[Guest]

here's a "puzzle" i came up with a while ago, let's see if you can get it.

you start of with a circle of radius 1. inside you put a equilateral triangle. inside the triangle, you inscribe another circle. inside that circle you inscribe a square. inside the square another circle, then regular pentagon, then circle, then regular hexagon, etc.

what would be the radius of the inner most circle?

[Guest]

Which one is the most inner circle? If you continue the sequence, there is no "inner most circle". Or do you wish to end the sequence at some point?

[Guest]

klmn, as the sequence approaches infinity, the regular shapes approach a circle.

[Guest]

The way I picture it, as the circle approaches infinity the so called circle will be a dot so 0 will be the radius

[Guest]

If I understood:

Yes they do approach the circle shape, but they don't reach it... in theory... there is always another regular shape between the "almost circle" regular shape and a true circle.

So, the sequence is converging to "point" or a "dot". I'm not sure of english math term... So the radius converges to 0.

Edited July 30, 2009 by klmn

[Guest]

what will the radius of the circle inside the triangle be?

what will the radius of the circle inside the square be?

what will the radius of the circle inside the pentagon be?

and no the limit as number of sides aproach infinity does not give a radius of zero.

[Guest]

I disagree with the previous posts.

this is a similar problem of a limit. As the "regular" shape approaches a sufficiently high number, the radius of the circle converges. It is similar to finding the limit of the sequence 1/n + 1/(n+1) + 1/(n+2)... Though the numbers will extend theoretically into infinity, the limit of this equation does not approach infinity. Now what the equationt to find the radius of the convergent sequence is, I'll leave to someone smarter than me.

[Guest]

to argue that the radius would be zero because the shapes continue to infinity would be like arguing that becuase you can inscribe a shape inside a circle with each side having near zero side length, that the circumference of the circle will be zero.

in the sequence you mention bekabah, i think you mean that just because each individual term approaches zero doesn't mean that the sequence itself approaches zero.

[Guest]

Sounds like an interesting problem. I'll see if I can come up with someting.

[superprismatic]

A quick calculation gives the radius of the innermost circle as Product{n=3 to infinity}cos(pi/n) which is about .115

[Guest]

A quick calculation gives the radius of the innermost circle as Product{n=3 to infinity}cos(pi/n) which is about .115

I agree.

0.1149.......

[Guest]

Letsee. The cirucumradius of a n sided polygon is the circumferance of the circle. This radius bisects the interior angle of the polygon. To find the next circumradius, you multiply the current circumradius by the sine of the bisected angle. For example, the angle of the triangle is 60. Therefore, the next circle will have a radius of 1*sin(30). This is .5 So now we have inscribe a square. The circumradius is .5, the angle is 1/2(90) so the next radius will be .5*sin(45)

So for n=6 we have sin(30)*sin(45)*sin(54)*sin(60).

To find the interior angle of the an n sided polygon is 180(n-2)/n. Each polygon we use is 1/2 that. So I come up with:

infinity

Product sin(90(n-2)/n)

n=3

I didn't know how to solve that. So I looked it up, and I found the Kepler-Bouwkamp Constant: which is

infinity

Product 1/cos(pi/n) ~= 8.7000366

n=3

The inverse of which (and equivilent to what i found above) is

infinity

Product cos(pi/n)

n=3

So the final answer ~= 1/8.7000366 ~= 0.11494205

[Guest]

Yes... I didn't put any math to my solution and certainly not enough attention to the problem (which it deserves)... I see my mistake... It converges to a value and not a point... nice work.