{lemma:Arc Midpoint
    as Centers}}
    \figdf{.9}{incircle_touchpoint_and_cevian}{\autoref{lemma:Incircle Touchpoint
    and Cevian}}
\end{minipage}

\newpage

\begin{minipage}{.5\linewidth}
    \lem{Apollonius Circle and Incenter, ISL 2002 G7}{
        Let $ \omega_a $ be the circle that goes through $ B, C $ and is
        tangent to $ (I) $ at $ X $. \hl{Then $ XD', EF, BC $ are concurrent and $
        X, D, I_a $ are collinear.} The same properties is held if the roles of
        incenter and excenter are swapped.
        \begin{itemize}[left=0pt, itemsep=0pt]
            \item The circle $ BXC $ is tangent to $ (I) $
            \item $ X $ lies on the Apollonius Circle of $ (B, C; D, G) $.
            \item $ XD $ bisects $ \angle BXC $.
        \end{itemize}
    }

    \lem{Line parallel to BC through I}{
        Let $E, F$ be the intersection of the $B, C$ angle bisectors with $AC,
        AB$. Then the tangent to $\odot ABC$ at $A$, $EF$ and the line through
        $I$ parallel to $BC$ are concurrent.
    }
    \lem{Midline Concurrency with Incircle Touchpoints}{
        $AI$, $ B, B_A $-tangent lines and $ C $-mid-line are concurrent. And,
        if the concurrency point is $ X $, then $ CS\perp AI $
    }
    \theo{}{Paul Yui Theorem}{
        $ B $-tangent line, $ C_A $-tangent line, and $ AH $ are concurrent.
    }

    \lem{Concurrent Lines in Incenter}{
        Let $ AD \cap (I) = G, AD' \cap (I) = H $. Let the line through $ D' $
        parallel to $ BC $ meet $ AB, AC $ at $ B', C' $. Then $ AM,\ EF,\
        GH,\ DD',\ BC',\ CB' $ are concurrent.
    }\label{lemma:concurrent_lines_in_incenter}
\end{minipage}\hfill%
\begin{minipage}{.48\linewidth}
    \figdf{.9}{InExLemma3}{\autoref{lemma:Apollonius Circle and Incenter, ISL 2002 G7}}

    \figdf{.9}{BC_parallel_through_I}{\autoref{lemma:Line parallel to BC
    through I}}

    \figdf{.8}{InExLemma4}{
        \autoref{lemma:Midline Concurrency with Incircle Touchpoints} \&
        \autoref{theorem:Paul Yui Theorem}
    }
\end{minipage}

\figdf{.5}{concurrent_lines_in_incenter}{
    \autoref{lemma:Concurrent Lines in Incenter}
    The lines are concurrent.
}

\begin{minipage}{.5\linewidth}
    \lem{Insimilicenter}{
        The \emph{insimilicenter} is the positive homothety center of the
        circumcircle and the incircle. It is also the \emph{isogonal conjugate of
        the Nagel Point} wrt $\triangle ABC$.
    }

    \begin{solution}
        Let $T$ be the $A$-mixtilinear touchpoint. If $AT\cap \left(I\right)
        =A'$, then if we can show that $A'B'$ arc has angle $\angle C$, where
        $A'B'\parallel AB$, we are done.
    \end{solution}
\end{minipage}\hfill%
\begin{minipage}{.49\linewidth}
    \figdf{.9}{insimilicenter}{}
\end{minipage}

\prob{}
{Application of Aollonius Circle and Incenter Lemma}{}{
    Let triangle $ABC$, incircle $(I)$, the $A$-excircle $(I_a)$ touches
    $BC$ at $M$. $IM$ intersects $(I_a)$ at the second point $X$. Similarly,
    we get $Y$, $Z$. Prove that $AX$, $BY$, $CZ$ are concurrent.\\

