15

There are geometrical diagrams in which it is evident* (to a skilled geometer) that two circles' radii have a certain integer ratio.

For example, for the following diagram, it is evident that the ratio of the brown circles' radius to the largest yellow circles' radius is $1:7$. The image, and proof, are found on pages 15 and 23 here, which has many other examples.

enter image description here

I have seen such diagrams with ratios $1:n$ for $n=1,2,3,\dots,10$, but I have never seen one with ratio $1:11$. So my question is:

Is there a geometrical diagram in which it is evident that two circles' radii have ratio $1:11$?

*By "evident", I mean:

  • The diagram can only show circles, circular arcs, lines, labels, and geometrical symbols such as right angle markers. No numbers.
  • Circles of the same color are congruent.
  • If circles look like they are tangent, then they are tangent.
  • If a circle has tangent points that look like they are diametrically opposite, then they are so.
  • No three circles of the same color can have collinear centres, either as shown, or upon rotation of circles and their contents.

(The last rule serves to avoid trivial answers such as a straight chain of say $11$ circles of the same color that span the diameter of a larger circle. I added the last rule after @Blue's answer, with their approval.)

Elegant answers are preferred. Proofs are optional.

Fun fact

There is a single diagram of circles in which it is evident that the radii have proportions $1,\dfrac12,\dfrac13,\dfrac14,\dfrac15,\dfrac16,\dfrac17,\dfrac18,\dfrac19,\dfrac{1}{10}$.

enter image description here

The equations of the circles are in this desmos graph.

Dan
  • 35,053
  • 1
    What about having 11 small (equal-sized) circles fit into a single large circle? To ensure collinearity of the centers, we can place $k$ of them into another circle $C_k$, for all $k$. – Calvin Lin Jun 20 '24 at 23:53
  • Just use binary? $11 = 8+2+1$ – Blue Jun 21 '24 at 01:24
  • @JyrkiLahtonen I fixed the link. – Dan Jun 21 '24 at 04:54
  • Thanks, @Dan. ${}$ – Jyrki Lahtonen Jun 21 '24 at 05:11
  • Would you consider Ford Circles to be in the spirit of the rules? Mind you, only the squares of rational numbers appear as ratios of radii within that family, so we won't get $1:11$ that way. Just listing a known infinite family (in case it qualifies). – Jyrki Lahtonen Jun 21 '24 at 05:24
  • @JyrkiLahtonen Yes, Ford Circles are in the spirit (and letter) of the rules. – Dan Jun 21 '24 at 05:45
  • A rule suggestion: remove the rule that "if two points of tangency appear to be diametrically opposit then they are", and instead use a rule "every line and circle must be tangent to at least three other circles/lines". I am aiming at a set of rules, where the diagram is, in some sense, "rigid". I did upvote Blue's post (for obviously clear reasons), but it is somehow unsatisfactory, because the smallish circles near the bottom can roll sideways, altering the size of the largest circle, but not disturbing tangency. – Jyrki Lahtonen Jun 21 '24 at 06:32
  • No, my suggested rule change won't capture everything I instinctively require :-( – Jyrki Lahtonen Jun 21 '24 at 06:33
  • 1
    @JyrkiLahtonen I have posted a "rigid" answer. It is a variation on Blue's answer. – Dan Jun 22 '24 at 13:21
  • "No three circles of the same color can have collinear centres, either as shown, or upon rotation of circles and their contents." This seems to be violated by the first example, in which rotation of one of the smaller yellow circles and the three blue circles in it can bring two of those blue circles into a line with the fourth blue circle. If the overlapping small yellow circles prevent this, then overlapping circles can be added to other figures to prevent unwanted rotations. – David K Jun 24 '24 at 05:48
  • @DavidK Yes, I hadn't noticed that. We should refine the rule about rotations: Rotations that extinguish tangencies are forbidden. – Dan Jun 24 '24 at 08:03
  • With that constraint on rotations, we cal easily convert any of the nonconformant constructions in the answers into conformant constructions by creating additional tangencies (including by rotation of circles and their contents). – David K Jun 24 '24 at 13:17
  • 1
    @DavidK Good point. Perhaps a better rule (than the one about three collinear) would be: the idea used to draw the diagram cannot be used to draw infinite diagrams that show arbitrary ratios. For example, in Blue's original answer (using binary), the idea there can be used to draw diagrams that show whatever ratio we want. But the drawback of such a rule is that it is not well-defined. I find that making these rules to match our intuition is quite challenging. Reminds me of the rules for geometric construction. – Dan Jun 24 '24 at 13:43
  • 1
    I think you are looking for constructions that are evident but not too obvious. It is like creating a good puzzle. Indeed this is very hard to pin down in terms of formal rules. In that vein, the binary construction is simply too obvious. The step that gives us $2n-1$ from $n$ may also be used in arbitrary constructions that are "too obvious" to anyone who knows how the $n\to 2n-1$ step works. – David K Jun 24 '24 at 14:09

6 Answers6

14

I believe this works:

enter image description here

I use (yellow) half-size circles and (red) third-of-half-size circles —with the "Y" arrangement conveniently allowing me to avoid a collinear chain of three— to determine (blue) five-sixths-size circles. These leave room for one-eleventh-size circles tangent to the boundary.

