Someone drew a circle on a chessboard around the center of one of the (inner) squares with a radius of 1.9 units (unit is the side of the squares), and found the total lengths of the arcs that fell on the light and dark squares to be approximately equal. Would increasing or decreasing the radius make the difference in these totals smaller? (However, the number of squares crossed should remain unchanged!)
I. solution. We choose the center of a light square (small square) as the center $O$ of our circle. The line of the circle intersects 16 squares, which alternate between light and dark, but it is sufficient to consider the arc of the circle that falls into one of the dark squares. Indeed, reflection in any symmetry axis of the square containing $O$ (the two midlines and the two diagonals) will map both the circle and the grid of the board that interests us onto itself, and the colors of the reflected pairs of squares will match. (The same applies to $90^{\circ}$ rotations around $O$, but we don't need this, as the result of a $90^{\circ}$ rotation can be achieved by two consecutive reflections, with the axes being one midline and one diagonal, in any order.)

![](https://cdn.mathpix.com/cropped/2024_05_02_b50c09ce5439a46c2873g-1.jpg?height=765&width=767&top_left_y=474&top_left_x=668)

Our circle does not intersect the dark squares adjacent to the sides of the square containing $O$, because the farthest vertices of these are $\sqrt{1.5^2 + 0.5^2} = \sqrt{2.5} \cdot 2 > 1.9$ from $O$. From these 4 light squares, the circle transitions into one dark square, and these 8 dark squares can be obtained from any of them by the mentioned symmetries. Finally, we mention the light squares that are the second neighbors of $O$ in a row or column; the circle passes close to the center of these (0.1 distance away), but does not include them.

We start drawing the circle in the second neighboring light square to the right of $O$ in the row containing $O$, and proceed in the positive rotational direction. Let the points of entry and exit into the first dark square be denoted by $A$ and $B$, and the angles that $OA$ and $OB$ make with the horizontal (row) axis be $\alpha$ and $\beta$. Thus, the central angle of the arc falling into the dark square is $\beta - \alpha$, the sum of the angles corresponding to the 8 dark arcs is $8(\beta - \alpha)$, and the sum of the angles corresponding to the light arcs is $360^\circ - 8(\beta - \alpha)$.

We show that the sum of the dark arcs is greater than the sum of the light arcs. For this, it is sufficient to show that $\beta - \alpha > 360^\circ / 16 = 22^\circ 30'$. For this,

\[
\begin{array}{ll}
\cos \beta = \frac{1.5}{1.9} = \frac{3}{3.8} = 0.7895 & \text{(rounded up)}, \\
\sin \alpha = \frac{0.5}{1.9} = \frac{1}{3.8} = 0.2632 & \text{(rounded up)},
\end{array}
\]

we chose the direction of rounding so that we write a slightly smaller value for $\beta$ and a slightly larger value for $\alpha$, thus we get a slightly smaller value for $\beta - \alpha$. In this sense,

\[
\beta > 37^\circ 51' \quad \text{and} \quad \alpha < 15^\circ 16',
\]

so

\[
\beta - \alpha > 37^\circ 51' - 15^\circ 16' = 22^\circ 35'.
\]

It is easy to see now that to get closer to the equality of the sums of the dark and light arcs—while maintaining the other conditions—we need to reduce the radius. Thus, $\cos \beta$ increases and $\beta$ decreases, while $\sin \alpha$ and $\alpha$ increase.

Remark. It was noted (at the special session announced for the III-IV. grade contestants on December 21, 1977) that if $O$ is chosen in one of the second rows of squares from the edge of the chessboard, on any internal square, then the sum of the two arcs is not unique. This is indeed the case. However, it is still stated before the word "internal" that the person "drew the circle" - or rather, drew the circle on the board, not an arc! Thus, the word "internal" was actually superfluous and even caused confusion, not what the editorial board intended. Our principle: do not complicate the problems!

II. solution. We calculate for what $r$ the following holds (using the notation from above):

\[
\beta - \alpha = 22^\circ 30'
\]

that is,

\[
\sin (\beta - \alpha) = \sin \beta \cos \alpha - \cos \beta \sin \alpha = \sin 22^\circ 30' = \sqrt{\frac{1 - \cos 45^\circ}{2}} = \sqrt{\frac{2 - \sqrt{2}}{4}}
\]

Let $\sin \alpha = \frac{1}{2r} = z$, so $\cos \beta = 3z$, and our equation becomes

\[
\sqrt{1 - 9z^2} \cdot \sqrt{1 - z^2} - 3z^2 = \sqrt{\frac{2 - \sqrt{2}}{4}}
\]

from which

\[
\begin{gathered}
1 - 10z^2 = \frac{2 - \sqrt{2}}{4} + 3 \sqrt{2 - \sqrt{2}} z^2 \\
\frac{1}{4z^2} = r^2 = \frac{1}{2}(2 - \sqrt{2})(10 + \sqrt{18 - \sqrt{162}})
\end{gathered}
\]

and using school function tables, always to 4 significant figures,

\[
r = 1.898 \quad (<1.9)
\]

This is in agreement with the first solution's conclusion that $r$ needs to be reduced.