Seven Deadly Sin-tors Sangaku

Let the sides of the square be $2$ , so that the radius the of the green semicircle is $r_g = 1$ and the radius of the violet circle is $r_v = \frac{1}{2}$ , and label some more points as follows:

$\triangle DMC \cong \triangle DML$ by HL congruency, so let $\theta = \angle LDM = \angle MDC$ .

From $\triangle DMC$ , $\tan \theta = \cfrac{1}{2}$ , so $\tan 2\theta = \cfrac{2\tan\theta}{1 - \tan^2 \theta} = \cfrac{2 \cdot \frac{1}{2}}{1 - (\frac{1}{2})^2} = \cfrac{4}{3}$ , and since $\angle KDF = 2\theta$ and $DF = 1$ , $KF = \cfrac{4}{3}$ .

By the Pythagorean Theorem on $\triangle KDF$ , $KD = \sqrt{DF^2 + KF^2} = \sqrt{1^2 + (\cfrac{4}{3})^2} = \cfrac{5}{3}$ , so the radius of the yellow circle is $r_y = \cfrac{1}{2}(1 + \cfrac{4}{3} - \cfrac{5}{3}) = \cfrac{1}{3}$ .

$EK = EF - KF = 2 - \cfrac{4}{3} = \cfrac{2}{3}$ . Since $\triangle EKG \sim \triangle FKD$ by AA similarity, the radius of the blue circle is $r_b = \cfrac{EK}{KF} \cdot r_y = \cfrac{\frac{2}{3}}{\frac{4}{3}} \cdot \cfrac{1}{3} = \cfrac{1}{6}$ .

Since $\triangle EKG \sim \triangle JKN$ by AA similarity and the blue incircles are identical, $\triangle EKG \cong \triangle JKN$ and $JK = KE = \cfrac{2}{3}$ .

That means $DJ = DK - JK = \cfrac{5}{3} - \cfrac{2}{3} = 1$ .

Since $\triangle JDH \sim \triangle FKD$ by AA similarity, the radius of the red circle is $r_r = \cfrac{DJ}{KF}\cdot r_y = \cfrac{1}{\frac{4}{3}} \cdot \cfrac{1}{3} = \cfrac{1}{4}$ .

That means the ratio of the sum of all six smaller radii to the radius of the green semicircle is $\cfrac{3r_b + r_v + r_y + r_r}{r_g} = \cfrac{3 \cdot \frac{1}{6} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}}{1} = \cfrac{19}{12}$ .

Therefore, $a = 19$ , $b = 12$ , and $a + b = \boxed{31}$ .

By observation, we note that all the triangles are similar right triangles. Let the radius of the green semicircle be $r_{\rm semicircle}= 1$ . Then the side length of the square is $2$ and the radius of the violet circle is $r_{\rm violet} =\frac 12$ . Since the violet circle is the incircle of $\triangle AGD$ , from $\Delta = sr$ , where $\Delta$ is the area of the triangle, $s$ , its semiperimeter and $r$ , the inradius, we have:

$\begin{aligned} \frac {AD \cdot AG}2 & = \frac {AD + AG + GD}2 \cdot \frac 12 & \small \blue{\text{Let }AG=x} \\ 2x & = \frac {2+x+\sqrt{4+x^2}}2 \\ 3x - 2 & = \sqrt{4+x^2} & \small \blue{\text{Squaring both sides}} \\ 9x^2 - 12x + 4 & = 4+x^2 \\ 8x^2 - 12 x & = 0 & \small \blue{\text{Since }x > 0} \\ 2x - 3 & = 0 \\ \implies x & = \frac 32 = AG \end{aligned}$

Now we note that $\triangle AGD$ is a $3:4:5$ right triangle. This means that all the other similar triangles are also $3:4:5$ right triangles, and this makes calculations easy. And the ratio of the inradius to the side lengths of the triangle is $1:3:4:5$ .

$\begin{aligned} r_{\rm blue} & = \frac {EG}3 = \frac {AG-AE}3 = \frac {\frac 32 -1}3 = \frac 16 \\ r_{\rm yellow} & = \frac {DI}5 = \frac {DF+FI}5 = \frac {1+\frac 43\cdot EG}5 = \frac {1+\frac 43 \cdot \frac 12}5 = \frac 13 \\ r_{\rm red} & = \frac {HD}5 = \frac {\frac 34 \cdot DI}5 = \frac {\frac 34 \cdot \frac 53} 5 = \frac 14 \end{aligned}$

Therefore $\dfrac {r_{\rm violet}+r_{\rm yellow} + r_{\rm red}+3r_{\rm blue}}{r_{\rm semicircle}} = \dfrac {\frac 12 + \frac 13 + \frac 14 + 3 \cdot \frac 16}1 = \dfrac {19}{12}$ , $\implies a + b = 19 + 12 = \boxed{31}$ .