Balti Tee 1995 · Ülesanne 20

Solution:

There are two vertices $A_{1}$ and $A_{2}$ of the pentagon that have their first coordinates of the same parity, and their second coordinates of the same parity. Therefore the midpoint $M$ of $A_{1}A_{2}$ has integer coordinates. There are two possibilities:

(i) The considered vertices are not consecutive. Then $M$ lies inside the pentagon (because it is convex) and is the common vertex of five triangles having as their bases the sides of the pentagon. The area of any one of these triangles is not less than $\frac{1}{2}$, so the area of the pentagon is at least $\frac{5}{2}$.

(ii) The considered vertices are consecutive. Since the pentagon is convex, the side $A_{1}A_{2}$ is not simultaneously parallel to $A_{3}A_{4}$ and $A_{4}A_{5}$. Suppose that the segments $A_{1}A_{2}$ and $A_{3}A_{4}$ are not parallel. Then the triangles $A_{2}A_{3}A_{4}$, $MA_{3}A_{4}$ and $A_{1}A_{3}A_{4}$ have different areas, since their altitudes dropped onto the side $A_{3}A_{4}$ form a monotone sequence. At least one of these triangles has area not less than $\frac{3}{2}$, and the pentagon has area not less than $\frac{5}{2}$.