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

\begin{document}
\title{RMM 2016/5}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 38}
\maketitle

\section*{Problem}
A convex hexagon $A_1B_1A_2B_2A_3B_3$
is inscribed in a circle $\Omega$ with radius $R$.
The diagonals $A_1B_2$, $A_2B_3$, $A_3B_1$ are concurrent in $X$.
For each $i=1,2,3$ let $\omega_i$ tangent to the segments $XA_i$ and $XB_i$
and tangent to the arc $A_iB_i$ of $\Omega$
that does not contain the other vertices of the hexagon;
let $r_i$ the radius of $\omega_i$.
\begin{enumerate}
  \ii[(a)] Prove that $R\geq r_1+r_2+r_3$.
  \ii[(b)] If $R= r_1+r_2+r_3$,
  prove that the six points of tangency of the circumferences $\omega_i$
  with the diagonals $A_1B_2$, $A_2B_3$, $A_3B_1$ are concyclic.
\end{enumerate}

\section*{Video}
\href{https://www.youtube.com/watch?v=ZqJNTYGwl8A&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/ZqJNTYGwl8A}}

\section*{External Link}
\url{https://aops.com/community/p5942512}

\newpage

\section*{Solution}
We let $M_1$ be the arc midpoint of the arc $A_2B_3$
containing $A_1B_1$, and so on.
The tangents to $\Omega$ at $M_1$, $M_2$, $M_3$
determine a triangle $ABC$ with incircle $\Omega$.
We also let $\omega_i$ be tangent to $\Omega$ at $T_i$,
and let $U_i$ and $V_i$ be tangency points as in the figure below.
\begin{center}
\begin{asy}
import graph; size(13cm);
pen qqwuqq = rgb(0.,0.39215,0.);
pair O = (0.,0.), X = (-0.09676,0.19152), A_1 = (-0.67013,0.74223), A_2 = (0.98123,0.19283), A_3 = (-0.6,-0.8), B_2 = (0.76864,-0.63967), B_3 = (-0.98169,0.19044), B_1 = (0.29152,0.95656), T_1 = (-0.25520,0.96688), T_2 = (0.98478,-0.17375), T_3 = (-0.97531,-0.22083), M_1 = (-0.00121,0.99999), M_2 = (0.89172,-0.45258), M_3 = (-0.69271,-0.72121), A = (0.28081,-1.65625), B = (1.62997,1.00198), C = (-2.48160,0.99697);
fill((-0.93731,0.99885)--X--(0.31376,1.00038)--cycle, palegreen);
fill((0.78754,-0.65783)--X--(1.21944,0.19313)--cycle, palegreen);
fill((-0.60350,-0.80689)--X--(-1.64105,0.18964)--cycle, palegreen);
draw(A--B--C--cycle, linewidth(0.6) + blue);
draw(circle(O, 1.), linewidth(1.2));
draw(circle((-0.17976,0.68109), 0.29558), linewidth(0.6) + red);
draw(circle((0.67752,-0.11954), 0.31201), linewidth(0.6) + red);
draw(circle((-0.64641,-0.14636), 0.33721), linewidth(0.6) + red);
draw(A--B, linewidth(0.6) + blue);
draw(B--C, linewidth(0.6) + blue);
draw(C--A, linewidth(0.6) + blue);
draw((-0.93731,0.99885)--(0.78754,-0.65783), linewidth(0.6) + qqwuqq);
draw((1.21944,0.19313)--(-1.64105,0.18964), linewidth(0.6) + qqwuqq);
draw((0.31376,1.00038)--(-0.60350,-0.80689), linewidth(0.6) + qqwuqq);
dot("$O$", O, dir(-90));
dot("$X$", X, dir(45));
dot("$A_1$", A_1, dir((1.312, 2.987)));
dot("$A_2$", A_2, dir((1.325, 3.074)));
dot("$A_3$", A_3, dir(10));
dot("$B_2$", B_2, dir(180));
dot("$B_3$", B_3, dir(135));
dot("$B_1$", B_1, dir(-90));
dot("$T_1$", T_1, dir(-T_1));
dot("$T_2$", T_2, dir(-T_2));
dot("$T_3$", T_3, dir(-T_3));
dot("$M_1$", M_1, dir(M_1));
dot("$M_2$", M_2, dir(M_2));
dot("$M_3$", M_3, dir(M_3));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$U_1$", (-0.38452,0.46790), dir(M_1-T_3));
dot("$V_1$", (0.08381,0.54731), dir(T_1-M_2));
dot("$U_2$", (0.67714,0.19246), dir(90));
dot("$V_2$", (0.46139,-0.34456), dir(225));
dot("$U_3$", (-0.34571,-0.29898), dir(M_2-T_3));
dot("$V_3$", (-0.64683,0.19085), dir(90));
\end{asy}
\end{center}
The three green segments drawn
(the extension of $A_1B_2$, $A_2B_3$, $A_3B_1$)
together with the sides of triangle $ABC$
determine three smaller triangles, shaded light green in the diagram above.
Notice that
\begin{itemize}
  \ii The red circles have radius smaller than the inradius
  of the green triangles they are contained in.
  \ii The sum of the three inradii of the green triangles equals $R$,
  because the homothety factors between the green triangles sum to $1$.
\end{itemize}
This immediately implies that $R \ge r_1 + r_2 + r_3$, so (a) is solved.
Moreover, equality holds if and only if $M_i = T_i$ for each $i$.

To finish (b), by shooting lemma twice we have
$M_3 = \ol{T_1 U_1} \cap \ol{T_2V_2}$ and
\[ M_3 T_1 \cdot M_3 U_1 = M_3 V_2 \cdot M_3 T_2. \]
(both equal to $M_3 A_1^2$).
However, by homothety from $T_3$ mapping $\omega_3$ to $\Omega$,
we also have
\[ \frac{T_3V_3}{T_3M_1} = \frac{T_3U_3}{T_3M_2}.  \]
So if $M_i = T_i$ for all $i$, then by power of a point from $T_3=M_3$
we are able to conclude that $U_1$, $V_3$, $U_3$, $V_2$ are cyclic.

Analogous statements imply that unless $U_1V_1$, $U_2V_2$, $U_3V_3$ are
parallel  --- which they obviously are not,
because they are perpendicular
to angle bisectors of $\angle A_i X B_i$ ---
then the six points must be cyclic, as desired.

\end{document}
