Sum of circumferences

[Barcallica]

What is the sum of circumferences of the infinite stack of circles in the pitcure?

What is the sum of areas?

 

Edited August 28, 2014 by Barcallica

[bonanova]

Diameter D₁ of the incircle is some fraction f of the triangle's height h.

The next circle has a diameter D₂ that is the same fraction f of the remaining height h₂ = h - D₁.

And so forth. Here, since h = 12 and D₁ = 20/3, we have f = 5/9.

The successive diameters are:

f h                          =                       h f      =  6.666666667

f (1 - h f)                =                  f - h f²     =  2.962962963

f (1 - f + h f² )         =             f -  f² + h f³   =  1.316872428

f (1 - f + f² - h f³)   =    f  -  f²+ f³ - h f⁴    =  0.585276635

  

and so on.

 

The corresponding areas are 34.90658504, 6.895127909, 1.362000575, 0.269037151 ...

Taking the first 50 terms, the diameters sum to 12.00000000

And the areas sum to 43.4989752.

 

If we note that successive diameters decrease by a factor of (1 - f)

Then the areas of successive circles decrease by a factor of r = (1 - f)².

Then the areas sum to A₁/(1 - r).

 

That's just pi (10/3)²/[1 - (4/9)²] = 34.9066/.80247 = 43.4989

 

[DeGe]

Circumference is quite straight forward

Circumference of each circle is 2.pi.r
So for all the circles it will be 24pi
Because the length of altitude is 12 (with sides 13 & 5)

[DeGe]

For area, had to check the web on how to calculate the radius of incircle
So, the first circle has radius 10/3
So the first circle covers 2/3 of the altitude length
Without going into details, I would say by symmetry that each circle covers the next 2/3rd
So the second circle would be of radius = 1/3(10 - 20/3) = 10/9
ratio of each consecutive circle radius = 1/3
So sum of areas = pi(r1² + r2² + ...)
= pi.(10/3)² / (1 - 1/9) = (25/4)pi

[Barcallica]

For area, had to check the web on how to calculate the radius of incircle

So, the first circle has radius 10/3

So the first circle covers 2/3 of the altitude length

Without going into details, I would say by symmetry that each circle covers the next 2/3rd

So the second circle would be of radius = 1/3(10 - 20/3) = 10/9

ratio of each consecutive circle radius = 1/3

So sum of areas = pi(r1² + r2² + ...)

= pi.(10/3)² / (1 - 1/9) = (25/4)pi

The first circle has radius 10/3 but doesn't cover 2/3 of the altitude length

[bonanova]

For each circle,

C_i = pi D_i

Sum(C_i) = pi Sum(D_i) = 12 pi