Taking the boundary to have radius $66$, here's a numerical verification:

enter image description here

Previous Solution

Note: OP has (reasonably) changed the rules to disallow three or more circles of the same size that can be rotated into collinearity, making this answer invalid ... but also making the question more interesting, so I'm okay with that.

It seems to me that you can achieve any $1:n$ ratio by just expressing $n$ in binary and making a stack of appropriate power-of-$2$ circles (rotating sub-circles thereof to prevent any three of the same size from being collinear; with only finitely-many centers to avoid, it's always possible to find a suitable rotation).

Here's $1:11$, noting that $11 = 1 + 2^1 + 2^3$ ...

enter image description here

Blue
  • 83,939
  • Thank you, this answers my question as stated. Now I realize that my last rule should have said "But there cannot be a straight chain of more than two circles of the same color, either as shown, or upon rotation of circles (and their contents)." Then the question would be more challenging (I've adjusted my "fun fact" accordingly). I would like to edit this new rule into my question, but you've already answered my original question, and I don't want to make my question a "moving target". Would you suggest that I start a new question, or edit the current one and make a note of the change? – Dan Jun 21 '24 at 03:51
  • 1
    @Dan: I'm generally against "moving target" edits, especially after an answer has been received, so I appreciate your hesitancy. In this case, I don't mind, since my figure didn't take much effort to make and the method I described felt like a bit of a cheat anyway. :) ... I wonder, though, if the adjusted rule should be that no three circles of the same color are ---or can be rotated to be--- collinear (not just "no straight chain ..."). With this stronger restriction, the two horizontal rows of circles in the revised Fun Fact Figure would be problematic. Challenge: Can you fix them? :) – Blue Jun 21 '24 at 04:15
  • 1
    My Fun Fact Figure now satisfies the stronger restriction. – Dan Jun 21 '24 at 15:13
  • 1
    I like the new solution better. I, too, was trying to create that $1, 1/11, 5/6$ configuration, but failed. Even though I spotted the need to triple a radius! I still think the configuration of 4 circles forming that Y is wobbly in the sense that I'm not seeing the 120 degree rotational symmetry forced on it. May be use a ring of 6 circles instead? – Jyrki Lahtonen Jun 22 '24 at 09:59
  • 1
    @JyrkiLahtonen: A ring of six circles would provide rigidity, which might well be a preferred feature overall. That said, I kinda like the "Y": the visual symmetry seems "evident enough" ... and the arrangement offers a nice "Ah-ha!" moment with the realization that it's equivalent to the hexagonal cluster of seven circles. Plus, four circles are simply lighter than six, and here they avoid being inelegantly sliced by the circumference of the blue circle; so, the composition looks a bit cleaner. These seem to me to be reasonable trade-offs for the lack of rigidity. – Blue Jun 22 '24 at 10:34
  • 1
    @Blue Great answer! I have posted an answer that is a variation of yours. – Dan Jun 22 '24 at 13:34
11

I found a simpler construction for $r = 1/11$ that seems to have been hiding under our noses all this time:

enter image description here

The ratio of the small red circle to the unit circle is $1:11$, which we can calculate via Descartes' circle theorem. The center coordinates and radii of the blue, cyan, and red circles are given by: $$\begin{array}{c|cc} \text{circle} & (x, y) & r \\ \hline \text{purple} & (\pm 1/2, 0) & 1/2 \\ \text{blue} & (0, \pm 2/3) & 1/3 \\ \text{cyan} & (1/2, 2/3) & 1/6 \\ \text{red} & (8/11, 6/11) & 1/11 \end{array}$$

More radii can be calculated with successive levels of inscribing, as well as with other initial arrangements of the largest circles in the packing. Notably, if the four initial circles of the packing have integer curvatures, then all subsequent circles in the packing will also have integer curvature. In the case shown above, the initial curvatures are $-1$ for the unit circle, $2$ for the purple circles, and $3$ for the blue.

