Sangaku problem involving eight circles

Question

I made the following sangaku problem.

$\dfrac{\text{Area of the orange circle}}{\text{Area of a blue circle}}=\space ?$

Description of diagram. In this question, circles of the same color are congruent. A chain of five tangent blue circles lies on top of a larger orange circle, such that each circle in the chain is tangent to the orange circle. Another blue circle sits on top of the chain's middle circle. A black circle encloses all of the orange and blue circles, such that the black circle is tangent to the orange circle, the left and right circles in the chain, and the circle on top of the chain.

I used a computer to work out that the answer is:

5

but I would like to know how to do this without a computer.

My attempt

Assume the blue circles have radius $1$ and the orange circle radius has $r$, where $r$ is a constant to be found. Superimpose a cartesian coordinate system with the origin at the orange circle's centre.

The black circle has equation

$$x^2+(y-2)^2=(r+2)^2\tag1$$

The rightmost circle in the chain has equation

$$\left(x-(r+1)\sin\left(4\arcsin\frac{1}{r+1}\right)\right)^2+\left(y-(r+1)\cos\left(4\arcsin\frac{1}{r+1}\right)\right)^2=1\tag2$$

In $(2)$ we can use the compound angle formulas $\cos(4\arcsin\alpha)=8\alpha^4-8\alpha^2+1$ and $\sin(4\arcsin\alpha)=\sqrt{1-\alpha^2}(4\alpha-8\alpha^3)$.

We want to find $r$ such that $(1)$ and $(2)$ together has exactly one solution. $(1)-(2)$ yields an equation that can be rearranged to get $y=f(x)$, which can be put into $(1)$ to get a quadratic in $x$. Then set the determinant equal to $0$ and solve for $r$. Then the answer to the OP is $r^2$. But this approach is very difficult without a computer.

Fun facts

Two more blue circles fit perfectly in the spaces above the chain's second-from-end circles (I asked about this in a Geometry question about a six-pack of beer).

And there are many sets of collinear points, as shown.

Edit

As shown in the Fun Fact diagram, an extended horizontal diameter of the orange circle is tangent to the bottom two blue circles. Here I will explain why this is true, in order to support @Edward Porcella's answer.

Start with a closed chain of ten tangent congruent blue circles, all internally tangent to a black circle, with one of the blue circles touching the top point of the black circle. Draw a horizontal diameter of the black circle (which must be tangent to four blue circles). Draw an orange circle inside the black circle, touching all the blue circles.

Remove the bottom five blue circles. Move the top five blue circles, and the orange circle, down, by a distance equal to the diameter of a blue circle. Add a blue circle at the top.

This is the diagram in the OP, showing an extended horizontal diameter of the orange circle, which is tangent to the bottom two blue circles.

Context

I have been asking questions about circles here on MSE, and also on Mathematics Educators and Puzzling.

Answer 1

Let the radii of the large (red) circle and small (blue) circles be $r$ and $s$, so that the radius of the enclosing circle is $r+2s$.

Let the center of the enclosing circle, the large circle, and the right-most small circle be $O$, $R$, $S$, with $S'$ and $S''$ the centers of other small circles as shown. Extend $OS$ to the point of tangency $T$ on the enclosing circle, and note that $$|OR|=(r+2s)-r=2s \qquad |OS|=|OT|-|ST|=r+s=|RS|$$ Thus, $\triangle SOR$ is isosceles with leg $r+s$ and base $2s$. But so are $\triangle RSS'$ and $\triangle RS'S''$. Since $\angle SRS'=\frac12\angle SRO$, we deduce that these are $36^\circ$-$72^\circ$-$72^\circ$ triangles, and can leverage the fact that $2\cos36^\circ=\phi=\frac12(\sqrt{5}+1)$ to show that $r=s\sqrt{5}$. But we can show this a bit more directly.

Writing $P$ for the point where $OT$ meets $RS'$, we observe that $|SP|=|SS'|=2s$, so that $|OP|=|OS|-|SP|=r-s$. Also, $\triangle ROP$ is similar to $\triangle SOR$ (and the others), and we can write $$\frac{|OP|}{|OR|}=\frac{|OR|}{|RS|} \quad\to\quad \frac{r-s}{2s}=\frac{2s}{r+s}\quad\to\quad r^2=5s^2 \quad\to\quad r = \sqrt{5}s$$

I suspect that an even-more-direct demonstration is possible, perhaps one that embeds a regular pentagon or other golden figure. I'll leave the search for such to OP and the reader.

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Addendum. Here's another look at the configuration, with larger circle $\bigcirc R$ (with radius $r$), top-most and right-most smaller circles $\bigcirc S$ and $\bigcirc S'$ (with radius $s$), and enclosing circle $\bigcirc O$ (with radius $r+2s$).

One can show that $\triangle S'OR$ and $\triangle SRS'$ are similar ($36^\circ$-$72^\circ$-$72^\circ$) isosceles triangles. Defining $p:=r+s=|OS|=|OS'|=|RS'|$ and $q:=2s=|OR|$, we have $$\frac{|RS'|}{|OR|}=\frac{|RS|}{|RS'|} \quad\to\quad \frac{p}{q}=\frac{q}{p+q}$$ which reveals $p/q$ to be the Golden ratio. A bit of algebraic manipulation then gets us to $r=\sqrt{5}s$, but that's mostly an afterthought here.

Answer 2

Done by hand

Using the steps you gave and a bunch of simplifications, the discriminant of the quadratic equation in $x$ is simply $$\Delta=-\frac { 32 r (r+3) \left(r^2-5\right)\left(r^2+2 r-1\right)^2} {\left(r^4+2 r^3-8 r^2-18 r-1\right)^2 }$$ So, the only possible conditates are $$r_1=\sqrt 2 -1 \qquad \text{and} \qquad r_2=\sqrt 5$$ Since $r_1$ is ridiculous, just remains

$$\huge \color{blue}{r=\sqrt 5}$$

Answer 3

Following the suggestion of @Blue, this approach uses the regular decagon.

Assuming, as indicated in OP's second figure, that the left and right small circles are tangent to the extended diameter of the larger circle at $E$ and $F$, then $CB$ is the side of a regular decagon, $\triangle ABC$ is $36^o-72^o-72^o$, and$$\frac{AB}{BC}=\phi=\frac{1+\sqrt 5}{2}$$

Letting radius $BD=1$, then$$\frac{AB}{BD}=\frac{1+\sqrt 5}{1}$$and$$\frac{AD}{BD}=\frac{1+\sqrt 5-1}{1}=\frac{\sqrt 5}{1}$$Thus the larger circle has five times the area of the smaller circle.