Dynamic Geometry: P21

Label the triangle $ABC$ , the center of the semicircle $O$ , the tangent points of the purple and red circles to the semicircle as $D$ and $E$ respectively. Let the centers and radii of the cyan, purple, and red circles be $P$ , $Q$ , and $R$ , and $r_1$ , $r_2$ , and $r_3$ respectively, and $\angle CAB = \theta$ . Note that $\angle ACB = 90^\circ$ , and $\angle CBA = 90^\circ - \theta$ . Then

$\begin{aligned} r_1 \cot \frac {\angle CAB}2 + r_1 \cot \frac {\angle CBA}2 & = AB \\ r_1 \cot \frac \theta 2 + r_1 \cot \left(45^\circ - \frac \theta 2\right) & = 2 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ \frac {r_1}t + \frac {1+t}{1-t}r_1 & = 2 \\ \implies r_1 & = \frac 2{\frac 1t + \frac {1+t}{1-t}} = \frac {2t(1-t)}{1+t^2} \end{aligned}$

Note that $O$ , $Q$ , and $D$ are colinear, $\angle COB = 2\theta$ , and $\angle DOB = \theta$ . Then

$\begin{aligned} OD & = OB \cos \angle DOB + 2r_2 \\ 1 & = \cos \theta + 2r_2 \\ \implies r_2 & = \frac {1-\cos \theta}2 = \frac {1-\frac {1-t^2}{1+t^2}}2 = \frac {t^2}{1+t^2} \end{aligned}$

Similarly,

$\begin{aligned} r_3 & = \frac {1-\sin \theta}2 = \frac {1-\frac {2t}{1+t^2}}2 = \frac {(1-t)^2}{2(1+t^2)} \end{aligned}$

The areas of the three circles are in a geometric progression, means that $r_1^2$ , $r_2^2$ , and $r_3^2$ are in a geometric progression. Since $r_2$ and $r_3$ are replaceable, and are not independent, we assume either $r_2^2$ or $r_3^2$ as the middle term of the geometric progression. Let us assume:

$\begin{aligned} r_1^2r_2^2 & = r_3^4 \\ r_1r_2 & = r_3^2 \\ \frac {2t^3(1-t)}{(1+t^2)^2} & = \frac {(1-t)^4}{4(1+t^2)^2} \\ 8t^3 & = (1-t)^3 \\ 2t & = 1-t \\ \implies t & = \frac 13 \end{aligned}$

Then $r_1 = \dfrac 25$ , $r_2 = \dfrac 1{10}$ , and $r_3 = \dfrac 15$ . Let $O$ be the origin $(0,0)$ of the $xy$ -plane. Then the coordinates of $P$ , $Q$ , and $R$ :

$\begin{aligned} P & = \left(\frac {r_1}t-1, r_1 \right) = (0.2, 0.4) \\ Q & = \left((1-r_2)\cos \theta, (1-r_2)\sin \theta \right) = \left((1-r_2)\frac {1-t^2}{1+t^2}, (1-r_2)\frac {2t}{1+t^2} \right) = (0.72, 0.54) \\ R & = \left(-(1-r_3)\sin \theta, (1-r_3)\cos \theta \right) = \left(-(1-r_3)\frac {2t}{1+t^2}, (1-r_3)\frac {1-t^2}{1+t^2} \right) = (-0.48, 0.64) \end{aligned}$

The area of $\triangle PQR$ ,

$[PQR] = (0.72+0.48)\frac {0.54+0.64}2 - (0.2+0.48)\frac {0.4+0.64}2 - (0.72-0.2) \frac {0.54+0.4}2 = \frac {11}{100}$

Therefore $a+b = 11+100 = \boxed{111}$ .

Label the diagram as follows, and let $p = AG = GC = OH$ :

Then by the Pythagorean Theorem on $\triangle OHB$ , $BE = \sqrt{1 - p^2} = EC = GO$ .

The diameter of the red circle is then $IG = IO - GO = 1 - \sqrt{1 - p^2}$ , and its radius is $r_r = \frac{1}{2}(1 - \sqrt{1 - p^2})$ .

Likewise, the diameter of the purple circle is $JH = OJ - HJ = 1 - p$ , and its radius is $r_p = \frac{1}{2}(1 - p)$ .

The radius of the cyan circle is $r_c = \frac{1}{2}(AC + BC - AC) = \frac{1}{2}(2p + 2\sqrt{1 - p^2 - 2}) = p + \sqrt{1 - p^2} + 1$ .

For the circles' areas to be in a geometric progression, either $\pi r_p^2 \cdot \pi r_c^2 = (\pi r_r^2)^2$ (which after substituting the above radii equations solves to $p = \frac{4}{5}$ and $\sqrt{1 - p^2} = \frac{3}{5}$ ), or $\pi r_r^2 \cdot \pi r_c^2 = (\pi r_p^2)^2$ (which solves to $p = \frac{4}{5}$ and $\sqrt{1 - p^2} = \frac{3}{5}$ ), or $\pi r_r^2 \cdot \pi r_p^2 = (\pi r_c^2)^2$ (which can be rejected because it solves to $p = 0$ or $p = 1$ , a degenerate triangle).

Without loss of generality, let $p = \frac{4}{5}$ and $\sqrt{1 - p^2} = \frac{3}{5}$ . Then $r_r = \frac{1}{2}(1 - \sqrt{1 - p^2}) = \frac{1}{5}$ , $r_p = \frac{1}{2}(1 - p) = \frac{1}{10}$ , and $r_c = p + \sqrt{1 - p^2} + 1 = \frac{2}{5}$ .

Also, $DO = DG + GO = \frac{1}{5} + \frac{3}{5} = \frac{4}{5}$ and $EO = OH + HE= \frac{4}{5} + \frac{1}{10} = \frac{9}{10}$ .

The height of $\triangle DOF$ with base $DO$ is $h_{\triangle DOF} = OJ - 2r_p - r_c = 1 - 2 \cdot \frac{1}{10} - \frac{2}{5} = \frac{2}{5}$ .

The height of $\triangle EOF$ with base $EO$ is $h_{\triangle EOF} = OI - 2r_r - r_c = 1 - 2 \cdot \frac{1}{5} - \frac{2}{5} = \frac{1}{5}$ .

The area of $\triangle DOF$ is then $A_{\triangle DOF} = \frac{1}{2} \cdot DO \cdot h_{\triangle DOF} = \frac{1}{2} \cdot \frac{4}{5} \cdot \frac{2}{5} = \frac{4}{25}$ .

The area of $\triangle EOF$ is then $A_{\triangle EOF} = \frac{1}{2} \cdot EO \cdot h_{\triangle EOF} = \frac{1}{2} \cdot \frac{9}{10} \cdot \frac{1}{5} = \frac{9}{100}$ .

The area of $\triangle DOE$ is then $A_{\triangle DOE} = \frac{1}{2} \cdot DO \cdot EO = \frac{1}{2} \cdot \frac{4}{5} \cdot \frac{9}{10} = \frac{9}{25}$ .

Finally, the area of $\triangle DFE$ is $A_{\triangle DFE} = A_{\triangle DOE} - A_{\triangle DOF} - A_{\triangle EOF} = \frac{9}{25} - \frac{4}{25} - \frac{9}{100} = \frac{11}{100}$ .

Therefore, $a = 11$ , $b = 100$ , and $a + b = \boxed{111}$ .