# Problem 7. (4 points)

$O A B C$ is a rectangle on the Cartesian plane, with sides parallel to the coordinate axes. Point $O$ is the origin, and point $B$ has coordinates $(9 ; 8)$. Inside the rectangle, a point $X$ with integer coordinates is taken. What is the smallest value that the area of triangle $O B X$ can take?

Answer: $\frac{1}{2}$
Solution:

Since the picture is symmetric with respect to the center of the rectangle, it is sufficient to consider the case when point $X$ lies below the diagonal $O B$. Let $Y$ and $Z$ be the projections of $X$ onto the x-axis and segment $B C$ (here $C(9 ; 0)$).

Then $S_{O B X}=S_{O A B C}-S_{O A B}-S_{O X Y}-S_{B X Z}-S_{X Y C Z}$. All these are areas of rectangles and right triangles, the sides and, respectively, the legs of which are parallel to the coordinate axes, so their areas are integers or half-integers. Therefore, $S_{O B X}$ is also an integer or half-integer, that is, at least $\frac{1}{2}$.

The number $\frac{1}{2}$ is obtained, for example, by taking the point $X(8 ; 7)$.

$S_{O B X}=72-36-28-\frac{1}{2}-7=\frac{1}{2}$.

In general, three points on the same line do not form a triangle, so degenerate triangles with an area of 0 can be ignored. However, in this problem, this is not important, as there are no points inside the segment $O B$.