Quick and cute summer Sangaku #12

I couldn’t resist using Archimedes’ book of Lemmas .
Let $AB$ , $AC$ , $CB$ be the diameters of the red, the pink and the green semicircles. Let $DE$ be a diameter of the blue circle. If we denote $x$ the radius of the pink semicircle, then the radius of the green semicircle is $1-x$ .

In Proposition 6 of the Book of Lemmas, we see that if $\dfrac{AC}{CB}=r$ then $\dfrac{DE}{AB}=\dfrac{r}{{{r}^{2}}+r+1} \ \ \ \ (1)$

In our case, $r=\dfrac{x}{1-x}$ and $AB=2$ , so $\left( 1 \right)\Rightarrow DE=\ldots =\dfrac{2x-2{{x}^{2}}}{{{x}^{2}}-x+1}$ .

Now, the white area, in terms of $x$ is: $A\left( x \right)=\frac{\pi }{2}-\pi \left[ \frac{1}{2}{{x}^{2}}+\frac{1}{2}{{\left( 1-x \right)}^{2}}+\frac{1}{4}{{\left( \frac{2x-2{{x}^{2}}}{{{x}^{2}}-x+1} \right)}^{2}} \right],\text{ }x\in \left( 0,1 \right)$ The maximum of $A\left( x \right)$ is $0.4422765949…$ , hence, the answer is $\left\lfloor {{10}^{7}}{{A}_{\max }} \right\rfloor =\boxed{4422765}$ .

• To solve this problem, we shall denote the radius of the green semi-circle as $\boxed{x}$ , the radius of the pink semi-circle as $\boxed{1-x}$ and the radius of the blue circle as $\boxed{R(x)}$ .
• We shall use the Pythagorean theorem to express $\boxed{R(x)}$ in terms of $\boxed{x}$
• We shall write three equations using the diagram:
• $\boxed{(1-R(x))^{2}=y^{2}+h^{2}}$
• $\boxed{(1-x+R(x))^{2}=(x-y)^{2}+h^{2}}$
• $\boxed{(x+R(x))^{2}=(1-x+y)^{2}+h^{2}}$

---

• After combining the equation and using minor algebra skills we find:
• $\boxed{R(x)=\frac{x-x^{2}}{x^{2}-x+1}}$
• Now we can express the white area in terms of $\boxed{x}$ :
• $\boxed{A(x)=\frac{\pi }{2}-\pi \cdot \left(\left(\frac{x-x^2}{x^2-x+1}\right)^2+\frac{x^2}{2}+\frac{\left(1-x\right)^2}{2}\right)}$
• The maximum of the function is $\boxed{0.44227659...}$
• $\boxed{\lfloor 10^7A_{max} \rfloor=4422765}$