Trapped Circles

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In this solution we use the fact that, in the case where the two black incircles are congruent, the following relation holds: $BF=\sqrt{s\cdot \left( s-b \right)} \ \ \ \ \ (1)$ where $b=CA$ and $s$ is the semiperimeter of $\triangle ABC$ .

This is problem 2.2.5 in Fukagawa, Hidetoshi, and Dan Pedhoe. 1989, Japanese temple geometry problems, Winnipeg: Charles Babbage Research Centre. In this book no solution is given, but you can find one here (problem 8).

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Reffering to the figure, let $CH\bot AB$ . Let $G$ be the intersection of lines $CD$ and $EB$ . Denote by $r$ and $R$ the radii of the red and black circles respectively.
Since $HC=\dfrac{AB-DC}{2}=\dfrac{1}{2}=\dfrac{BC}{2}$ , we conclude that $\angle CBH=60{}^\circ$ .

By cosine rule on $\triangle ABC$ ,

$A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2AC\cdot BC\cdot \cos B=4+1-2\times 2\times 1\times \dfrac{1}{2}=3\Rightarrow AC=\sqrt{3}$ This means that $\triangle ABC$ is a right triangle, with $\angle C=90{}^\circ$ .

Using proposition $(1)$ , we have $B{{F}^{2}}=s\left( s-b \right)=\dfrac{2+1+\sqrt{3}}{2}\cdot \dfrac{2+1-\sqrt{3}}{2}=\dfrac{3}{2}\Rightarrow BF=\sqrt{\dfrac{3}{2}}$ By Pythagorean Theorem on $\triangle BCF$ ,

$C{{F}^{2}}=B{{F}^{2}}-B{{C}^{2}}=\frac{3}{2}-1=\frac{1}{2}\Rightarrow CF=\frac{\sqrt{2}}{2}$ Furthermore, $\triangle CGF\sim \triangle ABF$ , thus,

$\dfrac{GF}{FB}=\dfrac{CF}{FA}=\dfrac{GC}{AB}\Rightarrow \left\{ \begin{matrix} GF=FB\cdot \dfrac{CF}{FA}=\sqrt{\dfrac{3}{2}}\cdot \dfrac{\dfrac{\sqrt{2}}{2}}{\sqrt{3}-\dfrac{\sqrt{2}}{2}}=\dfrac{6+\sqrt{6}}{10} \\ GC=AB\cdot \dfrac{CF}{FA}=2\cdot \dfrac{\dfrac{\sqrt{2}}{2}}{\sqrt{3}-\dfrac{\sqrt{2}}{2}}=\dfrac{\sqrt{6}+1}{5} \\ \end{matrix} \right.$ For the inradius of the right triangle $BCF$ we have

$R=s-BF=\frac{1}{2}\left( BC+CF-BF \right)=\frac{1}{2}\left( 1+\frac{\sqrt{2}}{2}-\sqrt{\frac{3}{2}} \right)$ Again, due to the similarity of $\triangle CGF$ and $\triangle ABF$ ,

$\dfrac{r}{R}=\dfrac{GC}{AB}\Rightarrow r=R\cdot \dfrac{GC}{AB}=\dfrac{1}{2}\left( 1+\dfrac{\sqrt{2}}{2}-\sqrt{\dfrac{3}{2}} \right)\cdot \dfrac{\dfrac{\sqrt{6}+1}{5}}{2}=\dfrac{1}{10}\left( -2+\sqrt{3}+\sqrt{2+\sqrt{3}} \right)$ For the answer, $a=2$ , $b=3$ , $c=10$ , thus, $a+b+c=\boxed{15}$ .

Let $F$ be the intersection of $BE$ and $AC$ , and extend $CD$ and $BE$ to meet at $G$ .

Since the trapezoid is a combination of three unit equilateral triangles, $AC = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$ , and since $AC^2 + BC^2 = AB^2$ , by the converse of the Pythagorean Theorem $\angle ACB$ is a right triangle.

Let $k = CF$ . Then $AF = AC - CF = \sqrt{3} - k$ , and by the Pythagorean Theorem, $BF = \sqrt{CF^2 + BC^2} = \sqrt{k^2 + 1}$ .

The area of $\triangle AFB$ is $A_{\triangle AFB} = \frac{1}{2} \cdot AF \cdot BC = \frac{1}{2} \cdot (\sqrt{3} - k) \cdot 1 = \frac{1}{2}(\sqrt{3} - k)$ , and the perimeter of $\triangle AFB$ is $P_{\triangle AFB} = AF + BF + AB = \sqrt{3} - k + \sqrt{k^2 + 1} + 2$ .

