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This lovely pattern comes from a show about arabian patterns. Thus the title.

enter image description here

It includes circles touching 8, 7, 6, 5 and 4 other circles.

The question: what are the exact sizes of the circles in decreasing order?

I know the numerical solution, I am interested in the exact expressions. I am not sure whether such an expression exists.

For the discussion purpose, lets call the radii a, b, c, d and e for the circles with resp. 8, 7, 6, 5 and 4 neighbors. Let's assume the period of the pattern is 2 units.

There is no catch in the picture. Where 2 circles seem to touch, they do touch. And you can assume perfect symmetry horizontal, vertical and diagonal.

PS: sorry if it is too easy for this forum, I believe it is too difficult and too mathematical for the puzzling forum.

Florian F
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2 Answers2

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"Just" solve this system:

enter image description here

$$\begin{align} |BC| &= b+c \\ |BD_1| &= b+d \\ |CD_2| &= c+d \\ |D_1 D_2| &= 2 d \\ |D_2 E| &= d + e \end{align}$$ where, say, $$A = (0,0) \qquad B = (a+b,0) \qquad C = \frac{(a + c )\sqrt{2}}{2} (1,1)$$ $$D_1 = (a+2b,x) \qquad D_2 = (x,a+2b) \qquad E = (a+2b)(1,1)$$

Edit. Numerically, taking $a=1$, we have $$\begin{align} b &= 0.7037139\dots \\ c &= 0.5493113\dots \\ d &= 0.7792658\dots \\ e &= 0.3227824\dots \\ x &= 1.3053795\dots \end{align} $$

which, according to Mathematica, is the only solution in positive real values.

Symbolically, well ... There's an unattractive degree-6 polynomial involved, and I don't really want to type it in right now. I'll revisit this later.

Blue
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  • Congratulations, that is how far I also got. The numbers are obviously correct if the picture is made using them. Anyway, they match mine. In my case, I end up with too many square roots to be able to solve the equation. – Florian F Sep 21 '15 at 05:11
  • @FlorianF: If you had gotten so far as the equations, then you should've given them and narrowed your question to ask for their exact solution. While I enjoyed thinking about this problem, I have to admit that I'm somewhat annoyed to learn that I spent a non-trivial chunk of my time covering ground you'd already passed. (I take some blame for not asking the traditional "What have you tried?".) Please be more considerate in the future. (And please pardon my tone. I dozed-off while pondering this problem, and just woke up to discover that all my work was for nothing. I'm a bit grumpy.) – Blue Sep 21 '15 at 06:13
  • Sorry, I said I know the solution numerically. I wanted to share the problem to let people have their own try. And my approach didn't give a symbolic solution so I was interested to see whether there is a different approach, maybe more elegant. – Florian F Sep 21 '15 at 19:08
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Do we know that the pattern really fits like it looks? Two starts, but too long to comment: Looking at the smallest circle and two of its neighbors, you have a $90-45$ triangle joining the centers with legs $b+e$ and hypotenuse $2b$. This gives $e=(\sqrt 2-1)b$ Looking at the largest circle, you have a triangle with sides $a+c,a+d,c+d$ with the angle opposite $c+d$ being $45^\circ$. This gives $(c+d)^2=(a+c)^2+(a+d)^2-2(a+c)(a+d)\frac {\sqrt 2}2$ from the law of cosines. Unfortunately, the other circles do not have nicely symmetric surrounds, so getting the other two equations will not be as easy.

Ross Millikan
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  • Yes, there are no "near misses" in the contact. Where they seem to touch, or almost touch, they do touch. – Florian F Sep 20 '15 at 20:04
  • It also looks like there are some squares, whose circumradius and side can be computed separately. For instance, consider the square with the four large circles: if the centers are indeed collinear, then its side is 2a+4c while the circumradius is $a+\sqrt{(b+d)^2-b^2} + \sqrt{(b+e)^2-b^2}$. – Aravind Sep 20 '15 at 20:09