2010-I-17

Since $\angle BAD=90^\circ$, then $BD$ is a diameter of the circle $ABCD$. Hence, the centre of the circle $ABCD$

$\begin{array}{cl}
= & \left(\dfrac{-6+8}{2}, \dfrac{8+6}{2} \right) \\
= & (1,7)
\end{array}$

$\begin{array}{cl}
= & \dfrac{1}{2} \sqrt{(-6-8)^2 +(8-6)^2} \\
= & \dfrac{1}{2} \sqrt{200} \\
= & \dfrac{1}{2} \times 10 \sqrt{2} \\
= & 5\sqrt{2}
\end{array}$

$\begin{array}{cl}
= & \sqrt{(8-0)^2 + (6-0)^2} \\
= & 10
\end{array}$

Since the circles $ABCD$ and $A_1B_1C_1D_1$ are similar, then we have

$\begin{array}{cl}
& \dfrac{\text{area of circle }A_1B_1C_1D_1}{\text{area of circle }ABCD} \\
= & \left( \dfrac{\text{diameter of circle }A_1B_1C_1D_1}{\text{diameter of circle }ABCD} \right)^2 \\
= & \left(\dfrac{10}{2\times5\sqrt{2}} \right)^2 \\
= & \dfrac{1}{2} \\
= & 1: 2
\end{array}$

$\begin{array}{cl}
= & 10^2 – \pi(5)^2 \\
= & 100-25\pi
\end{array}$

Therefore, the area of the region between square $A_1B_1C_1D_1$ and circle $A_2B_2C_2D_2$

$\begin{array}{cl}
= & \dfrac{1}{2} \times (100-25\pi) \\
= & \dfrac{100-25\pi}{2}
\end{array}$

Hence, the total area of all shaded region

$\begin{array}{cl}
= & 100-25\pi + \dfrac{100-25\pi}{2} + \dfrac{100-25\pi}{2^9} \\
= & (100-25\pi)(1+\dfrac{1}{2} + \cdots +\dfrac{1}{2^9}) \\
= & (100-25\pi) \times\dfrac{1[1-(\frac{1}{2})^{10}]}{1-\frac{1}{2}} \\
= & 42.878~452~9
\end{array}$

Hence, the total area of all shaded region to the area of circle $ABCD$

$\begin{array}{cl}
= & \dfrac{42.878~452~9}{\pi (5\sqrt{2})^2 } \\
= & \dfrac{0.272~972~709}{1}
\end{array}$

Therefore, $p=0.272~972~709$ which is lying between $0.2$ and $0.3$. Hence, the design of the logo is good.