jmc

The actual size of the diagram doesn't matter. To make calculation easier, we discard the original area of the circle, $1$, and assume the side length of the octagon is $2$. Let $r$ denote the radius of the circle, $O$ be the center of the circle. Then $r^2=1^2+(\sqrt{2}+1{)}^2=4+2\sqrt{2}$. Now, we need to find the "D"shape, the small area enclosed by one side of the octagon and 1/8 of the circumference of the circle:$$D=\frac{1}{8}\pi r^2-[A_1{A}_{2}O]=\frac{1}{8}\pi (4+2\sqrt{2})-(\sqrt{2}+1)$$ Let $PU$ be the height of $△A_1{A}_{2}P$, $PV$ be the height of $△A_3{A}_{4}P$, $PW$ be the height of $△A_6{A}_{7}P$. From the $1/7$ and $1/9$ condition we have$$△P{A}_1{A}_2=\frac{\pi r^2}{7}-D=\frac{1}{7}\pi (4+2\sqrt{2})-(\frac{1}{8}\pi (4+2\sqrt{2})-(\sqrt{2}+1))$$ $$△P{A}_3{A}_4=\frac{\pi r^2}{9}-D=\frac{1}{9}\pi (4+2\sqrt{2})-(\frac{1}{8}\pi (4+2\sqrt{2})-(\sqrt{2}+1))$$which gives $PU=(\frac{1}{7}-\frac{1}{8})\pi (4+2\sqrt{2})+\sqrt{2}+1$ and $PV=(\frac{1}{9}-\frac{1}{8})\pi (4+2\sqrt{2})+\sqrt{2}+1$. Now, let $A_1{A}_2$ intersects $A_3{A}_4$ at $X$, $A_1{A}_2$ intersects $A_6{A}_7$ at $Y$,$A_6{A}_7$ intersects $A_3{A}_4$ at $Z$. Clearly, $△XYZ$ is an isosceles right triangle, with right angle at $X$ and the height with regard to which shall be $3+2\sqrt{2}$. Now $\frac{PU}{\sqrt{2}}+\frac{PV}{\sqrt{2}}+PW=3+2\sqrt{2}$ which gives $PW=3+2\sqrt{2}-\frac{PU}{\sqrt{2}}-\frac{PV}{\sqrt{2}}$ $=3+2\sqrt{2}-\frac{1}{\sqrt{2}}((\frac{1}{7}-\frac{1}{8})\pi (4+2\sqrt{2})+\sqrt{2}+1+(\frac{1}{9}-\frac{1}{8})\pi (4+2\sqrt{2})+\sqrt{2}+1))$ $=1+\sqrt{2}-\frac{1}{\sqrt{2}}(\frac{1}{7}+\frac{1}{9}-\frac{1}{4})\pi (4+2\sqrt{2})$ Now, we have the area for $D$ and the area for $△P{A}_6{A}_7$, so we add them together: $\text{Target Area}=\frac{1}{8}\pi (4+2\sqrt{2})-(\sqrt{2}+1)+(1+\sqrt{2})-\frac{1}{\sqrt{2}}(\frac{1}{7}+\frac{1}{9}-\frac{1}{4})\pi (4+2\sqrt{2})$ $=(\frac{1}{8}-\frac{1}{\sqrt{2}}(\frac{1}{7}+\frac{1}{9}-\frac{1}{4}))\text{Total Area}$ The answer should therefore be $\frac{1}{8}-\frac{\sqrt{2}}{2}(\frac{16}{63}-\frac{16}{64})=\frac{1}{8}-\frac{\sqrt{2}}{504}$. The answer is $\boxed{504}$.