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In the diagram, circles (or disks, if you like) of the same color have the same radius. (For an explicit description of the diagram, see below.)

enter image description here

Let $g=$ radius of the green circles, $r=$ radius of the red circles. Find the value of $g/r$.

I have used a computer to find that the answer is exactly $5$. I'm looking for a solution that does not require a computer (like my previous self-made Sangaku-style problem).

About my computer-assisted solution: I used desmos to manually draw the circles, zooming in to get close approximations of their centres and radii. Then I discovered that the value in question was extremely close to $5$. Then I put my approximations into Wolfram, and it gave me suggested closed forms. Then I put these closed forms back into the equations of the circles, and they seem to be "perfect fits" (when I zoom in, there are no gaps or overlaps between neighboring circles or chords), with the value in question being exactly $5$.

Fun facts: Another blue circle could fit perfectly in the centre. The ratio of black to green (radii) is $\phi+1$, and the ratio of green to blue is $\phi$, where $\phi=\frac{1+\sqrt5}{2}$ is the golden ratio.

Explicit description of the diagram: In a black circle, two chords of equal length meet at a point on the circle. In each of the two segments thus formed, a largest possible green circle is inscribed, followed by a pair of largest possible blue circles (one on each side of the green circle), followed by a pair of largest possible red circles (so each blue circle is between the green circle and a red circle). A third green circle touches the black circle and the two chords. Circles of the same color have the same radius.

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  • $\begingroup$ two chords have equal lengths right ? Also, arbitrary choices of these two chords will not guaranteed that the three green circles will be of same size, so this is also an extra condition right? $\endgroup$
    – dezdichado
    Commented May 28, 2023 at 22:42
  • $\begingroup$ Wouldn't you return back the first picture? I liked it more :) Also, draw please a horizontal line tangent to the upper green and two blue circles. It can help. $\endgroup$
    – user376343
    Commented May 28, 2023 at 22:44
  • 3
    $\begingroup$ Please provide the solution you have. It may serve as a jumping-off point for the type of solution you seek, without everyone having to spend time duplicating your effort. (It may also clarify such things as the congruence of the chords.) Moreover, knowing the final value can help people create accurate images for study. (Sending people to a third-party site to hunt through your equations for clues isn't the most considerate way to get that information across. A good rule of thumb I've seen: Always treat your reader's time as more important than your own.) $\endgroup$
    – Blue
    Commented May 28, 2023 at 22:46
  • 1
    $\begingroup$ @Blue Please see my edit. $\endgroup$
    – Dan
    Commented May 28, 2023 at 22:59
  • 1
    $\begingroup$ If we add in the picture to more tiny disks, a brown one, and a pink one, so that we have a chain of touching circles: green, blue, red, brow, pink... can you guess which is the radius proportion green versus pink? $\endgroup$
    – dan_fulea
    Commented May 29, 2023 at 16:59

2 Answers 2

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A symmetry argument shows that the top two blue circles are tangent not just to the green circles on the same side of the chords, but also to the top green circle. This is because a reflection about a diameter that passes through, say, the center of the top right blue circle, must map that blue circle to itself, and the rightmost green circle to the topmost green circle, and vice versa.

Consequently, there is a horizontal tangent line that can be drawn so that the top five circles forms a third sector with all the corresponding tangencies preserved. This also implies that a circle centered at the origin that is externally tangent to all three green circles will have the same radius as the blue circle, since a reflection about a line joining the centers of the top green circle with the rightmost green circle will map the center of the top right blue circle to the origin.

Next, it is easy to determine the radius of the green cirles. Assuming without loss of generality that the large circle has unit radius. Then the radius of the green circle must satisfy

$$\frac{g}{2-g} = 1 - 2g, \implies g = \frac{3 - \sqrt{5}}{2}. \tag{1}$$

This in turn means that the blue circle has radius $$b = 1 - 2g = \sqrt{5} - 2. \tag{2}$$

The radius of the red circle $r$ is the most difficult to determine. One way is to place the figure on the coordinate plane and compute the coordinate of the center of the top right blue circle, which is $$(x_b, y_b) = \left(\sqrt{10 \sqrt{5} - 22}, 2(\sqrt{5} - 2) \right), \tag{3}$$ the proof of which I leave as an exercise (hint: compute the coordinate of the intersection of the horizontal chord with the chord on the right). Then $r$ must satisfy

$$\begin{align} x_r^2 &= (1-r)^2 - (b+r)^2, \\ (x_r - x_b)^2 + (b - r)^2 &= (b + r)^2, \end{align} \tag{3}$$ where $x_r$ is a nuisance variable representing the $x$-coordinate of the top right red circle. This gives us the unique nontrivial solution $$x_r = 4 \sqrt{\frac{\sqrt{5}-2}{5}}, \quad r = \frac{3 - \sqrt{5}}{10}. \tag{4}$$ hence $g/r = 5$.

