Sangaku Folding B

Label the figure as above. We note that $\triangle ABC$ , $\triangle CDE$ , and $\triangle EFG$ are similar. Let $AC=p$ and $CA=q$ . Then $p+q$ is the side length of the square paper. Then

$\begin{aligned} \frac {CD}{AB} & = \frac {r_g}{r_b} = \frac 45 \times \frac 34 = \frac 53 & \small \blue{\text{where }r_g \text{ and }r_b \text{ are radii of green}} \\ \implies CD & = \frac 53 q & \small \blue{\text{and blue circles respectively.}} \\ p+q - \sqrt{p^2-q^2} & = \frac 53 q \\ 3p - 2q & = 3\sqrt{p^2-q^2} & \small \blue{\text{Square both sides and rearrange.}} \\ q(13q-12p) & = 0 \\ \implies \frac pq & = \frac {13}{12} \end{aligned}$

For ease of computation, let $p=13$ and $q=12$ and hence $p+q = 25$ instead of $10$ . The $BC=\sqrt{13^2-12^2}=5$ , $CD = 25-5 = 20$ , $CE=20 \times \dfrac {13}{12} = \dfrac {65}3$ , and $EG=25-\dfrac {65}3 = \dfrac {10}3$ . Then the radius of the red circle:

$\begin{aligned} \frac {r_r}{r_b} & = \frac {\frac {10}3}5 = \frac 23 \\ \implies r_r & = \frac 23 r_b = \frac 23 \times \frac 45 = \frac 8{15}\end{aligned}$ .

Therefore $a+b = 8 + 15 = \boxed{23}$ .

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Bonus: General solution as mentioned by @Michael Mendrin

Since $\triangle ABC$ , $\triangle CDE$ , and $\triangle EFG$ are similar, the ratio of the radii of the three circle $r_b: r_g : r_r = BC:DE:EG$ . Let $\angle BAC = \theta$ and $p=1$ ; then $q=\cos \theta$ and the side length of the square paper is $1+\cos \theta$ . Then $BC = \sin \theta$ , $CD = BD-BC = 1+\cos \theta - \sin \theta$ , $CE= \frac {CD}{\cos \theta} = \sec \theta + 1 - \tan \theta$ , $EG = CG-CE = \cos \theta - \sec \theta + \tan \theta$ , and

$\begin{aligned} DE & = CE\sin \theta \\ & = \tan \theta + \sin \theta - \sec \theta + \cos \theta \\ & = EG + BC \\ \implies DE - BC & = EG \end{aligned}$

Therefore $\boxed{r_r = r_g - r_b}$ .

In general, for a square piece of paper, $R = G - B$ , where $(R,G,B)$ are the radii of the red, green, blue circles.

In particular,

$\dfrac{8}{15} = \dfrac{4}{3} - \dfrac{4}{5}$

Let the triangle whose incircle is the blue circle be $\triangle {ABC}$ right angled at $A$ . Let $|\overline {BA}|=x, \angle {ABC}=α$ . Then $x=\dfrac{10\cos α}{1+\cos α}$ , and radius of the blue circle is $\dfrac{4}{5}=\dfrac{x\tan α}{1+\sec α+\tan α}$ . Substituting for $x$ , we get $\cos α=\dfrac{12}{13}, \sin α=\dfrac{5}{13}, \tan α=\dfrac{5}{12}, \sec α=\dfrac{13}{12}$ . Let the triangle whose incircle is the red circle be $\triangle {EFG}$ right angled at $F$ and $|\overline {GF}|=y$ . Then $y=\dfrac{10(1-\sin α)}{1+\cos α}=\dfrac{16}{5}$ . Then radius of the red circle is $\dfrac{y\tan α}{1+\sec α+\tan α}=\dfrac{8}{15}$ . So $a=8, b=15$ and $a+b=\boxed {23}$