[bonanova]

A moment's reflection will verify that in the plane a set of circles,

each of which touches all the others at unique, single points, numbers at most four.

Let three circles have radii 1, 2 and 3, respectively, and arrange them so each touches the other two externally.

That is, the center of each circle is outside both of the other circles. Like a dime, nickel and quarter might look.

There are now two ways a 4^th circle can be in contact with each of these circles.

Make a sketch of each:

.

1. It may lie in the space between the first three.
2. It may enclose the first three, so that their outer edges each touch the inside of the 4^th circle at different points.

.

Determine the radius of the 4^th circle in each case.

[Guest]

if you draw a line from one radius to another for the three initial circles, you get a right triangle. (3,4,5)

the radius of the large outer circle will be the diameter of the large inner circle. (large inner = 3, large outer = 6)

the radius of the small inner circle will be 1/3 the diameter of the small outer one. (small outer = 1, small inner = 1/3)

[Guest]

A moment's reflection will verify that in the plane a set of circles,

each of which touches all the others at unique, single points, numbers at most four.

Let three circles have radii 1, 2 and 3, respectively, and arrange them so each touches the other two externally.

That is, the center of each circle is outside both of the other circles. Like a dime, nickel and quarter might look.

There are now two ways a 4^th circle can be in contact with each of these circles.

Make a sketch of each:

.

1. It may lie in the space between the first three.
2. It may enclose the first three, so that their outer edges each touch the inside of the 4^th circle at different points.

.

Determine the radius of the 4^th circle in each case.

Does each circle touch each other once and only once?

[bonanova]

Does each circle touch each other once and only once?

Yes. In the words of the OP, "... unique, single points ...."

i.e. tangents, not intersections.

[Guest]

OK. In that case:

I thought this looked familiar...from what I remember about Appolonius(sp) problem with four tangental circles (Descartes):

(1/r₁ + 1/r₂ + 1/r₃ + 1/k)² = 2(1/r₁² + 1/r₂² + 1/r₃² + 1/k²)

(1 + 1/2 + 1/3 + 1/k)² = 2(1 + 1/4 + 1/9 + 1/k²)

(11/6 + 1/k)² = 2 + 1/2 + 2/9 + 2/(k²)

121/36 + 1/k² + 22/(6k) = 49/18 + 2/k²

121/36 - 98/36 + 22/(*k) = 1/k²

23/36 + 22/6k = 1/k²

23k² + 132k = 36

k = -6 (outer) or 6/23 (inner).

I also remember that there is an observation that if you take a set of 4 circles (say the set with the inner), then there exists a second set of circles that also have their tangent points at the same set of tangent points of the original four circles...and that the second set of circles' tangents are perpendicular to the first set. (did i say that right???) I don't remember who observed that...I do remember the name of the mathematician was very similar to a small plane manufacturer (and I only remember THAT because my roommate was taking flying lessons at the same time that I was studying this in school).

[bonanova]

if you draw a line from one radius to another for the three initial circles, you get a right triangle. (3,4,5)

the radius of the large outer circle will be the diameter of the large inner circle. (large inner = 3, large outer = 6)

the radius of the small inner circle will be 1/3 the diameter of the small outer one. (small outer = 1, small inner = 1/3)

Correct for the outer circle; close but not quite for the inner circle.

[bonanova]

OK. In that case:

I thought this looked familiar...from what I remember about Appolonius(sp) problem with four tangental circles (Descartes):

(1/r₁ + 1/r₂ + 1/r₃ + 1/k)² = 2(1/r₁² + 1/r₂² + 1/r₃² + 1/k²)

(1 + 1/2 + 1/3 + 1/k)² = 2(1 + 1/4 + 1/9 + 1/k²)

(11/6 + 1/k)² = 2 + 1/2 + 2/9 + 2/(k²)

121/36 + 1/k² + 22/(6k) = 49/18 + 2/k²

121/36 - 98/36 + 22/(*k) = 1/k²

23/36 + 22/6k = 1/k²

23k² + 132k = 36

k = -6 (outer) or 6/23 (inner).

I also remember that there is an observation that if you take a set of 4 circles (say the set with the inner), then there exists a second set of circles that also have their tangent points at the same set of tangent points of the original four circles...and that the second set of circles' tangents are perpendicular to the first set. (did i say that right???) I don't remember who observed that...I do remember the name of the mathematician was very similar to a small plane manufacturer (and I only remember THAT because my roommate was taking flying lessons at the same time that I was studying this in school).

Bingo.

Yeah, the things we remember sometimes.. .