enter image description here

For fun, I also made a little animation.

enter image description here

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  • $\begingroup$ Interesting approach. Are you sure $x_r=4\sqrt{\frac{\sqrt{5}-2}{5}}$ is correct? That's approximately $0.869$. Looking at the diagram (and assuming the black circle is a unit circle), that seems far too large. Anyway, without a computer, I would have great difficulty solving your equations labelled (3), which would entail solving a quartic equation. $\endgroup$
    – Dan
    Commented May 29, 2023 at 3:39
  • $\begingroup$ @Dan $x_r$ is the $x$-coordinate of the red circle, and as you can clearly see from the diagram, it is located between $0.8$ and $0.9$. It is not the radius, which, as you have denoted in your question, is $r$. As for the solution of $(3)$, it is tractable by hand, as the solution itself demonstrates. You should try to solve it yourself, as an exercise. $\endgroup$
    – heropup
    Commented May 29, 2023 at 4:44
  • $\begingroup$ @Dan You need to look at the diagram I have included in my answer, because what I have done is exploited the symmetry argument and considered the red circle that is not the one originally in your question, but instead the reflected one in the equivalent circular sector cut off by the horizontal chord. I did this to make the calculation easier. $\endgroup$
    – heropup
    Commented May 29, 2023 at 5:13
  • $\begingroup$ @Dan your top-right red circle and heropup's red circle are congruent. This is why I suggested to draw the horizontal line. $\endgroup$
    – user376343
    Commented May 29, 2023 at 5:26
  • $\begingroup$ @heropup Oh OK. My computer here at work doesn't show stack exchange images due to local internet restriction. Sorry for my misunderstanding. $\endgroup$
    – Dan
    Commented May 29, 2023 at 5:31
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Assume that the radius of the bigger circle is $R=1$, we compute first the radius $G$ of the green circle from the picture:

Appolonius like configuration of circles

There are two similar triangles with the same marked violet angle, the similarity (or the same sine of this angle) gives: $$ \frac{R-2G}R=\frac G{2R-G}\ ,\qquad\text{ so }\qquad \bbox[lightgreen]{\color{green}{\ G=\frac 12(3-\sqrt 5)\ }}\ . $$ Now we have a special situation to be arranged inside the smaller part of the disk with radius $R=1$ delimited by the pink chord in the picture. We need to fit inside one of the remained empty white space the blue, and the red circle. We use inversive geometry, centered in the point $\Omega$ in the picture below.


In the second part we compute the proportions $G:b:r$ between the sizes of the green, the blue, and the red circle. The picture had a lot to do with the details, but the inversive argument is simple:

Appolonius like configuration of circles

Let $\color{magenta}{\Omega}$ be the point on the pink chord and the violet arc delimiting the region where the green, blue, red circles live in. The other end of the chord is $\color{magenta}{\Upsilon}$. Let $\color{magenta}{\Psi}$ be the other end of the diameter through $\color{magenta}{\Omega}$. Let $A,B,C$ be the tangency points of the green, blue, red circles shown in the picture with the line $\color{magenta}{\Omega\Upsilon}$. The picture shows, but we do not need, the points $S,T,U$ of tangency of the green, blue, red circles with the big initial violet circle of radius one.

We are performing an inversion, denoted by star, $$X\to X^*\ $$ with center in $\color{magenta}{\Omega}$ and with power $\color{magenta}{\Omega\Upsilon^2} =4\;\Omega A^2=4(1^2-(1-2G)^2)=16(\sqrt 5-2)$.