    \href{http://artofproblemsolving.com/community/c6h1595900p9909100}{Extension,
    by buratinogigle}: Triangle $ABC$ and $XYZ$ are homothetic with center $I$ is
    incenter of $ABC.$ Excircles touches $BC,$ $CA,$ $AB$ at $D,$ $E,$ $F.$ $XD,$
    $YE,$ $FZ$ meets excircles again at $U,$ $V,$ $W.$ Prove that $AU,$ $BV,$ $CW$
    are concurrent.
}

\newpage

\den{Isodynamic Points}{
    Let $ ABC $ be a triangle, and let the angle bisectors
    of $ \angle A $ meet $ BC $ at $ X, Y $. Call $ \omega_a $ the
    circumcircle of $ \triangle AXY $. Define $ \omega_b, \omega_c $
    similarly. The first and second isodynamic points are the points where the
    three circles $ \omega_a, \omega_b, \omega_c $ meet. I.e. these two points
    are the intersections of the three Apollonius circles. These two points
    satisfy the following relations:
    \begin{enumerate}
        \item $PA\sin A = PB\sin B = PC\sin C$
        \item They are the isogonal congugates of the
            Fermat Points, and they lie on the `Brocard Axis'
    \end{enumerate}
    \figdf{.6}{Isodynamic_Points}{}
}

\theo{https://en.wikipedia.org/wiki/Isodynamic_point}{Pedal Triangles of
    Isodynamic Points}{
    Prove that the pedal triangles of the isodynamic points are
    equilateral triangles. Also, Inverting around the Isodynamic Points trasnform
    $\triangle ABC $ into an equilateral triangle.
}

\prob{https://artofproblemsolving.com/community/c6h1568534p9617561}{China TST
    2018 T1P3}{EM}{
    Circle $\omega$ is tangent to sides $AB$,$AC$ of triangle $ABC$
    at $D$,$E$ respectively, such that $D\neq B$, $E\neq C$ and $BD+CE<BC$.
    $F$,$G$ lies on $BC$ such that $BF=BD$, $CG=CE$. Let $DG$ and $EF$ meet at
    $K$. $L$ lies on minor arc $DE$ of $\omega$, such that the tangent of $L$ to
    $\omega$ is parallel to $BC$. Prove that the incenter of $\triangle ABC$ lies
    on $KL$.
}

\solu{
    Using \autoref{lemma:Collinearity with antipode and center}, in the
    touch triangle of $\omega$.
}

\prob{}
{}{E}{
    Given a triangle $A B C$ with circumcircle $\Gamma$. Points $E$ and $F$
    are the foot of angle bisectors of $B$ and $C, I$ is incenter and $K$ is
    the intersection of $A I$ and $E F$. Suppose that $N$ be the midpoint of
    arc $B A C$. Circle $\Gamma$ intersects the $A$ -median and circumcircle
    of $A E F$ for the second time at $X$ and $S$. Let $S^{\prime}$ be the
    reflection of $S$ across $A I$ and $J'$ be the second intersection of
    circumcircle of $A S^{\prime} K$ and $A X$. Prove that quadrilateral $T J'
    I X$ is cyclic.
}

\begin{solution}[Reim and lemmas]
    Since $NI\cap \odot ABC=T$, the mixtilinear touchpoint, if we can show
    that $AT||IJ'$, we will be done.  Instead of working with $S'$ and $J'$,
    we reflect them back and work with $S, J$. Then we need to prove that
    $IJ||AD'$ where $D'$ is the reflection of $D$ over $I$.

