## From Heron's formula to Descartes' circle theorem

Descartes' Circle Theorem is a very remarkable and simple statement and, to prove it, I had to use three other surprising theorems of the triangle, including Heron's formula. While I had been aware of Heron's formula before, it was during my research on Descartes' theorem that I discovered the inradius and exradius formulas. The circle theorem was first described by Descartes in 1643 and then rediscovered by Philip Beecroft in 1842, Frederick Soddy in 1936, and M. E. Wise in 1960. To be exact, Wise actually discovered the three dimension version of the theorem unaware of the plane version. Descartes' circle theorem has a history that was as interesting to read as its various proofs. Without further ado...

Fig 1. Finding the area

Heron's formula. Given a triangle with sides $${a, b, c}$$, then its area is given by $Area = \sqrt{s(s-a)(s-b)(s-c)}$ where $${s = \frac{1}{2}(a+b+c)}$$

Proof: The basic area formula is $${Area = \frac{1}{2} \cdot base \cdot height}$$. Without loss of generality, let $${base = a}$$ and $${height = c \sin(B) }$$.Then \begin{aligned} Area &= \frac{1}{2} a c \sin(B) \\ &= \frac{1}{2} a c \sqrt{1 - \cos^2(B)} \\ &= \frac{1}{2} a c \sqrt{1 - \left(\frac{a^2 + c^2 - b^2}{2 a c}\right)^2} \\ &= \frac{1}{4}\sqrt{(a+b+c)(a-b+c)(a+b-c)(-a+b+c)} \\ &= \sqrt{s(s-a)(s-b)(s-c)} \end{aligned}

$$\Box$$

Inradius formula. Given a triangle with sides $${a, b, c}$$, then the radius of the inscribed circle is given by $r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$

Proof: (Cliff suggested this proof to me.) Let $${E}$$ be center of the circle, then the areas of the triangles $${ABC, AEC, AEB, BEC}$$ are given by $${ \sqrt{s(s-a)(s-b)(s-c)}, \frac{br}{2}, \frac{cr}{2}, \frac{ar}{2} }$$. So { \begin{aligned} \sqrt{s(s-a)(s-b)(s-c)} &= \frac{br}{2} + \frac{cr}{2} + \frac{ar}{2} \\ &= sr \end{aligned} }

$$\Box$$

Exradius formula. Given a triangle with sides $${a, b, c}$$, then the radius of the exterior circle tangent to $${a}$$ and extensions of the other sides is given by $r_a = \sqrt{\frac{s(s-b)(s-c)}{s-a}}$

Proof: Let $${ E }$$ be the center of the circle, then the areas of the triangles $${ABC, AEP, AEN, CEP, BEC, BEN}$$ are given by $${ \sqrt{s(s-a)(s-b)(s-c)}, \frac{s r_a}{2}, \frac{s r_a}{2}, \frac{(s-b)r_a}{2}, \frac{a r_a}{2}, \frac{(s-c)r_a}{2} }$$. So \begin{aligned} \sqrt{s(s-a)(s-b)(s-c)} &= s r_a - \frac{(s-b)r_a}{2} - \frac{a r_a}{2} - \frac{(s-c)r_a}{2} \\ &= (s-a)r_a \end{aligned}

$$\Box$$

Fig 4a. Interiors of the three circles are disjoint

Fig 4b. Interiors are not disjoint

Descartes' Circle Theorem. Given four circles $${{ C_i : 1 \le i \le 4}}$$ that are mutually tangent to each other, then their curvatures (reciprocal of radius) satisfy the following relation: $2\left(k_1^2 + k_2^2 + k_3^2 + k_4^2\right) = \left(k_1+k_2+k_3+k_4\right)^2$

Coxeter gave this proof in 1968 and it is a simplified version of Beecroft's original proof; I have made a few changes to make it clearer

Proof: For any three of the four given circles, say $${C_2, C_3, C_4}$$, define a new circle that passes through their three points of contact.