  • So $\color{magenta}{\Upsilon}$ is invariated, $\color{magenta}{\Upsilon^*=\Upsilon}$.
  • The half line $\Omega ABC\Upsilon$ is mapped into itself, and the points $A^*,B^*,C^*$ are beyond $\Upsilon$. Since $A$ is the mid point of $\Omega\Upsilon$ the transformed $A^*$ is the reflection of $\Omega$ in $\Upsilon$. The points $\Omega,A,B,C,\Upsilon$ are transformed in $\infty,A^*,B^*,C^*,\Upsilon$ so that the cross ratios of any four points tuple are preserved.
  • The violet big circle of radius $R$ is transformed in the line through $\Upsilon=\Upsilon^*$ which is perpendicular on the diameter $\Omega\Psi$. (So the arc $\overset\frown{\Omega\Upsilon}$ through $\Omega,S,T,U,\Upsilon$ is mapped in a half-line $\infty S^* T^*U^*\Upsilon $.
  • Which is the angle $\bbox[orchid]{2y}$ between the two lines $\Omega ABC\Upsilon C^*B^*A^*$ and $\Upsilon U^*T^*S^*$? We have $$ \sin \bbox[orchid]{2y}=\sin \widehat{\Upsilon\Psi\Omega}=\frac{\Omega\Upsilon}{\Omega\Psi} =\frac{4\sqrt{\sqrt 5-2}}2=2\sqrt{\sqrt 5-2}\ . $$Here, we can easily compute a trigonometric function of $2y$, however we will use in the sequel expresions in $y$, the angle between the orange line of the centers and either of $\Upsilon A^*$ or $\Upsilon S^*$. Let us compute $\bbox[lightblue]{s:=\sin y}$ by solving $2s\sqrt{1-s^2}=2\sin y\cos y=\sin 2y=2\sqrt{\sqrt 5-2}$. After dividing by $2$ and squaring, $s^2(1-s^2)=\sqrt 5-2$. So $0=s^4-s^2+\sqrt 5-2$, i.e. $$ s^2=\frac 12(1\pm \sqrt{1-4\sqrt 5 + 8})=\frac 12(1\pm(\sqrt 5-2))\ , $$ and we pick the value with minus, to get an angle less $45^\circ$, so $s^2=G=\frac 12(3-\sqrt 5)=\frac 14(6-2\sqrt 5)=\left(\frac 12(\sqrt 5-1)^2\right)$. This gives: $$ \bbox[lightblue]{ \ s=\sin y=\frac 12(\sqrt 5-1)\ }\ . $$ There must be a geometric argument, which i am missing now, it would simplify considerably the whole exposition.
  • The given green, blue, red circles have a complicated placement. However, their inversive transforms are in a similarity, being tangent to the two lines through $\Psi$. Let us denote by $G_i,b_i,r_i$, respectively $G_p,b_p,r_p$ the radius lengths of the transformed circles, respectively the projection of these lengths from the common orange line of their centers on either of the lines $\Upsilon C^*B^*A^*$ or $\Upsilon U^*T^*S^*$. There are also a similarity relations for the many pairs of circles, when seen from $\Omega$. So we have $$ \begin{aligned} G &=\frac{\Omega A}{\Omega A^*}G_i\ ,& b &=\frac{\Omega B}{\Omega B^*}b_i\ ,& r &=\frac{\Omega C}{\Omega C^*}r_i\ , \end{aligned} $$ and $\displaystyle\frac{G_p}{G_i} = \frac{b_p}{b_i} = \frac{r_p}{r_i} = \cos y$, and $\displaystyle\frac{G_i}{\Upsilon A^*} = \frac{b_i}{\Upsilon B^*} = \frac{r_i}{\Upsilon C^*} = \tan y$. Considering $\Upsilon A^*$, $G_p$ as known, we determine the "next level" from the system $$ \left\{ \begin{aligned} \Upsilon B^*+b_p&=\Upsilon A^*-G_p \ ,\\ \frac {b_p}{G_p} &=\frac{b_i}{G_i}=\frac{\Upsilon B^*}{\Upsilon A^*}\ , \end{aligned} \right. $$ so we insert $b_p=\frac{\Upsilon B^*}{\Upsilon A^*}G_p$ from the second equation into the first one, getting: $$ \Upsilon B^* =\Upsilon A^*\cdot\frac{\Upsilon A^*-G_p}{\Upsilon A^*+G_p} =\Upsilon A^*\cdot\underbrace{\bbox[lightblue]{\frac{1-\sin y}{1+\sin y}}}_{\bbox[lightblue]{\text{Notation: }k }} =\Upsilon A^*\cdot \underbrace{\bbox[lightblue]{\ (\sqrt 5-2)\ }}_{\bbox[lightblue]{\ =k\ }}\ , $$ and for $b_p$, $b_i$ we have the same factor of proportionality: $$ b_p = G_p\cdot k\ ,\ b_i=G_i\cdot k\ . $$ We can now conclude: $$ \begin{aligned} \frac rG &= \frac {\displaystyle \frac{\Omega C}{\Omega C^*}\cdot r_i} {\displaystyle \frac{\Omega A}{\Omega A^*}\cdot G_i} = \frac {\Omega C\cdot\Omega C^*} {\Omega C^*\cdot\Omega C^*} \cdot\underbrace{ \frac{\Omega A^*}{\Omega A}}_{=4} \cdot\underbrace{ \frac{r_i}{G_i}}_{=k^2} = \frac{\Omega\Upsilon^2}{{\Omega C^*}^2}\cdot 4k^2 = \left(\frac{\Omega\Upsilon}{\Omega C^*}\right)^2\cdot 4k^2 \\ &= \left(\frac{\Upsilon A^*}{\Upsilon A^* +\Upsilon C^*}\right)^2\cdot 4k^2 = \left(\frac{1}{ \displaystyle 1+ \frac{\Upsilon C^*}{\Upsilon A^*}}\right)^2\cdot 4k^2 = \left(\frac 1{1+k^2}\right)^2\cdot 4k^2 \\ &= \left(\frac {2k}{1+k^2}\right)^2 %= %\left(\frac {2(\sqrt 5-2)}{1+(9-2\sqrt 5)}\right)^2 = \left(\frac {2(\sqrt 5-2)}{10-2\sqrt 5}\right)^2 = \left(\frac {2(\sqrt 5-2)}{2\sqrt 5(\sqrt 5-2)}\right)^2 \\ &= \left(\frac 1{\sqrt 5}\right)^2 =\bbox[yellow]{\ \frac 15\ }\ . \end{aligned} $$ $\square$