    \figdf{.7}{mixti_symmedian}{}

    From \autoref{lemma:Line parallel to BC through I} we know that the $A$
    symmedian, $EF$, $IM$ are concurrent at a point $P$. We prove that $P
    \equiv J$. For that we need to show that $P$ lies on $\odot AKS$.\\

    If $\odot AKP\cap AB, AC = U, V$, it is sufficient to prove that
    \[\frac{UF}{FB} = \frac{VE}{EC}\]

    We have:
    \begin{align*}
        \frac{UF}{KU} = \frac{\sin FAP}{\sin KFA} &\quad\frac{VE}{KV} =
        \frac{\sin EAP}{\sin KEA}\\[1em]
        \therefore \frac{UF}{VE} &=\frac{\sin FAP}{\sin EAP} \frac{\sin KEA}{\sin KFA}
        =\frac{\sin CAM}{\sin BAM}\frac{AF}{AE}\\[1em]
        &=\frac{BA}{CA}\frac{AF}{AE}=\frac{BA}{AE}\frac{AF}{CA}
        =\frac{BC}{EC}\frac{CF}{BC}\\[1em]
        &=\frac{BF}{EC}
    \end{align*}
\end{solution}

\begin{minipage}{.55\linewidth}
    \prob{https://artofproblemsolving.com/community/c6t45786f6h1618685}{Vietnamese
        TST 2018 P6.a}{M}{
        Triangle $ABC$ circumscribed $(O)$ has $A$-excircle
        $(I_a)$ that touches $AB,\ BC,\ AC$ at $F,\ D,\ E$, resp. $M$ is the
        midpoint of $BC$. Circle with diameter $MI_a$ cuts $DE,\ DF$ at $K,\ H$.
        Prove that $(BDK),\ (CDH)$ have an intersecting point on $(I_a)$.
    }

    \vspace{2em}

    \prob{}
    {After Inverting Around D}{}{
        $MD$ is a line, $ I_a $ is an arbitrary point such that $ DI_a\perp
        MD$. $l$ is the perpendicular bisector of $ DI_a $. $ F, E $ are
        arbitrary points on $ l $. $ B=I_aF\cap MD, C=I_aE\cap MD\, H=FD\cap
        MI_a, K=DE\cap MI_a $. Then $ BK, CH, l $ are concurrent.
    }

    \begin{solution}
        It is straightforward using Puppus's Theorem on lines $ BDC $
        and $ HI_aK $.
    \end{solution}
\end{minipage}\hfill%
\begin{minipage}{.4\linewidth}
    \figdf{}{Vietnamese_TST_2018_P6_a_problem}{}
    \figdf{}{Vietnamese_TST_2018_P6_a_inv}{After inverting around $ D $}
\end{minipage}

\begin{solution}[Synthetic: Length Chase]
    \sollem{
        Let $ G, H, B', C' $ be
        defined the same way in Lemma 3.2. Prove that $ F $ lies on the radical
        axis of $ \odot D'GI, D'C'H $. By extension prove that $ B $ lies on the
        radical axis of $ \odot D'B'I, D'C'H $
    }\label{problem:vietTST2018P6.a}

    \figdf{.7}{Vietnamese_TST_2018_P6_a_modified}{\hrf{problem:vietTST2018P6.a}{Vietnamese
    TST 2018 P6.a}}

    We prove the first part, and the second part follows using spiral
    similarity.

    Suppose $ K\in FD\cap \odot KDI $. Due to spiral similarity on $ \odot
    KDI, \odot (I) $, we have $ \triangle GFK \sim \triangle GD'I $. Which
    implies: \[\frac{FK}{GF}=\frac{ID}{GD'} \implies FK = ID\frac{GF}{GD'}\]
    Now, if $ KDCE $ is to be cyclic, we need to have $ \triangle HFK \sim
    \triangle HDC $. So we need,

    \[\frac{FK}{HF}=\frac{DC}{HD}\implies FK=DC\frac{HF}{HD}\]
    Combining two equations:
    \[\frac{GF}{GD'}\cdot \frac{ID}{DC}=\frac{HF}{HD}\]

