2006 IMO Problems/Problem 2
Problem
Let be a regular 2006-gon. A diagonal of is called good if its endpoints divide the boundary of into two parts, each composed of an odd number of sides of . The sides of are also called good. Suppose has been dissected into triangles by 2003 diagonals, no two of which have a common point in the interior of . Find the maximum number of isosceles triangles having two good sides that could appear in such a configuration.
Solution
Call an isosceles triangle odd if it has two odd sides. Suppose we are given a dissection as in the problem statement. A triangle in the dissection which is odd and isosceles will be called iso-odd for brevity. Let ABC be an iso-odd triangle, with AB and BC odd sides. This means that there are an odd number of sides of the 2006-gon between A and B and also between B and C. We say that these sides belong to the iso-odd triangle ABC. At least one side in each of these groups does not belong to any other iso-odd triangle. This is so because any odd triangle whose vertices are among the points between A and B has two sides of equal length and therefore has an even number of sides belonging to it in total. Eliminating all sides belonging to any other iso-odd triangle in this area must therefore leave one side that belongs to no other iso-odd triangle. Let us assign these two sides (one in each group) to the triangle ABC. To each iso-odd triangle we have thus assigned a pair of sides, with no two triangles sharing an assigned side. It follows that at most 1003 iso-odd triangles can appear in the dissection. This value can be attained, as shows the example from the ﬁrst solution.
See Also
2006 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |