Start with a circle O of radius 1 and let a and b be arbitrary real numbers.

geometry/six.tangent.circles New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

Start with a circle O of radius 1 and let a and b be arbitrary real
numbers.
Construct a first circle of radius a tangent to circle O.
Construct a 2nd circle of radius b tangent to O and the a-circle.
Construct a 3rd circle of radius b/a tangent to O and the b-circle.
Construct a fourth circle of radius 1/a tangent to O and the b/a-circle.
Construct a fifth circle of radius 1/b tangent to O and the 1/a-circle.
Construct a sixth circle of radius a/b tangent to O and the 1/b-circle.
...
Prove that the sixth circle is tangent to the first circle.

-- Frank Pinto <rolls@cis.umassd.edu>

[ Editor: the "..." is in the problem as quoted in the only collected
reply. ]

Show solution

Here's a more serious effort.  This time I assume the six circles are
constructed around the circumference of the unit circle.

Let O be the center of the unit circle, A be the center of the circle
with radius a, B be the center of the circle of radius b, C be the
center of the circle with radius b/a, ..., and let F be the center of
the circle of radius a/b.   In order to show tangency of the last
circle with the first, construct a seventh circle centered at G of
radius a, tangent to the unit circle and to the a/b-circle.  We want to
show that G and A are coincident.

Note the lengths of the various line segments:

OA=a+1   OB=b+1     OC=b/a+1   OD=1/a+1    OE=1/b+1    OF=a/b+1
OG=a+1
    AB=a+b    BC=b+b/a  CD=b/a+1/a  DE=1/a+1/b  EF=1/b+a/b  FG=a/b+a

Triangle OAB is similar to triangles DOC and EFO, since
OA/DO=AB/OC=BO/CD=a and OA/EF=AB/FO=BO/OE=b.

Therefore the sum of the angles AOB, COD, and EOF is equal to 180
degrees.

By the same argument, the sum of the angles BOC, DOE, and FOG is 180
degrees.

Therefore the sum of the six central angles AOB, BOC, COD, DOE, EOF,
FOG [Editor: printed as EOG] is 360 degrees, which means that G is
coincident with A as desired.

Clever puzzle.

-- Ken Smith <ken@jpl354.jpl.nasa.gov>