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.

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Circle $A$ has curvature 2; circle $B$ has curvature 1 .

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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 three mutually tangent circles with curvatures $a, b, c>0$, suppose that $(a, b, c, 0)$ does not satisfy Descartes' Circle Formula. Show that there are two distinct values of $r$ such that there is a circle of radius $r$ tangent to the given circles.

Solution:
Apply Descartes' Circle Formula to yield

$$
(a+b+c+x)^{2}=2 \cdot\left(a^{2}+b^{2}+c^{2}+x^{2}\right),
$$

a quadratic equation in $x$. Expanding and rewriting in standard form yields the equation

$$
x^{2}-p x+q=0
$$

where $p=2(a+b+c)$ and $q=2\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2}$.

The discriminant of this quadratic is

$$
\begin{aligned}
p^{2}-4 q & =8(a+b+c)^{2}-8\left(a^{2}+b^{2}+c^{2}\right) \\
& =16(a b+a c+b c)
\end{aligned}
$$

This last expression is positive because it is given that $a, b, c>0$. Therefore the quadratic has two distinct real roots, say $d_{1}$ and $d_{2}$. These usually correspond to two distinct radii, $r_{1}=\frac{1}{\left|d_{1}\right|}$ and $r_{2}=\frac{1}{\left|d_{2}\right|}$.

One possible exception case to consider is if $r_{2}=r_{1}$, which would occur if $d_{2}=-d_{1}$. This case can be ruled out because $d_{1}+d_{2}=p=2(a+b+c)$, which must be positive if $a, b, c>0$. (Notice too that this inequality rules out the possibility that both circles have negative curvature, so that there cannot be two distinct circles to which the given circles are internally tangent.)

When both roots $d_{1}$ and $d_{2}$ are positive, the three given circles are externally tangent to both fourth circles. When one is positive and one is negative, the three given circles are internally tangent to one circle and externally tangent to the other.



While the foregoing answers the question posed, it is interesting to examine the result from a geometric perspective: why are there normally two possible fourth circles? Consider the case when one of $a, b$, and $c$ is negative (i.e., two circles are internally tangent to a third). Let $A$ and $B$ be circles internally tangent to $C$. Then $A$ and $B$ partition the remaining area of $C$ into two c-triangles, each of which has an incircle, providing the two solutions.

If, as in the given problem, $a, b, c>0$, then all three circles $A, B$, and $C$, are mutually externally tangent. In this case, the given circles bound a c-triangle, which has an incircle, corresponding to one of the two roots. The complementary arcs of the given circles bound an infinite region, and this region normally contains a second circle tangent to the given circles. To demonstrate this fact geometrically, consider shrink-wrapping the circles: the shrink-wrap is the border of the smallest convex region containing all three circles. (This region is called the convex hull of the circles). There are two cases to address. If only two circles are touched by the shrink-wrap, then one circle is wedged between two larger ones and completely enclosed by their common tangents. In such a case, a circle can be drawn so that it is tangent to all three circles as shown in the diagram below (shrink-wrap in bold; locations of fourth circle marked at $D_{1}$ and $D_{2}$ ).

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On the other hand, if the shrink-wrap touches all three circles, then it can be expanded to make a circle tangent to and containing $A, B$, and $C$, as shown below.



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