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.

<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 $d$.
Find the areas of the circles removed on day 3.

Solution:
Day 3 begins with two circles of curvature 15 from the configuration $(2,2,3,15)$, and four circles of curvature 6 from the configuration $(-1,2,3,6)$. Consider the following two cases:

Case 1: $(a, b, c, d)=(2,2,3,15), s=22$

- $a=2: a^{\prime}=2 s-3 a=\mathbf{3 8}$
- $b=2: b^{\prime}=2 s-3 b=\mathbf{3 8}$
- $c=3: c^{\prime}=2 s-3 c=\mathbf{3 5}$
- $d=15: d^{\prime}=2 s-3 d=-1$, which is the configuration from day 1 .

Case 2: $(a, b, c, d)=(-1,2,3,6), s=10$

- $a=-1: a^{\prime}=2 s-3 a=\mathbf{2 3}$
- $b=2: b^{\prime}=2 s-3 b=\mathbf{1 4}$
- $c=3: c^{\prime}=2 s-3 c=\mathbf{1 1}$
- $d=6: d^{\prime}=2 s-3 d=2$, which is the configuration from day 1 .



So the areas of the plots removed on day 3 are:

$$
\frac{\pi}{38^{2}}, \frac{\pi}{35^{2}}, \frac{\pi}{23^{2}}, \frac{\pi}{14^{2}}, \text { and } \frac{\pi}{11^{2}}
$$

There are two circles with area $\frac{\pi}{35^{2}}$, and four circles with each of the other areas, for a total of 18 plots.

Final answer: $\frac{\pi}{38^{2}}, \frac{\pi}{35^{2}}, \frac{\pi}{23^{2}}, \frac{\pi}{14^{2}}, \frac{\pi}{11^{2}}$