A rectangle has side lengths $a$ and $b; a \leq \sqrt{3}$. How many unit-radius circles are needed to cover the rectangle?
We will show that the number of circles required for the coverage is $\left\lceil\frac{b}{\sqrt{4-a^{2}}}\right\rceil$ (where $\lceil x\rceil-$ denotes the ceiling of $x$ - the smallest integer not less than $x$).

![](https://cdn.mathpix.com/cropped/2024_05_02_550d08d31ae2efc3b5d3g-1.jpg?height=442&width=439&top_left_y=302&top_left_x=824)

Figure 1

First, we will show that the combined length of two parallel chords of a unit circle, which are at a distance of $a \leq \sqrt{3}$ from each other, is at most $2 \sqrt{4-a^{2}}$. If the diameters parallel to $h_{1}$ and $h_{2}$ do not separate them, then reflecting the line of the farther chord from the center ($h_{2}$) over the line of the other chord results in a combined length of $h_{1}$ and $h_{2}^{\prime}$ that is clearly greater than the sum of the lengths of $h_{1}$ and $h_{2}$ (Figure 1).

By repeating this step a finite number of times, we can ensure that the center of the circle lies between the two chords. During this process, the distance of the chords from the center does not change. In this case, if the distance from the center of the circle to one of the chords is $c$, then by the Pythagorean theorem, the combined length of the two chords (Figure 2) is: $2\left(\sqrt{1-c^{2}}+\sqrt{1-(a-c)^{2}}\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_02_550d08d31ae2efc3b5d3g-1.jpg?height=499&width=461&top_left_y=1171&top_left_x=821)

It is sufficient to show that

$$
\sqrt{1-c^{2}}+\sqrt{1-(a-c)^{2}} \leq 2 \sqrt{1-\left(\frac{a}{2}\right)^{2}}
$$

Squaring both sides and rearranging:

$$
\sqrt{\left(1-c^{2}\right)\left(1-(a-c)^{2}\right)} \leq 1+c^{2}-a c
$$

Squaring both sides again and rearranging: $0 \leq(2 c-a)^{2}$.

This is clearly true, and since we squared positive numbers, (1) also holds. Therefore, a unit circle can cover at most $2 \sqrt{4-a^{2}}$ units of the combined length of the two $b$-length sides of the rectangle. If, however, the circles cover the entire rectangle, they must also cover the $2b$-length sides. Therefore, at least $\left\lceil\frac{b}{\sqrt{4-a^{2}}}\right\rceil$ circles are required for the coverage.

![](https://cdn.mathpix.com/cropped/2024_05_02_550d08d31ae2efc3b5d3g-1.jpg?height=426&width=1116&top_left_y=2237&top_left_x=491)

Figure 3

It is easy to see that this number is sufficient. If the circles are placed as shown in Figure 3 - their centers on one of the symmetry axes of the rectangle, and their intersection points on the perimeter of the rectangle - they will cover the rectangle. This completes the proof of our statement.

Remark. The equality (1) can more simply be derived from the fact that the function $y=\sqrt{1-x^{2}}$ (whose graph is a quarter circle) is convex from below (Figure 4).

![](https://cdn.mathpix.com/cropped/2024_05_02_550d08d31ae2efc3b5d3g-2.jpg?height=521&width=458&top_left_y=409&top_left_x=823)