Solution. Since the sum of the interior angles of an $n$-sided polygon is $(n-2) \cdot 180^{\circ}$, the sum of the interior angles of a 2006-sided polygon is $2004 \cdot 180^{\circ}$. Therefore, the polygon cannot have 2004 or more angles that are greater than $180^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b6c2e892e78512fc380cg-1.jpg?height=317&width=458&top_left_y=264&top_left_x=823)

We will show that 2003 such angles are possible. Take three points in the plane, $A$, $B$, and $C$, and draw two sides of the triangle they define ($AC$ and $BC$). Over the third side ($AB$), draw an arc that passes inside the triangle defined by the three points. Such an arc can be easily drawn. Then, place 2003 points on the arc and connect the adjacent ones (the two endpoints with $A$ and $B$, respectively). This way, we get a polygon where the angles at the three vertices, $A$, $B$, and $C$, are less than $180^{\circ}$, and the interior angles at the other two thousand and three vertices are greater than $180^{\circ}$.
What is the maximum number of interior angles greater than $180^{\circ}$ that a 2006-sided polygon can have?

 

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