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

\begin{document}
\title{USEMO 2020/5}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 35}
\maketitle

\section*{Problem}
The sides of a convex $200$-gon $A_1 A_2 \dots A_{200}$ are
colored red and blue in an alternating fashion.
Suppose the extensions of the red sides determine a regular $100$-gon,
as do the extensions of the blue sides.

Prove that the $50$ diagonals $\ol{A_1A_{101}},\ \ol{A_3A_{103}},\ \dots,
\ \ol{A_{99}A_{199}}$ are concurrent.

\section*{Video}
\href{https://www.youtube.com/watch?v=5a_XCGKiXnI}{\texttt{https://youtu.be/5a\_XCGKiXnI}}

\section*{External Link}
\url{https://aops.com/community/p18486857}

\newpage

\section*{Solution}
We present a diagram (with $100$ replaced by $6$, for simplicity).
\begin{center}
  \begin{asy}
    unitsize(50);
    int m = 3, n = 2*m;
    pair[] B = new pair[n], R = new pair[n], X = new pair[n];
    pair XX;
    B.cyclic = true;
    R.cyclic = true;
    X.cyclic = true;

    B[1] = (-1.49, 0.34);
    B[2] = (0.52, -0.09);
    R[1] = (-0.82, 0.05);
    R[2] = (1.19, 0.36);

    for (int i = 1; i < n-1; ++i) {
      B[i+2] = rotate(-180*(n-2)/n, B[i+1]) * B[i];
      R[i+2] = rotate(-180*(n-2)/n, R[i+1]) * R[i];
    }

    for (int i = 0; i < n; ++i) {
      X[i] = extension(B[i], B[i+1], R[i], R[i+1]);
    }

    XX = extension(X[0], X[m], X[1], X[m+1]);

    for (int i = 0; i < m; ++i) {
      draw(X[i]--X[i+m], dotted);
    }
    draw(circumcircle(X[4], B[4], R[4]), lightred);
    draw(circumcircle(X[4], B[5], R[5]), lightred);

    for (int i = 1; i <= n; ++i) {
      draw(B[i]--B[i+1], heavycyan);
      draw(R[i]--R[i+1], heavymagenta);
      dot("$B_{"+(string)i+"}$", B[i], dir(2B[i]-B[i+1]-B[i-1]), heavycyan);
      dot("$R_{"+(string)i+"}$", R[i], dir(2R[i]-R[i+1]-R[i-1]), heavymagenta);
      dot("$X_{"+(string)i+"}$", X[i], dir(dir(B[i+1]-B[i])+dir(R[i]-R[i+1])));
    }
    dot("$O$", XX);
  \end{asy}
\end{center}
Let $B_1 \dots B_{100}$ and $R_1 \dots R_{100}$ be the regular $100$-gons
(oriented counterclockwise),
and define $X_i = A_{2i+1} = \ol{B_iB_{i+1}} \cap \ol{R_iR_{i+1}}$ for all $i$,
where all indices are taken modulo $100$.
We wish to show that $\ol{X_1X_{51}},\ \dots,\ \ol{X_{50}X_{100}}$ are concurrent.

We now present two approaches.

\paragraph{First approach (by spiral similarity).}
Let $O$ be the spiral center taking $B_1 \dots B_{100} \to R_1 \dots R_{100}$
(it exists since the $100$-gons are not homothetic).
We claim that $O$ is the desired concurrency point.
\begin{claim*}
  $\dang X_iOX_{i+1} = \tfrac\pi{50}$ for all $i$.
\end{claim*}
\begin{proof}
  Since $\triangle OB_iB_{i+1} \stackrel{+}{\sim} \triangle OR_iR_{i+1}$,
  we have $\triangle OB_iR_i \stackrel{+}{\sim} \triangle OB_{i+1}R_{i+1}$,
  so $O$, $X_i$, $B_{i+1}$, $R_{i+1}$ are concyclic.
  Similarly $O$, $X_{i+1}$, $B_{i+1}$, $R_{i+1}$ are concyclic, so
  \[ \dang X_iOX_{i+1} = \dang X_iB_{i+1}X_{i+1} = \frac\pi{50} \]
  as wanted.
\end{proof}
It immediately follows that $O$ lies on all $50$ diagonals $\ol{X_iX_{i+50}}$, as desired.

\paragraph{Second approach (by complex numbers).}
Let $\omega$ be a primitive $100$th root of unity.
We will impose complex coordinates so that $R_k = \omega^k$,
while $B_k = p \omega^k + q$, where $m$ and $b$ are given constant complex numbers.

In general for $|z| = 1$, we will define $f(z)$
as the intersection of the line through $z$ and $\omega z$,
and the line through $pz + q$ and $p \cdot \omega z + q$.
\begin{center}
\begin{asy}
  size(13cm);
  defaultpen(fontsize(11pt));
  pair C = (3,1);
  pair D = (-6,-2);
  pair A = (4,-1);
  pair B = (-4,1.3);
  dot("$z$", A, dir(A), heavymagenta);
  dot("$\omega z$", B, dir(B), heavymagenta);
  dot("$p z + q$", C, dir(C), heavycyan);
  dot("$p \cdot \omega z + q$", D, dir(D), heavycyan);
  draw(A--B, heavymagenta);
  draw(C--D, heavycyan);
  pair X = extension(A,B,C,D);
  dot("$f(z)$", X, dir(-90));
\end{asy}
\end{center}
In particular, $X_k$ is $f(\omega^k)$.

\begin{claim*}
  There exist complex numbers $a,b,c$ such that $f(z) = a + bz + cz^2$,
  for every $|z| = 1$.
\end{claim*}
\begin{proof}
  Since $f(z)$ and $\frac{f(z)-q}{p}$ both lie on the chord joining $z$ to $\omega z$ we have
  \begin{align*}
    z + \omega z  &= f(z) + \omega z^2 \cdot \ol{f(z)} \\
    z + \omega z  &= \frac{f(z)-q}{p} + \omega z^2 \cdot \frac{\ol{f(z)}-\ol q}{\ol p}.
  \end{align*}
  Subtracting the first equation from the $\ol p$ times the second to eliminate $\ol{f(z)}$,
  we get that $f(z)$ should be a degree-two polynomial in $z$
  (where $p$ and $q$ are fixed constants).
\end{proof}
\begin{claim*}
  Let $f(z) = a + bz + cz^2$ as before.
  Then the locus of lines through $f(z)$ and $f(-z)$,
  as $|z|=1$ varies, passes through a fixed point.
\end{claim*}
\begin{proof}
  By shifting we may assume $a=0$, and by scaling
  we may assume $b$ is real (i.e. $\ol b = b$).
  Then the point $-\ol c$ works, since
  \[ \frac{f(z) + \ol c}{f(-z) + \ol c}
    = \frac{\ol c + bz + cz^2}{\ol c - bz + cz^2} \]
  is real --- it obviously equals its own conjugate.
  (Alternatively, without the assumptions $a=0$ and $b \in \RR$,
  the fixed point is $a - \frac{b \ol c}{\ol b}$.)
\end{proof}
\begin{remark*}
  [We know a priori the exact coefficients shouldn't matter]
  In fact, the exact value is
  \[ f(z) = \frac{-\omega \ol q z^2 + (1-\ol p)(1+\omega) z
    - \frac{\ol p}{p} q}{1 - \frac{\ol p}{p}}. \]
  Since $p$ and $q$ could be any complex numbers,
  the quantity $c/b$ (which is all that matters for concurrence)
  could be made to be equal to any value.
  For this reason, we know \emph{a priori} the exact coefficients should be irrelevant.
\end{remark*}

\end{document}
