\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{asydef}
	defaultpen(fontsize(9pt));
	pointfontsize = 9;
	real[] default_ticks = {-4,-3,-2,-1,0,1,2,3,4};
	void cplane(real theta=-45, real[] T = default_ticks, real xmin = -5, real xmax = 5, real S = 3) {
		graph.xaxis("Re", xmin, xmax, graph.Ticks(format="%", Ticks=T, Size=S), Arrows);
		graph.yaxis("Im", xmin, xmax, graph.Ticks(format="%", Ticks=T, Size=S), Arrows);
		Drawing("0", origin, dir(theta));
		markscalefactor *= xmax;
	}
	markscalefactor=0.01;
\end{asydef}

\begin{document}
\title{Bashing Geometry with Complex Numbers}
\author{Evan Chen}
\date{August 29, 2015}
\maketitle

\begin{abstract}
	This is a (quick) English translation of the complex numbers note
	I wrote for Taiwan IMO 2014 training.
	Incidentally I was also working on an airplane.
\end{abstract}


\section{The Complex Plane}
Let $\CC$ and $\RR$ denote the set of complex and real numbers, respectively.

Each $z \in \CC$ can be expressed as
\[ z = a+bi = r \left( \cos \theta + i \sin \theta \right) = re^{i \theta} \]
where $a,b,r,\theta \in \RR$ and $0 \le \theta < 2\pi$.
We write $\left\lvert z \right\rvert = r = \sqrt{a^2+b^2}$ and $\arg z = \theta$.

More importantly, each $z$ is associated with a conjugate $\ol z = a-bi$.
It satisfies the properties
\begin{align*}
	\ol {w \pm z} &= \ol w \pm \ol z \\
	\ol {w \cdot z} &= \ol w \cdot \ol z  \\
	\ol {w / z} &= \ol w / \ol z \\
	\left\lvert z \right\rvert^2 &= z \cdot \ol z
\end{align*}
Note that $z \in \RR \iff z = \ol z$ and $z \in i \RR \iff z + \ol z = 0$.

\begin{figure}[ht]
	\centering
	\begin{asy}
		size(7cm);
		cplane(135);
		pair z = (3,4);
		pair p = (z.x,0);
		pair x = (z.x,-z.y);
		Drawing("z = 3+4i", z, dir(45));
		Drawing("\overline z = 3-4i", x, dir(-45));
		Drawing("-1-2i", (-1,-2), dir(225));
		draw(x--origin--z);
		MP(" \left\lvert z \right\rvert = 5", 0.7*z, rotate(90)*dir(z));
		draw(z--x, dotted);
		MarkAngle("\theta", p, origin, z, 1);
		draw(rightanglemark(z,p,origin));
	\end{asy}
	\caption{Points $z = 3+4i$ and $-1-2i$; $\ol z = 3-4i$ is the conjugate.}
	\label{fig:explain_complex_plane}
\end{figure}

We represent every point in the plane by a complex number.
In particular, we'll use a capital letter (like $Z$) to denote the point associated to a complex number (like $z$).


Complex numbers add in the same way as vectors.
The multiplication is more interesting: for each $z_1, z_2 \in \CC$ we have
\[
	\left\lvert z_1z_2 \right\rvert = \left\lvert z_1 \right\rvert \left\lvert z_2 \right\rvert 
	\text{ and }
	\arg z_1z_2 = \arg z_1 + \arg z_2.
\]
This multiplication lets us capture a geometric structure.
For example, for any points $Z$ and $W$ we can express rotation of $Z$ at $W$ by $90\dg$ as
\[ z \mapsto i(z-w) + w. \]

\begin{figure}[ht]
	\centering
	\begin{asy}
		size(11cm);
		cplane();
		pair z = (3,4);
		pair iz = rotate(90)*z;
		Drawing("z = 3+4i", z, dir(90));
		Drawing("iz = 4-3i", iz, dir(90));
		draw(z--origin--iz,Arrows(SimpleHead,9));
		draw(rightanglemark(z,origin,iz));
		picture pic = CC();
		cplane();
		pair z = Drawing("z", (-1,-2), dir(0));
		pair w = Drawing("w", (-2,-4), dir(-90));
		pair f = Drawing("i(z-w)+w", rotate(90)*(z-w)+w, dir(100));
		pair zw = Drawing("z-w", z-w, dir(45));
		pair izw = Drawing("i(z-w)", rotate(90)*(z-w), dir(135));
		draw(f--w--z, Arrows(SimpleHead,9));
		draw(izw--origin--(z-w), Arrows(SimpleHead,9));
		add(shift((-11,0))*pic);
	\end{asy}
	\caption{$z \mapsto i(z-w) + w$.}
	\label{fig:complex_rotate}
\end{figure}
	

