1821 Japanese Sangaku

Using the notation of the referred-to paper, we have $a=5$ , $b=3$ , $c=4$ , $h = \frac{c^2-b^2}{2a}$ , $k = \frac{bc}{a}$ , $m = \frac{b}{c}$ . If $\beta$ is the semi-major axis of the ellipse, then the coordinates $(x_0,y_0)$ of the centre of the ellipse are given by $x_0 \; = \; \frac{k(h-2\beta)}{h-\beta} \hspace{2cm} y_0 \; = \; h - \beta$ while the semi-major axis $\alpha$ of the ellipse, and its eccentricity $e$ are given by $\alpha \; = \; \frac{1}{m}\sqrt{(mx_0 - y_0)^2 - \beta^2} \hspace{2cm} e \; =\; \sqrt{\frac{(mx_0 - y_0)^2 - (m^2+1)\beta^2}{(mx_0 - y_0)^2 - \beta^2}}$ while the radius of the three circles is $r \; = \; e\beta$ while the coordinates of the centre of the bottom circle are $(x_1,y_1)$ where $x_1 \; = \; \frac{r(c-b)}{a} \hspace{2cm} y_1 \; = \; \frac{r(c+b)}{a}$ These conditions (for a suitable range of values of $\beta$ ) ensure that the ellipse is tangent to all three sides of the triangle, and that the two circles of radius $r$ are mutually tangent, and both tangent to the ellipse, and also that the bottom circle is tangent to the bottom two sides of the triangle. We now need to choose $\beta$ so that the bottom circle is tangent to the ellipse. The important point is that all the parameters of the problem have been expressed in terms of $\beta$ (and $a,b,c$ ). There is only one degree of freedom for any right-angled triangle.

My aim is to find a way of solving this problem that does not descend into the involved manipulation of three different quartics required in the paper. If a simpler approach can be found, a numerical solution will be relatively easy.

We want to determine the value of $\beta$ such that the shortest distance from the ellipse $\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1$ to the point $(x_0-x_1,y_0-y_1)$ is equal to $r$ . Applying the method of Lagrange multipliers, the shortest distance from the point to the ellipse is equal to $X \; = \; \sqrt{\frac{\lambda^2(x_0-x_1)^2}{(\alpha^2 + \lambda)^2} + \frac{\lambda^2(y_0-y_1)^2}{(\beta^2 + \lambda)^2}}$ where $\lambda$ is the positive value such that $\frac{\alpha^2(x_0-x_1)^2}{(\alpha^2 + \lambda)^2} + \frac{\beta^2(y_0-y_1)^2}{(\beta^2 + \lambda)^2} = 1$ (the maximum distance from the point to the ellipse is obtained by finding the negative solution of $\lambda$ ).

The equation $X = r$ can be solved numerically, and we obtain $\beta = 0.5631982331252968$ , $\alpha = 1.8144501500345467$ , $e = 0.950607314800319$ , $r = 0.5353803600915225$ , and hence $D = 1.070760720183045$ , so that $\lfloor 1000 D\rfloor = \boxed{1070}$ .

The diameter of the red circles works out to $1.070760720...$ , or radius of $0.535380360...$

The major and minor axes of the blue ellipse works out to $1.814450150...$ , and $0.563198233...$

Edit: After seeing Mark's posted numbers, I pushed my approximation to $D=1.070760720183043...$

Refer to the figure below, showing a right triangle of sides in proportions of $3, 4, 5$ , a blue ellipse centered at the origin $(0,0)$ , and red and green unit circles. The center of the green unit circle is $(x_0,y_0)$ . The major axis of the blue ellipse is $p$ while the minor axis is $pq$ . Then if the blue ellipse is tangent to the right triangle, the red unit circles tangent to it and each other, and the green unit circle is tangent to the right triangle, then the following must be true

$p=\dfrac { 1 }{ q\sqrt { 1-{ q }^{ 2 } } }$

$x_0=\dfrac { 1 }{ 5 } +\dfrac { 1 }{ 25q\sqrt { 1-{ q }^{ 2 } } } \left( \dfrac { 27+48q^{ 2 } }{ \sqrt { 9+16{ q }^{ 2 } } } -\dfrac { 64+36q^{ 2 } }{ \sqrt { 16+9{ q }^{ 2 } } } \right)$

$y_0=-\dfrac { 7 }{ 5 } +\dfrac { 1 }{ 25q\sqrt { 1-{ q }^{ 2 } } } \left( \dfrac { 48+27q^{ 2 } }{ \sqrt { 16+9{ q }^{ 2 } } } +\dfrac { 36+64q^{ 2 } }{ \sqrt { 9+16{ q }^{ 2 } } } \right)$