Optical check:

I insisted against my will to do from the start an exact picture. Now it partially pays back the cost of the time. Let us see if the result matches the exact picture. We magnify the last picture at the right place...

An Appolonius configuration of points final check

And yes, the parallel to $\Omega\Upsilon$ drawn through the center of the red circle was used to move the red radius $r$ onto a fraction of the green radius $G$. Then taking successive reflections there is a match after the fifth step.

Mission accomplished.




LATER EDIT: (Just the story to the solution, and a bonus generalization.)

As mentioned in the comments, this solution grew out from the first picture with the big circle, and the position of the three green disks. Having this we can remove two of the disks and keep only one of them, the one that will get an appended tangent blue disk, which will be also accompanied on the other side by the tangent red disk. My initial intention was then to construct these two further disks geometrically, and check the proportionality red vs. green. (Somehow i doubted there will be a rational number in there as result, instead of a number in the field $\Bbb Q(\sqrt 5)$, i saw no (Galois) reason for such a coincidence, and i wanted to see if there is a coincidence, and if yes, then why. This was my question inside the question.) Now the geometric construction is easy and natural, use inversion to go

  • from the $D$-shape of the region where all five, well only three needed disks (green, blue, red) live in,
  • to an angular shape.

The transformed circles will be now similar to each other, and we need to get once for all times the factor of similarity, which is determined by the angular shape. Furthermore, each transformed disk is similar to the initial, original disk. Using this geometric argument, i could quickly construct in geogebra one by one

  • the transformed green disk, tangent to the legs of the angle in $\Upsilon$,
  • the "orange" angle bisector from $\Upsilon$ which will carry now the centers of the next disks,
  • the centers of the next transformed disks, blue and red, and their diameters on the orange ray, one by one in geometric progression,
  • and finally the initial disks.

And indeed, the red radius was i proportion $1:5$ to the green radius. I decided to make the computation, which is dictated by the construction. Going cartesian is easier, when the red disk is the final target, but the idea cannot be generalized. What if we are adding some two more circles in brown, and then in pink to the green-blue-red chain? So i did the calculus, it is always simple to do it on paper, but then name notations and display formulas...

