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 $\rho=\phi+\sqrt{\phi}$.
Prove that $\rho^{4}=2 \rho^{3}+2 \rho^{2}+2 \rho-1$.

Solution:
Note that

$$
\begin{aligned}
\frac{1}{\rho} & =\frac{\phi^{2}-\phi}{\phi+\sqrt{\phi}} \\
& =\phi-\sqrt{\phi}
\end{aligned}
$$



Therefore

$$
\begin{aligned}
\left(\rho-\frac{1}{\rho}\right)^{2} & =(2 \sqrt{\phi})^{2}=4 \phi \\
& =2\left(\rho+\frac{1}{\rho}\right) .
\end{aligned}
$$

Multiplying both sides of the equation by $\rho^{2}$ gives $\left(\rho^{2}-1\right)^{2}=2\left(\rho^{3}+\rho\right)$. Expand and isolate $\rho^{4}$ to obtain $\rho^{4}=2 \rho^{3}+2 \rho^{2}+2 \rho-1$.

Alternate Proof: Because $\phi^{2}=\phi+1$, any power of $\rho$ can be expressed as an integer plus integer multiples of $\sqrt{\phi}, \phi$, and $\phi \sqrt{\phi}$. In particular,

$$
\begin{aligned}
\rho^{2} & =\phi^{2}+2 \phi \sqrt{\phi}+\phi \\
& =2 \phi \sqrt{\phi}+2 \phi+1, \\
\rho^{3} & =4 \phi \sqrt{\phi}+5 \phi+3 \sqrt{\phi}+4, \text { and } \\
\rho^{4} & =12 \phi \sqrt{\phi}+16 \phi+8 \sqrt{\phi}+9 .
\end{aligned}
$$

Therefore

$$
\begin{aligned}
2 \rho^{3}+2 \rho^{2}+2 \rho-1 & =2(4 \phi \sqrt{\phi}+5 \phi+3 \sqrt{\phi}+4)+2(2 \phi \sqrt{\phi}+2 \phi+1)+2 \rho-1 \\
& =12 \phi \sqrt{\phi}+16 \phi+8 \sqrt{\phi}+9 \\
& =\rho^{4}
\end{aligned}
$$