How many circles of radius $r$ can be placed around a circle of radius $R$ (close to it)? $r$ can be bigger, equal or smaller than $R$.
Geometry – How Many Equal Circles Can Be Placed Around a Circle?
circlesgeometry
Related Solutions
Imagine there are $n \geq 3$ circles surrounding your unit circle. Then the situation would look like:
Whence $\cos(x)=\frac{r}{r+1}$. The angle $x$ is half of the interior angle of the corresponding regular polygon, so $x=\frac{n-2}{2n} \cdot 180^\circ$. You can then solve for $r=\dfrac{\cos(x)}{1-\cos(x)}$.
For example
- when $n=4$ we have $r=\dfrac{\cos(45^\circ)}{1-\cos(45^\circ)}=1+\sqrt{2}=2.41421\ldots$.
- when $n=6$ we have $r=\dfrac{\cos(60^\circ)}{1-\cos(60^\circ)}=1$.
- when $n=8$ we have $r=\dfrac{\cos(67.5^\circ)}{1-\cos(67.5^\circ)}=0.619914\ldots$.
If the radius of circle is $R$, then I get the length of the triangle as $(6+2\sqrt{3})R$.
Drop a perpendicular from the center of the rightmost bottom circle to the bottom side.
The distance from the bottom right vertex of the triangle to the base of the perpendicular is $D = \sqrt{3}\ R$, as we get a triangle of angle $30^{\circ}$ and we get $\tan(30^{\circ}) = \frac{R}{D}$. Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$, we get $D = \sqrt{3}\ R$.
Now drop perpendiculars from the centers of the bottom circles to the bottom side and add up the lengths of the segments formed on the bottom side.
Area of an equilateral triangle of side $a$ is $\sqrt{3}\ a^2/4$.
Best Answer
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 $\theta$.