Soddy-Descartes relation

The Kiss Precise by Frederick Soddy
In Nature #137, 1936

For pairs of lips to kiss maybe
Involves no trigonometry.
'Tis not so when four circles kiss
Each one the other three.
To bring this off the four must be
As three in one or one in three.
If one in three, beyond a doubt
Each gets three kisses from without.
If three in one, then is that one
Thrice kissed internally.

Four circles to the kissing come.
The smaller are the benter.
The bend is just the inverse of
The distance form the center.
Though their intrigue left Euclid dumb
There's now no need for rule of thumb.
Since zero bend's a dead straight line
And concave bends have minus sign,
The sum of the squares of all four bends
Is half the square of their sum.

Expanded to spheres : (Nature #139, 1937)

To spy out spherical affairs
An oscular surveyor
Might find the task laborious,
The sphere is much the gayer,
And now besides the pair of pairs
A fifth sphere in the kissing shares.
Yet, signs and zero as before,
For each to kiss the other four
The square of the sum of all five bends
Is thrice the sum of their squares.

This formula relates the radii of 4 circles mutually tangent.
Let k = 1/r be the "bend" or curvature. Then :

 2(k1² + k2² + k3² + k4²) = (k1 + k2 + k3 + k4)² 

When one circle is around the others, its curvature is taken < 0.
When one circle is a straight line, its curvature is 0.

This formula was first established by Descartes, but brought to fame by Soddy with a poem "The Kiss Precise", and extended to spheres, and n dimensions.

It gives the radius of the 4th circle tangent to three given tangent circles :

  r4 = r1r2r3 / ( r1r2 + r1r3 + r2r3 ± 2√(r1r2r3(r1 + r2 + r3)) ) 
(A script)

We'll give here the proof by Philip Beecroft (1842).
First of all, consider two sets of 4 circles mutually tangent :
Circles (C1) to (C4) with radii r1 to r4 and curvatures k1 to k4
Circles (Γ1) to (Γ4) with radii R1 to R4 and curvatures m1 to m4

4) is the incircle of triangle ABC with sides a, b, c.

Let s the semiperimeter s = (a + b + c)/2, from Heron formulas result :
R4² = (s-a)(s-b)(s-c)/s.
Also r1 = s - a, r2 = s - b, r3 = s - c and r1 + r2 + r3 = s
Hence results :  m4² = k1k2k2(1/k1 + 1/k2 + 1/k3) = k1k2 + k2k3 + k1k3  
 

Considering in the same way circles (Γ1) to (Γ3) with centers O1 to O3 and incircle C4 in O1O2O2 :
 k4² = m1m2 + m2m3 + m1m3 
As well as all formulas from swapping the indexes (considering excircles).

We get then the sum of the squares of ki's and mi's :
 ∑ki² = 2∑mimj and ∑mi² = 2 ∑kikj 

Calculate then the square of the sum of ki's : (∑ki)² = ∑ki² + 2∑kikj = ∑ki² + ∑mi²
And similarily (∑mi)² = ∑mi² + ∑ki²
Hence (∑ki)² = (∑mi)², and as they are >0 :  ∑ki = ∑mi 

Calculate then the value :
(k1 + k2 + k3 + k4)(k1 + k2 + k3 - k4) = (k1 + k2 + k3)² - k4² = k1² + k2² + k3² + 2m4² - k4² =
2,3,4mimj + ∑1,3,4mimj + ∑1,2,4mimj - ∑1,2,3mimj + 2m4² = 2(m1m4 + m2m4 + m3m4) + 2m4² = 2m4(∑mi)

And, as ∑mi = ∑ki :  k1 + k2 + k3 - k4 = 2m4 
Squaring the 4 similar relations and summing up, all terms as kikj cancel and remains :
 ∑ki² = ∑mi² 

Hence (∑ki)² = ∑ki² + 2∑kikj = ∑ki² + ∑mi² = 2∑ki²
That is finally the Descartes relation :

 (∑ki)² = 2∑ki² 

 

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