heropup
  • 143,828
  • 2
    Nice! I found that this is called a Pappus chain of a symmetric arbelos. The radii are $\frac{1}{n^2+2}$. – Dan Jun 23 '24 at 03:17
  • 1
    Very nice! I would another blue circle below the two purple ones. Just to force the centers of the purple circles on the $x$-axis. I know that Dan's question specified that if two points of tangency appear to be diametrically opposite, you can assume that to be the case. But I would rather do without that rule even though it seems to often be easy to force it to be the case (like here). – Jyrki Lahtonen Jun 23 '24 at 03:56
  • 1
    @JyrkiLahtonen Thanks! I have added the circle to the diagram. – heropup Jun 23 '24 at 04:05
9

The following would probably get me booted out of a shinto monastery, but posting it anyway. Alas, I need two pictures to make it clear, but I think the configuration is more "rigid" than the one in Blue's first solution — I don't need to assume any pair of points of tangency on a circle to be diametrically opposite (unless the rest of picture forces that to be the case).

The ingredients are:

  • The example Sangaku realizing the $1:7$ ratio.
  • When we have a hexagonal ($\approx$honeycomb) arrangement of seven circles of equal radii $R$ (blue in the image below), the central one tangent to the other six, forming a bracelet around the central one, then the bigger circle (black) circumscribing the whole lot necessarily has radius $3R$.
  • In the following arrangement of circles of three radii $R_1$(black), $R_2$(red) and $R_3$(blue) we have the relation $$ R_3=\frac{R_1 (R_1-R_2)}{R_1+3R_2}.\qquad(*) $$

Proof. Look at the triangle with vertices at the centers of the obvious circles. Two of its sides have lengths $R_1-R_2$ and $R_1-R_3$ with the angle between equal to $\alpha=\pi/3$. The third side has length $R_2+R_3$. Apply the law of cosines to that triangle. Plug in $\cos\alpha=1/2$, and solve for $R_3$ from the resulting equation. QED

Observe that the roles of $R_2$ and $R_3$ are interchangeable (or red vs. blue circles), and we can equally well calculate $R_2$ when knowing $R_1$ and $R_3$. In particular, when $R_2=R_1/3$ the formula ($*$) gives $R_3=R_2$ as in the hexagonal layout of the second bullet.

Observe that by selecting $R_1=1$, $R_2=1/7$ we get $R_3=3/5$. With those choices the above diagram is a key part of the given Sangaku yielding that precious $1:7$ ratio. It is still crucial that the four circles of radius $R_4=1/5$ (blue in Dan's image) fit exactly as prescribed. If you take a peek at Dan's link you will learn that those blue circles of radius $R_4$ are a part of the hexagonal picture (see my second bullet).

The remaining piece is that by selecting $R_1=1$, $R_2=1/11$, the formula (*) yields $R_3=5/7$. This gives us the following solution consisting of two pictures:

The picture on the left is essentially the $1:7$ Sangaku Dan posted. I recap the argument. Assume that the radius of the black circles is $R_{\text{black}}=1$. The four brown circles form a part of that hexagonal pattern, and this quickly leads to the relation $$R_{\text{cyan}}=3 R_{\text{brown}}.$$ But the diameter of a cyan circle is clearly the sum of the radii of the brown and black circles, so we have $$2R_{\text{cyan}}=R_{\text{brown}}+R_{\text{black}}.$$ From this pair of equations and the known value of $R_{\text{black}}$ we easily solve $R_{\text{brown}}=1/5$. Hence $R_{\text{cyan}}=3/5$ matching with our example use of the formula $(*)$. So the left diagram, indeed, yields $R_{\text{blue}}=1/7$.

Moving to the diagram on the right. In its center we have a hexagonal arrangement of blue circles circumscribed by the gray circle, so we have $$ R_{\text{gray}}=3R_{\text{blue}}=3/7. $$ But, analogously to the second step in our handling of the left diagram, the diameter of a red circle is the sum of radii of grey and black circles, so $$ R_{\text{red}}=\frac12\left(R_{\text{grey}}+R_{\text{black}}\right)=5/7. $$ But this matches perfectly our second example application of the formula $(*)$, and we can conclude that $$ R_{\text{green}}=\frac1{11}. $$

Jyrki Lahtonen
  • 140,891
  • I'm more than a bit unhappy with this. Reverse engineering was to be expected, but I could (should!) do a better job in removing some of the scaffolding, after the building was completed. Also. Two diagrams. Really? – Jyrki Lahtonen Jun 21 '24 at 18:22
  • A similar application of the law of cosines gives a nice formula relating the radii of Black-Red-Blue circles also in the case where there are only two red+blue circles similarly osculating the black circle. This time the diagram has only 180 degree rotational symmetry instead of the order three (=120 degree rotation) symmetry of my diagrams. – Jyrki Lahtonen Jun 21 '24 at 18:26
8

This is a variation of @Blue's excellent answer.

enter image description here

$\dfrac{\text{radius of blue}}{\text{radius of orange}}=\boxed{\dfrac{1}{11}}$

Here are the equations I used:

  • Orange: $\space x^2+y^2=66^2$
  • Black: $\space x^2+(y\pm11)^2=55^2$
  • Yellow: $\space x^2+(y\pm33)^2=33^2$
  • Green: $\space x^2+(y\pm22)^2=22^2$
  • Blue: $\space (x\pm60)^2+y^2=6^2$ and $(x\pm12\sqrt5)^2+(y\pm30)^2=6^2$

Here is the desmos graph.

