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 1 of 5: Parity and the geometric facts

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In plain words

The three boundary arcs cut off by a triangle have total length $2006$ , so an even number of them are odd. A special triangle therefore has exactly two good sides, and its equal sides are those two good sides.

\text{each triangle has }0\text{ or }2\text{ good sides; special triangles have two congruent good sides}

Detailed analysis

For a triangle with vertices on the regular $2006$ -gon, let the three boundary-arc lengths between consecutive vertices be $u,v,w$ . Then $u+v+w=2006$ , and a side is good exactly when its corresponding arc length is odd; hence the number of good sides is $0$ or $2$ . In an isosceles triangle with two good sides, the equal sides must be the good pair: otherwise the third side has the same parity as an equal side, contradicting that the number of good sides is even. The standard equal-chord and non-crossing argument also gives that two special triangles cannot share a good side.