Sangaku problem with seven circles in an isosceles triangle

Self-answering. This solution is not easy without a computer.

Assume the radius of the orange circles is $1$. Then the radius of the blue circles is $2$.

Let $g=$ radius of the green circles.

Pythagorus gives the two lengths labelled in red below:

Then Pythagorus with the dashed line right triangle gives:

$$\left(2\sqrt2-g\right)^2+\left(1+\sqrt{(g+1)^2-g^2}\right)^2=(g+2)^2$$

$$\implies g=\frac{4+6\sqrt2+4\sqrt{1+\sqrt2}}{9+4\sqrt2}$$

Superimpose cartesian axes with the origin at the bottom of the lower orange circle.

The equations of the right blue circle and right green circle are, respectively:

$$\left(x+2\sqrt2\right)^2+\left(y-2\right)^2=2^2$$

$$\left(x+g\right)^2+\left(y-3-\sqrt{2g+1}\right)^2=g^2$$

Let the equation of the right side of the triangle be $y=mx+k$. The right side of the triangle is tangent to both the right blue and right green circles. By setting discriminants equal to $0$, we get two equations with $m$ and $k$. We solve for $m$ and $k$. There are three pairs of solutions (corresponding to the three lines that are tangent to both circles); we choose the negative value of $m$ with smaller magnitude. We get that the equation of the right side of the triangle is:

$$y=\left(-\frac27\sqrt{2+10\sqrt2}\right)x+\frac87\left(3+\sqrt2+\sqrt{1+5\sqrt2}\right)$$

Now let $C$ be a circle of radius $1$ that is above and tangent to the two green circles. (We will show that $C$ is tangent to the sides of the triangle, which means that $C$ is the red circle.) The equation of $C$ is:

$$x^2+\left(y-3-2\sqrt{2g+1}\right)^2=1^2$$

It can be shown that $C$ is tangent to the right side of the triangle, by calculating the discriminant and finding that it equals $0$.

Thus, $C$ is the red circle, so the red circle has radius $1$, so the red circle is congruent with the orange circles.

Here is a Desmos graph of the triangle and circles.