It's All Tangents!

Extend radii $\overline{OB}$ and $\overline{CP}$ so that $\overrightarrow{\rm OB}$ and $\overrightarrow{\rm CP}$ intersect at $D$ .

$\overline{BC}$ is tangent to circle $C_{2}$ at $B \implies \angle{OBC}$ and $\angle{DBC}$ are right angles $\implies \triangle{DBC}$ is a right triangle and $\overline{AC}$ is tangent to circle $C_{3}$ at $C \implies \angle{DCA}$ is a right angle and $\triangle{ABC}$ is an equilateral triangle $\implies m\angle{BCA} = 60^{\circ} \implies m\angle{DCB} = 30^{\circ} \implies \triangle{DBC}$ is a $30 - 60 - 90$ right triangle.

Let $\overline{DB} = x \implies \overline{BC} = \sqrt{3}x$ and $\overline{CD} = 2x$

$C_{2}$ and $C_{3}$ are tangent to each other as shown above $\implies \overline{OP} = 2r$ and each radii $\overline{OB} = \overline{PC} = r \implies \overline{PD} = 2x - r$ and $\overline{OD} = x + r$ .

Using the law of cosines on $\triangle{DOP}$ formed we have:

$4r^2 = (2x - r)^2 + (x + r)^2 - 2(2x - r)(x + r)\cos(60^{\circ}) = 3x^2 - 3rx + 3r^2 \implies 3x^2 - 3rx - r^2 = 0 \implies$

$x = \dfrac{3 + \sqrt{21}}{6}r$ dropping the negative root

$\implies \overline{BC} = \sqrt{3}x = \dfrac{3 + \sqrt{21}}{2\sqrt{3}}r \implies$ the area of isosceles $\triangle{ABC}$ is

$A_{\triangle{ABC}} = \dfrac{\sqrt{3}}{4}(\dfrac{3 + \sqrt{21}}{2\sqrt{3}})^2 r^2 =$ $\dfrac{\sqrt{3}}{8}(5 + \sqrt{21})r^2 \implies$

$\dfrac{A_{\triangle{ABC}}}{r^2} = \dfrac{\sqrt{3}}{8}(5 + \sqrt{21}) = \dfrac{\sqrt{a}}{b}(c + \sqrt{d})$ $\implies a + b + c + d = \boxed{37}$ .

Label the diagram as follows, with $r = CD = DE$ :

Since $\angle DFE = 60°$ , $\angle DEF = 90°$ , and $DE = r$ , $DF = \frac{DE}{\sin 60°} = \frac{2\sqrt{3}}{3}r$ .

Since $\angle FCA = 30°$ and $\angle DCA = 90°$ , $\angle DCF = \angle DCA - \angle FCA = 90° - 30° = 60°$ .

By the law of cosines on $\triangle CDF$ , $DF^2 = CD^2 + CF^2 - 2 \cdot CD \cdot CF \cdot \cos \angle DCF$ , or $(\frac{2\sqrt{3}}{3}r)^2 = r^2 + CF^2 - 2 \cdot r \cdot CF \cdot \cos 60°$ , which solves to $CF = (\frac{1}{2} + \frac{\sqrt{21}}{6})r$ for $CF > 0$ .

The area of $\triangle ABC$ is then $A_{\triangle ABC} = 3 \cdot A_{\triangle CFA} = 3 \cdot \frac{1}{2} \cdot CF^2 \cdot \sin 120° = 3 \cdot \frac{1}{2} \cdot (\frac{1}{2} + \frac{\sqrt{21}}{6})^2r^2 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}(5 + \sqrt{21})r^2$ .

Therefore, $a = 3$ , $b = 8$ , $c = 5$ , $d = 21$ , and $a + b + c + d = \boxed{37}$ .

Denote by ${{O}_{1}}$ , ${{O}_{2}}$ and ${{O}_{3}}$ the centers of circles ${{C}_{1}}$ , ${{C}_{2}}$ and ${{C}_{3}}$ respectively. These three points form an equilateral triangle. Let $D$ be the point of intersection of $BC$ and ${{O}_{2}} {{O}_{3}}$ . Let $E$ be the point of intersection of $AB$ and ${{O}_{1}} {{O}_{2}}$ . Let $F$ be the point of intersection of $AB$ and ${{O}_{2}} {{O}_{3}}$ . Let $G$ be the point of intersection of $BC$ and ${{O}_{1}} {{O}_{3}}$ , as shown on the figure.

Since $\angle DBE$ = $\angle D{{O}_{2}}E$ =60{}^\circ), points $B$ , $D$ , $E$ and ${{O}_{2}}$ are concyclic. $BC$ is tangent to ${{C}_{2}}$ , thus $\angle {{O}_{2}}BD=90{}^\circ$ and hence, $\angle {{O}_{2}}ED=180{}^\circ-\angle {{O}_{2}}BD=90{}^\circ$ .

This means that $\triangle {{O}_{2}}DE$ is a $90{}^\circ$ - $60{}^\circ$ - $30{}^\circ$ triangle, therefore, ${{O}_{2}}D=2 {{O}_{2}}E$ .