\section{Elementary Propositions}
First, some fundamental formulas:

\begin{proposition}
	Let $A$, $B$, $C$, $D$ be pairwise distinct points.
	Then $\ol{AB} \perp \ol{CD}$ if and only if $\frac{d-c}{b-a} \in i \RR$; i.e.\
	\[ \frac{d-c}{b-a} + \ol{\left( \frac{d-c}{b-a} \right)} = 0. \]
	\label{lem:complex_perp}
\end{proposition}
\begin{proof}
	It's equivalent to 
	$\frac{d-c}{b-a} \in i \RR
	\iff \arg \left( \frac{d-c}{b-a} \right) \equiv \pm 90\dg \iff \ol{AB} \perp \ol{CD}$。
\end{proof}

\begin{figure}[ht]
	\centering
	\begin{asy}
		size(12cm);
		cplane();
		pair a = Drawing("a", (2,2), dir(120));
		pair b = Drawing("b", (3,4), dir(120));
		pair c = Drawing("c", (-1,1), dir(90));
		pair d = Drawing("d", (-4,2.5), dir(90));
		draw(a--b); draw(c--d);
		add(shift(-11)*CC());
		cplane();
		pair x = Drawing("b-a", b-a, dir(90));
		pair y = Drawing("d-c", d-c, dir(90));
		draw(x--origin--y);
	\end{asy}
	\caption{$\ol{AB} \perp \ol{CD} \iff \frac{d-c}{b-a} \in i \RR$.}
\end{figure}

\begin{proposition}
	Let $A$, $B$, $C$ be pairwise distinct points.
	Then $A$, $B$, $C$ are collinear if and only if $\frac{c-a}{c-b} \in \RR$; i.e.
	\[ \frac{c-a}{c-b} = \ol{\left( \frac{c-a}{c-b} \right)}. \]
	\label{prop:complex_collin}
\end{proposition}
\begin{proof}
	Similar to the previous one.
\end{proof}

\begin{proposition}
	Let $A$, $B$, $C$, $D$ be pairwise distinct points.
	Then $A$, $B$, $C$, $D$ are concyclic if and only if
	\[ \frac{c-a}{c-b} : \frac{d-a}{d-b} \in \RR. \]
\end{proposition}
\begin{proof}
	It's not hard to see that
	$\arg \left( \frac{c-a}{c-b} \right) = \angle ACB$
	and $\arg \left( \frac{d-a}{d-b} \right) = \angle ADB$.
	(Here angles are directed).
\end{proof}

Now, let's state a more commonly used formula.

\begin{lemma}[Reflection About a Segment]
	Let $W$ be the reflection of $Z$ across $\ol{AB}$. Then
	\[ w = \frac{(a-b) \ol z + \ol{a}b - a\ol{b}}{\ol a - \ol b}. \]
	\label{lem:complex_reflect}
\end{lemma}
Of course, it then follows that the foot from $Z$ to $\ol{AB}$ is exactly $\half (w+z)$.

\begin{figure}[ht]
	\centering
	\begin{asy}
		size(4cm);
		real[] T = {0,2,4};
		cplane(T=T, xmin=-1.9,xmax=5);
		Drawing("1", (2,0), dir(-90));
		pair a = Drawing("a", (1,1), dir(-90));
		pair b = Drawing("b", (4,3), dir(10));
		pair z = Drawing("z", (2,3), dir(90));
		pair w = Drawing("w", 2*foot(z,a,b)-z, dir(-45));
		draw(a--b);
		draw(z--w, dotted);
	\end{asy}
	\quad
	\begin{asy}
		size(4cm);
		real[] T = {0,2,4};
		cplane(T=T, xmin=-1.9,xmax=5);
		Drawing("1", (2,0), dir(-90));
		pair a = origin;
		pair b = Drawing("b-a", (3,2), dir(10));
		pair z = Drawing("z-a", (1,2), dir(90));
		pair w = Drawing("w-a", 2*foot(z,a,b)-z, dir(45));
		draw(a--b);
		draw(z--w, dotted);
	\end{asy}
	\quad
	\begin{asy}
		size(4cm);
		real[] T = {0,1,2};
		cplane(T=T, xmin=-0.95,xmax=2.50);
		pair a = origin;
		pair x = (3,2);
		pair b = Drawing("1", (3,2)/x, dir(-90));
		pair z = Drawing("\frac{z-a}{b-a}", (1,2)/x, dir(70));
		pair w = Drawing("\frac{w-a}{b-a}", 2*foot(z,a,b)-z, dir(-70));
		draw(a--b);
		draw(z--w, dotted);
		draw(origin--(1,0), black+1);
	\end{asy}
	\caption{The reflection of $Z$ across $\ol{AB}$.}
	\label{fig:proof_reflect}
\end{figure}

