Dynamic Geometry: P55

Let the leg of the orange triangle on the left side be $a$ , and the other leg be $b$ , and let the radius of the pink incircle be $r_a$ , the radius of the cyan incircle be $r_b$ , and the radius of the purple incircle be $r_c$ .

By Thales's Theorem, the angle at the red point is right, so the orange triangles and the smaller triangles are similar to each other, which means $r_a = \cfrac{ar_c}{2}$ and $r_b = \cfrac{br_c}{2}$ .

Since the three radii are in an arithmetic progression, $r_a + r_c = 2r_b$ . Substituting $r_a = \cfrac{ar_c}{2}$ and $r_b = \cfrac{br_c}{2}$ and solving gives $a + 2 = 2b$ .

By the Pythagorean Theorem on the orange triangle, $a^2 + b^2 = 4$ . Combining this with $a + 2 = 2b$ and solving gives $a = \cfrac{6}{5}$ and $b = \cfrac{8}{5}$ .

As an inradius of a right triangle, $r_c = \cfrac{1}{2}(a + b - 2) = \cfrac{1}{2}\bigg(\cfrac{6}{5} + \cfrac{8}{5} - 2\bigg) = \cfrac{2}{5}$ .

That means $r_a = \cfrac{ar_c}{2} = \cfrac{\frac{6}{5} \cdot \frac{2}{5}}{2} = \cfrac{6}{25}$ and $r_b = \cfrac{br_c}{2} = \cfrac{\frac{8}{5} \cdot \frac{2}{5}}{2} = \cfrac{8}{25}$ .

The sum of the areas of the three incircles is $S = \pi r_a^2 + \pi r_b^2 + \pi r_c^2 = \pi \bigg(\cfrac{6}{25}\bigg)^2 + \pi \bigg(\cfrac{8}{25}\bigg)^2 + \pi \bigg(\cfrac{2}{5}\bigg)^2 = \cfrac{8\pi}{25}$ , so $p = 8$ , $q = 25$ , and $p + q = \boxed{33}$ .

Let the radius of the largest incircle (purple) be $r_0$ and the left angle of the orange triangle be $\theta$ . Then the angle on the right is $90^\circ - \theta$ . And $r_0 cot \dfrac \theta 2 + r_0 \cot \left(\dfrac {90^\circ - \theta}2\right) = 2$ , where $2$ is the diameter of the semicircle. By half-angle tangent substitution and let $t = \tan \dfrac \theta 2$ ,

$\begin{aligned} \frac {r_0}t + \frac {1+t}{1-t}r_0 & = 2 \\ \frac {1+t^2}{t(1-t)}r_0 & = 2 \\ \implies r_0 & = \frac {2t(1-t)}{1+t^2} \end{aligned}$

Note that the length of the green segment is $2\cos \theta \sin \theta$ . Let the radius of the cyan circle be $r_1$ . Then we have:

$\begin{aligned} r_1 \cot \frac \theta 2 + r_1 & = 2 \cos \theta \sin \theta \\ \left(\frac {1+t}t\right)r_1 & = \frac {4t(1-t^2)}{(1+t^2)^2} \\ \implies r_1 & = \frac {4t^2(1-t)}{(1+t^2)^2} \end{aligned}$

Similarly, let the radius of the pink circle be $r_2$ . Then

$\begin{aligned} r_2 \cot \left(45^\circ - \frac \theta 2\right) + r_2 & = 2 \cos \theta \sin \theta \\ \left(\frac 2{1-t} \right)r_2 & = \frac {4t(1-t^2)}{(1+t^2)^2} \\ \implies r_2 & = \frac {2t(1-t^2)(1-t)}{(1+t^2)^2} \end{aligned}$

Since $r_0 > r_1, r_2$ for $t >0$ , and $r_1$ and $r_2$ are interchangeable, we can assume $r_1$ be the middle term of the arithmetic progression. Then we have:

$\begin{aligned} r_0 + r_2 & = 2r_1 \\ \frac {2t(1-t)}{1+t^2} + \frac {2t(1-t^2)(1-t)}{(1+t^2)^2} & = \frac {8t^2(1-t)}{(1+t^2)^2} & \small \blue{\text{Multiply both sides by }\frac {(1+t^2)^2}{2t(1-t)}.} \\ 1 + t^2 + 1 - t^2 & = 4t \\ \implies t & = \frac 12 \\ r_0 & = \frac {2t(1-t)}{1+t^2} = \frac 25 \\ r_1 & = \frac {4t^2(1-t)}{(1+t^2)^2} = \frac 8{25} \\ r_2 & = \frac {2t(1-t^2)(1-t)}{(1+t^2)^2} = \frac 6{25} \\ \implies \pi r_0^2 + \pi r_1^2 + \pi r_2^2 & = \frac {200}{625}\pi = \frac 8{25} \pi \end{aligned}$

Therefore $a+b = 8 + 25 = \boxed{33}$ .