Dynamic Geometry: P25

• To solve this problem, we shall denote the radius of the cyan semi-circle as $a$ , the radius of the green semi-circle as $1-a$ and the radius of the yellow circle as $R(a)$ , $x(a)$ is the $x$ coordinate of the yellow circle's center and $y(a)$ is the $y$ coordinate of the yellow circle's center.

• We shall use the Pythagorean theorem to express $R(a)$ , $x(a)$ and $y(a)$ in terms of $a$

• We shall write three equations using the diagram:

• $(1-R(a))^{2}=x(a)^{2}+y(a)^{2}$
• $(1-a+R(a))^{2}=(a-x(a))^{2}+y(a)^{2}$
• $(a+R(a))^{2}=(1-a+x(a))^{2}+y(a)^{2}$

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• After combining the equation and using minor algebra skills we find:
• $R(a)=\dfrac{a-a^{2}}{a^{2}-a+1}$
• $x\left(a\right)=\dfrac{2a-1}{a^2-a+1}$
• $y\left(a\right)=2\cdot \dfrac{\left(a^2-a\right)}{-a^2+a-1}$

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• Now we can use $x(a)$ and $y(a)$ to express $y$ in terms of $x$ to get the equation of the locus.
• We find : $y=\dfrac{2}{3}\left(-1+\sqrt{4-3x^2}\right)$
• We now evaluate the required area : $\int _{-1}^1\dfrac{2}{3}\left(-1+\sqrt{4-3x^2}\right)\:=\dfrac{8\pi \:\sqrt{3}-18}{27}$
• $8+3+18+27=56$

Let the centers of the large, green, and cyan semicircle, and the yellow circle be $O$ , $A$ , $B$ , and $C$ , and the radii of the green and cyan semicircles, and the yellow circle be $r_1$ , $r_2$ , and $r$ respectively. Then $r_1 + r_2 = 1$ . Since $AO = 1-r_1 = r_2$ and $OB = 1 - r_2 = r_1$ . By cosine rule ,

$\begin{aligned} BC^2 & = OB^2 + OC^2 - 2\cdot OB \cdot OC \cdot \cos \blue{\angle BOC} & \small \blue{\text{Let }\angle BOC = \theta.} \\ (r_2+r)^2 & = r_1^2 + (1-r)^2 - 2 r_1 (1-r) \cos \blue \theta \quad ...(1) & \small \blue{\text{Similarly for }CA^2} \\ (r_1+r)^2 & = r_2^2 + (1-r)^2 + 2 r_2 (1-r) \cos \theta \quad ...(2) \end{aligned}$

From $r_2 \cdot (1) + r_1 \cdot (2)$ :

$\begin{aligned}r_2(r_2+r)^2 + r_1(r_1+r)^2 & = r_2r_1^2 + r_1r_2^2 + (r_2+r_1)(1-r)^2 \\ r_1^3 + r_2^3 + 2r(r_1^2+r_2^2) + \blue{(r_1+r_2)}r^2 & = r_1r_2\blue{(r_1+r_2)} + \blue{(r_1+r_2)}(1 - 2r + r^2) & \small \blue{\text{Note that }r_1+r_2 = 1} \\ (r_1+r_2)((r_1+r_2)^2 - 3r_1r_2) + 2r((r_1+r_2)^2 - 2r_1r_2) + r^2 & = r_1r_2 + 1 - 2r + r^2 \\ 1 - 3r_1r_2 + 2r - 4r_1r_2r + r^2 & = r_1r_2 + 1 - 2r + r^2 \\ 4r - 4r_1r_2r & = 4r_1r_2 \\ \implies r_1r_2 & = \frac r{1+r} \end{aligned}$

To find what form the locus is, let $O=(0,0)$ , the origin of the $xy$ -plane, and an arbitrary point on the locus $C=(x,y)$ . By Heron's formula , the area of $\triangle ABC$ ,

$\begin{aligned} A_\triangle & = \sqrt{(r_1+r_2+r)\blue{r_1r_2}r} = \sqrt{(1+r) \cdot \frac {r^2}{1+r}} = r & \small \blue{\text{Note that }r_1r_2 = \frac r{1+r}} \end{aligned}$

Since the base of $\triangle ABC$ , $AB=r_1+r_2=1$ , the height $y = 2r$ . By Pythagorean theorem ,

$\begin{aligned} x^2 + y^2 & = OC^2 = (1-r)^2 = 1 -2r + r^2 & \small \blue{\text{Note that }y = 2r} \\ x^2 + \frac 34 y^2 + y & = 1 \\ x^2 + \frac 34 \left(y+\frac 23\right)^2 & = \frac 43 \\ \frac {x^2}{\frac 43} + \frac {\left(y+\frac 23\right)^2}{\frac {16}9} & = 1 \end{aligned}$

Therefore the locus is part of an ellipse with center at $\left(0, -\frac 23\right)$ , minor-axis of length $\frac 2{\sqrt 3}$ , and major-axis $\frac 43$ . Since it is an elliptical segment, we can find its area with geometrical means without using integration. The area under the locus is a $120^\circ$ -elliptical segment and its area is given by:

$A_e = \frac ab\left(\frac {\pi b^2}3 - \frac {b^2\sin 120^\circ}2 \right) = \frac {\sqrt 3}2 \left(\frac {16}{27}\pi - \frac {4\sqrt 3}9 \right) = \frac {8\sqrt 3 \pi -18}{27}$

Therefore the required answer is $8+3+18+27 = \boxed{56}$ .