This is one of fifty problems on a local math test, so it shouldn't be real difficult, but neither I nor my friends have been able to solve it.

Three circles, two with radius a, and one with radius b, are externally tangent to one another. A fourth circle is circumscribed around the other three so that the first three are all internally tangent to the fourth circle. Find the radius of the fourth circle in terms of a and b.

The problem actually told you a and b, but I forgot them. I'd be interested if someone could find a solution where the three circles in the middle all have different radii.

Thanks

I'm afraid you're not going to like this ...

There does exist a very neat solution. This is via a formula of Descartes, which states that for four circles, all tangent to one another, we have

$({\kappa }_1+{\kappa }_2+{\kappa }_3+{\kappa }_4{)}^2=2({\kappa }_1^2+{\kappa }_2^2+{\kappa }_3^2+{\kappa }_4^2)$.

Here the curvature values ${\kappa }_i$ are defined in terms of the radii $r_i$ of the circles by

1. $\left|{\kappa }_i\right|=1/r_i$
2. If the circles with radius $r_i$ and $r_j$ touch externally, then ${\kappa }_i$ and ${\kappa }_j$ have the same sign. If they touch internally, the curvatures have opposite signs.

David,
I suppose you use inversion here as well....

love arun

Sorry I took so long to acknowledge your reply, I read your solution before 1st period in the library, but didn't have enough time to post. On Friday, Saturday, Sunday, I got web site not responding messages.

Your solution interests me, since we just covered curvature in math class a week or so ago. I asked my math teacher what the relation between curvature and the four tangencies is, and he said it was simply that curvature and coordinates determine a circle (so that it was a formula of radii). I guess this might be so, I was hoping there was some proof involving curvature. He also said something about repeated roots.

Thanks for your answer!

PS-I've heard of inversion, but I'm not exactly sure what it is.

Inversion!!!I don't seem to get that idea very fast...

You may like to have a look at this site...
http://www.maths.gla.ac.uk/~wws/cabripages/inversive/inversive0.html

love arun