Dynamic Geometry: P38

See the diagram below: - We use the Pythagorean theorem to express $x_D$ and $y_D$ in terms of $r$ :

• $\begin{cases}\left(1+y_D\right)^2=\left(y_D-1\right)^2+x_D^2\\\left(r+y_D\right)^2=\left(2+r-y_D\right)^2+x_D^2\end{cases}$

• $\begin{cases}x_D=\dfrac{2\sqrt{1+r}}{\sqrt{r}}\\y_D=\dfrac{1+r}{r}\end{cases}$

• Now we can express $y_D$ in terms of $y_D$ and we get the equation of the locus: $y=\dfrac{x^2}{4}$

• Since the radius of each circle is: $1$ , $r$ , $\dfrac{1+r}{r}$ . We can find the value of $r$ for which the radii are in arithmetic progression, with $1<r<\dfrac{1+r}{r}$ . We can solve this system of equation, where $k$ is the common difference between the radii:

• $\begin{cases}r=1+k\\\dfrac{1+r}{r}=r+k\end{cases}$

• $\begin{cases}k=\dfrac{\sqrt{3}-1}{2}\\r=\dfrac{\sqrt{3}+1}{2}\end{cases}$

• So we can find the corresponding value of $x_D$ for $r=\dfrac{\sqrt{3}+1}{2}$ and we get $x_D=2\cdot \sqrt[4]{3}$ .

• Now we need to solve $x_D=y_D$ . We use the equation of the locus and solve $x=\frac{x^2}{4}<=>x=4$

• We can evaluate the area: $\int _{2\sqrt[4]{3}}^4\dfrac{x^2\:}{4}=\frac{2}{3}\cdot \left(8-3^{\frac{3}{4}}\right)$

• So $a=2$ and $b=3$ , then $a+b=5$

Label the origin as $O$ , the centers of the green, cyan, and yellow circles be $A$ , $B$ , and $C$ . Let $CN$ be perpendicular to the $x$ -axis, and $AN'$ and $Bn"$ be parallel to $ON$ . We note that $ON = AN' = BN" = \ell$ . Let the radius of the yellow circle be $R$ . By Pythagorean theorem ,

$\begin{aligned} AC^2-CN'^2 & = AN'^2 \\ (R+1)^2 - (R-1)^2 & = \ell^2 \\ \implies 4R & = \ell^2 \end{aligned}$

SImilarly,

$\begin{aligned} BC^2 - CN"^2 & = BN"^2 \\ (R+r)^2 - (R-(2+r))^2 & = \ell^2 \\ 4rR + 4R - 4r - 4 & = \ell^2 \end{aligned}$

Therefore, $4rR + 4R - 4r - 4 = 4R \implies R = \dfrac {r+1}r$ . Now consider any point on the locus $C(x,y)$ . Then $x = ON = \ell = 2\sqrt R$ and $y = CN = R = \dfrac {x^2}4$ . The locus is a parabola. And the locus under the parabola is given by:

$\int_a^b y \ dx = \int_a^b \frac {x^2}4 \ dx = \frac {b^3-a^3}{12}$

where $a$ is the value of $x$ when $1$ , $r$ , and $R$ are in arithmetic progression, and $b$ is the value of $x$ when $x=y$ . When $1$ , $r$ , and $R$ are in a arithmetic progression, we have:

$\begin{aligned} 1 + \blue R & = 2r & \small \blue{\text{Note that }R = \frac {r+1}r} \\ 1 + \blue{\frac {r+1}r} & = 2r \\ 2r^2 - 2r - 1 & = 0 \\ \implies r & = \frac {\sqrt 3 + 1}2 & \small \blue{\text{As }r > 0} \\ \implies R & = \frac {r+1}r = \frac {3+\sqrt 3}{\sqrt 3 +1} = \sqrt 3 & \small \blue{\text{Since }x = 2\sqrt R} \\ \implies a & = 2 \cdot 3^\frac 14 \end{aligned}$

Since $y = \dfrac {x^2}4$ , when $y=x$ $\implies x = 4 = b$ . Therefore the area under the locus is

$\frac {4^3 - 2^3 \cdot 3^\frac 34}{12} = \frac 23 \left(2^3 - 3^\frac 34\right)$

Therefore the required answer is $2+3 = \boxed 5$ .