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"When you draw a circle in a plane of radius 1 you can perfectly surround it with 6 other circles of the same radius."

BUT when you draw a circle in a plane of radius 1 and try to perfectly surround the central circle with 7 circles you have to change the radius of the surround circles.

how can i find the radius of the surround circles if i want too use more that 6 cirles?

ex : 7 circles = radius 0.4

8 circles = radius 0.2

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have to chance or have to change ? –  xport Nov 28 '10 at 4:03
    
"have to change" sorry. –  Mateuszk Nov 28 '10 at 4:06
    
This is related. –  J. M. Nov 28 '10 at 14:52
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2 Answers

Imagine there are $n \geq 3$ circles surrounding your unit circle. Then the situation would look like:

alt text

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 $r=\dfrac{\cos(67.5^\circ)}{1-\cos(67.5^\circ)}=0.619914\ldots$.
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Let's recollect a very simple fact about the situation when having n identical circles, $n\geq3$ perfectly surrounding the unit circle. If we connect all intersection points of these circles with the unit circle we get a regular polygon inscribed to the unit circle. The side length formula of the regular polygon inscribed to a circle is $l=2r\sin(\frac{\pi}{n})$, $n$ - the number of the sides of the polygon and is equal to the number of the surrounding circles. In our case $l=2\sin(\frac{\pi}{n})$. Now imagine that in all the picture you have there remains only the unit circle and two of the surrounding circles that are against each other. Connect the three centers of these circles and also consider the side of the polygon inside to the unit circle and get similar triangles that by Thales theorem yield:

$$ \frac{l}{2r}=\frac{1}{1+r};r=\frac{2\sin(\frac{\pi}{n})}{2-2\sin(\frac{\pi}{n})}=\frac{\sin(\frac{\pi}{n})}{1-\sin(\frac{\pi}{n})}.$$

The proof is complete.

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