For short, the main merit of this approach is the possibility to immediately generalize. So assume we use general notations in the following spirit: $$ \begin{array}{|r|l|r|l|c|r|l||} \hline \scriptstyle\text{Disk} & & \scriptstyle\text{Its radius} & & \scriptstyle\text{Disk}^* & \scriptstyle\text{Its radius} & & \scriptstyle\text{Point on } \Omega\Upsilon & \\\hline % \text{green} & = \Gamma(0) & G & = r(0) & \Gamma(0)^* & G_i & = r_i(0) & A & = A(0) \\\hline % \text{blue} & = \Gamma(1) & b & = r(1) & \Gamma(0)^* & b_i & = r_i(1) & B & = A(1) \\\hline % \text{red} & = \Gamma(2) & r & = r(2) & \Gamma(2)^* & r_i & = r_i(2) & C & = A(2) \\\hline % \text{brown} & = \Gamma(3) & br & = r(3) & \Gamma(3)^* & br_i & = r_i(3) & D & = A(3) \\\hline % \text{pink} & = \Gamma(4) & p & = r(4) & \Gamma(4)^* & p_i & = r_i(4) & E & = A(4) \\\hline % \vdots & \ \vdots & \vdots & \ \vdots & \vdots & \vdots & \ \vdots & \vdots & \ \vdots \\\hline % \text{?} & = \Gamma(N) & ? & = r(N) & \Gamma(4)^* & ?_i & = r_i(N) & ? & = A(N) \\\hline % \vdots & \ \vdots & \vdots & \ \vdots & \vdots & \vdots & \ \vdots & \vdots & \ \vdots \end{array} $$ Then the natural question is what is the proportion : $$ \bbox[lightyellow]{\ \frac {r(N)}{r(0)} \ } $$ for a general $N\ge 0$? Well, the same computation delivers, using the more general notations: $$ \begin{aligned} \bbox[lightyellow]{\ \frac {r(N)}{r(0)} \ } &= \frac {\displaystyle \frac{\Omega A(N)}{\Omega A(N)^*}\cdot r_i(N)} {\displaystyle \frac{\Omega A(0)}{\Omega A(0)^*}\cdot r_i(0)} = \frac {\Omega A(N)\cdot\Omega A(N)^*} {\Omega A(N)^*\cdot\Omega A(N)^*} \cdot\underbrace{ \frac{\Omega A(0)^*}{\Omega A(0)}}_{=4} \cdot\underbrace{ \frac{r_i(N)}{r_i(0)}}_{=k^N} \\ &= \frac{\Omega\Upsilon^2}{{\Omega A(N)^*}^2}\cdot 4k^N = \left(\frac{\Omega\Upsilon}{\Omega A(N)^*}\right)^2\cdot 4k^N \\ &= \left(\frac{\Upsilon A(0)^*}{\Upsilon A(0)^* +\Upsilon A(N)^*}\right)^2\cdot 4k^N = \left(\frac{1}{ \displaystyle 1+ \frac{\Upsilon A(N)^*}{\Upsilon A(0)^*}}\right)^2\cdot 4k^N \\ &= \left(\frac 1{1+k^N}\right)^2\cdot 4k^N =\bbox[lightyellow]{\ \frac{4k^N}{(1+k^N)^2}\ }\ . \end{aligned} $$ Recall the value: $$ \bbox[lightblue]{\ k = \sqrt 5-2\ }\ . $$ Now for even values of $N$, the above expression is Galois-invariant w.r.t. the conjugation (denoted by a bar) $a+b\sqrt 5\to \overline{(a+b\sqrt 5)}:=a-b\sqrt 5$, $a,b\in\Bbb Q$, since $k\cdot \bar k= (-k)\overline{(-k)}=(2-\sqrt 5)(2+\sqrt 5)=4-5=-1$, i.e. $\bar k=-1/k$, so $\bar k^2=1/k^2$, and thus for an even $N$, since conjugation is a field automorphism (i.e. compatible with all field operations), $$ \overline{\ \left(\frac{4k^N}{(1+k^N)^2}\right) \ } = \left(\frac{4(1/k)^N}{(1+(1/k)^N)^2}\right) = \frac{4k^N}{(1+k^N)^2}\ . $$ So the fraction is rational! (This was may question inside the question, we win rationally, since the involved polynomials are reciprocal in some sense.) In fact, taking the inverse, $r(0):r(N)$ for an even $N=2n$, since the norm of $k^2$ is one, we get an integer! $$ \frac{(1+k^N)^2}{4k^N} =\frac 14(k^{n} + k^{-n})^2 =\left( \frac 12(\sqrt 5-2)^{n} + (\sqrt 5+2)^n\right)^2 \ . $$ (And this integer is five times a square for $n=N/2$ odd, and a true square for $n=N/2$ even. Use binomial expansion above to obtain this information.)

