Dynamic Geometry: P39

Define the center of the circular sector as $O(0,0)$ and center of the pink circle as $P \left (\dfrac{24r}{7},r \right)$ , which implies that the radius of the circle is $r.$

In right-triangle $ODP$ , $\begin{aligned} OD^2+PD^2&=OP^2 \\ \\ \left(\frac{24r}{7}\right)^2+r^2&=(1-r)^2 \\ \\ \frac{576r^2}{49}+r^2&=1-2r+r^2 \\ \\ \implies 576r^2+98r-49&=0 \end{aligned}$ Solving the above quadratic, we have, $r=\dfrac{7}{32}\implies |OP| = 1-r=\dfrac{25}{32}$ .

Therefore, $a=25,\; b=32\implies \sqrt[2]{a}-\sqrt[5]{b}=5-2=\boxed{3}$

$\begin{aligned} y &= ( 1-r ) \sin\beta \quad\quad \text{Eq1} \\ \quad \\ d &= \sqrt{x^2 + y^2} \quad\quad \text{Eq2} \end{aligned}$

We are given:

$\frac{x}{y} = \frac{24}{7} \quad\quad \text{Eq3}$

We can deduce from the geometry and the given coordinate ratio locating the center of the circle:

$\begin{aligned} y &= r \quad\quad \text{Eq4}\\ \quad \\ x &= \frac{24}{7} r \quad\quad \text{Eq5} \\ \quad \\ \sin\beta &= \frac{ 1 }{ \sqrt{ \left( \frac{24}{7} \right)^2 + 1 } } \quad\quad \text{Eq6} \end{aligned}$

Let:

$\varphi = \sqrt{ \left( \frac{24}{7} \right)^2 + 1 } \quad\quad \text{Eq7}$

Next , substitute $\text{Eq4, Eq5} \to \text{Eq2}$ , then $\text{Eq7} \to \text{Eq2}$ and let the result be $\text{Eq2'}$ :

$\begin{aligned} d &= \sqrt{ \left( \frac{24}{7}\right)^2 r^2 + r^2} \\ \quad \\ &= r \sqrt{ \left( \frac{24}{7} \right)^2 + 1 } \\ \quad \\ &= r \varphi \quad\quad \text{Eq2'} \end{aligned}$

Substitute $\text{Eq7} \to \text{Eq6} \to \text{Eq1}$ and followed by $\text{Eq4} \to \text{Eq1}$ and let the result be $\text{Eq1'}$ :

$r = \left( 1 - r \right) \frac{1}{ \varphi}$

$\Rightarrow$

$r = \frac{1}{1+\varphi} \quad\quad \text{Eq8}$

Finally, substitute $\text{Eq8} \to \text{Eq2'}$

$d = \frac{\varphi}{ \varphi + 1} = \frac{25}{32} = \frac{a}{b}$

Thus;

$\sqrt{a} - \sqrt[5]{b} = 5 - 2 = 3$