From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Very difficult geometry problem...
Date: 10 Jan 2000 03:53:11 -0500
Newsgroups: sci.math
Summary: [missing]

In article <slya4yt7q7ru@forum.swarthmore.edu>,
Bill Kinnersley <bkinners@hotmail.com> wrote:
:Okay, here's a geometry problem that I never quite solved.  I worked
:it down to the point where I simply had an equation that needed
:solving, so I resorted to using a TI-92, and found the answer. 
:Unfortunately, I was supposed to be able to solve the problem without
:a calculator.  :(
:
:Three mutually externally tangent circles have radii of lengths 3, 4,
:and 5.  A fourth circle is drawn which is externally tangent to these
:three circles.  What is its radius?
:
:Any ideas?  

There is a planar transformation called inversion with respect to a
circle (centre C, radius r): every point X with CX = u>0 is mapped to a
point Y on the ray CX, but with CY = r^2/u . The point C has no image in
the plane, but this is no big problem.

Inversion turns most circles into circles, except those which pass through
C; they get mapped into straight lines (except for one point, of course).

Taking advantage of the external tangency: select an inversion circle so
that its centre is at one of the points of tangency. The inversion will
turn the two circles into parallel lines, and the third circle into a
circle. So, the transformed problem reads:

Find a circle which touches two parallel lines and a given circle. That is
an easy ruler-and-compass problem (draw an auxiliary circle touching the
two lines, and perform a suitable parallel translation).

Then re-invert the solution.

If you place the coordinate system suitably, you can base the calculations
on this construction.

Good luck, ZVK(Slavek).
==============================================================================

From: Jim Ferry <"jferry"@[delete_this]uiuc.edu>
Subject: Re: Very difficult geometry problem...
Date: Mon, 10 Jan 2000 14:03:54 -0600
Newsgroups: sci.math

Zdislav V. Kovarik wrote:
> 
> There is a planar transformation called inversion

There is an even more elegant way to do the Soddy circle
problem.  You can treat the circle
a (x^2 + y^2) - 2 p x -2 q y + r = 0
as the vector (a,p,q,r), then define the inner product
[c1.c2] = p1 p2 + q1 q2 - (r1 a2 - r2 a1)/2.

See Chapter 40 in _Geometry: A Comprehensive Course_ by Dan Pedoe.

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