    Now, using Ptolemy's theorem in $ \square FDEH $, we have,

    \begin{align*}
        FD\cdot EH + DE\cdot FH &= DH\cdot EF\\
        EH \cdot \frac{FD}{FH} + DE &= EF \cdot\frac{DH}{FH}\\
        2\ \frac{DE}{EF} &= \frac{DH}{FH}
    \end{align*}

    Similarly from $ \square FGED' $ we get, \[2\ \frac{D'E}{EF} =
    \frac{GD'}{FG}\] Combining these two equations gives us the desired result.
\end{solution}

\begin{minipage}{.45\linewidth}
    \gene{https://artofproblemsolving.com/community/c374081h1619335}{Vietnamese
        TST 2018 P6.a Generalization}{
        Let $ABC$ be a triangle. The points $D,$
        $E,$ $F$ are on the lines $BC,$ $CA,$ $AB$ respectively. The circles $(AEF),$
        $(CFD),$ $(CDE)$ have a common point $P.$ A circle $(K)$ passes through $P,$
        $D$ meet $DE,$ $DF$ again at $Q,$ $R$ respectively. Prove that the circles
        $(DBQ),$ $(DCR)$ and $(DEF)$ are coaxial.
    }

    \begin{solution}[Inversion]
        Invert around $ D $, and use Pappu's Theorem as in\\
        \autoref{problem:Vietnamese TST 2018 P6.a}.
    \end{solution}
\end{minipage}\hfill%
\begin{minipage}{.52\linewidth}
    \figdf{}{Vietnamese_TST_2018_P6_a_gene}{\hrf{VNTST2018P6a_Gene}{Vietnamese TST
    2018 P6.a Generalization}}
\end{minipage}

\rem{
    The synthetic solution of \autoref{problem:Vietnamese TST 2018 P6.a}
    can't be reproduced here maybe because here we don't have $ A, P, D $
    collinear, and we can't have harmonic quadrilaterals either.
}

\theo{https://en.wikipedia.org/wiki/Poncelet's_closure_theorem}{Poncelet's
    Porism}{
    Poncelet's porism (sometimes referred to as Poncelet's closure
    theorem) states that whenever a polygon is inscribed in one conic section and
    circumscribes another one, the polygon must be part of an infinite family of
    polygons that are all inscribed in and circumscribe the same two conics.
}

\prob{https://artofproblemsolving.com/community/c6h1181536p5720184}{IMO 2013
    P3}{M}{
    Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent
    to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$
    on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively.
    Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle
    of triangle $ABC$. Prove that triangle $ABC$ is right-angled.
}

\solu{
    Straightforward use of \autoref{lemma:excenter_touchpoint_bigarc-midpoints}
}

\prob{https://artofproblemsolving.com/community/c74453h1225408_some_geometric_problems}{buratinogigle's
    proposed probs for Arab Saudi team 2015}{E}{
    Let $ABC$ be acute triangle with
    $AB < AC$ inscribed circle $(O)$. Bisector of $\angle BAC$ cuts $(O)$ again at
    $D$. $E$ is reflection of $B$ through $AD$. $DE$ cuts $BC$ at $F$. Let $(K)$
    be circumcircle of triangle $BEF$. $BD, EA$ cut $(K)$ again at $M, N$, reps.
    Prove that $\angle BMN = \angle KFM$.
}

\fig{.5}{SATST2015proposed_by_bura/derakynay1134-8}{}

\prob{https://artofproblemsolving.com/community/c6h54506p340041}{USAMO 1999
    P6}{E}{
    Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$. The
    inscribed circle $\omega$ of triangle $BCD$ meets $CD$ at $E$. Let $F$ be a
    point on the (internal) angle bisector of $\angle DAC$ such that $EF \perp
    CD$. Let the circumscribed circle of triangle $ACF$ meet line $CD$ at $C$ and
    $G$. Prove that the triangle $AFG$ is isosceles.
}