\begin{proof}
	According to \Cref{fig:proof_reflect} we obtain
	\[ \frac{w-a}{b-a} = \ol{\left( \frac{z-a}{b-a} \right)} = \frac{\ol z - \ol a}{\ol b - \ol a}. \]
	From this we derive $w = \frac{(a-b)\ol z + \ol a b - a \ol b}{\ol a-  \ol b}$.
\end{proof}

Here are two more formulas.

\begin{theorem}
	[Complex Shoelace] Let $A$, $B$, $C$ be points.
	Then $\triangle ABC$ has signed area
	\[
		\frac{i}{4}
		\left\lvert
		\begin{array}{ccc}
			a & \ol a & 1 \\
			b & \ol b & 1 \\
			c & \ol c & 1 \\
		\end{array}
		\right\rvert.
	\]
	\textbf{In particular, $A$, $B$, $C$ are collinear if and only if this determinant vanishes.}
	\label{thm:complex_shoelace}
\end{theorem}
\begin{proof}
	Cartesian coordinates.
\end{proof}
Often, \Cref{thm:complex_shoelace} is easier to use than \Cref{prop:complex_collin}.

Actually, we can even write down the formula for an arbitrary intersection of lines.
\begin{proposition}
	\label{prop:complex_intersect}
	Let $A$, $B$, $C$, $D$ be points. Then lines $AB$ and $CD$ intersect at
	\[ \frac{ (\bar a b - a \bar b )(c-d) - (a-b)(\bar c d - c \bar d) }{(\bar a - \bar b )(c-d) - (a-b)(\bar c - \bar d)}. \]
\end{proposition}
But unless $d=0$ or $a$, $b$, $c$, $d$ are on the unit circle, this formula is often too messy to use.


\section{The Unit Circle, and Triangle Centers}
\label{sec:unitcircle}
On the complex plane, the \textbf{unit circle} is of critical importance.
Indeed if $\left\lvert z \right\rvert = 1$ we have \[ \ol z = \frac 1z. \]
Using the above, we can derive the following lemmas.

\begin{lemma}
	If $ \left\lvert a \right\rvert = \left\lvert b \right\rvert = 1$ and $z \in \CC$,
	then the reflection of $Z$ across $\ol{AB}$ is
	$a+b-ab \ol z$,
	and the foot from $Z$ to $\ol{AB}$ is
	\[ \half \left( z+a+b -ab \ol z \right). \]
\end{lemma}
\begin{lemma}
	If $A$, $B$, $C$, $D$ lie on the unit circle then
	the intersection of $\ol{AB}$ and $\ol{CD}$ is given by
	\[ \frac{ab(c+d)-cd(a+b)}{ab-cd}. \]
\end{lemma}

These are much easier to work with than the corresponding formulas in general.
We can also obtain the triangle centers immediately:
\begin{theorem}
	Let $ABC$ be a triangle center, and assume that the circumcircle of
	$ABC$ coincides with the unit circle of the complex plane.
	Then the circumcenter, centroid, and orthocenter of $ABC$
	are given by $0$, $\frac 13 (a+b+c)$, $a+b+c$, respectively.
\end{theorem}
Observe that the Euler line follows from this.

\begin{proof}
	The results for the circumcenter and centroid are immediate.
	Let $h = a+b+c$.
	By symmetry it suffices to prove $\ol{AH} \perp \ol{BC}$.
	We may set
	\[ z  = \frac{h-a}{b-c} = \frac{b+c}{b-c}. \]
	Then
	\[ \ol z = \ol{\left( \frac{b+c}{b-c} \right)}
		= \frac{\ol b + \ol c}{\ol b - \ol c}
		= \frac{\frac 1b + \frac 1c}{\frac 1b - \frac 1c}
		= \frac{c+b}{c-b} = -z \]
	so $z \in i \RR$ as desired.
\end{proof}

