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

\begin{document}
\title{ToT Spring 2004 S-A5}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 106}
\maketitle

\section*{Problem}
A circle and a parabola share exactly two points, $A$ and $B$.
Suppose they are tangent at $A$
(that is, the tangent lines to the circle and parabola at $A$ coincide).
Does it follow that they are also tangent at $B$?

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

\section*{External Link}
\url{https://aops.com/community/p14147724}

\newpage

\section*{Solution}
It does not follow; for a counterexample use the construction
\[
  y = x^2 \qquad\text{and}\qquad
  (x+4)^2 + \left( y - \frac 72  \right)^2 = \frac{125}{4}.
\]
See figure below.
They meet at $A = (1,1)$ and $B = (-3,9)$.
\begin{center}
\begin{asy}
  import graph;
  real f(real x) { return x*x; }
  draw(graph(f,-3.5,3,operator ..), blue, Arrows);
  draw(circle((-4, 7/2), (125/4.0)**.5), red);
  dot("$A$", (1,1), dir(315));
  dot("$B$", (-3,9), dir(45));
\end{asy}
\end{center}

\begin{remark*}
[Motivation for these numbers]
If we write the parabola as $y=x^2$ WLOG,
and the circle as $(x-\alpha)^2 + (y-\beta)^2 = \gamma^2$,
then the intersections are controlled by a quartic in $x$, namely:
\begin{align*}
  0 &= \left( x-\alpha \right)^2 + \left( x^2 - \beta \right)^2 - \gamma^2 \\
  &= x^4 + (1-2\beta) x^2  - 2\alpha x + (\alpha^2+\beta^2-\gamma^2).
\end{align*}
To fail the problem, we need this quartic to have a
``triple root'' at $A$.
The above example is constructed by choosing
$(x-1)^3(x+3)$ as the quartic and equating coefficients.
\end{remark*}

\end{document}
