Revenge of the progressive Sangaku

Let us place a point M on the circle with diameter [AC], we obtain a right triangle AMC with $\boxed{AC=2}$ , $\boxed{MC=x}$ and $\boxed{MA=\sqrt{4-x²}}$

• Now we will be able to express the radii of the three circles in terms of $\boxed{x}$ . We denote $\boxed{R_1}$ the radius of the incircle of the right triangle AMC. We denote $\boxed{R_2}$ the radius of the circle tangent to (MC) and $\boxed{R_3}$ the radius of the circle tangent to (MA)

• $\boxed{R_1(x)=\frac{2Area}{Perimeter}=\frac{x\sqrt{4-x²}}{2+x+\sqrt{4-x²}}}$

• We shall use the product of chords equality in a circle to express $\boxed{R_2}$ and $\boxed{R_3}$ in terms of $\boxed{x}$ . Since the two circles inscribed in the circular segments are the largest possible, their centers are located on the perpendicular bisectors of [AM] and [CM] respectively

• This gives us $\boxed{2R_2(4-2R_2)=\frac{x²}{4}}$ <=> $\boxed{R_2(x)=\frac{2-\sqrt{4-x²}}{4}}$

• This gives us also $\boxed{2R_3(4-2R_3)=\frac{4-x²}{4}}$ <=> $\boxed{R_3(x)=\frac{2-x}{4}}$

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• Since the radii have to be in an arithmetic progression we can write : $\boxed{R_1-R_2=k_1}$ and $\boxed{R_2-R_3=k_1}$ where $\boxed{k_1}$ is the common difference of successive members and $\boxed{R_1>R_2>R_3}$

• But we can also consider the possibility of the incircle not being the largest circle, this gives us $\boxed{R_2-R_1=k_2}$ and $\boxed{R_2-R_3=k_2}$ where $\boxed{R_2>R_1>R_3}$

• We solve $\boxed{R_1-R_2=k_1}$ and $\boxed{R_2-R_3=k_1}$ in terms of $\boxed{x}$ where $\boxed{R_1>R_2>R_3}$ , we find $\boxed{x_1=\frac{6+16\sqrt{2}}{17}}$ and $\boxed{k_1=\frac{10\sqrt{2}-9}{34}}$

• We solve $\boxed{R_2-R_1=k_2}$ and $\boxed{R_2-R_3=k_2}$ in terms of $\boxed{x}$ where $\boxed{R_2>R_1>R_3}$ , we find $\boxed{x_2=\frac{6+\sqrt{14}}{5}}$ and $\boxed{k_2=\frac{\sqrt{\frac{7}{2}}}{10}}$

• After calculating $\boxed{R_1}$ , $\boxed{R_2}$ and $\boxed{R_3}$ for both possible values of $\boxed{x}$ we find that the area of the three circles is smaller for $\boxed{x_2}$ and equals : $\boxed{\frac{19π}{100}}$

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• Now let's determine the area of each rectangle. One way to do it consists in drawing a coordinate system, the center of the incircle being the origin. Let's solve it for $\boxed{x_2}$ . We place a point $\boxed{W(x;y)}$ . The height of the rectangle will be $\boxed{\frac{1}{5}-y}$

• Let's determine the length of the rectangle, the diagram below shows the different parts of the diameter, the first is determined using basic trigonometry in the triangle. The second part is equal to the x coordinate of point W, and since point W is on the circle with equation $\boxed{x²+y²-\frac{1}{25}=0}$ , the length x can be expressed in terms of y, thus $\boxed{x=\frac{\sqrt{1-25y²}}{5}}$ . Next we have the length of the rectangle. At last we use trigonometry to express the last portion of the diameter in terms of $\boxed{y}$ , we get $\boxed{\frac{11(\frac{1}{5}-y)}{25-6\sqrt{14}}}$

• Since the diameter is 2, we find $\boxed{L=-\frac{\sqrt{1-25y²}}{5}-\frac{11(\frac{1}{5}-y)}{25-6\sqrt{14}}-\frac{11}{5\sqrt{39+10\sqrt{14}}}+2}$

• Now we express the area in terms of y $\boxed{A(y)=(-\frac{\sqrt{1-25y²}}{5}-\frac{11(\frac{1}{5}-y)}{25-6\sqrt{14}}-\frac{11}{5\sqrt{39+10\sqrt{14}}}+2)(\frac{1}{5}-y)}$

• The maximum of the function is $\boxed{0.13914807...}$ . We use a similar method to solve the maximum area of the rectangle for $\boxed{x_1}$ and we find $\boxed{β=0.13941851...}$ and this is the larger rectangle.

• $\boxed{\frac{\lfloor{10^{7}β}\rfloor-\lceil{π}\rceil+a}{b}=\frac{1394185-4+19}{100}=13942}$

@Valentin Duringer , thanks for inviting to solve this problem. I have been put off long problem questions that was why I didn't solve your earlier problems. I couldn't come up with solution that is as accurate as yours. But nonetheless I got it done though less dramatically.

Let the radii of the small circle and big circle of the two circular segments be $r_1$ and $r_2$ respective. Let the measure of the smaller angle of the right triangle be $\alpha$ . Then we note that $r_2 = \dfrac {1-\sin \alpha}2$ and $r_1 = \dfrac {1-\cos \alpha}2$ . Let the inradius of the right triangle be $r_3$ . From $\text{semiperimeter} \times \text{inradius} = \text{area of triangle}$ , we have $r_3 = \dfrac {2\sin \alpha \cos \alpha}{1+ \sin \alpha + \cos \alpha}$ . Since $r_1$ , $r_2$ , and $r_3$ are in an arithmetic progression, we have for the upper semicircle $r_1+r_2=2r_3$ and for lower semicircle $r_1+r_3=2r_2$ . Solving for the two $\alpha$ 's, we get:

$\begin{array} {r|c|c|c|c|c} \text{Semicircle} & \alpha & r_1 & r_2 & r_3 & \text{Area of 3 circles} \\ \hline \text{Upper} & 0.227799337 & 0.012917131 & 0.387082869 & 0.2 & 0.19\pi \\ \text{Lower} & 0.569848253 & 0.079008574 & 0.230247857 & 0.38148714 & 0.204788868\pi \end{array}$

Since the smaller sum of areas of three circles is $0.19\pi$ , then $a=19$ and $b=100$ .

To find the area of the largest inscribed rectangle, let the measure of the angle extended at the incenter between the vertical and the point the corner of the rectangle touches the circle be $\theta$ . Then the area of the rectangle is given by:

$A=r_3^2(1-\cos \theta)\left(\cot \frac \alpha 2 - \sin \theta - \cot \alpha(1- \cos \theta)\right)$

Solve for the $\theta$ that gives the largest area then calculate $A_{\max}$ . We get:

$\begin{array} {r|c} \text{Semicircle} & A_{\max} \\ \hline \text{Upper} & 0.139148066 \\ \text{Lower} & 0.139418508 \end{array}$

Therefore $\beta \approx 0.139418508$ and $\dfrac {\lfloor 10^7\beta\rfloor - \lceil \pi \rceil}b = \boxed{13942}$ .