My ultimate Sangaku part 1

Let the centers of the purple circle, the yellow semicircle and right blue semicircle be $A$ , $B$ , and $C$ and their radii be $r_1$ , $r_2$ , and $r_3$ respectively. Points $P$ and $Q$ are tangent points of the purple circle and the left blue semicircle with the black semicircle. Then $PB = 2r_1 + r_2 = 1$ and $QB = 2r_3 + r_2 = 1$ . This means that $r_1 = r_3$ . Then we note that $\triangle ABC$ is an isosceles right triangle with $\angle B = 90^\circ$ and

$\begin{aligned} CA & = \sqrt 2 AB \\ 2r_1 & = \sqrt 2(1 - r_1) \\ \sqrt 2 r_1 & = 1 - r_1 \\ \implies r_1 & = \frac 1{\sqrt 2+1} = \sqrt 2 -1 \\ \implies r_2 & = 1 - 2r_1 = 3 - 2\sqrt 2 \end{aligned}$

The area of the white region is given by:

$\begin{aligned} A & = \frac \pi 2 - 2 \pi r_1^2 - \frac {\pi r_2^2}2 = \pi \left(\frac {1-4(\sqrt 2-1)^2 - (3-2\sqrt2)^2}2 \right) = \pi \left(10\sqrt 2 - 14 \right) \end{aligned}$

Therefore, $\sqrt[b]{ac + \lceil \pi \rceil} = \sqrt {140+4} = \boxed {12}$ .

Let the radius of the yellow semicircle be $r_1$ , blue semicircle be $r_2$ and pink circle be $r_3$ . Then

$r_1+2r_2=r_1+2r_3=1\implies r_2=r_3$

Also, $(r_2+r_3)^2=(r_1+r_2)^2+(r_1+r_3)^2\implies 4r_2^2=2(r_1+r_2)^2$

$\implies r_1=(\sqrt 2 -1)r_2$

$\implies r_2=\dfrac {1}{\sqrt 2 +1}=\sqrt 2 -1,r_1=(\sqrt 2 -1)^2$

Area of all the colored figures is

$2πr_2^2+\dfrac π2 r_1^2=\dfrac π2 (29-20\sqrt 2)$

Hence, area of the required region is

$\dfrac π2 (20\sqrt 2-28)=π(10\sqrt 2-14)$

So, $a=10,b=2,c=14$ , and

$\sqrt[b] {a.c+\lceil π\rceil }=\sqrt {10\times 14+4}=\boxed {12}$ .

We can easily solve this problem by using the pythagorean theorem : - $\begin{cases}\left(x+R\right)^2=\left(R+1-2x\right)^2+\left(x+1-2x\right)^2\\R+1-2x=1-R\end{cases}$

• We solve this system of equation and get $\boxed{R=\sqrt{2}-1}$ and $\boxed{x=\sqrt{2}-1}$
• Then the white area is : $\boxed{\frac{\pi }{2}-\pi \:\left(2\left(\sqrt{2}-1\right)^2+\frac{\left(1-2\sqrt{2}+2\right)^2}{2}\right)=\pi \left(10\sqrt{2}-14\right)}$
• $\boxed{\sqrt[2]{10\cdot 14+\lceil \pi \rceil }=12}$