Problem: Let $n > 1$ be an integer.
Suppose $2n$ points are given in the plane, no three of which are collinear.
Suppose $n$ of the given $2n$ points are colored blue and the other $n$ colored red.
A line in the plane is called a \emph{balancing line}
if it passes through one blue and one red point
and, for each side of the line, the number of blue points on that
side is equal to the number of red points on the same side.
Prove that there exist at least two balancing lines.

Solution: Consider the convex hull $\mathcal H$ of the polygon. There are two cases.

\paragraph{Easy case: the convex hull has both colors.}
If the convex hull $\mathcal H$ is not all the same color,
there exist two edges of $\mathcal H$ (at least) which have differently colored endpoints.
The extensions of those sides form balancing lines;
indeed given any such line $\ell$ one side of $\ell$ has no points,
the other has $n-1$ red and $n-1$ blue points.

\paragraph{Harder case: the convex hull is all one color.}
Now assume $\mathcal H$ is all blue (WLOG).
We will prove there are at least $|\mathcal H|$ balancing lines in the following way.
\begin{claim*}
For any vertex $B$ of $\mathcal H$ there is a balancing line through it.
\end{claim*}
\begin{proof}
Assume $A$, $B$, $C$ are three consecutive
blue vertices of $\mathcal H$.
Imagine starting with line $\ell$ passing through $B$ and $A$,
then rotating it through $B$ until it coincides with line $BC$,
through the polygon.

\begin{center}
\begin{asy}
size(5cm);
filldraw(dir(90)--dir(210)--dir(330)--cycle,
  rgb(1,1,0.9), blue);
label(scale(2)*"$\mathcal H$", midpoint(dir(90)--dir(330)), dir(30), blue);
dot("$B$", dir(90), dir(130), blue);
dot("$A$", dir(210), dir(210), blue);
dot("$C$", dir(330), dir(330), blue);
dot( (0.2,-0.4), red);
dot( (-0.1, 0.3), red);
dot( (0.4, 0.1), red);
dot( (-0.5, -0.1), red);
dot( (0.3, -0.2), blue);
draw(dir(90)--dir(234), dashed, EndArrow);
label("$\ell$", dir(234), dir(234));
\end{asy}
\end{center}

During this process, we consider the set of points on the same side of $\ell$ as $C$,
and let $x$ be the number of such red points minus the number of such blue points.
Note that:
\begin{itemize}
  \ii Every time $\ell$ touches a blue point, $x$ increases by $1$.
  \ii Every time $\ell$ touches a red point, $x$ decreases by $1$.
  \ii Initially, $x = +1$.
  \ii Just before reaching the end we have $x = -1$.
\end{itemize}
So at the moment where $x$ first equals zero, we have found our balancing line.
\end{proof}

\begin{remark*}
  Some students may hope that in fact for every red point there is balancing line through it.
  This claim is false even for $n=2$.
  For example, if $ABC$ is an equilateral triangle with center $O$,
  if we color $A$ and $O$ red but $B$ and $C$ blue,
  then no balancing line passes through $O$.

  On the other hand, in \emph{On the number of balanced lines},
  by J.\ Pach and R.\ Pinchasi, (Discrete and Computational Geometry, 2001, p.\ 611-628)
  it is shown that in fact there are at least $n$ balancing lines.
\end{remark*}