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Since I botched my first attempt at posing this puzzle, let me try again, this time hopefully closing all loopholes by dropping the packing angle and making this a purely geometric puzzle:

enter image description here

The figure is E-W symmetric, all circles have the same diameter d, the centres of the red circles form a regular 18-gon with side length d and the centres of the circles numbered 10 to 15 form a regular hexagon with side length d. Circle 16 is at the centre of both the hexagon and the 18-gon. Circles 1 to 9 each touch two red circles.

It looks like a weird coincidence that six of them also touch the hexagon, i.e. 1 touches 10, 2 touches 11, 4 touches 12, etc.

Can you prove or disprove that the apparent contacts are indeed precise touches?

Hint

Two more will do no harm, in fact, they may be quite helpful in providing an outside angle.

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Might this be the intended solution?

enter image description here

We will call the centres of the circles by their label and the circles by prepending C. Draw two more equal circles CA1 and CA2 each touching two circles of the 18-gon from the outside at a rotational offset of 3 circles. Do it in such a way that A1 is centred on the axis of E-W symmetry. 3 out of 18 is 1/6, so the angle subtended by A1 and A2 over 16 is 60°. Because of rotational symmetry triangle T1 is, in fact, equilateral.

Further, by bilateral symmetry the line 16-A1 runs precisely between C14,C15 and between CR0,CR1. By rotational symmetry a similar statement is true for the line 16-A2 and cirlces C16,C15,C10 and CA2,CR3,CR4. For the moment, we only need 16-A2 touches C15. From this it follows by the N-S symmetry of triangle T1 which extends to C16,C14,C15 vs. A1,CR0,CR1 that the line A1-A2 touches CR1.

By hexagonal symmetry we get a similar statement for line 16-A2 and adjacent circles. In particular, the three circles CR1,CR2 and C15 are equal size and stuffed into the corners of an equilateral triangle. Consequently, the purple triangle T2 formed by their centres is itself equilateral.

The orange triangle T3 is equilateral by construction of C8. Therefore rotating by 60° ccw around the shared corner R1 of T2,T3 maps the red arc CR1,CR2,CR3 onto the arc formed by CR1,C8 and C15. In particular, C8 and C15 must touch.

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  • $\begingroup$ Yep, that's very similar to my reference solution. Well done. $\endgroup$ – loopy walt Apr 29 at 2:01
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Two circles are touching if and only if the length of the line segment formed by connecting their centers has length $1$ (well technically, $d$, but I'm just going to set $d=1$ because the scale doesn't really matter).

Let's only consider circles 16 (let this be $A$), 14 ($B$), 15 ($C$), 7 ($D$), the circle below it to the left ($E$), the other circle touching $D$ and $E$ ($F$), 8 $(G)$, the circle below it to the left ($H$), and the other circle touching $G$ and $H$ ($I$).

Thus, $AB=BC=AC=DE=EF=DF=GH=HI=GI=1$ because we know they are all touching. Also, $FH=1$.

The internal angle of an 18-gon is $\angle EFH=\angle FHI=\frac{16\pi}{18}$. The internal angle of an equilateral triangle is $\frac\pi3$ so we can determine that $\angle DFH=\angle EFH-\angle EFD=\frac{16\pi}{18}-\frac\pi3=\frac{5\pi}9$.

Connect $DG$ and draw an altitude from $F$ to this line. Then, label the connecting point $J$. $\angle DFJ=\angle DFH-\angle JFH=\frac{5\pi}9-\frac\pi2=\frac\pi{18}$. Note that $\cos\frac\pi{18}=\frac{\text{adjacent}}{\text{hypotenuse}}$. In this case, $\text{adjacent}=JF,\text{hypotenuse}=DF=1$, so $JF=\cos\frac\pi{18}$.

Now, using the apothem formula, $a=\frac s{2\tan\frac\pi n}$. The side length is $s=1$ and the number of sides is $n=18$ (this is for the entire 18-gon), therefore $a=\frac1{2\tan\frac\pi{18}}=\frac12\cot\frac\pi{18}$. This is the length of the altitude if we draw one down from $A$ to $FH$.

Now, from that, subtract the altitude from $A$ to $BC$ (this and the above altitude are clearly both vertical and thus if we subtract them, we get the height of the hexagon $BCGHFD$, which is true because $BC$ and $FH$ must be horizontal for this to be E-W symmetric). Using simple Pythagorean formula, that length is $\frac{\sqrt3}2$. Therefore, the height of this hexagon is $\frac12\cot\frac\pi{18}-\frac{\sqrt3}2$.

Note that due to E-W symmetry and that $BC=FH$, $B$ must be vertical to (above) $F$, and $C$ to $H$. Also, we have $JF$, which is the height of the lower half of the hexagon. Now it remains to prove that the upper half is the same height, and we can prove that this hexagon is symmetric.

$\begin{align*}\frac{\frac12\cot\frac\pi{18}-\frac{\sqrt3}2}{\cos\frac\pi{18}}&=\frac{\frac{\cos\frac\pi{18}}{\sin\frac\pi{18}}-\sqrt3}{2\cos\frac\pi{18}}\\&=\frac{\csc\frac\pi{18}-\sqrt3\sec\frac\pi{18}}2\\&=2\end{align*}$

(According to Wolfram Alpha. Hopefully that's acceptable :P)

Therefore, this proves that the vertical distance from $BC$ to $DG$ is the same as from $FH$ to $DG$. Also, $B$ and $F$, and $C$ and $H$ are vertical to each other. Therefore, hexagon $BDGHFD$ is symmetric across $DG$. Therefore, $BD=DF$ and $CG=GH$. These equal $1$ (or $d$, technically).