We can actually even get the formula for the incenter.
\begin{theorem}
	Let triangle $ABC$ have incenter $I$ and circumcircle $\Gamma$.
	Lines $AI$, $BI$, $CI$ meet $\Gamma$ again at $D$, $E$, $F$.
	If $\Gamma$ is the unit circle of the complex plane
	then there exists $x,y,z \in \CC$ satisfying
	\[ a=x^2, b=y^2, c=z^2 \text{ and } d=-yz, e=-zx, f=-xy. \]
	Note that $\left\lvert x \right\rvert = \left\lvert y \right\rvert = \left\lvert z \right\rvert = 1$.
	Moreover, the incenter $I$ is given by $-(xy+yz+zx)$.
\end{theorem}
\begin{proof}
	Show that $I$ is the orthocenter of $\triangle DEF$.
\end{proof}

\section{Some Other Lemmas}
\begin{lemma}
	Let $A$, $B$ be on the unit circle
	and select $P$ so that $\ol{PA}$, $\ol{PB}$ are tangents.
	Then \[ p = \frac{2}{\ol a + \ol b} = \frac{2ab}{a+b}. \]
\end{lemma}
\begin{proof}
	Let $M$ be the midpoint of $\ol{AB}$ and set $O = 0$.
	One can show $OM \cdot OP = 1$ and that $O$, $M$, $P$ are collinear;
	the result follows from this.
\end{proof}
\begin{figure}[ht]
	\centering
	\begin{asy}
		size(4cm);
		draw(unitcircle);
		pair A = Drawing("a", dir(110), dir(110));
		pair B = Drawing("b", dir(30), dir(30));
		pair P = 2/conj(A+B);
		Drawing("\frac{2ab}{a+b}", P, dir(P));
		draw(A--P--B);
		draw(A--B, dashed);
		draw(D(origin)--P, dashed);
		dot( (A+B)/2 );
	\end{asy}
	\caption{Two tangents. $p = \frac{2}{\ol a + \ol b}$.}
\end{figure}

\begin{lemma}
	For any $x$, $y$, $z$, the circumcenter of $\triangle XYZ$ is given by
	\[ \left\lvert
		\begin{array}{ccc}
			x & x\bar x & 1 \\
			y & y\bar y & 1 \\
			z & z\bar z & 1
		\end{array}
	\right\rvert
	\div
	\left\lvert
		\begin{array}{ccc}
			x & \bar x & 1 \\
			y & \bar y & 1 \\
			z & \bar z & 1
		\end{array}
	\right\rvert. \]
\end{lemma}
This formula is often easier to apply if we shift $z$ to the point $0$ first, then shift back afterwards.

\section{Examples}
\begin{example}
	[MOP 2006]
	Let $H$ be the orthocenter of triangle $ABC$.
	Let $D$, $E$, $F$ lie on the circumcircle of $ABC$ such that $\ol{AD} \parallel \ol{BE} \parallel \ol{CF}$.
	Let $S$, $T$, $U$ respectively denote the reflections of $D$, $E$, $F$ across $\ol{BC}$, $\ol{CA}$, $\ol{AB}$.
	Prove that points $S$, $T$, $U$, $H$ are concyclic.
\end{example}
\begin{proof}
	Let $(ABC)$ be the unit circle and $h = a+b+c$.
	WLOG, $\ol{AD}$, $\ol{BE}$,  $\ol{CF}$ are perpendicular to the real axis (rotate appropriately); thus $d = \ol a$ and so on.
	Thus $s = b+c - bc \ol d = b+c-abc$ and so on; we now have
	\[ \frac{s-t}{s-u}=\frac{b-a}{c-a}\quad\text{and}\quad\frac{h-t}{h-u}=\frac{b+abc}{c+abc}. \]
	Compute \[ \frac{s-t}{s-u} : \frac{h-t}{h-u} = \frac{(b-a)(c+abc)}{(c-a)(b+abc)}=\frac{\left(\frac{1}{b}-\frac{1}{a}\right)\left(\frac{1}{c}+\frac{1}{abc}\right)}{\left(\frac{1}{c}-\frac{1}{a}\right)\left(\frac{1}{b}+\frac{1}{abc}\right)} \implies \frac{s-t}{s-u}:\frac{h-t}{h-u} \in \RR \]
	as desired.
\end{proof}