\prob{https://artofproblemsolving.com/community/c6h1619730p10134424}{Serbia
    2018 P1}{E}{
    Let $\triangle ABC$ be a triangle with incenter $I$. Points $P$
    and $Q$ are chosen on segments $BI$ and $CI$ such that $2\angle  PAQ=\angle
    BAC$. If $D$ is the touch point of incircle and side $BC$ prove that $\angle
    PDQ=90$.
}

\solu{Straightforward Trig application.}

\prob{https://artofproblemsolving.com/community/c6h1628676p10217476}{Iran TST
    T2P5}{E}{
    Let $\omega$ be the circumcircle of isosceles triangle $ABC$
    ($AB=AC$). Points $P$ and $Q$ lie on $\omega$ and $BC$ respectively such that
    $AP=AQ$ .$AP$ and $BC$ intersect at $R$. Prove that the tangents from $B$ and
    $C$ to the incircle of $\triangle AQR$ (different from $BC$) are concurrent on
    $\omega$.
}

\prob{}{}{M}{
    Let a point $ P $ inside of $ \triangle ABC $ be such that the
    following condition is satisfied \[\frac{AP+BP}{AB} = \frac{BP+CP}{BC} =
    \frac{CP+AP}{CA}\]	Lines $ AP, BP, CP $ intesect the circumcirle again at $
    A', B', C' $. Prove that $ ABC $ and $ A', B', C' $ have the same incircle.
}

\solu{
    After finiding the point $ P $, we get a lot of ideas.
    \figdf{.8}{itti_same_incircle}{two lines are parallel}
}

\prob{https://artofproblemsolving.com/community/c6h1623012p10163453}{Iran TST
    2018 P3}{EM}{
    In triangle $ABC$ let $M$ be the midpoint of $BC$. Let $\omega$
    be a circle inside of $ABC$ and is tangent to $AB,AC$ at $E,F$, respectively.
    The tangents from $M$ to $\omega$ meet $\omega$ at $P,Q$ such that $P$ and $B$
    lie on the same side of $AM$. Let $X \equiv PM \cap BF $ and $Y \equiv QM \cap
    CE $. If $2PM=BC$ prove that $XY$ is tangent to $\omega$.
}

\solu{Work backwards}

\prob{https://artofproblemsolving.com/community/c6h1623417p10167655}{Iran TST
    2018 P4}{E}{
    Let $ABC$ be a triangle ($\angle A\neq 90^\circ$). $BE,CF$ are the
    altitudes of the triangle. The bisector of $\angle A$ intersects $EF,BC$ at
    $M,N$. Let $P$ be a point such that $MP\perp EF$ and $NP\perp BC$. Prove that
    $AP$ passes through the midpoint of $BC$.
}

\prob{https://artofproblemsolving.com/community/c6h1662902p10561154}{APMO 2018
    P1}{E}{
    Let $H$ be the orthocenter of the triangle $ABC$. Let $M$ and $N$ be
    the midpoints of the sides $AB$ and $AC$, respectively. Assume that $H$ lies
    inside the quadrilateral $BMNC$ and that the circumcircles of triangles $BMH$
    and $CNH$ are tangent to each other. The line through $H$ parallel to $BC$
    intersects the circumcircles of the triangles $BMH$ and $CNH$ in the points
    $K$ and $L$, respectively. Let $F$ be the intersection point of $MK$ and $NL$
    and let $J$ be the incenter of triangle $MHN$. Prove that $F J = F A$.
}

\prob{https://artofproblemsolving.com/community/c6h155710p875026}{ISL 2006
    G6}{E}{
    Circles $ w_{1}$ and $ w_{2}$ with centres $ O_{1}$ and $ O_{2}$ are
    externally tangent at point $ D$ and internally tangent to a circle $ w$ at
    points $ E$ and $ F$ respectively. Line $ t$ is the common tangent of $ w_{1}$
    and $ w_{2}$ at $ D$. Let $ AB$ be the diameter of $ w$ perpendicular to $ t$,
    so that $ A, E, O_{1}$ are on the same side of $ t$. Prove that lines $
    AO_{1}$, $ BO_{2}$, $ EF$ and $ t$ are concurrent.
}

