96. On a grid paper with a cell side of 1 cm, a circle with a radius of $100 \mathrm{~cm}$ is drawn, not passing through the vertices of the cells and not touching the sides of the cells. How many cells can this circle intersect?
96. Answer: 800 or 799.

One of the directions of the grid lines will be called horizontal, and the other - vertical. A circle with a diameter of 200, not passing through the nodes and not touching the grid lines, intersects 200 horizontal lines and 200 vertical lines, and each of them twice. Thus, the maximum number of intersection points is 800. These 800 points divide the circle into 800 parts. Each of these parts lies within one cell. Therefore, the maximum possible number of cells is 800.

However, it may happen that some two parts belong to one cell, i.e., that the circle intersects some cell twice (Fig. 54). We will prove that there are no more than one such "special" cell.

Let's see where the center O of a circle with a diameter of 200, intersecting some side AB of a cell twice, can lie.

The distance from the center O of such a circle to points A and B is more than 100, and from the line AB - less than 100, so O lies outside the circles of radius 100 with centers at points A and B, between two vertical grid lines passing through A and B, and within a strip 200 wide with the horizontal axis AB. All such points fill the interior of two curvilinear triangles (one of them is shaded in the figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_7b8a6b2625543c3dae75g-142.jpg?height=607&width=600&top_left_y=256&top_left_x=286)

Fig. 54

Obviously, for different segments AB, such sets do not have common points, and therefore there is no more than one special cell.