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

\begin{document}
\title{IGO 2020/A4}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 52}
\maketitle

\section*{Problem}
Convex circumscribed quadrilateral $ABCD$
with its incenter $I$ is given such that
its incircle is tangent to $\ol{AD}$, $\ol{DC}$,
$\ol{CB}$, and $\ol{BA}$ at
$K$, $L$, $M$, and $N$.
Let $E = \ol{AD} \cap \ol{BC}$
and $F = \ol{AB} \cap \ol{CD}$.
Let $X = \ol{KM} \cap \ol{AB}$ and $Y = \ol{KM} \cap \ol{CD}$.
Let $Z = \ol{LN} \cap \ol{AD}$ and $T = \ol{LN} \cap \ol{BC}$.

Prove that the circumcircle of triangle $\triangle XFY$
and the circle with diameter $EI$ are tangent
if and only if the circumcircle of triangle $\triangle TEZ$
and the circle with diameter $FI$ are tangent.

\section*{Video}
\href{https://www.youtube.com/watch?v=hlP-tSi3Eb8&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/hlP-tSi3Eb8}}



\newpage

\section*{Solution}
We are going to prove both conditions
are equivalent to $\ol{KM} \perp \ol{LN}$.

We introduce the following notation.
\begin{itemize}
  \ii $\Omega$ denotes the circumcircle of $KLMN$.
  \ii Rays $EN$ and $EL$ meet $\Omega$ again at $N'$ and $L'$.
  \ii Points $X'$, $Y'$, $F'$, $W$ are the midpoints
  of $\ol{NN'}$, $\ol{LL'}$, $\ol{NL}$, $\ol{N'L'}$.
  Note that $X'$ and $Y'$ lie on the circle with diameter $\ol{IE}$.
  \ii Let $G = \ol{LN} \cap \ol{KM}$ and $V = \ol{X'Y'} \cap \ol{MK}$.
\end{itemize}

The relevance of $F'$, $X'$, $Y'$ is explained as follows:
\begin{claim*}
  The points $F'$, $X'$, $Y'$ are the inverses
  of $F$, $X$, $Y$ with respect to $\Omega$.
\end{claim*}
\begin{proof}
  For $F$ it's clear.
  For $X$, the inverses of $\ol{MK}$ and $\ol{NN}$
  are the circles with diameter $\ol{IE}$ and $\ol{IN}$.
\end{proof}
Hence, we are interested in when $(F'X'Y')$ is tangent to $\ol{MK}$,
the latter being the inverse of the circle with diameter $\ol{EI}$.

