Subfield: Geometry


Question: A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.

This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.

Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.

<img_4001>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<img_3649>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Let $\phi=\frac{1+\sqrt{5}}{2}$, and let $\rho=\phi+\sqrt{\phi}$.
Show that four pairwise externally tangent circles with nonequal radii in geometric progression must have common ratio $\rho$.

Solution:
If the radii are in geometric progression, then so are their reciprocals (i.e., curvatures). Without loss of generality, let $(a, b, c, d)=\left(a, a r, a r^{2}, a r^{3}\right)$ for $r>1$. By Descartes' Circle Formula,

$$
\left(a+a r+a r^{2}+a r^{3}\right)^{2}=2\left(a^{2}+a^{2} r^{2}+a^{2} r^{4}+a^{2} r^{6}\right)
$$

Cancel $a^{2}$ from both sides of the equation to obtain

$$
\left(1+r+r^{2}+r^{3}\right)^{2}=2\left(1+r^{2}+r^{4}+r^{6}\right) .
$$

Because $1+r+r^{2}+r^{3}=(1+r)\left(1+r^{2}\right)$ and $1+r^{2}+r^{4}+r^{6}=\left(1+r^{2}\right)\left(1+r^{4}\right)$, the equation can be rewritten as follows:

$$
\begin{aligned}
(1+r)^{2}\left(1+r^{2}\right)^{2} & =2\left(1+r^{2}\right)\left(1+r^{4}\right) \\
(1+r)^{2}\left(1+r^{2}\right) & =2\left(1+r^{4}\right) \\
r^{4}-2 r^{3}-2 r^{2}-2 r+1 & =0 .
\end{aligned}
$$

Using the identity from 4a, $r=\rho$ is one solution; because the polynomial is palindromic, another real solution is $r=\rho^{-1}=\phi-\sqrt{\phi}$, but this value is less than 1 . The product of the corresponding linear factors is $r^{2}-2 \phi r+\phi^{2}-\phi=r^{2}-2 \phi r+1$. Division verifies that the other quadratic factor of the polynomial is $r^{2}+(2 \phi-2) r+1$, which has no real roots because $(\sqrt{5}-1)^{2}<1$.