20160324, 00:27  #1 
"Rashid Naimi"
Oct 2015
Remote to Here/There
7·283 Posts 
Inscribed Circles
Hi,
Find the trigonometric formula for the radius r1 of n equal, tangent and inscribed circles inside a larger circle of radius r2 . To clarify all the circles are tangent to their 2 equal neighbors and also to the large circle. 
20160324, 17:43  #2 
"Rashid Naimi"
Oct 2015
Remote to Here/There
7·283 Posts 
[B]r1 = (r2 sin (180/n))/(1+sin (180/n))[/B]
Follow up questions: * For what values of n can the above equal, tangent and and inscribed circles inside a larger circle be drawn using a compass and straight edge? (Hint: this is a prime numbers related question) * Describe the process of drawing n=5 such circles using a compass and a straight edge. 
20160325, 19:19  #3  
"Rashid Naimi"
Oct 2015
Remote to Here/There
7×283 Posts 
Quote:
Quote:
Hint 2: This is a Fermat Primes related question. 

20160325, 23:06  #4 
"Rashid Naimi"
Oct 2015
Remote to Here/There
7·283 Posts 
Here are the detailed instructions for drawing the pentagons:
http://www.mathopenref.com/constinpentagon.html 
20160326, 18:41  #5 
"Rashid Naimi"
Oct 2015
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7×283 Posts 
In order to draw n inscribed circle using a compass and a straightedge, It must be possible to draw a regular ngon using a compass and a straightedge. With the exception of the special cases n=1 and n=2, which are possible to draw, n must be the product of any number of distinct Fermat Primes m times 2^{i}, where i is any natural number including 0:
n = 2^{i} ยท m So n=6 has a solution but n=9 does not, n=17 has a solution but n=19 does not. https://en.m.wikipedia.org/wiki/Prim...tible_polygons 
20160327, 17:03  #6  
"Rashid Naimi"
Oct 2015
Remote to Here/There
7·283 Posts 
Due to overwhelming interest in this subject matter here are some more details:
Quote:
About proofs and animations: Quote:


20170625, 19:46  #7 
"Rashid Naimi"
Oct 2015
Remote to Here/There
7·283 Posts 
For the record:
The trigonometric formula for the radius r1 of n equal, tangent and inscribed circles inside a larger circle of radius r2 is: r1 = (r2 sin (180/n))/(1+sin (180/n)) 
20170626, 13:48  #8  
Feb 2017
Nowhere
2^{4}×3^{2}×29 Posts 
Quote:
In any case, the radii of the large circle passing through the centers of the smaller circles, intersect the large circle at the vertices of a regular ngon. Now, let O be the center of the large circle, and let C1 and C2 be the centers of two adjacent small circles. let T be the point of tangency between these two small circles. Let R be the length of the radius of the large circle, and r the common radius of the small circles. Then O, C1, and T and O, C2 and T form congruent right triangles*. The angles C1OT and C2OT are both (2*pi/(2*n) = pi/n. The hypotenuses OC1 and OC2 have length Rr. The legs C1T and C2T have length r, which is also (R  r)*sin(pi/n). Thus r = (R  r)*sin(pi/n) or r = R*sin(pi/n)/(1 + sin(pi/n)). *If n = 2, the "right triangles" collapse; in this case O = T. Last fiddled with by Dr Sardonicus on 20170626 at 13:55 

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