Packing Sangaku

The ellipse and the pink circles are just red herrings. Leave them out and the thing is very simple. That can be done once you realize that fixing the square fixes everything else. Leave the left dashed circle and the left violet circle away as well.

Let the square have vertices $(1,1),(-1,1),(-1,-1),(1,-1)$ .

Then the dashed circle must be centred at $(x,0)$ , such that its distance to (-1,0) and (1,1) is equal. We find $x=\frac{1}{4}$ and the radius of the dotted circle equals $R=\frac{5}{4}$ .

The rightmost point of both circles is at $x+R=\frac{3}{2}$ .

The radius of the small circle is $r=\frac{1}{2} (\frac{3}{2}-1)=\frac{1}{4}$ .

The ratio of the radii equals $\frac{1/4}{5/4}=0.2$ and the answer is $\lfloor0.2×10^5\rfloor=20000$ .

Let the center of symmetry of the figure be the origin $O(0,0)$ of the $xy$ -plane, the radius of the six congruent circles be $1$ , and the ellipse be $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$ . Then the four sides of the square are $x = \pm a$ and $y = \pm a$ .

As the figure is symmetrical about the origin, we can just consider the first quadrant. Let the tangent point of the two unit circles and the ellipse in the first quadrant be $P(x_0, y_0)$ . Then the lower unit circle centered along the $x$ -axis is $(x-1)^2 + y^2 = 1$ . Let the center of the upper circle be $Q(x_1, y_1)$ , then $x_1 = 1 + 2(x_0-1) = 2x_0 - 1$ . But $x_1 = a - 1$ , Then we have $2x_0 - 1 = a - 1 \implies x_0 = \dfrac a2$ . From

$\begin{cases} \dfrac {x_0^2}{a^2} + \dfrac {y_0^2}{b^2} = 1 & \implies y_0 = \dfrac {\sqrt 3}2 b \\ (x_0-1)^2 + y_0^2 = 1 & \implies \dfrac {a^2}4 - a + \dfrac {3b^2}4 = 0 & ...(1) \end{cases}$

We note that at $P(x_0, y_0)$ the gradients of the unit circle and the ellipse are equal.

$\begin{cases} \dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1 & \implies \dfrac {dy}{dx} = - \dfrac {b^2x}{a_2y} \\ (x-1)^2 + y^2 = 1 & \implies \dfrac {dy}{dx} = - \dfrac {x-1}y \end{cases} \implies \frac {b^2x_0}{a^2y_0} = \frac {x_0-1}{y_0} \implies \blue{b^2 = a^2 - 2a}$

And from $(1)$ :

$\begin{aligned} \frac {a^2}4 - a + \frac {3\blue{b^2}}4 & = 0 & \small \blue{\text{Note that }b^2 = a^2 - 2a} \\ \frac {a^2}4 - a + \frac {3\blue{(a^2 - 2a)}}4 & = 0 \\ a \left(a - \frac 52\right) & = 0 & \small \blue{\text{Since }a > 0} \\ \implies a & = \frac 52 \end{aligned}$

Then $b = \dfrac {\sqrt 5}2$ , $x_0 = \dfrac 54$ , $y_0 = \dfrac {\sqrt{15}}4$ , $x_1 = \dfrac 32$ , and $y_1 = 2y_0 = \dfrac {\sqrt{15}}2$ .

Let the center of the right dash circle be $S(x_2, 0)$ and its radius $r$ . Then $x_2 + a = x_2 + \dfrac 52 = r$ and

$\begin{aligned} \sqrt{(x_1-x_2)^2+y_1^2} + 1 & = r \\ \sqrt{\left(\frac 32 - r + \frac 52 \right)^2+\frac {15}4} & = r - 1 & \small \blue{\text{Squaring both sides}} \\ r^2 - 8r + 16 + \frac {15}4 & = r^2 - 2r + 1 \\ \implies r & = \frac {25}8 \end{aligned}$

Let the radius of the violet circle be $r_1$ . Then we have:

$\begin{aligned} x_2 + r - a & = 2r_1 \\ r - \frac 52 + r - \frac 52 & = 2r_1 \\ \frac {25}4 - 5 & = 2r_1 \\ \implies r_1 & = \frac {5}8 \end{aligned}$

