Problem: Let $P$ be a regular $2006$-gon.
A diagonal is called \emph{good} if its endpoints
divide the boundary of $P$ into two parts,
each composed of an odd number of sides of $P$.
The sides of $P$ are also called \emph{good}.
Suppose $P$ has been dissected into triangles by $2003$ diagonals,
no two of which have a common point in the interior of $P$.
Find the maximum number of isosceles triangles having two good
sides that could appear in such a configuration.

Solution: Call a triangle with the desired property \emph{special}.
We prove the maximum number of special triangles is $1003$,
achieved by paring up the sides of the polygon.

We present two solutions for the upper bound.
Both of them rely first on two geometric notes:
\begin{itemize}
  \ii In a special triangle, the good sides are congruent
  (and not congruent to the third side).
  \ii No two isosceles triangles share a good side.
\end{itemize}

\paragraph{Solution using bijections.}
Call a good diagonal \textbf{special} if it's part of a special triangle;
special diagonals come in pairs.
Consider the minor arc cut out by a special diagonal $d$,
which has an odd number of sides.
Since special diagonals come in pairs,
one can associate to $d$ a side of the polygon
not covered by any special diagonals from $d$.
Hence there are at most $2006$ special diagonals,
so at most $1003$ special triangles.

\paragraph{Solution using graph theory.}
Consider the tree $T$ formed by the $2004$
triangles in the dissection, with obvious adjacency.
Let $F$ be the forest obtained by deleting
any edge corresponding to a good diagonal.
Then the resulting graph $F$ has only degrees $1$ and $3$,
with special triangles only occurring at degree $1$ vertices.

If there are $k$ good diagonals drawn,
then this forest consists of $k+1$ trees.
A tree with $n_i$ vertices ($0 \le i \le k$)
consequently has $\frac{n_i+2}{2}$ leaves.
However by the earlier remark at least $k$ leaves
don't give special triangles
(one on each side of a special diagonal);
so the number of leaves that do give good triangles is at most
\[ -k + \sum_i \frac{n_i+2}{2}
  = -k + \frac{2004 + 2(k+1)}{2} = 1003. \]