How many circles (radius $r$) are needed to cover circle whose radius is $2$ times bigger (radius $2r$).
I think we need to use area which is $S=\pi R^2$ but I don't really know what to do.
Geometry – How Many Circles to Cover a Circle Twice as Large
circlesgeometry
Related Solutions
If you connect the centers of two adjacent little circles and the center of the big one, you'll get a triangle. The sides of this triangle have lengths $r+R, r+R$ and $2r$. A little trigonometry will get you that the top angle is
$$\theta=2\arcsin\left(\frac{r}{r+R}\right) \; .$$
Since you want the small circles to form a closed ring around the big circle, this angle should enter an integer amount of times in $360°$ (or $2\pi$ if you work in radians). Thus,
$$\theta=360°/n \; .$$
From this, you can compute that
$$r=\frac{R \sin(180°/n)}{1-\sin(180°/n)} \; .$$
Here's a plot of the result for $n=14$:
Here's the code in Scilab:
n=14;
R=1;
r=R*sin(%pi/n)/(1-sin(%pi/n));
theta=2*%pi*(0:999)/1000;
plot(Rcos(theta),Rsin(theta));
hold on;
for k=0:(n-1),
plot((r+R)*cos(2*k*%pi/n)+r*cos(theta),(r+R)*sin(2*k*%pi/n)+r*sin(theta));
end;
hold off;
Draw the two lines through the center of the central circle that are tangent to one of the touching circles. Call the angle between them $\theta$. The question then is how many times will $\theta$ go into the full circle? You need $\left\lfloor\dfrac{2\pi}{\theta}\right\rfloor$, or if you're using degrees, $\left\lfloor\dfrac{360}{\theta}\right\rfloor$.
Now draw the line through the center of the two circles. The angle between that line and one of the lines you drew earlier is $\theta/2$. The distance between the centers is $R+r$. Now draw the segment from the center of the touching circle to the tangent line. Its length is $r$. Now you have a right triangle in which the hypotenuse has length $R+r$ and one of the legs has length $r$ and the angle opposite that leg is $\theta/2$. Therefore $$ \sin\frac\theta2 = \frac{r}{R+r}. $$ Take arcsines and multiply by $2$ and you've got $\theta$.
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Best Answer
This is a disk covering problem. As you have stated it, it is not quite as difficult as some others.
The first task is to find the minimum number of small circles which cover the circumference of the bigger circle rather than the whole area. If this is $m$ then it will be impossible to have a regular $m$-gon with edges of length $2r$ fitting strictly inside the circle of radius $2r$. This implies $m \ge 6$.
It turns out to be just possible to cover the circumference of the bigger circle with 6 smaller circles, and ignoring symmetries there is only one way to do it (you get a regular hexagon whose vertices lie on the bigger circle and whose edges are diameters of the smaller circles). But this leaves an uncovered area in the middle, which needs at least one (and in fact exactly one) more.
So the answer is seven smaller circles.