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Prove: I define the radius of three mutually externally tangent to be $d,e,f$ respectively. The circle with radius $x$ is internally tangent to all three circles. Then

$$ddeeff+ddeexx+ddffxx+eeffxx = \\2(deffxx+ddeffx+deefxx+ddeefx+ddefxx+deeffx)$$

[Reference: p.189-190 of The Changing Shape of Geometry.]

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The phrase "a very original version" perplexes me. Do you simply want some proof of Descartes' own statement of the theorem? Are you looking for Descartes' own proof? In any case, a Google search of "proof of Descartes Circle Theorem" generates 2.5 million results; what are you seeking here that can't be found among those? –  Blue May 2 '12 at 0:01
    
@DayLateDon - i think this is most basic version of the theorem, that is, even a HS student could understand its statement before it is converted into more complicated form. In addition i think the early version may easier to prove. –  Victor May 2 '12 at 0:08
    
@DayLateDon - Any proof would work for me, but more alternate way tend to be better. –  Victor May 2 '12 at 0:09

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