The three circles of radii 1, 2 and 3 tangent to each other contemplated last week will have their centers at the vertices of a triangle with sides 3, 4 and 5 (1+2, 1+3 and 2+3), that is, of a right triangle, so it is not difficult to draw them without help other than a compass (and a non-graduated ruler to draw straight lines, although if you have a good hand it is not essential). We draw two perpendicular lines and, starting from their point of intersection, we mark with the compass on one of them a point located three arbitrary units from it and, on the other, a point located four units away, and we now have the centers of the three tangent circles: the point of intersection is the center of the circle of radius 1 and the other two are the centers, respectively, of the circle of radius 2 and that of radius 3. Now it is easier (or not?) to find the radius of the two circles tangent to the other three, one exterior and one interior.

The key lines of Soddy's poem, as far as the statement of Descartes' theorem is concerned, are those that say “it is the addition of its squares/half a square of the sum.” That is, the sum of the squares of the curvatures is equal to half the square of the sum of said curvatures (remember that the curvature of a circle is the inverse of its radius). If we call Q, R, S and T the respective curvatures of the four circles tangent to each other:

Q² + R² + S² + T² = 1/2 (Q + R+ S + T)²

In the case of circles of radius 1, 2 and 3, if we call r the radius of the fourth circle tangent to those three, we will have:

Q = 1

R = 1/2

S = 1/3

T = 1/r

Therefore:

1 + 1/4 + 1/9 + 1/r² = 1/2 (1 + 1/2 + 1/3 + 1/r)²

From which it is easy to deduce the value of r (as it is a second degree equation, we will obtain two values, one for the inner circle and another for the outer circle). Easy, but cumbersome, so it is advisable to resort to a simple formula that allows you to calculate the value of the fourth radius based on the other three (can you find it?).

In the second part of Soddy's erotic-mathematical poem, which extends the theorem to the three-dimensional case of five spheres tangent to each other, the key lines are “is the square of the sum / three times the sum of squares,” which means ( passing the factor 3 as a divisor to the other side of the equation to homologate it to the previous one):

Q² + R² + S² + T² + V² = 1/3(Q + R+ S + T + V)²

#### Philosopher vs. lawyer

We cannot say goodbye to Descartes, and even less so after talking about analytical geometry, without mentioning Pierre de Fermat, the other great French mathematician of the time (who, by the way, was actually a lawyer who in his free time was distracted by doing math, for which he was called “prince of amateurs”), since he discovered analytical geometry before Descartes, and if today we talk about Cartesian coordinates instead of Fermatian, it is only because of Descartes' greater prestige and because he presented his works in a way clearer and more systematic.

Annoyed by the fact that Fermat had beaten him to it, Descartes tried to discredit the “incompetent lawyer” by calling his methods lax. But given the good results obtained by his rival, at one point he wrote to him: “In view of the latest method that you use to find tangents of curved lines, I can only say that it is very good and that if you had explained it this way way from the beginning, I wouldn't have questioned it at all.”

A detail of *fair play* somewhat deceptive, since, secretly, both heavyweights of mathematics turned each other on at the slightest opportunity (as Mersenne, the third great mathematician of the time, who, as the involuntary arbiter of the dispute, had than listening to the complaints of both). It was a combat over several rounds that, even though Descartes ended up naming the coordinates after him, neither of them actually won. Among other things because, as we saw, the trophy of analytical geometry had already been awarded to Apollonius of Perga and Omar Jayam long before.

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