The idea of Archimedean twin circles is that if we have three semi circles such that the semicircle with diameter AB and center O, the semicircle with diameter AG and center O1, and the semicircle with diameter GB and center O2, then the inscribed circles with radius x and y are same, it means they both have the same radius such that x=y. Also this theorem includes that x=y= a.b/a+b, this means radius of these circles are equal to harmonic mean of a and b.
A picture for this theorem
The link for this problem:
http://www.gogeometry.com/problem/p295_archimedes_twins_harmonic_mean_radius.htm
GSP for above picture
I solve this problem using Pythagorean theorem. First, I am going to show x= a.b/a+b. To show this we can draw a perpendicular from center K to base AB. I called h to perpendicular segment from K to H. Since O is the center of larger semicircle with radius a+b and GB= 2b then OG= a+b - (b+b) = a-b. Since x parallel to AB then HG= x. OG = a-b and HG = x then HO = x-(a-b)= x+b-a and O1H= O1G- HG = a-x. Since HK = radius = x then OK = a+b-x. It will be clear if we look at below picture.
Picture for above eplanations.
Now, if we apply Pythagorean theorem, then we will get
HK² + O1H² = O1K²
therefore (a-x)² + h² = (a+x)² (1) and
HK² + OH² = OK²
thus h² + (x+b-a)² = (a+b-x)² (2)
If we do calculation for (1) then we get a² - 2ax + x² + h² = a²+ 2ax + x² <----> h² = 4ax. If we replace h² = 4ax in the second equation then we can get
4ax + ( x+b-a)² = ( a+b-x)² <-----> 4ax + (x+b)² - 2a(x+b) + a² = (a+b)² - 2x(a+b) +x² <----> 4ax + x² + 2bx + b² - 2ax -2ab + a² = a²+ 2ab + b² - 2ax- 2bx +x²
<----> 4ax + 4bx = 4ab <----> x(a+b) = ab <----> x = a.b / a+b.
If we apply the same method for y then we will get
JM² + OM² = OJ² therefore k² + (a-b+y)² = (a+b-y)² (3) and
JM² + MO2² = JO2² (4) thus k² +(b-y)²= (b+y)²
Therefore solving (3) and (4) will give us y = a.b / a+b. The below picture explains how to find y value.
Since x = y then the circles ( yellow and purple ) have same radius so these two circles are congruent.