\begin{example}
	[Taiwan TST 2014]
	In $\triangle ABC$ with incenter $I$,
	the incircle is tangent to $\ol{CA}$, $\ol{AB}$ at $E$, $F$.
	The reflections of $E$, $F$ across $I$ are $G$, $H$.
	Let $Q$ be the intersection of $\ol{GH}$ and $\ol{BC}$, and let $M$ be the midpoint of $\ol{BC}$.
	Prove that $\ol{IQ}$ and $\ol{IM}$ are perpendicular.
\end{example}
\begin{figure}[ht]
	\centering
	\begin{asy}
		pair A = dir(110), B = dir(210), C = dir(330);
		Drawing("A", A, A);
		Drawing("B", B, dir(-90));
		Drawing("C", C, dir(-90));
		pair I = Drawing("I", incenter(A,B,C), dir(90));
		pair E = Drawing("E", foot(I,C,A), dir(C+A-I));
		pair F = Drawing("F", foot(I,A,B), dir(A+B-I));
		pair D = Drawing("D", foot(I,B,C), dir(B+C-I));
		pair G = Drawing("G", 2*I-E, dir(I-E));
		pair H = Drawing("H", 2*I-F, dir(I-F));
		pair Q = Drawing("Q", extension(B,C,G,H), dir(-90));
		pair M = Drawing("M", midpoint(B--C), dir(-90));
		draw(A--B--C--cycle);
		draw(H--Q);
		draw(incircle(A,B,C));
		draw(B--Q--I--M);
	\end{asy}
\end{figure}
\begin{soln}
	Let $D$ be the foot from $I$ to $\ol{BC}$, and set $(DEF)$ as the unit circle.
	(This lets us exploit the results of \Cref{sec:unitcircle}.)
	Thus $\left\lvert d \right\rvert = \left\lvert e \right\rvert = \left\lvert f \right\rvert = 1$,
	and moreover $g = -e$, $h = -f$.
	Let $x = \ol d = \frac 1d$ and define $y$, $z$ similarly.
	Then
	\[ b = \frac{2}{\ol d + \ol f} = \frac{2}{x+z}. \]
	Similarly, $c = \frac{2}{x+y}$, so
	\[ m = \half(b+c) = \frac{1}{x+y} + \frac{1}{x+z} = \frac{2x+y+z}{(x+y)(x+z)}. \]
	Next, we have $Q = DD \cap GH$, which implies
	\[ q = \frac{dd(g+h)-gh(d+d)}{d^2-gh}
		= \frac{\frac{1}{x^2} \left( -\frac1y-\frac1z \right) - \frac{1}{yz} \frac{2}{x}}{\frac{1}{x^2} - \frac{1}{yz}}
		= \frac{2x+y+z}{x^2-yz}. \]
	so
	\[ m/q = \frac{x^2-yz}{(x+y)(x+z)}. \]
	Now,
	\[ \ol{m/q} = \frac{\frac{1}{x^2}-\frac{1}{yz}}{\left( \frac1x+\frac1y \right)\left( \frac1x+\frac1z \right)} = \frac{yz-x^2}{(x+y)(x+z)} = -m/q \]
	thus $m/q \in i \RR$, as desired.
\end{soln}

\begin{example}
	[USAMO 2012]
	Let $P$ be a point in the plane of $\triangle ABC$, and $\gamma$ a line through $P$.
	Let $A', B', C'$ be the points where the reflections of lines $PA, PB, PC$ with
	respect to $\gamma$ intersect lines $BC, AC, AB$ respectively.
	Prove that $A', B', C'$ are collinear.
\end{example}

\begin{figure}[ht]
	\centering
	\begin{asy}
		size(4cm);
		pair A = dir(110), B = dir(210), C = dir(330);
		Drawing("A", A, A);
		Drawing("B", B, dir(-90));
		Drawing("C", C, dir(-90));
		draw(A--B--C--cycle);
		pair P = Drawing("P", 0.1*dir(40), dir(-35));
		pair X1 = P + .5 * dir(10);
		pair X2 = P - .5 * dir(10);
		Drawing(L(X1,X2), dashed, Arrows);
		pair T = 2*foot(A,X1,X2)-A;
		pair A1 = Drawing("A'", extension(P,T,B,C), dir(-40));
		draw(A--P--A1);
	\end{asy}
\end{figure}

