Inscribed Circles
 

Hi,
Find the trigonometric formula for the radius [B]r1[/B] of [B]n equal[/B], [B]tangent [/B]and [B]inscribed [/B]circles inside a larger circle of radius [B]r2 [/B].
To clarify all the circles are tangent to their 2 equal neighbors and also to the large circle.

[SPOILER][B]r1 = (r2 sin (180/n))/(1+sin (180/n))[/B][/SPOILER]

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 [B]prime numbers[/B] related question)

* Describe the process of drawing n=5 such circles using a compass and a straight edge.

[QUOTE=a1call;429980][SPOILER]
* Describe the process of drawing n=5 such circles using a compass and a straight edge.[/QUOTE]

Please see the attached drawing.

[QUOTE=a1call;429980][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 [B]prime numbers[/B] related question)

[/QUOTE]
Still an open question.
Hint 2: This is a [B]Fermat Primes[/B] related question.

Here are the detailed instructions for drawing the pentagons:
[url]http://www.mathopenref.com/constinpentagon.html[/url]

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 2[SUP]i[/SUP], where i is any natural number including 0:
n = 2[SUP]i[/SUP] · m

So n=6 has a solution but n=9 does not, n=17 has a solution but n=19 does not.:smile:

[url]https://en.m.wikipedia.org/wiki/Prime_number#Fermat_primes_and_constructible_polygons[/url]

Due to overwhelming interest :smile: 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).[/QUOTE]

[url]http://mathworld.wolfram.com/ConstructiblePolygon.html[/url]

About proofs and animations:
[QUOTE] constructions of a 17-gon, 257-gon and 65537-gon. Only the first stage of the 65537-gon construction
[/QUOTE]

[url]https://en.wikipedia.org/wiki/Constructible_polygon[/url]

For the record:

The trigonometric formula for the radius [B]r1[/B] of [B]n[/B] equal, tangent and inscribed circles inside a larger circle of radius [B]r2[/B] is:
[B]r1 = (r2 sin (180/n))/(1+sin (180/n))[/B]

[QUOTE=a1call;429903]Hi,
Find the trigonometric formula for the radius [B]r1[/B] of [B]n equal[/B], [B]tangent [/B]and [B]inscribed [/B]circles inside a larger circle of radius [B]r2 [/B].
To clarify all the circles are tangent to their 2 equal neighbors and also to the large circle.[/QUOTE][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.