Quick and cute summer Sangaku #2

We note that the center of red circle and the center of blue circle has a horizontal distance of $4$ . If the height of the green rectangle is $h$ , then the width of the green rectangle $w$ is given by:

$\begin{aligned} w & = 4 - \sqrt{1^2 - (1-h)^2} - \sqrt{4^2-(4-h)^2} \\ & = 4 - \sqrt{2h-h^2} - \sqrt{8h-h^2} \end{aligned}$

Then the area of the green rectangle is $hw = h(4 - \sqrt{2h-h^2} - \sqrt{8h-h^2})$ . And the maximum area $A \approx 0.598745976$ $\implies \lfloor 10^7 A \rfloor = \boxed{5987459}$ .

• We can draw a coordinate system and give the pink circle the equation : $\boxed{x^{2}+(y-1)^{2}-1=0}$ and the red circle this equation : $\boxed{(x-4)^{2}+(y-4)^{2}-16=0}$
• There must be a point $\boxed{M(x;y)}$ on the pink circle and a point $\boxed{N(x';y)}$ on the blue circle to construct the rectangle.
• Since point $\boxed{M(x;y)}$ is on the red circle, we can express $\boxed{x}$ in terms of $\boxed{y}$ using the equation $\boxed{x^{2}+(y-1)^{2}-1=0}$ <=> $\boxed{x=\sqrt{2y-y^{2}}}$
• Since point $\boxed{N(x';y)}$ is on the blue circle, we can express $\boxed{x}$ in terms of $\boxed{y}$ using the equation $\boxed{(x-4)^{2}+(y-4)^{2}-16=0}$ <=> $\boxed{x'=4-\sqrt{8y-y^{2}}}$
• Then the width of the rectangle is $\boxed{y}$
• The length of the rectangle is $\boxed{4-\sqrt{8y-y^2}-\sqrt{2y-y^2}}$
• We can now express the rectangle's area in terms of $\boxed{y}$
• $\boxed{A(y)=y(4-\sqrt{8y-y^2}-\sqrt{2y-y^2)}}$
• The maximum of the function is $\boxed{0.58974597...}$
• Finally, $\boxed{\lfloor{10^{7}A}\rfloor=5987459}$