If the circles are all externally tangent to each other like in Figure 4a, then this circle is the inscribed circle of the triangle defined by their centres. Using $${s-a = r_2, s-b = r_3, s-c = r_4}$$, and the inradius formula, then the radius of the new circle, $${\rho_1}$$ is $\rho_1 = \sqrt{\frac{r_2 r_3 r_4}{r_2 + r_3+r_4}}$ If we let its curvature be $${\kappa_1 = \rho_1^{-1}}$$, then we get the following relation $\kappa_1^2 = k_3 k_4 + k_2 k_4 + k_2 k_3$

If one circle, say $${C_2}$$, contains the other two circles like in Figure 4b, then the newly defined circle is the exterior circle of the triangle defined by the three centres and opposite $${C_2}$$. Letting $${s = r_2, s-c = r_3, s-b = r_4}$$, and the exradius formula, then the radius of the new circle, $${\rho_1}$$ is $\rho_1 = \sqrt{\frac{r_2 r_3 r_4}{r_2 - r_3 - r_4}}$ Similarly, we get the following relation of curvatures $\kappa_1^2 = k_3 k_4 - k_2 k_4 - k_2 k_3$ To keep the two relations identical: let's use $${k_2 = - r_2^{-1}}$$ and we get as before $\kappa_1^2 = k_3 k_4 + k_2 k_4 + k_2 k_3$

So we adopt the convention that the curvature is negative if the circle contains the other circles. Since the choice of $${C_2, C_3, C_4}$$ was arbitrary, we can permute the subscripts to get three other relations

\begin{aligned} \kappa_2^2 &= k_1 k_3 + k_1 k_4 + k_3 k_4 \\ \kappa_3^2 &= k_1 k_2 + k_1 k_4 + k_2 k_4 \\ \kappa_4^2 &= k_1 k_2 + k_1 k_3 + k_2 k_3 \end{aligned}

The four new circles are also mutually tangent to each other and this leads to the following relations

\begin{aligned} k_1^2 &= \kappa_2 \kappa_3 + \kappa_2 \kappa_4 + \kappa_3 \kappa_4 \\ k_2^2 &= \kappa_1 \kappa_3 + \kappa_1 \kappa_4 + \kappa_3 \kappa_4 \\ k_3^2 &= \kappa_1 \kappa_2 + \kappa_1 \kappa_4 + \kappa_2 \kappa_4 \\ k_4^2 &= \kappa_1 \kappa_2 + \kappa_1 \kappa_3 + \kappa_2 \kappa_3 \end{aligned}

If two circles of the same radius were contained in a circle with twice their radius, then the "circle" joining their points of contact is actually a straight line with curvature of $$0$$. This is a degenerate case and it is easy to verify that the above relations still hold.

Noting that there is at most one negative $${k_i}$$ and $${\kappa_i}$$ and their corresponding circles must have the largest radius (to contain the other three circles): then $${\sum k_i, \sum \kappa_i > 0}$$. \begin{aligned} \left(\sum k_i\right)^2 = \sum k_i^2 &+ \sum \kappa_i^2 = \left(\sum \kappa_i\right)^2 \\ \implies \sum k_i &= \sum \kappa_i \\ \ -k_1^2 + (k_2 + k_3 +k_4)^2 &= -k_1^2 + k_2^2 + k_3^2 + k_4^2 + 2 \kappa_1^2 \\ &= 2 \kappa_1 (\kappa_1 + \kappa_2 + \kappa_3 + \kappa_4 ) \\ &= 2 \kappa_1 (k_1 + k_2 + k_3 + k_4) \\ \implies -k_1 + k_2 + k_3 + k_4 &= 2 \kappa_1 \end{aligned}

Permute the subscripts to get \begin{aligned} k_1 - k_2 + k_3 + k_4 &= 2 \kappa_2 \\ k_1 + k_2 - k_3 + k_4 &= 2 \kappa_3 \\ k_1 + k_2 + k_3 - k_4 &= 2 \kappa_4 \end{aligned}

Square each of these four equations and add them up to get $\sum k_i^2 = \sum \kappa_i^2$ Finally, $2\sum k_i^2 = \sum k_i^2 + \sum \kappa_i^2 = \left(\sum k_i \right)^2$

$$\Box$$

Bibliography

Coxeter, H. S. M. "The Problem of Apollonius." Amer. Math. Monthly 75.1 (1968): 5-15.

Pedoe, Daniel. "On a Theorem in Geometry." Amer. Math. Monthly 74.6 (1967): 627-640.