Due to the threefold symmetry of the compound shape of the two equilateral triangles, segments ${{O}_{2}}E$ and ${{O}_{3}}D$ are congruent. Let ${{O}_{2}}E={{O}_{3}}D=y$ . Then ${{O}_{2}}D=2y$ , so we have ${{O}_{2}}{{O}_{3}}={{O}_{2}}D+D{{O}_{3}}\Rightarrow 2r=2y+y\Rightarrow y=\frac{2r}{3}$ Using Pythagorean theorem on $\triangle {{O}_{2}}DE$

$D{{E}^{2}}={{O}_{2}}{{D}^{2}}-{{O}_{2}}{{E}^{2}}={{\left( 2y \right)}^{2}}-{{y}^{2}}=3{{y}^{2}}\Rightarrow DE=y\sqrt{3}\Rightarrow DE=\frac{2r}{\sqrt{3}}$ Again, because of the symmetry, $\left( BD,\ AE \right)$ , $\left( DG,\ EF \right)$ and $\left( GC,\ BF \right)$ are pairs of congruent segments. Label $BD$ , $AE$ by $k$ , $DG$ , $EF$ by $l$ and $GC$ , $BF$ by $m$ .

By Pythagorean theorem on $\triangle {{O}_{2}}BD$

$\begin{aligned} B{{D}^{2}}+{{O}_{2}}{{B}^{2}}={{O}_{2}}{{D}^{2}} & \Rightarrow {{k}^{2}}+{{r}^{2}}={{\left( 2y \right)}^{2}} \\ & \Rightarrow {{k}^{2}}+{{r}^{2}}=\frac{16{{r}^{2}}}{9} \\ & \Rightarrow k=\frac{r\sqrt{7}}{3} \ \ \ \ \ (1)\\ \end{aligned}$

From the similarity of triangles $\triangle {{O}_{2}}EF$ and $\triangle BDF$ we get

$\dfrac{{{O}_{2}}F}{BF}=\dfrac{{{O}_{2}}E}{BD}\Rightarrow \dfrac{x}{m}=\dfrac{y}{k}\overset{\left( 1 \right)}{\mathop{=}}\,\dfrac{\frac{2r}{3}}{\frac{r\sqrt{7}}{3}}=\dfrac{2}{\sqrt{7}}\Rightarrow x=\dfrac{2m}{\sqrt{7}} \ \ \ \ \ (2)$ Likewise, from the similarity of triangles $\triangle {{O}_{2}}BF$ and $\triangle EDF$ we have

$\dfrac{{{O}_{2}}F}{EF}=\dfrac{BF}{DF}=\dfrac{{{O}_{2}}B}{DE}\Rightarrow \dfrac{x}{l}=\dfrac{m}{\dfrac{3r}{3}-x}=\dfrac{r}{\dfrac{2r\sqrt{3}}{3}}=\dfrac{\sqrt{3}}{2}\Rightarrow \left\{ \begin{matrix} x=\dfrac{\sqrt{3}}{2}l \ \ \ \ \ (3) \\ \phantom{0} \\ lm=x\left( \dfrac{3r}{3}-x \right) \ \ \ \ \ (4) \\ \end{matrix} \right.$

$\left( 2 \right),\left( 3 \right)\Rightarrow {{x}^{2}}=\dfrac{\sqrt{3}}{\sqrt{7}}lm\overset{\left( 4 \right)}{\mathop{\Rightarrow }}\,{{x}^{2}}=\dfrac{\sqrt{3}}{\sqrt{7}}x\left( \dfrac{4r}{3}-x \right)\overset{x\ne 0}{\mathop{\Rightarrow }}\,x=\dfrac{\sqrt{3}}{\sqrt{7}}\left( \dfrac{4r}{3}-x \right)$ which solves to $x=\frac{\sqrt{21}-3}{3}r \ \ \ \ \ (5)$ Now,

$\left( 2 \right)\overset{\left( 5 \right)}{\mathop{\Rightarrow }}\,m=\dfrac{\sqrt{7}}{2}\cdot ~\dfrac{\sqrt{21}-3}{3}r\Rightarrow m=\dfrac{7\sqrt{3}-3\sqrt{7}}{6}r \ \ \ \ \ (6)$

$\left( 3 \right)\overset{\left( 5 \right)}{\mathop{\Rightarrow }}\,l=\dfrac{2}{\sqrt{3}}\cdot ~\dfrac{\sqrt{21}-3}{3}r\Rightarrow l=\dfrac{2}{3}\left( \sqrt{7}-\sqrt{3} \right)r \ \ \ \ \ (7)$

Calculating the length of the side of the red triangle,

$\left( 1 \right),\left( 6 \right),\left( 7 \right)\Rightarrow BC=k+l+m=\dfrac{\sqrt{7}}{3}r+\dfrac{2}{3}\left( \sqrt{7}-\sqrt{3} \right)r+\dfrac{7\sqrt{3}-3\sqrt{7}}{6}r\Rightarrow BC=\dfrac{\sqrt{7}+\sqrt{3}}{2}r$ Finally,

$\begin{aligned} & {{A}_{ \triangle ABC}}=\dfrac{\sqrt{3}}{4}B{{C}^{2}}=\dfrac{\sqrt{3}}{4}{{\left( \dfrac{\sqrt{7}+\sqrt{3}}{2} \right)}^{2}}{{r}^{2}}=\dfrac{\sqrt{3}}{8}\left( 5+\sqrt{21} \right){{r}^{2}} \\ & \Rightarrow \dfrac{{{A}_{\triangle ABC}}}{{{r}^{2}}}=\dfrac{\sqrt{3}}{8}\left( 5+\sqrt{21} \right) \\ \end{aligned}$ .

For the answer, $a=3$ , $b=8$ , $c=5$ , $d=21$ , thus $a+b+c+d=\boxed{37}$ .