\solu{\hrf{lemma:concurrent_lines_in_incenter}{This}}

\lem{Tangential Quadrilateral Incenters}{
    Let $ ABCD $ be a tangential
    quatrilateral. Let $ I_1, I_2 $ be the incenters of $ \triangle ABD, \triangle
    BCD $. Then $ (I_1), (I_2) $ is tangent to $ BD $ at the same point.
    \fig{.5}{tangential_quad_incenters}{}
}

\prob{https://artofproblemsolving.com/community/c6h21758p140322}{Four
    Incenters in a Tangential Quadrilateral}{E}{
    Let $ABCD$ be a quadrilateral.
    Denote by $X$ the point of intersection of the lines $AC$ and $BD$. Let
    $I_{1}$, $I_{2}$, $I_{3}$, $I_{4}$ be the centers of the incircles of the
    triangles $XAB$, $XBC$, $XCD$, $XDA$, respectively. Prove that the
    quadrilateral $I_{1}I_{2}I_{3}I_{4}$ has a circumscribed circle if and only if
    the quadrilateral $ABCD$ has an inscribed circle.
}

\solu{
    There is a lot going on in this figure, firstly, the $ J_1, J_2 $ and $
    M $, then $ K $, then $ \angle I_4ME = \angle I_3ME $. Connecting them
    with the \hyperref[Incircle Touchpoint and Cevian]{lemma}.

    \figdf{1}{tangential_quad_four_incenters}{}
}

\prob{}{Geodip}{E}{
    Let $ G $ be the centeroid. Dilate $ \odot I $ from $ G $
    with constant $ -2 $ to get $ I'$. Then $ I' $ is tangent to the circumcircle.
    \figdf{.5}{nice_prob_by_geodip}{}
}

\theo{http://mathworld.wolfram.com/FuhrmannCircle.html}{Fuhrmann Circle}{
    Let $ X', Y', Z' $ be the midpoints of the arcs not containing $ A, B, C $ of $
    \odot ABC $. Let $ X, Y, Z $ be the reflections of these points on the
    sides. Then $ \odot XYZ $ is called the \textbf{Fuhrmann Circle}. The
    orthocenter $ H $ and the nagel point $N$ lies on this circle, and $ HN $
    is a diameter of this circle.

    Furthermore, $ AH, BH, CH $ cut the circle for the second time at a
    distance $ 2r $ from the vertices.

    \figdf{1}{fuhrmann_circle}{Fuhrmann Circle}
}

\prob{https://artofproblemsolving.com/community/c6h213443p1178421}{Iran TST
    2008 P12}{E}{
    In the acute-angled triangle $ ABC$, $ D$ is the intersection of
    the altitude passing through $ A$ with $ BC$ and $ I_a$ is the excenter of the
    triangle with respect to $ A$. $ K$ is a point on the extension of $ AB$ from
    $ B$, for which $ \angle AKI_a=90^\circ+\frac 34\angle C$. $ I_aK$ intersects
    the extension of $ AD$ at $ L$. Prove that $ DI_a$ bisects the angle $ \angle
    AI_aB$ iff $ AL=2R$. ($ R$ is the circumradius of $ ABC$)
}

\solu{}

\lem{Polars in Incircle}{
    In the acute angled triangle $ABC$, $I$ is the incenter and $DEF$ is the
    touch triangle. Let $EF$ meet $\odot ABC$ at $P, Q$ such that $E$ lies
    inside $F, Q$. If $QD$ meets $\odot ABC$ for the second time at $U$, prove
    that $AU$ is the polar line of $P$ wrt $\left(I\right)$.
}

\proof{[Projective]
    We have:
    \begin{align*}
        \left(B, C; D, EF\cap BC\right) &= Q\left(B, C; P, U\right) \\
                                        &= A(E, F; P, U)\\
                                        &= -1
    \end{align*}

