# Archimedean Twin Circles:

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

GSP for above picture

## Solution:

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.

## Conclusion:

Since x = y then the circles ( yellow and purple ) have same radius so these two circles are congruent.