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.

<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 $d$.

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 $P^{(3)}$ dominates $Q^{(3)}$.

Solution:
Because $\mathcal{C}\left(P^{(3)}\right)=\left(a, b, d, c^{\prime}\right)$ and $\mathcal{C}\left(Q^{(3)}\right)=\left(w, x, z, y^{\prime}\right)$, it suffices to show that $c^{\prime} \leq y^{\prime}$. If $a \geq 0$, but if $a<0$, then there is more to be done.

Arguing as in 9b, $c, c^{\prime}=a+b+d \pm 2 \sqrt{a b+a d+b d}$. If $a<0$, then the other three circles are internally tangent to the circle of curvature $a$, so this circle has the largest radius. In particular, $\frac{1}{|a|}>\frac{1}{b}$. Thus $b>|a|=-a$, which shows that $a+b>0$. Therefore $c$ must get the minus sign, and $c^{\prime}$ gets the plus sign. The same argument applies to $y$ and $y^{\prime}$.

When $a<0$, it is also worth considering whether the square roots are defined (and real). In fact, they are. Consider the diameters of the circles with curvatures $b$ and $d$ along the line through the centers of these circles. These two diameters form a single segment inside the circle with curvature $a$, so the sum of the diameters is at most the diameter of that circle: $\frac{2}{b}+\frac{2}{d} \leq \frac{2}{|a|}$. It follows that $-a d-a b=|a| d+|a| b \leq b d$, or $a b+a d+b d \geq 0$. This is the argument of the square root in the expressions for $c$ and $c^{\prime}$. An analogous argument shows that the radicands are nonnegative in the expressions for $b$ and $b^{\prime}$.

The foregoing shows that

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

and, by an analogous argument for $w<0$,

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

It remains to prove that $c^{\prime} \leq y^{\prime}$. Note that only $a$ and $w$ may be negative; $b, c, d, x, y$, and $z$ are all positive. There are three cases.



(i) If $0 \leq a \leq w$, then $a b \leq w x, a d \leq w z$, and $b d \leq x z$, so $c^{\prime} \leq y^{\prime}$.

(ii) If $a<0 \leq w$, then $a b+a d+b d \leq b d$, and $b d \leq x z \leq w x+w z+x z$, so $c^{\prime} \leq y^{\prime}$. (As noted above, both radicands are nonnegative.)

(iii) If $a \leq w<0$, then it has already been established that $a+b$ is positive. Analogously, $a+d, w+x$, and $w+z$ are positive. Furthermore, $a^{2} \geq w^{2}$. Thus $(a+b)(a+d)-a^{2} \leq$ $(w+x)(w+z)-w^{2}$, which establishes that $a b+a d+b d \leq w x+w z+x z$, so $c^{\prime} \leq y^{\prime}$.