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.

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347)

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

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386)

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 $d$.
Two unit circles and a circle of radius $\frac{2}{3}$ are mutually externally tangent. Compute all possible values of $r$ such that a circle of radius $r$ is tangent to all three circles.

Solution:
Use Descartes' Circle Formula with $a=b=1$ and $c=\frac{3}{2}$ to solve for $d$ :

$$
\begin{aligned}
2 \cdot\left(1^{2}+1^{2}+\left(\frac{3}{2}\right)^{2}+d^{2}\right) & =\left(1+1+\frac{3}{2}+d\right)^{2} \\
\frac{17}{2}+2 d^{2} & =\frac{49}{4}+7 d+d^{2} \\
d^{2}-7 d-\frac{15}{4} & =0
\end{aligned}
$$

from which $d=\frac{15}{2}$ or $d=-\frac{1}{2}$. These values correspond to radii of $\frac{2}{15}$, a small circle nestled between the other three, or 2 , a large circle enclosing the other three.

Alternatively, start by scaling the kingdom with the first four circles removed to match the situation given. Thus the three given circles are internally tangent to a circle of radius $r=2$ and curvature $d=-\frac{1}{2}$. Descartes' Circle Formula gives a quadratic equation for $d$, and the sum of the roots is $2 \cdot\left(1+1+\frac{3}{2}\right)=7$, so the second root is $7+\frac{1}{2}=\frac{15}{2}$, corresponding to a circle of radius $r=\frac{2}{15}$.

Final answer: $2$, $\frac{2}{15}$