\begin{soln}
	Let $p=0$ and set $\gamma$ as the real line.
	Then $A'$ is the intersection of $bc$ and $p\bar a$.
	So, using \Cref{prop:complex_intersect} we get
	\[ a' = \frac{\bar a(\bar b c - b \bar c)}{(\bar b - \bar c)\bar a-(b-c) a}. \]
	Note that
	\[ \bar a' = \frac{a (b \bar c - \bar b c)}{(b-c) a - (\bar b - \bar c)\bar a}. \]
	Thus by \Cref{thm:complex_shoelace}, it suffices to prove
	\[ 0 = \left\lvert
	\begin{array}{ccc}
		\frac{\bar a(\bar b c-b \bar c)}{(\bar b - \bar c)\bar a - (b-c) a} & \frac{a (b \bar c - \bar b c)}{(b-c) a - (\bar b - \bar c) \bar a} & 1 \\
		\frac{\bar b(\bar c a-c \bar a)}{(\bar c - \bar a)\bar b - (c-a) b} & \frac{b (c \bar a - \bar c a)}{(c-a) b - (\bar c - \bar a) \bar b} & 1 \\
		\frac{\bar c(\bar a b-a \bar b)}{(\bar a - \bar b)\bar c - (a-b) c} & \frac{c (a \bar b - \bar a b)}{(a-b) c - (\bar a - \bar b)\bar c} & 1
	\end{array}
	\right\rvert.
	\]
	This is equivalent to
	\[ 0 = \left\lvert
	\begin{array}{ccc}
		\bar a(\bar b c -b \bar c) & a (\bar b c - b \bar c) & (\bar b - \bar c) \bar a - (b-c) a \\
		\bar b(\bar c a -c \bar a) & b (\bar c a - c \bar a) & (\bar c - \bar a) \bar b - (c-a) b \\
		\bar c(\bar a b -a \bar b) & c (\bar a b - a \bar b) & (\bar a - \bar b) \bar c - (a-b) c \\
	\end{array}
	\right\rvert. \] 
	Evaluating the determinant gives
	\[
	\sum_{\text{cyc}} ((\bar b - \bar c) \bar a - (b-c) a) \cdot - \left\lvert
		\begin{array}{cc}
			b & \bar b \\
			c & \bar c
		\end{array}
		\right\rvert \cdot \left( \bar c a - c \bar a \right)\left( \bar a b - a \bar b \right)
	\] or, noting the determinant is $b\bar c - \bar b c$ and factoring it out,
	\[
		(\bar b c - c \bar b)(\bar c a - c \bar a)(\bar a b - a \bar b) \sum_{\text{cyc}} \left( ab - ac + \bar c \bar a - \bar b \bar a \right) = 0.
	\qedhere \]
\end{soln}

\begin{example}
	[Taiwan TST Quiz 2014]
	Let $I$ and $O$ be the incenter and circumcenter of $ABC$.
	A line $\ell$ is drawn parallel to $\ol{BC}$ and tangent to the incircle of $ABC$.
	Let $X$, $Y$ be on $\ell$ so that $I$, $O$, $X$ are collinear and $\angle XIY = 90\dg$.
	Show that $A$, $X$, $O$, $Y$ are concyclic.
\end{example}
\begin{figure}[ht]
	\centering
	\begin{asy}
		pair A = dir(110), B = dir(220), C = dir(320);
		pair A = Drawing("A", A, A);
		pair B = Drawing("B", B, B);
		pair C = Drawing("C", C, C);
		draw(A--B--C--cycle);
		pair O = Drawing("O", origin, dir(-90));
		draw(unitcircle);
		draw(incircle(A,B,C), dotted);
		pair I = Drawing("I", incenter(A,B,C), dir(-90));
		pair B1 = 2*I-B, C1 = 2*I-C;
		pair X = Drawing("X", extension(B1,C1,O,I), dir(20));
		pair Y = Drawing("Y", 2*foot(circumcenter(A,X,O),X,2*I-foot(I,B,C))-X, dir(190));
		draw(circumcircle(A,X,Y));
		draw(X--Y--I--cycle);
		pair P = IP(L(I,Y),unitcircle);
		pair Q = 2*foot(O,I,P)-P;
		Drawing("P", P, dir(P-Q));
		Drawing("Q", Q, dir(Q-P));
		draw(P--Y,dashed); draw(I--Q,dashed);
		Drawing("Y'", IP(I--Q,B--C), dir(-110));
		pair X1 = extension(I,O,B,C);
		Drawing("X'", X1, dir(-75));
		draw(X1--I, dashed);
	\end{asy}
\end{figure}

\begin{soln}
	Let $X'$ and $Y'$ respectively denote the reflections of $X$ and $Y$ across $I$.
	Note that $X$, $Y$ lie on $\ol{BC}$.
	Also, let $P$, $Q$ be the intersections of $\ol{IY}$ with the circumcircle.

