Semi-Circles

First, we need to express $\boxed{R}$ in terms of $\boxed{r}$ . We draw 3 right triangles, a pink one, a blue one and a green one

• The blue triangle gives us this equation with the pythagorean theorem : $\boxed{(R+r)²=(R-r)²+(a+b)²}$
• The green triangle gives us this equation with the pythagorean theorem : $\boxed{(1-R)²=R²+b²}$
• The pink triangle gives us this equation with the pythagorean theorem : $\boxed{(1-r)²=r²+a²}$
• Next, we use the second and the third equation to express $\boxed{a}$ and $\boxed{b}$ in terms of $\boxed{r}$ and $\boxed{R}$ . Then we use substitution in the first equation to get an equation only in terms of $\boxed{R}$ and $\boxed{r}$ .
• The first equation becomes : $\boxed{2Rr=1-r-R+\sqrt{(1-2r)(1-2R}}$
• Now we express $\boxed{R}$ in terms of $\boxed{r}$ and get : $\boxed{R=\frac{-2r²+2\sqrt{2}\sqrt{r²-2r^{3}}+3r}{4r²+4r+1}}$
• We can now use this fonction to express the sum $\boxed{R+r}$ in terms of $\boxed{r}$ :
• $\boxed{f(r)=r+\frac{-2r²+2\sqrt{2}\sqrt{r²-2r^{3}}+3r}{4r²+4r+1}}$
• Now set the derivative equal to 0 to find for which value of r, the function is maximum, we find $\boxed{r=\sqrt{2}-1}$
• Finally we calculate $\boxed{f(\sqrt{2}-1)=2(\sqrt{2}-1)}$ so $\boxed{a=2}$ , $\boxed{b=2}$ , $\boxed{c=1}$
• Then, $\boxed{(2+2 \times 2+3 \times 1)! =362880}$