4 Circles in Love

Relevant wiki: Descartes' Circle Theorem

Let $r> 0$ denote the radius that we're looking for, then using the statement given in the aforementioned theorem:

The curvature of $C_i$ is defined to be $k_i = \pm 1/r_i$ , where the sign of $k_i$ is chosen depending on whether or not $C_i$ is internally or externally tangent to the other circles. For example, in the above picture, the large green circle has negative curvature, since the other three circles are internally tangent to it. On the other hand, the small red circle has positive curvature, since the other three circles are externally tangent to it.

The curvatures of all these inner circles are the positive reciprocal of their respective radius, whereas the curvature of the external circle has a negative curvature.

Thus, $k_1 = r, k_2 = \dfrac12, k_3 = \dfrac12, k_4 =-\dfrac13$ .

And the theorem states that

$(k_1 + k_2 + k_3 + k_4)^2 = 2(k_{1}^{2} + k_{2}^{2} + k_{3}^{2} + k_{4}^{2})$

Plugging in these values and simplify gives

$\left( \dfrac76 - r\right)^2 = 2\left( \dfrac{49}{36} + r^2\right) \;$

so solving for $r$ gives $r = \dfrac mn = \dfrac76$ . The answer is $m+n=\boxed{13}$ .

In general, if three circles of radii a, b & c are touching each other externally then the radius ( $R$ ) of circle which touch all three circles internally, is given by a general formula (for detailed analysis of three tangent circles go through: Geometry of three externally touching circles by H.C. Rajpoot )

\boxed{\color{dark black}{R=\frac{abc}{2\sqrt{abc(a+b+c)}-ab-bc-ca}}\quad \underbrace{\{\forall \ 0<c\le a, b<R \quad \text{\&}\quad c>\frac{ab}{(\sqrt a+\sqrt b)^2}}_{\text{conditions for validity of formula}\}}}

As per given problem, the given values of radii are $R=3, a=1, b=2, c=\text{?}$ , substituting these values in above general formula

$3=\frac{1\cdot 2\cdot c}{2\sqrt{1\cdot 2\cdot c(1+2+c)}-1\cdot 2-2\cdot c-c\cdot 2}$

$6\sqrt{2c^2+6c}=11c+6$

$(6\sqrt{2c^2+6c})^2=(11c+6)^2$

$49c^2-84c+36=0$

$c=\frac{-(-84)\pm\sqrt{(-84)^2-4(36)(49)}}{2(49)}=\frac{84}{98}=\frac{6}{7}=\frac mn$

$\therefore m+n=6+7=\boxed{13}$

We can try Descartes' kissing circles.