OK, then let us plot the first few values for the proportion which is $>1$, for $N$ even we see no denominators: $$ \begin{array}{|r|c|c|} \hline N & r(0):r(N) & \text{Factorization of } r(0):r(N)\\\hline 0 & 1 & 1\\\hline 1 & \frac{1}{2} \sqrt{5} + \frac{1}{2} & \\\hline 2 & 5 & 5\\\hline 3 & \frac{17}{2} \sqrt{5} + \frac{1}{2} & \\\hline 4 & 81 & 3^{4}\\\hline 5 & \frac{305}{2} \sqrt{5} + \frac{1}{2} & \\\hline 6 & 1445 & 5 \cdot 17^{2}\\\hline 7 & \frac{5473}{2} \sqrt{5} + \frac{1}{2} & \\\hline 8 & 25921 & 7^{2} \cdot 23^{2}\\\hline 9 & \frac{98209}{2} \sqrt{5} + \frac{1}{2} & \\\hline 10 & 465125 & 5^{3} \cdot 61^{2}\\\hline 11 & \frac{1762289}{2} \sqrt{5} + \frac{1}{2} & \\\hline 12 & 8346321 & 3^{6} \cdot 107^{2}\\\hline 13 & \frac{31622993}{2} \sqrt{5} + \frac{1}{2} & \\\hline 14 & 149768645 & 5 \cdot 13^{2} \cdot 421^{2}\\\hline 15 & \frac{567451585}{2} \sqrt{5} + \frac{1}{2} & \\\hline 16 & 2687489281 & 47^{2} \cdot 1103^{2}\\\hline 17 & \frac{10182505537}{2} \sqrt{5} + \frac{1}{2} & \\\hline 18 & 48225038405 & 5 \cdot 17^{2} \cdot 53^{2} \cdot 109^{2}\\\hline 19 & \frac{182717648081}{2} \sqrt{5} + \frac{1}{2} & \\\hline 20 & 865363202001 & 3^{4} \cdot 41^{2} \cdot 2521^{2}\\\hline \end{array} $$ (... and we may also want to say something about the case of an odd $N$?!)

The above table was easily produced:

K.<a> = QuadraticField(5, latex_name=r'\sqrt{5}')
k = a - 2
for N in [0..20]:
    f = (1 + k^N)^2 / (4*k^N)
    print(f"{N} & {latex(f)} & "
          f"{latex(QQ(f).factor()) if f in QQ else ''}"
          r"\\\hline")

So, to have a final short message, if we further add to the green-blue-red-chain a further brown-pink appendix, then the pink radius is $1/81$ of the green one.

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  • 1
    $\begingroup$ The given solution should be further applicable for the next small, then tiny, then microscopic disks that can still be squashed in the small region near $\color{\magenta}\Upsilon$ between the pink chord and the violet arc of the big circle... $\endgroup$
    – dan_fulea
    Commented May 29, 2023 at 6:15
  • $\begingroup$ this is monster solution. $\endgroup$
    – dezdichado
    Commented May 29, 2023 at 14:45
  • $\begingroup$ @dezdichado I have to confess that the story above is also not my favorite way to present and reread mathematics. The notations are indeed monstrous, but caused by the initial interest to have a construction. Which is the first picture. There i considered the radius $R=1$, a bad decision to occupy $R$ with a letter. But ok, the diameter was $2R=2$, we paste one more unit, get $3$, and subtract from $3$ the diagonal of the rectangle with sides $1$ ad $2$, get the point determining the diameter of the upper green disk. No cartesian coordinates so far. So i decided to not insert any, and... $\endgroup$
    – dan_fulea
    Commented May 29, 2023 at 15:04
  • $\begingroup$ @dezdichado ... and construct geometrically the other two disks, the blue one, and the red one. This is usually done by a convenient inversion. The computations may get wild, and going in the details may make them dirty an undigestible, but the idea is simple: Transform the disks inside the $D$-shape region, tangent to its boundaries, into disks inside an angle, tangent to its legs, so a similitude is transparent. Each inversed disk comes with its own similarity to its original cousin. The angle determines the figure, and one and the same homothety factor $k$. We can generalize. I'll add it. $\endgroup$
    – dan_fulea
    Commented May 29, 2023 at 15:11
  • $\begingroup$ I meant monster in a good way. An informal adjective to describe something outstanding. $\endgroup$
    – dezdichado
    Commented May 29, 2023 at 15:20

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