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 2016-03-24, 00:27 #1 a1call     "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.
 2016-03-24, 17:43 #2 a1call     "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.
2016-03-25, 19:19   #3
a1call

"Rashid Naimi"
Oct 2015
Remote to Here/There

7×283 Posts

Quote:
 Originally Posted by a1call [SPOILER] * Describe the process of drawing n=5 such circles using a compass and a straight edge.
Please see the attached drawing.

Quote:
 Originally Posted by a1call [SPOILER] * 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)
Still an open question.
Hint 2: This is a Fermat Primes related question.
Attached Thumbnails

 2016-03-25, 23:06 #4 a1call     "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
 2016-03-26, 18:41 #5 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 7×283 Posts In order to draw n inscribed circle using a compass and a straightedge, It must be possible to draw a regular n-gon 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 2i, where i is any natural number including 0: n = 2i 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
2016-03-27, 17:03   #6
a1call

"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:
 In 1796 (when he was 19 years old), Gauss gave a sufficient condition for a regular n-gon to be constructible, which he also conjectured (but did not prove) to be necessary, thus showing that regular n-gons were constructible for n=3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, 48, 51, 60, 64, ... (OEIS A003401).
http://mathworld.wolfram.com/ConstructiblePolygon.html

About proofs and animations:
Quote:
 constructions of a 17-gon, 257-gon and 65537-gon. Only the first stage of the 65537-gon construction
https://en.wikipedia.org/wiki/Constructible_polygon

 2017-06-25, 19:46 #7 a1call     "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))
2017-06-26, 13:48   #8
Dr Sardonicus

Feb 2017
Nowhere

24×32×29 Posts

Quote:
 Originally Posted by a1call 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.
[nit-pick]When n = 2, there are only two circles, each having only one neighbor[/nit-pick]

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 n-gon.

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 C1-O-T and C2-O-T are both (2*pi/(2*n) = pi/n. The hypotenuses OC1 and OC2 have length R-r. 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 2017-06-26 at 13:55

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