\begin{center}
\begin{asy}
import graph; size(12cm);
pen qqwuqq = rgb(0.,0.39215,0.); pen xfqqff = rgb(0.49803,0.,1.); pen dcrutc = rgb(0.86274,0.07843,0.23529); pen qqffff = rgb(0.,1.,1.); pen ffxfqq = rgb(1.,0.49803,0.); pen yqqqqq = rgb(0.50196,0.,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw(circle((0.,0.), 1.), linewidth(0.6) + qqwuqq);
draw((-0.32917,0.94426)--(-0.31711,-0.94838), linewidth(0.6) + red);
draw(circle((0.,-0.83333), 0.83333), linewidth(0.6) + xfqqff);
draw(circle((-0.31933,-0.30201), 0.29798), linewidth(0.6) + dcrutc);
draw(circle((-0.15855,-0.47419), 0.5), linewidth(0.6) + qqffff);
draw(circle((-0.16458,0.47213), 0.5), linewidth(0.6) + qqffff);
draw((0.,-1.66666)--(-0.91446,0.40466), linewidth(0.6) + ffxfqq);
draw((0.,-1.66666)--(-0.32917,0.94426), linewidth(0.6) + ffxfqq);
draw((-0.08450,-0.99642)--(-0.91446,0.40466), linewidth(0.6) + ffxfqq);
draw((-1.16175,-0.6)--(-0.20684,-0.02607), linewidth(0.6) + red);
draw((-0.31933,-0.6)--(-0.61578,-0.27186), linewidth(0.6) + yqqqqq);
draw((-0.31933,-0.6)--(-0.20684,-0.02607), linewidth(0.6) + yqqqqq);
draw((-1.16175,-0.6)--(0.8,-0.6), linewidth(0.6) + red);
draw((0.,-1.66666)--(-0.8,-0.6), linewidth(0.6) + qqwuqq);
draw((0.,-1.66666)--(0.8,-0.6), linewidth(0.6) + qqwuqq);
dot("$I$", (0.,0.), dir((1.160, 2.848)));
dot("$K$", (0.8,-0.6), dir((4.948, -3.148)));
dot("$M$", (-0.8,-0.6), dir((-11.235, -8.026)));
dot("$E$", (0.,-1.66666), dir((-3.143, -12.694)));
dot("$N$", (-0.31711,-0.94838), dir((-9.882, -9.342)));
dot("$N'$", (-0.91446,0.40466), dir((-12.414, 5.711)));
dot("$L$", (-0.32917,0.94426), dir((1.080, 2.827)));
dot("$X'$", (-0.61578,-0.27186), dir((-13.014, 0.767)));
dot("$Y'$", (-0.20684,-0.02607), dir((1.185, 2.300)));
dot("$F'$", (-0.32314,-0.00206), dir((1.051, 2.194)));
dot("$G$", (-0.31933,-0.6), dir((-9.086, -8.599)));
dot("$L'$", (-0.08450,-0.99642), dir((2.725, -9.703)));
dot("$V$", (-1.16175,-0.6), dir((1.124, 2.304)));
dot("$W$", (-0.49948,-0.29588), dir((1.181, 2.308)));
\end{asy}
\end{center}

The first critical lemma is the following:
\begin{claim*}
  [A generalization of butterfly theorem]
  Unconditionally, we have that $(GX'Y')$ is tangent to $\ol{MK}$.
\end{claim*}
\begin{proof}
  Suppose $\ol{NN'} \cap \ol{MK} = X_1$
  and $\ol{LL'} \cap \ol{SMK} = Y_1$.
  Then by shooting lemma from $E$, the quadrilateral $X'X_1Y_1Y'$ is cyclic.

  Also, by the dual of Desargues involution theorem (DDIT)
  on $N'NLL'$, there is an involution swapping $M \leftrightarrow K$,
  $X_1 \leftrightarrow Y_1$, $G \leftrightarrow G$.
  Combined with power of a point from $V$, we are able to get
  \[ VX_1 \cdot VY_1 = VX' \cdot VY' = VM \cdot VK
    \implies VX' \cdot VY' = VG^2. \]
  This proves the desired tangency.
\end{proof}

Thus, the tangency takes place exactly when $F' \in (GX'Y')$.

To prove the result, we now need:
\begin{claim*}
  Unconditionally, $\ol{GF'}$ and $\ol{GW}$ are isogonal.
\end{claim*}
\begin{proof}
  Since $\triangle GN'N \overset{-}{\sim} \triangle GLL'$,
  we get $\triangle GN'X' \overset{-}{\sim} \triangle GLY'$.
\end{proof}

Now, $WX'F'Y'$ is always a parallelogram,
because these four points are the midpoints of $NN'L'L$.

So we focus entirely on $\triangle GX'Y'$ now,
and consider the following four statements:
\begin{itemize}
  \ii $\ol{GW}$ and $\ol{GF'}$ are isogonal wrt $\angle G$
  (and they are distinct, provided $X' \neq Y' \iff X \neq Y$,
  which I think should be assumed for the problem to make sense).
  \ii $W$ and $F'$ are reflections are the midpoint of $\ol{X'Y'}$.
  \ii $F'$ lies on $(GX'Y')$.
  \ii $\ol{GW}$ and $\ol{GF'}$ are altitude and diameter respectively.
\end{itemize}
The first two statements are always known;
under these two assumptions,
one can then show the third and fourth statements are equivalent.

Finally, the last bullet is equivalent to $\ol{GF'} \perp \ol{MK}$,
because $\ol{MK}$ is a tangent,
and hence equivalent to $\ol{LN} \perp \ol{MK}$.

\end{document}