Therefore, circles 7 and 14 touch, and 8 and 15 touch (the second could be given by symmetry based on the first, anyway). Since this is rotationally symmetric at 120 degrees, therefore 1/10 and 2/11 touch, and finally, 4/12 and 5/13 touch.

QED

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    $\begingroup$ This is probably a really stupid way of going about this. But it works (I think). $\endgroup$ – hyper-neutrino Apr 26 at 2:46
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    $\begingroup$ Not pretty, but +1 for the sheer will power. $\endgroup$ – loopy walt Apr 26 at 3:21
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    $\begingroup$ @loopywalt Haha, thanks :P Surely there's a more elegant solution. I'll keep thinking about it, but just bashing it with geometric formulas was my first idea so I stuck with it to see if I'd get a better idea halfway in, and it just ended up being a working solution, lol. $\endgroup$ – hyper-neutrino Apr 26 at 3:22
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Here is the proverbial picture worth a 1000 words.

enter image description here

You can see how it works for disks 2, 5, 8. Disks 4, 1, 7 are mirror images of 2, 5, 8.

And here are the 1000 words explaining the picture.

The illustration below shows 3 hexagons. Each hexagon extends a center point in 6 directions.

enter image description here

You can "multiply" these hexagons. You get a figure in 3 layers, each layer extending the central figure according to the direction defined by the hexagons.
Each layer adds to the perimeter the sides of one hexagon. You can see that the perimeter of the figure combines the sides of all the hexagons preserving the orientation of the originals.

enter image description here

By carefully rotating the hexagons by 1/18 of a turn relative to each other, as I did, you get a prefectly regular 18-gon.
The lengths of all segments are one unit. If you draw a circle centered on each node, with a radius of 1/2 unit, you get that the circles touch perfectly and never overlap.

See? No math!

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    $\begingroup$ Very nice. And with your argument it can be easily generalised. Perhaps you want to mention that? You can use this slightly megalomaniac picture if you like: i.stack.imgur.com/YRDFD.png $\endgroup$ – loopy walt May 2 at 2:38
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This is probably not pretty, but it relates to the concept of circumcenter which I like.

The triangle ABC

We establish a coordinate system with the origin $O$ at the center of the shape. Assume $d=1$ for convenience. Let $A$ be the center of the circle 10 at $(0,1)$, and $B,C$ be the centers of the other two circles touching circle 1 (with $B$ the lower one). Then by symmetry of the setting, it suffices to check there exists a point $X$ (the center of circle 1) whose distance to $A,B,C$ are all 1. It's equivalent to saying $X$ is the circumcenter of $\Delta ABC$, and the circumradius is 1. So first we solve for the coordinate of the circumcenter $X$, and then verify the circumradius is 1.

First, we define $R=|OB|$, and solve for it. The chord $BC$ with length 1 corresponds to a central angle of $\frac{2\pi}{18}$, so $2R\sin(\frac{2\pi}{18}/2)=1$, i.e. $R=\frac1{2s}$ where we define $s=\sin\frac{\pi}{18}, c=\cos\frac{\pi}{18}$ (these will come up later).

Let $X=(x,y)$ be the circumcenter (i.e. center of circle 1). Then it lies on the perpendicular bisector of $AB$, so $x=\frac{R+1}2$. It also lies on the perpendicular bisector of $BC$, but this passes through $O$ and is characterized by $\frac yx=\tan\frac{\pi}{18}$ (in polar coordinate this is $\theta=\frac{\pi}{18}$). Hence $y=\frac{R+1}2\tan\frac{\pi}{18}=\frac{R+1}2\cdot\frac{s}{c}$.

The remaining is purely not-so-pretty algebraic manipulation. Preparation: first we have $c^2=1-s^2$. And due to $\sin3\theta=3\sin\theta-4\sin^3\theta$, taking $\theta=\frac{\pi}{18}$ and noticing $\sin\frac{\pi}6=\frac12$, we get the equation $$8s^3-6s+1=0.$$

Using this equation, we can verify the circumradius is indeed 1. Its square equal to $|AX|^2$, which is $$(\frac{R+1}2-1)^2+(\frac{R+1}2\cdot\frac{s}{c})^2=(\frac{1-2s}{4s})^2+(\frac{1+2s}{4c})^2=\frac{4s^2+2s}{16s^2c^2}=\frac{4s^2+2s}{16s^2(1-s^2)},$$

so it suffices to check $16s^2(1-s^2)=4s^2+2s$. But by our cubic equation above, $16s^2(1-s^2)=16s^2-2s(6s-1)=4s^2+2s$, so we're done. This algebraic manipulation is cumbersome (and probably can be simplified), but it does not need calculator. (If we have calculator, the part of "not-so-pretty algebraic manipulation" can be done immediately).

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    $\begingroup$ Whoa, another impressive "stare-it-down" solution. Thanks for that. $\endgroup$ – loopy walt Apr 28 at 7:56
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Not an answer, but I wanted to share a picture. It looks like you can squeeze in 3 more circles: enter image description here

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