Dynamic Geometry: P18

Label the diagram as follows, and let $p$ , $q$ , $r$ , $t$ be the radii of the cyan, orange, red, and purple circles, respectively, and let $\theta = \angle OAE$ :

Since the centers of the cyan and red circles lie on the bisector of the green angle, $\tan 2\theta = \cfrac{336}{527} = \cfrac{2\tan\theta}{1 - \tan^2 \theta}$ , which solves to $\tan \theta = \cfrac{7}{24}$ .

Then from $\triangle OAE$ and $\triangle OBD$ , $OA = \cfrac{24}{7}p$ and $OB = \cfrac{24}{7}r$ .

Since $ED = r + p$ and $DC = r - p$ , by the Pythagorean Theorem on $\triangle EDC$ we have $EC = \sqrt{(r + p)^2 - (r - p)^2} = 2\sqrt{pr} = AB$ .

Since $OA + AB = OB$ , $\cfrac{24}{7}p + 2\sqrt{pr} = \cfrac{24}{7}r$ , which rearranges to $r = \cfrac{16}{9}p$ .

By Descartes' Theorem on the cyan, orange, and red circles and the line, $\cfrac{1}{q} = \cfrac{1}{p} + \cfrac{1}{r} + 2\sqrt{\cfrac{1}{pr}} = \cfrac{1}{p} + \cfrac{9}{16p} + 2\sqrt{\cfrac{9}{16p^2}} = \cfrac{49}{16p}$ , which solves to $q = \cfrac{16}{49}p$ .

The semiperimeter $s$ of the black triangle is then $p + q + r$ , and $s - a$ , $s - b$ , and $s - c$ are $p$ , $q$ , and $r$ , so that by Heron's formula the area of the triangle is $T = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{(p + q + r)pqr} = \sqrt{(p + \cfrac{16}{49}p + \cfrac{16}{9}p)\cdot p \cdot \cfrac{16}{49} p \cdot \cfrac{16}{9}p} = \cfrac{592}{441}p^2 = \cfrac{37}{16}$ , which solves to $p = \cfrac{21}{16}$ .

That means $q = \cfrac{16}{49}p = \cfrac{16}{49} \cdot \cfrac{21}{16} = \cfrac{21}{49}$ and $r = \cfrac{16}{9}p = \cfrac{16}{9} \cdot \cfrac{21}{16} = \cfrac{7}{3}$ .

Finally, by Descartes' Theorem on all four circles, $\cfrac{1}{t} = \cfrac{16}{21} + \cfrac{49}{21} + \cfrac{3}{7} + 2\sqrt{\cfrac{16}{21} \cdot \cfrac{49}{21} + \cfrac{16}{21} \cdot \cfrac{3}{7} + \cfrac{49}{21} \cdot \cfrac{3}{7}} = \cfrac{148}{21}$ , which solves to $t = \cfrac{21}{148}$ .

Therefore, $a = 21$ , $b = 148$ , and $\sqrt{a + b} = \boxed{13}$ .

Let the radius of the red, cyan, orange, and purple circles be $r_1$ , $r_2$ , $r_3$ , and $r_4$ respectively. Let the green angle to be $\theta$ , then we note that the line joining the centers of the red and cyan circles has a gradient of $\tan \dfrac \theta 2$ . SInce $\tan \theta = \dfrac {336}{527}$ , then $\tan \dfrac \theta 2 = \dfrac {1-\cos \theta}{\sin \theta} = \dfrac 7{24} \implies \sin \dfrac \theta 2 = \dfrac 7{25}$ and

$\frac {r_1-r_2}{r_1+r_2} = \sin \frac \theta 2 = \frac 7{25} \implies r_2 = \frac 9{16}r_1$

We note that the horizontal distance between the centers of the red can cyan circles is given by $\sqrt{(r_1+r_2)^2-(r_1-r_2)^2} = 2\sqrt{r_1r_2}$ . This formula is applicable to any two circles. So we have:

$\begin{aligned} 2\sqrt{r_2r_3} + 2\sqrt{r_3r_1} & = 2\sqrt{r_1r_2} \implies \sqrt{r_3} = \frac {\sqrt{r_1r_2}}{\sqrt{r_1}+\sqrt{r_2}} = \frac 37 \sqrt{r_1} \implies r_3 = \frac 9{49}r_1 \end{aligned}$

Then the three sides of the triangle is $r_1+r_2$ , $r_2+r_3$ , and $r_3+r_1$ . By Heron's formula , the area of the triangle is

$\begin{aligned}: \sqrt{(r_1+r_2+r_3)r_1r_2r_3} = \frac {333}{784}r_1^2 = \frac {37}{16} \implies r_1 = \frac 73 \end{aligned}$

Then $r_2 = \dfrac 9{16}r_1 = \dfrac {21}{16}$ and $r_3 = \dfrac 9{49}r_1 = \dfrac 37$ . Thanks to @David Vreken and of course Descartes' theorem , with three know radii we can find the radius of the fourth circle which is tangent to the other three circles as follows:

$\begin{aligned} \frac 1{r_4} & = \frac 1{r_1} + \frac 1{r_2} + \frac 1{r_3} + 2\sqrt{\frac 1{r_1} \frac 1{r_2} +\frac 1{r_2} \frac 1{r_3} + \frac 1{r_3} \frac 1{r_1}} = \frac {148}{21} \\ \implies r_4 & = \frac {21}{148} \end{aligned}$

Therefore $\sqrt{a+b} = \sqrt{21+148} = \sqrt{169} = \boxed{13}$