Subfield: Combinatorics


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 $d$.

Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states

$$
\left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) .
$$
Prove that the center of each circular plot has coordinates $\left(\frac{u}{c}, \frac{v}{c}\right)$ where $u$ and $v$ are integers, and $c$ is the curvature of the plot.

Solution:
It suffices to show that for each circle $C$ with curvature $c$ and center $z_{C}$ (in the complex plane), $c z_{C}$ is of the form $u+i v$ where $u$ and $v$ are integers. If this is the case, then each center is of the form $\left(\frac{u}{c}, \frac{v}{c}\right)$.

Proceed by induction. To check the base case, check the original kingdom and the first four plots: $U, C, C^{\prime}, P_{1}$, and $P_{2}$. Circle $U$ is centered at $(0,0)$, yielding $-1 \cdot z_{U}=0+0 i$. Circles $C$ and $C^{\prime}$ are symmetric about the $y$-axis, so it suffices to check just one of them. Circle $C$ has radius $\frac{1}{2}$ and therefore curvature 2 . It is centered at $\left(\frac{1}{2}, \frac{0}{2}\right)$, yielding $2 z_{C}=2\left(\frac{1}{2}+0 i\right)=1$. Circles $P_{1}$ and $P_{2}$ are symmetric about the $x$-axis, so it suffices to check just one of them. Circle $P_{1}$ has radius $\frac{1}{3}$ and therefore curvature 3 . It is centered at $\left(\frac{0}{3}, \frac{2}{3}\right)$, yielding $3 z_{P_{3}}=0+2 i$.

For the inductive step, suppose that $a z_{A}, b z_{B}, c z_{C}, d z_{D}$ have integer real and imaginary parts. Then by closure of addition and multiplication in the integers, $a^{\prime} z_{A^{\prime}}=2 \hat{s}-3 a z_{A}$ also has integer real and imaginary parts, and similarly for $b^{\prime} z_{B^{\prime}}, c^{\prime} z_{C^{\prime}}, d^{\prime} z_{D^{\prime}}$.

So for all plots $A$ sold, $a z_{A}$ has integer real and imaginary parts, so each is centered at $\left(\frac{u}{c}, \frac{v}{c}\right)$ where $u$ and $v$ are integers, and $c$ is the curvature.

Final answer: 证明题，略