# 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 $(11; 8)$. 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 relative 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 $C(11 ; 0)$).

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 of 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 $X(7 ; 5)$.

$S_{O B X}=88-44-\frac{35}{2}-20-6=\frac{1}{2}$.

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 $O B$.