Therefore $R = \dfrac {r_1}r = \dfrac 5{25} = \dfrac 15 \implies \lfloor 10^5R \rfloor = \boxed{20000}$

Let the radius of the pink circles be $1$ , and side length of the square by $s$ , and the let the angle that the line segment joining the centers of the right central pink circle and the upper right pink circle, makes with the horizontal direction be $\theta$ . Note that the ellipse has a semi-major axis of $\dfrac{s}{2}$ , and let's call its semi-minor axis $b$ . Finally, let the eccentric angle of the point of tangency between the two aforementioned pink circles be $t$ , then we can write the following relations between these four variables, which are $s, \theta, t, b$ :

Firstly, taking the horizontal distances along the path indicated in the figure above, we get,

$2 + 2 \cos \theta = s / 2 \hspace{24pt} (1)$

Next, the tangency point satisfies:

$T = (1 + \cos \theta, \sin \theta) = ( \frac{1}{2} s \cos t , b \sin t )$

so that,

$1 + \cos \theta = \frac{1}{2} \cos t \hspace{24pt} (2)$

and

$\sin \theta = b \sin t \hspace{24pt} (3)$

and finally, since the parametric equation of the ellipse is $(\frac{1}{2} \cos t , b \sin t)$ , then its tangent vector is along $(-\frac{1}{2} \sin t , b \cos t)$

and this tangent vector is perpendicular to the line joining the two centers of the two pink circles whose unit vector is $(\cos \theta, \sin \theta)$ , therefore,

$- \frac{1}{2} s \cos \theta \sin t + b \sin \theta \cos t = 0 \hspace{24pt} (4)$

Applying the Newton-Raphson method, we define our vector function to be

$\mathbf{f} = \begin{bmatrix} f_1 \\ f_2 \\ f_3 \\ f_4 \end{bmatrix}$

where

$f_1 = 2 + 2 \cos \theta - s / 2 \hspace{24pt} (5)$

$f_2 = 1 + \cos \theta - \frac{1}{2} \cos t \hspace{24pt} (6)$

$f_3 = \sin \theta - b \sin t \hspace{24pt} (7)$

$f_4 = - \frac{1}{2} s \cos \theta \sin t + b \sin \theta \cos t \hspace{24pt} (8)$

Let our variable vector be

$\mathbf{x} = \begin{bmatrix} s \\ \theta \\ t \\ b \end{bmatrix} \hspace{24pt} (9)$

Then according to the Newton-Raphson iteration, we start with an initial guess $\mathbf{x_0}$ , and update our guess according to the following

recursive formula:

$\mathbf{x_{k+1} } = \mathbf{x_k} - \mathbf{J}^{-1} \mathbf{f_k} \hspace{24pt} (10)$

where the matrix $\mathbf{J }$ is the jacobian defined as follows:

$\mathbf{J} = [ J_{ij} ] = [ \dfrac{\partial f_i }{\partial x_j} ]$

Implementing this method, led to the solution in just $4$ iterations. The solution is

$\mathbf{x } = \begin{bmatrix} 5 \\ 1.318116071 \\ 1.047197551 \\ 1.118033989 \end{bmatrix}$

Now, we need to find $R$ the radius of the big circles, and $r$ the radius of the small circles. Passing a line through the center of the upper right pink circle, making an angle $\phi$ with the horizontal direction, and noting that the y-coordinate of the center of this circle is $y = 2 \sin \theta$ , then we want to find $\phi$ such that

$R = 1 + y \csc \phi = s - 1 - y \cot \phi$

Multiplying through by $\sin \phi$ ,

$\sin \phi + y = \sin \phi (s - 1) - y \cos \phi$

whose solution is $\phi = 1.146406617$ , from which it follows that $R = 3.125 =\dfrac{25}{8}$

Finally, the radius of the small circle is $r = \dfrac{1}{2} (2 R - s ) = 0.625 = \dfrac{5}{8}$

Hence, $\dfrac{r}{R} = \dfrac{5}{25} = \dfrac{1}{5} = 0.2$

And this makes the answer $\lfloor 10^5 (0.2) \rfloor = \boxed{20000}$