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

$$
\left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) .
$$

Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper.

Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term "dominates" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.)

Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations

$$
\begin{aligned}
\mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\
\mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\
\mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right),
\end{aligned}
$$
Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$.

Prove that $P^{(1)}$ dominates $Q^{(1)}$.

Solution:
Because $\mathcal{C}\left(P^{(1)}\right)=\left(b, c, d, a^{\prime}\right)$ and $\mathcal{C}\left(Q^{(1)}\right)=\left(x, y, z, w^{\prime}\right)$, it is enough to show that $a^{\prime} \leq w^{\prime}$. $a$ and $a^{\prime}$ are the two roots of the quadratic given by Descartes' Circle Formula:

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

Solve by completing the square:

$$
\begin{aligned}
X^{2}-2(b+c+d) X+2\left(b^{2}+c^{2}+d^{2}\right) & =(b+c+d)^{2} \\
(X-(b+c+d))^{2} & =2(b+c+d)^{2}-2\left(b^{2}+c^{2}+d^{2}\right) \\
& =4(b c+b d+c d) .
\end{aligned}
$$

Thus $a, a^{\prime}=b+c+d \pm 2 \sqrt{b c+b d+c d}$.

Because $a \leq b \leq c \leq d$, and only $a$ can be less than zero, $a$ must get the minus sign, and $a^{\prime}$ gets the plus sign:

$$
a^{\prime}=b+c+d+2 \sqrt{b c+b d+c d} \text {. }
$$

Similarly,

$$
w^{\prime}=x+y+z+2 \sqrt{x y+x z+y z} .
$$

Because $P$ dominates $Q$, each term in the expression for $a^{\prime}$ is less than or equal to the corresponding term in the expression for $w^{\prime}$, thus $a^{\prime} \leq w^{\prime}$.

Final answer: 证明题，略