Dan
  • 35,053
  • 2
    +1. Very nice! ... There seemed to be a danger of the figure corresponding to more than one ratio, but I have confirmed uniqueness. With only the outer blue circles, the various radii are constrained by these relations: $$\begin{align}R_{orange}&=2R_{yellow} \[4pt] R_{black} &=R_{yellow}+R_{green} \[4pt] 2R_{blue}R_{black} &=(R_{orange}-R_{blue}),(R_{orange}-R_{black})\end{align}$$ (The first two are trivial; the third follows from a triangle argument.) Then, ignoring degeneracies, the tangencies of the central blues with yellow, green, black force $R_{orange}=11R_{blue}$. – Blue Jun 22 '24 at 16:28
  • 1
    I think this is the best solution. Good job, both of you. – Jyrki Lahtonen Jun 22 '24 at 17:54
3

If you can show $\space 1:n\space$ then you can show $\space 1:(2n-1)\space$

Suppose there is a diagram showing a unit circle, and inside there are two circles with radii $\frac{1}{n}$ and $\frac{n-1}{n}$, so that the circle with radius $\frac{1}{n}$ is internally tangent to the unit circle at their top points, and the circle with radius $\frac{n-1}{n}$ is internally tangent to the unit circle at their bottom points.

Then, starting from the top circle, fill the gap between the unit circle and the circle with radius $\frac{n-1}{n}$ with a chain of $n$ tangent circles on each side (or just one side, if you prefer).

The last ($n^{\text{th}}$) circle in the chain will have radius $\dfrac{1}{2n-1}$. (Fun fact: its centre, and the centre of the unit circle, lie on the same horizontal line.)

Here is an example with $n=6$.

enter image description here

White circle has radius $1$.

$\dfrac{\text{radius of black}}{\text{radius of orange}}=\dfrac15$ (proof left as an exercise)

$\therefore\dfrac{\text{radius of black}}{\text{radius of white}}=\dfrac16$

$\therefore\dfrac{\text{radius of red}}{\text{radius of white}}=\dfrac{1}{2(6)-1}=\boxed{\dfrac{1}{11}}$

I got the general equation of the circles in the chain from Mathworld's page on Pappus Chain. (Note: They start with a circle of diameter $1$, but I start with a unit circle. They condsider the top circle in the chain to be $n=0$, but I consider it to be $n=1$.).

Dan
  • 35,053
2

enter image description here

$\dfrac{\text{radius of yellow}}{\text{radius of grey}}=\boxed{\dfrac{1}{11}}$

Equations:

  • Grey: $\space x^2+y^2=1$
  • Light blue: $\space x^2+(y\pm\frac12)^2=\left(\frac12\right)^2$
  • Orange: $\space\left(x+\frac{\sqrt6}{5}\right)^2+y^2=\left(\frac{1}{5}\right)^2$ and $\space \left(x-\frac{2\sqrt3}{5}\right)^2+\left(y+\frac{2}{5}\right)^2=\left(\frac{1}{5}\right)^2$
  • Black: $\space \left(x-\frac{2\sqrt3}{5}\right)^2+\left(y-\frac{1}{10}\right)^2=\left(\frac{3}{10}\right)^2$
  • Red: $\space\left(x+\frac{2\sqrt6}{7}\right)^2+\left(y\pm\frac{5}{14}\right)^2=\left(\frac{3}{14}\right)^2$
  • Dark blue: $\space \left(x+\frac{2\sqrt6}{7}\pm\frac{3}{28}\right)^2+\left(x+\frac{5}{14}\right)^2=\left(\frac{3}{28}\right)^2$
  • Green: $\space\left(x+\frac{2\sqrt6}{7}\right)^2+\left(y+\frac{3}{14}\right)^2=\left(\frac{1}{14}\right)$ and $\space\left(x-\frac{2\sqrt3}{7}\right)^2+\left(y+\frac{3}{14}\right)^2=\left(\frac{1}{14}\right)^2$
  • Yellow: $\space \left(x-\frac{2\sqrt3}{11}\right)^2+y^2=\left(\frac{1}{11}\right)^2$

Here is the desmos graph.

Dan
  • 35,053