	Of course, $(ABC)$ is the unit circle.
	Let $j$ be the complex number corresponding to $I$ (to avoid confusion with $i = \sqrt{-1}$).
	Thus,
	\[ x' = \frac{\left( \ol b c - b \ol c \right)(j-0) - \left( \ol j 0 - j \ol 0 \right)(b-c)}{(\ol b - \ol c)(j-0) - (b-c) (\ol j - \ol 0)}
		= \frac{j \cdot \frac{c^2-b^2}{bc}}{j \cdot \frac{c-b}{bc} - (b-c) \ol j}
		= \frac{j(b+c)}{j + bc \ol j} . \]
	We seek $y'$ now. Consider the quadratic equation in $z$ given by
	\[ \frac{z-j}{j} + \frac{\frac1z - \ol j}{\ol j} = 0
		\iff z^2 - 2j z + j / \ol j = 0.\]
	Its zeros in $z$ are $p$ and $q$,
	which implies that $p+q = 2j$ and $pq = j / \ol j$
	(by Vieta!).
	From this we can compute
	\[ y' = \frac{pq(b+c)-bc(p+q)}{pq-bc}
		= \frac{j(b+c)-2bcj \ol j}{j - bc \ol j} = \frac{j(b+c)-2bcj \ol j}{j-bc\ol j}. \]
	which gives
	\[
		x = 2j - x' = \frac{j(2j-b-c+2bc \ol j)}{j + bc \ol j}
		\quad\text{and}\quad
		y = 2j - y' = \frac{j(2j-b-c)}{j - bc \ol j}.
	\]
	From this we can obtain
	\begin{align*}
		y-x &= j \cdot \frac{(2j-b-c)(j+bc\ol j)-(2j-b-c+2bc\ol j)(j-bc \ol j)}{(j-bc \ol j)(j+bc \ol j)} \\
		&= j \cdot \frac{2bc \ol j(2j-b-c)-2bc \ol j(j  - bc \ol j)}{(j-bc\ol j)(j+bc\ol j)} \\
		&= j \cdot \frac{2bc \ol j \left( j-b-c+bc \ol j \right)}{(j-bc\ol j)(j+bc\ol j)} \\
		X = \frac{y-x}{x} &= \frac{2bc \ol j \left( j-b-c+bc \ol j \right)}{(j-bc\ol j)(2j-b-c+2bc\ol j)} \\
		A = \frac{y-a}{a} &= \frac{j(2j-b-c-a)+abc\ol j}{a(j-bc\ol j)}
	\end{align*}
	We need to prove $X/A = \ol{X/A}$.
	Now set $a=x^2$, $b=y^2$, $c=z^2$, $j=-(xy+yz+zx)$, $\ol j = -\frac{x+y+z}{xyz}$
	(this is a different $x$, $y$ than the points $X$ and $Y$.)
	So, the above rewrites as
	\begin{align*}
		X &= \frac{2\frac{yz}{x}(x+y+z)(\frac{yz}{x}(x+y+z)+y^2+z^2+xy+yz+zx)}{\left( -\frac{yz}{x}(x+y+z)+xy+yz+zx \right)\left( y^2+z^2+2(xy+yz+zx)+2\frac{yz}{x}(x+y+z) \right)} \\
		&= \frac{2yz(x+y+z)\left( 2xyz+\sum_{\text{sym}} x^2y \right)}{(y+z)(x^2-yz)\left( x(y+z)(2x+y+z)+2yz(x+y+z) \right)} \\
		&= \frac{2yz(x+y+z)(x+y)(x+z)}{(x^2-yz)\left( (x^2+yz)(y+z)+(xy+yz+zx)(x+y+z) \right)} \\
		\intertext{and}
		A &= \frac{(xy+yz+zx)(x+y+z)^2-xyz(x+y+z)}{x^2(-(xy+yz+zx)+\frac{yz}{x}(x+y+z))}
		= \frac{(x+y+z)(x+y)(y+z)(z+x)}{x(yz-x^2)(y+z)} \\
		\intertext{thus}
		X/A &= \frac{-2xyz}{(x^2+yz)(y+z)+(x+y+z)(xy+yz+zx)} \\
		&= \frac{-\frac{2}{xyz}}%
		{(\frac{1}{x^2}+\frac1{yz})(\frac1y+\frac1z)  +  (\frac1x+\frac1y+\frac1z)(\frac1{xy}+\frac1{yz}+\frac1{zx})} = \ol{X/A}. \qedhere
	\end{align*}
\end{soln}


