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On a hot summer day like today, I like to put a six-pack of beer in my cooler and enjoy some cold ones outdoors.

My cooler is in the shape of a cylinder. When I place the six-pack in the cooler against the wall, with three beer cans touching the wall, it seems that I can always draw an (imaginary) circle that is tangent to the other three beer cans and the opposite side of the wall, like this:

enter image description here

This seems to be true no matter what size the beer cans are (as long as they are the same size as each other).

Is this true?

I made this animation:

enter image description here

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    $\begingroup$ What an excellent little puzzle! Perfect for elementary school geometrists. $\endgroup$
    – Prime Mover
    Commented Jun 30 at 14:50
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    $\begingroup$ You got me with the growing beer cans. Awesome, +1. Too bad they had to shrink again. $\endgroup$
    – Oscar Lanzi
    Commented Jul 1 at 12:30
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    $\begingroup$ Congratulations on the cylindrical cooler; it has less surface area and therefore less heat leaking through than the usual rectangular cooler. For the best possible taste on a hot Nevada desert day, add salt to the ice to reduce the temperature even further. $\endgroup$
    – richard1941
    Commented Jul 4 at 3:47
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    $\begingroup$ Epic circles only question. $\endgroup$
    – mick
    Commented Jul 5 at 19:03
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    $\begingroup$ I am curious what did you use to make the animation? $\endgroup$
    – jw013
    Commented Jul 11 at 14:23

3 Answers 3

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Step 1

Remove the lower three beers from the six-pack. Put one of them at the bottom of the cooler. Obviously, we can draw a circle inside the cooler touching all four beers.

enter image description here

Step 2

Remove the bottom beer. Move the top three beers, and the circle, down, by a distance equal to the diameter of a beer.

enter image description here

Step 3

Put three beers back at the top.

enter image description here

Conclusion

The answer to the OP is: Yes.

It's not just six-packs

In the OP, if we replace "six-pack" with "$2n$-pack", the answer is still yes. In Step 1, instead of saying "Remove the lower three beers from the six-pack", say "Remove the lower $n$ beers from the $2n$-pack", etc.

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    $\begingroup$ I just thought of this solution to my own question. I am surprised that my question had 700+ views and 25+ upvotes before someone posted this solution. $\endgroup$
    – Dan
    Commented Jul 1 at 1:41
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    $\begingroup$ but I guess it is easy to prove that those four points are equidistant ($R-2r$) from the center of the large circle. $\endgroup$
    – dezdichado
    Commented Jul 1 at 1:54
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    $\begingroup$ @dezdichado Slide a beer along the wall of the cooler. By symmetry, the envelope curve must be a circle. $\endgroup$
    – Dan
    Commented Jul 1 at 1:59
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    $\begingroup$ I can't see how step 2 is obvious. If I start with 5 beers at the top in your step 1 for example, I can still shift all beers down as per step 2, but there wouldn't be enough space for 5 more beers after doing so. What makes 3 special? $\endgroup$
    – hiccups
    Commented Jul 1 at 3:32
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    $\begingroup$ +1. Excellent. ... To @hiccups's point, there is a limitation to the can size in order for a second row of (however-many) cans to "physically" fit inside the cooler. That said, if we abandon the cans-and-cooler conceit, the idea of your argument shows generally that a "row" of circles (of diameter at most half that of the enclosing circle) can be translated in any direction by their diameter to obtain a second "row" that has a common tangent circle with the enclosing one. This, even though the rows may overlap each other or the boundary of the enclosure. $\endgroup$
    – Blue
    Commented Jul 1 at 3:50
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This one is fairly self-evident:

enter image description here

(Too-)Verbosely ... Let the enclosing circle have radius $r$ and center $O$. Let the small circles have radius $s$, and let the left-most four of these have centers (going counter-clockwise from top-left) $S$, $S'$, $S''$, $S'''$. Extend $OS$ to point of tangency $T$.

Now, note that $\square SS'S''S'''$ is a rhombus with sides $2s$. Let $O'$ complete parallelogram $\square SS'O'O$, so that $|OO'|=2s$, making $O'$ bisect the lower portion of the "vertical" diameter into lengths $r-2s$.

Also, $|O'S'|=|OS|=|OT|-|ST|=r-s$. Letting $O'S'$ meet $\bigcirc S'$ at $T'$, we conclude that $|O'T'|=|O'S'|-|S'T'|=r-2s$, the same distance as above. Consequently, $O'$ is the center of a circle with that radius, which is tangent to the lower three small circles (for any $s$ up to $r/3$), as suspected by OP.

(Of course, when $s=r/3$, we have a hexagonal cluster of seven equal circles within the enclosing one.)

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    $\begingroup$ Is your point that due to symmetry we can remove the left or right hand beer can and then we just need to find a circle tangent to three others (middle can, side can, enclosing circle) which is en.wikipedia.org/wiki/Problem_of_Apollonius ? $\endgroup$
    – Cong Chen
    Commented Jun 30 at 13:30
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    $\begingroup$ @CongChen: Interesting thought. However, as an Apollonian problem, there's no apparent reason for the circle tangent to middle-can, side-can, enclosing-circle to be tangent to the other side-can. The tangency properties among the cans and enclosure ultimately impose the desired symmetry. (They also threaten to over-determine the system and make a symmetry-respecting fourth circle impossible. That they don't over-determine the system is what makes the configuration interesting.) $\endgroup$
    – Blue
    Commented Jun 30 at 14:50
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    $\begingroup$ Not self-evident at all, if you ask me! $\endgroup$
    – TonyK
    Commented Jun 30 at 14:52
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    $\begingroup$ What tool was used to create this figure? $\endgroup$
    – Jacob Ivanov
    Commented Jun 30 at 19:32
  • $\begingroup$ @JacobIvanov: I use GeoGebra (specifically, the GeoGebra Classic 5 desktop app) for my figures. $\endgroup$
    – Blue
    Commented Jun 30 at 22:24
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Not an answer, but the program to control the group of scaling and rotations

 Manipulate[
  Graphics[
   {Circle[],
    Circle[{0,-q},1-q],
    Circle[{0,1-q},q],
   tc={
      Circle[{0,1-q/2},q/2],
      Circle[{0,1-3q/2},q/2]},
   Rotate[Rotate[tc, -\[Alpha]  \[Pi], {0, p}], \[Alpha] \[Pi]]},
   PlotRange->If[zoom==1,{{-0.6,1/2},{0.2,1}},All]],
{{q,1/3},0,1},
{{\[Alpha],0.12},-1,1},
{{p,-0.26},-1,1},
{zoom,{0,1}} ]

4 packed circles between two tangent circles

Here the position of the circles is translated by a counter rotation around the origin following a rotation by $\alpha $ around a point $(0,p)$ on the y-axis.

The tangent conditions for $F(q,p,\alpha)=0$ have to be found, if possible.

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