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Ha, finally no knot theory :-)

First of all, let's define the "power line" of three circles. (Very probably, someone had the idea before me, but no math forum ever came up with something .) Call the circles c1,c2,c3 and "draw" some circle c0 such that the intersection angle @ between c1,c0 = that between c2,c0 = c3,c0. I use " because c0 might be not drawable due to negative or imaginary radius; it surely is in the Gergonne case below. But you can as well use cos(@)=(r1^2+r2^2-d12^2)/(2*r1*r2) instead. The locus of all c0 centers is (for general c1,c2,c3) a line. This is immediately clear in analytic form, just subtract the circle equations. I call this line "power line" p. The center of the in-in-in and out-out-out Apollonius circles as well as the radical center obviously lie on p. (cos(@)=-1,1,0)

It is fun to draw the power line for three prominent circles of a triangle, e.g. the excircles or Soddy circles. In the latter case p is the Soddy line, of course. Now the Gergonne point has an unique property: It's value of cos(@) is just a number (i.e. independent of the triangle shape). (Unique in the sense cos(@) != 0,1,-1 and for not too obscure circles of a triangle. I checked many.)

I won't spoil your fun to compute cos(@). In any computer algebra system this is done in a jiffy, but although I'm just an amateur, I'd like a purely geometric proof of the theorem much better as probably anyone else. Can you come up with one?

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Added tag "geometry". Someone already voted to close. Perhaps you should try a site like artofproblemsolving.com/Forum since there is an aversion to "puzzle" type problems here. –  Gerald Edgar May 9 '11 at 16:41
    
To whoever votes to close this: This is more interesting than most of the standard triangle geometry riddles (already because not many people have considered power lines before, and because this might lead to $n$-dimensional generalizations of the Gergonne point), and I don't believe posting it on artofproblemsolving.com/Forum would be of much help nowadays (it certainly would have done so a few years ago). But have you tried Hyacinthos, Hauke? –  darij grinberg May 11 '11 at 7:22
    
Hi Darij, where have you been? :-) Yes, I mentioned this somewhen on Hyacinthos, but neither was the "power line" known there nor did someone attempt a geometric proof (as I said, drawing imaginary circles is a bit complicated :-) In any case, I don't object to closing the question; the solution BTW is 1/Sqrt[-3]. –  Hauke Reddmann May 13 '11 at 13:20
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