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 $d$.
Given $A, B, C, D$ as above with $s=a+b+c+d$, if there is a second circle $A^{\prime}$ with curvature $a^{\prime}$ also tangent to $B, C$, and $D$. We can describe $A$ and $A^{\prime}$ as conjugate circles. Use Descartes' Circle Formula to show that $a^{\prime}=2 s-3 a$ and therefore $s^{\prime}=a^{\prime}+b+c+d=$ $3 s-4 a$.

Solution:
The equation $(x+b+c+d)^{2}=2\left(x^{2}+b^{2}+c^{2}+d^{2}\right)$ is quadratic with two solutions. Call them $a$ and $a^{\prime}$. These are the curvatures of the two circles which are tangent to circles with curvatures $b, c$, and $d$. Rewrite the equation in standard form to obtain $x^{2}-2(b+c+d) x+$ $\ldots=0$. Using the sum of the roots formula, $a+a^{\prime}=2(b+c+d)=2(s-a)$. So $a^{\prime}=2 s-3 a$, and therefore

$$
\begin{aligned}
s^{\prime} & =a^{\prime}+b+c+d \\
& =2 s-3 a+s-a \\
& =3 s-4 a .
\end{aligned}
$$

Final answer: 证明题，略