A unit-radius closed circular disk is covered with smaller, congruent closed circular disks.

a) What is the minimum number of covering disks needed?

b) What is the minimum radius of the covering disks if the smallest number of disks is used?
First, we show that two circles are not sufficient for the covering. A circle with a radius smaller than 1, if it intersects the unit circle at two points, can cover less than half of its circumference, since the distance between the two intersection points of the circles is less than the diameter of the unit circle. However, if the two (non-overlapping) circles have at most one common point, then the smaller circle can cover at most one point.

Three circles, however, are sufficient. The sides of an equilateral triangle inscribed in the unit circle, as diameters, cover the unit circle (Figure 1). Therefore, at least three covering circles are needed.

In the case of Figure 1, the radius of the covering circles is $\frac{\sqrt{3}}{2}$. If three circles with a radius smaller than this are taken, the covering fails because each circle covers less than a third of the unit circle's circumference. Therefore, the radius of the three congruent covering circles must be at least $\frac{\sqrt{3}}{2}$.

Rudolf Gábor (Fazekas M. Főv. Gyak. Gimn., III. o.t.) and Terék Zsolt (Fazekas M. Főv. Gyak. Gimn., II. o.t.)

Remarks. 1. It can be shown that not only a 2-unit diameter circle, but any planar point set with a diameter of at most 2 units can be covered by three circles with a radius of $\frac{\sqrt{3}}{2}$. To prove this, we use the following theorem: Any planar point set with a diameter of at most $d$ can be covered by a regular hexagon with a side length of $\frac{d}{\sqrt{3}}$. The proof of this theorem for convex domains, known as Gyula Pál's theorems, can be found on pages 295-296 of István Reiman's book "Geometry and Its Boundary Areas." Given this theorem, we only need to show that a regular hexagon with a side length of $\frac{2}{\sqrt{3}}$ can be covered by three circles with a radius of $\frac{\sqrt{3}}{2}$. In Figure 2, points $E, F, G$ are the midpoints of three sides of a regular hexagon with a side length of $\frac{2}{\sqrt{3}}$. Clearly, triangle $EFG$ is equilateral, and its side length is $\sqrt{3}$. Based on Figure 2, we can see that the circles with diameters equal to the sides of this triangle cover the hexagon, and thus also cover any 2-unit diameter shape covered by the hexagon.

2. Gergely Bérczi (Szeged, Ságvári E. Gyak. Gimn., II. o.t.) pointed out the following generalization possibility: Consider a bounded and convex planar figure $F$ instead of a circle. We want to cover $F$ with $m$ smaller, not necessarily congruent, centrally similar figures. What is the smallest positive integer value of $m$? Our problem is a special case of the problem posed by I. C. Gohberg and A. Sz. Markus in 1960, when the figure $F$ is a circle. In this case, the minimum value of $m$ is 3.

The mentioned authors proved that the smallest value of $m$ is 4 for a parallelogram and 3 for any other bounded convex figure $F$. The elementary but rather lengthy proof can be found on pages 53-64 of V. G. Boltyanski and I. C. Gohberg's book "Decomposition of Figures into Smaller Parts."

For the second part of the problem, i.e., how large the covering figures must be when $m$ is the smallest, there is no simple answer, as the answer depends on the shape of the convex figure.
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