\section{Practice Problems}
\begin{enumerate}
	\ii Let $ABCD$ be cyclic.
	Let $H_A$, $H_B$, $H_C$, $H_D$ denote the orthocenters of $BCD$, $CDA$, $DAB$, $ABC$.
	Show that $\ol{AH_A}$, $\ol{BH_B}$, $\ol{CH_C}$, $\ol{DH_D}$ are concurrent.
	\ii (China TST 2011) 
	Let $\Gamma$ be the circumcircle of a triangle $ABC$.
	Assume $\ol{AA'}$, $\ol{BB'}$, $\ol{CC'}$ are diameters of $\Gamma$.
	Let $P$ be a point inside $ABC$ and let $D$, $E$, $F$ be the feet from $P$ to $\ol{BC}$, $\ol{CA}$, $\ol{AB}$.
	Let $X$ be the reflection of $A'$ across $D$; define $Y$ and $Z$ similarly.
	Prove that $\triangle XYZ \sim \triangle ABC$.
	\ii In circumscribed quadrilateral $ABCD$ with incircle $\omega$,
	Prove that the midpoint of $\ol{AC}$ and the midpoint of $\ol{BD}$ are collinear with the center of $\omega$.
	\ii (Simson Line) Let $ABC$ be a triangle and $P$ a point on its circumcircle.
	\begin{enumerate}[(a)]
		\ii Let $D$, $E$, $F$ be the feet from $P$ to $\ol{BC}$, $\ol{CA}$, $\ol{AB}$. Show that $D$, $E$, $F$ are collinear.
		\ii Moreover, prove that the line through these points bisects $\ol{PH}$, where $H$ is the orthocenter of $ABC$.
	\end{enumerate}
	\ii (PUMaC Finals) Let $\gamma$ and $I$ be the incircle and incenter of triangle $ABC$.
	Let $D$, $E$, $F$ be the tangency points of $\gamma$ to $\ol{BC}$, $\ol{CA}$, $\ol{AB}$ and let $D'$ be the reflection of $D$ about $I$.
	Assume $EF$ intersects the tangents to $\gamma$ at $D$ and $D'$ at points $P$ and $Q$.
	Show that $\angle DAD' + \angle PIQ = 180\dg$.
	\ii (Schiffler Point) Let triangle $ABC$ have incenter $I$.
	Prove that the Euler lines of $\triangle AIB$, $\triangle BIC$, $\triangle CIA$, $\triangle ABC$ are concurrent.
	\ii (USA TST 2014) Let $ABCD$ be a cyclic quadrilateral and let 
	$E$, $F$, $G$, $H$ be the midpoinst of $\ol{AB}$, $\ol{BC}$, $\ol{CD}$, $\ol{DA}$.
	Call $W$, $X$, $Y$, $Z$ the orthocenters of $AHE$, $BEF$, $CFG$, $DGH$.
	Prove that $ABCD$ and $WXYZ$ have the same area.
	\ii Let $O$ be the circumcenter of $ABC$.
	A line $\ell$ through $O$ cuts $\ol{AB}$ and $\ol{AC}$ at points $X$ and $Y$.
	Let $M$ and $N$ be the midpoints of $\ol{BY}$, $\ol{CX}$. Show that $\angle MON = \angle BAC$.
	\ii (APMO 2010) Let $ABC$ be an acute triangle, where $AB > BC$ and $AC > BC$.
	Denote by $O$ and $H$ the circumcenter and orthocenter.
	The circumcircle of $AHC$ intersects $AB$ again at $M$;
	the circumcircle of $AHB$ intersects $AC$ again at $N$.
	Prove that the circumcenter of triangle $MNH$ lies on line $OH$.
	\ii (Iran 2013) Let $ABC$ be acute, and $M$ the midpoint of minor arc $\widehat{BC}$.
	Let $N$ be on the circumcircle of $ABC$ such that $\ol{AN} \perp \ol{BC}$,
	and let $K$, $L$ lie on $AB$, $AC$ so that $\ol{OK} \parallel \ol{MB}$, $\ol{OL} \parallel \ol{MC}$.
	(Here $O$ is the circumcenter of $ABC$). Prove that $NK = NL$.
	\ii (MOP 2006) Cyclic quadrilateral $ABCD$ has circumcenter $O$.
	Let $P$ be a point in the plane and let $O_1, O_2, O_3, O_4$ be the circumcenters of $PAB,PBC,PCD,PDA$.
	Show that $\ol{O_1 O_3}$, $\ol{O_2O_4}$, $\ol{OP}$ are concurrent.
\end{enumerate}

 
\end{document}
