Problem 2

Let

P

be a regular

2006

-gon. A diagonal of

P

is called 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 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.

Step 5 of 5: Pair boundary sides to attain the bound

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In plain words

Cut off one isosceles triangle from every adjacent pair of boundary sides; the remaining polygon can be triangulated arbitrarily.

\triangle P_{2i-1}P_{2i}P_{2i+1}\ (i=1,\ldots,1003)\quad\Longrightarrow\quad1003\text{ special triangles}

Detailed analysis

Label the vertices $P_1,\ldots,P_{2006}$ cyclically and draw $P_1P_3,P_3P_5,\ldots,P_{2005}P_1$ . These $1003$ non-crossing diagonals cut off the $1003$ triangles $P_{2i-1}P_{2i}P_{2i+1}$ , each with two polygon sides and hence two good sides. Triangulate the remaining $1003$ -gon with $1000$ diagonals. This is a valid dissection with $1003$ special triangles, so the upper bound is attained.