The congruent inradii of $\triangle BFC$ and $\triangle AFB$ is $R = \frac{1}{2}(CF + BC - BF) = \cfrac{2A_{\triangle AFB}}{P_{\triangle AFB}}$ , or $R = \frac{1}{2}(k + 1 - \sqrt{k^2 + 1}) = \cfrac{\sqrt{3} - k}{\sqrt{3} - k + \sqrt{k^2 + 1} + 2}$ , which solves to $k = \frac{\sqrt{2}}{2}$ and $R = \frac{1}{4}(2 + \sqrt{2} - \sqrt{6})$ .

By alternate interior angles of parallel segments $CG$ and $AB$ , $\angle GCA = \angle FBA$ and $\angle CGB = \angle ABG$ , so $\triangle CFG \sim \triangle AFB$ by AA similarity, which means $\cfrac{r}{R} = \cfrac{CF}{AF}$ , or $\cfrac{r}{\frac{1}{4}(2 + \sqrt{2} - \sqrt{6})} = \cfrac{\frac{\sqrt{2}}{2}}{3 - \frac{\sqrt{2}}{2}}$ , which solves to $r = \cfrac{-2 + \sqrt{3} + \sqrt{2 + \sqrt{3}}}{10}$ .

Therefore, $a = 2$ , $b = 3$ , $c = 10$ , and $a + b + c = \boxed{15}$ .

Let the radius of the black circles be $R$ and that of the red circle be $r$ , $\angle EBA = \theta$ and $AC$ and $BE$ intersect at $F$ .. Using the fact that the line joining the incircle and a vertex of the triangle bisects the vertex angle, we have:

$\begin{aligned} R \cot \frac {\angle CAB}2 + R \cot \frac {\angle EBA}2 & = AB \\ R \cot 15^\circ + R \cot \frac \theta 2 & = 2 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ R(2 + \sqrt 3) + \frac Rt & = 2 \\ \implies R & = \frac {2t}{(2+\sqrt 3)t + 1} \end{aligned}$

Similarly,

$\begin{aligned} R \cot \frac {\angle CBE}2 + R \cot \frac {\angle ACB}2 & = BC \\ R \cot \left(30^\circ - \frac \theta 2\right) + R \cot 45^\circ & = 1 \\ R \left(\frac {\sqrt 3 + t}{1-\sqrt 3 t}\right) + R & = 1 \\ \implies R & = \frac {1-\sqrt 3 t}{1+\sqrt 3 + (1-\sqrt 3)t} \end{aligned}$

Therefore

$\begin{aligned} \frac {2t}{(2+\sqrt 3)t + 1} & = \frac {1-\sqrt 3 t}{1+\sqrt 3 + (1-\sqrt 3)t} \\ 2(1+\sqrt 3)t + 2(1-\sqrt 3)t^2 & = 1 + 2t - (2\sqrt 3 + 3)t^2 \\ 5t^2 + 2\sqrt 3t - 1 & = 0 \\ \implies t & = \frac {2\sqrt 2 - \sqrt 3}5 \\ \implies \tan \theta & - \frac {2t}{1-t^2} = 4\sqrt 2 - 3\sqrt 3 \end{aligned}$

Similarly for the red circle,

$\begin{aligned} r \cot \frac {\angle EFC}2 + r \cot \frac {\angle DCF}2 & = FC \\ r \cot \left(75^\circ - \frac \theta 2\right) + r \cot 15^\circ & = \tan (60^\circ - \theta) \\ r \left(\frac {2-\sqrt 3 + t}{1-(2-\sqrt 3)t} + 2 + \sqrt 3 \right) & = \frac {\sqrt 3 - 4\sqrt 2 + 3\sqrt 3}{1+\sqrt 3(4\sqrt 2 - 3\sqrt 3)} \\ \frac {4r}{1-(2-\sqrt 3)t} & = \frac 1{\sqrt 2} \\ \implies r & = \frac {1-(2-\sqrt 3)t}{4\sqrt 2} \\ & = \frac {5-(2-\sqrt 3)(2\sqrt 2 - \sqrt 3)}{20\sqrt 2} \\ & = \frac {1-2\sqrt 2 + \sqrt 3 + \sqrt 6}{10\sqrt 2} \\ & = \frac {-2+\sqrt 3}{10} + \frac 1{10} \cdot \frac {1+\sqrt 3}{\sqrt 2} \\ & = \frac {-2+\sqrt 3}{10} + \frac 1{10} \sqrt{\left(\frac {1+\sqrt 3}{\sqrt 2}\right)^2} \\ & = \frac {-2+\sqrt 3 + \sqrt{2+\sqrt 3}}{10} \end{aligned}$

Therefore (a+b+c = 2+3+10 = \boxed {15}}.