Dynamic Geometry: P34

Label the semicircle $ABC$ , where $AB$ is its diameter and $C$ its highest point; the centers of the semicircle, yellow, cyan, and pink circles, $O$ , $P$ , $Q$ , $S$ and $T$ respectively; and the red point $R$ . Let the radii of the yellow, cyan, and pink circles be $r_1$ , $r_2$ , and $r_3$ respectively, and $\angle RAO = \angle RBO = \theta$ .

We note that $\dfrac {OP}{AO} = r_1 = \tan \dfrac \theta 2 = t$ (my favorite half-angle tangent substitution ). Let $OR = v = \tan \theta = \dfrac {2t}{1-t^2}$ . If we extend the two green segments to meet the tangent to the semicircle at $C$ , we note that the cyan and yellow circles are inscribed in two similar triangles. The ratio of their radii is the ratio of their heights. Therefore $r_2 = \dfrac {1-v}v\cdot r_1 = \dfrac {1-2t-t^2}2.$

Because of symmetry, $S$ , $R$ , and $T$ are colinear and $ST || AB$ . By Pythagorean theorem, we note that:

$\begin{aligned} OR^2+\blue{RT}^2 & = OT^2 & \small \blue{\text{Note that }RT=RS=r_3 \csc \theta} \\ v^2 + (r_3 \csc \theta)^2 & = (1-r_3)^2 \\ v^2 + r_3^2 (1+ \blue{\cot^2 \theta}) & = 1 - 2r_3 + r_3^2 & \small \blue{\text{Note that }v = \tan \theta} \\ v^2 + \frac {r_3^2}{v^2} & = 1 - 2r_3 \\ \frac {r_3^2}{v^2} + 2r_3+v^2 & = 1 \\ \left(\frac {r_3}v + v \right)^2 & = 1 \\ \implies r_3 & = v(1-v) = \frac {2t(1-2t-t^2)}{(1-t^2)^2} \end{aligned}$

The area of the quadrilateral $PSQT$ is given by $PQ \cdot RS$ and when it is $\dfrac {25}{144}$ , we have:

$\begin{aligned} (1-r_1-r_2) r_3 \csc \theta & = \frac {25}{144} \\ \left(1 - t - \frac {1-2t-t^2}2 \right) \cdot \frac {2t(1-2t-t^2)}{(1-t^2)^2} \cdot \frac {1+t^2}{2t} & = \frac {25}{144} \\ \frac {(1+t^2)^2(1-2t-t^2)}{2(1-t^2)^2} & = \frac {25}{144} \\ \frac {-(t^6 + 2t^5 + t^4+4t^3- t^2 + 2t-1)}{t^4 - 2t^2 + 1} & = \frac {25}{72} \\ 72t^6+144t^5+97t^4+288t^3-122t^2+144t - 47 & = 0 \\ (3t-1)(24t^5+56t^4+51t^3+113t^2-3t+47 & = 0 \end{aligned}$

There is a positive root $t = \dfrac 13$ . Then the ratio $\dfrac {r_2}{r_1} = \dfrac {1-2t-t^2}{2t} = \dfrac {9-6-1}{6} = \dfrac 13$ . Therefore the required answer is $\sqrt{1+3}= \boxed 2$ .

See the diagram below:

• Point $A\left(0;a\right)$ is our parameter in this problem. First, we shall evaluate $r_1$ , the inradius of triangle $ABC$ .

• We get $r_1=\dfrac{2\cdot Area}{perimeter}=\dfrac{AO\cdot OB}{AC+AB+BC}=\dfrac{2\cdot a}{2+2\cdot \sqrt{a²+1}}=\dfrac{a}{1+\sqrt{a²+1}}$

• Now we use the pytagorean theorem and basic trigonometry to get $r_2$ , with $\alpha =90°-\beta$

• We get $\sin \left(\alpha \right)=\cos \left(90°-\alpha \right)=\cos \left(\beta \right)=\dfrac{a}{\sqrt{a^2+1}}$ and $\sin \left(\alpha \right)=\dfrac{HF}{AF}$

• We combine those two equations and get $\dfrac{a}{\sqrt{a^2+1}}=\dfrac{r_2}{AF}\:<=>AF=\dfrac{\sqrt{a^2+1}\cdot r_2}{a}$

• Now by the Pythagorean theorem: $OF^2=AF^2+FM^2\:<=>\left(1-r_2\right)^2=AF^2+a^2\:<=>AF=\sqrt{\left(1-r_2\right)^2-a^2}$

• We combine the two expressions for $AF$ and get: $r_2=a-a^2$

• Let's get $r_3$ . We use a similar method, it's even easier since the center is on the $y$ axis.

• We get $r_3=\dfrac{1-a}{1+\sqrt{1+a^2}}$

• Diving the black quadrilateral into two isosceles triangles, its area, after some algebra, can be expressed as: $AF\cdot \left(AJ+AI\right)=\dfrac{\left|-1+a\right|\left(1+a^2\right)}{1+\sqrt{1+a^2}}$

• Now we solve $\dfrac{\left|-1+a\right|\left(1+a^2\right)}{1+\sqrt{1+a^2}}=\dfrac{25}{144}$

• We get $a=\dfrac{3}{4}$ and evaluate $\dfrac{r_3}{r_2}$ and we obtain $\dfrac{1}{3}$

• Finally $\left|\sqrt{c+b}\right|=\left|\sqrt{1+3}\right|=2$