At least how many convex pentagons are needed to construct a convex 2011-gon?
Solution. The sum of the interior angles of a 2011-gon is $(2011-2) \cdot 180^{\circ}$. The sum of the interior angles of a pentagon is $(5-2) \cdot 180^{\circ}$. Since every angle of the 2011-gon will be covered by some pentagon, at least $\left\lceil\frac{(2011-2) \cdot 180^{\circ}}{(5-2) \cdot 180^{\circ}}\right\rceil=670$ pentagons are needed. This is achievable. Choose a vertex and connect it to the 4th, 7th, 10th, ..., 2005th vertex that follows it. This creates 668 convex pentagons and leaves a heptagon that still needs to be covered. Connect the chosen vertex to a point on the edge between the 2007th and 2008th vertices. This creates two more pentagons that cover the heptagon.