    Which means $\left(P, AU\cap EF; E, F\right)$ is harmonic.
    \figdf{.5}{ISL2019G6_lem}{}
}

\prob{}{ISL 2019 G6}{HM}{
    In the acute angled triangle $ABC$, $I$ is the incenter and $DEF$ is the
    touch triangle. Let $EF$ meet $\odot ABC$ at $P, Q$ such that $E$ lies
    inside $F, Q$. Prove that \[\angle APD + \angle AQD = \angle PIQ\]
}

\begin{solution}
    Since $PI \cap AU$ at $X$ from \autoref{lemma:Polars in Incircle}, we have
    $AFPX$ is cyclic. And so
    \[\angle FAX = \angle FIX\]
    \figdf{.5}{ISL2019G6}{}
    After some more angle chasing, we reach our goal.
\end{solution}

\newpage
\subsection{Feurbach Point}

\den{Feurbach Point}{
    The point where the nine point circle touches the incircle is called the
    \emph{Feurbach Point}.
}

\thmbox{}
{It Exists!}{
    The nine point circle touches the incircle and the excircles.
}

\begin{prooof}[Inversion]
    Let $D, D'$ be the incircle and the $A$-excircle touchpoints with $BC$.
    Let $M, N, P$ be the midpoints of $BC, CA, AB$ resp. Also let $B'C'$ be
    the reflection of $BC$ on $AI$. Now let $N', P'$ be the intersection
    points of $MN, MP$ with $B'C'$. \\

    We invert around $M$ with radius $MD = \frac{b-c}{2}$. We prove that the
    image of $\odot MNP$ after the inversion is $B'C'$. And since $(I)$ and
    $(I_a)$ are orthogonal to $(M)$, we will be done.
    \figdf{.8}{feurbach_inversion}{}
    Wlog, assume that $b \ge c$.
    \[\begin{aligned}
        B'N &= AB' - AN = c - \frac{b}{2} &\quad NN' &= B'N \cdot\frac{AC'}{AB'}\\[.5em]
        MN' &= MN - NN' &\quad &= \frac{2c-b}{2}\cdot \frac{b}{2}\\[1em]
        MN'\cdot MN &= \frac{c}{2}\left(\frac{c}{2} -
        \frac{b}{c}\cdot\frac{2c-b}{2}\right) &= \frac{b-c}{2}^2
    \end{aligned}\]
    Which concludes the proof.
\end{prooof}

\thmbox{}
{Construction of Feurbach Point}{
    Let $D$ be the incenter touch point with $BC$. Let $M, L$ be the midpoints
    of $BC$ and $AI$. Let $D_1, D'$ be the reflections of $D$ over $I$ and
    $M$. Let $K, P$ be the refecltions of $D_1, D$ over $L$ and $AI$. Let $Q$
    be the intersection of $AD_1$ with the incircle.\\

    Then $D_1K$ and $MP$ meet at $F$ on the incircle, which is the Feurbach
    Point of $\triangle ABC$.
}

\figdf{.7}{feurbach_construction}{}

\begin{prooof}
    It is easy to see that the tangents at $P$ and $M$ to the incircle and the
    nine point circle are parallel. So if we let $F = MP\cap \left(I\right)$,
    then we have $F$ is the Feurbach point. \\

    And since $MQ$ is tangent to $(I)$, we also have $\left(F, P; D,
    Q\right)=-1$. But notice that
    \[\begin{aligned}
        D_1(A, I; L, P) &= D_1(K, P; Q, I)\\
                        &=-1
    \end{aligned}\]
    So $D_1K$ passes through $F$.
\end{prooof}

\newpage
\subsection{Assorted Diagrams}

\figdf{.7}{circles_with_arc_midpoints}{The smaller circles touches the side
and the circumcircle}