54. What is the smallest number of circles with which a circle of twice the radius can be completely covered?
54. Since the diameter of the smaller circles is equal to the radius $R$ of the larger circle, each such circle intersects the circumference of the larger circle at points no more than $R$ apart, and thus covers an arc of this circumference no greater than $60^{\circ}$. It follows that at least six smaller circles are required to cover the entire circumference of the larger circle; in this case, six circles are sufficient only when these circles have diameters equal to the six sides of a regular hexagon inscribed in the larger circle, and in all other cases, at least seven circles are required. However, if the circumference of the larger circle is covered by six circles, then at least one more small circle is required to cover the remaining part of the larger circle (since the first six circles do not cover, for example, the center of the larger circle). In this case, one circle is sufficient to cover the remaining part of the larger circle (Fig. 143; from the fact that $O_{1} M=\frac{1}{2} R$, it follows that $O_{1} M$ is the median of the right triangle $O O_{1} A$ with an angle of $30^{\circ}$ and, therefore, $\left.O M=\frac{1}{2} R\right)$. Thus, we see that the smallest number of circles with radius $\frac{1}{2} R$ required to cover a circle with radius $R$ is 7.