## Tuesday, January 26, 2010

### Fox 235

#### 8 comments:

1. This one is really interesting:)

2. perpendiculars to tangent points will give the solution

3. We can use the well-known result of the length of common external tangent to two touching circles as being 4√(product of radii). If radii of the bottom three semi-circles are denoted as r1,r2 & r3 then (r2)^2 = r1*r3 by similar triangles. The second rung circles have radii of √(r1r2) and √(r2r3) which means radius of B is √√(r1r2)(r2r3) or √(r2(√r1r3))=√(r2r2)=r2=Radius of A.
Ajit

4. Typo - Please read: the length of common external tangent to two touching circles as being 2√(product of radii).
Ajit

5. where can u get the info about "the length of common external tangent to two touching circles as being 2√(product of radii)."

any link u suggest?
i cant find one....:(

6. The following can help:
http://www.8foxes.com/Home/238
See the discussion, it is solved by Phytagoras:
http://8foxes.blogspot.com/2010/02/fox-238.html

Now is |BC|=0 => Comman tangent line:
TU^2=AB*CD
TU^2=2r1*2r2, where r1 and r2 are radiuses
TU^2=4r1r2
TU=2Vr1r2
Actually, TU is the geometric mean of the diameters (2r1 and 2r2).
Hope this helps.
-8foxes

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