Then we solve the following simple equation for $x$

$y_0-\sqrt{1-(x-x_0)^2}=q\sqrt{p^2-x^2}$

...and that's when things explode. But $q$ can still be worked out, $q = 0.310396090581256029$

This is how bad it gets when it's done this way. Let

$a=\left(q^2-1\right)^2$

$b=\dfrac{4 }{25 q}\left(5 q^3+\sqrt{1-q^2} \left(4 \sqrt{9 q^2+16}-3 \sqrt{16 q^2+9}\right)-5 q\right)$

$c=\dfrac{1}{625 q^2 \left(q^2-1\right)} \left( 3650 q^6-1148 q^4-8 \left(12 \sqrt{9 q^2+16} \sqrt{16 q^2+9}+613\right) q^2+20 \sqrt{1-q^2} \left(33 \sqrt{9 q^2+16}+19 \sqrt{16 q^2+9}\right) q+96 \sqrt{9 q^2+16} \sqrt{16 q^2+9}+20 \sqrt{1-q^2} \left(17 \sqrt{9 q^2+16}+31 \sqrt{16 q^2+9}\right) q^3-2598 \right)$

$d=-\dfrac{4}{125 q^3 \left(q^2-1\right)^2} \left(25 q^7-76 q^5+35 \sqrt{1-q^2} \left(\sqrt{16 q^2+9}-2 \sqrt{9 q^2+16}\right) q^2+\left(2 \sqrt{9 q^2+16} \sqrt{16 q^2+9}+99\right) q+5 \sqrt{1-q^2} \left(3 \sqrt{16 q^2+9}-4 \sqrt{9 q^2+16}\right)+5 \sqrt{1-q^2} \left(6 \sqrt{9 q^2+16}-\sqrt{16 q^2+9}\right) q^4-2 \left(\sqrt{9 q^2+16} \sqrt{16 q^2+9}+24\right) q^3\right)$

$e= \dfrac{1}{625 q^4 \left(q^2-1\right)^2} \left( 625 q^8-1350 q^6+2 \left(52 \sqrt{9 q^2+16} \sqrt{16 q^2+9}+2549\right) q^2-500 \sqrt{1-q^2} \left(\sqrt{9 q^2+16}+\sqrt{16 q^2+9}\right) q+500 \sqrt{1-q^2} \left(\sqrt{9 q^2+16}+\sqrt{16 q^2+9}\right) q^5-\left(200 \sqrt{9 q^2+16} \sqrt{16 q^2+9}+1873\right) q^4-20 \sqrt{1-q^2} \left(33 \sqrt{9 q^2+16}+19 \sqrt{16 q^2+9}\right) q^3+625 \right)$

then solve for $q$ such that

$0= 2^{2/3} a^2 \sqrt[3]{\sqrt{\left(27 \left(a d^2+b^2 e\right)-9 c (8 a e+b d)+2 c^3\right)^2-4 \left(12 a e-3 b d+c^2\right)^3}-72 a c e+27 a d^2+27 b^2 e-9 b c d+2 c^3}+\frac{2 \sqrt[3]{2} a^2 \left(12 a e-3 b d+c^2\right)}{\sqrt[3]{\sqrt{\left(27 \left(a d^2+b^2 e\right)-9 c (8 a e+b d)+2 c^3\right)^2-4 \left(12 a e-3 b d+c^2\right)^3}-72 a c e+27 a d^2+27 b^2 e-9 b c d+2 c^3}}+\frac{3 \sqrt{3} a \left(8 a^2 d-4 a b c+b^3\right)}{\sqrt{2\ 2^{2/3} a \sqrt[3]{\sqrt{\left(27 \left(a d^2+b^2 e\right)-9 c (8 a e+b d)+2 c^3\right)^2-4 \left(12 a e-3 b d+c^2\right)^3}-72 a c e+27 a d^2+27 b^2 e-9 b c d+2 c^3}+\frac{4 \sqrt[3]{2} a \left(12 a e-3 b d+c^2\right)}{\sqrt[3]{\sqrt{\left(27 \left(a d^2+b^2 e\right)-9 c (8 a e+b d)+2 c^3\right)^2-4 \left(12 a e-3 b d+c^2\right)^3}-72 a c e+27 a d^2+27 b^2 e-9 b c d+2 c^3}}-8 a c+3 b^2}}+8 a^2 c-3 a b^2$